Remedial Mathematics

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REMEDIAL MATHEMATICS

Sudhir Kumar Pundir

Asian Books Private Limited

REMEDIAL MATHEMATICS

Sudhir Kumar Pundir Reader, Department of Mathematics and Computer Science S.D. (P.G.) College, Muzaffar Nagar (UP)

7/28, Mahavir Lane, Vardan House, Ansari Road, Darya Ganj, New Delhi -110002.

"This page is Intentionally Left Blank"

Remedial Mathematics S.K. Pundir

Registered and Editorial Office 7/28, Mahavir Lane, Vardan House, Ansari Road, Darya Ganj, New Delhi - 110 002. E-Mail: [email protected] .. World Wide Web: http://www.asianbooksindia.com Phones: 23287577,23282098,23271887.23259161 Fax: 91 1123262021 ©Publisher First Published 2010 ISBN 978-81-8412-112-4 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording and/or otherwise, without the prior written permission of the publishers.

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Preface

The book "Remedial Mathematics" has been written according to the latest syllabus of B. Pharma and many other courses related to Bioscience of different technical colleges of India. The book is furnished with well defined theoretical background of the subject followed by well graded set of examples. At the end of each chapter, a set of unsolved problems as an exercise and chapter revision and learning fundamentals with the name "Refresher" to revise the chapter quickly to the students has been given. Various concepts !lnd related theories have been given in simple manner. The language used in this book is simple and a fairly large number of solved and unsolved problems with "hints have been added. I believe that the subject matter of this book will be very helpful to each of B. Pharma students in getting high percentage of marks in this paper. I gratefully acknowledge my indebtedness to various authors and publishers, whose books have been freely consulted during the preparation of the book. However, 'No Claim' is made of originality of results but presentation is my own. I have extended a conscious effort to make the book student friendly. I have tried my best to keep the book free from misprints. The author shall be grateful to the readers who point out errors and omissions which in spite of all cares, might have been there. I wish to sincerely thank Smt. Purobi Biswas, Production Manager, Asian Books Private Limited, New Delhi, whose encouragement and support enabled me to complete my book timely. I must also record my appreciation due to my wife Dr. Rimple, daughter Rijuta and son Shrish for their understanding and love during the long period that I have taken to complete this book. Mr. S.P. Singh, MIET Meerut; Mr. Aftab Alam, SDCMS, Muzaffar Nagar; MI'. Nishutosh, SDCET, Muzaffar Nagar; deserve special attention and thanks for their kind help and support. Above all, I am thankful to Almighty, without whose grace nothing is possible for anyone. Suggestions for further improvement will be gratefully acknowledged and accepted. Dr. Sudhir Kumar Pundir Email: [email protected]

Contents I.

QUADRATIC EQUATIONS ..................................................................................,.......... 1-60 1.1

Introduction .................................................................................................................... I

1.2

Linear Equation ......................................................................................... :.................... I

1.3 1.4

Solution of Quadratic Equations .................................................................................. 4 Equations Reducible to Quadratics ................................................................................ 12

I.S 1.6

Nature of Roots of a Quadratic Equation ................................................................ 36 Symmetric Function of The Roots ........................................................................... 36

1. 7

Relation Between Roots and Coefficients ................................................................ 38

1.8

Formation of Equations ...................................................... ........................................ 39

1.9

Common Roots ........................................................................................................... 39

1.10 Application of Quadratic equations ............................................................................. 48 Objective Evaluation ..................................................................................................... S6 Refresher ....................................................................................................................... S9

2.

SIMULTANEOUS LINEAR EQUATION ................................................................. 61-89 2.1

3.

Linear Equation in Two Variables ................................................................................. 61

2.2

Solution ofa Given System of Simultaneous Equations ............................................... 61

2.3

Graphical Method for Solving Simultaneous Linear Equations ............................... 62

2.4

Algebraic Methods of Solving a Pair of Linear Equations ............................................. 71

2.5

Word Problems on Simultaneous Linear Equations ....................................................... 81 Objective Evaluation ..................................................................................................... 86 Refresher ....................................................................................................................... 88

DETERMINANTS ........................................................................................................

~1-126

3.1 3.2 3.3

Introduction .................................................................................................................. 90 Determinant of Order Two ...................................................... ...................................... 90 Determinant of Order Three ...................................................... .................................... 91

3.4 3.5

Co-factors and Minors of an Element ...................................................... ................ 92 Properties of Determinants ........................................................................................ 9S

3.6

Cramer's Rule ............................................................................................................. 117 Refresher ............................................. ............................. ... ......... ............. ......... ......... 12S

(vi) 4.

Contents

MATRICES ............................................................................................................. ;•. 127-210 4.1 Introduction ............................................................................................... ,................ 127 4.2 Types of Matrices .................................................................................................... 127 4.3 Determinant of Square Matrix .................................................................................. 129 4.4 SingularandNon-SingularMatrix ...........................................................~ .................... 129 4.5 Sub- Matrix of a Matrix ............................................................................................ 130 4.6 Minors of a Matrix ................................................................................................... 130 4.7 Transpose of Matrix ....................................................................... 130 4.8 Symmetric and Skew-Symmetric Matrices ................................................................... 130 4.9 Complex Matrix ......................................................................................................... 131 4.10 Algebra of Matrices ................................................................................................. 132 4.11 Properties of Matrices Addition ............................................................................ 133 4.12 Multiplication of Matrices ........................................................................................ 136 4.13 Adjoint of a Matrix ............................................................................................... 149 4.14 Inverse or Reciprocal of Matrix ................................................... ,........................ 150 4.15 Orthogonal and Unitary Matrices ................................................... ~ .................... 151 4: 16 Solution of Equations Using Inverse of a Matrix ................................................ 165 4.17 Rank of a Matrix .................................................................................................... 172 4.18 Echelon Form of a Matrix ..................................................................................... 172 4.19 Elementary Transformations of a Matrix ............................................................. 173 4.20 Elementary Matrices ................................................................................................ 173 4.21 Invariance of Rank Under E-transformation .......................................................... 174 4.22 Normal Form ........................................................................................................... 174 4.23 Equivalence of Matrices ........................................................................................... 174 4.24 Rank of Product of Matrices ..................................................................................... 175 4.25 Linear Equations ....................................................................................................... 189 4.26 Nature of the Solution of Equation Ax = 0 ................................................................. 190 4.27 Non-homogeneous Equations .................................................................................. 193 4.28 Conditions for Consistency .................................................................................... 193 4.29 Conditions for a System of N-equations in N-unknowns to have a ......................... . Unique Solution ........................................................................................................ 194 Objective Evaluation ................................................................................................. 206 Refresher ................................................................................................................... 209 ¥

5.

••••••••••••••••••••••••

MEASURE OF CENTRAL TENDENCY ............................................................. 211-254 5.1 Introduction ................................................................................................................ 211 5.2 Kinds of Statistical Averages .................................................................................. 211 5.3 Arithmetic Mean (A.M.) ............................................................................................ 211 5.4 Methods of Calculating Arithmetic Mean in Individual Series ............................... 212 5.5 Properties of Arithmetic Mean ................................................................................ 220 5.6 Combined Mean ......................................................................................................... 225 5.7 Geometric Mean ......................................................................................................... 227 5.8 Properties of Geometric Mean ............................................................................... 230 5.9 Harmonic Mean .......................................................................................................... 233

Contents (vii) 5.10 Properties of Harmonic Mean .................................................................................. 237 5.11 Median ........................................................................................................................ 242 5.12 Mode ........................................................................................................................... 248 6.

TRIGONOMETRY ..................................................................................................... 255-321 6.1 Introduction ................................................................................................................ 255 6.2 Angles and Quadrants ............................................................................................... 255 6.3 Measurement of Angles ............................................................................................ 255 6.4 Various Types of Angeles ......................................................................................... 256 6.5 Trigonometric Ratios or Functions .......................................................................... 264 6.6 TRigonometric Identities and Equations .................................................................. 265 6.7 Fundamental Trigonometric Identities ...................................................................... 265 6.8 Signs of Trigonometric Functions ............................................................................. 275 6.9 Trigonometrical Ratio of Compound Angles (to be Used Directl) .......................... 292 6.10 Trigonometrical Ratio for Half Angles

(Obtained by Replacing A by

~

in the above Formulae) ...................................... 293

6.11 Conditional Identities ................................................................................................ 308 Objective Evaluation ................................................................................................... 315 Refresher ..................................................................................................................... 318 7.

LOGARITHMS ........................................................................................................... 322-345 7.1 Introduction ................................................................................................................. 322 7.2 Properties of Logarithms ........................................................................................... 323 7.3 System of Logarithms ............................................................................................... 323 7.4 Standard Form of Decimal ........................................................................................ 324 7.5 Characteristic and Mantissa ...................................................................................... 324 7.6 Method to Determine The Characteristic and Mantissa ......................................... 332 7.7 Antilogarithm .............................................................................................................. 335 7.8 ,Application of Logarithm in Pharmaceutical Problems ........................................... 339 €lbjective Evaluation ................................................................................................... 342 Refi;esher ..................................................................................................................... 344

8.

SYSTEMS OF COORDINATES ............................................................................. 346-374 8.1 Introduction ................................................................................................................. 346 8.2 Distance between Two Points ................................................................................. 349 8.3 Collinear Points ......................................................................................................... 352 8.4 Section Formula ........................................................................................................... 354 8.5 Mid Point Formula ..................................................................................................... 355 8.6 Area of a Triangle ..................................................................................................... 360 8.7 Locus and Equation to a Locus ................................................................................ 365 Objective Evaluation ................................................................................................... 370 Refresher ..................................................................................................................... 372

9.

THE STRAIGHT LINES ........................................................................................... 375-411 9.1 Introduction ................................................................................................................. 375 9.2 Slope or Gradient of a Line ..................................................................................... 375

(viiO

Contents

9.3 9.4 9.5 9.6 9.7 9.8

Slope of a Line Through Two Points ..................................................................... 376 Equation of Lines is Standard Form ...................................................................... 380 Transformation of General Equation in Different Standard Forms ........................ 394 Point of Intersection of Two Lines ......................................................................... 397 Condition of Concurrency of Three Given Lines ................................................... 397 Angle between Two Intersecting Lines .................................................................... 400 Objective Evaluation ................................................................................................... 408

10. FUNCTIONS AND LIMITS .................................................................................... 412-458 10.1 Introduction ............................................................................................................... 412 10.2 Type of Functions .................................................................................................. 413 10.3 Some Particular Functions ...................................................................................... 423 10.4 Algebra of Functions .............................................................................................. 428 10.5 Composition of Functions ....................................................................................... 429 10.6 Concept of Limit ..................................................................................................... 439 10.7 One Side Limits ....................................................................................................... 447 10.8 Limit at Infinity and Infinite Limits ..................................................................... 447 10.9 Continuity ................................................................................................................ 448 Objective Evaluation ................................................................................................. 454 Refresher .............................................~..................................................................... 457 11. DIFFERENTIATION ................................................................................................... 459-514 11.1 Introduction ............................................................................................................... 459 11.2 Method for Finding the Derivative Using First Principle .................................... 459 11.3 Derivative of the Sum of Two Functions ............................................................. 464 11.4 Derivative of the Difference of Two Functions ................................................... 465 11.5 Derivative of the Product of Two Functions ....................................................... 468 11.6 Derivative of the Quotient of Two Functions ..................................................... 469 11.7 Derivative of Functions of aFunction (Chain Rule) ............................................ 472 11.8 Differentiation of Implicit Functions ..................................................................... 483 11.9 Logarithmic Differentiation ...................................................................................... 486 11.10 Second Order Derivatives ........................................................................................ 505 Objective Evaluation ................................................................................................. 510 Refresher ................................................................................................................... 513 12. INTEGRATION ........................................................................................................... 515-586 12.1 Introduction ............................................................................................................... 515 12.2 Indefinite Integral .................................................................................................... 515 12.3 Methods of Integration ........................................................................................... 520 12.4 Integration by Parts ................................................................................................ 530 12.5 Integration by Partial Fractions ............................................................................. 537 12.6 Definite Integral ....................................................................................................... 564 12.7 Properties of Definite Integrals ............................................................................. 564 APPENDIX .................................................................................................................... 587-594

QUADRATIC EQUATIONS

•

INTRODUCTION (i) Polynomial: A function/defined by

lex) where ao' a p a2, ... , an

E

ra +a x+azx2+ ... +a/,x O

1

E

R

R is called a polynomial of a real variable with real coefficients.

Remarks

*

• If an 0 then the degree of the polynomial is n. • If ao' aI' ... , an E C, the set of complex number and x called complex polynomial.

E

R, then the polynomial is

(ii) Polynomial Equation: Let/(x) be a polynomial, then/ex) = 0 is called the polynomial equation. Generally, a polynomial equation of degree two is called quadratic equation. (iii) Degree of an Equation: The degree of an equation is the index of the highest power of variable quantity involved in the equation, when the equation has been expressed to the rational integral form (radical free form). (iv) Roots of an Equation: Let/(x) = 0 be a quadratic equation. A real or complex numbers a is said to be a root or solution of a quadratic equation/ex) = ax2 + bx + c = 0, if/(a) = aa2 + ba + c = 0 i.e. a satisfies the given quadratic equation. (v) Solution Set: The set of all roots of an equation, is called the solution set of

the given equation. (vi) Identity: An expression involving equality and a variable is called an identity, if it is satisfied by every value of the variable.

Remark • A root of an equation is also called zero.

III

LINEAR EQUATION

An equation of the form ax + b = 0 is called a linear equation of x, where x is unknown variable (quantity) and a and b are any constants. !fere, a 0 because if a ~ 0 then the equation gives b = 0, which has no unknown.

*

2

Remedial Mathematics

For example, x - 2 = 0, 3x - 9 = 0, x +

~ = 0, x = 2, x = 0 are the linear equation in x.

Remark • To find the degree of an equation, the unknown variable x must be in the numerator only in the equation and power of x must be a positive integer.

Solution of a Linear Equation All those values of the unknown variable, which is involved in the equation, for which equation is satisfied, are called the solutions (or roots) of the equation. To find the solution ofa linear equation we write the equation in standard form (ax + b = 0) ifit is not. Method: consider a linear equation

ax + b = 0, a*O Here, we want to find the value of x, So, add - b to both sides in (1.1) we get ax+b-b = O-b => ax=-b Now, dividing both sides by a, we get

...(1.1)

b x=--,a*O a Hence, the solution of the given equation is x =

~, a * 0 a

For example: Consider the equation 2x + 3 = 0. First adding - 3 in both sides of the given equation, we get

0-3 2x =-3

2x+3-3

=>

=

Now, dividing both sides by 2, we get

x = -

3 2'

which is the required solution.

SOLVED EXAMPLES Example 1: Solve the following equation 10 11_1::._ y =y

10 5 Solution: Here, the given equation is

y - 10 =y 10 5 and y is the unknown variable in this equation. So, multiplying the equation (1) by 10, we get 11_1::._

10 10(11-1::._ Y ) = lOy

IO

5

...(1 )

QJladratic Equations

110_y_1O(y-1O) = lOy 5 11O-y-2y+20 = lOy => 130-3y = lOy => Now, adding 3y on both sides of (2), we get 130-3y+3y = lOy+3y => 130 = 13y

y Hence,

=

3

... (2)

130 -=10 13

y=1O

Example 2: Solve the equation 3 (x + 2) = 5 Solution: Here, the given equation is 3 (x + 2) = 5 and x is unknown variable in the given equation. 3 (x+2) = 5 So, => 3x+3 x2 = 5

=>

~+6=5

Now, adding -6 on both sides of the above equation, we get 3x+6-6 = 5-6 3x =-1 1 x = -3 Example 3: Solve the equation x + 2 = 2x - 8. Solution: Here, the given equation is x + 2 = 2x - 8, adding 8 on both sides of the above equation, we get x+2+8 = 2x-8+8 x+IO =2x => 10 = 2x-x => 10 = x or x= 10 =>

Example 4: Solve

x -+3 2

x-6 8

=--

H ere, th · . .IS -x + 3 =x-6 · e gIven equatIOn -S o IutIOn: 2 8

=>

8(~+ 3)

=x-6

8x -+8x3 =x-6 2 => 4x+24 = x-6 Now, adding -24 on both sides, of the above equation, we get 4x+24-24 = x-6-24 4x = x-30

=>

4


4x-x = -30 3x = -30 x = - 30 =-10 3 x = -10

Theorem 1: A quadratic equation can not have more than two roots. Proof: Let, if possible, a, p, y be the three distinct roots of the quadratic equation ax2 + bx + c = 0, then we have aa2 +ba+c = 0 ...(Ll ) ap2+bp+c = 0 ...(1.2) and ar+by+c = 0 ... (103) Using (Ll) and (1.2), we get a(a2-p2)+b(a-p) = 0 => a(a-p)(a+p)+b(a-p) = 0 => (a-p)[a(a+p)+b] = 0 a(a+p)+b = 0

a+p

b

= --

a

... (1.4)

Similarly, using (1.2) and (1.3), we get b P+y = - ...(1.5 ) a Equation (1.4) and (1.5) gives a+p = P+y => a=y which is a contradiction (:. a and yboth are distinct) Hence, the quadratic equation cannot have more than two roots

III

SOLUTION OF QUADRATIC EQUATIONS

There are three methods to solve a complete quadratic equation: (a) Method offactorization (b) Method of Completing the square (c) Method offormula SOLVED EXAMPLES

Type I: Method of Factorization : Example 1: Solve Xl - 4x + 3 = 0 S"ution: Here, we have x 2 - 4x + 3 = => x 2 - 3x - x + 3 = => x(x-3)-1 (x-3) = => (x-3)(x-1) = x =

0 0 0 0

1,3

Quadratic Equations

5

E~a mple 2. Solve x2 - 2x - 3 = 0

Solution: Here, we havex2-2x-3 = 0 ~ x2-3x+x-3 = 0 ~ x(x-3)+1(x-3)=0 ~ (x-3)(x+ I) = 0 x=-1,3 Type II: Method of Completing the Square:

Remark • This method can be employed when the factorization method fails i. e. when the quadratic expression is not factorize or difficult to factorize. Example 3: Solve 4x2 - 3x - 1 = 0 Solution: Here, wehave4x2-3x-1 =0

~ ~

4( x2 - ~x) = I 3 x 2 --x

4

=

I

4

~ + (-~J

[By adding both sides (1/2 the coeff. of x)]

1. + ~ = 25 = (~y 4

64

64

8)

[By taking square root of both sides 1

x= l±~ 8 8 x= l+~ 8 8 x=

or

3 5 or x= - - 8 8 4

Hence, the solution set of the given equation is

{I, - ±}

Example 4: Solve 2x2 - 7x + 6 = 0 Solution: Here, the given equation is 2x2 -7x + 6 = 0 ~ a2-7x =-6 7 =-3 x2 - -x 2

~

x2_7....x+(_7....)2 = -3 + ( 2 -4 7 49 x 2 --x+2 16

=

49 16

-3+-

~y [BY adding both sides,H the coeff. Ofxt

6


=>

16 7 4

x-- =±4

=>

(By taking square root of both sides)

Therefore x = 2. ± 1.. = 2. + 1.. = ! = 2 and x = 2. - 1.. = .§. = .l , 44444 4442 Hence the solution set of the given equation is

{%' 2}.

Type III: Method offormula Consider a quadratic equation ax 2 + bx + c = 0, a,* 0 The solution of equation (l) is given by

... (1)

2

-b± Jb -4ac 2a Example 5. Solve each a/the/allowing equation by quadratic/ormula. (i) 5x2 - 15x + 11 = 0 Oi) ~ - 3x + 5 = 0 Solution: (i) Comparing the given equation with ax 2 + bx + c = 0, we get, a =S,b=-IS,c=ll x=

Therefore,

x=

- b± Jb 2

-

4ac

2a -C-1S)±J(-IS)2 -4xSxll

IS ± ~22S - 220

2xS

10

lS±)5

IS+)5 IS-)5 ---or---

10

10

10

. set IS . [IS+)5 , S-)5] Hence, the solutlOn -10 10 2 (ii) Comparing the given equation x - 3x + S = 0 with the standard quadratic equation ax 2 + bx + c = 0, we get a = l,b=-3,c=S Therefore,

-b ± Jb 2 - 4ac 2a _ (_ 3) ± ~r-(--3-i---4-x-l-x-S

x=

2xI

3±~ =3±FU 2

3±i,Ji! 2

2 (:.P=-l)

Quadratic Equations

7

m m}

. d so IuttOn ' set IS " gIven by {3--2--' +i 3 - i2 H ence, th e reqUIre

Example 6: Solve the following equations by quadratic method. (i) 2~ - 4x + 3 = 0 (ii) 25~ - 30x + 11 = 0 Solution: (i) Here, the given equation is 2x2 - 4x - 3 = 0 Comparing with the standard quadratic equation ax2 + bx + c = 0, we get a = 2, b = - 4, c = 3 x = -

Now,

b± Jb

2

4ac = 4±.J16=24

-

2x2

2a

= 4±R = 4±W = 4±i)8 4 4 4

4 ± 2i.fi

=

I .

=I±-l

J2

4

I. I - -II. = I +-Ior .fi .fi Hence, the solution set is {I +

~ i, 1- ~ i}

(ii) Comparing the given equation 25x2 - 30x + 11 = 0 with the standard quadratic equation, we get a = 25, b = - 30, c = II

Therefore,

x=

-b±~b2 -4ac

30±J900-1100

2a

50

30 ±

ROO

30 ±

50

J200ii 50

30±IOJi =l±.fii 50 5 5

1 + .fi i or 1_ J2 i 5

5

Hence, the required solution set is

5

{1

5

+

5

J2 i,l- J2 5

5

Example 7: Solve the following equations: I 2 4 (i) - - + - - - - x+1 x+2 - x+4 I 2 Solution: (i) Here, - - + - x+1 x+2

(ii) 2x Jl3 + 2x-i / 3

4 x+4

= --

(x + 2) + 2 (x + 1)

4

(x + I)(x + 2)

x+4

=

5

5

i}

8


(3x+4)(x+4) 3.x2+ 16x+ 16 x2 -4x-8 Put all these values in x

=

=4 (x+ 1)(x+2) = =

4 (.x2 + 3x+2) 0

_b±~b2 -4ac 2a

x =

' we get

-(-4) ± ~(-4)2 - 4·1(-8)

--'---'-----'--'--'----~

2 ·1 =

4±~16+32 2

=

2

(1

± v3{:;)

(ii) Here, the given equation is 2xl/3 + 2x- I / 3 = 5

So, Put

2 x l/3 = y, we get2y+ - = 5 y 21-5y+2 = 0 (y-2)(2y-l) = 0

Y = 2,

x

Now,

=

Hence,

x = 8 or

Example 8: Solve the equation x

::::> ::::>

Now,

or Hence,

= y=(2)3=8

(tY i

or

So, put

1

2'

+..Ix =

8

;5

..Ix = y in the given equation, we get 1 + Y = 265 251-25y-6 = 0 (5y+ 6)(5y-l) = 0 6

y

= --

x

=1= (-~f = ~~

(~f

5

=

or

5

1 25 36

x = 25

1 or x=25

Quadratic Equations

1. Solve the following equations by factorization method (/) .x2+x+I=O (ii) x2-x-12=0 (iii) x2+1=0 (iv) x2-4x+3=0 2. Solve the following equations by quadratic method. (l) x2-7x+12=0 (ii) x2-4x+7=0 2 (iii) 2x -3x+ 1 =0 (iv) 27.x2-10x+ I =0 (v) 3x2-x-1O=0 (vi) 9.x2+12x+4=0

.

3. Solve the equatlOn

x x 6"x -"5x = I5 - 3" + 7

4. Solve the following equations: (i)

x 2 -3x ---+2x =6

'it) 2.x2-lOx=3x-15

7 5. Solve the following equations:

(ii)

4x + 9

2X2 - 3x - 5 3

=--

5

6. Solve the following equations:

(/)

3x -7 16 --=-

5

x-5

7. Solve the equation: pqx2 - (p2 + 8. Solve:

~{2(x2 -

x + I)}

=

l) x + pq = 0

3x-4

9. Solve: 52x -126· 5x +3 =0 10. Solve the following equations: (I)

J(3x + 10)

=

9-

J(9x + 7)

2

(ii) J3x +1 +

~ 2 3x + I

. . vlpx) Fl-x) ~ ) vl--;--x-) ="6 +

(1/1)

13

HINTS TO THE SELECTED PROBLEMS 2. (I) .x2-7x+12=0. Here a= I, b =-7, c= 12

x=

b ± ~b2 -4ac 2a

x

7 ± ~(- 7)2 - 4 x I x 12 2x 1

=

(ii) x2-4x+7=0.

x= l,b=-4,c=7. -b ± ~r-:b2:--_-4a-c x =

2a 4 ± ~(- 4)2 - 4 x I x 7 2xl

9

10 Remedial Mathematics 7±.J49-48

7±1

2

2

7+1 - --

7-1

or

2

=4±..jl6=2s 2 = 4±..FI2 2

2

= 4±2J)i =2± J) i 2

x = 4or3.

3.

~- ~

=

1~ -

5x -6x --30 -x 30 4x x --15 30 8x-x 30 -7x => 30 x =>

j +7

4. (i)

x 2 -3x 7

+ 2x = 6

x 2 -3x + 14x

x -5x =--+7 15 -4x = --+7 15

7

~+ 14x-3x-42 = 0

=7

=>

=7

=> x(x+ 14)-3 (x+ 14) = 0

=7

=>

(x+ 14)(x-3)

=

0

= 30

=>

x x

=

-14 +3

or

5.

~ll)

2X2 - 3x - 5

4x+9 5

3

=> => =>

10~-15x-25 = 12x+27 1O~-27x-52 = 0

a = 10, b =-27, c=-52 x =

-b± ~b2 -4ac 2a

27 ± ~(- 27)2 - 4 x 10 x (- 52) 2 x 10 27 ± .J729 + 2080 27 ± .J2809 20 20 27 - 53 27 +53 -13 or =4or20 20 10

6. (I)

=>

3x-7 5

--

= 6

16 x-5 (3x-7)(x- 5)

=

80

27 ±53 20

=

Quadratic Equations

3~-15x-7x+35 = 80

3~-22x-45 = 0

3~-27x+ 5x-45 = 0 3x(x-9)+5(x-9) = 0 (3x+ 5)(x-9) = 0 3x+5 = 0

x

or x-9=O

-5

= -

3

or x=9.

7. pq~_(p2+q2)x+ pq= 0 => pqx2 _ p 2x - q2x + pq = 0 => px (qx-p)-q (qx-p) = 0 => (qx-p) (px-q) = 0 either qx - p = 0 or px - q = 0 p q x=or x=q P 2 8. )2(x - x + 1) = 3x-4 Squaring both sides. 2 (~-x + 1) = (3x-4)2

=> =>

Y-2x+2 = 9~+ 16-24x. 7~-22x+ 14 = 0 x= =

-~x7xI4

+22±) (22)2

2x7

22±~484-392 14 22 ±

J92

22 ± 2

14 let

=>

53

14

11 ±

53

7

5X = y 1-126y+3 = 0

y= =

126 ± )(126)2 - 4 x 1 x 3 2xl ±126

ANSWERU (ii) 4,-3

(iii) (i, - i)

11

12


2. (I) 3,4 (iv)

(ii) 2 +

~±J2i 27

27

3. 30

p q

!L p

10. (I) 2

III

5 3

(v) 2 --

'

,

3

-5

10

6. (z) 9'"3

8. 11±m

(ii) +3,-3

9. OJ

7

(ii) 0,

3

(ii) 2,5

(") 4 -13 II

2

(vi)

4. (I) 3,-14

5. (z) 2,-3 7.

J3 i, 2 - J3 i

(")

±,fS

III

9

4

13' 13

EQUATIONS REDUCIBLE TO QUADRATICS

Type I Equation of the form aX' +bX'+ c=O, whereXis an expression inx.

STEP KNOWLEDGE

In order to solve such type of equation we use the following steps: Step 1. Putx.' = y, and obtain the equation. ai + by + c = O. Step 2. Now solve the obtained equation for y. Step 3. Finally get the value of x, by using the relation Xn = y.

~~~~~~~I SOLVED

EXAMPLES

~I~~~~~~

Example 1: Solve x4 - 9x2 - 10 = 0 Solution: The given equation is x4 - 9x2 - 10= 0 Put, x 2 = yin (1), we get 9y - 10 = 0, which is a quadratic equation iny. i-9y-l0 = 0 => ;-10y+y-l0 = 0 => y(y-1J)+ 1(y-l0) = 0 => (y-lO) (y+ 1) = 0 => y = -1, lO Now Y = 10 => x 2 = 10

i -

=> and

=> =>

x=±J1O y =-1 x 2 =-1

x =±i

Hence, the solution set is given by

(JlQ, - JlQ, i, -i)

...(1)

Quadratic Equations

2x+I Example 2: Solve - x-I

(

2x+I )2 4- 10 (- + 9 = O.

J

x-I

=> => => =>

x-I

y-lOy+9

=>

= 0

= 0

y-9y-y+ 9 = y(y-9)-1 (y-9) = (y-1)(y-9) = y =

0 0 0

1 or y=9

2x + 1)2 y -_ 1 => ( -

Now

rUPTU B. Pharma 20051

x-I

. 2x + 1 2x + I SolutIOn: We have (- -)4- 10 (- -J2 + 9

x-I

2x + I

2x + 1 _ I = 1=>---± x-I

2x + I - - =-1 x-I 2x+l =x-I or 2x+ 1 =-x+ I x = -2 or x=O.

=>

x-I

=> =>

= I

or

2

2x + I 2x + 1 =9 =>-- =±3. ) x-I x-I

y = 9 => - -

Again

(

2x + I =3 x-I

=>

or

2x+ I = 3x-3

=>

x = 4

or

2x+ I --=-3 x-I

or

2x + I =

-

3x + 3

2 x= 5'

Hence, the solution of the given equation is {-2, 0, 4,~}

~

Example 3: Solve the equation (x _ _ x_)2 + 2x = 3 x+1 x+1

Solution: Here, the given equation is

(x __x+x_)2 + 2x ~ = 3 I x+ I

x2)2 x2 which can be written as - - + 2 - - = 3 ( x+1 x+1 2

Now putting _x_ x +1

=> =>

13

= y, the equation (I) reduces to y + 2y = 3

y+2y-3 =0=>y+3y-y-3=0 y(y+3)-I(y+3) =0=>(v+3)()'-I)=0

14 Remedial Mathematics y = 1,-3 y = 1

=> Now,

-

x2

=1

x +1 x 2 = x+ 1

x 2 -x-l =0 x=

1 ± .ji+4

2

1±

.J5

=--

2

y =-3

Also, 2

x =-3 x +1 x 2 = -3x-3 ~+3x+3 = 0

-

x=

. · set IS Hence, th e so IutlOn

- 3 ± ~9 -12

2

- 3 ± i J3 =---2

{1±J5 -2-' -3±i.J3} 2-

Example 4: Solve the following equation 4x - 5. :]X + 4 Solution: 4x-5.2x+4 = 0 which can be written as (2 Xi - 5 (2X) + 4 = 0 Put '? = y, we get y-5y+4 = 0 which is quardrat in y. 'So1ving (l) for y, we get (y-4)(y-l) = 0 y = 1,4 => Now, y =1

=

0

'? = 1 '? = 2°

=> => =>

x = 0

Also,

y =4

=> => =>

'? = 4 '? = 22

x=2 Hence, the solution set of the given equation is (0,,2). Example 5: Solve the equation ~/3 + x l13 - 2 = 0 Solution: Here the given equation is x 2/3 + X 113 - 2 = 0

i

(x ll3 + x l/3 - 2 = 0 Putx l/3 = y, we gety + y-2 = 0, which is quadratic iny,

=>

Quadratic Equations

15

Solving for y, we get (y+2)(y-l) = 0

=>

y

= 1,-2

Now,y= 1 X l/3 =

=> =>

I

x = 1 Y =-2 x l/3 =-2 x = (_2)3

also,

=> =>

x =-8 Hence, the solution set of the given equation is (1, - 8).

1)rpell Equation of the form aX + !!.- + C = 0, where X is an expression in x. X

STEP KNOWLEDGE In order to solve, such type of equations, we use the following steps Step I. Put X = Y and obtain the quadratic equation in y. Step 2. Solve the quadratic equation for y. Step 3. Finally get the values of x by using the relation X = y.

~~~~~~~I Example I: Solve

~

x x-I

SOLVED EXAMPLES

+JX-l x

Put

~

x

x-I

13, (.'(*I,x*O) 6

=

· H ' . .IS ere, t I 1e gIven equatIOn So I utlOD:

~I~~~~~~

~ - - + Fxl-l - - = -13 x-I

x

6

= y, then the above equation reduces to

I 13 7 Y + - = - => 6y- - 13y + 6 = 0

6 Solving for y, we get y

(2y-3)(3y-2) = 0

3 2

y =

2'3

Now,

y

12 => ~ x-Ix = 12 => _x_ = (1)2 => x = 2. 2 5

also,

y=~=>~ x =~3 3 x -I

=

x-I

16


Hence, the solution set is given by Example 2: Solve 8x

3/2

-

8 -m

x

{~, -~}

= 63

H ere t h ' . .IS 8 x 3/2 · e given equatIOn S o IutJon: 8

Putx3/2 = y, we get 8y - -

=> => => =>

8 = 63 -m x

= 63

y 1-63y-8 81-64y+y-8 8y(y-8)+ 1 (y-8) (y- 8) (8y+ I)

= 0

=0 =0 =0

y =

8,-i

=>

y = 8,x312 =8 x = (8)2/3 = (2 3 xl/3i = 22 = 4

also,

Y = _k=>x3/2 = -k=>x =

Now,

Hence, the solution set is given by

{4, i}

Example 3: Solve the equation 7' + x + 7' -x = 50 Solution: Here, the given equation can be written as 7·7 x +7·T x = 50

IUPTU B. Pharma 2008]

7.7 x +2.=50 x

7

Put

=>

r

7y+1. y 71 - 50y + 7

= y, we get =

50

= 0, which is quadratic

Solving for y, we get

71-49y-y+7 = 0 7y(y-7)-1 (y-7) = 0 (y-7) (7y-l) = 0 1 Y = 7,7

Now,

=>

y = 7 7x = 7

(_ky/3 = ±

iny.

{,)uadratic Equations

17

x = 1 (By putting the value ofy)

1 x 1 y = -::::::>7 =-::::::>x =-1

also,

7

7

Hence, the solution set is given by {I, - I} Example 4:

4x + 1 x + 1 5 Jf-+ - - = - = 0 ,jindthevalue a/x. x +I

4x + 1

4x + 1 x+I

2

x +I 4x + 1

IUPTU B. Pharma 20081

5 2

Solution: We have - - + - - = -

4x+1

Let

~=y

then

I 5 y+- = y 2 5

1+ 1 = -y 2 21+2 = 5y 21-5y+2 = 0 21-4y-y+2 = 0 2y(y-2)- I(y-2) = 0 (y-2)(2y-l) = 0 y = 2, ory= 1/2 Ify= 2 then 4x+ 1 - - =2 x+l 4x+ I = 2(x+ 1)::::::>4x+ 1 =2x+2 2x = I ::::::>x= 112. EXERCISE 1.2

~I~~~~~~~~~

Solve thefollowing equations: 1. x 4 -S.x2-9=0

4. 5. (.x2-3x+3i-(x-l)(x-2)=7 7. 3x-2 + 7x- 1 +5=0

(~)2 -5(~)+6=0 x+a x+a

6. (.x2-5xi-30(.x2-5x)-216=0 8. (.x2-5x+7)2_(x-2)(x-3)= 1 10. 3x + 1 + ~ = ~ (x E R) x + 1 3x+ 1 2

x l+x 13 11. - - + - - = l+x x 6

18


13. 5x+ 1 +5 2- x = 53 + 1

14. 3x+rX-2 =0

15. 2 2t +8 _8. 2x+2+1 =0

4x -I 4x + 1 10 16. - - + - - = 4x + 1 4x -I 3

17.

8~-f¥=2 x+3 x

18. )3x 2 +1+

~=5 2 3x + 1

HINTS TO THE SELECTED PROBLEMS 1.

x4 -

8~ - 9 = O.

x 2 =y =>;-8y-9=0 => ;-9y+y-9 = 0 =>y(y-9) + 1 (y-9)=0 => (y-9)(y+ I) = 0 =>y-9=0 or y+ 1 =0 => x 2 = 9 or x 2 = - 1 => y = 9 or y = - 1 => x = ± 3 or x = ± i 3. substitute x-I = y. to obtain 3; + 7y + 5 = O.

Let

a)

4. substitute (x = y. to obtain; - 5y + 6 = O. x+a 5. (~-3x+3i-(x-I)(x-2)=7 => (x 2 -3x+3)2_(x2-3x+2)-7=0 => (~-3x + 3)2_(~_ 3x+ 3)-6 = O. => substitute ~ - 3x + 3 = y. to obtain; - y - 6 = O. 9.2x =42x _1

=> Let 11. substitute x-I

Then

=> => => => =>

=

y.

;-12=-y ;+y-12=0 ;+4y-3y-12=0 (y+4)(y-3) = 0 y = 3 or y=-4 x-I = 3 or x-I =-4 x=- orx=-l 3

ANSWERS l. (± 3, ± i)

1 2. ±-,± 1 2

7 Jli. 3. - - ± - l

6

10

Quadratic Equations

3±

iJII

4. -2a,-3a

5.0,3, - - 2 -

7. ± 2,± 3

8. 2,3,

I -- I 5'

5 ±iJ] 2

6. 2,3, - 4,9 9

1 1 --. 3' 4

11. -3,2

12. -1,1

13. -1,2

14. 0

15. -4

1 16. ±2

17. 1

18. 0,0, ±

10.

19

J5

Typem Equation ofthe form (x + b) (x + b) (x + c)(x + d) = e, where a, b, c, d are constant such that a+b=c+d.

STEP KNOWLEDGE In order to solve such type of equations, we use the following steps Step 1. Put~ + (a + b) x = y and obtain the quadratic equation iny. Step 2. Solve this quadratic equation for y. Step 3. Finally get the value of x by putting ~ + (a + b) x = y

~~~~~~~I

SOLVED EXAMPLES

~I~~~~~~

Example 1: Solve (x + 1) (x + 2) (x +3) (x + 4) = 120 Solution: Here, the given equation is (x+ I)(x + 2)(x + 3)(x+ 4) = 120 which can be written as [(x + I) (x + 4)][(x + 2)(x + 2)] = 120 => (~+5x+4)(~+5x+6) = 120 2 put x + 5x = y, then equation (I) reduces to (y+4)(y+6) = 120 => y+ lOy+24-120 = 0 => y+ 10y-96 = O. which is quadratic iny, solving fory, we get y (y + 16) - 6 (y + 16) = 0 => (y-6) ()It 16) = 0 Y =6,-16 => Now, Y = 6 x 2 +5x = 6 => ~+5x-6 = 0 => x = 1,-6 => y = -16 Also, x 2 + 5x = -16 =>

20 Remedial Mathematics xl+5x+ 16

=>

=

x =

=>

0

-5±iJ39 2

· set 0 f th · . .IS {I , -6, - 5 ±2i Hence, the so IutIOn e given equatIOn

J39}

Example 2: Solve (2x - 7) (~ - 9) (2x + 5) = 91 Solution: Here, the given equation is (2x - 7)(xl - 9)(2x + 5) = 91 which can be written as (2x-7)(x-3)(x+ 3)(2x+ 5) = 91 => [(2x-7)(x + 3)][(x-3)(2x+ 5)] =91 => [2.x2-x - 21] [2x 2 -x - 15] = 91 Put 2x2 -x = y, we get (y-21) (y-15) = 91 ;-36y+315-91 = 0 => ;-36y+224 = 0 => (y-8)(y-28) = 0 => Y = 8,28 => Now, Y = 8 2x2 _x = 8 => 2x2 -x-8 = 0 =>

1±J65

=>

x =

also,

y = 28

4

2xl-x = 28 2 2x -x-28 = 0

=> =>

x = 1 ± 15 = 4 _2 4 '2

=>

M}

· set 0 f th · . .IS {4 , - "2' 7 l± 4 Hence, the so Iutlon e given equatIOn

7l-

Example 3: Solve (~ - 5x + (x - 2) (x - 3) = 1 Solution: Here, the given equation can be written as (x 2 - 5x + 7i- (xl - 5x + 6) = 1 Put xl - 5x = y, we get (y + 7)2 - (y + 6) = 1 => 14y+49-y-6-1 = 0 => 13y+42 = 0 => ;+6y+7y+42 = 0 => y(y+6)+7(y+6) = 0 => (y + 6) (y + 7) = 0

;+

;+

Quadratic Equations

21

y = -6,-7

=:>

Now,

Y = 6 xl-5x+6 = 0 (x-2) (x-3) = 0 x = 2,3 y = - 7 =:> x 2 - 5x + 7 = 0

=:> =:> =:>

also

5±iJ3

x=--2

. set of the given . equation . .IS Hence, the solution

{2,3, -5±iJ3} 2-

'JYpeIV Equation of the type ~ax + b + ~cx + d = ~ex + f where a, b, c, d, e,fare constant. STEP KNOWLEDGE In order to solve such Jype of equation we use the following steps. Step 1. Square both the sides of the given equation. Step 2. Put the rational terms on one side and irrational terms on other side. Step 3. Again squaring and obtain the quadratic equation. Step 4. Solve the obtained quadratic equation.

~~~~~~~ISOLVED

EXAMPLES

~I~~~~~~

Example 1: Solve ~x + 5 + ~x + 21 = ~6x + 40 Solution: Here the given equation is ~ x + 5 + ~ x + 21

=

.J6x + 40

On squaring both the sides, we get

(x+5)+(x+21)+2 =:>

"Jx+5. ~x + 21 =6x+40

2~(x + 5) (x + 21)

=

4x+ 14

~(x + 5)(x + 21) = 2x+7 Again squaring, we get (x+ 5)(x+21) = (2x+7i =:> xl+26x+ 105 = 4x2+28x+49 =:> 3x2+2x-56 = 0 (3x+ 14)(x-4) = 0 (3x + 14) or (x-4) = 0 =:>

14 x = --orx=4 3

Now, we check, whether the obtained values x = 4 and x = -

1~ .)

equation or not.

satisfy the given

22


x =4,wehave, ~ +~ = J24+40

When

=>

3 + 5 = 8, which is true Hence, x = 4 is solution of the given equation. 14 x = - then

when

3'

)_1; +5 +)_1; +21

6(_1;)'+40

J¥

Jf+f!

=>

=> 1 + 7 *- 6, which is not true

Therefore, x = - 1; is not a solution. It is an extraneous root and so reject it. Hence, the solution is 4.

Remark A root which is obtained by solving an equation but does not satisfy it, is called an extraneous root. Such roots enter the equation in the process of squaring because this process is irreversible.

Example 2. Solve = J(x + 5) + J(x + 12) = J2x + 41 Solution: Here, the given equation is J(x+5)+Jx+12 = J2x+41

Squaring both the sides, we get (x+5)+(x+ 12)+ 2J(x + 5)(x + 12) =2x+41

=>

2J(x + 5)(x + 12) = 24

=>

Jex + 5)(x + 12) = 12

Again squaring, we get (x+5)(r+12) = 144 => .?+ 17x + 60 = 144 x 2 + 17x-84 = 0 => x = 4,-21 => Therefore x = 4 is a root, because it satisfy the given equation. Also x = - 21 is an extraneous root, because it does not satisfy the given equation. Example 3: Solve

~5x2

- 6x + 8 -

~5x2

- 6x - 7

Solution: Here, the given equation reduce to

~5x2

Let 5.? - 6x = y, then given equation reduce to JY+8-JY-7 =\

=~ 1 - 6x + 8 -

~5x2

- 6x - 7 = 1

Quadratic Equations

23

Squaring both the sides, we get

(y + S) + (y - 7) - 2,Jr-(y-+-S"--')(y---7-) = 1

~y2 + Y _ 56

y =

=> Again squaring, we get

y = y+y-56 Y = 56 5x2 -6x = 56 5~-6x-56 = 0 (5x+ 14)(x-4) = 0

14

x=4-, 5 Since, both the obtained values

(x

= 4, and -

1;)

satisfies the given equation.

Hence, the solution set of the given equation is {4, _ I;} Example 4. Solve ~x + 4 + ~x + 20 = 2F+li Solution: Here, the given equation is ~x + 4 + ~x + 20

=

2 F+1i

Squaring both the sides, we get

(x + 4) + (x + 20) + 2 ~(x + 4)(x + 20) =4 (x+ 11)

=>

2 ~(x + 4) (x + 20) = 2x+20

=>

~(x + 4)(x + 20) = x+ 10

Again squaring, we get (x+4)(x+20) = (x+ 1O)2=~+20x+ 100 ~+24x+SO = ~+20x+ 100

=> => Clearly, x equation.

=

4x = 20 x =5 5, satisfy the given equation. Hence, x = 5 is the required root of the given

~~~~~~~I

EXERCISE 1.3 ~I~~~~~~~

Solve the following equations: (ii) (2x+ 3) (2x+ 5)(x-l)(x-2) = 30 1. (i)x(x+ l)(x+3)(x+4) = ISO (iii) (x-5)(x-7)(x+4)(x + 6) = 504 (iv) x(2x+ l)(x-2)(2x-3)=63 (v) (~-3x-1O)(~-5x-6)= 144 (vi) (x+ 2)(3x + 4)(3x + 7)(x+ 3) = 2600

2. (i)

~3x - 1 -

F=l = 2

(iii)

~x+4 +~x+20 = 2F+1i

(iv)

F+1- F=l =

~4x - 1

(ii) ~2x + S +

F+5 = 7

(v) ~5x + 7 -- ~3x + 1 = ~x + 3

24 Remedial Mathematics HINTS TO THE SELECTED PROBLEMS 1. (2x+3)(2x+5)(x-l)(x-2)=30 ~ [(2x + 3)(x-l)[(2x+ 5)(x-2)] = 30 ~ (:zx2+x-3)(2~+x-lO) = 30 ~Let 2x2+x = y. ~

(y-3)(y-lO)=30

~

Y-13y+30 =30

~

y-13y y(y-13)

~

= 0 =

0

y = 0 ot Y - 13 ~ 0

~

:zx2+x = 0 or 2xl+x-13=0

~

-1±)(1)2 -4x2x(-13) x (2x + 1) = 0 or x = ----'--2-x-2----

~

1

x =0 or = - 2'

~

- 1 ± .JI 4

2.(ii) .J2x+8+.Jx+5 =7. squaring both sides.

(2x + 8) + (x + 5) + 2 .J(2x + 8)

F+5 =49

3x + 13 + 2 .J(2x + 8)(x + 5) =49 3x-36 = - .J(2x + 8) (x + 5) squaring again, we get

(3x-36i =4 (2x+ 8) (x+ 5) ~ 9 (x-12)2 =4 [:zx2 + 10x+ 8x+ 40] ~ ·9 (xl + 144-24x) =4[:zx2+18x+40] ~ 9xl + 144 x 9 - 24 x 9x = 8xl + 160 + 72x. ~ xl-144x+ 1136 = 0 on solving we get the required result.

.J5x + 7 - .J3x + I = .Jx + 3

2. (v)

squaring both sides

(5x + 7) + (3x + 1) - 2.J5x + 7 .J3x + 1 = x + 3

~ 8x + 8 - 2 .J5x + 7 .J3x + 1 = x + 3 ~ ~

7x + 5 = 2 . .J5x + 7 .J3x + 1 squaring both sides.

(7x+5i = 4 (5x+7)(3x+l)

+ 104

-I ± ft05 = -----'-4

Quadratic Equatiolls

=> => =>

25

49X3 +25 +70x = 4 (15~ + 5x+2Ix+7) 49.x2 + 25 + 70x = 60.x2 + lOOx + 28 Ilx 2 +30x+3 = 0

x=

- 30 ± ~900 - 4 x II x 3 2 x II

- 30 ± ~900 - 132 22 - 30 ± J768

- 30 ± 16 F3

22

22

-15±8F3 11

ANSWERS

.ftl

I

(ii)

0, - 2'

(iii) -7,-2,3,8

(iv)

_i

(v) - 3,2,2, 7

(vi)

1. (/) - 6, 2, - 2 i

-1±OJiQ5 4

3 3±iJ47

2"

4

19 13±~-599 -- 2 3" 6 (ii) 4 (iii) 5

2. (i) 1,5

1 11 4 (b) Removal of Common Factor Throughout in an Irrational Equation. 5 (iv) -

(v) - -

STEP KNOWLEDGE In this method, we use the following steps. Step 1. Factorize each given expression. Step 2. Put common factor equal to zero and find one value ofx. Step 3. Solve the remaining equation by the method discussed in (a).

~~~~~~~I SOLVED Example 1: Solve

~x 2 -

16 -

~x 2 -


EXAMPLES

~I~~~~~~

5x + 4 = x - 4

~x 2 -

=> ~(x - 4)(x + 4) - ~(x -I) (x - 4)

16 =

~x 2 -

(x-4)

5x + 4 = x - 4

26 Remedial Mathematics

:::::> .J(x-4) [.Jx+4 -..r;=I-.Jx-4] =0 Now,either .Jx-4 =0:::::>x-4=0:::::>x=4 or[.Jx +4 -~ - .Jx- 4]

=

0

.Jx+4-..r;=I = ~x-4

:::::>

On squaring, we get. 4(~+3x-4) = (x+7)2=~+ 14x+49

4~+ 12x-16 = ~+ 14x+49 3~-12x-65 = 0

x=

2±.J4+780 2±28 13 6 =-6-=5'-3

Here, it is clear that x = 5, satisfy the given equation. Although x = -

1:

-13

satisfy the given equation, therefore x = -3- is an extraneous root. Hence, the solutions set of the given equation is {4, 5}.

Example 2: Solve

~x2 + 2x - 3 + ~


= .J5 (x -1)

~x 2 + 2x - 3 + ~(x2

- x) =

:::::> ~(x -1)(x + 3) + Jx (x -1) = J5 (x -1) :::::>

JX-1[.Jx+3+Fx-J5] = 0

~ = 0 :::::>x-l =0 :::::>x= I

Now, either or

.Jx + 3 + Fx - J5 = 0:::::> J x + 3 +.Jx = J5

Again, squaring, we get (x+3)+x+2.Jx :::::>

F+3

= 5

2.Jx(x+3) =2-2x

Again squaring 4x (x + 3) = (2 - 2x)2 :::::> 4~+ 12x = 4-8x+4~:::::>20x=4 4

Therefore,

x

1

= 20 ="5

Hence, the solution set of the given equation is {I. ~} 2

(c) Equation ofthe form ax + bx + c + p

~ax2 + bx + c

=

q

~5 (x -1)

does not

Qlladratic Equatiolls

27

STEP KNOWLEDGE

In order to solve such type of equation, we use the following steps. Step 1. Assume

~ ax 2 + bx + c

=

y and obtain the quadratic equation in y.

Step 2. Solve the obtained equation for y. Step 3. Finally obtain the ~alue of x by putting

~ax 2 + bx + C

=Y

~~~~~~~I SOLVED EXAMPLES I~~~~~~~ Example 1: Solve x 2 - 4x -12 ~x2 - 4x + 19 =-51 Solution: Here, the given equation can be written as (x 2 - 4x + 19) -12 ~x2 - 4x + 19 + (51-19) =0

~

(x 2 -4x+19)-12~x2 -4x+19 +32 =0

~x 2 -

Put

4x + 19

;-12y+ 32

=

y is (\), we get

=

0, which is a quadratic equation iny.

Solving for y, we get, y = 4~

~x 2 -

4x + 19 = 4 2 x - 4x + 19 = 16 ~ ~ - 4x + 3 = 0 (x-l)(x-3) = O~x= 1,3

Now, ~ ~

Also

y

=

8 ~ ~r-x-2---4-x-+-}-9 = 8

~

~-4x+ 19 = 64

~

~-4x-45 = 0

~

~+5x-9x+45 = 0 x (x + 5) - 9 (x + 5) = 0 ~ (x+ 5)(x-9) = 0 ~ x = 9,-5 Hence, the solution set of the given equation is {I, 3, - 5, 9} ~

(d) Method ofIdentity : Equation of the form

~ ax 2 + bx + c + ~dx + ex + f

=

k

In order to solve such type of equation we proceed as the following example:

~~~~~~~I SOLVED EXAMPLES ~I~~~~~~ Example 1: Solve

~5x2

- 6x + 8 -

~5x2

- 6x -7 = 1


~5x2

- 6x +·8 -

~5x2

- 6x - 7 = 1

...(1)

28


Let

~5x2

- 6x + 8 = A and J5x 2 - 6x - 7 = B, then given equation reduces to

Also

=> =>

A -B = 1 A2_B2 = (5x 2-6x+8)-(5x2 -6x-7)= 15 (A-B)(A+B) = 15(\) l.(A+B) = 15 ... (2)

...(1)

[Using(l)]

Solving (1) and (2), we get 2A = 16

=>

A =8

=>

J5x 2 -6x + 8 =8

=> =>

5~-5x-56 = 0

5~-6x+ 8 = 64

6 ± ~36 + 1120

=>

x=

=>

x=4-, 5

10 14

6±34 10

=--

Hence, the solution set of the given equation is {4, _

Example 2: Solve

~2x2 - 3x - 5 - ~x2 - 3x + 4 =x + 3

Solution: Here, the given equation is J2x 2 - 3x - 5 Let J2x 2 - 3x - 5 = A and Also,

=>

I;}

Jx 2 - 3x + 4

=

~x 2 -

3x + 4

A - B = x+3 A2_B2 = (2x2-3x-5)-(x2-3x+4)=~-9 (A - B)(A + B) = x2 - 9 (x+3)(A+B) = (x-3)(x+ 3)[using (1)] A+B =x+3

A =x

On squaring,

=>

(x + 3)

B, then given equation reduces to

2A=2x

Therefore,

=

~2x2 ---3x -

5 =x ~-3x-5,= x 2 ~-3x-5 = 0

x=

3 ± ~9 2

+ 20

3 ± 59 =--2

.J29}

' . .IS {3 ± 2 · set 0 f th Hence, t he so IutlOn e gIven equatIOn

...(2)

... (2)

Quadratic Eqnations

Solve tile following equations: 1.

~x2-5x+6+~x2-9 = ~2x2

2.

~x2 +2x-3 +~

=

-llx+15

~5(x-1)

3. ~2x2 - 5x - 2 - ~2x2 - 5x - 9 = 1 4.

~4x2 -7x-15 _~x2 -3x = ~x2 -9

3x - 2 I 2 (x + 1)2 5. - - + V2x - 5x + 3 = -'------'-2 3 6.

~3x2 - 4x + 34 + ~3x2 - 4x - 11 = 9

7.

~x2

+ ax + b -

~x

Jb +..Jc

+ 9x + 6 =

8. x (x + 3) + 3 ~2x2 + 6x + 5

=

25

HINTS TO THE SELECTED PROBLEMS 1.

~x2

- 5x + 6 + ~x2 - 9 =

~2x2 -llx + 15

=> ~(x - 2)(x - 3) + ~(x - 3)(x + 3) => ~x - 3 [ ~x - 2 + ~x + 3]

~x-3[~x,-2+~x+3J

=

=

=

~2x2 - 6x - 5x + 15

~2x (x - 3) - 5 (x - 3)

~(x-3)(2x-5)

=>~x-3[~x-2+~x+3-~2x-5J =0 either

~ x - 3 = 0 => x = 3.

or ~ x - 2 + ~ x + 3 - ~2x - 5 squaring both sides.

x-2+x+3+2~x-2 ~x+3 =2x-5 1+5 = -2~x-2 ~x+3

6= -2F=2F+3 =>

3 = - ~(x -2)(x + 3)

squaring again, we get 9 = (x-2)(x+ 3) 9 = ~-x-6

29

30 Remedial Mathematics x2-x- 15

=

x= x= 4.

~4x2 :::)

-7x-15

_~x2 -3x

=

0 1 ± ~(1)2 + 4 x 1 x 15

2xl

I±KI 2

~

~4x2 -12x + 5x -15 - ~x (x -

3) =

:::) ~4x (x - 3) + 5(x - 3) - ~x (x - 3) :::) ~(x-3)(4x+5) -~x(x-3)

=>

~(x - 3)(x + 3)

~(x-3)(x+3)

~x-3)[~4x+5)-J;-~x+3J =0 ~x-3

either or

=

=

~(x - 3)(x + 3)

=

0,x=3

~4x+5-J;-~x+3. =0

:::)

~4x+5 -J; = ~x+3

squaring both sides.

4x+5+x-2J; ~4x+5

=

2+3.

4x+5-3 = 2J; ~4x + 5 :::)

:::) :::) :::) :::) :::)

4x+2 = 2J; ~4x + 5

2x+l=J;~4x+5 2 4x + 1 +4x = x(4x-15) 4x2 + 1 + 4x = 4x2 + 5x 5x-4x = 1 x =1

ANSWERS 1 3 I±KI . , 2

2. 1, 1. 5

3. -2, -4

4. 3

5. 2, 1. 2

5 6. 3, - ,.,

7. O,a

8.2,-5

9

-'

Type V: Reciprocal Equations 1

An equation which remains unchanged when x is changed to - is called reciprocal equation.

-

x

Quadratic Equations

31

Remarks 1. The roots of a reciprocal equations occurs in pairs. 2. If a is a root of reciprocal equation, then

1..a

also a root of the given equation.

To solve such type of reciprocal equation, we use the following steps :

STEP KNOWLEDGE (A) For Even Degree (say, degree = 4)

(I)

Divide both side by x?-

l I (ii) Put x + - or x - -

x

x

=

y and solve for y.

.

1

1

(iii) Finally obtained the value ofx by putting x + - = y or x - - = y. x x (B) For Odd Degree (i) If the coefficient of the terms equidistant from the starting and end are equal in magnitude as well as in sign. Then by inspection, we have that - 1 is a root and then taking (x + I) as a common factor and get even degree equation. (ii) If the coefficients of the terms equidistant from the starting and end aye equal in magnitu~e but opposite in sign. Then, by inspection take x = 1 is awot'and take (x - I) as a common factor and get even degree equation, which can be easily solved.

~~~~~~~ISOLVED

EXAMPLES

I~~~~~~~

Example 1: Solve x4 - x 3 + 2J? - x + 1 =0 Solution: Here, the given equation is x4 -X3 + 2x? -x + 1 = 0 Divide throughout by x?-, we get

x2

_

x + 2 _1.. + _1_ = 0 x x2 ...(1)

2 Put (x+';) =yi.e. x +

=> => Now, if

x~

=1-2in(l),weget

1-2-y+2 = 0 l-y=o y(y-I) = 0 y = 0,1 1 Y = o=>x+- =0 x

32


1 y=l=>x+-=1

or

x

x2 +1 =x ~-x+1 =0 x

=

l±i.J3 2

· set 0 f th ' . .IS {±l,. 1± 2i .J3} Hence, the so IutlOn e gIven equatIon Example 1: Solve (x

r-

+';

Solution: We have ( x + .; =>

%(x -.;)

y-%(

=

4, x

~O

[UPTU B. Pharma 2004)

x - .;) = 4

[(x-.;y +4]-%(X-';) =4 2 3 Y +4-- Y =4 2

i-%Y = O=>Y(Y-%)=O 3 Y = 0 or Y - - =0 2 3 Y = 0 or y=Now

y=O

2

x-!

= 0 x x2 -1 =0 x2 = 1 X

Again

=> => => =>

=± 1.

3

1

3

y=-=>x--=2 x 2 2x2_2 = 3x ~ = 3x-2 2x(x-2) + 1 (x-2) = 0 (x-2)(2x+ 1) = 0 x-2=00r2x+1=0 1 x = 2 or x = - 2

Hence, The solution set of the given equation is

{I, -I,

2,

-~}.

Quadratic Equations

33

f-

Example 2: Solve (x + .;

2 ( x - .; + 4 ) - /1 = 0


(X+;-f -2(X-;-+4)-11 Put x -

~ = Y i e. ( x + ;-

f

=

=

... (1 )

0

(x - ;-

y

+ 4=l +4

Then, equation(l) becomes ul+4)-2 (y+4)-11 = 0 or l-2y-15 = 0

=> Now, if

=>

(y-5)(y+3) =0 y = -3,5 1

y=5=>x--=5 x .x2-5x-l = 0 x= x

orif

.x2 + 3x-l x

5±m 2

1 x

= -3, then x - - =-3 = 0

=

- 3±

J9 + 4 = --"-- 3 ± v'i3

2

2

· set 0 f th ' . .IS Hence, the so1utlOn e given equatIOn

Solve the fol/owing equations: 1. 2x4 -x3 -11.x2-x+2=0

{5±m -3±v'i3} 2'

2

2. ~ -3.x2-3x+2 =0

r- x-;-

(x + ;-

3. 4x4 -4~ -7.x2-4x+4 = 0

4.

5. 2x4_~+ 14.x2-9x+2 =0 7. x 4-4x3 -3.x2-4x+ 1 =0

6. x4+1-3(~+x)=2.x2 8. x 5 - 4x4 + ~ + .x2 - 4x + 1 = 0

9. (x+;-f -%(x-;-)-4 =0 11. x 6 - x 5 + x4 -.x2 - x-I

=

0

.

6(

+ 1) - 5

=

0

34 Remedial Mathematics HINTS TO THE SELECTED PROBLEMS 1. 2x4-x3_11~-x+2=0 divide by~, we get

2x2 - x-II

_l + ~2 x

=0

x

~

2 (x2 + xI2) - ( x + ) - 11 = 0 2

=> 2 ( x + => 2(x2 Let

:2 +

2) - 4 - ( x +

~ ) - 11 = 0

+~J -(X+~)-I5 =0 (x+~) 21-y-I5 21-6y+5y-15 2y(y-3)+5(y-3) (y-3)(2y+ 5) Y- 3

=y =0 =0 =0 = 0 = 0 or 2y + 5 = 0

1 5 x + - = 3 or y = - x 2

~ + 1 = 3x

or x + 1 x

= _1 2

2

5 x +1 x 2 - 3x + 1 = 0 or - - = -x 2 X

=

+ 3 ± ~9 - 4 x 1 x l ? 2xI tj

± 15

= --

2

2x- + 2 = - 5x

or z.x2+5x+2

=

0

2~+4x+x+2 = 0

2x(x+ 12)+ 1(x+2)

=

0

1 x=-2x=-, 2

Quadratic Equations

35

~(X2+ X~-2}-6(X-~)-7=0

~ (x-~f -6(X-~)-7 =0 1 x

x--

Let

6.

=y

~ y-6y-7 = x4+ 1-3 (x 3 +x)=al x4+ l-al-3x3 -3x =

0 0

Divide by x?, we get.

x 2 + _I _ 2 - 3 x x2

~

(x2 + ~2

_1..

2) - 3 ( x +

-

= 0

X

~)

=

0

~(x2+ ~2+2)-3(X+~)-4=0

~(x+~f -3(X+~)-4

=

...(1)

0

x+-1 =y

~Let

x

Then eqn (I) reduced to

8.

x 4x4 + x 5-

3+

y-3y-4 = 0 x2- 4x + 1 = 0

putting x = 1 ~ (-1)5-4 (_1)4 + (- 1)3 + (_1)1 -4 x (-I) + 1

=-1-4-1+1+4+1=0 so (x + 1) is a factor of given equation. (x+ l)(x4-5x3 +6x?-5x+ 1)= 0

5 1)

(x+l) ( x2.-5x+6--+-2 =0 x

x

2 (X+l)[(X + ~2 )-5(X+~)+6] =0

~ (x+l)[(x+~r -2-5(X+~)+6l =0 ~ (x+l)[(x+~f -5(X+~)+4l =0 ~ Consider { x+~f

- (x +~) 5

+4

= 0

36 Remedial Mathematics 1

x+- = y

Take

x y-5y+4 = 0

On solving we get required Answers.

ANSWERS 1. 4.

7. 10.

III

_1-

2'

-2

,

1 ' '2

3±J5 2

7±J53 1 ±J5 2

2

5 ±51 -±F3 , 2 6

1-

2

-3±R

2. -1 2 -

3.

1 5. 1,2,2

6. 2 ± .)3,

8. -1,

1±.,F3 ,2 ±.)3 2

2"

4

t

(l ± -)'- 3) .

1 9. -1--12 ' 2"

1±.,F3 11. +. _I, 2

3 ±J5 1_2 , , 2

NATURE OF ROOTS OF A QUADRATIC EQUATION

The roots of the quadratic equation ax 2 + bx + c = 0 are

-b ±~b2 -4ac 2a Here, the expression D = b2 - 4ac is called discriminant. The nature of the roots, depend upon the value of D as given below: (a) If b2 - 4ac ~ 0, then roots are real (I) If b2 - 4ac > 0, then roots are real and distinct (ii) If b2 - 4ac = 0, then the roots of the equation are real and equal.

-b+O

... (5)

-b

In this case, each root = ---- = 2a 2a (iii) Also if b2 - 4ac is a perfect square, then the roots are rational and in case it can't be a perfect square, then the roots are irrational.

(b) Ifb 2 -4ac O,x-b>Oi.e.x>a,x>b Therefore, x > b (:. a> b) ...(1) Therefore, x b. In other words, we mean that x does not lie between a and b (a < b). (x-a)(x-b) = - ve CaseU It is possible if one factor is positive and the other is negative. Let x-a = +ve>O,xb=-ve a, x < b or a < x < b i. e. x lies between a and b (a < b) or x-a = -veO i.e. x < a and x> which is not possible. Therefore, (x - a) (x - b) = Positive if x does not lie between a and b and is negative if x lies between a and b. For Example: Consider the expression. (x+3)(x-5) = [x-(-3)] (x-5)

a =-3,b=5=:::>a O. 2 . ( b)2 4ac _ b . Hence, the expressIOn x + + 2 is positlve for all real values of x. 2a 4a Therefore, ax2 + bx - c has same sign for all real value ofx. Case II Let the roots of the equations (1) are real and distinct, denoted by a and ~. Let a > ~ Then we have the identity. ...(3) If~ <x < a, thenx-a > 0 andx- ~ < 0 so that (x-a) (x-~) < O. It follows that the sign ofax 2 - bx + c is opposite to that of a Ifx> a or x 0 since the factor (x-a) and (x-~) are either both positive or both negative. Hence, in this case the sign ofax2 + bx + c is the same as that of a. Case ill Let the roots a, b be equal. Then ax2 + bx + c = (x - a)2 and (x - a)2 is positive for all real values of x and therefore ax 2 + bx + c has the same sign as a.

Remark • From, above three cases, we conclude that, for all real values of x, the expression ax2 + bx + c has the same sign as a except when the roots ofthe equation ax2 + bx + c = 0 are real and unequal, and x has a value lying between them.

III

RELATION BETWEEN ROOTS AND COEFFICIENTS

Let us consider the quadratic equation ax2 + bx + c = 0, a 0, a, b, c E R To find the sum and product ofthe roots in terms ofthe coefficients a, b, and c. Consider the quadratic equation ax 2 +bx+c = O,(a*O) If a, ~ be the roots, then by the theory of equation, we have

*

a

=

-b+~b2-4ac 2a

and ~ =

-b-~b2-4ac --...!.---2a

... (1)

Quadratic Equations

39

(i) The sum of the roots

=a+p=

J

2 - b + b - 4ac 2a

+

J

- b + b 2 - 4ac

b =-a

2a

(ii) The product of the roots

~ a~{b+J:: -4ac)( -b- J:: -4ac 1 (_b)2 -(Jb 2 -4ac)2 4a Hence, we have sum of the roots =

b

2

-- =a

2

2

b - b + 4ac

4a

2

c a

= -

coefficient of x 2 coefficentofx

c Constant term 2 and product of the roots = - = a Coefficent of x

III

FORMATION OF EQUATIONS

To find the equation whose roots are a, 13: Let the equation be ax2 +bx+c = O,a:;eO Then by theory of equation, a+ (3 =

Sum of roots

... (1)

b

a

c Product of the roots a(3 = a Now equation (1) can be written as 2 b c x +-x+- = 0

a

=> =>

a

.x2-(a+(3)x+a(3 = 0 or x(x-a)-(3(x-a)=O (x-a)(x-(3) =0

Remark • Let S = a + (3, P = a(3, then the required equation is x 2 - Sx + P = O.

III

COMMON ROOTS

(l) Condition for One Common Root Consider the two quadratic equations, such that ax 2 +bx+c = 0 and a'x2 + b'x + c' = 0 Let a be the common root then equations (1) and (2) gives aa2 + ba+c = 0

a'a 2 + b'a + c' = 0

... (1) ... (2)

...(3) ... (4)

40


Solving (3) and (4), we get

a2

a

be' - b'c

cd - e'a

=

... (5)

ab' - db

\

Now taking first two members, we get ,

a

, :::>a=

ea -ea

be' - b'e

be' -b'e cd -e'a

...(6)

Taking last two members, we get

cd - e'a ex = ab' - db Now (5) and (6) gives.

be' - b'e cd - e'a

:::>

(ab' - d b) (be' c)

ea' -e'a ab' - db =

(cd - e'ai

which is the required condition for one common root. (il) Condition for Both Roots Common Here, the given equations are ax2 +bx+e = 0

...(1)

dx 2

and + b'x + e' = 0 Let ex, ~ be the common roots, then from (1), we have

•

...(2)

b a

ex+~ = - -

e a

ex~ = -

and From (2), we have

...(3)

ex+~

...(4)

b' a'

= --

...(5)

and

...(6)

Now (3) and (5) gives

b a

b' d

b a

b' a' -

a a'

b b'

=--:::>-=-:::>-=-

...(7)

Equation (4) and (5) gives

e e' a e =-:::>-=add e' Combining (7) and (8), we have

a d

b b'

e

e'

which is the required condition for both common roots

...(8)

Quadratic Equations

~~~~~~ISOLVED

41

I~~~~~~~

EXAMPLES

Example 1: !fa, 13 are the roots ofaX! + bx + c = O,findthefollowing: ~

1 1 (a) ao. + b + a~ + b (e) (ao. + br3 + (a~ + br3

a

(b) ao. + b +

a~ + b

(d) (ao. + br2 + (a~ + br2

Solution: Since a, ~ are the roots of d + bx + e = 0

e a

ao.2 + bo. + e = 0 => ao. + b + -

Then,

b

e

o.+~ = --;o.~ = a a

also, 1

1

~

a

- - + - - = ---ao. + b a~ + b e e

(a)

= --1 (e

b) b

ao. + b

a~

ae

= _ o.~ _ o.~ +bee 3

a + 13 3 e

(ao.+~r3+(a~+br3 = -

1

e a

a

3

[(a + 13)

c3

(ao.+br2+(o.~+br2 =

_~ . .:. =_~

=

3

= - -

(d)

=-

a

_~_+_o._

(b) (e)

1 e

= --(o.+~)

- 3a13 (a + 13)] =

3 b - 3abe 3 3

a e

+ ~2

0.2

e

2

1

= "2 [(a +~)

2

e

- 2o.~]

=

b 2 - 2ae 2 2

a e

Example 2: !fa and 13 are the roots ofaX! + bx + e = O. Find the equation whose roofs are as given below. I

I

(/) a + ~ , a

I

(ii) a

+"i3

2

+~

2

I

'-2 +

a

I

2" ~

Solution: Let a,~ be the roots ofthe given quadratic equation ax2 +bx+e = 0

Then, we have

-b

e

a

a

a+~ =-ando.~ = -

o.+~ a b (ae+b 2 ) 1 Sum = - - + - - = - - - - = - -'-------'a +~ a~ b e be

42 Remedial Mathematics 1 o,+p 1 a Product = a + P . ~ = a P = -;;Now, consider the equation 2

x 2 -xS + P = 0 => x 2 - x (-a c + b ) bc bc . x 2 + (b 2 + ac)x+ ab =

=>

(ii) S = a

2

+ -a = 0 c

0

2 1 1 + P + -~ + 2" o,~

P

_ (2

A2) (0,2 + p2) _ (b 2 - 2ac)2 22 22

p-o,+~.

ex,

pac

2

x - Sx + P = 0 gives ~c2~ - (b 2 - 2ac) (~+ c2 ) x + (b 2 - 2aci = 0 Example 3: If a, fJ are the roots o/x2 - p (x + J) - c = 0, show that (a + J) (fJ + J) = J - c.

Therefore,

Hence show that

a 2 + 2a + 1 fJ2 + 2fJ + 1 2 2 + fJ2 2fJ = J a + a+c + +c

Solution: Here, the given equation is x 2 - P (x + 1) - c = 0 Therefore, 0,+ P = p, and o,p = - (p + c) Now (O,+l)(p+l) = O,p+(o'+p)+1 =

Also,

... (1)

-rt-c+p+ 1 = l-c

0,2 + 20, + 1

p2 + 2P + 1

0,2 + 20, + C

p2 + 2P + c

... (2)

---::----+-'-::---'---

(0,+1)2

(P+l)2

(o'+l)2-0-c)

(p+l)2-0-c)

---'--::---':--- +

[using (1)]

a +1 P+1 (a + 1) - (P + 1) = --+--= =1 O,-p p-O, O,-p

Example 4: If a be a root a/the equation 4x2 + 2x - J = 0, prove that 4 cJ - 3 a is the other root. (UPTU B. Pharma 20021 Solution: Given a be a root ofthe equation 4x2 + 2x - 1 = o. Then Let

40,2+20,-1 = 0

p be the other root of the given equation

Then

1 1 0,+ P = - => p=---O, 2 2_

... (1)

Quadratic: Equations

43

We have to show 3

1

2

1

2

Now4a -3a=a(4a +2a-I)- -(4a +2a-I)-a--

2

2

1 1 =a·O--O-a·-

2

2

1

= -a--=~.

Hence 4a3 -

2 3a is the other root of the gives equation.

Example 5: Two students solve an equation. In solving, one commits a mistake in constant term andfind the roots 8 and 2. Other commits a mistake in the coefficient ofx andfind the roots - 9 and - 1. Find the correct roots.

Solution: Let the correct equation be x 2 +ax+b = 0

...(1)

Roots found by first student are 8, 2 i. e. S = 10, P = 16 :. Equation is

~-IOx+ 16 = 0

...(2)

Since, the committed mistake only in constant term :. a = - 10 Roots found by second student are - 9, -1, i. e. S = - 10, P = 9 Therefore equation is ~ + lOx + 9 = 0 Since the committed mistake is in the coefficient ofx. :. b = 9 Having found a = - 10, b = 9, the required equation is x 2 -IOx+9

=

0 or (x-9)(x-l)=0

1,9 Hence, the correct roots are 1 and 9. Example 6: If a, fl be the roots ofax2 + bx + c = 0 and y, 8those oflx2 + mx + n = 0, then find the equation whose roots are ar + b8 and ao + flr Solution: Here the given equations are ax 2 +bx+c = 0 ...(1) and ...(2) i.e.

Now, we have Now,

and Now,

x

=

a+~

=

bam

a2+~2 = (a+~)2-2a~= .2

2

n

-~,a~=~'Y+O=-T'YO=1 b2

-

2ac

-""'7"-

a

2

2

r +0 =

m - 2nl

P

S = (ay+~o)(ao+By)=a(y+o)+~(y+o) bm =(a+ ~)(y+o)=al

44


p

and

= (ay+~o)(ao+~y)

= a 2yo + a~o2 + a~r + ~2yO = (a2+~2yo+a~(r+02)

=

T(b

2

2aC :2 )

+ ; (m2

~ 2nl )

nl (b 2 - 2ac) + ac (m 2 - 2nl) a

2

z2

b 2 nl + m 2ac - 4acnl

a2 12 Therefore the required equation is given by x?-xS+ P = 0 2 2 (bm) b nl + m ac - 4acnl + =0 al a 212 => ~t2x? - xalbm + b2nl + m2ac - 4acnl = 0 Example 7: Jfp and q be the roots of2x2 - 6x + 3 (p3 + q3) _ 3pq (p2 + q2) _ 3pq (p + q)

=>

2

x -x -

= 0, find the value of

Solution: Since p and q are roots of2x? - 6x + 3 = 0 Therefore, sum of the roots p + q = and product of the roots pq =

-6 -2 =3

3

2"

Now, (p3 + q\_ 3pq (p2 + q2)- 3pq (p + q) = (p + q)3_ 3pq (p + q)- 3pq [(P + q)2 - 2pq]- 3pq (p + q) = 27-3.%(3)-3.%[9-2.%]-3.%.3 =

27 27 27 ---27 - - =-27

2

2

Example 8: Solve the equation x 2 + px + 45 = 0, given the squared differences ofits roots is equal to 144. Solution: Here, the equation x? + px + 45 = 0 ...(i) Let a, ~ be its roots such that a > ~ .. a + ~ = - p, ex.~ = 45 From the given condition (ex. - ~)2 = 144 .. (ex. + ~)2-4ex.~ = 144 ~ p2-4.45 = 144 => p2 = 324

=>

p = ± 18

Quadratic Equations

45

When p = 18, equation (1) becomes x 2 + 18x + 45 = 0

=> =>

(x+3)(x+ 15) = 0 x =-3,-15 When p = - 18, equation (1) becomes x 2-18x+45 = 0 => (x-3)(x-15) = 0 => x = 3,15 Hence, the roots of the given equation is 3 and 15 or - 3 and - 15. 1 1 1 Example 9: If the sum of the roots ofthe equation x + a + x + b = ~ is zero. Show that the product of roots is

-"21(a2+2 b ).


1 x+a

1 x+b

--+--

1 c

=-

=> =>

c(x+b)+c(x+a) = (x+a)(x+b) a+~+a+w=~+~+fu+~ =>x2 + (a + b -2c)x + (ab - bc-ca) = 0 Given that, the sum of its roots is equal to zero. _ (a + b - 2c) = 0 1 2c-a-b = 0

or

a+b c= - 2 Product of roots =

=

... (1)

ab -bc -ca 1

=ab-c(a+b)

ab_(a~b») ~+x2+ 1 +2x = 61 => ~+2x-60 = 0

=> => =>

(x + 6)(x - 5)

=

0

x = -6, or x= 5 x is a whole number

x = 5 And required numbers are x and x + 1 as 5 and 5 + 1 = 6. Example 3: The sum ofthe squares oftwo consecutive odd integer is 394. Find the integers. Solution: Let the required integers be (2x + 1) and (2x + 3) (2x+ Ii + (2x +3)2 = 394 => ~+ 16x-384 = 0 => x2+2x-48 = 0

=> => => => =>

~+8x-6x-48=O

x (x + 8) - 6(x + 8) = 0 (x+8)(x-6) = 0 x + 8 = 0 or x - 6 = 0 x =-8 or x=6. When x = - 8, the required integers are - 15 and - 13 when x = 6, the required integers are 13 and 15 Hence, the required integers are {-15, - 13} or {13, 15}. Example 4: The sum of two natural numbers in 8. find the two numbers. Solution: Let the numbers be x and (8 - x).

If the sum of their reciprocals in

8

15"


1 8-x

8 15

8-x+x x (8 - x)

8 15

1

-+-X

8

8

~8x - x 2

15 64x - 8x2 0

•

120 = 8x2-64x+ 120 = x2-8x+ 15 = 0 (x-5)(x-3) = 0 x = 5 or x=3 when x = 5, the Numbers are x and 8 - x = 5 and 3 when x = 3, The numbers are x and 8 - x = 3 and 5 i.e,

:. Required numbers are 5 and 3. Example 5: A number consists of two digits whose product in 18. When 27 in subtracted from the number, the digits interchange their places. Find the numbers.

Solution: Let the tens digit be x. Then the units digit = Number formed =

(lOX +

18

-

x

1:)

(lOX 1: + x ) (lOx + I:) _27 (10 x1: + x) 162 9x- --27 = 0 x

Number formed on reversing the digits =

.. => => => => => => =>

=

9x2-27x-162 = 0 x2-3x-18 = 0 x2 - 6x + 3x - 18 = 0 x (x - 6) + 3 (x - 6) = 0 (x-6)(x+ 3) = 0 x = 6 or x = - 3. [ Since a digit can never be negative] x =6

Thus we have the tens digit

=

6, the unit digit =

18 (; =

3

Hence, the required number = 63. Example 6: Two positive numbers are in the ratio 2:5. JfdifJerence between the squares of these numbers is 189. Find the numbers.

Quadratic Equations 51 Solution: Let the numbers be 2x and 5x. .. (5xi-(2x)2 = 189 =:> =:>

25~-4~

=

189

21~ = 189

189 x2 = =9 21 x = ±3 Since the required numbers are positive. :. x = 3. And required numbers = 2x and 5x = 2 x 3 and 5 x 3 =6andI5. Example 7: One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his son s age. Find their present age. Solution: Let the son's age one year ago be x years. Then, the man's age one year ago = (8x) years. Present age of the son = (x + 1) years. Present age of the man = (8x + 1) years. (8x+ 1) = (x+ 1)2 =:> ~-6x = 0 =:> x = 0 or x = 6 x =6 [Son's age can not be zero] Present age of the son = (x + 1) years = 7 years. and the present age of the man = (8x + 1) years = 49 years. Example 8: The sides (in cm) of a right triangle are x - 1, x and x + 1. Find the sides of triangle. Solution: It is clear that the largest side x + 1 is hypotenuse of the right triangle. According to Pythagoras Theorem, we have ~ + (x - 1i = (x + 1)2 =:> ~+~-2x+ 1 = x2+2x+ 1 x 2 -4x = 0 x+1 x(x-4) = 0 x i.e., x = 0 or x=4 ;, Since with x = 0 the triangle is not possible, hence x = 4 x-1 x-I, x and x + 1 = 4 - 1,4 and 4 + 1 sides are Fig. 1.1 i.e., 3 em, 4 cm and 5 cm. Example 9: Rs 250. was divided equally among a certain number of children. If there were 25 more children, each would have received 50 paise less. Find the number of children. Solution: Let the number of required children be x. 25000 The share of each = - - paise x


25000J

It there were (x + 25) children, share of each = ( x + 25

..

25000

paise.

25000

- - - - - =50 x x+ 25

=>

I ---- x x+25 500

=>

x+ 25-x x (x + 25)

500 x (x + 25) = 12500 => ~+25x-12500 = 0 => (x+ 125)(x-100) = 0 => x = -125 or x= 100 => => x = 100 [since number of children can not be negative] Hence, the number of children = 100. Example 10: The Hypotenuse ofright triangle is J m less then twice the shortest side.lfthe third side is J m more than the shortest side, find the sides of the triangle. Solution: Let the shortest side be x m. .. Hypotenuse = (2x-I) m and third side = (x + l)m. Applying Pythagoras Theorem, we get (2x-li =x2 +(x+li => 4~-4x+ 1 = x 2 +x2 +2x+ 1 x 2x2 -6x = 0 i.e., x 2 -3x = 0 => x(x-3) = 0 x+1 => Fig. 1.2 x = 0 or x=3 => Since x = 0 makes the triangle impossible. There fore x = 3 and sides of the triangle are = x, 2x - 1 and x + I = 3,2 x 3-1,and3+1 = 3m, 5 mand I m Example 11: A passenger train takes 3 hours less for journey of 360 km if its speed is Increased by 10 kmlhr. What is the usual speed? Solution: Let the usual speed of the train be x km/hr . k en to cover 360 km atxkm/hr= -360 hr. Tlmeta x New speed = (x + 10) km/hr 360 Tune taken to hour 360 km at (x + 10) km/m = ) hr (x + 10 360 360 - - - - =3 x x + 10

Quadratic Equations

x + 10

x

53

20

x+lO-x x (x +10)

120

x(x+ 10) = 1200 => ~+ IOx-1200 = 0 => .x2+40x-30x-1200 = 0 => x(x+40)-30(x+40) = 0 => (~+40)(x-30) = 0 => x = -40 or x=30. => [ .: speed cannot be negative] => x=30 Hence, the usual speed of the train is 30 kmlhr. Example 12: A plane left 30 minutes later than the scheduled time and in order to reach its destination 1500 km awcry in time, it has to increase its speed by 250 kml from its usual speed Find its usual speed Solution: Let the usual speed to the plane = x km/hr => The increased speed of the plane = (x + 250) km/hr usual time taken by the plane to cover 1500 km =

1500 --

x

hrs

1500 and the new time taken to cover 1500 km = x + 250 hrs From the given statement, it is clear that the new time taken is 30 minutes i.e., less than the usual time.

i.e.,

=>

1500

1500

x

x+250

2

1500 (x + 250) -1500x x (x + 250)

2

1500 x + 375000 -1500x x 2 + 250 x 375000 2

x + 250x

2 1

=2

On cross mUltiplying we get.x2 + 250x = 750000. .x2+250x-750000 = 0 On factorising we get (x + 1000) (x-750) = 0 x = -10000rx-750

=>

Rejecting the negative values of x, we get x = 750 The usual speed of the plane = 750km/hr

"21 hrs


~~~~~~~I

EXERCISE 1.8

~I~~~~~~

1. Two numbers differ by 3 and their product in 504. Find the numbers. 2. The sum of two numbers in 18 and their product in 56. Find the numbers.

3. The sum of two numbers in 15 and the sum of their reciprocals in 4. 5.

6. 7.

8. 9.

10.

11.

12. 13.

3

10. Find the

numbers. The sum of the squares ofthree consecutive positive integers is 50. Find the integers. Find two consecutive even integers whose squares have the sum 340. Find two consecutive positive odd integers whose squares have the sum 290. A two digit numbers is 5 times the sum of its digits and is also equal to 5 more then twice the product of its digits. Find the number. A two digit number is such that product of the digits in 35. When 18 is added to this number the digits interchange their places. Determine the number. The sides (in cm ) of a right triangle containing the right angle are 5x and 3x - 1. If the area of the triangle is 60 cm2, find the sides of the triangle. The Hypotenuse of a right angle triangle is 6 meters more than thrice the shortest side. If the third side is 2 meters less than the hypotenuse, find the sides of the triangle. The area ofa right angle triangle is 600 sq. cm. If the base of the triangle exceeds the altitude by 10 cm, find the dimensions of the triangle. If the perimeter of a rectangular plot is 68 m and its diagonal is 26 m, find its area. A fast train takes 3 hours less than a slow train for ajourney of600 km. If the speed of the slow train is 10 kmlhr less than that of the fast train, find the speeds of the two trains.

HINTS TO THE ,SELECTED PROBLEMS 1. Let the required no. are x and x - 3 Their product = x (x - 3) x(x-3) =504 On solving, we get x = 24 thenx- 3 = -21 and x =- 21 thenx- 3 =- 24 3. Let the required numbers be x and (15 - x) 1 3 =15-x 10 x = 10 and 5. on solving we get 6. Let the required two consecutive positive odd integers be x and x + 2. Then x2 + (x + 2)2 = 290 7. Let the numbers is (10 x + y) Then (lOx+y) =5(x+y) 10x+y =2xy+ 5 and Then

1 x

-+--

...(1) ...(2)

Quadratic Equations

55

From (1), we get 5 x=4y 4 x= -y 5

substituting

x=

4 sy in (2), we get

(lOX 4;)+ Y = (2X 4; xy )+5 8Y-45y+25 =0 Y =5 or

since Hence Then 9. It is clear from the figure,

5 8

Y "# 5/8

Y =5 x=4

1

-x5xx(3x-l) =60 2

3x2 -x-24 =0

8 x= -- or-3 3 since, sides of the triangle can not be negative x = 3 and sides =5xand3x-l = 3 x 5 and 3 x 3 -1 = 15 cm and 6 cm. Applying Pythagoras Theorem, we get

Ii

5x

Fig. 1.3

= 152 +82 =289=(l7i

H = 17cm.

10. Let the length of the shortest side be x meters Then, the hypotenuse = (2 x + 6) metres. and, the third side = (2x+4) By Pythagoras theorem we have (2x+6i = x 2 + (2x+4i => x2- 8x-20 = 0 x = 10 or x=-2 x = 10 [since side ofa triangle can never negative] Shortest side = 10 m Hypotenuse = 2 x + 6 = 26 m third side = 24 m 13. Let the speed of the train = x Kmlhr and speed of the slow train = (x-l0) Kmlhr.


Time taken by fast train to cover 600 Km = 600 hrs

x

Time taken by slow train to cover 600 Km = 600 hrs x-l0 600 _ 600 = 3 x-l0 x 600x - 600(x -10) x(x-l0) =3

=> =>

x'l-10x-2000 x Neglecting x Speed to fast train speed of slow train

0 = 50 or x=-40 = -40, we getx = 50 andx-lO = 50-10 =40 = 50 Km/hr = 40 KmIhr.

=

ANSWERS 1. 5. 9. 11. 13.

(24 and21)or (-21 and-24) 12 and 14 6. 11 and 13 15 em, 8 em and 17 em Altitude = 30 em, base = 40 em 50 kmlhr and 40 kmlhr.

2. 7. 10. 12.

4 and 14 3. 10 and 5 45 8. 57 10m,26mand24m 240m2

OBJECTIVE EVALUATION MULTIPLE CHOICE QUESTIONS (Choose the appropriate answer) 1. Which of the following is linear equation (a)

b =0 x

ax2 + bx + C = 0

(b) ax+ -

(c) ax+2=0 (d) None of these 2. Which of the following are quadratic equation

1 x (c) 2x+3=0 3. The degree of the equation 2x - 3 = 0 is (a) 1 (b) 2 4. Solution of the equation 2x - 3 = 0 is (a)

(a)

x-- =0

2 5

-

3 (b) -

2

(b) x 3 + 3x2 + x-I = 0 (d) None of these (c) 3

(d) None ofthese

(c) 1

(d) 0

Quadratic Equations

57

5. The root of the quadratic equation 2.x2 - 3 x-I = 0 is (a) 2 (b) 1 (c) 3 ,6. The roots of the equation 3x2 - 4x -7 = 0 is

7 (a)

3

'30rl

(d) None of these

7 (c) '3 or - 1

(b) '70r2

(d) None of these

*

7. Ifax2 + bx + c = 0 (a 0) is a quadratic equation and b2 - 4 ac = 0 then the roots are (a) real and distinct (b) real and equal (c) imaginary (d) None of these 8. Solution of the equation 16x2 = 25 is 3 4 (b) ±~ (a) ±(c) 4 4 5 9. lithe discriminant ofax2 + bx + c = 0 (a 0) is greater then zero then root are (a) real and distinct (b) real and equal (c) imaginary (d) None of these 10. The root of the equation x 2 + ax + b = 0 are equal if (a) a2 =3b (b) a2 =4b (c) a2 =b (d) None of these 2 11. Solution of the equation x + 16x + 60 = 0 is (a) -IOor6 (b) lOor-6 (c) -lOor-6 (d) None of these 12. The sum and product of the roots of the equation ax2 + bx + c = 0 is

*

(a)

be a a

be a a

be a, a

(b) --, - -

-, -

(c) - - -

(d) None of these

13. The sum ofthe roots of the equation 6x 2 + x - 2 = 0 is (a)

1

"6

(b)

1

-"6

(c)

5

4"

14. If a, 13 are the roots of the equation d + bx + c (a

3abc - c

3

3abc - b3 (b)

(a)

15. The equation whose roots are (a)

x 2 + 4.J3x + 9

(c)

x2-4.J3x-9 =0

=

(c)

d

J3

and

(d) 6

* 0), then the value of a 3 + 13 3 is

3abc - a 3 (d) None of these

3J3 is (b) x 2 - 4.J3x + 9 = 0

0

(d) none ofthese

16. The quadratic equation whose roots are 4 and 5 is (a) x2-9x-20=0 (b) x2-9x+20-0

..z

(c) + 9x + 20 = 0 (d) none of these 17. lithe sum and product ofthe roots ofthe equation d - 5x + c = 0 is 10 then the value of a and cis

1

(a)

a=

1

2'c='5

(b) a

18. The roots of the equation

G>;2

= 5. c =

'21

+ bx + c, a

(c) a =

1 2

-,C

=5

* 0 is

(a)

b+ Jb 2 -4ac 2a

(b)

-b± Jb 2 +4ac 2a

(c)

-b± Jb 2 -4ac 2a

(d) None of these

(d) None of these

58


19. The discriminant of the equation 5~ + I6x + 3 = 0 is (a) 195 (b) 196 (c) 197 20. The discriminant of the equation 3J7x2 + 4x (a)

98

(b) 99

(d) none of these

J7 = 0 is

(c) 100

(d) 1001

TRUE I FALSE (Write T for True and F for False) 1. The solution of the linear equation 3x + 6 = 0 is - 2 2. The solution of the equation x 3. The degree of the equation x -

~

= 0 is ± 1.

(T/F)

- 3 = 0 is.

(T/F)

x

~ 2x

(T/F)

4. The sum of the roots of the equation 6x 2 + x - 20 = 0 is

.! . 6

(TIP)

5. The roots of the equation x 2 - 5x + 6 = 0 is 2,3.

(T/F)

. 6 • ax + -b - 2 = O·IS a I'mear equatIOn.

(T/F)

. x2 7. x=-5Isarootof -+2x-1O =0

(T/F)

3x

-5

8. The discriminant of (x - I) (x + 2) = 0 is 9. 9. If the roots of the equations x 2 + 2x + ab = 0 are real and unequal then the equation. x 2 - 2 (z + b) x + a 2 + b2 + 2c 2 = 0 has no real roots. 10. The roots of J2x 2 - 5x - 3 = 0 will be equal. 11. For a quadratic equation exist. 12.

J2.x + ~ =.J3 x

ax2 + bx

+ c = 0, a*,O it

(T/F)

(TIP) (T/F)

b2 -

4 ac < 0 the two distinct real roots will (T/F)

is not a quadratic equations.

(TIF)

13. The roots of the quadratic equation ax2 + bx + c = 0, a*,O can be found by using the quadratic

+ b ± ~b2 - 4ac

. 2 , provided b - 4ac ~ O. 2a 14. The equations (x - 1)3 = x 3 - 2x + 1 is not a quadratic equation.

formula

15. _1- + _1_ = 3 has two distinct real roots. x-I 2x+ 1

(T/F) (T/F)

(TIF)

FILL IN THE BLANK 1. If x = 1 is a common root of the equations px 2 + px + 3 = 0 and x 2 + x + q = 0 the pq=---

2. [;iscriminate for x +

3. - I = 0 is _ __

x 3. If the roots of the equations 4x 2 - Kx + 9 = 0 are not real this K should be _ __ 4. The roots of a quadratic equations can be found by using the method of _ __ 5. If a,

p, are roots of the equations x2 -

2x + I = O. then a -

Pis _ _ _ .

Quadratic Equations

59

ANSWERS MULTIPLE CHOICE QUESTIONS 1. (c) 5. (b) 9. (a) 13. (b) 17. (c)

3. 7.

(b) (b)

H.

(c)

15. 19.

(b) (b)

2. (a) 6. (c) 10. (b) 14. (b) 18. (c)

2. 6. 10. 14.

3. False 7. False H. False 15. True

4. 8. 12. 16. 20.

(b) (b) (c)

(b) (c)

TRUE/FALSE 1. 5. 9. 13.

True True True False

True False False False

4. False. 8. True. 12. False.

FILL IN THE BLANK 2. -7

1. 3

3. -6J2 O. Can we do? (Frequently Asked Questions)

1. Solve (2X+I)4 _IO(2X+I)2 +9 =0. x-I

x-I

[UPTUB.Pharma2005j

60


[UPTU B.Pharma 2003]

3. If a. be a root of the equation 4~ + 2x + 1 = 0, prove that 4a.3 - 3a. is the other root. [UPTU. B Pharma 2002] 4. Solve

(x+~r -%(x-~) =4,x,tO,

5. Divide 33 into two parts whose product is 342. 6. Solve 7x+ 1 + 7 1- x = 50

4x

x

+1 +1 5 7. If - - + - - - - =0 findthevalueofx x + 1 4x + 1 2 '

[UPTU B.Pharma 2004] [UPTU B.Pharma 2001] [UPTU B. Pharma 2008] [UPTU B. Pharma 2008]

DOD

SIMULTANEOUS LINEAR EQUATION

•

LINEAR EQUATION IN TWO VARIABLES

If a, b, and c are three real numbers with a"* 0 and x and yare two variables, then the equation ofthe type ax + by + c = 0 or ax + by = c is called a linear equation in two variables.

For Example: (i) 3x+ 5y-7 =0 (ii) 5x-8y= 15.

Solution of a Linear Equation We say that x = a and y =

~

is a solution ofax + by + c = 0

aa+b~+c

if

=0.

Simultaneous Linear Equations in Two Variables Two linear equations in two unknowns x and yare said to form a system of simultaneous linear equations if each of them is satisfied by the same pair of values of x and y.

For Example: The pair of linear equations 3x+ 2y = 7 and4x- 8y-2 = 0 in two variables x andy forms a system of simultaneous linear equations.

•

SOLUTION OF A GIVEN SYSTEM OF SIMULTANEOUS EQUATIONS

The values of x and y which satisfies each equation of the given system of linear equations is called its solution.

For Example: Show that x = 5, y = 2 is a solution of the system oflinear equations 2x+3y = 16,x-2y=1. Solution: The given equations are and putting

2x+3y = 16

...(1)

x-2y =1

... (2)

x = 5 andy = 2 ip(l) we get IRS =2x5+3x2=16=RHS.

62


x = 5 andy = 2 in (2), we get

Putting

LHS = 5 - 2 x 2 = 1 = RHS. Thus x

=

5 and y = 2 satisfy both (1) and (2)

Hence x = 5, y = 2 is a solution of the given system of equations

Consistent / Inconsistent Pair of Linear Equations A pair oflinear equations in two variables which has a solution is called a consistent pair of linear equations. A pair oflinear equations is two variables which has no solution is called an inconsistent pair of linear equations.

Remarks 1. A system of simultaneous linear equations in two variables will have either (I) only one solution

(il) no solution (iii) an infinite number of solutions.

2. It must be taken into memory that no system of linear equations in two variables will have only two solutions, only three solutions, only four solutions, etc. lnfact, if any system of linear equations has two or more solution, it will always have an infinite number of solutions .

•

GRAPHICAL METHOD FOR SOLVING SIMULTANEOUS LINEAR EQUATIONS

To solve the system of linear equations in two variables adopt the following steps.

STEP KNOWLEDGE

l. On the same graph paper, draw graph (straight line) for each given equation. 2. If the lines drawn intersect each other at a unique point; read the values of x and y for this point. The values of x and y so obtained, gives the required solution of the given system of equations.

Method Let the given system of linear equations be

a/x + b\y + c\ =0 and

a;X + bzY + c2

=0

... (1) I

••.

(2)

We draw the graph of e~ch of: the 'given linear equations on the same graph paper. Let the lines L\ and L2 represent these graphs.

Casel: When the lines L\ and L2 intersect at a point. Let the graph lines L\ and L2 intersect at a point P (a, I)). Then x = a, y = I) is the unique solution of the given system of equations.

Simultaneous Linear Equation

63

y

----~----~~------~---.x

o

Fig. 2.1

Case 2: When the lines L) and L2 are coincident. In this case, the given system has infinitely many solutions. y

x'

-----+--------------x o y'

Fig. 2.2

Case 3: When the lines L) and L2 are parallel. In this case, there is no common solution of the given equation, as shown in figure, i.e., the given system of equations has no solution. Thus in this case, the system of given equations is inconsistent. Y

~L' L1

x'

x

0 y'

Fig. 2.3

64


Remarks • A system of two linear equations in x and y has (;) a unique solution if the graph lines intersect at point.

(ii) infinitely many solutions if the two graph lines coincide. (iii) no solution if the two graph lines are parallel.

• Straight lines as graphs oflinear equations alx + bly + c I = 0 and alx + bLY + c2 = 0 intersect each other at a point if

b., -a2al *-. b 2

• Straight lines as graphs of linear equations alx + bly+ c I are parallel to each (i.e. do not intersect) if

=

0 and a

r

+ bLY + c 2 = 0

r

+ bLY + c2 = 0

b., -a2al -_ -*-. b2 C2 cI

•

Straight lines as graphs of Iinear equations a IX + b IY + c I = 0 and a coincide (i.e. becomes single line) if.

b.,

al

cI =-=-

-

b2

a2

c2

Here one equation is a constant multiple of the other equation.

~~~~~~I

SOLVED EXAMPLES

~I~~~~~~

Example 1: Solve graphically the system ofequations: x

+ 2y = 3, 4x + 3y = 2.

Solution: We have x + 2y = 3 => y = Now

.!. (3 - x) 2

1

= "2 (3 - 1) = 1.

x = 1

=>

y

x =3

=>

y="2 (3-3)=0.

1

Thus we have the following table:

x

3

1

y

o

Now, plot the point A (1,1) andB (3,0) on a graph paper. JoinAB and produce it on both sides. Now we have

4x+3y

=

2 1

y = -;; (2-4x). -'

Simultaneous Linear Equatioll 65

Now

1

x =-1

y=

3"

x =2

y=

3"

1

[2-4(-1)]=2. (2-4x2)]=-2.

Thus, we have the following table:

x

-I

2

y

2

-2

Now, plot the point C (-I, 2) and D (2, - 2) on the same graph paper. y

5 4

3

-5

-4

-3

-2

-3 -4

-5 y'

Fig. 2.4

Joint CD and produce it on both sides. The two graph lines AB and CD intersect at the point C (-I, 2). :. x = - 1, Y = 2 is the solution of given system of equation.

Example 2: Show graphically that the system ofequations: 3x - y = 2, 9x - 3y = 6. has an infinite number of solutions.

Solutions: Graph 3x - y = 2. 3x-y =2 Now

y=3x-2.

x =1

y=3 x l-2=1.

x =2

y=3 x2-2=4.

66


Thus, we have the following table: 2

x y

Now plot the points A (1, 1) and B (2, 4) on the graph paper. Join AB and produce it on both sides.

Graph of 9x - 3y = 6: 1

9x-3y = 6

~

y= -(9x-6) 3

x =0

~

Y = - (9 x 0 - 6) =-2 3

x =3

~

y= 3(9 x 3-6)=7.

Now

1 1

Thus, we have the following table x

0

3

y

-2

7.

Now plot the points C (0, - 2) and D (3, 7) on the same graph paper y

D (3, 7)

x' ---+----f--+--I----f--+--++-+---+--+------i"-- x -5 -4 -3 -2 4 5

Fig. 2.5

We find that both the points C and D lie on the line AB. Since the graph lines AB and CD coincide, the given system has an infinite number of solutions.


67

Example 3: Solve the following system of linear equations graphically: 2x + y = 6, x - 2y + 2 = o. Find the vertices of the triangle formed by the above two lines and the x - axis. Also, find the area of the triangle. 2x+y =6 Solution: ... ( I) Y =6-2x x Y

4

I

4

2

-2

2

x-2y+2 =0

... (2)

x+2 y=-2

x

o

y

4

2

-3

2

The two straight lines intersect at A (2, 2). y

-4

-5 y'

Fig. 2.6

:. x = 2,y = 2 is the solution. Vertices of the triangle are A (2,2), B (3,0) and C (-2, 0) AL .1 BC

BC = 5 units, AL = 2 units. Area of 11 ABC =

.!. x (5 x 2) = 5 square units. 2

Example 4: Show graphically that the system of linear equations 2x - 3xy = 5, 6y - 4x = 3. has no solution.

68


Solution: Graphof2x-3y= 5. 2x-3y = 5

'3I (2x-5).

=>

y =

Now

x = I 1

'3 (2x 1-5)=-1.

Y =

x = 4 y

'I3 (2 x 4 - 5) = - 1.

=

Thus, we have the following table r-------r-------~----~

4

x

Y

-1

Now plot the point A (1, - 1) and B (1, 4) on a graph paper. y Join AB and produce it on both sides. Graphof6y-4x=3 6y-4x = 3 I

=>

y

=

'6 (3 + 4x).

x =0 I

Y =

1

·o---+--+-~~+--+--+?~--+--+----

'6 (3 + 4 x 0) = '2'

x

x

x =3

I 5 Y = - (3+4 x 3)=-. 6 2 Thus we have the following table: x y

o 112

yO

3

Fig. 2.7

512

On the same paper, plot the points C (0, 112) and D (3,5/2). Join CD and produce it on both sides. Then line CD is the graph of6y-4x = 3. Since the graph lines AB and CD of the given equations are parallel, it follows ~hat the given system of equations has no solution. Example 5: Which of the following pair of linear equation are consistent/in consistent? U"e the algebraic conditions for ratios

~

!l and.:L.

b2 (i) 5x - y = 7, x - y + 1 = O. (ii) 3x + 2y - 4 = 0, 3x + 2y + I = O. (iii) 3x + 4y + 5 = 0, 6x + 8y + 10 = O. a2

0

c2

Simultaneous Linear Equatioll

Solution: (I) and

~

5x-y-7 = 0 x-y+ 1 =0

69 ...(1) ...(2)

~;t;!!L

a2 b2 So, the pair of linear equations has unique solution and therefore, the equations are consistent. (ii) 3x+2y-4 =0 ... ( 1) 3x+2y+ 1 =0 ...(2)

~ =~=I·!!L=~=I·.:L=-4--4 a2

3

' , b2

2

1 -

'C2

.

~=!!L;t;.:L. a2 b2 c2 So, the linear equations are in consistent because the pair of linear equations has no solutions. (iii) 3x+4y+5 =0 ...(1) 6x+ 8y+ 10 =0 ... (2)

~ = ~ = _. !!L = i = L a2

6

2' b2

8

Ct

-

2. - ~

2' ~ - 10 - 2 .

~=!!L=.:L a2 b2 C2 There are infinitely many solution for the given pair of lines. Hence, the equations are consistent.

Example 6: Find the values ofkfor which the system ofequation. x - 2y = 3, 3x + ky = 1. has a unique solution. Solution: The given system of equations are x-2y =3, 3x+ky= 1. These equations are ofthe form: atx+bty+c t =0,a2x+bY'+c2 =0. where at = -I, b t = -2, c t = -3 and a2 = 3, b2 = k, c2 = -I. for a unique solution we must have ~

a2 This happens when k;t; - 6. Thus, for all real values of k other than - 6; the given system of equations will have a unique solution.

70


Example 7: Find the values ofkfor which the system ofequations

3x + Y = 1 (2k - 1)x + (k - 1)y = (2k + 1) has no solution. Solution: The given equations are 3x +y-l =0 (2k - 1) x + (k - 1) Y - (2k + 1) = O. These equations are of the form.

atx + bty+ c t = 0, a2 x + b2Y+ c2 = O. at =3,b J =I,c]=1 a 2 = (2k-l), b2 = (k-l), c2 =-(2k+ 1).

where

for no solution, we must have

~

=

a2

!!L *.:L . b2

c2

1 -1 3 =--* k-l -(2k+l)' 2k -1

3 I 1 1 = - - and - - * - 2k - 1 k- 1 k - 1 2k + 1 (3k-3 =2k-l)and(2k+l*k-l) k =2 and k*-2

¢:> ¢:>

~

!!L

= *.:L holds when k= 2. a2 b2 c2 Hence, the given system of equation has no solution when k = 2.

Thus

~~~~~~~I

EXERCISE 2.1

I~~~~~~~

Solve the following system of equations graphically. 1. x+y=3,2x+5y=12 2. 2x+y=3,2x-3y=7. 3. 2x+ 3y=4, 3x-y=-5. 4. 4x+3y=5,2y-x=7. 5. 3x+y+ 1 =0,2x-3y*8=0. 6. Solve the following system of linear equations graphically. 2x-3y-17=0,4x+y-13 =0. Shade the region between the lines and the x-axis. 7. Solve the following system of linear equations grapJ,;r- l l 2x+y-5 =0,x+y-3 =0. fmd the points where th~ graph meet the x-axis. 8. Solve the following system of linear equations graphically 2x-y-4=0;x+y+ 1 =0,.

Simultaneous Lilleal" Equation

71

9. Solve the following system of linear equations graphically. x-y+ I =0, 3x+2y-12=0. Calculate the area bounded by these lines and the x-axis. 10. In each of tte following systems of equations, find whether it has a unique solution, an infinite number of solution or no solution:

(I) 3x+5y=13,5x+3y=4. (il) 2x-3y= 5, 6x-9y= 15. (iii) 6x-lOy=3,3x-5y=7. ( iv)

~+.,[ =3 x-2y=2.

3 2 ' 11. For what value of k, the system of equations kx+2y= 5, 3x-4y= 10. has (i) a unique solution (ii) no solution.

12. Find the value of k for which the following system of equations has no solution. 3x+ky= I,(2k-I)x+(k-l)y=(2k+ 1).

I

ANSWERS I

1. x = I,y= 2 5. x=-I,y=2.

2. x = 2,y=-1 3. x=-I,y=2 4. x=-I,y=3. 6. x=4,y=-3. 7. x=2,y=I,P(0,5)andQ(O,3). 8. x = I, y = -2, P (2, 0) and Q (-I, 0) 9. x = 2, y = 3, 75 sq units. 10. (i) Unique (ii) Infinite (iii) No solution (IV) Unique.

3

3

2

2

11. (i) All real values except - - (ii) k = - -.

•

12. k = 2,

ALGEBRAIC METHODS OF SOLVING A PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

It is difficult to ex'amine non-integer solutions of linear equations correctly by graphical method. In such cases only approximate values of x and y can be obtained from the graph. Algebraic methods provide exact and correct solution for each and every pair of linear equations

a1x + b1y+ c 1 = 0 and azX+ by> + c2 = O. The most commonly used algebraic methods of solving a pair of linear equations in two variables are as under. 1. By substitution method. 2. By elimination method 3. By cross-multiplication method.

72


Substitution Method STEP KNOWLEDGE Suppose we are given two linear equations in x and y. To solve these we proceed using the following steps: Step 1. Express y in terms of x, taking one of the given equations. Step 2. Substitute this value of y in terms of x in the other equation to solve it for x. Step 3. Substitute the value of x in the relation taken in step (i) solve it for y.

Remark • We may interchange the role of x and y in the above method. Example 1: Solve 2x - y - 3 = 0, 4x - y- 5 = 0 by substitution method. Solution: Given equations are

2x-y-3 =0 4x-y- 5 =0

... (1) ... (2)

y =2x-3 from(l): Substitute y from (3) in (2), we get 4x-(2x-3)-5 =0

... (3)

4x - 2x - 3 - 5 = 0 2x=2=:.x=1 Substituting x = I in (3), we get

y =2-3 The required solution in x = 1, y

y=-l

= - 1.

Elimination Method STEP KNOWLEDGE In this method, we eliminate one of the unknown variables and proceed using the following steps. Step 1. Multiply the given equ~tions by suitable numbers so as to make the coefficients of one of the unknown variables numerically equal. Step 2. Ifthe numerically equal cofficients are opposite in sign-then add the new equations. Otherwise subtract them. Step 3. The resulting equation is linear in one unknown variable. Solve it to get the value of one of the unknown variables. Step 4. Substitute this value in any of the given equations. Step 5. Solve it to get the value of the other unknown variable. Example 2: Solve lOx + 3y = 75

6x- 5y = 11 Solution: The given system of equations is 10x+3y =75

... (1 )


6x-5y=11

73 ... (2)

We multiply equation (1) by 3 and equation (2) by 5 so that the coefficients of x in both the equations become equal and we get the following equations

30x+9y =225 30x-25y =55

... (3) ... (4)

On subtracting (4) from (3) we get

34y = 170

170 y=-=5 34 .

=>

Substituting y = 5 in (1) we get

=>

lOx+3x5 =75

10x=60 60 x=1O=6.

=>

Hence, the solution is x = 6,y = 5. Example 3: Solve 2x + 3y - 9 = 0, 4x + 6y - 18 = 0 by substitution method. Solution: 2x+ 3y-9 =0 4x+6y-18 =0 3y =9-2x from(l).

9-2x y=--

... (1) ... (2)

... (3)

3

substitutingy from (3) in (2), we get.

=>

4x + 6 (

9

~2x ) -

18 = 0

4x+ 2 (9-2x)-18 = 0

=>

4x-4x + 18 -18 =0

0=0

Which is a true statement. Hence, the given pair of linear equations has infinitely mnay solutions. Let us find solutions putting x = k (any real constant) is (3), we get

y

=

9-2x -3-' 9-2x

Hence,x=k,

y = -3-'

Example 4: Solve the following system oflinear equations:

J3 y = 0,

J2 x -

J5 x + J2 y

= O.

Solution: The given system of equations is

J2 x-

J3 y

=0

... (1 )

J5x+J2y=O

... (2)

Multiplying (1) by J2 and (2) by

J6 y J15 x+ J6 y 2x-

J3 , we get

=0

...(3)

=0

... (4)

74


Adding (3) and (4), we get (2x +

J1s) x

=0

=>

x=O.

=0

=>

y=O

substituting x = 0 in (2), we get

J2 y Hence, the so lution in x

= 0, y = O.

Example 5: Solve 78 x + 91 Y = 53, 65 x + 117 Y = 60 by elimination method. Solution: 78x+9Iy=53 6x+ 117y =60 Multiplying (1) by 5 and (2) by 6. 390x+455y =265 We get 390x+702y =360

... (1) ... (2) ... (3) ...(4)

Substracting (3) from (4), we get

702y-455y =360 -265 247y =95 Substitutingy =

135

95 y- 247

5

5

-13 =>y-13

in (1) we get 5

=>

78x+ 91x =53 13 78x+35 =53

=>

18 3 3 x = - = - =>x= - . 78 13 13

Hence,

3 5 x =13,y=13.

I I I I Example 6: Solve---=-l, - + - =8. 2x y x 2y I I Solution: Taking - = u, and - = v, the given equation becomes x y

~-v =-1 2

u+ ~ =8

=>

u-2v=-2

...(1)

=>

2u+v= 16

... (2)

=>

30 u= - 5 =6 .

=> =>

2v=8 v=u.

2 MUltiplying (2) by and adding the result with (I) we get

5u =30 substituting u = 6 in (I) we get

6-2v =-2

Simultaneolts Linear Equation

Now

-

=6

•

v =4

- =4

x 1

y

x= -

6' I

y=-. z

[

x =-

Hence, the solution is

I

u =6

75

6' y"" -. 4

Method of Cross Multiplication Theorem: The system a/two linear equations

has a unique solution, given by

Proof: The given equations are a,x+b,y+c, = 0 a,x+b;V'+c2 = 0 Multiplying equation (I) by b2 and (2) by b, and subtracting, we get (a,b 2 -a2b,)x = (b,c 2 -b 2c,).

q C2 -b2c, [ .. ~;otll :=;,a b -a b ;otO] '22' a, b2 -a2 b, · a2 b2 Multiplying equation (2) by al' (I) by a 2 and subtracting, we get (a 1b2 -a2b 1)y = (c,a 2 -c2a 1)· x =

C1 a2 -c2 a1 [ .,' (a,b 2 -a2b,);otO]. a1 b2 -a2 b, Hence, a unique solution exists, which is given by y =

This can be written as

x

y

Remark • The following diagram helps in remembering the above solution. x y 1

b b'Xc'Xa'X ' ~~~

b2

c2

a2

b2

... (1 ) ... (2)

76


Numbers with downward arrows are mUltiplied fIrst and from this product, the product of numbers with upward arrows is subtracted.

~~~~~~~~I

SOLVED EXAMPLES

Example 1: Solve the system ofequations 2x + 3y = J 7, 3x -2y = 6 by the method ofcross multiplication. Solution: The given equations may be written as 2x+3y-17=O ... (1) 3x-2y-6 = 0 ...(2) By cross multiplication, we have: 1 x y ~~~

3

-17

2

3

-2XX X2 3

x {3 x (-6)-(-2) x (-17)}

y {(-17) x 3 -(-6) x 2} 1

{2 x (-2)-3 x 3} x

=>

-18-34

-.::....y-=--51+12 -4-9·

x

y

-52

-39

-=-=-

-13

-52 -39 x = -13 =4,y= -13 =3.

=> Hence x = 4, y

=

3 is the required solution.

Example 2: Solve 3x - y - 2 = 0; 2x + Y - 8 = 0 by method ofcross multiplication. Solution: 3x+y-2 = 0

2x + Y - 8 = 0 are the given equations. By cross multiplication method

... (2)

x -t t

X

Y -2 -8

X

{( -1)(-8) - (1)( -2)}

=>

x -8+2

...(1)

-2

3

8X 2

3

2X

-t t

1 Y = {(-2)(2) - (-8)(3)} {(3)(1) - (2)( -I)} y

-4+ 24

=--

3+2

Simultaneous Linear Equatioll

77

x =L=.!.. 10

20

5

I

y 1 = - and - =10 5 20 5 X

20 10 x = - andy=-

2

=>

2

x=5andy=4.

Hence x= 5,y=4 is the required solution. x y x y Example3:Solve -+-b =a+b; 2"+2"=2. a a b Solution: The given equations may be written as

1 1 -x+-y-(a+b) =0

...(1)

1 1 -x+-y-2 =0 2 a b2

... (2)

a

b

By cross multiplication x

~ X-(a+b)

Il

x

y

-2

-(a + b)

X

-;;

-2

x

a I

1

1

~X

b 1

-;;

li

y

[-(aa;b)

+~] [a~2 - )b]

y =--=--

a-b

and

(a-b) a

a-b

a 2b 2 (a-b)

2

y = -x - - =b. 2

Hence, x = ~, and y = b2 in the required solution.

a b ab 2 a 2b Example4: So/ve---=O, - + - =(cI+b2);wherex~O,y~O. . x y x Y 1 1 Solution: Taking - = u and - = v the given equations become. x y au-ab *0 =0 ab 2u + ~bv- (~+ b2) =0

... (1) ... (2)

78 Remedial Mathematics By cross multiplication, we have

v

u

o

a

0

_(a 2 +b 2 ) ab 2 u

v

[b(a 2 + b 2 )-0]

[0 + a(a 2 + b 2 )]

2 2 u = b(a + b ) a 3 b + ab 3

1

u=- v=-

=>

a'

a 3 b + ab 3

a(a 2 +_ b 2_'_) v= ----0.-_ a 3 b + ab 3 .

'

1

b·

-b

ab 2 a 2 b 2

1

1

=>-=- -=x a' y b

=>

x =a,y=b. Hence, x = a, y = b is the solution. Hence, x = 4, Y = 3 is the required solution.

Solve each of the following systems of equations by using the method of cross multiplication.

°

2. 3x - 2y + 3 =

1. 2x - y - 3 = 4x+y-3 =0

3. 2x+y=35 3x+4y=65 5. 4x-7y+28=0 5y-7x+9=0. 2 x

3

5 x

4

6.

Solve - + - = 13 and ---=-2, wherex*O,y *0.

7.

ax+by=(a-b)

y

y

8. ~= l.

~ + l. = (0 + b) a

b

a

b

1

x

1 y

2 x

y

ax + by= if + b2 .

bx-ay= (a + b) 9.

°

4x+3y-47=0. 4. 3x-5y+25 =0 2x+y+ 10=0.

10. -+ - =7

2.+1::..=2 a2 b2 •

3

-+ - = 17. (x* O,y*O).

ANSWERS 1. x=l,y=-l

2. x=5,y=9

3. x= 15,y= 5

4. x= 41,y=TI

5. x=7,y= 8.

6.x=- v=-. 2 '. 3

7. x= l,y=-l

8. x = a,y = b

9. x=a 2,y=b2

125

-89

I I 10. x= 4'y= 3".

I

1

Simultaneolls Linear Equation

79

HINTS TO THE SELECTED PROBLEMS 5. Given equations are:

4x-7y+28 = 0 -7x+5y+9=0

... (1 )

... (2)

By cross multiplication, we have x -7 x 9-5 x 28

=>

y

28 x (-7) - 9 x 4

x

y

-63 -140

-196-36

x

=>

4 x 5 - ( - 7)( -7)

=--

20-49·

=--L=_I_

-203

-232 -29 -203 -232 x = --=7andy= - - =8. -29 -29

=>

Hence, x = 7, y = 8 is the required solution. 1 1 - = uand - =v.

6. Taking

x 2u+3v-13 = 0 5u-4v+2 = 0

We get

y

...(1) ... (2)

8. The given equations are bx-ay = 0 ax+by-(el+b 2) = o. 9. The given equations are bx+ ay-ab (a+ b) = 0

b2 X + ely - 2el b2 = O. 10. Put

1

x

= u and - =v. y

EXERCISE 2.3 ~~~~~~~~~ Solve 1. x-y=3

3x-2y= 10 3. 2x+y=7 4x-3y=-1 5. 4x-3y=8 29 6x-y= -

3 8 7.2x+5y=3 5 3x-2y== -. 6

2. 2x+ 3y=0 3x+4y=5.

4. l1x+ 15y+23=0 7x-2y-20=0 6. 0.4x-1.5y=6.5 O.3x + O.2y = 0.9

8. 31x+23y=39 23x+31y= 15.

80


9 4 10. - - - = 8.

9. ax+ by=a- b

x Y 13 7 - + - = 101, (x;toO,y;t:O) x y

bx-ay= a+ b. I 7x

6y

I

I 2x

I 3y

It. -+-=3

12. 4x+ 6y= 3xy

- - - =5 (x;toO,y;toO).

8x+9y=5xy. x+y

x-y xy

14. --=2, - - =6, (x;toO,y;toO)

13. 3(x+3y)=11xy

xy

3 (2x+y) = 7xy 12 7(3x - 2y)

15. - - - + - - 2(2x + 3y)

7

4

2x+3y

3x-2y

-

2

- - + - - =2 where (2x+ 3y);toO, (3x-2y);toO.

I

ANSWERS

1. x =4,y= I

2. x = 15, Y

4. x=2,y=-3.

5. x= 2,y=-3

6. x=5,y=-3.

1 1 7. x= -,y=-.

8. x= 2,y=-1

9. x=l,y=-1

10. x=

2

3

1

1

-,y=-. 4 7

= -

3

1

11. x=

14'y =

10.

2

1 6'

3. x= 2,y= 3

12. (x=O,y=O)or(x=3,y=4)

13. (x=O,y=O)(x= l,y=3/2) 15. x = 2, Y

=

1.

HINTS TO THE SELECTED PROBLEMS

3. Wehave(l)y=(7-2x) substitutingy = (7 - 2x) in (2), we get 4x-3(7 -2x) = -1 4x-21 +6x =-1 => lOx = 20 => x=2 => from (1): 2 x 2+y=7 y=3. 4. The given equations are l1x + 15 y = - 23 7x-2y = 20 Multiplying (1) by 2 and (2) by 15 we get 22x+30y = -46 105x-30y = 300

... (1 )

... (2)

...(3) ...(4)


81

Adding (3) and (4), we get

127x = 254

=>

254 x= 127 =2.

Substituting x = 2 is (1) we get 11 x2+ 15y = -23

=>

15y = -45,y=-3. 1

1

1

1

10. put - =u, - =v. x Y

11. put - =u, - =v. x Y 12. Divide these equations by xy. 15. Let __1_ =uand __1_ =v. 2x+3y 3x+2y

•

WORD PROBLEMS ON SIMULTANEOUS LINEAR EQUATIONS

Example 1: 7 audio cassettes and 3 video cassettes cost Rs 1395, while 5 audio cassetts and 4 video cassettes cost Rs 1665. Find the cost of an audio cassette and that of a video cassette. Solution: Let the cost of each audio cassette be Rs x and that of each video cassette be Rs y. Then 7x+3y = 1395 ...(1) 5x+4y = 1665 ...(2) Multiplying (1) by 4 and (2) by 3 and subtracting, we get 13x = 585 => x=45. substituting x = 45 in (1) we get (7 x 45) + 3y = 1395 => 3y = (1395-315)= 1080 y = 360 cost of 1 audio cassette = Rs 45 and cost of 1 video cassette = Rs 360. Example 2: Five years ago, a man was seven times as old as his son, andfive years hence, the man s age will be three times his son s age. Find their present ages. Solution: Let the present ages of the man and his son be x years and y years respectively. The man's age 5 years ago = (x-5) years. The son's age 5 years ago = (y-5) years. ... (1) .. (x-5) = 7 (y-5) => x-7y=-30. The man's age 5 years hence = (x + 5) years. The son's age 5 years hence = (y + 5) years. (x + 5) = 3 (y + 5)

82


x-3y = 10

=:>

...(2)

on subtracting (1) from (2)

4y =40

=:>

y= 10.

x-3 x 10 = 10 :. the man's present age = 40 years. :. the son's present age = 10 years.

=:>

x =40.

puttingy = lOin (2), we get

Example 3: Four years ago mother was four times as old as her daughter. Six years later, the mother will be two and half times as old as her daughter, form the pair of linear equations for the situation and determine the present ages of mother and her daughter in years, solving the linear equations by substitution method Solution: Present age of mother = x years. Present age of daughter = y years. Four years ago. The age of mother = (x - 4) years. = (y - 4) years. (x-4) = 4 (y-4) x-4y+ 12 = 0

The age of daughter =:>

...(1)

Six years later: The age of mother

=

(x + 6) years.

The age of daughter = (y + 6) years. (x + 6) =

2 (y + 6) 2

2x-5y-18 =0 x =4y-12. from (1) Substituting x from (3) in (2), we get 2(4y-12)-15y-18 =0 or 8y-24-5y-18 =0 or 3y =42 or

y

=

... (2) ... (3)

14

substitutingy = 14 in (2) we get

x=4 x 14-12=:>x=44 Therefore age to mother = 44 years and the age of daughter = 14 years. Example 4: A two digit number is 4 more than 6 times the sum of its digits. tracted from the number, the digits are reversed. Find the number.

If 18 is sub-

Solution: Let the tens digit of the required number be x and let its units digit be y. Then the number =:>

=

(lOx + y).

10x+y =6(x+y)+4. 4x-5y =4

...(1)


Number formed on reversing the digits = (lOy + x). .. IOx+y-18 = lOy+x => 9 (x-y) =18 => x-y=2. Multiplying (2) by 5 and subtracting (1) from the result. We get x = 6. Substituting x = 6 in (2), we get y = (6-2)=4 Thus x = 6 and y=4. tens digit = 6 and units digit = 4 Hence, the required number = 64.

83

... (2)

Example 5: Seven times a two digit number is equal to four times the number obtained by reversing the order of its digits. If the difference the digits is 3, find the number. Solution: Let the tens and units digits of the required number be x andy respectively. Then, the number = 10 x + y. The number obtained reversing the digits = (10 y + x). .. 7(10x+y) =4(10y+x) 33 (2x-y) = 0 => 2x-y= 0 => y=2x ... (1) => Thus, unit digit = 2 x tens digit (unit digit) > (tens digit) so y > x y-x =3 ...(2) using (1) is (2), we get (2x-x) =3,x=3 On substituting x = 3 in (I), we gety = 2 x 3 = 6. Hence, the required number = 36. Example 6: The sum of the present ages of Kamal and his mothers is 89 years. After 11 years, mother s will be Kamal s age. Find their present ages. rUPTU B. Pharma 20011 Solution: Let the present age of Kamal be x years and that of his mother be y years. Then according to the question, we have x+y =89 ...(1 ) ...(2) and (y + 11) = 2 (x + 11) or 2x - y = - 11 Adding (1) and (2) we get 3x =78 x =26. putting x = 26 in (1), we get y = 89 -26 = 63. Hence, Kamal's present age is 26 years and his mother's present age is 63 years. Example 7: In a given fraction, if the number is multiplied by 3 and the denominator is

reduced by 1, we get

~, 2

multiplied by 3, we get

1

3"

but if the numerator is increased by 12 and the denominator is find the fraction.

Solution: Let the fraction be ~. Then according to question, y

IUPTU B. Pharma 2007j

84


3x y-l

=> => and

3 2 6x-3y =-3

=>

6x=3y-3

2x-3y =-1 x + 12 3y

1 = 3

...(1)

=>

3x+36=3y

3x-3y = -36 x-y=-12 Subtracting (2) from (1), we get x = 11. putting x = 11 in (1), we get => Y = - 1- 22 2 x 11 - y = - 1 => Y = 23.

=> =>

...(2)

11 Hence, the required fraction is 23 .

Example 8: If the numerator of a fraction is multiplied by 2 and its denominator is increased by I, it becomes I . However, if the numerator is increased by 4 and denominator is multiplied by 2, then the ratio of the numerator and denominator is I : 2. from a pair of linear equations for the problem and solve by substitution method and hence find the fraction. x Solution: Let the given fraction be - . According to the given conditions. y

2x x +4 1 = 1 and - - =-. y +1 2y 2

=>

2x = y+ 1 and x+4=y.

Thus we get the required pair of linear equations: 2x-y-1 = 0

x-y+4 = o. Y = 2x-l

...(3)

from (1) substitutingy from (3) is (2), we get

x-(2x-I)+4 =0

... (1)

... (2)

=>

x=5.

substituting x = 5 in (3), we get

y =10-1=9. Hence, the required fraction is

5

"9'

~~~~~~~I EXERCISE 2.4~1~~~~~~~ 1. Find two numbers such that the sum of twice the first and thrice the second is 92; and four times the first exceeds seven times the second by 2. 2. If2 is added to each of two given numbers, their ratio becomes I: 2. However, if 4 is subtracted from each of the given numbers, the ratio becomes 5 : II. Find the numbers.


85

3. The monthly incomes of A and B are in the ratio 8 : 7 and their expenditures are in the ratio 19 : 16. If each saves Rs 2500 per month, find the monthly income of each. 4. A mother is three times as old as her daughter five years later, the mother will be two and a halftimes as old as her daughter, find the age of the daughter and that of her mother in years. 5. A fraction becomes

.!. if 5 is subtracted from its numerator and 3 is subtracted from its

2 denominator. However, if we divide the numerator by 2 and add 7 to the denominator, the fraction becomes

.!.. Determine the fraction. 4

6. Of the numbers, one number is greater than thrice the other number by 2 and 4 times the smaller number exceeds the larger number by 5. Find the numbers. 7. The sum ofthe digits of a two digit number is 93. The number obtained by interchanging the digits of the given number exceeds that number by 27, find the number. 8. Harsh purchased 4 chairs and 3 tables for Rs 1650. From the same place and at the same rates_Sunit purchased 3 chair~ and 2 tables for Rs 1150. Find the cost per chair and per table. 9. The sum of the digits of a two digit number is 12. Ifthe digits are reversed, the new number is 12 less than twice the original number. Find the original number. 10. If 1 is added to both the numerator and the denominator of again fraction, it becomes

~.

If however, 5 is subtracted from both the numerator and the denominator, the

fraction becomes 112. Find the fraction.

HINTS TO THE SELECTED PROBLEMS 3. Let the monthly income of A and B be Rs 8x and Rs 7x respectively and let their expenditures be Rs 19y and Rs 16y respectively. Then A's monthly saving = Rs (8x-19y). and B's monthly saving = Rs (7x - 16y).

..

8x-19y =2500 7x-16y =2500 6. Let the larger number = x and the smaller number = y. Then according to the gives conditions, we have (largernumber)-3 x (smallernumber) = 2 and 4 x (small number) - (larger number) = 5 x-3y =2 -x+4y =5 Adding (1) and (2) we gety = 7 substituting y = 7 in (i) we get x-3 x 7 =2 ~ x=23. the larger number = 23 and the smaller number = 7.

... (1) ...(2)

...(1) ...(2)

86


9. Let the tens digit and units digits be x and y respectively.

x+y =12

Then

... (1)

original number = lOx + y. Number obtained on reversing the digits = (lOy + x).

lOy+x =2(lOx+y)-12 19x-8y =12

... (2)

10. Let the required fraction be .::.. The according to the given condition, we have

y

x +1 y +1

4

x-5 y-5

1

- - = - and - - = -. 5

2

5 (x + 1) = 4 (y + 1) and 2 (x - 5) = (y - 5). -5x-4y =-1 2x-y =5 ANSWERS

2. 34, 70 1. 25, 14 3. Rs 1200 and Rs 10500 4. Age of daughter = 15 years, Age of mother = 45 years. 5.

3 7

6. 23 and 7.

7. 58

8. cost of chair = Rs 150, costoftable=Rs350.

9. 48.

7

10.

'9' OBJECTIVE EVALUATION

MULTIPLE CHOICE QUESTIONS . Choose the most appropriate Answers: 1. Solution of3x- 4y= I, 4x- 3y-6 = 0 is (a) x=3,y=2 (b) x=2,y=3 (c) x=2,y=2 (d) x=3,y=3 2. The value ofk for which x - 2y = 3, Icy + 3x = I represent paralleilines is (a) k= 6, (b) k=-6, (c) k-ci' q, (d) k *-6. 3. The system.of linear equations 21'. - 5 = 3y and 4x - 6y = 3 represents a (a) Intersecting lines. (b) Paralieilines. (c-) Coincident lines. (d) None of these.

... (1) ...(2)

:Simultaneous Linear Equatioll

87

4. The system of linear equations 3x + 2y - 4 = 0 and ax -y - 3 = 0, will represent intersecting lines if (a)

a =-3/2

(b)

*- 3/2

a

2 (d) a*- -. 3 5. If the system of equations 2" + 3y = 7 and (a + b) x + (2a - b) y = 21 has infinitely many solutions then (c)

a=-2/3

a = 1, b = 5 (b) a=5,b=1 a = -I, b = 5 (d) a=5,b=-1. 6. The system of linear equations ax + by + e = 0 and xd + ey + f = 0 will represent coincident lines if (a) ae = bd and bf= ec. (b) ae = bd and bi ee. (c) ab=deandbc=ef (d) ab = de and be ef (a)

(c)

* *

7. Equation of line which is prallel to (a)

J2 x - fj y = 5 is

J8x-2 fjy=5

J8x-2 fjy= I

(b)

(d) -J8x+2 fjy=l. (a) -J8x+2.Jjy=5 8. The system of linear equations b = 3 (y - 3) and 6x - 9y = 5 represents a

(a) parallel line

(b) coincident lines

(a) intersecting lines

(d) none of these.

9. Equation of a line which is parallel to (a)

3. 3

2 - x+ 3y+ 1 = 0 3

x + 3y + I (b)

=

0 is 2 - x-3y+5=0 3

3 3 3 -x+ 3y+ 1 = 0 (d) -x-y+)=O 2 2 10. Value of k for which 2 (k - 1) x + Y = I, 3x - y = I represents parallel lines is (c)

(a)

(c)

k* 5/2 k= 112

(b) k= 5/2 (d) k*-1I2.

ANSWERS 1. (a) 8. (a)

2. (b)

3. (a)

9. (a)

10. (c)

4. (b)

5. (b)

6. (a)

7. (a)

7. (b)

FILL IN THE BLANKS 1. The lines b - 3y = 9 and kx - 9y = 18 will be parallel if k is - - 2. Value of k for which

x - 2y = 5, 3x + ky + 15 = 0 is incoincident is - - 3. The value ofp for which the syStem of equations 3x + 5y = 0 and px + lOy = 0 has a non zero solution is - - 4. The value of k for which the system of equations kx - 2 = y and - 2y 3 = 6x is consistent with unique solution is - - -

88


J3

J7

5. Solution of.J2 x- ../5y= 0,2 xy=O i s - - - . 6. ax + by = c, Ix + my = n has a unique solution. i f - - 7. The equation

~x- 5y+ 1 = 0 and ~

3 8. The system of equations

2

x + 3 = 5yare---

4x + py + 8 = 0 and 2x + 2y + 2 = 0 will have a unique solution i f - - -

ANSWERS 1. k= 6 5. x= O,y= 0

2. k= 6. 6. am

3. 6*p. 7. intersecting

* bl.

4. k* 6. 8. p*4.

TRUE/FALSE State whether given statement is true or false. 1. The lines 2x - 3y = 9 and 8 - 2y = x are parallel. 2. The lines 5x - 1= 2y and y = -

(T/F)

.!. + ~ x are c~incident. 2

(T/F)

2

3. Equation ofline which is coincident to 5 (x - y) = 3 is lOx - 10 y - 3 = O. (T/F) 4. The lines 2 (x - 3) = 5y and 4x -1 = lOy represents intersecting lines. (TIF) 5. If 2x - 3y = 7 and (a + b)x - (a + b - 3)y = - 4a + b represent coincident lines then a+5b=0.

(T/F)

6. The solution of 2x - '"3y = 5 and ~ - 3y = 1 is x 4

= 4, y = 1.

(T/F)

ANSWERS 1. False. 5. False.

2. True. 6. False.

3. False

4. False.

REFRESHER Do you know? After reading this chapter you should be able to know the following concepts: • The equation ofthe type ax + by + c = 0 or ax + by = c is called a linear equation in two variables. • A pair of linear equations in two variables which has a solution is called a consistent pair oflinear equations. • A pair oflinear equations in two variables which has no solution is called an inconsistent pair of linear equations. • A system of two linear equations in x and y has (I) a unique solution if the graph lines intersect at a point. (il) Infinitely many solutions if the two graph lines coincide. (iii) No solution if the two graph lines are parallel.


89

• Straight lines as graph of linear equations alx + bly + c l = 0 and ai' + b2y + c2 = 0 intersect each other at a point if at q -*-a2 b2 • Straight lines as graph of linear equations atx+ bty+ c l = 0 and ai' + b2Y+ c 2 = 0 and parallel to each other if al _ q Ct ---*--. a2 b2 c2 • Straight lines as graph of linear equations atx + bty + c t = 0 and ai' + b2y + c2 = 0 coincide if a2

b2

C2

Can we do? (Frequently Asked Questions) 1. Calculate the value of x and y when 4x-5y=3 andxy=2. [UPTU B. Pharma 2001] 2. In a given fraction, if the numerator is multiplied by 3 and the dinominator is reduced by I, we get

~. 2

But if the numerator is increased by 12 and the denominator is

multiplied by 3, we get

1

"3' Find the fraction.

[UPTU B. Pharma 2007]

3. The sum of the present ages of Kamal and his mother is 89 years. After 11 years, mother's age will be twice Kamal's age. Find their present ages. [UPTU B. Pharma 2007]

4. Solve the following system of linear equations by using the method of cross multiplication. x a

Y

-+- =(a+b) b

""::"'+L =2 2 2 a

b

.

5. Show Graphically that the system of linear equations 2x-3y = 5, 6y-4x= 3. has no solution.

DDD

DETERMINANTS

III

INTRODUCTION

Consider two homogeneous linear equations

alx+b,y =0, a2!+b7Y =0; Multiplying the first equation by b2 , the second by bI' subtracting and dividing by x, we obtained

a l b2 - a2b, =0 This result is sometimes written as

I: ~I

=0

and the expression on the left is called the determinant. A determinant also is an arrangement of numbers in rows and columns but it always has a square form and can be reduced to a single value. Therefore, a determinant is distinct from matrix in the sense that the determinant is always in square shape and it has a numerical value. The arrangement of the numbers of a determinant is enclosed within two vertical parallel lines.

Order of a Determinant The determinant of a square matrix of order n is known as determinant of order n. DETERNDNANTOFORDERTWO Let all' a 12 , a21' a22 be any four number (real or complex). Then

IA 1= la,1

bl2l

a21 ~2 represent the number all a22 - a21 a l2 and is called a determinant of order two. For example

IA I =

I~ ~71 =(5)(-7)-(3)(2)

=-35-6=-41

Determinants

•

91

DETERMINANT OF ORDER THREE

IA I

Let

aIJ

a12

al3

= a21

a22

a23

a31

a32

a33

is called a determinant of order 3 and its value can be obtained as follows:

IAI

=alll::~ :::I-a

12

=a

I: : :::I+ I: :

::~I

a13

ll (a22a33-a32a23)-a 12 (a2Ia33-a31a23)+ a 13 (a21a32-a3Ia22)

2

IA I

For example,

3

= -I

5

2 3 -2

4

1 31_ 31- 31+41-1 21 -2 I 4 I 4-2 2

=21

=2(2 +6)-3(-1-12)+4(2-8) = 16+39-24=31

Remarks • The value of a determinant is not changed if it is expanded a!~ng any row or column. • When no reference of the corresponding matrix is needed, we may denote a determinant by D. • The determinant of a square zero matrix is zero.

~~~~~~~I SOLVED Example 1: Find the value ofl A

I ifA is given by

sa IA I _leo. sma

Solution:

EXAMPLES cosa . sma

I

~I~~~~~~

-sinal cosa

-sinal cosa

= cos2 ex - (- sin 2 ex) = cos 2 ex + sin 2 ex = 1.

I

Example 2: Find the value of J (J)

Solution:

IA 1=

II

(J)

(J) -(J)

(J) -(J)

I.

I =-ro-{02=_(ro+ro2)=(-I)= 1.

Example 3: Solve for x: x

I5

31 = 155 -413

2x

92


I~ :xl I~ ~41

Solution: We have

=

2~ - IS = IS + 20 2x2=SO => x 2 =2S=> x=±S.

=> =>

a

h

g

Example 4: Find the value of h

b

f.

g

f

e

Solution: Let

a

h

g

/1 = h

b

f

g

f

e

We expand /1 along first row, we get

/1=a(-1)21;

~1+h(-l)31; ~1+g(-1)41;;1

= a (be- f2)-h (he -fg) + g (hf- bg) = abc - af2 - eh 2 + fgh + fgh + fgh - bi = abc + 2fgh - a/- bi - eh2

o

1

sec 8

Example 5: Find the value of tan8 -seeS tan8 101

o Solution: Let/1= tanS

sec8 -secS

tanS

o expand along Rl /1=0+1(-1) 3 1tan 8 1

tan 81 +secS(-I) 4 \tan 8 1 1

- sec 8\ 0

= - (tan 8 - tan 8) + sec S (0 + sec S) = sec S (0 + sec S) = sec 2 S.

__

CO-FACTORS AND MINORS OF ANELEMENT

If in the expansion of a determinant I aijl, all the containing aij as a factor, are collected and their sum a is denoted by aijAi; then the factor Aij is called the co-factor of the element a , .. Hence, in a determinant of order n 1J n

laijl

=ai1AI1+ai2A,2+····+a,~in= 'L,aijAij j-l

Now, let ~J be the (n-l) x (n -1) sub-matrix ofl a y In xn obtained by deleting the ith row andjth column. Then IMijl is called the minor of the element aij the determinant layl of order n. Thus we can express the determinant as a linear combination ofthe minors of the elements of any row or any column.

Determinants

93

Remark • (- 1)1 +j is 1 or -1 according as t + j is even or odd :. Ai) and Mi} coincides if i +j is even and if i + j is odd then we have Ai} = -Mi)'

~~~~~~~I SOLVED

EXAMPLES

~I~~~~~~

Example 1: Find the minors and cofactors ofelements of the determinant Solution:

Minor of the element all is Mil Minor of the element a l2 is MI2 Minor of the element a 21 is M21 Minor of the elem~nt a 22 is M22 Hence, A II = (-1) I + I Mil Al2 = (_1)1+2 MI2 1 A21 =(-li+ M21

I~ ~ I

= 171 = 7 =3 =-2 =5 =7 =-3

=2 A22 = (-li+ M22 =5 Example 2: Find all the minors and cofactors ofthe elements infollowing determinants 2

4.3

J

J 3 2

2 J 5 Solution: Here

all = 4a l2 = 3al3 = 1 a 21 = la22 = 3a23 = 2 a 31 = - 2a32 = 1a 33 = 5

=

I~ ~I = 15-2= 13

M l2 =

Ml3 =

I~ ~I = 1 - 6 = - 5

M21 =

M22 =

I~ ~I = 20- 2 = 18

M23 =

M31 =

I~ ~I =6-3 =3

M32 =

Mil

M33

I~ ~I =5-4= 1 I~ ~I = 15 -1 = 14 I~ ~I =4-6=-2 I~ 211 =8-1=7 '

=I~ ~1=12-3=9.

The co-factors are All =(_1)1+1 Mil = 1 x 13 = 13 Al3 =(_1)1+3 Ml3 = Ix (-5) =-5 A22 =(_1)2+2 M 22 = Ix 18= 18 A31 =(-1)3+IM31 =l x 3=3 A33 =(-1)3+3M33 =l x 9=9.

Al2 =(-1) 1+2 Ml2=-lx 1 =-1 2+1 A21 = (-1 ) M21 =-lx 14 =-14 A 23 =(_1)2+3 M23=-lx (-2)=2 A32 = (_1)3+2 M32 =-1 x 7 =-7

94


Example 3: Find the minor and co-factors ofelements ofthe following determinant

2 -3

5

6

0

4

1

5 -7

Solution: We have

0

M1·1 --

41 = 0-20 =-20

5 1

-7

-7

M12

=

I~

Ml3

=

I~ ~I =30-0=30

M2l

3 51 =21-25 =-4 = 1-5 -7

M22

=

I~

M 23

=

I~

M31

3 = 1-0 :1 =-12-0=-12

M32

=

M33

=I~

=-20

Al3

=30

A2l

=4

41 =-42-4=-46

51 =-14-5 =-19

-7

-31 5 =10+3=13

I~

All

A22 =-19

A 23 =-13

A3l

=-12

:1 = 8-30= -22

A32

= 22

-31 0=0+18=18

A33 =-18.

Example 4: Write the co-factors ofelements ofthe second row ofthe following determinants and hence evaluate them

1 a be

Solution: Let

1 b

ca

1 c

ab

~ =

A 21

a

be

1 b

ca

1 c

ab

=(-l)2+1l ac 1

bel

2 =ab-bc A22 = C-l + 11 ab

i

Determinants

•

95

PROPERTIES OF DETERMINANTS

Theorem 1: The value of a determinant does not change when rows and columns are interchanged. bl

cI

Proof: Let I A I= a2

b2

c2 be a determinant of order three.

a3

~

c3

al

Expanding I A I along the first row, we get IA I = a l (b 2c 3 - b 3c 2) - b l (a 2c 3 - a 3c 2) + c I (a 2b 3 - a 3b 2) = a l (b 2c 3- b3c2) - a 2 (b l c3 - a3c l ) + a 3 (b l c2 - b 2c l ) (by rearrangement of terms)

c3 Hence, the theorem is proved. Theorem 2: Ifany two rows [or columnsJ ofa determinant are interchanged, the sign ofthe determinant is changed. C2

CI

bl

al

Proof: Let I A

cI

I~ lb._ j/2.,

t. I I

bJ

a3

c2 be a detenninant of order three. '

c3

Expanding A along the first row, we get IA I = a l (b 2c 3 - b3c2) - b l (a2c 3 - a 3c 2) + c I (a2b3 - a 3b2) = - {a 3 (b 2c l - b l c2) - b3 (a2c I - a l c2) + c 3 (a2b l - a l b2)} (by rearrangement ofterms)

=-

al

a2

a3

q

b2

~ =(-I)IA I

CI

c2

c3

Theorem 3: If two rows or two columns ofthe determinant are identical, then the value of the determinant vanishes, i.e.,

I A 1=

al

bl

a2

b2 c2

al

bl

cI

cI

=o.

96


Proof: We have I A I is a determinant of order 3 whose first and third row are identical. If we interchange the two identical rows, then obviously there will be no change in the value of 1A I. But by theorem 2, the value ofA ismu'ltiplied by-l if we interchange two rows. Therefore, we get

IAI=-IAI 21 A = 0 or A = 0 1

I

1

Theorem 4: Ifall the elements ofany row, or any column, ofa determinant are multiplied by the same number then the determinant is multiplied by that number.

Proof: Let IAI =

We have

all

a12

a21

a22

ani

a n2

mall

al2

ma21

a22

manl

a n2

be a determinant of order n

(where Ail' A i2 ...A m be the cofactor of elements ail' a i2 , ... am of ith row ofl A I)

If in the determinant, the elements of a row are added in and m times the corresponding elements ofthe another rows (or column), the value ofthe determinant does not change in particular, Theorem 5:

al

a2

+ mb.. + nCI b.. + mb2 + nC2 b2

cI

al

b..

cI

c2

a2

c2

~

c3

a3

b2 b3

b..

cl

al

bl

cI c2

a3 +m~ +nc3

c3

Proof: We have bl

cI

nCI

b..

cI

c2

+ nC2

b2

c2

c3

nC3

~

c3

cI

b..

cI

c2 +n c2

b2

c2

~

c3

+mbl +ncI a2 +m~ +nc2

b2

c2

a2

b2

a3 +m~ +nc3

~

c3

a3

b3

c3

m~

al

bl

cI

b..

b..

ci

a2

b2

c2 +m b2

b2

a3

b3

c3

~

b3

c3

al

mb..

+ mb2 b2 b3

c3

(By theorem 4) al

b..

cI

a2

b2

c2

a3

b3

c3

al

b..

cI

a2

b2

c2'

a3

b3

C3

+m (O)+n(O)

(By theorem 3)

Determinants

97

~~~~~~~I SOLVED EXAMPLES I~~~~~~~ Example 1: Evaluate the following determinant

Solution: We have IA I=

I! ~21

-:1

I~

= 3 x 5 - 4 x (-2) = 15 + 8 = 23

Example 2: Find the value ofthe determinant ofthe matrix

~ [~ ~l

A

;

123 Solution: We have

IAI=231

312 On expanding the detenninant along the first row, we get = 1

I~ ~I-21~ ~I + 31~ ~I

= 1. (6-1)-2. (4-3) + 3. (2 -9) =-18

4 J 4 Example 3: Evaluate the determinant of 0

J O.

J 2

Solution: We have I A

1=

4

4

0

0

J

1 2 On expanding the detenninants along first column, we get

=41~ ~I-ol~ ~I + ll~ ~I =4(1-0)-0+ I (0-4) =4-4=0. Example 4: Show that: J

x

0

cosx

siny = cos (x + y)

0

sinx

cosy

x

y

Solution: We have 0

cos x

siny

0

sinx

cosy

I

Y

98

Remedial Mathematics On expanding the determinant along first column, we get

I

y I+ 0 x 1 cosx sinyl -0 I x sinx siny cosx Isinx cosy = cos x cos y - sin x sin y = cos (x + y)

=

I I I Example 5: Show that I l+x I =xy l+y I I 1 .

I

Solution: We have L.H.S. =

I +x 1+ y

Applying C2 - C1 and C3 - C] in the given determinant, we get =

1 0 0 1 x 0 lOy

On expanding the determinant along the first row, we get

=ll~ ~I-ol; ~I-ol; ~I =xy=R.H.S. Example 6: Without expanding, show that

b-e e-a a-b e-a a-b b-e a-b b-e e-a

=

0

Solution: We have

b-e e-a a-b e-a a-b b-e a-b b-e e-a Example 7: Without expanding, show that

b 2 e2

0 e-a

a-b 0 a-b b-c 0 b-e c-a

(Operating C1 ~ C1 + C2 + C3 , we get) = 0

be b+e

2 2

ea e+a =0

2 2

ab a+b

e a

a b Solution: Consider

b2 c 2

be b + e

2 2

e a

ea e + a

a2 b 2

ab a + b

b2 e2

be b+e

abc 2 2 e a abc a2 b 2

ea e+a ab a+b (Multiplying

RJ

by a,

R2

by band

R3

by c)

Determinants

ab 2 c 2

abc

= _1_

bc 2 a2

abc bc + ab

abc

2 2

ca b

ab + ca (Take abc out from C I and C2 )

ca + bc

abc

bc abc.abc ca abc ca

ab+ca bc+ab ca+bc

bc

ab +bc +ca

= abc ca

ab +bc +ab

cb

ab +ca+bc bc

= abc (ab + bc + ca)

C{:J

cb = abc (ab + bc + ca) x 0 = 0 x+l

x+2

x+a

Example 8: Ifa, b, c are in A.P. prove that x + 2

x +3

x +b = 0

x+3

x+4

x+c

Solution: Given a, b, care inA.P. therefore a + c = 2b =>a+c-2b=0 Operating Rl ~ Rl + R3 - 2R2 , we get x+ 1 x+2

x+a

0

0

a+c-2b

x+2

x+3

x+b

x+2

x+3

x+b

x+3

x+4

x+c

x+3

x+4

x+c

0

0

0

=x+2

x+3

x+b =0

x+3

x+4

x+c

Example 9: Prove that a

b

c

1

1

J

a

2

b

2

c

2

= abc a

b

c

a

3

b

3

c

3

2

2

2

a

b

a

b

2

2

Solution: We have I A 1= a

a3

c

= abc (a - b) (b - c) (c - a)

c

1

c 2 = abc a b 3 c3 a2

b

b

c

2

c2

b

1

Now again

IAI

99

= abc a

b

c

2

2

c2

a

h

100


Applying C2 - C t and C3 - C t , we get

o = abc a

0

b- a

a2

b2 _

c- a

a2

c2 _

a2

On expanding along the first row, we get

= abc

~

- a2 b -a

1

~

- a 21 c-a

= abc [(b - a) (c 2 -~) - (b 2 - a2) (c - a)] = abc [(b -a)(c-a) {(c + a) - (b + a)}] = abc (b-a) (c-a) (c+ a- b-a) = abc (a- b)(b - c)(c-a) Example 10: Prove that

Solution: Let IAI =

a + b + 2c

a

b

c

b+c + 2a

b

c

a

c + a + 2b

2 (a + b + c/

=

a +b +2c

a

c

b +c + 2a

b b

c

a

c + a +2b

Adding C2 and C3 in Cl' we get 2(a +b +c)

a b =2(a+b+c) b+c+2a b 2(a +b +c) a c+a+2b

a =2(a+b+c) 1 b+c+2a _

b b

1

c + a + 2b

a

Applying (R2 - R t ) and (R3 - R t ), we get 1

b

a

=2(a+b+c) 0 b+c+a

o

O·

0

c+a+b

On expanding the determinant along the first column, we get

=2(a+b+C)\b+c+a

o

=2(a+b+c)(a+b+ci =2(a+b+d

0 \ a + b+'c

Determinallls

101

Example 11: Prove that

( 1 1 1)

1+a 1 1 1+b 1 1 = abc 1 + - + - +abc 1+e 1 1

[VPTV B. Pharma 2000, 06J

Solution: Operating C]~ C] - C3 and C2 ~ C2- C3 , we get I+a

I l+b

I

1

l+e

=

a

0

0

b

-e -e l+e

=a[b.(l +e)-(-e).l]+ 1 [0. (-c)-(-c)b] = a (b + be + c) + be = abc + be + ea + ab = abc ( 1 + -1 + -1 + -1 ) abc Example 12: Prove that

a-b-e 2b 2e

2a b-e-a 2e

2a 2b =(a+b+el e-a-b

Solution: Operating R] ~ R] + R2 + R3, we get

a- b- e 2a 2a 2b b-e-a 2b 2e 2e e-a-b

a + b +e a + b + e a + b + e 2b b-e-a 2b 2e 2e e-a-b [Take (a + b + c) out from Rd

=(a+b+e) 2b b-e-a 2b 2e 2e e-a-b (Operate C2 ~ C2 - C] and C3 ~ C3 - C])

o

0

=(a+b+e) 2b -b-e-a o (expand by R]) 2e 0 a-b-e =(a+b+e) 1 (-a-b-e)(-a-b-e)=(a+b+e)3 Example 13: Without expanding the determinant, show that 0 b -e -b 0 a =0. e -a 0 0

b

-e

Solution: Let.1 = -b

0

a

-a

0

e

[UP TV B. Pharma 20011

102


By changing columns into rows: 0

-b

A = b

0

-e

a

e 0 -a =(_1)3 -b e

0

0

-e a

-a

0

b

(taking (-I) Common from each column) = (_1)3 A =-A.

2 A =OorA=O.

Example 14: Without expanding the determinant, show that I

a be

I

a a2

I

b ca

I

b b2 and evaluate it.

I

e

ab

I

e

c2 [UP TU B. Pharma 2001,20081

1 a

be

Solution: Let A = 1 b

ea

I

c

ab

Multiplying the 1st, 2nd and 3rd rows by a, b, c respectively. We get

a I A=- b abc c

a2

abc

b2

bca

e2

abc

=

a a2 abc b b2 abc c e2

Taking abc common from 3rd column

a I

a2

b

b 2 applying C2 ~ C3

C

1 e2

1 a

I

a A = 0

a2

b

b 2 applying C 1 ~ C2.

c

e2

a2

b-a b 2 _a 2

0 e-a

e 2 _a 2

b-a b 2 _a 2 e-a

on expanding the determinant along C 1 =

(b-a)(e-a) 1 b + al 11 e + a

c2 _a 2

Delerminants

103

taking (b - a) common from R\ and (c - a) common from R2 = (b -a)(c-a) [c + a- (b + a)] = (b -a)(c-a) (c- b)

= (a- b}(b -c)(c-a). Example 15: Without expanding the determinant show that (a + b + c) is a factor oj [ollowing determinant. abc A= b

c

a

[UPTU B. Pharma 20031

cab

If a, b, and c are positive and unequal, show that the value of A is always negative. Solution: Applying C\

~

~ =

C\+ C 3, we get

a+b+c b

c

a+b+c

c

a =(a+b+c) 1 c

a+b+c

a b b

b c a

a b c

=(a+b+c) 0 c-b a-c

o

a-b b-c

Applying R2 ~ R2 - R\, R3 ~ R3 - R\

=(a+b+c) IC-b a-cl a-b b-c = (a+ b + c) {-(b-d-(a-b)(a-c)} = (a + b + c) (- ~ - b2 - c2 + ab + bc + ca) Thus (a + b + c) is 0 factor of~. Now we shall prove the next part. we have ~ = (a + b + c) (--A + (A + B) =-A + (A + C) adding-A both side. Then, by associative law of addition, we have (-A + A)+ B = (-A + A)+C O+B=O+C B=C

SOLVED EXAMPLES

Examplel:ifA=[~ ;lB=[-~ ~JFindA+B. Solution: We have

A+B Example 2: if A = [2

o

=

o =

If A

~[~ ~ -~l -/

A +2C=B.

=

=

:]

3 1] andB = [1 2 -1] find 3A -4B. -1 5 0 -1 3

Solution: Wehave3A-4B =3 [2

Example 3:

[~ ~] + [-~ _~] [~=~ ~:~] [~

/

1 3 1]_4[1 2 -3 ] -1 5 0-1

[~ -~ l~]-[~ -48 -4] 12

1 7]

6-4 = [ 0-0

9-8 3-(-4)] = [2 -3-(-4) 15-12 0 1 3

and B =

[~ ~ -~l· -2 0

-/

Find the matrix C such that

135

Matrices

Solution: Given A + 2C = B or 2C = B - A Now,

2C

=

[~ -!]-[~ ~ -~] 1 0 -1

-2

1 -1

1

=[~=~ ;=~ -24-_(~3)]=[_~-1~] -2-1 C =

~[_~

0-(-1)

-1-1

-2

~] = [_5 /2 ~~~2 1~2]. 1

-1

-3 -2 Example 4: Find,the additive inverse o/the matrix A =

-3

-3/2

1/2

-1

[~ =~ -~ ~] 287

J

Solution: The additive inverse of matrix A is the matrix each of whose elements is tlie negative of the corresponding element ofA. Hence, if we denote the additive inverse of A by - A then we have -A

=

-2 3 1-I] [ -3

I -2

-I -2

-2

-8 -7

Example 5: Solve the/ollowing equations/or A and B, 2A-B =

. Solution: GIven 2A - B = [33

-3 3

[~ -33 20] and2B +A = [-J4

J

4

~]

~l

Multiplying both sides by 2, we get

-3 3

4A-2B = 2G Also given that 2B + A = [4 I -1 4 adding equation (I) and (2), we get 5A

=[:

~] =[:

-6 6

... (1)

-!J

... (2)

I -6 0] + [ 4 6 4 -1 4

= [6+4 -6+1 0+5] 6-1 6+4 4-4 A

~J

=.5!.fO5

-5 10

~] =[~

-!] =[105 -1 2

~]

-5 10

~]

136


Now substituting the value of A in equation (2), we get 2B

= =

Example 6:.if A = [ 1/3

sin

2

[4-2 1-(-1) 5-1] -1-1 4-2 -4-0

=

[2 2 -44] -2 2

[_~ ~ _~]

B sec 2 e

[_~ ~ _!]-[~ -~ ~]

e ] andB= [-tan 2 e

cosec 2e

2;'3

2

e]

cos 2 . Find A +B -cot e

sec2 e sin 2 e ]+[-tan 2 e cos 2 e ] Solution:A+B= [ 113 cosec 2e 2/3 -cot 2 e 2 2 2 2 = [sec e-tan e sin e+cos e] = [1 11] 1/3+2/3 cosec 2e-cot 2 e 1 Example 7: .if A

=

[~

;

:]

and B =

456

Solution: We have 3A - 4B = 3

[~ ~ ~] thenfind 3A - 4B . 001

[~ ~ ~1-4 [~ ~ ~l 456

=

001

[~ ! 1~1-[~ ~ ~l 12 15 18

=

•

0-4 3-0 6-0 9-4 [ 12-0 15-0

0 0 4

1~]

14

MULTIPLICATION OF MATRICES

Let A = [aij] m x nand B = [bjk] n x p be two matrices such that the number of columns in A is equal to the number of rows in B then the product of A and B denoted by A 13 is defined as a matrix C = [c ik] mx p where cik = '£aij bjk or The product AB is defined as the matrix whose element in the ith row and! kth column is ail b1k + ai2 b2k + ai3 b3k ... + ain bnk' thus we conclude that: if A is an m x n matrix and B is an n x k matrix then the product matrix AB, is an m x k matrix. In the product AB, the matrix A is called the pre-factor and the matrix, B is called the postfactor. Also we saythatthe matrix A has been post -multiplied by the matrix B and the matrix

137

Matrices

B has Deen pre multiplied by the matrix A. The product in both the above cases AB and BA mayor may not exist and maybe equal or different. The productAB can be calculated only if the number of columns in A is equal to the number of rows in B.

Remarks • If AB = BA, then the matrices A and B are called commutative and if AB =-BA then the matrices A and B are called anticommutative. • The product oftwo non- zero matrices may be a zero matrix. • The product of matrices generally does not obey the law of cancellation. Theorem 1: Let A and B are symmetric matrices, show that AB is symmetric if and only if AB=BA. Proof: It is given A and B are symmetric matrices, therefore

A ' = A and B '= B

Let us first suppose To prove AB is symmetric. We know that

... (1)

AB =BA

...(2)

(AB) , = B' A' =B.A =AB (AB)' =AB

(R,eversallaw) [using (1)] [using (2)]

Hence, AB is symmetric. Conversely, Let AB is symmetric, i. e., (AB)' =AB = (AB)' = B' A' =BA AB = BA

Consider

L.H.S.

Hence,

Examplel:lfA

~[~ 3] I

[1

andB= 0

[using (1)]

0 2] 1 2.FindBA,canwefindAB?

0 2 3

4

Solution: We cannot finds AB since the number of columns of A is not equal to the number of rows of B. i.e., column of A is 2 and rows of B is 3. They are not equal. Since the matrix B has 3 columns and matrix A has 3 rows in BA, so product BA is defined.

Now,

I 0 2] [o 2 3

[I 3]

BA=012

x21 33 x

_[1.1 + 0.2 + 2.0 -

0.1+1.2+2.0 0.1+2.2+3.0

0 4

3 x2

1.3 + 0.1 + 2.4] 0.3+1.1+2.4

0.3+2.1+3.4

= 3x2

[I 2 4

11]

~4

3x2

138


2

-2 3 ] and B = 4 Example 2: IfA = [ -4 2 5 [ 2 J

Solution: We have AB

n

Find AB and shaw that A B '" BA.

:]

~ [~ -~ ~l'l~

= [1.2+(-2).4+3.2

-4.2+2.4+5.2

L~ ~]

=

Now

BA

1.3+(-2).5+3.1] -4.3+2.5+5.1

~ [~

l [-~ -2 3] 2 5

2.1 + 3(-4)

2( -2) + 3(2) 2(3) +3(5)] 4(-2) + 5(2) 4(3)+ 5(5) 2(-2)+1(2) 2(3) + 1(5)

4.1 + 5(-4)

=

[

2.1+1(-4) -10

=

[

2

21] -16 2 37 -2 -2 11

Hence, AB ~ BA.

Example 3: IfA

=

[~ -3

oJ

-;

J

~

BA.

2 0

Solution: Since A and B both are 3

AB

2]2. Find AB and show that AB

=

x

3 type square matrices, therefore

[~ -~ -~] [~ ~ ~] x

-3

2

1.1 + (-2).0 + 3.1

2.1+3.0+(-1).1 [

=

-3.1+1.0+2.1

[~ ~ ~~] -1

5 -4

1 2 0

1.0 + (-2).1 + 3.2 1.2 + (-2).2 + 3.0] 2.0+3.1+(-1).2 2.2+3.2+(-1).0 -3.0+1.1+2.2 -3.2+ 1.2 +2.0

Matrices

Now

BA =

[~ ~ ~] [~ -~ -~] x

1 2 0

-3

2

_ [1.1+0.2+2.(-3) 1.(-2)+0.3+2.1 - 0.1+1.2+2.(-3) 0.(-2)+1.3+2.1 1.1 + 2.2 + 0.(-3) 1.(-2) + 2.3 + 0.1

=

5 4

U

~ ~ ~]

Solution: WehaveAB=

and B

~

1

H~ ~lt~n

2.3+3:2'+4.0 1.3+2.2+3.0

1.1+2(-1)+3.0

=

[

(-1).1 + 1.(-1) + 2.0 (-1).3+1.2+2.0

12 [ -I = -1 7 -2 -1

BA~H

BA

2.0+3.1+4.2] .1.0+2.1+3.2 (-1).0+1.1+2.2

I:]

... (1)

2

0][ 1 x 21 23 41 3

0 2

-1

2 1.3+3.2+0.1

1.2+3.1+0.(-1) (-1).2+2.1+2.1+1(-1)

0.3+0.2+2.1

[ 2+3+0

3+6+0

"" -2+2-1 0+0-2

-3+4+1

-4+6+2

0+0+2

0+0+4

4+9+0]

From (1) and (2) , we conclude thatAB7f!: BA.

~ [ ~ 1 3] 2

-J

6. ShowthatA 2 = 0

-3

1.4+3.3+02]

(-1) . 3 + 22 + 1.1 (-1).4+2.3+1.2

0.2+0.1+2.(-1)

-J

AB~

3

[

EnmpleS, if A

prow that

1

2.1+3(-1)+4.0

=

1.3+2.(-1)+0.2

[~ ~ -1

Now

1.3 + 0(-1) + 2.2] 0.3+1.(-1)+2.2

[~ ~ ~]

HenceAB 7f!:BA

Example 4, ifA

139

[5 I~]

0.4 + 0.3 + 2.2

9

=

-1

2

-2

2

...(2)

140 Remedial Mathematics -Solution: We have A2

=A

xA

=[

~.

-1

-1

=

[

2

1.1+l.2+3(-1)

1.1 + l.2 + 3( -1)

2.l + 2.2 + 6( -1)

2.1+2.2+6(-1)

1.3+l.6+3(-3) ] 2.3+2.6+6(-3)

-1.1 + -l.2 + -3( -1) -1.1-1.2 - 3(-1) -1.3 -l.6 - 3(-3)

[~o ~0 ~] ~ 0, where 0 is 3 , 3 null matrix.

=

A2 = 0

Hence,

Example 6: Find the product ofthe following matrices

o A

2

c -b

= [-c

0 b -a

a

a] and B 0

= ab [

ac

Solution: We have c

0

AB=-c? [ b -a o.a 2 +c.ab+(-b).ac = (-c).a

[

2

+ O.ab + a.ac

O.ab+c.b 2 -b.(bc) 2

-c(ab)+O.b +a.(bc)

-c.(ac) + O(bc) + a.c

2

2

2

b.a +(-a).ab+O.(ac) b(ab)+(-a).b +O.(bc) b(ac) + (-a)bc + O.c 2 2

~[~ ~

n

Example 7: Prove that the product oftwo matrices 2 2 cos8sin8] and [ cos ~ cos 8 [ cos8sin8 sin 2 8 cos~sin~ is zero when

O.ac + c.(bc) + (-b).C

(J and

cos~sin~]

7t

¢ differ by an odd multiple of "'2

Solution: The required product

2 cos2 8 cos 8 sin 8] x [ cos ~ cos ~ sin~] 2 [ cos8sin 8 sin 8 cos~sin~ sin 2 ~

sin2~

j

Matrices

=

=

=

Now if8 -

~=

Then cos (9 -

2 2 [COS 9cos ~+cos 9sin 9cos ~sin ~COS2 9cos ~sin ~ +cos 9sin 9sin2 ~] cos 9sin 9cos 2 ~+ sin 2 9cos ~sin ~cos 9sin 9cos ~sin ~+sin 2 9sin2 ~ [cos 9cos ~(cos 9 cos ~ + sin 9sin ~)cos 8 sin ~(cos8 cos ~ + sin 8 sin~] sin 9cos ~(cos9 cos ~ + sin 9sin ~sin 9sin ~(cos 9cos ~ + sin 9sin ~ [cos 9 cos ~ cos(8 - ~) cos 9 sin ~ cos(8 - ~)] sin 8cos ~cos(8 -~)sin 8sin ~cos(8 -~) an odd multiple of

~)

>=

h

Example 8: IfA = [xyz}, B = h

b

[g

f

Solution: We have AB

'21t .

0 and consequently the above product is zero .

a

~ [xyzl' [;

~ h

] and C

~ ~] [

be three maMces. thenfind ABC

g]

b

f

f

e

= [x. a + y.h +z.g x.h + y.b + z.J x.g +y.J + z.e] Now

ABC

~[x.a +

y.h +zg'" + yb + ifxg+yf+ zc]

x[~]

= [x(ax + by +gz) + y(hx + by + fz) + Z (g:x +fy = [ax 2 + by + e~ + 2hxy + 2gzx + 2fyz]

Example 9: JfA

=

Solution: We have

141

[~ 2 2] 1

2 ,show that A2 - 4A -51 = 0

2

1

+ ez)]

142


9-4-5

8-8+0

8-8+0] = 8-8+0 9-4-5 8-8+0 [ 8-8+0 8-8+0 9-4-5

~ [~ ~ ~] Hence, A2 - 4A - 51 = 0 Example 10: Find the value o/x, y, z in the/ollowing equation

SoloUon: We have

[~ 2 3]

Y [X] [lX+2. +3.Z] 1 2 x Y = 3x+1.y+2.z

31 4 -2] o -6 x [ -1 2

and

[~]

Z

2x+3.y+1.z

[4.2+(-2).1] [6] = 0.2+(-6).1 = -6 -1.2+2.1 0

With the help of (1) and (2) the given equation reduces to

[;::;:~:] [-!]. =

0

2x+3y+z

On comparing the corresponding elements on both sides, we get x+2y+3z = 6 3x+y+2z =-6 and 2x+3y+z = 0 Solving these, we get ; : ;4} z=2 Example 11: IfA =

[4-1 2]1 Find (A - 21) (A - 31).

Solution: We have A _ 21 =;0

[

4

-1

= Also

2] _ 2 [1 1

0

0] 1

[~1 ~]-[~ -~]=[~1 ~1]

Matrices

143

[_~ _~] [_~ _~] = [~ ~] =0

(A - 21) (A -31) =

Hence (A - 21) (A - 31) = O. Example 12:.lf I =

[~ ~] C = [~ ~] show that (aI + bc)3 = a3I +3clbc.

[~ ~] + b[~ ~]

Solution:We have aI + bc = a

=[~ ~]+[~ ~]=[~ ~] 2

(aI +bci = [a b] [a b] = [a OaOa 0 2

(aI + bc)3

a [0

=

2ab] [a . b] a2 0 a =

2ab] a2 3 2 3a b] [a 0 a3

Verify that (AB),= B'A' where A' and B' are the transpose ofA and B respectively. [RGPV B. Pharma 2001)

Solution: We have AB =

[~ ~ ~1] [~ 0 ~] 4

5

=

Now

0

0

3

[~ ~ ~] [5 3 14] 14 5

0

L. H.. S = (AB)' =

1 -3

[1 0 0]

B = 2

1 0

o

1 3

2 6

5 0

~B'=

[1

2

0

0 0

:]

I

144


A{

and

3 2 -I] [ 1 o 2 ::::::>A '= 2 0 5

0

-1 2

~]

[5I 32 14]5

B'A' =

-3 6

0

L. H.S = RH.S. Example 14: IfA =

[~ =~] show that Ak =

[l+/k

1~4;k] where k is any positive integer.

Solution: We have A 2 = [3

-4] [3 -4] I -I I -I

=

[5 -8] [1 =

2 -3

+ 2.2 -4.2] 2 1-2.2

... (1 )

A3=[~ =~][~ =~]=G ~:]=[1+32.3 1~~~3] Thus the result is true for k

=

2,3.

Now assume that result is true for integer k i.e Ak = [ k [1+2k A k+/ -_ A·A k

-4k 1-2k

=[O+2k).3-4k.1 k.3+(l-2k).1 =

1+2k

][3 -4] I

k.

-4k] then 1-2k

-I

(1+2k).-4+ C-4k)C-I)] = [3+2k -4Ck-1)] k.(-4) + (l-2k).(-I) k+1 -1-2k

[I + 2(k + 1) -4(k + 1) ] k +1 1- 2(k + 1)

Hence, the result is true for Ak . Then it is also true for Ak+ / . Hence by induction the required result follows. Example 15: IfA = [ ;

Where

~l thenprovethatA 2 -6A +51= 1=

Solution: We have A2 = A.A =

Now

A2 -6A +51

=

=

[~ ~l [~

[1918 6]7 + [-24 -18 [~ ~]


Matrices

145

Example 16: Find the value o/x such that

[IIxJ

[~ ~ ~][}o

[UPTU B. Pharma 2001, 061

Solution: The given matrix is

[I Ix]

=>

;

m:]

= 0

[1+0+2x 0+2+x 2+1+0{] =0

[I

=>

[~

+2x2+X{] =0

[1 +2x+2+x+3] =0 =>3x+6=0 x=-2

Enmplel7'ifA =

[~

:]andB = [-;

=n FindD

= [; :] ,"chthalA + B-D = 0

Solution: We have A + B - D = 0 or D = A+B

1 2] [-3 -2] [1-3 2-2] [ -2 0] [P q] = [-2: - ~0] . = [: -~

D = 3 4 +

or

5 6

D

Hence

= 3 + 1 4-5

1 -5

4

or

5+4 6+3

3

;

:

P = -2 ,q=O ,r=4 ,s=-I, t=9, u=9.

Example: 18: IfA

=

I] . Find number a, b, so that (al +bA /

[0 -I 0

- °1 +

Solution: We have al + bA - a [1

=[

~

°

0] b [0 1 ] -1

: ] +[

~b ~] = [:b ~]

=

A.

146


(aJ +bA/

=

[a b] [a b] -b a x -b a

a 2 _b 2

= [ -2ab ... If(aJ + bA)2 = A then, we have 2

a _b2 [ -2ab

2ab a 2 _b 2

1 [0 1] -1 ° . =

Equating the corresponding elements, we have ~ a2-b 2 =0,2ab =1

a =b Example 19: A =

[~

=;1

[1I.J2].

=

B = [:

~l]

and(A +Bi =A2 + B2. Find a and b.

Solution: We have

-I] [1-2 -1+1] [-1 0] =

-1

=

2-2 -2+1

0-1

a +b a-I] ~I] = [ab-b b+ I 2

and

A2+B2

Also

A

=[-1°

2 0]+[a +b a-I] -I ab - b b + I

B -

1+a

0]

[I+a

-2 x 2+b

(1+a)2 40] = [ (2+b)(a-l) Now given that (A + B/ = A2 + 8 2. Hence from (I) and (2) , we get 2 [a +b-l (1+a)2 [ (2+b)(a-l) 4 = ab-b

0]

a -I

Hence

ab-I]

+ G=;]+[: ~I] -G:: =;:;]=G::

(A+ Bi = [ 2+b

or

=[a 2ab+b-l -b

°

= and b = 4 .

a=l,b=4.

0] [

-2

(1+ai +0 (2+b)(1+a)-2(2+b) ...(2)

Matrices

~~~~~~~I 1. If A = [20

2. If A =

3 -1

l]andB=[1 5 0

2 -I

~I~~~~~~~

-I]. Find2A-3B. 3

[~ ~ ~]andB=[~ ~ ~].Find3A-2B. 689

3. If A =

EXERCISE 4.1

[~ ~]

570

andB =

4 1

[~ ~ ~l'

FindBA.

2 3 0

4. Find the product ofthe matrices A = [; ~7 ~ 2 4

1

~8] , [~3 ~l' B=

-3 1

I -3 2] [1 4 ~ ~] [

3

6. If A = [ 1 -1 =

8. IfI=

9. [fA

10. If A

-1] 1

Showthat

-1

ShowthatA 2 =2A andA 3 =4A.

[~ ~lB= [~ ~lc= [~ ~lShowthatA(B+C)=AB+AC. [~ ~] andE= [~ ~] [;

cosh u . [ smh u

3

provethat(aI+bE)3=a I+3a 2 bE.

~ ~l ~lFindA2-5A =

1

[~ ~2 =; -o-~]

5. IfA = 2 1 -3, B = 2 1 andC, = 4 -3 -1 1-2 1 2 2 -5 AB =AC.

7. If A

147

+61.

sinh u] [COSh nu then show that An = . cosh u smh nu

sinh nu ] , where n in any cosh nu

positive integer. 11. Show that multiplication of matrices is not commutative.

148 Remedial Mathematics HINTS TO THE SELECTED PROBLEMS

to. Let A

=

cosh U sinh U] [ sinhu coshu

A2=A.A = [COShU SinhU][COShU SinhU] = [COSh2U sinh U cosh U sinh U cosh U sinh 2u In equation (1) and given value of A. Let us assume that the An = [COSh na sinh na A n+ 1 = An.A

Now to show

= [COSh na sinh na

...(2)

sinh na] [COSh a cosh na sinh a

sinh a] cosh a

sinh(n + I)a] cosh(n + I)a

i.e., equation (2) holds forn+ 1 ifittrue fonn. Hence by mathematical Induction, we have An = [COSh na sinh no.

sinh na] cish no.

-23 -I3] andB= [I0 11. Let us take A ~ [ ~ ]

-3

2

AB~p -23 -I3]['0 -3 2 1 [1.1 + (-2).0 + 3.1 ]

] 0

1 2

0

1 2

~l

~]

1.0+(-2).1+3.2

1.2 + (-2).2 + 3.0]

2.1+3.0+(-1).1

2.0+3.1+(-1).2

2.2 + 3.2 + (-1).0

-3.1+1.0+2.1

-3.0+1.1+2.2

-3.2 + 1.2 + 2.0

[~ ~ ~~l -1

Now BA = [0]1

5 -4

~ ~l [~ ~2 ~]l = [~ ~ ~l x

2 0

-3

... (1)

sinh na] cosh na

[COSh(n + I)a sinh(n + I)a

=

Sinh2U] cosh 2u

2

5

4 ]

Hence AB ;If BA Clearly, the ~ultiplication of matrix is not commutative.

Matrices

149

'ANSWERS

1.

4.

Co 10

AB~ [:

6 3] [~O

[ -I

5] 1

2.

15 11 10 27

~6

: ] BA j, undefined

9. [

~]

~I

-1 -1 -~O

-5

-8 -8 •

11

3.

-3]

4

ADJOINT OF A MATRIX

Definition: Let A = [aij] n x n be a square matrix of order n x n. Then the adjoint ofA is a matrix of the same order n x n which is obtained by the transposing of a matrix whose elements are cofactor ofthe element ofA in the determinant A . That is ifB = [A ti] n x n where Aij are the cofactors ofthe elements aij in the determinant IAI. Then B'is called the adjoint ofA. It is denoted by ad) A.

Remark • Sometimes the adjoint of a matrix is also called the adjugate of the matrix.

Theorem 1: IfA is a square matrix oforder n then A. (adj A) = (adj A) .A = IAl.ln where I is the unit matrix ofthe same order as A. "'-

Proof: We have the (i,j) th element of the product A (adj. A) = Product of the ith row of A and jth column of adj A I = aliAjl +a2i Aj2 · .. +aniA jn = 0, when i;f:j =

IAI when i = j.

Thus, is the product, only the diagonal elements exist and each is equal to IAI while all other elements are zero , so that

A .(adjA) =

or

IAI

0

0

o

IAI

0

0

0

IAI

o o o

o

0

o

IAI

A.(~AHA{t Similarly, (ad) A). A = IAI . I. Hence the theorem is proved.

:

. ~.l ... I

=

". IAI

.1

150


~~~~~~~I SOLVED

EXAMPLES

~I~~~~~~

Example 1: Find the adjoint ofthe matrix A =

[~

: :]_ 9 10 12 Solution: For the given matrix A, we have

7 [170 1~] =4; A12=-[~ 1~] [~ 10] =-13; =- [1~ 1:] = G1:] =-24; [~ 1~] =8; =[27 4]8 =-12- =_[15 4]8 =12- =[15 72] =-3

AJI=

A21

A

=12;A J3 =

16;A 22 =

A

31

A23=-

A

'32

'33

Therefore the matrix B formed by the cofactors of the elements of IAI is: B =

[1: ~;4 ~~3] -13

8 Nowadj A = transpose ofthe matrix B =

•

-3

[1:-12 ~;412

INVERSE OR RECIPROCAL OF MATRIX

Definition: Let A be a square matrix of order n x n and there exists a square matrix ofthe same order such that AB = BA = 1, where In is a unit matrix oforder n x n _ Then the matrix B is called the inverse of a matrix A_

Remarks • A matrix 'A' is invertible ifit is non - singular. adj.A I I -lA!' A :;1;0

-1_

• A

Theorem 1: The inverse ofa matrix is unique. Proof: Ifpossible, let Band C be two inverses of the matrix A, then by definition AB =BA =1 and AC =CA =/. From (l) and (2) , we get AB = AC , each being equal to I or B(AB) = B(AC) or (BA) B = (BA) C or IB =IC B=C. or Hence, the inverse of a matrix, if exists is unique.

• ...(1) ... (2)

Matrices

151

Theorem 2: A square matrix A has an inverse if and only ifA is non - singular Proof: The condition is necessary; LetB be the inverse of the matrix A, thenAB = I Therefore,

IAIIBI IAI

=

III

=1

¢O

The condition is sufficient: Let

IAI

¢

0, we assume that

B = adjA

IAI (adjA) AB =AliAif

IAII =TAT1 (A ad]. A) =,lAI =I . Similarly

BA =1 AB =BA=I

Hence, A has an inverse. Theorem 3: IfA and B are two non - singular matrices of the same order, then AB is also non - singular and (AB ri = S-i A-i Proof: Let A -i and S-I exist. Since A and B are non - singular, therefore, (AB) (B- 1 A-I) = A (BS-I) A-I, (By associative law) (By inverse property)

= AlA-I =AA- I =1

Similarly, (S-I A-I) (AB) = I (S-I A-I) (AB) = (AB)(S-I A-I)=I i.e., S-I A-I is tIl1e inverse of AB or (AB)-I = S-I A-I and as such AB is also non - singular Theorem 4: IfA is a non - singular matrix, then (A-ir i =A. Proof: Let A-I be the given matrix instead of A, then (A-I)A =A(A-I)=l.

This show that A is the unique inverse of A-I i.e., A =(A-irl Theorem 5: The inverse of the transpose ofa matrix A is the transpose ofthe inverse ofA, i.e., (AT i = (A-i) '. Proof: We have AA-i = A-i A .. (AA- I ), =f=(A-IA),or(A-I),A'=I=A'(A- I ), Hence, (A"I)' is the inverse of A' (~.4')-1 = (A-I),

•

ORTHOGONAL AND UNITARY MATRICES

(A) Orthogonal Matrix Definition: A matrix A is called an orthogonal matrix. matrix .

IfAA '=1 = A 'A, where I is an identio

152 Remedial Mathematics Theorem 1: ifA is an orthogonal matrix, then A-I is also an orthogonal matrix. Proof: By definition, if A is orthogonal, then AA'=A'A =1 or (AAT I =(A'Arl =]' [:.1' =1] or (A- 1),A-1 =A-1 (A- 1),=1. Hence A-I is orthogonal.

i.e., Inverse of an orthogonal matrix is also orthogonal. Theorem 2: IfA and Bare n- square orthogonal matrices then AB and BA are orthogonal matrices. Proof: Since A and B are orthogonal matrices, we have AA' =1andBB' =1 ... (1) Now (AB)(AB)' = (AB)(B'A') =A (BBJA' by associative law =AIA' since BB'= 1 = AA' Hence AB is an orthogonal matrix. Similarly we can show that BA is also orthogonal.

(8) Unitary Matrix Definition: A square matrix A is said to be unitary if AB A = 1 where 1 is an identity matrix and A Bis the transposed conjugate ofA . The elements ofA are complex numbers. SinceAs=IAland IAAsl = IAIIA S I,thereforeifAA s =1,wehave IAI IAI = I. Thus determinant of a unitary matrix is of unit modulus. For a matrix to be unitary, it must be non- singular. Hence AA s =/::::;.AsA = 1 i. e., AA s =1=A sA.

~~~~~~~I SOLVED

EXAMPLES

Example 1: find the adjoint ofthe matrix A =

[1 2] 3 -5

~I~~~~~~

and verify the theorem

A(adj A) = (adjA)A = IAIl. Solution: We have IAI = [1

2] = 1(-5) - (3) 2 =- 5 - 6 =- 11 .

3 -5

The cofactors of the elements of the first row of IAI are - 5 , - 3 respectively. The cofactors ofthe elements of second row of IAI are - 2 ,1 respectively. Therefore the matrix B formed by the cofactors of the elements of IAI is

B

=

[-5 -3] -2

1

adj A = transpose of the matrix B =

[=~ ~2]

Matrices

Now

A(adj A) = [1 2] [-5 -2] 3 -5 -3 1 =

[-5-6 -15+15

=(-11)

Also

(adj A)A =

-2+2] = [-II -6-5 0

0] -11

[~ ~] = IAI I

[=~ ~2] [~ ~5]

= [-5-6

-3+3 = [

-10+10] -6-5

-~ 1 _~ 1] = (-11) [~ ~ ]

HenceA (adj A) = (adj A)A = IAII.

Example 2: Find the adjoint a/the matrix B

A~[~ iJ ~l Solution: For the given matrix A, we have All

=1-12 -313 =3, 21=_5 ' -1

A 12

=_1 21

A 21

=_1 -11 311 =-4 '

11 3

A 23

=_1 2

11 -1

=3 '

A

=

_III

11 -3

=4 '

= I

1

'

A31

=I~ ~31 =-5,

A33

=I~ ~I =1.

32

-31 3 =-9 '

Therefore the matrix B formed by the cofactors ofthe elements of A is B=

[~4

-9 ~5l

-5

4

Adj A = B'=

[~9 -5

1

: ~l

153

154


Example 3: ifA =

[~ ~ ~]

,find A2 - 2A + Acij A .

243 Solution: We have A2 =

[~ ~ ~] [~ ~ ~] x

24324 3

[1+0+6 = 0+0+0

2+10+12

3+0+9]

0+25+0 0+0+0 2+0+6 4+20+12 6+0+9

[7 24 12] 0

= 0 25

Also

All =

I~ ~I = 15, !I =- 10

A21

=-I~

A22

=I~

31 =-3 3 '

A 23

=_1 2

21 =0 4 '

A31

=I~

o31

A32

=-I~

o31 =0 '

=-15

1

'

31 =6 3 '

I~ ~I =5.

[ 15 B= 6

0 -3

AdjA =B'=

0

A'-2A+AdjA

~ [~

-n

[150 -36 -15] 0 o

-10 .,

=-I~ ~I =0.

=I~

-15

..

AI2

AI3

A33 =

.,

...(1)

36 15

8

24 25

...(2)

5

12] 3] [15 6 -15] o -2 [I0 2 5 0 + 0 -3 0

36 15

2 4 3

-to

o

5

(Using(l) and (2»

4 6]

[15 10 0 + 0 8 6 -to

Matrices

7 -2 + 15 24-4+6 12-6-15] 0-0-0 25-10-3 0-0-0

= [

=

[

Example 4: Find the inverse ofA

8-4-10

36-8+0

20 0

28 12

-9] 0

-6

28

14

=

[~

15-6+5

:

3 5 Solution: For the given matrix A ,we have A

11

=14 51=_1 5 6 '

A 12

=_1 23

51 =3 6 '

AJ3 =

I~

;1 =-2,

A 21

=_1 25

31 =3 6 '

=

I~

!I

=- 3,

A 23

=_1 31

21 =-1 5 '

31 =-2 5 '

A 32

=_1 21

31 = 1 5 '

A22

A31 = A33 =

..

I! I~

!I

=0. 3

B= [-I3

-2] 1 .

-3

-2

..

0

[-I AdjA =B'= 3

3 -3

-2

Also

[I 2

IAI = 2 4 3 5

-2] 1 . 0

:]

= 1(24-25)-2(12-15)+3(10-12) =-1+6-6 =-1. Now

A-I

~ 01A {~ -~

-J [~ ~: ~I]

ISS

156


Example 5: ifA =

[

9

7

5

-J

6

8

n

Findadjaint AandA-/

Solution: The given matrix is A =

[RGPV B. Pharma 20021

[~ ~1 ~l 682

IAI

=

[~ ~1 ~] =9(-2-32)-7(10 -24) +3(40+6)

682 =-70. as IAI :#: 0 so A-I will exist. The cofactors of the given matrix are We have All =-34 A21 = 10 A31 =31 A22 =0 A32 =21 AI2 = 14 A23 = 30 AJ3 = 46 A33=-44. Adj A =

[-34 10 31

..

A-I =

a~~

46] = [-34 10

10 0 30 21 -44 1 =_ 70

46

[-34 10 46

31 ]

10 0 21 30 -44

31 ]

10 0 21 30 -44

1 [34 -10 -31] = --- -10 0 -21 70 -46 -30 44

Example 6, Find the in"",e afthe matrix

[~

;

:J

Solution: The given matrix is

A=

[~ ! ~] 2 7 11

IAI as IAI :#:, so A-I exists.

= 1(44-35)-2(33+10)+3(21-8) =2.

[RGPV B. pharma 2003]

Matrices

157

The cofactors ofthe given matrix A are: All = 9 A21 =-1 AI2 =-23 A 22 =5 Al3 = 13 A 23 =-3 AdjA =

[~1

-:3 ~~11

-2 A-I =

A31 =-2 A32=4 A33=-2

4

=

13-3 -2

-2

AI~IA ~ [-~3 ~I =

Example 7: Find the inverse ofthe matrix A =

~l

-3 -2

13

[~ ~ 3

Solution: We have

[-~3 ~I ~21

1

~l


IAI = 1(6-1)-2(4-3)+3(2-9) = 5 -2-21 =-18

Since IAI :t; 0, therefore £1 exists. Let Ai} be the cofactor of ai} in IAI, Then we have All =5 AI2=-1 Al3=-7

Now

adjA =

A31=-7 A 32 =5 A33=-1

[~1 =~ ~71 [~1 =~ ~71 =

-7

A-I

A21=-1 A22=-7 A 23 =5

5

-1

-7

-1

=1~ladjA=-I~ [~1 =~ ~71 -7

Example 8: Find the inverse ofthe matrix A = [; 2 Solution: We have

5

5

-1

~1 ~11 3

-1

IAI = (1 +3)-2(-1 +2) +5 (3 +2) = 4-2+25 =27

Since IAI:t; 0, therefore A-I exists. Let Ai} be the cofactor of ai} in IAI , then we have . A I1 =4, AI2=-I, A 13 =5, A21 = 17 A 31 =3,

A22=-ll, A23= 1 A 32 =+1, A33=-3

[U PTU B. Pharma 2001)

158

Re~dilll Mathematics

adjA ~ [I;

Now

A-I

-1

-11

T[4 1

=-1

-3

5

~ I~I adj ~ 12~1 [ ~I

1

JJ

17

JJ

1

~ [~ 33 4]1 2

IAI

-11

-II

A

Example " Find the invem ofthe mo"ix A Solution: We have

17

[UPTU B. Pharma 2002J

4

=2(12-2)-3(16-1)+4(8-3) = 20-45 +20 =-5.

IAI * 0, Therefore A-I exists. Let Aij be the cofactor of aij in IAI Since

Then we have

Now

Example 10, ifA "

[~ ~ ~],

using A' - 4A - 51

~

0, ond hence find A-I

[UPTU B. Pharma 20003-2008) Solution: We have

[~1

22

2~]

Matrices

159

! :]-4[~ 2~]-5[~ ~ ~]

2 A -4A-5/= [: 889221001

9 4 5 8-8-0

8-8-0] = 8=8=0 9-4-5 8-8-0 [ 8-8-0 8-8-0 9-4-5

n

~ [~ ~

Now we have to find A-I Multiplying both sides of A2 - 4A - 51 = 0 by A-I, we get A-I (A 2 -4A -51) = O. 2 A-I A -4A- IA-5A- I 1=0 or

5A-

1

~A-41~ [~

~3 ~]

= [-;

2

2

-3

5[ ~ ~3

~J

2

-J

-3

2

1

-I

A

=

Example 11: Show that the matrix A

=

-

[ A3 -

6A 2

:

J

2

J -J

~

] satisfi., the equation

+ 9A - 41 = O. Hencejind A-I.

[UPTU B. pharma 20041

2-I I] [

Solution: We have A = -I

1

2 -I -I 2 2

-I

A2 = A.A= -I [ 1

2

~I] [~I ~I ~I] [~5

-I

2

=

1

-I

2

5

-: ~5]

-5

6

~I ~Il [~5 ~

51 [22 -21 -5 = -21 22

-I

6

2

5

-5

21

-21

. 21] -21

22

160


Now A 3 -6A 2 +9A-41 =

[

22 -21 -21 22

21; -6-5 [ 6 -21

21

22

-21

5

0 0 0] = 0 0 0 =0. [ 000 Now we have to find A-I Multiplying both sides of A 3_ 6A 2 + 9A - 41 = 0 by A-I, we get A-I (A 3 -6A 2 +9A-,4J) = 0 £1 A 3-6A- I A 2 +9A- I A-4A-I 1 = 0 A 2 -6A +91-4A- = 0 4A- I = A2-6A +91.

-[~ ~]-f' ~}9[~ ~] -5

-1

0

6

2 -1

0

-5

1

-[: A-I -

3

1

:']

' -I]

±[:

3 -1 1

1 3

1 2 1] [112

Example 12: Given that A = 3 2 3 ,compute (i) det A (ii) Adj A (iii) A-I. Solution:We have

(i)

IAI

=

[~ ~ ~] I

I

2

1 (4-3)-2(6-3)+ I (3-2) =1-6+1 =-4. =

(ii) Now the cofactors of the elements of the first row of

IAI

I~ ~I'I~ :21'1~ ~1'i.e.,arel,-3,t.respectivelY.

are

Matrices

The cofactors of the elements of the second row of

IAI

are

-I~ ~I' I~ ~I' -I~ ~I i. e., are - 3 , 1 , 1 respectively. The cofactors of the elements of the third row of

IAI

are

I~ ~I' -I~ ~I'I~ ~I i. e., are 4 , 0 , - 4 respectively. B

~ [~

AdjA = B'=

-3 0

[

~3

~l

~l

-3

-4

[,

.... Adj A 1 (m) The mverse of A = = - "4 ~3

lAI

-3 1 I

[0 01

Example 12: IfA = 0

1 0

So'ution. W,have IAI .

~

[: 1

il

,how thot A-I

o1 0'] = [~ o 0

~

1 3 - -1 4 4 3 1 -0 4 4

--

l

-4

4

4

A.

~] =-1.

Now for the given matrix A, we have All

AI3

A22 A31 A33

[~ o0] =0' [~ o1] =-1 ' [~ o1] =-1 ' [~ o1] =-1 ' [~ ~] =0.

AJ2 =-

[~

o0]

A21 =-

[~

o1] =0 '

A23

=-[~

A32 =

-[~

o0]

=0

=0

'

'

o1] =0 '

161

162


Therefore, the matrix B formed the cofactor of the elements of IAI is B

0

0 -1] -1 0 .

-I

0

= 0 [

0

Acij A = B' = transpose of the matrix B

=

[~ ~l ~l]. -1

£1

0

0

= AcijA = _1_ [

IAI

1 . Example 13: Show that the matrzx A

Solution: We have A =

Then

We have

~

1

="3

[

2

~1 ~1]=[~ ~ ~]=A o

~ ~2]

0

1 0 0

is orthogonal.

-2 2 -1

1 21 -22] .

2 -2

[

~

(-1)_1

2 -1

A' =

~

[

1

2

2

1

2 -2

AA'=.!.X.!.[~-2 3 3

9 0

=

io

0 9

[

0

Hence, the matrix A is orthogonal.

cose Sine] -sme cose is orthogonal.

Example 14: Show that the matrix. [ cose Sine] . Solution:LetA = [ -sine cose Then

A' = [cose

sine

-Sine] cose

Matrices sinO] [COSO cosO sinO

cosO AA'= [ -sinO

We have

2

I

2

-sinO] cosO

cos 0+sin 0

cososino-sinocosoj

[ sin Ocos O-cos Osin 0

sin 2 O+cos 2 0

=[~ ~]=I. Hence, the condition of orthogonality is satisfied. Therefore, the given matrix is orthogonal.

+i -1+i] 2 2.

1 Example 15: Prove that the matrix

[ 1+ I 2

.

I-I

.

IS

. umtary.

2

+i -1 +i] 2 2 I

'Solution: LetA = 1+; [. 2

1-; 2

Then

Now

AA=[ 1;; l;i][I;i -12+i] -1-; l+i l+i l-i - -- -- -9

2

Hence, the matrix A is unitary .

2

2

2

163


1. Find the adjoint of matrix A

~ [~

0

1 2

2. Find the adj oint of the matrix A

~ [i

3. Find the inverse ofthe matrix A =

[~

~] 2

-1

~l

~

~l

0

4. Find the inverne of the matrix A [:

1 0 -2

5. Find the inverse ofthe matrix A = [ 22

n

3

2

3 6. Show that the matrix

2 3 2 3

3

2

3

3

2

2

3

3

is orthogonal.

1 3

. A =.fi 1 [1 . 7. Show that the matnx

i ] is unitary. -1

2 -/

8. Find the inve"e of the matrix A

~l

-1

-1

~ [~ ~

:]

9. Find the inverse ofthe matrix A , where

A

~ [~ =: ~l

10. Find the inverne of A

~ ~ ~l [;

11. Find the inverne of the matrix A

~

[! : ~1

[RGPV B. Pharma 2004]


Matrices

165

I ANSWERS

[~2 ~2] 4

1.

4.

9.

11.

2.

-5 -2

[~I-1

0 0

[~2

-1 3 -2 3

67 11

•

5.

10.

~ [-5 0 10 5

~[~

~]

1 3

-5

~] ~]

__I[~

[~9

-4

2

-~6]

0

10

2

2 2 -3

3.

8.

[

~7 ~2]

[~I

0 -1

7]

7]

-37 -1 8 18 -19

13]

SOLUTION OF EQUATIONS USING INVERSE OF A MATRIX

In this section, we shall express the system of linear equations as matrix equation and solve them using inverse ofthe coefficient matrix. Consider the system of equations.

alx+bly+ clz = + b2Y+ c2z = a 3x + bJY + c 3z =

azx

Lcl

A

dl d2 d3 .

f : ].

= [::

X= [;].

B=

[~:]

Then, the system of equations can be expressed in the form

[~ ~ .

a3

i.e.,

v3

: : ] [ : ] = [::] C3 Z d3

AX=B. If A is non-singular matrix, then its inverse exists. Hence, we have A-I (AX) = £1 B

X=A-1B This matrix equation provides solution for the variables x, y and z .


ISOLVED

EXAMPLES

I

Example 1: Solvefor x andy by inverting the matrix in the following IUPTU B. Pharma 20011

Solution: We have

IAI

= 6 - 1 = 5 :# 0 . Therefore A-I exists.

Cofactors ofthe elements ofthe first row of IAI are 3 ,-I . Cofactors of the elements of the second row of IAI are -I ,2. Now

..

adj A = [

A-I =

3 -I] 3 -I]. T

-I

2

[;J

=

Hence

-I

2

I~I adjA= ~[~I ~Il

I X=A- B= or

=[

~[~1 ~IJ [~J

=

GJ

[~J

x = l,y=2

Example 2: Solve using matrix method, the following equations x+y=O, y+z=l, z+x=3, [UPTUB.Pharma2005] Solution: The given system of equations can be written is matrix form as AX=B. where

IAI = 1 (1-0)-1(0-1)= 1 + 1 =2:#0. Let Aij be the cofactor of aij is IAI . Then we have We have

A\1 = 1, A21 = -1, A31 = 1 A12=1 A22=1 A32=-1 A\3 = - 1 A 23 = 1 A33 = 1. Now

adjA =

[~1 1

A-I

~ I~I

:

~llT

= [:

-1

-1

1

-I

adj A

~~

[I -I ~Il

~ll 1

Matrices

1

~[~

We have

X=A- B=

Hence,

-1 x = 1, y=-I,z=2.

Example 3: IfA

~ [~ ~1 ~ 1

,FindA- 1 .

1

-1

Hence solve the equations x + y + z = 6, x - y + z = 2, Solution: We have

IAI Since

IAI

=1(1-1)-1(-1-2)+1(1+1)=0+3+3=6.

:1= 0, therefore A

is invertible.

Let Aij be the cofactor of aij in

A A

11

12

A13

=(-1)1+21 =3, 0,

adj A =

A21

II I-I

-I

11

2

=2,

-1

=

= 0.

1=3.

A 22 =-3,

A 23 = I,

A31

=2,

A33 =-2.

[~ ~3 ~ 1= [~ ~3 ~ 1 2

A-I

IAI . Then we have

= (_1)1 + 1 1-1 1

A32 =

Now

2x + Y - z = 1.

~ I~I

°

adj A

-2

3

-2

~ ~[: ~3 ~J

Writing the given system as a single matrix equation, AX = B. we get

~[~ ~I ]l[~l

~3 ~J [~l or Hence,

x = 1, y=2,z=3.

167'

168 Remedial Mathematics Example 4: Solve the following equations by matrix method: 2x-y+3z=l, x+2y-z=2, 5y-5z=3, [UPTUB.Pharma2003,06] Solution: The given system can be written as a single matrix equation AX=B. where IAI = 2(- 10 + 5) + 1(-5 + 0)+ 3(5 -0)=-10-5 + 15 =0.

We have

Since IAI = 0. Therefore, to judge whether the given system is consistent or inconsistent, we shall calculate (adj A) B. Let Aij be the cofactor of aii' then we have All =-5, A 12 =5, A 13 =5, A 21 =0, A22=-10, A 23 =-10, A31 =-5, A 32 =5, A 33 =5.

++H: ~:~ -!l ~:: :ml {:~~E::H~l

adjA~ [~~

Now

.

(adjA)B

~

[:

Since IAI = 0, (adj A) B = 0, therefore the system of linear equations is consistent and possesses infinite number of solutions. Taking Z = K where K is real Substituting for z in the first and second equations of the given system, we get 2x-y= 1-3k, x+2y=2+k. Writing these equations as a single matrix equation AX = B, we get

[~ ~1] [;] = [~-:: Here

l

IAI =4+1=5.

Cofactors of the elements of the first row of IAI are 2, - I. Cofactors of the element of the second row of IAI are 1,2. Now

adj A

=

[21

_I]T = [ 2 2

A-I=~(adjA)=.!.[2 IAI

We have

=>

I]

-I 2

5 -1

_I 1 [2 X=A B="5 -1

I]

2

1][1-3k] 2 2+k

1 1 x = - [4-5kl, y= - [3+5kl,

5

5

1 [2-6k+2+k] -1+3k+4+2k

="5

I [4-5k] 3+5k .

="5

Matrices

169

1 1 Since x = - [4 - Sk], y = - [3 + Sk], z = k, k is real, also satisfy the third equation of S

S

the given system for all real values of x. Therefore,

1 x = - [4 - Sk] S

1 Y = -S [3+Sk] '

z=k form infinite solutions of the given system. Example 5: Solve the following system ofequations by matrix method: x + 2y + z = 4 2x + Y = 3 x +z = 2 [RGPV B. Pharma 2004] Solution: The given systems is equivalent to AX = B :. A-I B.

Where

Now

IAI

~ [~ : ~] ~

10-0)-2(2-0)+

1(0-1)~

The cofactors of A are All = 1

Adj A =

A21 =-2

A31 =-1

A22 =0

A32 =2

A 23 =2

A33

[~2 ~ ~1]1 [~2 ~2 ~1]. =

-1

2 1

Then

A

-I

=-3

1 =-"4-2 [ -1

-3

-1

-2o -1] 2

x=l,y=l, z=l.

2 -3

2

-3

1-4-\

~-4

170


Example 6: Using matrices solve the equations 2x-y + 3z =9 x+y+z=6 x-y+z=2 Solution: Given system is equivalent to AX = B.

[RGPV B. Pharma 2001)

where

X=A- I B.

(A)

~

-1 [:

1

:]

~2

(1+ 1)+ 1 (1-1)+3

(-1-1)~4-6~-2.

-1

The cofactors ofthe elements of IAI are

2 AI2 = 0 A\3 =-2

All =

A21

A31

=-1 A 23 = 1

A-I

0 -1

A33 =3

-2) [2

1 = 0 3 -2

~ ~~IA ~ ~IP-2

X=A- I B

=-4

A32 = 1

A22

AdjA+~-4 Then

=-2

-2 -1

-2 -1 1

~]

~]

~~~~~~~I EXERCISE 4.31~~~~~~~~ Solve the following system of equations by matrix method 1. 5x+2y=4 2. 2x-y=-2 7x+3y=5 3x+4y=3 3. 4x-3y=3 4. 2x+5y= 1 3x - 5y = 7 3x + 2y = 7

Matrices

5. 2x+y+z= 1 x-2y-z= 3/2 5y-5z=9 7. 2x+3y+5z=5 x-2y+z=-4 3x-y-2z=3

171

6. x-y+z=4

2x+y-3z=O x+y+z=O 8. x-y+2z=7 3x+4y-5z=-5 2x-y+3z= 12.

9. Solve the system of equations using matrix method

x+y+2z=4 x+3z= 5 2y+3x= 7 10. Solve2x+3y+z=9 x+2y+3z=6 3x+y+2z=8.


By finding inverse of coefficient matrix.


HINTS TO THE SELECTED PROBLEMS 4. The system of equations can be written in the form AX = B, where

Now

Hence

IAI

=

All

=

-11 ,.to. 2,A I2 =-3,A 21 =-5,A 22 =2.

A-I = -

/1 [~3

~]

1[2

X=A -I B=-II -3 = -

1\ [~313]

=

[~1]

x = 3,y=-1. 8.

IAI

=-17,.tO. All =-IA I2 =-8A 13 =-1O A21 = -5A 22 =-6A 23 = 1 A31 =-IA 32 =9A 33 =7.

Hence,

1

A-

~ ~ ~~o ~ ~ll -

1 [

172

Remedial Mathematics -I

X =

we have

[IB=_

I~

-8 [

-10

x =1,y=2,z=3.

I ANSWERS 5

12

x=-ll ,y= II

6

1. x=2,y=-3

2.

4. x= 3,y=-1

1 3 5. x= \,y= 2'z=- 2

3.

-19

x=-ll ,y= I t

6. x=2,y=-1,z= 1 7. x= I,y=2,z=-I 8. x= 1,y=2,z=3. •

RANK OF A MATRIX

Definition: A positive integer r is said to be the rank ofa matrix A if it contains at least one square submatrix oforder r x r, whose determinant is non-zero while any square submatrix ofA of order (r + 1) x (r + I) or greater is singular i.e., having determinant zero. The rank ofa matrix A is denoted by p (A). It is obvious that the rank r of a matrix of order m x n may at most be equal to the smaller of the numbers m and n, but it may be less. I f the rank of a square matrix A of order n x n is rand r < n, then matrix A is said to be singular, on the other hand if r = n, then the matrix is said to be non-singular.

Remarks • If the rank of a matrix is zero, then matrix is a null matrix • The rank of every non-zero matrix must be greater than or equal to 1. • The rank of a unit matrix is equal to the order of the unit matrix.

III

ECHELON FORM OF A MATRIX

Definition: A matrix A is said to be in Echelon form ifit satisfies following conditions: (i) Every row of A has all its entries zero which occurs below the every row having a non-zero entry. (ii) The number of zeros before the first non-zero entry in the same row is less than the number of zeros in the next row.

Matrices

173

Remark • The rank ofa matrix is equal to the number of non-zero rows in Echelon form of the given matrix.

For example: Let

0 2 33 25] [0 0 0 0

A= 0 0

This matrix A is in Echelon form and it has two non-zero rows since rank ofA is equal to the number of non-zero rows. Hence rank of A = 2.

Theorem 1: The rank ofthe transpose ofa matrix is the same as that ofthe original matrix. Proof: Let us suppose A is any matrix and A I is its transpose and let rank of A = r. This implies that A contains at least one r-rowed square matrix whose determinant is non-zero, let it be B. Obviously B' is a submatrix of A' but we know that det B' = det B and since det B *- 0 => det B' *- O. Thus the rank of A' ?r. Now if A contains a (r + 1) -rowed square submatrix C, then det C = 0 because rank of A = r. Obviously C' is a submatrix ofA' and det C' = det C = 0, it follows that A does not contain (r + 1)-rowed square submatrix with non-zero determinant. Hence rank of A'::; r and consequently we obtained rank of A' = rank of A . I

•

ELEMENTARY TRANSFORMATIONS OF A MATRIX

Definition: A transformation is said to be elementary transformation ~r it is one of the following: (1) Interchanging of any two rows (or columns). (ii) Multiplying any row (or column) by any non-zero number. (iii) Addition of any row to K times the other row, where K is any non-zero number.

Remarks • lfthe elementary transformation (or E- transformation) is performed on rows, then it is called row-transformation. • If the E-transformation is performed on column, it is called column-transformation .

•

ELEMENTARY MATRICES

Definition: A matrix which is obtained by a single E-transformation is called an elementary matrix. For example

[:

A~ [~

0 1 o , 0 0I :] cto '][' 0 o 0 0 -10 8 -4

~']

-2

174


Here first E- matrix is obtained from 13 by interchanging C 1 and C3 columns and the second E-matrix is obtained by RI~R2 + 2R2.

Remarks • All the elementary matrices ar~ non - singular. • Each elementary matrix possesses its inverse.

III

INVARIANCE OF RANK UNDERE-TRANSFORMATION

Casel. Elementary transformation (E- transformation) do not change the rank ofa matrix. Case ll. Multiplication of the elements of a row by a non-zero number does not change the rank. Case ID. Addition of any row to the product of any number K and other row does not change the rank.

Remarks • The rank of a matrix does not change by a series of E-transformation. • The rank ofa matrix does not change by a column-transformation.

III

NORMAL FORM

Definition: If a matrix is reduced to the form (I;

~) . Then this form is called normalform

of the given matrix.

Theorem 3: Every matrix oforder m x n ofrank r can be reduced to the form

(

Ir 0

~) bya

finite number ofE-transformations, where Ir is the unit matrix of order r x r.

Remark • The form (

~ ~) of A is also called first canonical form.

Corollary 1. The rank ofmatrix oforder m x n is r ifitcan be reduced to

(~ ~) byafinite

number ofE-transformations.

Corollary 2. IfA is a matrix oforder m x n ofrank r, then there exist non-singular matrices P and Q such that

PAQ =

III

(~ ~)

EQUIVALENCE OF MATRICES

Definition: Let A be a matrix oforder m x n. Ifa matrix B oforder m x n is obtainedfrom A by afinite number ofE-transformations, then A is called equivalent to B. It is denoted by A - B (Read as A is equivalent to B).

Matrices

175

Theorem 4: The relation "-" in the set ofall m x n matrices is an equivalence relation. Proof: (l) Reflexivity. IfA is a matrix of order m x n, then A is equivalent to A i.e., A-A. (il) Symmetry. Let A and B be two matrices of order m x n and A-B. This implies if B is obtained from A by a finite number of E-transformation, then A can also be obtained from B by a finite E-transformations. Hence B - A. (iii) Transitivity: LetA, B, C, be three matrices of order m x n and A - B, B -CO This implies that of B is obtained from A by a finite number of E-transformation and C is obtained from B by a finite number of E - transformations, then C can also be obtained from A by a finite number of E-transformations. Hence A - C.

Hence the relation "-" is an equivalence relation.

Remarks • An equivalence relation is a relation which is reflexive, symmetric and transitive. • Two equivalent matrices have the same rank. • Two matrices of same order and of same rank are always equivalent.

_ _ RANK OF PRODUCT OF MATRICES Theorem 7: The rank ofa product oftwo matrices cannot exceed the rank ofeither matrix. Proof: Let A and B be two matrices of order m x nand n x p respectively. Let r 1 and r2 be the ranks of A and B respectively and let r be the rank ofAB. We shall prove that r :S;r l and r:S; r2 . Since the rank of A is rl' then there exist a non-singular matrix P such that PA =

[~]

... (i)

where G is a matrix of order r1 x n ofrankrl and 0 is a zero matrix oforder(m-r l ) n. Now post multiplying both sides of (i) by B, we get PAB =

[~] B.

...(ii)

Since we know that (PAB) = rank of (AB) rank of rank of(PAB) = r [:. rank of (A B) = r] or

rank of

[~] B = r.

Since rank ofG is r l so it has only r l non-zero rows, therefore the matrix.

[~] B Cannot have more than r 1 non-zero rows. Thus we have rank of

[~] B :s; r l

...(iii)

176 Remedial Mathematics or or

r:::; r l [from(iii)]

Rank of (A B) :::; RankofA. Further since we have rank (AB) = rank (AB)' and (AB)' = B' A'. ..

rank of (A B)

=

... (iv)

Rank of (B'A'):::; rank of B' using (iv)

rank of (A B) :::; rank of B' = rank of B rank of (A B) :::; rank of B

or or

r :::;r2 .

or

Remark • The rank of matrix does change by pre (post) multiplication with a non-singular matrix.

~~~~~~~I SOLVED

EXAMPLES

~I~~~~~~

Example 1: Determine the rank a/the/allowing matrices

Solution: (i) The square submatrices of the given matrix are

AI =

[~ ~ 3

[~ ~

:] A2 = 6 9 3

6

[~

:] A3 = : 12 3 9

:1] 12

A4 =

[~

6

: 9

: ] 12

det AI = 1 (36-36) +2 (18-18) + 3 (12 - 12)= 0

detA 2 = 1 (48-48)+2(24-24)+4(12-12)=0 detA 3 = 1(72-72)+3(24-24)+4(18-18)=0 det A4 = 2(72-72)+3 (48-48)+4(36-36)=0. Therefore, determinant of all square submatrices of the given matrix of order 3 x 3 are . zero so the rank of the given matrix is less than 3. Now the square submatrices of the given matrix of order 2 x 2 are

[~ ~l[~ l~l[: ~l[: l~l[~ l~l[~ ~l[~ ~] [~ l~l[~ ~l[~ l~l[! l~l Obviously, the determinant of all square submatrices of order 2 x 2 are zero. Thus the rank ofthe given matrix is less than 2. Since the given matrix is non-zero matrix. Hence, the rank of the given matrix is 1.

Matrices

(il)

A

=

177

[~ ! ~l 456

det A = 1(24-25)+2 (20-18) + 3 (15 -16)

=-1+4-3=-2+2=0 Therefore the rank A #:-3. Now the square submatrices of A of order 2 x 2 are Al

=

G

!J.A2=

G ~]

etc.

detA I = 4-6 =-2 #:-0.

A

Hence the rank

Example 2: IfA =

=

2.

l~ ~ ~ ~1' 000

o

1

Find the rank ofA and A2.

000

Solution: Since the matrix A is in Echelonform and there are three non-zero rows. Therefore, rank of A is equal to the number of non-zero rows. Hence, rank of A = 3. NextfindA 2

A2 =

l~ m~ ll~ ~J 1 0

1 0

0

0

0

0 0

0 0

0 0 0 0

0 0

0 0

0 0

1

1

Obviously, A2 is an Echelonform and having two non-zero rows. Hence the rank of A2=2.

Example 3: Use E-transformation to reduce the following matrix A to triangular form and hence find the rank ofA.

A~U

36]

1 3 2 2 -1 -3 4

Solution: Since we have

A~[j

-[ ~

I

3

3

2

-I

-3 3

!l

3 2 ;] byC,-> -I -3

i [i C,-

1 3 3 2

~l by

R3 -> R3 + R,o

-I 0 0 10 This matrix is a triangular matrix (Echelon form) and it contains three non-zero rows. Hence, the rank of A = 3.

178


Example 4: Reduce the ma"ix A

l

~~

-1 2 1 0 3 0 1 0

hence, find the rank ofA. Solution: Since the given matrix is

A

~ l~ -~ ~ -~1

Performing C 2 ~ C 2 + C j , C 3 ~ C 3 -2C j , C4 ~ C 4 + 3C j

Performing R2 ~ ~ - 4R j

-l~ 1~1 -l~ 1~1 -l~ 1~1 0

0

5

-8

3

0 0

0

0

5

-8

3

0

0

Performing R2 B R4

Performing C 4 ~ C4 -2C2

0

0

3

0 0

5

-8

{ -~1 0

0 0

3

0

5

-8

Performing R3 ~ R3 - 3R2, R4 ~ R4 - 5R2

-r~ -~] -l~ -~1 0

0

0

0

LO

0 -8 0

Performing C 3 B C4

0

0 0

0

-2

0

4

Matrices

Performing R3 ~ -

-[~

.!.2 R3

179

0 0 0 0 1 0 4

lj

[~ lj [~ ~] 0 0 1 0

-

0

1

0 0 0 0

.

Perfiormmg R4 ~-

1

8" R4

-

0 0 1 0 0

Hence the rank of A = 4.

Example 5: Find two non-singular matrix P and Q such that PAQ is in the normal form where 1

4 =

1

1 -1 [

3

1

Solution: Since we have

i.e.,

... (1)

Now applying E-transformation on the matrix A on the L.H.S. of(1) until A reduced to the normal form. In this process we apply E-row transformation to pre-factor 13 ofR.H.S. of(1) and E-column transformation to post-factor 13 ofR.H.S. of(l). Now performingR2 ~ R2 -RI' R3 ~ R3 -3R J , we get

[ ~o ~2

-2

~2l [~1 ~ ~lA[~ 0 ~l =

-2

-3

0

1

0 0

1

180


Performing R2 ~ -

!2 R2

[~

~J[! ;H~ 0

0

1

--

1

2 0

-2

[~ r] "[i [~ ~] "[i 0

1

Performing R3 ~ R3 + 2R2

0 0

1

Performing C3 ~ C3 - C2

-1

0

..

(I~ ~) =PAQ

where

P"[~

-2

0

1 2

-1 0 2

-1

0 2

-1

~H~ ~H~ ~H~

0

-1 1 0

-1 1 0

-1 1 0

~ll ~1]

~1] ~1]

Hence rank of A = 2. Example 6: With the help of E-transformation, find the rank ofthe following matrix 1

li

Solution: Let

3

0

-2

-3 2

1

A"

fi

2

-i1

2

3

0

-2

-3 2

-i1

181

Matrices

1 1 Perfonning R3 ~ R2 + 3R2

~

o o [

o

1

0 0

This is an Echelon form and having three non-zero rows. Hence, the rank of the given matrix = 3. 1 1 1] a b c a, b, c being all real numbers.

Example 7: Find the rank ofthe matrix A =

[ 3

a

Solution: Let

IAI+ I] a3

b

c ,

b3

c3

a

=

=

3

C3~C3-Cl

0

b-a b3

_a 3

(b-a)(c-a)

.. (b-a)(c-a)

c3

C,-+C,-C,

0

a

b3

c-a , expanding vector R, c3

1 2

_a3

1

2

~ C2- C

1 21 C2 +ca+a

b +ab+a 2,

C

I b 2 +ab+a2 ,

I 1 c 2 +ca-b 2 -ab

1

I

= (J-a)(c-a)[(c 2 +ca-b 2 -ab)-O] = (b - a) (c- a) [(c 2 - b2 ) + a (c- b)] = (b - a)(c-a) [(c- b)(c + b + a)].

IAI = (a-b)(b-c)(c-a)(a+b+c).

...(A)

Now following cases arise. Case I. Let a = b = c, then

Therefore all minors of order 3 and 2 of A are zero. Also as no element of A is zero, so A has non-zero minors of order 1. Hence rank (A) = 1.

182


Case II. Let a = b :#= c, then A=

l~

3

a

a a Also A have a minor of order 2 viz.

3

~ 1 °as C and C are identical. =

c

1

2

3

I~ ~I =c-a:#=O. Hence, rank (A) = 2. Similarly, we can discuss the cases b = c :#= a and c = a :#= b. Case III. Let a, b, c be all different such that a + b + c = 0. By above discussion IAI = O. Also, A has a minor of order 2

I~ ~I =b-a:#=O Case IV, Let a, b, c be all different but a + b + C :#= 0. In this case from (case I), it is evident that

IAI

i.e., Ahas a non-zero minor of order 3. Also A has no minor of order greater than 3. Hence, r(A) = 3.

:#= 0,

I EXERCISE 4.41 Determine the rank of the following matrices:

Matrices

10.

[~ -~ -~ 5

11.

13.

1

-3

[9

3

3

-!] 11

~ ~]

[I~ ~ :-~] 16 4

12

15 1

15. (a)

~1

2 _:

-2

I

2

0

2311]

[ 16. Reduce the following matrix to its Echelon form and find its rank:

I

(a) [

3

4

3 9 12 -1 -3 -4

(b)

[~ ~ l~ I -1

401]

2

17. Reduce the following matrix to normal form and find its rank:

(a) [;

~

! -1]

(b)

I 24 3] [3 6 107 2

1 -2 0 18. Reduce the following rltatrix to normal form nI)O find its rank:

[~ ;~ ~ ~J

19. Find the ranks of A, B, A + B and AB, where

1 I -1] [

A= 2

-3

3 -2

4,

3

B=

[-1 -2 -IJ' 6

12

6 .

5

10

5

20. If A and B are two equivalent matrices, then show that rank A = rank B.

183

184


21. Change the following matrix A into normal form and find its rank

WA

~ U_~ -! -~l

(i,) A

~ [~ ~ ~l :


1. A =

[~ ~ ~l 012

detA = 1 (2-0)-2(4-0)+3(2-0) =2-8+6=0 A Rank A < 4 Also determinant values of all square sub matrices order 3 x 3 and 2 x 2 are also zero. Thus Rank < 2. Since A is not a null matrix. Hence Rank (A) = 1

3. A =

[~ ! ~l 456

Now

detA = 1 (24-25)-2 (18-20)+3(15-16) =-1+4-3 =0 Rank(A) < 3.

The square submatrices of A are Al = and

[~ !~], A2 =[!

det(A,) = 4-6=-2*0 Rank(A) = 2

!]

etc.

Matrices

4. LetA =

185

1 25 -70 85] [o 0 0-8 0

Since the matrix A is of Echelon form and contains three non-zero rows. Hence Rank ofA is3.

2 -1 5. A=

[

3 4]

o

3

4

1

2

3

7

5

2

5 11

6

.

Performmg R2 ~

1

"3 R2

4]

1/3

-1I~

Thus A is converted into Echelon form, having 3 non-zero rows. Hence Rank of A=3.

6. LetA =

[-~ -~ -~ =~] 1

o

0 1

1 I 1-1

Then use the following steps: (1) Performing R2 B R]

(2) Performing R2 ~ R2 + 2R], R3 ~ R3 - R J (3) Performing R4 B R2 (4) Performing R3 ~ R3 + 2R2, R4 ~ R4 - 3R2, we get

186


[~ ; i =~] Then A is converted to Echelon form having two non-zero rows. Hence Rank of A = 2.

7. A=

2 3 -1 -4-I] [6 3 0-7 1 -1 -2

3

1

3-2

Then using the following steps: (1) Performing R2 ~ Rl (2) PerformingR2 ~ R2 -2Rp R3 ~ R3 -3R p R4 ~ R4 -6R t .

(3) Performmg R2 ~

'51 R2

(4) Performing R3 ~ R3 - 4R2, R4 ~ R4 - 9R2 (5) Finally performing R4 ~ R4 - R3

1 -1

A-

o o [ o

-2

-4]

3/5

7/5

0 33/5

22/~

1 0

0

Thus A is converted into Echelon form and having 3 Non-zero rows. Hence the Rank of A = 3.

:~ ~l abO c d 1

Here Rl = R3 and R2 = R4 A = 0 => Rank of A C4 ~ C4 - C2 (6) PerformingR3 ~R4 . 1 (7) Performmg R3 ~ 6' R3

.

8

(8) Performmg C4 ~ C4 + 3' C3

(9) Finally perl'onning R, -+ H) R ,. we get A - [

~

or Thus A is reduced to normal form and hence the rank of A = 3. 19. Rank of A: det A = 1(-9+ 8)-1 (6-12)-1 (-4 + 9) =-1 + 6-5 =0

=> Rank of (A) < 3 The square submatrices of A are

A = t

and Hence Now

[12 -31] [1-3 -I]4 etc. ,A 2 =

detA t = -3 -2 =-5 *0 => Rank (A) =2. RankA = 2 B

-1 -2 -Ij [

= 6

12 6 5105

188


Performing R2 ~ R2 + 6R], R3 ~ R3 + 5R]

_[~1 ~ ~1] :. B is reduced to Echelon form, which having one non-zero row. Hence, Rank of B = 1.

Further

A+B =

[

1 1 -1] [-1 2 -3 4 + 6

-2 12

-1] [0 6

= 8

-19 -2] 10

3 -2

10

5

8

8

3

5

8

det (A + B) = -1 (80-64)-2 (64-72) =-16+16=0 Rank of (A + B) < 3 Now the square submatrices of (A + B) are (A + B)] =

etc. [~ ~Il [~ -2] 10

det(A +B)] = 0+8=8;t:0 ~Rankof(A

Next,

+ B)=2. AB

=

[~ ~3 ~I][~1 ~~ ~ll = [~ ~ ~l 3 -2

3

5

10

5

0 0 0

:. AB is Null matrix, then Rank of (AB) = 0 20. Since A and B are equivalent. Then by definition, B is obtained by a finite chain E-transformation applied on 'A' and vise-versa and further since we know that E-transformation do not change the Rank. Hence Rank A = Rank B 21. Since we have

Performing R2 ~ R2 - 2R], R3 ~ R3 + R] 1 2 -1 A

~

[

0

0

5

o

0

5

Matrices

189

~ [~ ~ ~ -~l . 0 000

Perfonning R2 ~

1

"5 R2

-

or

[~

o o o

0 0

-3/~l

A~ (I~ ~J

Thus A reduced to nonnal form and hence rank of A = 2.

ANSWERS 1. 5. 9. 13. 17. 19. 21.

•

4. 8. 12. 16.

2. 3. 2 2 7. 3 3 6. 2 10. 3 4 11. 2 14. 4 15. (a) 3 (b) 3 2 18. 3 (a) 3 (b)2 Rank A = 2, Rank B = 1, Rank (A + B) = 2 RankAB= 0 (i) 2.

3 2 3 (a) 2 (b)3

LINEAR EQUATIONS

In this section we shall study of two type of linear equations: (1) Homogeneous linear equations (2) Non-Homogeneous linear equation.

1. Homogeneous Linear Equations Let us consider a system of linear homogeneous equation as follows:

=O} a2l~1:~~~'.~:::~'.:.~:~.~ a\\x\ +al2x2 +···+alnxn

amlxj +an12 x 2 +···+amnxn-O

.

... (1)

190


These are m equation in n unknowns. Any set ofnumbersx l ,x2 , ... xn that satisfies all the equations (1) is called a solution of(1). Trivial Solution: The solution xI = 0, x 2 = 0, ... xn = 0 of the equation (1) are called trivial solution. Non -trivial solution: Any other solution other than trivial, if exist, is called a non-trivial solution of equations (1). Let the coefficient matrix be

A=

l

o o o

al2

all

a~.I.

a22

amI

am2 Xn

nxl

o mxl

Then the equation (1) can also be written as AX=O.

...(2)

This equation (2) is called a matrix equation. Theorem 1: If Xj and X 2 are two non-trivial solutions of (2), then kjX j solution of (2), where k j and k2 are any arbitrary numbers.

+ k72 is also a

Proof: Since the equation (2) is AX = 0 and AXI= 0, AX2 = 0 are given. Now consider A (kIXI + kzX2) = kl (AXI) + k2 (AX2) = k (0) + k2 (0). Hence, klXI + kzX2 is the solution of (2).

III

NATURE OF THE SOLUTION OF EQUATION AX = 0

Since AX = 0 is a matrix equation of a system of m homogeneoll,,') linear equations in n unknowns and A is a coefficient matrix of order m x n. Let the rank of A be r. Then obviously r can not be greater than n. So, that either r is n or r is less than n. Therefore these are following cases: Case I: Ifr= n, then the equationAX=O will have no linearly independent solution. So in this case only trivial solution will exist. CaseII: Ifr < n, then there will be (n-r) linearly independent solution ofAX= 0 and thus in this case we shall have infinite solutions. Case III: Suppose the number of equations are less than number of unknowns i.e., m < nand since r :s; m, then obviously r < n thus in this case a non - zero solution will exist. Therefore the equation AX = 0 will have infinite solution.

Matrices

~~~~~~~I SOLVED

EXAMPLES

~I~~~~~~

Exanple 1: Find the solution of the following system of linear homogeneous equation 2x\-x2 +x3 =0 3x\ +2x2 +x3 =0 x\-3x2 + 5x3 =0. Solution: The given equation in mattix form AX = B can be written as

[: ~: m:} [~] A

~ ~: ~] [:

=> Performing R2 ~ 2R2 - 3R\, R3 ~ 2R3 - R\

[~ ~: ~,] Now R3 ~ 7R3 + 5R2

0>

[~ ~J[~] ~ m :'

Hence rank of A =3, i.e., r= 3 So, trivial solution exists which.is given by

=0 7x2 -x3 = 0 58x3 = 0

2X\-X2 +X 3

2x\ =0 7x2 =0

o

= - =0 58 x\ =0,x2=0,x3 =0. Interchanging R\, R2 and R3, we get X

3

~

1 0 0] =/3' [o 0 -1 0

1 0

Hence, the rank of A is 3 and equal to the number of unknowns x\, x 2 and x 3 • Hence,

191

192


Example 2: Find the solution o/the/ollowing system o/linear homogeneous equations: x+y+z=O 2x-y-3z = 0 3x-5y+4z =0 x+ 17y+4z = O. Solution: The coefficient matrix is given by

A=

[

~ ~1 ~31

3 -5 4 1 17 4

First reduce A into Echelon form Performing R2 ~ R2 - 2R 1, R3 ~ R3 - 3R 1, R4 ~ R4 - R1

Performing R2 ~

1

-[i ~ d

"3 R2, we get 1 1

1 5 0 1 3 0 -8 1 0 16 3

Performing R3 ~ R3 + 8R2, R4 ~ R4 - 16R4, we get

111

o

1

5 3

o

0

43

3

o .

0

71 3

3

Performmg R 3 ~ 43 R 3, we get

1 1 1 o 1 5 3

001

o

0

71 3

Matrices

193

11

Performing R4 ~ R4 + "3 R3, we get

1 1 1 0 1

5

3 0 0 1 0 0 0 This is an Echelon form and having three non-zero rows so A has the rank 3. Since there are 3 unknowns, hence a trivial solution exists here i.e., x = 0, y = 0, Z = is the only solution .

°

•

NON-HOMOGENEOUS EQUATIONS

Let us consider a system of equations which are non-homogeneous as follows: aUxl +a12 x 2 +···+alnxn

=ht }

:,l;:;:::~~~;.~::~;:::':~?~,:

...(1)

There are m equations in n unknowns. Let

A = [:::

:::

Cltn] a2n

amI

a m2

a mn

X= IIIxn

Xl] [ X2

Xn nxl

B=

[htb2 ] bm

1IIxi

Then the system of equations (1) can also be written as AX=B. ...(2) This equation is called matrix equation. If Xl' x 2 , .. .xn simultaneously satisfy the equation (2), then (xI' x 2' ...x n) is called the solution of(2). Consistency and inconsistency: When there will exist one or more than one solution of the equation AX = B, then the equations are said to be consistent otherwise said to be inconsistent. Augmented matrix: The matrix ofthe type

[AlB] = [ ; : amI

~: a m2

Cltn a2n

ht

b2

1

~:I/n ~,:!

is called the Augmented matrix of the equations .

•

CONDITIONS FOR CONSISTENCY

Theorem 1: The equation AX = B is consistent ifand only ifthe rank ofA and the rank ofthe augmented matrix [AlBJ are same.

194


Proof: Since the equation is AX = B

...(1)

The matrix A can be written asA = [Cl' C2 ... Cn ] Where Cl' C2 ",Cn are column vectors. Then the equation (1) can be written as

xlC l + x 2C2 + '" + xnCn = B. ... (2) Suppose the rank of A is r, then A has r linearly independent columns. Let these columns be C l , C 2 ""Cr and C l , C2""Cr are linearly independent and remaining (n - r) columns are in linear combination ofC I, C2 ... Cr .

or

Necessary condition: Suppose the equations are consistent, there must exist kl k2 ... kr such that

kl C l + k2C2 + ... + knCn = B. ...(3) But Cr + l' Cr +2' ... , Cn is a linear combination ofC l , C2 ... Cr, then from (2) it is obvious that B is also a linear combination of C l , C2, ",Cr and thus [AlB] has the rank r. Hence, the rank of A is same as the rank of [AlB]. Sufficient condition: Suppose rank A = rank [AlB] = r. This implies that [AlB] has r linearly independent columns. But Cl' C2 , ",Cr of [AlB] are already linearly independent. Thus B can be expressed as B = klC I + k2C2 + .. , + krCr.

...(4)

where k]l ~ ... kr are scalars. Now, equation (4) becomes

B = klC I + k2C2 + ... + krCr + O.Cr + I + ... + O.Cn· Comparing (2) and (5), we get

...(5)

xl =k l ,x2 =k2 , ... xr=kr,xr + l =0, ... xn=O and these values of xI' x 2 ...xn are the solution of AX = B. Hence, the equation are consistent .

•

CONDITIONS FOR ASYSTEM OF N-EQUATIONSIN N-UNKNOWNS TO HAVE A UNIQUE SOLUTION

if A be an n-rowed non-singular matrix, X be an n x 1 matrix B be an n matrix, then the system of equation AX = B has a unique solution.

Theorem 1:

x

I

Proof: If A be an n-rowed non-singular,matrix, the ranks of both the matrices A and [AlB] is Therefore the system of equations AX = B is consistent. Multiplying both sides ofAX= B by A-I, we have A-I AX=A- l BorIX=A- 1 B

17.

or

X = A-I B is a solution of the equation AX = B.

To show that the solution in unique, let us suppose that Xl and X 2 be two solution of AX=B.

Matrices

195

AXI = E, AX2 = B => AXI = AX2

Then

A-I AX = A-I AX I

2

IXI = IX2 => XI = X 1·

=>

Hence, the solution is unique.

Remarks • IfrankA"# rank of[AIB], then there is no solution. • If r = n, then there will be a unique solution. • If r < n, then (n - r) variables can be assigned arbitrary values. Thus there will be infinite solution and (n - r + 1) solution will be linearly independent. • If m < nand r ~ m < n, then equation will have infinite solutions.

I SOLVED EXAMPLES ~I~~~~~~ Example 1: Express the following system ofequations in matrix form

9x+ 7y+3z = 6; 5x+y+4z= I; 6x+ 8y+2z=4. Solution: The given equation are 9x+7y+3x = 6 5x+y+4z = I 6x+ 8y+2z = 4. :. The required matrix form of these equations is given by AX= B

when

A" [:

~ ~l;X" m

MdB"

Example 2: Solve by matrix method x + y + Z = 6, x - y + Z Solution: The given equation are

m

= 2, 2x + Y - z = l.

x+y+z=6 x-y+z=2 2x+y-z=l. Let

and assume that the... exists a matrix X"

Then

[~ ~1

iJ[l[H

[~l

such that AX" B.

196


Performing R2 ~ R2 - R\ and R3 ~ R3 - 2R\

we get

[~~2 ~] [~] o -1 -3

= [

z

:4] -11

Performing R \ ~ R2 + R3

~ [~;~ ~2][~] ~ [~l:J => => => =>

x-2z =-5,-2y=-4,-y-3z= 11 x-2z = -5,y=2, 2 + 3z= 11 x =2z-5,y=2,z=3 x = I,y=2,z=3.

Example 3: Solve the/allowing equation by matrix method:

x+y+z=9 2x+5y+7z=52 2x + y-z =0. Solution: The given equation can be written as

i.e.,

AX=B. :. Augmented matrix is

[AI~ ~

1 1

[:

5 7 1 -1 :

H

Performing R2 ~ R2 - 2R\ and R3 ~ R3 - 2R\, we get

[A~ [~ ~1 ~al[l [~:l -

Performing R2 ~ ~ + 3R3

~ [~o ~ ~4][~] [-~o]. =

-1 -3

x

-18

The matrix equation is equivalent to the equations

which gives x =

x+y+z=9 -4z =-20 -y-3z =-18 1,y= 3,z= 5.

Matrices

197

Example 4: Solve the following equations by matrix method x-2y + 3z =6 3x+y-4z =-7 5x-3y+2z = 5. Solution: The given equation can be written as

i.e.,

AX=B ,'. Augmented matrix is

[A~l ~

-2 3 : -4 [: 1 -3 2

;7]

This is an Echelon form and having two non-zero rows and rank A = rank [A I B] = 2. Thus, the equation are consistent.

i.e., Let

x-2y+3z = 6 7y-13z = -25. z = c then 25 13 8 5 y =- - + -candx=- - +-c 7 7 7 7'

Hence, the solution is 8 5 x = -;:; + ;:; c,y =

25 13 h' b' -7 + "7 c, z = c, were c IS an ar Itrary constant.

Example 5: Solve the following equation by matrix method: 2x+3y+z=9 x+2y+3z = 6 3x+y+2z = 8.

198 Remedial Mathematics Solution: The given equation can be written as

[~ i.e.,

3 2 1

ml[:] AX=B

:. Augmented matrix is [AlB}

1 · 9] 2 3 : 6 ~ [~ 3 1 2 : 8

Performing RI ~ RI - 2R2 and R3 ~ R3 - 3R 3, we get

~ [~

;1;5

~3]

o

-5 -7

-10

Performing R3 -+ R3 - 5R I , we get

Which gives

Hence,

[~ ~1 ~

tl

-y-5z =-3 x+2y+3z =6 1& = 5

th 1·· 35 29 5 e so utlon IS x = 18 ' Y = 18 ' z = 18 .

Example 6: Show that the equation x + 2y-z = 3, 3x-y+2z= 1, 2x-2y+ 3z=2,x-y+ z=-1 are consistent and solve them. Solution: The given equation can be written as

i.e.,

AX=B. Therefore augmented matrix is

[A IB] = [11:

~~ t -1

1

Matrices

199

~81

-4 -4

-

[~o ~ t ~41 -3 2

-4 -4

Performing R3 --) R3 - 6R2 , R4 --) R4 - 3R2 we get

-

[~o ~1 t ' ~l. 0

2

: 8

. 1 1 Performmg R3 --) 5" R3, R4 --) 2" R4

Performing R4 --) R I - R3

This is an Echelon form and having three non-zero rows. Thus rank A = rank of [A IB] = 3. Therefore the equations are consistent

and

x-2y-z = 3,-y=-4,z=4. Hence, the solution is x = - I, Y = 4, z = 4. Example 7: If the system offollowing equations is consistent thenjind the solution x + y + 4z =6 3x + 2y-2z =9 5x + Y + 2z =13

200 Remedial Mathematics Solution: The equations can be written as

i.e .•

AX=B

.'. Augmented matrix is

1 4 [Aim" [: 2 -2

1 2 Perfonning ~ -+ R2 -3R\. R3 -+ R3 -R] If 1

4

- 0 -1 -14 [ 4 0 -2 Perfonning R3 -+ R3 - 4R\. we get -

[~ ~1 ~14 o

-4 -18 Perfonning R3 -+ R3 - 4R2• we get

-

: l· 13

~9l 6-9

1

-17

[~ ~1 ~14 ~9l. o

0

38

19

1 Perfonning R3 -+ 38 R3• we get

-

[~ ~1 ~14 ~ ~9l. o

0 1 : 112 PerfonningR\ -+ R\ -4R3' R2 -+ R2 + 14R3• we get -

[1 0 : 4]

0 0 -1 0 : -2 . o 0 1 : 112

Perfonning R\ -+ R\ + R2• we get -

[1 0

0 0 -1 0 o 0 1

2-2 ] 112

Matrices

201

Performing R2 ~ - R2, we get

r1 -l~

o

0

2 ].

1 0 o 1

1/2

This is an Echelon form and having three non-zero rows and rank A = rank [A IB] = 3. Thus the system of equations are consistent.

Hence, the solution is x = 2,y=2,z= 112. Example 8: Show that the equations

x+y+z=-3 3x + y-2z =-2 2x + 4y + 7z = 7 are not consistent. Solution: The given equations can be written as

[: i.e.,

1 1 4

~2][:H]

AX=B :. Augmented matrix is

[AI~~

, -3]

1 1 [: 1 -2 : -2 4

: 7

7

PerformingR2~R2-3Rl,R3 ~R3-2Rl'

[A IB]-

[~ ~2 ~5

we get

::]

Performing R3 ~ R2 + R3

-

[~

1

1

-2 -5 0

0

-3] 7 . 20

This is an Echelon form and having three non-zero rows therefore the rank [A I B] = 3 and we see that

A-

[~ ~l :

202


Obviously rank A = 2. Since, rank A :F- rank [A IB]. Therefore, the given equations are not consistent.Example 9: Investigate for what value of A, ",the simultaneous equations x+y+z=6

x + 2y + 3z =10

="

x + 2y + A.z have (i) No solution (ii) a unique solution (iii) an infinite solutions. Solution: The given equations can be written as

AX=B. i.e., Therefore, augmented matrix is

1 1

1 : 6]

1 2 3 : 10 . [ 1 2 A : 11 Performing R2 ~ R2 - R J , R3 ~ R3 - Rl' we get [AlB] =

-

[~

1 1 1 2 1 A-I

-

[~

1 1 1 2 1 A-3

6

4

1

11- 6

Performing R3 ~ R3 - R2

If A :F- 3, the rank A = rank [A I B]

'*

6

4

1

11- 10

= 3. Thus in this case a unique solution exists.

In. = 3 and 1l:F- 10 then rank A rank [A IB] is 3. Thus rank A :F- rank [A IB]. Hence, in this case equations are inconsistent. In. = 3 and 11 = 10, then rank A = rank [A IB] = 2. Thus, in this case infinite solutions exist.

1. Use matrix method to solve the equations 2x-y + 3z= 9,x +y +z = 6,x-y+ z=2. 2. Use matrix method to solve the equations x+2y+ z=2, 3x+ 5y+ 5z=4, 2x+4y+ 3z= 3. 3. Show that the equations are consistent and hence solve them ~ x-3y-8z+ 10=0, 3x+ y-4z =0, 2x+ 5y+ 6z- 13 = 0.

Matrices

203

4. 5x+3y+7z=4,3x+26y-2z=9, 7x+2y+ lOz=5. 5. 5x-6y+4z= 15, 7x+4y-3z= 19, 2x+y+ 6z=46. 6. 2x-y+3z= 8,-x+2y+z=4, 3x+y-4z=O. 7. Show that the following equations are not consistent.

2x-y+ z=4, 3x-y+ z= 6, 4x-y+ 2z= 7 -x+y-z= 9. 8. Prove that the following system of equations have a unique solution

5x+3y+ 14z=4,y+2z= l,x-y+2z=O. 9. Use matrix method to equation to solve 3x+y+2z=3, 2x-3y-z=-3,x+2y+z=4

HINTS TO THE SELECTED PROBLEMS 1. The given system of equation can be written in the form: AX=B -1

=>

[:

1 -1

:t] ~m

Consider the augmented matrix

[AI~~[:

-1 3 1 1 -1 1

:] (by R\ --> R,)

[~

1 1 -1 3 -1 1

~]

We get

-

Now R2~R2-2RI,R3 ~R3-RI

[A IH]

~ [~ -

[~

1 1

-:]

-3 1

-4

-2 0 1

1

-3 1 -2 -2/3

~] (bYR3~Ra-~~ )

-2

:. Rank [A IB] = The Number of Non-Zero rows in Echelon form = 3. 3. The given system of equation can be written in the form of AX=B such that

[~ ~3 =:][~] [-~O] =

2

5

-6

z

13

Now consider the augmented matrix.

[AlB]

~ [~

!~

-1~] 13

204


Now R2 ~R2-3R2,R3 ~R3-2RI We have

Now R3

[~ - [~

-3 -8 10 20 11 22

[~

-3 -8 1 2 0 0

-

-10] 30 33

-I:] (by

-3 -8 1 2 1 2

~ ~-..!..~,R3 ~-..!..Rl) 10 11 '

~R3-R2

We get

-

-I~]

The Rank of [A IB] = Number of Non-Zero rows = 2. Here we observe that Rank (A) = Rank [A IB] Therefore, the given system of equation is consistent Now Rank(A) = 2, which is less than the number of unknown Therefore, the given system of equation having infinite number of solutions. Now the given, system of equation reduces to

x-3y-8z =-10 y+2z = 3 Letz = c, theny = 3 -2c andx= 2c-1.

which gives

Hence the solution of the given system of equation isx= 2c-l,y = 3 -2c, andz= c. 7. Consider the augmented matrix of the given system

[A IB] = [ :

=:

-1

~

1 -1

;j 9

-n -~ ~] ~yR, ~R4) =i

Now R2 ~R2 +3Rl'R3 ~R3 +4R 1,R4 + R4 +2R 1

We get

[-~o =~ :::] :

1 -1 : 22

Matrices

205

1 -1 1 -1

2 -2 3 -2 1 -1

9]

1 -1

o o

22

0

-11 23

1

1 -1

9]

1 -1

o

22

23 -11

1

o 0 This is Echelon fonn. Here, we observe that Rank of [A I B] = 4 and Rank of A = 3 Here, the given system of equation is inconsistent. 8. The given system of equations can be written as AX=B

[~ ~1 ~][~] m Consider the augmented matrix

[AIBJ~[~ We get

-

[~

_:

14 : 4] 2 : 1

2 : 0

[~

14 2 1 -4 -12 3

3 14

4]

2 : 1 0 -4 : 0 1

J (by R3 ~R3 +4R 1)

Which is in Echelon form Rank of [A IB] = 3 = Rank of A. ~ System is inconsistent and have a unique solution such that x+3y+14z =4 y+2z = I -4z = 0 z=O y = I, andz=O Hence, the given system of equations have a unique solution given by x = 1,y= 1 andz=O.

206


ANSWERS 1. x= 1,y=2,z=3

2. x=3,y=0,z=-1

3. Consistent:x=2c-l,y=3-2c,z=c

4. x=

5. x=3,y=4,z=6 9. x= 1,y=2,z=-1

6. x=2;y=2, z=2.

7

3

11' y =11' z =0

OBJECTIVE EVALUATION MULTIPLE CHOICE QUESTIONS 1. The vertical lines in a matrix are called. (a) Rows (b) columns 2. The Horizontal lines in a matrix are called (a) Rows (b) Columns 3. The single entity in a matrix is called. (b) Column (a). Rows 4. The elements of a matrix are defined for (a) Only real no's

(c) elements

(d) None of these.

(c) Elements

(d) None of these.

(c) element

(d) None of these.

(b) Complex no's (c) Both real or complex (d) None of these. 5. The elements' aij' of a matrix A = [a,j]n x n for which i = j are called. (a) Row elements (b) Column elements (c) Diagonal elements (d) None of these. 6. A square matrix A = [aij] is a unit matrix if. (a) aij= I when i = j and aij = 0 when i ±j (b) aij = 1 when i *- j and aij = 0 when i =j (c) aij*- 1 when i = j and aij = 0 when i*-j (d) ai, *- 1 when i *- j and a ij = 0 when i *- j 7. If A = [a,j]m x n and B = [b'j]m x n then 'A + B' is a matrix of the type. (b) m x n (c) m x m (d) n x n. (a) n x m

8. A =

(a)

[3 2 -I], [I -2 7] 3 -3

1

B=

[~ ~3~]

3

(b)

2

-I

then A + B=

[~ ~I~]

(c)

[~I ~ ~]

(d)

[-I0 60 7]4

9. A row matrix has only (a) One element (e) One column with one or more rows

10. If [ 5 k+1 (a) 0

k+2] = [k+3 -2

3

4]

-k

(b) 2

(b) One row with one or more columns

(d)

One row with one element.

then k= (c) -2

(d) 1.

Matricc!s

11.

IfP

(a)

(c)

=[~

2

:],nd

5 8

4

37 18

[-4

10

4

-9

-3 -5

=[:

0 3l 0 ~ then 2P = 3Q is

J

7 -4

IS]

[:,

10

Q

1207

(b)

29

-J] -1

[:

-10 5

-:8]

(d) None of these.

18

12. If 'A' is a matrix of order 3 x 5, then each row of 'A' has. (a)

3 elements

(b) 5 elements

(c) 8 elements

(d) 2 elements

13. If'AB' is the product of matrix 'A' and matrix 'B'thenmatrix 'A'iscalled (a) Pre-factor

(b) Post-factor

(c) Co-factor

(d) None of these

;j. B= [~ ~]

A

14. If = [ ;

and A + B - C = 0, then C =

(a)

15.

[~

;]

IfA=[~4

-22

~]

MdB=[~ i]ili" (a) AB. BA exist and are equal

(c)

(b)

AB. exist and BA does not exists

AB. BA exist and are not equal

(d) AB. does not exist and BA exists

16. Which is true for the product of matrix A and adjoint A when A = [aij]n x n (a)

Oifi=j

17. If A =

(a)

[~

[;

:l

~]

(b) IAlifi;t)

(c) IAlifi=j

(d) None of these

then Acij. A (b)

[8

-y

-13] a

(c)

8 [a

-y]

-13

(d) None of these.

208


[' , 'l

19. IfA= 1

2

2

-1

-3 thenA(A4IA)= 3

(a)

(c)

[T

[" ~

20. If A = [;

(a)

[;

-11 0 0

11 0

[~ ~3l 1

0

,~J

-u

(b)

2

-1

(d) None of these.

;] then A (Ad) A) =

!]

(b)

[-pU ~]

(c)

[~ ~]

(d) None of these.

FILL IN THE BLANK 1. If AlB are square matrices of the same order such that _ _ _. then (ABt = AnBn .

2. The matrix A = [

~ ~ !1

is a impotent matrix of index __ . _.

-2 -1 -3 3. The sum of two idempotent matrices A and B is idempotent if AB= BA ___. 4. If A and B bare symmentric matrices, then AB is symmetric ~ _ _ _. S. If A and B are symmetric matrices of oreder n, then AB + BA is _ _ _ . 6. If A are singular matrix, then det (A-I) = - - 7. If A is a skew symmetic matrix of odd order then 1A S. If A is an orthogonal matrix, then 1A

1=

1= ___

_ __

9. If A and B are two invertible matrices such that AB = C, then det (B)

= ___

10. If A and B be two non-null square matrix such that AB is a null matrix, then A and B both are =

TRUE I FALSE 1. A diagonal matrix is both an upper triangular and a lower triangular.

(T/F)

2. If A, B are two matrices such that AB and A + B are both defined, then A, B are square matrices of different order. (T/F) 3. If A, B are square matrices of the same order, then (ABt = An Bn.

4. The matrix A = [

~ ~ ~1

is on nilpotent matrix of index 4.

-2 -1

(T/F)

(T/F)

-3

S. If A, Bare n -rowed square matrices, then AB = 0 implies that at least one of A and B is the null (T/F) matrix. 6. Every matrix can be written as then sum ofa symmetric matrix a skew symmetric matrix. (T/F) 7. All positive integral powers ofa skew symmetric matrix are skew symmetric.

(T/F)

Matrices 8. 9. 10. 11. 12. 13. 14.

Positive odd integral powers of a skew symmetric matrix are symmetric. Positive even integral powers of a skew symmetric matrix are symmetric. If A and B are two symmetic matrices of order n, then ABA is a symmetric matrix. Every invertible matrix is orthogonal. If A and B are two invertible matrices, then (ABt I = A-I Ir I. If A is a non-singular matrix, then (ATrl = (A-Il. The inverse of a non-singular diagonal matrix is a diagonal matrix.

15. The product of two diagonal matrices of the same order is a diagonal matrix.

209 (T/F) (T/F) (T/F) (T/F)

(TIF) (TIF) (T/F) (TIF)

I ANSWERS MULTIPLE CHOICE QUESTIONS 1. 5. 9. 13.

(b) (c) (b) (a)

17. (c)

(a) (a) (b) (a)

3. (c) 7. (b) 11. (c) 15. (b)

4. 8. 12. 16.

18. (a)

19. (a)

20. (d)

2. 6. 10. 14.

(c) (b) (b) (c)

FILL IN THE BLANKS 1. AB=BA

2. 3

5. Symmetric

6.

9.

Det(e)

Det(A)

3. I

--Det(A)

o(Null matrix)

7. 0

4. AB=BA.

8. ±J

10. Singular

TRUE OR FALSE 1. True 5. False 9. True

2. False 6. False 10. True

3. False 7. False 11. False

13. True

14. True

15. True

4. False 8 False 12. False

REFRESHER Can we know? (Frequently Asked Questions) 1. Find the value of x such that

[lIXJ[~ ; ~l[}o 2.

IUpruB.Ph..

ma2002,~1

IfA=[~ ~]thenprovethatA2=6A+5I=[~ ~lWhereI= [~ ~]. (UPTU B. Pharma 2002]

210


3. If A =

[~I

_13

~],

5

-5

5

~ 4 3] 4

B= [

show that A2 B2 = A2

6

-1 4 4 IUPTU B. Pharma 20051

4. Find the inverse ofthe matrix A "

[i ~I ~Il

IUPTUB. Pharma 2001]

3 -I

5. Find the inverse of the matrix A " [:

6. Find the inve"e ofthematrixA "[;

7. If A"

[~ ~ ~]

~ !] ~



:]

usingA'-4A - 51" 0 and hence find A-I. [UPTU B. Pharma 2003, 2008]

8. Show that the matrix A =

[~1 ~1 ~

I] satisfies the equation A

1

3

-

6A' + 9A + 41" 0

-1

and hence find A-i.


9. Find the inverse of the matrix [;

~

:]

IUPTU 8. Pharma 20061

10. Solve for x andy by inverting the matrix in the following

[21 31] [yx] __ [47] 11. If A =

[~2

1

-


~]

find A-i. Hence solve the equations x + y +

Z

= 6,

-1

x - y + z = 2 , 2x + Y - z = 1.

[UPTU B. Pharma 2005] 13. Solve using matrix method, the following equations x + y + z =O,y + z = 1, z + x= 3. [UPTU B. Pharma 2003, 06]

DOD

MEASURE OF CENTRAL TENDENCY

III

INTRODUCTION

In previous chapter, we have studied about the classification and tabulation of data. But our study about classification and tabulation is not enough to get all the desirable results since when two or more series of same type are under observations, we can not classify and tabulate them. So we need an arithmetical idea or characteristic about the distribution. These characteristics are measure of central tendency, measure of dispersion, skewness and the peakedness. In this chapter, we will study the measure of central tendency or'average. According to Dr. Bowley, "Statistics may rightly be called the science qf averages" and averages are "statistical constants which enable us to comprehend in a single effort the significance of the whole." "Average is a point about which all the values of the variate cluster."

III

KINDS OF STATISTICAL AVERAGES

The statistical averages can be divided into five parts:

1. Arithmetic mean (A.M.) 2. Median (Md) 3. Mode (Mo) 4. Geometric Mean (GM.) 5. Harmonic Mean (H.M.)

III

ARITHMETICMEAN (A.M.)

"The arithmetic mean is the amount obtained by dividing the sum of values ofthe items in a series by their number." The arithmetic mean of some observations is the value which we can obtain by pividing the sum of all the numbers by total number of terms.

x (A.M.)

= Sum of all the observations Total number oftenns Let xI' x 2' x 3' .... xn are the observations. Then arithmetic mean is given by

212


Xl

+ x2 + x3 + ... + xn

n LXi i=I

n

n

n If the variate xI' x 2' ... , xn occursjj, h, /3, ... ,In times, then the arithmetic mean is known as due arithmetic average and,given by

x

=

fixl + hX2 + ... + fnxn fi+f2+ .. ·+fn

:E'·x· X = -'-;'-'

:EJ,

If wI' w2' .... wn be the weights assigned to the n values xI' x 2' ... ~ xn respectively, then arithmetic mean is known as weighted mean and is defined by Weighted mean = wlxl + w2 x 2 + ... + wnxn 11'1 + w2 w3 + ... + wn :Ewx

:Ew

•

METHODS OF CALCULATING ARITHMETIC MEANIN INDIVIDUAL SERIES

We can calculate the arithmetic mean by following three methods: 1. Direct Method 2. Short-cut method 3. Step-deviation method 1. Direct Method: In this method, the mean is calculated using the following formula n LXi

XI + X2 +",+xn ;=1 =-n n

STEP KNOWLEDGE Step 1. Add all the observations to find :Exi Step 2. Divide this sum:Ex; by total number of observations, i.e., n. 2. Short-cut Method: In this method, we assumed a middle number as an assumed mean. Here, we use the following formula. :Ed n = arithmetic mean; A = assumed mean; d = (x - A)

x =A+where,

x

:Ed = sum of deviations

Measure of Central Tendency

213

STEP KNOWLEDGE

Step 1. Select the assumed mean i.e., A. Step 2. Calculate the deviation from A, i.e., d=x-A. Step 3. Find the sum of deviation as 'i.d.

x=

A + 'i.d, we get the required mean. n 3. Step Deviation Method. Let us assume a number h called scale then x d' = h

Step 4. Using tpe formula

'i.d'

= 'i.x LX = hLd'

'i,x

hLd' n = hd'

n

x

h '

AM. = hLd' n

~~~~~~~I

SOLVED EXAMPLES I

Example 1. Calculate the arithmetic mean of 129,117, 112,200, 172, 138, 183 Solution: Arithmetic mean ofthe above data can be given by: 129+ 117 + 112+200+ 172 + 138+ 183 x (AM) 7 = 1051 = 150.14 7 So, AM. = 150.14 Example 2. Hemoglobin percentage (Hb %) of a patients of a ward of hospital ~ " obtained as 6 mg, 7 mg, 5 mg, 4 mg, 8 mg, 7 mg, 9 mg, 6 mg and 8 mg. Find out the arithmetl" mean of the data. 'i,x

Solution:

N 6+ 7 +5+4+8+ 7 +9+6 + 8 = 60=6.66mg

9

9

Example 3: WBC's in Number of 10 malefrogs (Rana trigina) are 8.19, 9.21,10.40, 10.95, 12.14,12.52,13.41,13.92,14.78, and 15.74 laclmm 3. Findmean WBC'sNumber. Solution: ..Here

x=

'i,x

N

8.19 +9.21 + 10.40 + 10.95 + 12.14 + 12.52 + 13.41 + 13.92 + 14.78 + 15.74 10 =

121.26=12.13 10

214


Example 4: Find the arithmetic mean offirst n natural numbers. Solution: Arithmetic mean is given by: Sum of all the observations X (A.M.) Total number of terms 1 + 2 + 3 + 4 + ... + n ~n

n = ~

n(n + 1)

2

n 1

= - (n

2

n

+ 1)

So the arithmetic mean of first n natural numbers is given by

~ (n + 1) .

2 Example 5: Show that the arithmetic mean of the series 1,2,22,23 ,24 , ... , 2n is given by (2 n + 1 _ 1) / n + 1. · A'thm' M Sum of all the observations S o Iu t IOn: n ebc ean= - - - - - - - - Total number of terms

1 + 2 + 22 + 23 + ... + 2 n n+l Sum of the G.P.

n +1 (2n+l_1)

(2n+l_1)

(2-1)(n+l)

n+l

(2n+l_l)

Arithmetic mean of the given G.P. =

1 n+ Example 6: A candidate obtain the following marks in an examination in a paper of 100 marks each English Maths Physics Chemistry Biology Subjects 6) Marks (out of 100) 82 64 48 70 It is agreed to give double weight to physics and mathematics as compared to other subjects. What is the arithmetic mean? [UPTU B. Pharma 2005) Solution: Since weights are given, we shall calculate weighted arithmetic mean in place of simple arithmetic mean. Weights are given accordance with the statement given in the example Subjects

English Maths Physics Chemistry Biology

Marks (X)

Weight(W)

WX

48 82 70

1

2 2

48 164 140 64 60

64

1

60

1 DV=7

LWX=476


x w

215

= LWX = 476 =68 LW 7

Example 7: The mean ofn numbers ofa series is Find the value of the last number. Solution: Mean ofn numbers =

x and the sum offirst (n -

1) numbers is A

x

Sum of first (n - 1) numbers =;'" Sum of all the observations . hm . M Ant etIc ean = - - - - - - - - - Total number of terms Sum of all the observations = Mean x Total no. ofterm = xxn Sum of all the observations = nX" Now, sum of first (n - 1) numbers = A Then the last number = (Sum ofn terms) -(Sum of (n - 1) terms) = nX"-A The value of the last number = (nX" - A) 5. Calculation of arithmetic mean in a Discr~te frequency distribution 1. Direct Method: In case of discrete frequency distribution, we multiply the values of a variable (x) by their respective frequencies (f). Then, we use the following formula __ Lf x Lf x x - --=-Lf n

where, L.fx = The sum of the products of observations with their respective frequencies "Lf = n = Total number of frequencies STEP KNOWLEDGE

Step 1. Multiply the value of the variable x by the corresponding frequency fto find.fx. Step 2. Calculate sum L.fx ofthe product obtained in step (1). Step 3. Putting the values in the formula - _ L.fx x --

n 2. Short-cut Method: Firstly, we shall assume a mean and then take deviation of the variable from this assumed mean. In this metfiod, we use the following formula: . -x -- A + Lfd n where, A = assumed mean; d = x - A deviation;

f = frequency

STEP KNOWLEDGE

Step 1. Select the assumed mean, i.e., A. Step 2. Calculate the deviation from A, i.e., d=x -A. Step 3. Calculatef d.

216


Step 4. Sum all the deviation to obtain I:.fd.

x=

A + I:.fd n 3. Step-Deviation Method: In this method, we divide our deviation by the common factor h. Therefore Step 5. Using the formula

x

Lfd' = A+--xh n

where, h = Common factor of the deviation d

!!. = x - A h

h

STEP KNOWLEDGE Step 1. Choose the assumed mean A. Step 2. Calculate d = x - A. Step 3. Calculate d = d / h. Step 4. Calculate fd' Step 5. CalculateI:.fd' Step 6. Using the formula

x=

A + Lfd' x h n

6. Calculation of arithmetic mean in a continuous frequency distribution In this case, we have to calculate the mid point ofthe various class intervals and denote it by x . Then, proceed same as above.

~~~~~~~I

SOLVED EXAMPLES

I~~~~~~~

Example 1. Compute the mean of the following data by direct and short cut method. Height (cm)

195

198

201

204

207

210

213

4

5

7

11

10

6

Children

216 4

Solution: Height x (em)

I

Ix

d(x-A)

Id

195 198 201 204 207 210 213 216 219

1 4 5 7 11 10 6 4 2

195 792 1005 1428 2277 2100 1278 864 438

-12 -9

-12 -36 -30 -21 0 30 36 36 24

Total

50

10377

-6

-3 0 3 6 9 12

27

219 2

Measure afCentral Tendency

217

Here, A =207 By direct method, we can calculate the A.M. by the formula A.M. = 'Lfx = 10377 =207.54cm N 50 By short- cut method, the formula is given by

= 207 + 27 = 207 + 0.54 = 207.54

A.M. = A + 'Ljd

N

50 Example 2: Values of fecundity (Rate of reproduction) of 50 fishes of a species offish (Macrognathus aculatus) were obtained and on the basis of that, a frequency table is given below. Calculate the mean of fecundity by long method (Direct Method). 1-10 3

Class Interval Frequency

11-20 11

21-30 7

3140

41-50 15

4

51-60

61-70 7

o

71-80 3

Solution: Class Interval

Mid"aluex

Frequencyf

ft

1-10

5.5

3

16.5

11-20

15-5

11

170.5

21-30

25.5

7

178.5

31-40

4

142.5

41-50

35.5 45.5

15

682.5

51-60

55.5

0

0

61-70

65.5

7

458.5

71-80

75.5

x=

3

226.5

Lf=50

Lfx= 1815

'Lfx = 1875 =37.5 'Lf 50

Example 3: Compute the mean of the following distribution by step deviation method Class Frequency

O-ll

11-22

22-33

3344

44-55

55-66

9

17

28

26

15

8

Solution: Class

Mid value x

f

(x - 38 5)

d'=

x - 38.5

fd'

J1

-33 -22 -II

-3 -2 -I

26

0

15

II

0 I

8

22

2

5.5 16.5 27.5

9 17 28

33-44

38.5

44-55

49.5

55-66

60.5

0-11 11-22 22-33

Total

"iJ= 103

-27 -34 -28 0 15 16

Lfd' =- 58

218


Here, the assumed mean, A = 38.5 and h= 11. A.M. by step deviation method is given by M = A+h 'Ljd' =38.5+ llx(-58) =38.5-6.194=32.306 N 103 Hence, A.M. =32.306 Example 4: What is the arithmetic mean ofthefollowing data Variate Frequency

0

2

1 nC

nco

3 nC

nC

2

I

3

...

n

...

nC n

Solution: The arithmetic mean is given by = 'Lft 'Lj 'Lft = nCo.O + nCI xl + nC2 x 2 + nC3 x 3 + ... nCn X n r.f nCo + nCI + nC2 + nC3 + ... + nCn n(n-I) (n-2) n(n-I) 0+1'1.1+--.2+ x3 ... +1.n 2! 3! (1+It

=

n[1 + (n -1) + (n - I)(n - 2) + ... + 1] 2! n 2

n [n-1Co +

(n-I)C + 1

... + (n-l)C _ ] n 1

n

n. (1 + It -\

2

n

2 n . 2n -\

2

n

So the arithmetic mean of the given data =

n

2

!!.

2 Example 5. Compute the arithmetic mean of the following data by using step deviation method Age (below) 25 30 35 40 45 50 55 ro No. of employees 8 23 51 81 103 113 117 120 Solution: Since the given data of the age is given as the age below x, so each column will contain all the previous frequencies. So at first we will construct the frequency table. Age

Mid value (x)

Frequency

d= (x -A)

d d'=h

Id'

20-25 25-30 30-35 35-40 40-45 45-50 50-55 55-60

22.5 27.5 32.5 37.5 42.5 47.5 52.5 57.5

8 15 28 30 22 10 4 3

-15 -10 -5 0 5 10 15 20

-3 -2 -I 0 1 2 3 4

-24 -30 -28 0 22 20 12 12

Total

'£1= 120

'£Id' =-16


219

Here, assumed mean is taken asA = 37.5 and h = 5. Now, by the above table, we can calculate 'Lf= 120 and 'f.fd' = -16. Then the arithmetic mean by step deviation method is given by A.M. = A + h 'f.fd' N = 37.5+ 5x (-16) 120 = 37.5 - 80 = 37.5 - 0.667 120 A.M. = 36.833 years Thus, the required arithmetic mean is given by 36.833. Example 6: A student passed the examination with the marks stated as follows Mathematics = 59, Physics = 55, Chemistry = 63, English = 75, Statistics

= 60

Compute the simple mean and also find out the weighted mean if weights are 3,3,1,2,1, respectively.

Solution: We have Sub

Marks obtained (x)

Weight (w)

wx

Maths Physics Chemistry English Statistics

59 55 63 75 60

3 3 1 2 1

177 165 63 150 60

Total

~=312

LW= 10

LWX = 615

. r.x 312 Simple mean = - = -=62.4 n 5 Therefore, 62.4 marks are the simple mean for the given data . 'f.wx 615 Weighted mean = - - = = 61.5 LW 10 Hence, 61.5 are the weighted mean ofthe given data Example 7. Find the weighted mean offirst n natural numbers whose weights are equal to the squares of the corresponding number. Solution: The table for the natural number and their respective weight is as follows x

w

wx

1 2 3

12 22 32

13 23 33

n

n2

n3

220


Now, we have to calculate the value of~u and ~w.x. ~w = 12 + 22 + 3 2 + ... + r? = ~r?

n(n + 1) (2n + 1) 6 ~w = n(n + 1)(2n + 1)

~wx

=

6 13 + 23 + 33 + ... + n3

=~n3

= [n(n + 1) 2

r

. ~wx n(n + 1) Then, the welghted mean = - - = 2 ~w

[

2

6

n( n + 1)

3

J x n(n + 1) (2n + 1) = (2n + 1) x -2

. 3 n(n + 1) Welghted mean = 2" (2n + 1)

Remark • Weight is the numerical value which shows the relative importance of the corresponding variable.

III

PROPERTIES OF ARITHMETIC

MEAN

Therorem 1. If every variable is increased by a particular value a , then the arithmetic mean is also increased by a. Proof. Let Yl' Y2' Y3 ... ,Yn be n variables and the frequency for these variables is given by

ii, h, ... , In·

Then arithmetic mean .:... ~fy Y =~f

let us introduce a new variable z such that z =y+a and

z

= ~fz ~f

= ~f(y+a) ~f

_

~fy

a~f

~f

~f

--+--

z = y+a Thus, it shows that if every variable is increased by a particular value a, then the arithmetic mean is also increased by a . Theorem 2. The algebraic sum ofthe deviations ofall the variate values from their arithmetic mean is zero.


221

Proof. Let zp z2' z3' ... , zn be the variables and the frequency for these variables is given by

Ii ,/z, ····,In· Then the arithmetic mean will be _ Z

'Lfz

=-

'Lf 'Lfz ='Lf·z 'Lfz = z'Lf

or

... (1)

Now the deviation from arithmetic mean is given by d=z-z

then

fd = f(z-z) 'Lfd

=

'Lf(z - z)

= 'Lfz -'Lfi = z'Lf -z'Lf

[From(l)'Lfz= z'Lf]

'Lfd =0 So, we can say that the algebraic sum of the deviations of all the variate values from their arithmetic mean is zero. Theorem 3. The sum of the squares of the deviations of all the values taken about their arithmetic mean is minimum. Proof: Let zl' z2' z3' ... , zn be the variables andfi ,/z, .. ·,In be their respective frequencies. Then the arithmetic mean z is given by __ 'Lfz '7' -'Lf 'Lfz = z'Lf

or

... (1)

Now deviation from the arithmetic mean then

d = z-z U ='Lj(z_A)2

(Sum of squares of deviation) The condition of maximum and minimum is given by au =0 aA

a2u

and if --2 > 0, then Uwill be minimum. aA Now differentiating equation (2) w.r.t. A we get au aA = -2'Lf(z - A) = -2'Lfz + 2A'Lf = -2z'Lf + 2A'Lf au = -2zN +2AN aA

(Since 'Lf= N)

... (2)

222


But

au aA =0 -2ZN +2AN =0 -ZN+AN =0

A=z au

... (3)

2

Now, where,

- 2 =2N aA N>O

a2u

-->0 aA 2 So, Uwill be minimum when A = So,

z (which is mean).

Remark • If M1, M2, ." Mk be the arithmetic mean of k distributions with respective frequencies nl' n 2, ... , n k , then the mean M of the whole distribution with frequency N = (n l + n2 + ... + nk ) is given by k

M= lIN'L nr M r r =1

Theorem 4: Arithmetic mean is not independent ofthe change of origin and scale. Proof: Letx l , x2 ' ... , xn be the n variables andfi ,.12, ... ,In be their respective frequencies. Then arithmetic mean is given by

M = 'Lfx 'Lf Now, we will change the origin and scale. For this, let us assume a new variable ~, which is given by x-A ~=- n 'Lfd' M=Then Mean 'Lf Put the values of d', we get 'Lf(x - A) h 'Lf

=! h

'Lf (X-A) 'Lf

=!

'Lfx _ A'Lf h' 'Lf 'Lf

M

=

i(M-A)

Thus, we can say that the arithmetic mean is not independent of the change of origin and scale.


223

1. Compute the arithmetic mean of first n natural numbers whose weights are equal to the corresponding number

.!. (2n + 1) . 3

2. Compute the mean marks of a student from the following table: Mark

No.o/students

Above 0 Above 10 Above 20 Above 30 Above 40 Above 50 Above 60 Above 70 Above 80 Above 90 Above 100

80 77 72 65 55 43 28 16 10 8 0

3. The rainfall of a certain town in centimeters for the first six months of the year are 102, 103,95,98, 100, 105. Compute the average rainfall of the town. 4. Compute the arithmetic average in rupees from the data given below: Salary 100 150 200 250 300 500 No. of Labours 30 20 15 10 4 5. Find the missing frequency from the following data, it is being given that 19.92 is the average number of the given data: 4-8 8-12 12-16 16-20 20-24 24-28 28-32 32-36 36-40 Tablets 11 13 16 14 ? 9 17 6 4 No. of persons cured 6. Calculate the mean marks of a student from the given data: Marks

No.o/students

Below 10 Below 20 Below 30 Below 40 Below 50 Below 60 Below 70 Below 80

15 35 60 84 96 127 128 250

7. Find the combined average daily wages for the workers of two factories: 250 200 No. of workers 250 Average 2.00

224


8. If the arithmetic average of data given below be 165 rupees, compute the missing term:

100 150 200 300 500 Monthly salary No. of Labours 30 20 15 10 4 9. Compute the weighted arithmetic average rate of31 building trade workers from the following table: Kind of worker

Daily wages (Rs)

Frequency

Masons Labourers Carpenters Painters

15 8 12 10

4 20 5 2

10. Compute the arithmetic average of the marks obtained by 9 students in a test: 75,43,52,65,48,35,40,70,40 11. Compute the missing frequer;tcy term from the following data whose arithmetic average is given by 35.64: Class 20-25 25-30 30-35 35-40 40-45 45-50 44 Frequency 18 102 57 19 12. Compute the arithmetic average of the following data: 5-8 8-10 10-12 12-15 15-17 17-20 20-25 0-5 25-30 2 5 6 4 4 9 5 7 6 13. The arithmetic average of a group of 40 items is 100 and that of another group of 50 items is 70. Find the mean of the combined group of size 90. 14. Compute the arithmetic mean for the following data: 15-20 Class 0-5 5-10 10-15 20-25 15.

16.

17.

18.

Frequency 4 16 2 15 2 Ifthe arithmetic average of the following frequency distribution is 7.85. Calculate missing frequency term. 10 12 15 Salary 5 6 7 Labourers 10 13 8 5 4 The average salary of 500 workers in a factory running in two shifts of360 and 140 workers respectively is Rs. 70. The average salary of360 workers working in day shift is Rs. 75. Find the average salary of 140 workers working in the night shift. Find the mean of the following distribution: Height (cm) 65 66 67 68 (f) 71 70 72 73 Plants 4 5 6 11 10 4 2 7 Find the mean of the following distribution: Class

0-7

7-17

14-21

21-28

28-35

35-42

42-49

Frequency

19

25

36

72

51

43

28

19. IfP + q = 1, compute the mean ofthe following: x 0 1 2 qn f

3

Measure oJ Central Tendency

225

ANSWERS 5. 250 9. 9.68 15. 15.05

2. 51.75 8. 250 13. 83.33

•

6. 50.4 11. 160 19. np

7. 2.22 12. 15.417

COMBINED MEAN

If x,

and x2 are the mean oftwo groups ofsizes n J and n2' then the combined mean x is the mean of two groups, given by __ nixi + n2 x 2 x n1 +n2

Proof: Letx"x2 , ... ,xn be the variates ofa group of size n, andYI'Y2' ""Yn be the variates of a group of size n2 . Then

and

x2

Let

x

=

YI + Y2 + Y3 + ... + Y n

n,

::::>n2 x 2 =y, + Y2 + Y3 + ... + Yn

...(1)

be the mean of these two groups, then x = (XI +x2 +x3 + ... +xn ) + (y,,,:r Y2 + Y3 + ... + Yn)

n,

+~

x = nix, + n2 x 2 n, +n2

~~~~~~~I SOLVED

[Using (1) and (2)]

EXAMPLES

~I~~~~~~

Example 1: The mean ofthe marks secured by 25 students ofsection A ofclass BCA is 47. that of 35 students of section B is 51, and that of 30 students of section Cis 53. Find the mean of marks secured by 90 students ofclass BCA. Solution! Let n" n 2, n3 be the numbers of students respectively in section A, Band C and Xi, x2 and x3 be the mean of marks secured by them. n, =25,n2 =35,n3 =30

and

Xi

=47, x 2 =51 and x) =53

25 x 47 + 35 x 51 + 30 x 53 25 + 35 + 30 = 4550 = 50.56 90

226


Example 2: The school has two sections. The mean marks ofone section ofsize 40 is 60 and mean marks ofother section ofsize 60 is 80. Find the combined mean ofthe students ofthe school. Solution: Here given that, n l =40, n2 =60 xi =60, x2 = 80 .. Combined mean

x

= nlXi + n2 x2 nl +n2 40 x 60 + 60 x 80 40+60

x

= 2400 + 4800 = 7200 =72 100 100 Example 3: The average score ofboys in an examination ofschool is 71 and that ofgirls is 73. The average score of school in that examination is 71.8. Find the ratio of number of boys to the number of girls appeared in the examination. Solution: Let there be n l boys and n2 girls in the school. Here

xI

= 71, x2 = 73 and

. .. Combmed mean

x

=

x = 71.8

nlxl +n2 x2 nl +n2

--'--'------"=-=-

71.8 = nl x 71 + n2 x 73 nl +n2 71.8(n l +n2) =71nl +73~ 71.8n l + 71.8n2 =71nl + 73n2 0.8n l = 1.2n2 nl 12 3 -=-=n2 8 2 Hence, n l : n2 =3: 2. Example 4: Three teachers ofstatistics reported mean marks oftheir classes, consisting of 69, 64 and 71 students as 30, 26 and 18 respectively. Determine the mean marks of all the three classes. [UPTU B. Pharma 2008] Solution: Total marks ofIst class = 69 x 30 Total marks of 2nd class = 64 x 26 Total marks of3rd class =71 x 18 Total marks of all the three classes Total number of students = Mean marks of all the classes =

1. The mean wage of 150 workers of the first shift in a factory is Rs. 400. The mean wage of75 workers of the second shift is Rs. 600. Find the combined mean wage of the workers of the factory.

MeaslIre afCentral Tendency

227

2. There are 50 students in a class out of which 20 are girls. The average weight of20 girls is 45 Kg and that of30 boys is 52 Kg. Find the mean weight in Kg of the entire class. 3. The average marks obtained by 30 students of group I is 60 and average marks of 40 students of II is 55 and that of 30 students of group III is 70. Find the combined average of students of all three groups. 4. There are 100 students in a class. The mean height of the class is 150 cm. If the mean height of60 boys is 170 cm. Find the mean height of the girls. 5. The mean weight of 150 students in a class is 60. The ~ean weight of boys is 70 Kg and that of girls is 55 Kg. Find the number of boys and girls in the class.

ANSWERS 2. 49.2Kg

1. 466.67

I

3. 61

4. 120 em

5. 50,100

•

GEOMETRIC MEAN

LetxJ• xl' x J• .... xn be the n variates ofavariablex. then the geometric mean G ofn variables is defined by G -- (XI'

) lin

X 2 ' X 3 "'Xn

Iffi,fi,J3' ·.. ,fn be the frequency of these variables and N =fi +fi+f3+···+fn Then,

G

=

(xf'x{:xf

log G

= -

10gG

=

1

N

···xf,n)'IN

[fi 10gxI + h

logx2 + f3 10gx3 + ... + fn logxn]

~ [~/; 10gX

i]

1=1

Thus, we can say that the logarithm of the geometric mean can be calculated by taking weighted mean of the logorithm of the variables XI'

STEP KNOWLEDGE Step 1. Find the logarithm ofthe variable x Step 2. Obtain r,f log x Step 3. Obtain ....:'i:f,---Io=g_x r,f

. r,flogx Step 4. Calculate the antIlog of -"---=-r,f

228


~~~~~~~I

SOLVED EXAMPLES

~I~~~~~~

Example 1: Calculate the geometric mean 3, 7,8,5,2. Solution: The geometric mean of the terms 3, 7, 8, 5,2, is given by G.M.

=

(3 x 7 x 8 x 5 x 2)1/5

=(1680)115 =4.416 Example 2: On 1" March a baby weighted 14 lbs. On ]-" May it weighted 20 lbs. What was the approximate weight of the said baby ion 1" April. Solution: G.M. =(xlx2x 3.. x)lIn

G.M.

= (XIX/12

=(14 X 20Y.12 = (2 x 7 x 2 x 2 X 5)112 = (2J70)=2x8.36=16.72 The weight of the baby was 16.72lbs on 1st April. Example 3: Calculate the geometric mean ofthe following data: 2574,0.005,0.8,0.0009,5,75,475,0.08 Solution: We know that the geometric mean of the data can be calculated by the formula G.M. =

~ ~Iogxi. Then, we will solve it by this formula. N

x

logx

2574

3.4106

0.005

3.6990

0.8

1.9031

0.0009 5 75 475

4.9542 0.6990 1.8751 2.6767

0.08

2.9031 2.1208

I 10gG.M. = N

~logxi

1 = gx2.1208

log G.M. =0.2651 G.M = Anti log 0.2651 = 1.841 Example 4: Calculate the geometric mean ofthe given data Then,

8, 15, 36, 40, 45, 70, 75, 85, 250, 500

Measure o/Central Tendency

229

Solution:

x

logx

8 15 36 40 45

0.9031 1.1761 1.5563 1.6021 1.6532 1.8451 1.8751 1.9294 2.3979 2.6990

70 75 85 250 500

17.6373

Now, the geometric mean is given by 1 logG.M. = -l:logxj N

~xI7.6373

=

lO

log G.M. = 1.7637 GM. = Antilog 1.7637 GM. =58.03 Example 5: The number ofBasophiles (a kind ofWBe) in blood of30 patients ofa hospital and their frequency were recorded as [sources II, 14, 17, 19,22 andfrequencies 5, 6, 8, 7, 4]. Find out the G.M Solution:

Scores (x)

Frequencies (j)

logx

j10p

11 14

5 6

1.0414

5.2070

17 19 22

8 7 4 30

1.1461 1.2304 1.2788 1.3424

6.8766 9.8432 8.9516 5.3696 36.2480

. . = Ant I·1og (1:fIOgx) = antI·1og (36.2480) GM 1:f

30

= antilog (1.20826) = 16.15 Example 6: Calculate the weighted geometric mean of the data given below: Articles A B C D E

Price 125 150 100 122 75

Weight 40 25 5 20 10

230


Solution: Articles A B C D

Price

125 150 100 122 75

E

Weight (w) 40

25 5 20 10 100

log X

wlogx

log 125 = 2.0969 log 150=2.1761 log 100=2 log 122 = 2.0864 log 75 = 1.8751

40 x 2.0969 = 83.876 25 x 2.1761 = 54.4025 5x2=10 20 x 2.0864=41.728 10 x 1.8751 = 18.751 208.7575

Weighted geometric mean can be calculated as .

log (weighted G.M.) =

So,

•

~wlogx

_-'='_

~w

= _1_ x 208.7575 = 2.0875 100 Weighted G.M. = Anti log 2.0875 = 122.3 Weighted G.M. = 122.3

PROPERTIES OF GEOMETRIC MEAN

Property 1: Ifwe put the value ofgeometric mean in place ofthe each value ofa series, then the product of the value of the series will be unchanged Proof: Let us assume a series with geometric mean G whose values are given by xl' X2' X3' ...xn. Then G = (X I.xZ.x3 ••• x)lIn Ifwe replace each value by G, then G = (G. G.G. ... G)'ln.

(ntirnes)

=(Cl')lln=G Thus, it proves that if we put the value of geometric mean in place of every value of the series, then the product of the value of the series will remain unchanged. Property 2: If G 1 is the geometric mean ofthe series xI' XZ' X 3' ••. , xII' Gz is the geometric mean of the series Y 1' YZ' Y3' ••• , Yn and G is geometric mean of the series obtained by the ratios of

corresponding observations. Then G will be equal to G/G z' i.e., G = ~. G2 Proof: Here, G is the geometric mean of the series obtained by the ratios of corresponding observations,


231

G=§. G2 Property 3: Let us consider n series with frequencies N I, N 2, N 3, ... , Nil' respectively and geometric means G I , G 2, G 3, ... , Gn respectively. Then the combined geometric mean qfn series with frequency NI + N2 + N3 + ... + N n is given by N N N G -- (GI IG2 2G3 3 .,.GNn n )IIN Proof: Taking log of both sides, we get

Iog G -I - og (GNIGN2GN3 I 2 3 =

...GNn)I/N n

1 -[NllogGI +N210gG2 +N3 10g G3 + ... +NnlogGnl N n

NlogG = LN;logG; ;=1

Property 4: Let us consider n sets ofobservations whose geometric means are respectively G I , G 2, ... ;Gn. Now, ifG is the geometric mean afthe product ofthese n sets, then the product ofthe geometric means of these series will be equal to the value ofG. G =G I.G2.G3...Gn n

10gG = LlogG; ;=1

Property 5: Let us consider a series xI' x 2, ... , xp' xp+ I ... xn whose geometric mean is given by G. In -Which G is greater than from the each value xI' x 2' ... , xp and less thanfrom each of the values xp + I' xp + 2 ",xn then = (x l x 2 ... xp.xp+ I",xn) GP.Gn-p = (x l .x2 .. .xp)(xp + I".xn)

an

G G x2

xI

G _ Xp + I XP + 2 Xp -0'0-

~~~~~~~I SOLVED

Xn

G EXAMPLES

~I~~~~~~

Example 1: Let us consider two series with m and n number of items and whose geometric mean is given by G and g respectively. Then find out the geometric mean of combined distribution. " Solution: Let xI' x 2' ... , x"i. and Yl' Y2' ...Yn be the two series whose geometric means are respectively given by G and g. Then, G is given by G = (x l .x2,x3 .. .xm) 11m and g= (Y1'Y2'Y 3 ...yi So the geometric mean of the combined distribution is given by

'n

I

Gc = (XI,X2,X3,,,Xm·YI·Y2 .. ·Yn)m+n

232


mn

Gc

=

(G. g)m+n m

n

G = Gm+n . gm+n

c

Taking log of both sides, we get m

n

log Gc =log Gm+n . gm+n =

1 --[m log G + nlogg] m+n

Gc =antilog _1_[mlogG+nlogg] m+n Example 2: The rate ofa certain item increases 5/4 times in first year, II/8 times in second year and 5/6 times in third year. What is the annual average increment.

Solution: Let x be the rate of item. It is increased by S/4 times in first year means it is

~ x in 4

. t h'lrd year. x m secon d year an d -S x m filrst year, -11.

8

6

Now, we have to calculate the geometric average of the rate, so G=

(~x x!..!.x ~x)1/3

4 8 6 Taking log of both sides, we get 1 S 11 S 3 log G = - log (- x - x - x x )

3

4

I[ S

8

6

S

11 + log- + logx = - log- + log348 6

3]

1

= -[logS + 10g11 + logS + 310gx -log4 -log8 -log6] 3 1

= - [210gS + 10g11 + 310gx - 210g2 - 310g2 -log2 -log3] 3 1 = -[2 x 0·6990 + 1·0414 + 310gx -6x 0·3010 - 0·4771] 3 1 = - [1· 398 + 1· 0414 + 310gx -1· 806 - 0·4771 3 1

=-[310gx+0·lS63] 3 = 10gx[0,l] + O· OS21

log G-logx = 0·OS21 10g(Glx) = 0·OS21 (Glx) = Anti log 0.00S21 = 1.127 G = 1·127 x Hence, the average annual increment will be 1·127 times of the price in first year.

Measure a/Central Tendency

233

1. Find the geometric mean of the following data: 50,100,1920,143740,204980,1206740,154910 2. Find the geometric mean of the series:

1,2,2 2,2 3,24 , ... .2n 3. The price ofcertain article rises 5% in first year, 8% in second year and 77% in third year. What is the average change per year?

4. The geometric mean of 10 data are calculated as 16.2. It was later found that one of the data was wrongly read as 12.9, in fact it was 21.9 Calculate the correct geometric mean.

5. Calculate the geometric mean from the following frequency distribution Marks obtained 11 12 13 14 15 Frequency 3 7 5 8 2 6. Find the geometric mean of2, 6, 18,54, 162. 7. Find the geometric mean of the following series Class 0-10 10-20 20-30 30-40 40-50 Frequency 10 15 12 8 5 8. Find the geometric mean from the following table: Marks obtained 5 7 9 11 13 15 No. of students 2 3 5 11 9 9. Calculate the average rate of increment in population which is increased by 20% in first year, 25% in second year and 44% in third year. 10. Find out the geometric mean of the following distribution: Marks obtained 0-10 10-20 20-30 30-40 No. of students 5 8 3 4

ANSWERS 1. 12700 5. 12.79 9. 28.02%

III

2. 2n/2 6. 18 10. 14.64

3. 26% 7. 19.\0

4. 17.08 8. 11.86

HARMONIC MEAN

The harmonic mean of any series is given by the reciprocal of the arithmetic mean of the reciprocals of the variables. For different type afseries, it can be calculated by different method (i)

For individual series: Letx\, x 2 , x 3 .• .xn be the n variables, then the harmonic mean of these variables [s' given by

234


H

=

n I 1 1 -+-+ ... +xl

X2

Xn

(ii) For discrete series: Letx l , x2,x3.• .xn be n variables andJ;,/i, ..

1n be the frequency

of them. Then the harmonic mean is given by 1 H

-

=

I n (-/ ) ,where N = :L / -:L Nix

H= N~(;} (iii) For grouped series: When the grouped series are given, we take the mid value of

each group and named them as Xl' X2' x3 ...xn and if the frequencies of these groups are J; ,/i,fj.. In, then the harmonic mean can be calculated by

H= :LU/x) N ~~~~~~~I

SOLVED EXAMPLES

I~~~~~~

Example 1: Find the harmonic mean o/the/ollowing data 12,8,6,24 Solution: Harmonic mean for individual series is given by n

H

=

1 1 I -+-+ ... +Xl

X2

xn

4

4

~+..!.+..!.+~ 12 8 6 24

0.0833 + 0.1250+ 0.1666+ 0.0416

H = _4_ =9.6038 0.4165 Example 2: Find the harmonic mean o/the/ollowing data: 4, 8, 16 Solution: n H= I 1 1 -+-+Xl

X2

x3

3 .!.+..!.+~ 4 8 16

3 0.25+0.125+0.0625

= _3_ = 6.8571 0.4375 So the harmonic mean is given by 6.8571.


235

Example 3: Hemoglobin percentage offive persons were measured as 1,5,10,15, and 25. Find out the Harmonic mean. Solution: HM HM

=.!. (!+.!. + -.!...+-.!...+...!....) 5 1 5 10 15 25 =.!. [150+30+ 15 + 10+6] 5 150 1 211 211 =-x- = 5 150 750

HM = 750 = 3.55 211 Example 4: Find the harmonic mean ofthefollowingfrequency distribution: Class 0-10 10-20 20-30 3040 40-50 Frequency 4 5 11 6 4 Solution: Class

Mid value x

Frequency

J/x

j7x

0-10 10-20 20-30 30-40 40-50

5 15 25 35 45

4 5 11 6 4 30

0.2 0.0666 0.04 0.0285 0.0222

0.800 0.333 0.04 0.171 0.088 1.832

Now, harmonic mean for grouped series is given by N 30 H.M. = "( ) = - =1.63755 L..; f / x 1.832 Example 5: Find the harmonic mean ofthefollowingfrequency distribution: Class 0-10 10-20 20-30 3040 Frequency 5 8 3 4 Solution: We have Class

Mid value x

Frequency(f)

J/x

j7x

0-10 10-20 20-30 30-40

5 15 25 35

5 8 3 4

0.200 0.0666 0.0400 0.0285

1 0.5336 0.12 0.114

20

Total

Now, the harmonic mean for grouped series is given by N H.M = "( L..;

f

20 ) = - - =11.3147 / x 1.7676

1.7676

236


Example 6: Hb% and its frequencies in 10 members ofa family was studied andfollowing results were obtained. Find the HM of the given series. Hb% mg I 100ml

Frequencies

12mg l3mg 14mg 15mg 16mg

3

-

3 1 2 1

Solution: Hb% mg/JOOml

Frequencyf

l/x

jlx

12 13 14 15 16

3 3 I 2 I 10

0.083 0.076 0.071 0.066 0.0625

0.25 0.23 0.071 0.133 0.0625 0.7465

, ""-

:~:

. .

-

'i,f= 10

'i,j7x = 0.7465

_ ~)flx) _ 0.7465 HM --10-

If

=0.07465 HM =

0.07465

=13.39

Example 7: A man drives a car for three days by covering a distance of360 km per day. First day he drives for a time of 10 hours and drive with the speed of36kmlh. On the second day, he drives 15 hours at a speed of 24kmlh and on the third day, he drives for 12hours at a speed of 30 kmlh. Calculate the average speed of the car. Solution: It is given that he covers a constant distance of360 km per day. His speed on the first day is given by = 36 kmlh His speed on the second day is given by = 24 km/h His speed on the third day is given by = 30 km/h

Since the distance is given to be constant so the average speed can be calculated by taking harmonic mean of the speeds. n So, average speed = 1 I 1 -+-+VI

v2

v3

3 1 1 1 -+-+36 24 30

3

0.0277 +0.0416+0.0333 _3_=29.2397 kmIh 0.1026


237

Remark • Where the distance in each part of the journey is given to be constant, then average speed will be calculated by harmonic mean. In the case when time being constant, the average is given by arithmetic mean. Example 8: In a certain factory a unit of work is completed by A in 4 minutes, by B in 6 minutes, by C in 8 minutes and by Din 12 minutes. What is their average rate of working? At this rate, how many units will they complete in a 8 hour day? Solution: The average rate ofwo]rking can be calculated by harmonic mean. So n 4 H= 1 1 1 1 1 1 1 1

-+-+-+- -+-+-+Xl

X2

X3

x4

4

6

8

12

_4_ = 6.41 minute 4 0.25 + 0.166 + 0.125 + 0.083 0.624 So, 6.41 minute per unit is the average rate of working 8 hour = 8 x 60 = 480 minutes They will complete the units= _1_x 480 . 6.41 Example 9: A train travels first part of its journey of 100 kms at a speed of20 kmlh, second part of 100 kms at a speed of 2 5 kmlh and the third part of the same distance at 30 kmlh, Find its average speed Solution: Since the train travels a distance of 100kms in each part. So, its average speed can be calculated by the harmonic mean. n 3 3 H= 1 1 1 1 1 1 0.05 + 0.04 + 0.033 -+-+- - + - + xl X2 x3 20 25 30 = _3_ = 24.390 0.123

II1II PROPERTIES OF HARMONIC MEAN Theorem 1: Ifxl and x 2 are any two observations, then A.H =G2

A = Arithmetic mean H = Harmonic mean G = Geometric mean Proof: We know that for two observations, arithmetic mean is given by Where,

A =

xl

+x2 2

Harmonic mean is given by 2 H = 1

I

xl

X2

-+-

238

Remedial Mathematics 2

Now,

H = 1

1

xl

X2

-+-

H

=

2XIX2

,A = xl

Xl +x2

+X2 2

and geometric mean G is given by G =

AH

Then,

=

~XIX2 (Xl +X2)

x

2

2XIX2

--,--~=----

(Xl

+ X2)

AH =x l x 2 AH = G2

Hence,

Theorem 2: /fx l , x 2' x 3, ... ,xn be the n positive observations then A cG 2 H

The sign ofequality will hold if the values ofall observations under consideration are same. Proof: Let xI' x 2, x 3, ... ,xn be the n positive observations. For these observations

A =

Arithmetic mean,

xI +x2 +x3 + ... xn

n H = 1

Hannonic mean,

n 1

1

-+-+ ... xI

x2

xn

G = (x l ·x2·x3···xn)l/n

and Geometric mean,

Now, we have to prove A 2: G 2: H Firstly, we will prove that A 2: G We know that 2(;-;; - ~)2 2: 0

=>

(xl +X2

-2~XIX2) 2:0

... (1)

2

>

(xl +X2)

(X3 +X4)

-

2

2

2:

=>

(XI+ X2+ X3+ X4)

~{~[XIX2]X3X4}

2: (x x r3 x 4)

[by equation (1)]

1/4

... (2)

l 4 In the same way, we can show that xI

+ x2 + x3 + x4 + Xs + x6 + x7 + Xg 8

IIg 2: (xI x2x3x4xSX6X7Xg)

...(3)


By eqn. (I), (2) and (3), we see that A n=21 ,2 2 ,23, ... , 2m .

~

239

G for n = 2, 4, 8, ... means it is true for

Now, we have to prove it for all values ofn. For this, let us assume the n observations x l ,x2,x3' ... , xk,xk + I ...x 2n (k < 2 n ,n = 2m)

Let

A = xI +x2 +",+xk

1

... (4)

and G = (XIX2 ....\k)lIk and all Now, we can write XI

x k + 1 = xk+2=xk+3= .. .x2n=A

+ X2 + "'Xk + Xk+1 + ... x2m > ( m)I/2m 2m XI X2 "'XkXk +1 ... x 2 -

From equation (4), it can be written as (2m -k)A+KA 2m KA+2m A-KA

2

m

~

~

(Gk A2m-~1I2m (Gk A2n1 _k)1/2n1

2m.A

2m

... (5) Now, we will prove that G ~ H or

Now, put 1 1 1 Then ( - ' -... , YI Y2 Y n

X;=Y;

)l/n

>

n

- YI + Y2 +"'Yn

240

or


{, )lln ; G.M. >H.M. 37,32,36,35,43,39,41 4. Calculate the geometric mean of the following data

I M"ks obtained I Frequency

5 3

7 8

6 4

9 2

8 7

5. Calculate the harmonic mean of the following frequency distribution Class Frequency

40-50 19

504> 25

60-70

36

70-80 72

80-90 51

90-100 43

6.Calculate the harmonic mean of the following frequency distribution Class Frequency

04 4

4-8 12

8-12

20

12-16 9

16-20 5

7.Compute the harmonic mean of the following frequency distribution Class Frequency

40-50 12

504>

10

60-70 15

70-80 17

80-90 8

90-100 3

8. A car runs at the rate of 15 km/h during the first 30 km, at 20 km/h during the second 30 km and at the rate of25 kmlh during the third 30 km. Find out the average speed ofthe car.

242


9. A train starts from rest and travel a distance of 1 km in four parts each of 0.25 km with average speed 12, 16, 24 and 48 kmlh. Explain the statement that the average speed over the whole journey of 1 km is 19.2 kmlh and not 25 kmlh. to. A variate takes values 1, r, ?, ... f'I- 1 each with frequency unity. Show that

A=

~

G = r n- 1/2 H n(l- r)' ,

= n(l-r)r

n l -

1- rn

From the above observations, also show that AH = G2 and A> G> H. 11. Find out the average speed of a car running at the rate of20 km/h during the first 30 km; at 25 kmlh during the second 30 km and at 30 kmlh during the third 30 km.

t2. Calculate the average speed of a train running at the rate of20 km/h during the first 100 km, at 25 kmIh during the second 100 km and at 30 kmlh during the third 100km.

I ANSWERS I 1. 13.5030 6. 7.246

2. 0.0002095 7. 32.049

4. 6.84 8. 19.15

5. 82.5669 11. 24.32 km

12. 24.39 km/h

III MEDIAN If we

arrange the whole data in ascending or descending order, then the value of the middle variable is known as median. In case when the number of variables are odd, then the middle value is known as median. If the number of variables are even, i.e., (2n), the value of the mean of nth, (n + l)th variables will be median.

Computation of Median 1. Formula for individual series: When the data given are ungrouped, then firstly, we arrange them in ascending or descending order. Then, if number of data are odd number, then the value of the middle variable will be median. Ifnumber of data are even number (2n), then the value of the mean of the nth and (n + I)'h variable will give the median. 2. Formula for discrete series: Let us assume that xl' X2 ....Xn are the n observation In . To calculate the median of such series whose frequencies are given by fi, first of all we calculate the cumulative frequency and then calculate the sum of the frequency. Now, we calculate the median of series according to the N (sum of the frequency) is odd or even. \ 3. Formula for Continuous series: In these type of questions all the data are divided into particular classes and their respective frequencies are given. Firstly, we calculate the cumulative frequencies. Then, we calculate the sum of the frequencies (N). According to N is even or odd, we fmd out the median. The class which contain this median is known as median class.

ii, ..

Mea~ure

ofCemral Tendency

243

Now, the median for this series can be calculated by the formula

IN-F Median = 1+ 2 xi

f

I = lower limit of the median class N = sum of all the frequencies F = Sum of all the frequencies preceding the median class

f

= Frequency of median class

i = Width of the median class

~~~~~~~I SOLVED

EXAMPLES

~I~~~~~~

Example 1: The height of the 11 students in inches of a hockey team is given. Find the median of these data . 65,67,69,6160,65,66, 70, 71,62, 72, Solution: First of all, we arrange all the terms in ascending or descending order 60,61,62,65,65,66,6769,70,71,72 Number of terms are 11 which is odd. bythe term = (n+l) 11+1 6 · . . Some d lanlsglVen -=-- = 2 2 Hence, the value of 6th term will be the median ofthese data. Median = 66 inches. Example 2: RBC's Number of8 patients is 35, 44,38,36,39,40,42 and 41laclmm 3. Find out the median ofthis series. Solution: First of all data is processed in ascending order i.e.,35, 36, 38, 40, 41 42, and 44 (Iac/mm3) th

(

Median =

(

~ ) item+ ~+I 2

2

2

)th

item (Since n = 8 is even)

= (%f item+(%+lf item 2 4th item + 5 th item 2 = 39+40 =39.5 2 Example 3: The daily wages in Rupees often labourers ofafactory are 4,6,9,12,11,8,5, 10, 11,8 Calculate the median ofthese wages. Solution: Firstly we arrange the data in ascending order 4,5,6,8,8,9,10, II" 11, 12

244


Here, n = 10, i.e., number of data are even so the mean ofnl2th and

(% + 1)th value will

be actual median.

n =5 ~+1 =5+1=6 2 ' 2 5 term is given by = 8 6 term is given by = 9

th th

Value of 5th term+value of 6th term · = ------------M elan d 2 8+9 17 =-=-

2

2

Rs. 8.5 Example 4: Hb % ofan animal was recordep as 6, 7, 4, 5, 5, 3 and 4gmlJ 00 mi. Calculate the median. . =

Solution: First of all above data is arranged in an ascending order i.e., 3,4,5,5, 6 and 7. Total Number of scores is 7 (an odd number)

. (n+ l)th Item .

Median = -2-

l)th Item .

7+ = ( -2=

(i

r

item = 4 th item = 5

Median = 5gm1100ml. Example 5: Find the median in the following frequency distribution: 5 3

9

7 12

28

11 10

13 9

Solution: (f)

Cumulative frequency

3

7

5 7 9 11

3 12 28

7 10

(x)

13 15

10 9

6

N=75

Here. N =75, which is odd.

22 50 60 69 75

15 6


245

So, the median = Value of the (75 + I) th term. 2

= Value ofthe 38 th term. In the table, we see that the cumulative frequency 50 contain the 38th term. So the value of x for this column will be the median Median = 9 Example 6: Compute the median ofthefollowingfrequency distribution: Size 8 10 12 14 16 18 20 Frequency 3 7 12 28 10 9 6 Solution:

(x)

(f)


8

3

10 12 14 16 18 20

7 12 28 10 9 6

3 10

22 50 60 69 75

N=75

Here, N= 75, which is odd. So, median is given by the value of (N + 1) th term. 2 Median = Value ofthe (75 + 1) th term = Value of the 38 th term. 2 The 38 th term will fall in the cumulative frequency 50. So the median for this distribution is given by the value of x for this frequency. Median = 14. Example 7: Percentage of body water of J5 Fishes and their frequencies given as follows. Find median of the given data. Water'% 60 62 64 70 72 74 76 78 82 84 86 Frequency 1 2 1 2 1 3 Solution: Cumulative frequency table Water %

60 62 64 70 72

74 76 78 82 84 86

Frequency(/)

Cumulativefrequency

I

1 2

1 1 2 1 1 2 1 1

10 II

3 I

14 15

3

5 6 7 9

246


~)th item

Median will fall in (

= So median

C;Jh

item=7.5 th item

= 76.

Example 8: Calculate the median in the/ollowing/requency distribution: x 20-30 3040 40-50 50-60 60-70

/ Solution:

8

30

'2fj

16

(j)


8 26 30 20 16 N= 100

8 34

(x)

20-30 30-40 40-50 50-60 60-70

20

64

84 100

Here, N = 100. So the median number is given by N = 50. The median number 50 falls in 2 the class 40-50. So, 40-50 will be the median class. Lower limit of median class 1= 40 Sum of all the frequencies, N = 100 Sum of all the frequencies preceding the median class, F = 34. Frequency of median class/= 30. Width of the median class i = 10.

~N-F Then, median is given by = I + 2 /

xi

(~x 100-34) = 40+ 2

= 40+

30

xl0

(50-34) 30 xl0 =40+5.3333=45.3333

Example 9: From the/allowing table, calculate the median a/the cost a/living index: (UPTU B. Pharma 2005]

Costoflivingindex(Rs) No. of weeks

140-150 5

150-160 10

160-170 20

170-180 9

180-190 6

190-200 2


247

Solution: Calculation of median Cost o/living index

No.'o/weeks

Cf

140-150 150-160 160-170 170-180 180-190 190-200

5 10 20 9 6 2 N=52

5 15 35 44 50 52

Median number = (

~) th item = ( 5; ) th item = 26th item.

which lies in the class 160-170. Thus, 160-170 is the median class in which I) = 160,/=20, F= 15, i= 10. Using the formula

(N

i --F Median = f. +I f 2

)

= 160 +.!Q.(26 -15) = 160+.!..!. 20 2 = 160+5.5= 165.5. Example 10: Find the medium ofthe following data: Daily wages (in Rs) 22 24.50 28 31.50 34 36.50 No. of Workers 10 23 32 28 12 5 [UPTUB.PHarma20081 Solution: We have Daily wages x

Frequency


22 24.50 28 31.50 34 36.50

10 23 32 28 12 5

10 33 65 93 105 110

HereN= 110. 110 then =55.

2 which lies between 33 and 65. Hence medium = 28.

~~~~~~~I

EXERCISE 5.5 ~I~~~~~~~

1. Compute the median ofthe data: 9,10,15,7, 11,9,8,11,7,9,10 2. The marks obtained by the ten students of class 8th is as follows: 75,80,96,92,89,94,100,82,63.105 Find the median

248


3. In a factory the daily wages of labourers are given by the following frequency distribution. Find the median. 10 12 14 Wages (Rs) 6 8 No. of labourers 6 3 4 5 2 4. Compute the median for the following frequency distribution Age No. of students

5-7 7

8-10 12

11-13 19

14-16 10

17-19 2

5. Compute the median for the following frequency distribution Variable 45-50 SO-55 55-60 60-65 65-70 70-75 75-80 80-85 85-90 90-95 Frequency 2 3 5 7 9 11 7 2 3 6. Calculate the median for the following frequency distribution Variable 0-5 5-10 10-15 15-20 20-25 Frequency 4 16 15 2 2 7. Calculate the median for the following frequency distribution 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Variable Frequency

2

18

30

45

35

20

6

8. Compute the median ofthe following frequency distribution: Age 15-19 20-24 25-29 30-34 35-39 Frequency

4

20

38

24

10

3 40-44 4

9. Compute the median ofthe following frequency distribution: Class Frequency

0-4 4

4-6 6

6-8 8

8-12 12

12-18 7

18-20 2

ANSWERS 1. 9 5. 7.583

2. 90.5. 6. 36.559

3. Rs. 10 7. 34.45

4. 11.447 8. Median = 28.49

9. Median = 8.5.

_MODE The variable whose frequency is maximum, is known as the mode of the distribution. In other words, we can say that the value which occurs most frequently in a distribution is known as the mode of the distribution.

Computation of the mode: 1. For individual series: For individual series, we can find the mode by inspection only. If number of variables are very large, then we arrange the data into discrete series and then we check the frequency for each variable to know the mode of the series. Iffrequency for different variables are same in the frequency table, then we use the method of grouping to calculate the mode of the distribution.


2.

249

For discrete series: Firstly, we arrange all the data in the frequency table. If the maximum frequency has the unique value, then it will be the mode of the series and if the maximum frequency occurs more than once, then mode can be calculated by grouping of data.

3. For continuous series: The class with maximum frequency is known as the modal class and we can obtain the mode of this series by calculating the formula Mode = 1+

f - f-l xi 2f - f-1- It

I = Lower limit of the modal class,

where,

f= Frequency of the modal class, 1-1 = Frequency preceding the modal class,

fi

= Frequency succeeding the modal class,

i = Class width.

If1-1 andfj are both (or one) is greater than fthen we use the following formula Mode = 1+

It fi + f-1

xi

~~~~~~~I SOLVED

EXAMPLES

~I~~~~~~

Example 1: Find the mode ofthe data given below

0, 1,6,7,2,3,7,6,6,2,6,0,5,6,

°

Solution: In the given data, we see that the frequency of 6 is 5, which is maximum and no other frequency is equal to this frequency. So, 6 is the mode of the given data. Example 2: Find the mode ofthe data given below

25, 15.23,

4~ 2~ 2~23,

25, 20

Solution: In the given data, the frequency for 25 is given by 3 which is maximum, since no other data has the same frequency or more than 3. So the mode ofthe given data is given by 25. Example 3: Water percentage offtfteenfishes ofa species offish wave recorded as 60, 64. 62. 76. 70, 84, 82, 72, 76, 84, 78. 84 and 86. Find the mode of this series.

Solution: First of all, data is arranged in ascending order. Not even single observation is spared. It comes as 60, 62, 64, 70, 72, 76, 76, 78, 82, 84, 84,84 and 86. By simple observation one can s,ay that 84 is the mode, because 84 is repeated maximum times (three times) in the above series. Example 4: Find the mode of the following frequency distribution Mid value

15

20

25

30

35

40

45

50

55

Frequency 2 22 19 14 3 4 6 1 1 Solution: Here, the mid value for each classes is given. So, firstly, we have to convert the given data into grouped frequency. So,

250

Remedial Mathematics Mid value

Class

frequency

15

12.5-17.5

2

20

17.5-22.5

22

25

22.5-27.5

19

30

27.5-32.5

14

35 40 45

32.5-37.5 37.5-42.5 42.5-47.5 47.5-52.5 52.5-57.5

3 4 6 1

50 55

1

Here, by inspection, we see that the maximum frequency is given by = 22. So the class 17.5-22.5 will be the modal class. For the modal class f=22, 1= 17.5,1=22'£1 =2,fi = 19, i= 5

M

Then, mode

=

1+

=

17.5+

=

20 17.5+---x5 44-21

a

f - f-1 xi 2f - f -1 +fi 22-2 x5 2x22-2-19

20 17.5+-x5 23 = 17.5+4.3478 Hence, Mode Mo =21.8478 Example 5: The expenses of lOOfamilies are given below. 48, thenjind the missingfrequency. =

Solution:

Ifthe mode ofthis distribution is

Expenses (Rs.)

Frequency

0-20 20-40

14

40-60 60-80 80-100

x 27 Y 15 N=56+x+y

N = 100 56+x+y = 100

It is given that

So,

x + y =44 ...(1) Now, it is given that mode of this distribution = 48 which falls in the class 40-60. So for this

1=40,1=27'£1 =x,fi =y, i=20


251

Then mode, M

= 1+

o

1-1-1 21 - 1-1 - fi

48 = 40+

8

=

xi

27-x x20 54-x- y

27-x x20 54-x- y

8 (54-x-y) = 20 (27-x) 2 (54-x-y) = 5 (27-x) 108-2x-2y = 135-5x 3x-2y =27 Solving equation (1) and (2) 5x = 115

x =

... (2)

.!..!2. =23

5 Put this value in equation (1) 23 +y =44,y=21

Hence, the missing frequency is given by = 23, 21.

Note: Empirical Formula The empirical relationship between mean, mode and median is given by

Mode = 3 Median - 2Mean Example 6: Calculate the mode of the following data Wages (Rs.)

Below 100

100-200

200-300

300-400

400-500

500 and above

No. of weeks

8

12

25

15

10

5

[UPTUB. Pharma - 2005J Solution: As the frequencies are regular and highest frequency 25 belongs to the class 200300. Thus.200-300 is the modal class in which 1=200,J= 25,L 1 = 12,J= 15,;= 100. Using the formula Mode = 1+

1-1-1 21 - 1-1 - fi

= 200+

xi

25-12 xl00 50-12-15

= 200 + 1300 = 200 -r 56;52 23 =256.52

Example 7: Compute the mode ofthefollowing distribution Class Frequency

10-20 22

20-30 35

30-40 40

40-50 62

50-60 50

60-70 45

70-80 40

Solution: Here, the frequencies are regular and highest frequency 62 belongs to the class 4050. Thus 40-50 is ~emodalclass in which 1=40,J=62,1-1 =40,fi =50, i= 10.

252


Using the formula Mode = 1+ =

f - f-1 2f - f-1 -

40+

fi

xi

62-40 xlO 2x62-40-50

= 40 + 220 =40 + 6.47 34 =46.47 Example 8: Calculate the mean, median and mode of the following data of 120 articles Weight in gms 0-10 10-20 20-30 30-40 40-50 50-60 No. of articles 14 17 22 26 23 18 [MEERUT B.Se. BIOTECH-2005]

Solution: Table for Calculation of Mean Weight in 9 ms No ofartic.:lesf

0-10 10-20 20-30 30-40 40-50 50-60

Mid value x

d=x-35

5 15 25 35 45 55 100

-30 -20 -10 -0 10 20

14 17 22 26 23 18 N= 120

d' =

!l.10

-3 -2 -1 0 1 2

N

120

= 35 -3.25 = 31.75 Table for Calculation of Median Wt ingms

Number ofarticles


0-10 10-20 20-30 30-40 40-50 50-60

14 17 22 26 23 18

14 31 53 79 102 120

C~O

r

item = 60 th item.

It belongs to 30-40 class. Thus, using the formula Median

=

1+

~ (~ - F )

= 30+~(60-53) = 30+~x 7 = 30+ 70 = 32.69 26

-42 -34 -22 0 23 36 'Lfd' =-39

" fd' (-39) X = A+-L..--xi = 35+--xlO

Median No. =

Id'

26

26


253

Mode Since highest frequency is 26 which belongs to 30-40 class. Thus, using the formula Mode = 1+

I - I-I

21 - I-I - Ii

= 30+

X

h

26- 22 xlO 2x 26 -22 -23

= 30+~ =35,71. 52-45

Example 9: The rise in prices ofacertain commodity was 5% in 1954,8% in 1955 and 77% in 1956. It is said that the average price rise between 1954 and 1956 was 26% and 30%. JustifY this statement and show how you would explain it before a layman. Solution: Assuming the price in year 1954 is 180.

Arithmetic mean =

105 + 108 + 177 3 = 130

i.e., an average change 000%. However, if we use the average change of30% per annum of result would not tally with the actual change in prices. If we use the geometric mean of 105, 108 and 177, the results would reflect the real change in prices. G.M. ofthe prices would be % olpri~e

Price at the end olyearX

log X

5 8 77

105 108 177

2.0212 2.0334 2.2480 LlogX= 6.3026

6.3026 G. M. = antilog - 3 - = 126.2 Average price rise during the period was 26.2% per annum. Thus, if we use G. M. The % increase would be 105 for 1954 and 113.4 (105 + 8% of 105) for 1955 and200.7(113.4+77% of113.4)for 1956. But if we use A.M., it would be 130 (for 1954),169 for 1955 and219.7 for 1956. Thus A.M. does not give a correct value, while G.M. tally the price.

~~~~~~~I

EXERCISE

5.61~~~~~~~

1. Find the mode ofthe following data: 4,5,8,6,9,8,8,6,5,11,9,8 2. Find the mode of the following data: 0, 1,6,7,2,3,7,6,6,2,6,0,5,6, 3. Find the mode of the following data:

2 3

3 8

4 10

5 12

6 16

7 14

8 10

9 8

10 17

II 5

° 12

13

4

)

254 Remedial Mathematics 4. Find the mode of the following data:

Monthly rent 2040 40-60 60-8080-100 100-120120-140 140-160 160-180180-200 No. 0/students 6 9 11 14 20 15 10 8 7 5. Find the mode ofthe following frequency distribution:

Height Frequency

120-124 125-129 130-134 135-139 140-144 145-149 150-154 2 8 15 20 10 5 5

6. Find the mode of the following frequency distribution: Marks Frequency

10-25 6

25-40 20

55-70 26

40-55 44

70-85 3

85-100

7. Find the mode for the following frequency distribution: 20-25 Age No.o/persons 50

25-30 70

30-35 80

35-40 180

40-45 150

45-50 120

50-55 70

55.(j() 50·

8. Compute the mode for the following frequency distribution:

Age below (in years) No.o/persons

5 2

10 4

15 14

20 27

25 48

30

64

35 72

40 75

9. Find the missing frequency for given frequency distribution when the mode is given by 47.5:

Class Frequency

0-10 7

10-20 8

20-30 10

30-40

40-50 40

50-60 35

60-70 10

70-80 7

10. Find the missing frequency for the given frequency distribution:

Classes Frequency

0-10 14

10-20

20-30 27

30-40

40-50 15

The mode and the median for the distribution is given by 24 and 25. The empirical relationship between Mean, Mode and Median Mode = 3 Median - 2 Mean 11. In a moderately asymmetrical distribution, the mode and mean are 32.1 and 35.4. Calculate the median.

ANSWERS 1. 1.8 5. 191.25 9. 25

2. 6 6. 58.75 10. 25,24

3. 6 7. 42 years 11. 34.3

4. 110.9 8. 23.1

DOD

TRIGONOMETRY

•

INTRODUCTION

The word trigonometry is derived from two Greek word "trigon" and "metron ", means "triangle" and "to measure" respectively. Therefore, trigonometry means to measure a triangle, i.e.,

"Trigonometry is that branch of Mathematics which deals with angles, whether of a triangle or any other figure ",

•

ANGLES AND QUADRANTS

Consider the Fig. 6.1, the angle is obtained by rotating a given ray about its end points. The original ray is called the initial side and the ray into which the initial sides rotates is called the terminal side ofthe angle.

0 ~------- X Initial side

Fig. 6.1

Remarks 1. The measure of an angle is the amount of rotation required to get the terminal side from initial side. 2. Ifthe revolving line revolves in anticlockwise direction, then add the angle is positive and if revolving line revolves in clockwise direction, then it is called negative angle. This may be clear in the following Fig. 6.2. p

o~---'r-------x

+ ve angle

o "----'I..------x

p

Fig. 6.2

•

MEASUREMENT OF ANGLES Generally, we measure the angles in degrees or in radian, which are defmed as follows:

(a) Sexagesimal System or English System (Degree Measure): We can divide the right angle into 90 equal parts, and each small part is known as degree.

256


Thus a right angle is equal to 90 degrees. Similarly we can say that, the circumference of a circle can be divided into 360 equal parts. One degree is denoted by 1°. Again we can divide a degree into sixty equal parts. Each small part is known as a minute and is denoted by 1'.

i.e.,

1° = 60' (sixty minutes).

A minute can also be divided into sixty equal parts and each small part is known as second and is denoted by I" c

I' = 60" (sixty seconds).

i.e., (b) Radians (Circular System):

B

Let us take a circle of radius a then 'a radian' is an angle subtended at the center of a circle by an arc equal in length to the radius of the circle. One radian angle is denoted by as 1C in the fig 6.3, LBOC= 1c.

Fig. 6.3

Remark • Relation between degree and radian: n radian = 180 degree

Grade Measure: If we divide the right angle into 100 equal parts, then each small part is said to be grade. Ifwe subdivide each grade into 100 equal part, then each part is known as minute and when we divide the minute into 100 parts then each part is known as seconds. So,

I right angle = 10 BC = perpendicular (y) and AC = Base (x) AB = Hypotenuse (r) We define the following trigonometric ratios which are also known as trigonometric functions: c Perpendicular y (i) sine e = = - , and is written as sin e. Hypotenuse r y Base x (li) cosine e = = - , and is written as cos e. Hypotenuse r

A "'--'-"0_ _ _ _ _--'-' B

Perpendicular y . . (iii) tangent e = = -, and IS written as tan e. Base x Hypotenuse r (iv) coseant e = d' I = -, and is written as cosec e. Perpen ICU ar y

(v) secant 8 =

Hypotenuse Base

r x

.

x

.

= - , and IS written as sec 8.

Base x .. 8 . = - , and IS wntten as cot . perpendicular y From the above definitions, it follows some definitions: 1 1 sin 8 tan e = --e' cosec 8 = -:-8 ' sec 8 = --e ' cos Sin' cos 1 cose cot e = and or sine tan 8

(vi) cotangent 8 =

Remarks o

o o o

o

o

o

o

Sine and cosine functions are called primary functions whereas tangent, cotangent, secant and cosecant functions are called secondary trigonometric functions. When the terminal ray coincides with x-axis, cosec e and cot e are not defined. When the terminal ray coincides with y-axis, sec e and tan e are not defined. The domain of the sine function is the set of real number, whereas its range is the set of real numbers from -I to 1. The domain of the cosine function is the set of all real numbers and range is the set of real numbers from -I to I. The domain of the tangent function is the set of all real numbers except odd multiples of nl2 and its range is the set of all real numbers. The student should not commit the mistake of regarding sine as sin X8; sin e means the sine of angle e, it is absolutely wrong to perform such operations as: sin (A + B) = sin A + sin B sin 2x + sin x = sin (2x + x). Power notation for trigonometric function (sin is written as sin2 e and is read as sin ) square e, (sin e)3 is written as sin3e and is read as sin cube e.

ei

Trigonometry

•

265

TRIGONOMETRIC IDENTITIES AND EQUATIONS

An expression involving trigonometric function which is true for all those values of 8 for which the functions are defined is called a trigonometric identity. Otherwise, it is a trigonometric equation.

I 1 For Example (i) sec 8 = --9 ' cosec e = -:--e etc. are true for all values of8 except those cos Sin . for which cos e = 0, sin e = o. So these are called trigonometric identities. For Example (ii) sin 8 = cos 8 is an expression which does not hold for all values of8. So it is an equation and not identity.

•

FUNDAMENTAL TIUGONOMETRIC IDENTITIES

For any angle e, we have (i) sin2 8 + cos 2 8 =1

(iii) 1 + cot2 e

=

(ii) 1 + tan 2 8 = sec 2 8 sin8 (iv) tan 8 = - cose

cosec2 e

cos9 (iv) cot 8 = --:--9 Sin

Proof: Let a ray starting from OX; trace out any angle 8 in any ofthe four quadrants and let it take the final position OP. From P draw PM perpendicular to x-axis. Now, in right angled triangle 1'10Mp, we have OP2 = OM2 +PM2

PM 2 +OM 2 OP2 PM)2 PM2 1+ ( - - =1+-OM OM 2 OM 2 +PM 2 = Op2 = (OP)2 OM 2 OM OM 2

(iii) 1 + cot2 e = 1+

y

p

x

x'

y'

Fig.6.S

(~~f

Op2 Op2 OM 2 PM 2 + OM 2 1+ - = ------,,-- = - = cosec 2 8. 2 = -2 2 PM PM PM PM 2 sine PM OP PM (iv) - - = - - x - - = - - = tan 8. OP OM OM cos 8 =

(v) cos9 sin9 .

=

OM x OP OP PM 1

(VI) cot e = --8 tan

=>

= OM = cote PM tan 8 cot 8 = \.

266

Remedial Matheniatics

Trigonometric Ratios of Standard Angles:

The values of Trigonometric Ratio of trigonometric angles of .:: (30°),':: (45°), .:: (60°), .:: 6 4 3 2 (90°) given in the following table 1: Table 1 8

0

d5

1tftl.

sin

0

-

1

1

2

J2

cos

1

fj

1

2

J2

tan

0

cot

00

sec

1

cosec

00

-

1

1t1l

1t

31t1l

2n:

1

0

-1

0

1 -

0

-1

0

1

fj

00

0

00

0

0

00

0

00

00

-1

00

1

1

00

-1

00

1t/3

fj

-

2 2

fj

1

fj

1

2

J2

2

J2

fj

fj 2

1

fj

2

SOLVED EXAMPLES I + COS 8

Example 1: Show that

--- =

I-cos8

cosec 8 + cot 8.

Solution: Here, we have

L.H.S.

=

1 + cos 8 1- cos8 1 + cos 8 1 + cos 8 I-cos 1 + cos8

/---x---

1 +cose sine

1 sine

cos 8 sine

---=-+--

= cosec 8 + cot 8 = R.H.S. Example 2: Show that

tan8-cot8 . 8 8 = sec2 {}- cosec2 Sin cos

{}

Trigonometry

L.H.S. =

Solution:

267

tanS - cotS sinS cos S sin 2 S - cos 2 S sinS cosS sinS cosS

sinS cosS cosS sinS sinS. cosS

-----

sin 2 S - cos 2 S 2

sin 2 S

2

sin S cos S

I

1

cos S

sin S

cos 2 S 2

2

2

sin S cos 2 S

sin S cos S

-- - 2- = sec2 9 - cosec2 S = R.H.S. 2 Example 3: Show tNat (sec A - tan Ai = Solution: Consider

L.H.S.

=

1- sin A .. I'+sinA


(sec A - tan A)2

sin A)2 1 = ( cosA - cosA

=

(I-sin A)2 cosA

(1- sin A)2 = (1- sin A)2 = cos

2

A

(l-sin 2

A)

(1- sin A)2 (l-sinA)(l +sinA)

= I-sinA =R.H.S. 1+ sin A Example 4: Sh ow t hat

tanS + seeS -1 1 + sinS eosS = =- - tan-seeS + 1 easS I-sinS

[RCPV B. Pharma 20041 [UPTU B. Pharma 20041

Solution: Consider

tanS + seeS -1 S S· tan - sec + 1

Using 1 + tan2 S = sec2 S, i.e., sec 2 S - tan 2 S = 1, we get tanS + seeS -1 tan S - sec S + 1

tan S + sec S - ( sec 2 S - tan 2 S) tan S - sec S + 1

(tan 8 + sec8) -(sec8 - tan 8)(sec8 + tan 8) tan 8 - sec 8 + 1 (seeS + tanS) (1- seeS + tanS) 1- sec S + tanS

-'-------'-"'-------....:.. =

S S sec + tan

1 sinS 1 + sinS --+--=--cosS

cosS

(1 + sinS)

~--...:..

cos S cosS 1- sin. S

R.H.S.

x

cosS

(I-sinS)

(1 -

sin 8)

cos 2 S

=- - - - cos8(1-sin S)

268


Example 5: Show that 2 (sin 6() + cos 6 fJ) -3 (sin 4 ()+cos 4 ()) + 1 = 0

[UPTUB. Pharma2005\

L.H.S. = 2 (sin6 e + cos6 e) -3 (sin4 e + cos4e) + 1 [.,' a' + b3 = (a+ b)3-3ab (c?+ b2 ) = 2[(sin2 e)3 + (cos 2e)3] - 3 [(sin2e)2 + (cos 2 e) + 1 2 2 = 2 [(sin2e + cos e)3 - 3 sin2 e cos e (sin2e + cos 2e) - 3 [(sin2e + cos 2 e)2 - 2 sin2 cos2e] + 1 = 2 [(1)3 - 3 sin2 e cos 2 e '1] -3 [(1)2 - 2 sin2 cos2e] + 1 = 2 [1 - 3 sin 2e cos 2e] - 3 [1 - 2 sin 2e cos 2e] + 1 = 2 - 6 sin2e cos2 e - 3 + 6 sin2 e cos 2 e + 1

Solution: Consider

= O=R.H.S. Example 6: Show that Solution: Consider

sine sinS = 2 +----cot e + cosec S cot e - cosec e sine

sine cose cot S + cosec e - - + -1sine sine 2 sine sin e I-cos 2 e

L.H.S. =

cose + 1 sinS

1 + cose

1 + cose

(1- cosS) (1 + cosS) -'-----'----- = (1 - cos e). 1 + cosS

...(1)

sinS R.H.S. = 2 + - - - - cot S - cos ece

Now,

2 sine = 2 + sin S = 2 + sin S cosS cos e - 1 eos e - 1 sine sinS sine 1- eos 2 e (1- eose)(1 + cos e) 2+ = 2 - -'-----'--'---cos e - 1 1 - eos e 2+

2 -(1 + eos S) 2 -1- cos S = I-cos e From (1) and (2), we conclude that sine cot S + eos eeS

sine

2+----cot e - cos eee

111 1 Example 7' Show that - -- = -- - ---. seeS + tane case case sece - tane Solution:

seeS + tan e

+---seeS - tan S

1 cose

1 cose

--+--

... (2)

Trigonometry

If

--------+-------see8 + tan 8 see8 - tan 8

sec8 - tan 8 + sec8 + tan 8 i.e.,if (sec+tan8)(sec8-tan8)

i.e., if

I cos8

--+-2 cos8

2 cos8

2sec8

i.e., if

I eos8

269

I

2 see 8, whieh is always true.

Example 8: Prove that the expression 2 (sin 6 {}+ cos6 B) - 3 (sin4 {}+ cos 4 B) is independent ofB. Solution: We have sin68 + cos6 8 = (sin2 8 + cos2 8) (cos4 8 + sin4 8 - sin2 8 cos2 ) I. ( cos 48 + sin 4 8 + 2 sin2 8 . cos 2 8 - 2 sin2 8 cos 2 8 - sin2 8 cos2 8) (eos 2 8 + sin 2 8)2 - 3 sin 2 8 cos 2 8. Now using this result in L.H.S. we get 2(1 - 3 sin2 8 cos 2 8) - (sin4 8 + cos 4 8) (1- 3 sin 2 8 cos 2 8)-3 [sin 48 + cos4 8 + 2 sin2 8 cos 2 8 -2 sin2 8 cos 2 8] = 2 - 6 sin2 8 cos 2 - 3 [(sin2 8 + cos 2 8)2 - sin2 8 cos 2 8] = 2 - 6 sin2 8 cos2 8 - 3 + 6 sin2 8 cos2 8 = -I which is independent of8. (I+sin8-cos8)21-cos8 [Meerut B.Se Biotech 2006) 2 I + cos 8 . (I + sin 8 + cos 8) I + sin 2 8 + cos 2 8 + 2 sin 8 - 2 cos 8 - 2 sin S cos S L.H.S. = 1+ sin 2 8 + cos 2 8 + 2 sin8 + 2 cos S + 2sin8 cosS

Example 9: Prove that Solution:

I + I + 2 sin S - 2 cos S (1 + sin S) I + 1 + 2 sin S + 2 cos 8 (I + sin 8) 2 (1 + sin8) - 2 cosS (1 + sinS) 2 (1 + sinS) + 2 cos8 (1 + sinS) 2 (1 + sin S) (1 - cos 8) 2 (1 + sinS) (I + cosS) 1- cos8 =R.H.S. 1 + cos S Example 10: If tan {} + sin {} = m and tan {}- sin {} = n, show that (m 2 - n2) = 16 mn. Solution: Here the given equation are tan 8 + sin 8 = m ... (1) tan 8 - sin 8 = n ...(2) and

270


Adding (1) and (2), we get 2tan8 = m+n. tan 8 =

m+n 2

2 m+n

=> cot8= - - .

Subtracting (2) from (1), we get 2 sin 8 = m - n cosec e =

=> sin 8 = m - n . 2

2 m-n

Putting the values of cot e, cosec 8 in equation cosec2 8 - cot2 = 1, we get or

or

4

4

(m_n)2 (m+n)2 4(m+nyl-4(m-n)2 = [(m-n)(m+n)f Therefore, 4(m 2 + n2 + 2mn)-4 (m 2 + n2-2mn) = [(m-n)(m + n)f => 4m 2 + 4n 2 + 8mn - 4m 2- 4n2 + 8mn = (m 2- n2i => 16mn=(m2-n 2i (m 2 _n 2)2 = 16 mn.

Example 11: Show that ( J + cot () + tan ()) (sin ()- cos()) =

Solution: Consider

sec ()

---=cosec

2()

L.H.S. = (1 + cot e + tan 8) (sin e - cos e) sin e ) . COS e 1 + - . - + - - (SIn e - cos 8) sme cose 2 sinecose+cos +sin28) (. e e) SIn - cos ( sin e cos e [.: if _b 3 = (a-b)(~ + b2 + ab) (

(sin8 - cos8) (sin2 8 + cos 2 + sin 8 cos 8) sin 8 cos8 3

3

sin e - cos 8 sin e cose

... (1)

secS cosec8 --R.H.S. = - - 2 cosec sec 20 1

2 2 cose sin e sin cos e -----=----cose sine 2 2 sin e cos e sin 3 e - cos3 8 sine cose

... (2)

Trigonometry

271

From (1) and (2), we get

(sin e- cos e) (1 + cot e + tan e) = Example 12:

sec e 2

cosec e

cosec e sin 2 e

1

1

x

2x

Ifsec e = x + -4 ,show that sec e + tan e = 2x or - .

Solution: We have I

sec e = x+ => 4x tan2 e = sec2 e - I

Now,

4x 2 + I sece= - - 4x

(4X;x+lr -1= (4x2

...(1)

+1~~2-16X2 = (4~::21)2

4x 2 -I tane = ± - - 4x Adding (1) and (2), we get 4x 2 + I 4x 2 -1 sece+tane = - - - + - - 4x 4x

... (2)

2 4x2+1 4x2-1) (4x2+1 4X -1) or ---( ---+ 4x 4x 4x 4x

- - - - - - or - - - - - 4x 4x

8x 2

2

-or4x 4x· 1

sec e + tan e = 2x or - . 2x

Hence,

1. Show that (a)

(c) (d) (f) (h)

(i)

0)

1- sine 1- cose / - - - = sec e -tan e (b) / - - - = cosec e - cot e. I + sine 1+ cose 2 4 2 4 2 sec e - sec e - 2 cosec e + cosec e = cot4 e - tan4 e. sec6 e = tan6 e + 3 tan2 e sec2 e + 1. (e) sec4 e - sec2 e = tan4 e + tan2 e. 2 tan2 e _ sin2 e = tan2 e . sin2 e. (g) I + cos e = tan e . 1- cose (sece -Ii sece - tane e = I - 2 sec e tan e + 2 tan 2 e. sece + tan 1- cose sine sine 1 + cose· tan2e + co2 e + 2 = sec2 e . cosec2 e

IUPTU B. Pharma 2001)

272


2. Eliminate 8 from the following equations: (a) x = a cos 3 8. y = b sin 3 8 (b) x=asec 3 8,y=btan 3 8 (c) sec 8 + tan 8 = m, sec 8 - tan 8 = 7t (d) acot8+ bcosec8=x2 ,bcot8+dcosec8=/

3. Prove that 1

cos8 sin8 8 +1 8 = sin 8 + cos 8 - tan - cot

[RGPV B. Pharnia 20011

4. Prove the following: (a) sins 8 - cos s 8 = (sin2 - cos 2 8) (1 - 2 sin2 8 cos 2 8).

(b)

sinS + 1 tanS S = sec 8 cosec 8 + cot 8. 1- cosS + cos

(c)

sin8 1 + cos8 + sin8 1 + cos8

(d)

cosec8 + cot8 - -si-n-8 = -si-n-8 - cosec8 - cot 8

=

2 cosec 8.

21 5. (a) If cos 8 = 29 and 8 lies in the fourth quadrant, find sin 8 and tan 8.

(b) If cos 8 cosec 8 = - 1 and 8 lies in the fourth quadrant, find cos 8 and cosec 8. 6. Prove that (a) sin 8 cot 8 + sin 8 cosec 8 = 1 + cos 8. (b) sec 8 (1 - sin 8)(sec 8 + tan 8) = 1. tan 8 cot 8 (c) 8+ 8 = 1 + sec 8 cosec 8. I-cot I-tan (d) (1 + cot 8 - cosec 8)(1 + tan 8 + sec 8) = 2. (e) (cosec 8 - sin 8) (sec 8 - cos 8) (tan 8 + cot 8) = 1. (t) (sec 8 - cos 8) (cosec 8 -

(g)

. SIll

8) =

8

1

tan + cot

8

~sec2 + cosec 28 = tan 8 + cot 8 = sec 8 cosec 8.

(h) (sin 8 + cosec

8i + (cos 8 + sec 8i = tan2 8 + co~ 8 = 7

(i)

cosecS + cosecS = 2 sec2 8. cosecS - 1 cosecS + I

G)

sin A - sin B cos A - cos B + =0 cos A + cos B sin A + sin B

(k) sec8 - tan 8

sin A 7. If - . SIll B

=

1 + sin 8 cos8

--- =

cosA

m and - cos B

=

sec 8 + tan 8.

n, show that tan A

=

m~-n2 -2-'

± -

n

m - 1

Trigonometry

9. (a) If sin 8 =

~~,

show that sec 8 + tan 8 =

273

%, if 8 lies between 0 and ~,

(b) What will be the value of the expression when 8 lies between ~, and 1t. 2

HINTS TO THE SELECTED PROBLEMS (i) (e). sec4 8 - sec 2 8 = = = =

sec2 8 (sec 28 - 1) (l + tan2 8) [tan2 8] tan2 8 + tan4 8 tan 4 8 + tan2 8 .

(sec 8 -tan8) x (sec8 - tan 8)

sec 2 S -tanS (h) - - secS+tanS

(sec 8 + tan 8) (sec8 - tan 8) (sec8 - tan 8)2 sec 2 e - tan 2 8 sec 2e + tan 2e - 2 seeS tanS

1

1 + tan 2 8 + tan2 8 - 2 sec tan 8 1 - 2 sec 8 tan 8 + 2 tan2e. 2. (a) x = a cos3 8 ,Y = sin3 8. x - =cos 3 8

a

'

113

cos 8 = ( ; 2

sin8=

)

(

i

113 )

2

cos e + sin 8 = I

(~f/3 + (1;)2/3 = 1. (c) sec 8 + tan 8 = m sec8-tan8=n (l) x (2) (sec e + tan 8) (sec e - tan e) = m = sec2 e - tan2 8 = mn

... (1) ... (2) x

n

~mn=1

4. (a) sin8 8 - cos 8 8 = (sin4 8 + cos4 8) (sin4 e - cos 4 e) = [(sin2 e + cos 2 e) - 2 sin 2 8 cos 2 8)] [(sin 2 8 + cos 2 8) (sin2 8 - cos 2 8)] = (1- 2 sin2 8 cos 2 e) (sin2 e - cos 2 e) = (sin2 e - cos2 e) (1 - 2 sin2 8 cos2 8).

1

1

1

1

(d)

cosece + cot 8 - -si-n-e = -si-n-e - cosec8 - cot 8

or

----+----cosec e + cot e cosec e - cot e

2 sin e

274


S. (a) If cos 9 =

~!

and q lies in fourth quadrant.

Then in fourth quadrant sin and tan both are negative 20

sin 9 = - 29 20

tan 9 = _ 20

21 6. (c)

tan 9 cot 9 +I 9 1- cot 9 - tan sin9

L.H.S. =

coS9 + I_COS 9 sin a

=

21

I + seq e cosec 9.

cos9

Fig. 6.9

sin 2 9 cos9(sin9 - cos a)

cos 2 9 sin a(cos 9 - sin9)

------------+------------

sin.9

9 I-~ cosa

(sin9 - cos9) (sin2 9 + cos 2 + sina cos 9)

sin 3 9 - cos9 3 sin 9 cos a (sin a - cos a)

sin a cos a (sin a - cos a)

1+ sin9 cosa sin 9 cos9 (g)

~sec2 9 + cos ec 29 L.H.S.:

= tan 9 + cot 9 = sec a cosec 9

~sec2 a + cos ec 2a

=

I

I

cos 9

sin 9

----+---2 2 sin 2 a + cos 2 sin 2 9 cos 2 9 - - - - = sec 9 cosec 9 = R.H.S. sin a cos9

Middle Term: tan a + cot 9 sina cos9 sin 2 9 + cos 2 9 I . = . 9 9 =sec9cosec9=R.H.S = --+-- = cos 9 sin 9 sin 9 cos 9 Sin cos

(k)

1 I+sin9 = sec9 - tan a cos9

L.H.S. = cos 9 1 - sin 9

=

sec a + tan a

. 9 Sin seca--cos9 cos 9(1 + sin 9) (1 - sin 9)(1 + sin 9) cos a (1 + sin 9) (1 - sin 2 a)

Trigonometry

=> =>

275

cosa (1 + sin a) cos 2 a 1+ sina => cosa

Middle Tenn. 1 sina cosa cosa sec a + tan a =>

--+--

R.H.S

ANSWERS 2. (a)

X)2/3 (y)213_ (-a . + -b -1,

(c) mn

(b) ( ~ )

2/3

(

-

f

)2/3 = 1,

= 1, (d)x4 _y4 = b2 -c?

20 20 7. (a) - 29' -21'

III

SIGNS OF TRIGONOMETRIC FT)NCTIONS

(a) When the angle is-x radians (x> 0): Take a circle with center 0 at the origin and radius equal to 1 unit. Take L. AOP = x radian Y a = cos x, b = sinx. => L.AOQ = -x Let [clockwise rotation from OX] Co-ordinates of point Q are (a, - b). X' ---t----:=*"-'r--+-- X A x-eo-ordinates = cos ( -x) a = cos (-x) => cos(-x) = a= cosx [By()] => = sin (-x) :. y-coordinate Y' =>-b=sin(-x) Fig. 6.10 => sine-x) =- b =-sinx

Now tan (-x) =

sin ( - x) - sin x = - - =-tanx. cos( - x) cosx

Hence, sin (-x)=-sinx, cos (-x) = cos x, tan (-x)=-tanx. (b) Trigonometric Ratio for 0 <x < 1t/2 (First quadrant) : The terminal side OP lies in quadrant I and as such both a and b are positive i.e., > 0 cos x = x-eo-ordinate of P = a > 0 sin x = y-co-ordinate of P = b > 0

276

Remedial Mathematics Y

sinx 1 tan x = --=->0 cosx a Taking the reciprocals, we get 1

sec x = ->0

a

X' --t----::-f0 b a cot x = ->0

Y'

b

Fig. 6.11

Hence, for 0 <x < 2: (first quadrant) all six trigonometric ratios are positive i.e., > 0 2

0<x 0, cosx> 0, tan x > 0, cosec x > 0, secx> 0, cot x > 0 Also for every P (a, b) on the circle of radius 1 unit, we have - 1 ~ a ~ 1 and - 1 ~ b ~ I - 1 ~ cos x ~ 1 and - 1 ~ sin x ~ 1.

For,

Remark • For any angle xc,

-1 -1

and (c) Trigonometric Ratios for

For

~

sinx< 1 cosx ~ I.

~

2: <x < 31t 2

2

(second quadrant):

'21t <x < 1t i. e., 90° <x < 180°, the terminal side CQ of LA OQ lies in quadrant if, when

co-ordinates of Q are (-a, b) cos x = -aO· b tan x = - < 0 -a 1 sec x = - 0 b -a cot x = - < 0 b

~

1t

Hence, for - < x < 1t (quadrant II) 2

Y

x'

A

Y'

Fig. 6.12

X

Trigonometry

277

sin x and cosec x are positive and remaining four ratios are negative. For'::' , 2 <x O,cosx sin x = -b < 0

tanx =

sin x cosx

X'

-b b =->0 -a a cosec x

1 - 0 cosec x < 0; sec x < 0; cot x> O. (e) Trigonometric Ratios for 31t <x < 2 1t (Fourth quadrant) : 2 For 31t < x < 21t, the terminal side OS of angle x lies in quadrant IV and as such the 2 Y co-ordinates of point S are (a, - b) cos x = a>O sinx = -b 0; cot x For both angle x and angle 21t + x, the co-ordinates of point P are (a, b) cos x = a; cos (21t + x) = a sin x = b; sin (21t + x) = b. This may be followed up by any number of revolutions to give cos (41t + x) = a. sin(41t+x)=b...

cos (2n1t +x) = cos x = a,n e Z sin (2n1t + x) = sin x = b, n e Z sec (2n1t + x) = sec x, ne Z cosec (2n1t + x) = cosec x, n e Z. However in case of tan x and cot x tan(n1t+x) = tan x, neZ cot(n1t+x) = cot x, neZ

Y

Hence,

X'

_-I--"T"f~=-"...1---JI--_

A

Y' Fig. 6.15

X

Trigonometry

279

as shown below: cos x = a sin x = b tanx =

cotx

=

sec x = cosec x =

b

a

a b

a 1 b

cos (1t + x) = - a ¢ cos x sin(1t+x) = -b ¢sinx sec (1t +x)

=

1 ¢ cosec x -a

-

I

cosec (1t + x) = -b ¢ cosec x

-b

b

-a

a

-b

b

tan x = - = - =tanx -a a cot x

= - =-

= cot x

:. the values of tan x and cot x repeated a rotation of an angle 1t. Hence tan (n1t + x) = tan x, n E Z cot (n7t + x) = cot x, n E Z

Remarks • In trigonometric ratios of sine, cosine, secant, cosecant we may add or subtract a multiple of21t to the angle without changing the value of the T-ratios . • In trigonometric ratios of tangent and cotangent, we may add or subtract a multiple of 1t and this will not change the value of the T-ratios. For Example: (i) sin (765°) = sin (765° -2 x 360°) = sin 45° =

.1

[from tables]

J3

.. cos (231t) (11) -6 = cos (231t -6 + 2 (21t) ) = cos (1t) "6 = cos 30° = 2' (iii) tan (420°) = tan [420°-2(180°)] =tan600=

J3.

280


Some Important Results: 1. cos (2mt + 8) = cos 8 and sin (2mt + 8) = sin 8, where n is any integer. 2. cos (- 8) = cos 8 and sin (- 8) = - sin 8,for all values of8. Behaviour of cos e and sin e as e Varies From 0 to 21t : y II

x../y../cos 8..L-

cos 8../-

sin 8 . L-

sin 8 t

x..L-

As x=cos8

yt

y=cos8

t

(1,0)

(-1,0)

0

x' cos 8 t

III

x

-+

,j.. -+

lndicates for increment Indicates for decrement

cos 8 t IV

sin 8 . L-

xt

xt

y..L-

yt y'

3. cos 8 = x and cos (-8) = x cos (-8) = cos 8 for all real numbers x ; (a)

cos(~ -

x) = sinx

(b) cos(i + x)= - sin x

~-

x ) = cos x

(d) sin

(c) sin (

4. For all real number x (a) cos(1t-x)=-cosx

(i +

x ) = cos x.

(b) cos(1t+x)=-cosx

(d) sin(1t+x)=-sinx. (c) sin (1t-x) =sinx 5. For all real number x, (b) sin (x + 21t) = sin x. (a) cos (x + 21t) = cos x 6. For all real number x and y, (a) sin (x + y) = sin x cos y + cos x sin y (b) sin (x - y) = sin x cos y - cos x sin y. 7. For all real number x and y, (a) cos (x + y) = cos x cos y - sin x sin y (b) cos (x-y) = cos x cosy + sin x siny. 8. For all real x, (a)sin2x =2sinxcosx (b) cos 2x = cos2 X - sin2 x = 2 cos 2 x-I = 1 - 2 sin2 x. 9. For all real x (a) sin 3x = 3 sin x - 4 sin3 x (b) cos 3x = 4 cos 3 X - 3 cos x.

Trigonometry

281

10. For all real number x andy (a) 2 cos x cosy = cos (x + y) + cos (x-y) (b) 2 sin x cosy = sin (x + y) + sin (x-y) (c) 2 sin x siny = cos (x-y) -cos (x + y) (d) 2 cos x siny = sin (x + y)- sin (x-y). Now, subtracting (4) and (3) we get 2 cos x sin y= sin (x + y) - sin (x - y) this proves the result (d). 11. Forallx,y E R (a)sinx+siny=2sin (x; Y)cos( x; y)

(b) sinx-siny = 2 cos (x; y) sin ( x; y) (c) cos x + cos Y = 2 cos ( x ; y) cos ( x ; y) (d) cos x - cos y = 2 sin ( x ; y) sin ( y ; x)

=- 2

y . (x+y) . (xsIn -2- Sin -2- ) .

12. Prove for all real number x andy. (a) sin (x + y) sin (x - y) = sin 2 x - sin2 y [UPTU B. Pharma 2001] 2 (b) cos (x + y) cos (x-y) = coi x-sin y Proof. (a) We have sin (x + y) sin (x -y) = [sin x cos y + cos x siny] [sin x cos y - cos x siny] = sin2 x cos 2 Y - cos2 x sin2 y = sin2 x (I - sin2 y) - (I - sin2 x) sin2 y sin2 x - sin2 y. (b) We have cos (x + y) cos (x - y) = [cos x cos y - sin x siny] [cos x cos y + sin x sin y] cos 2 x cos2 Y - sin2 x sin2 y cos 2 x (1- sin2 y) - (I - cos2 x) sin2 y = cos 2 x - sin2 y. 13. For all real number x prove that tan (-x) = - tan x. sin (-x) -sinx Proof. We have tan (-x) = ( ) = - - =-"-tanx. cos -x cosx ' 14. For all x E R, prove that (a) tan

(~-x)""cotx (b)tan(~+x)=-cotx

(c) tan (1t -x) =- tan x

(d) tan (1t + x) = tanx.

sin(~2 1t) Proof. We have tan (- - x = ( 2 1t

x)

cos --x 2

cosx ) "" - . - = cot x. smx

282


(b) We have tan

sin(~+ x) 2

(% + X)

1t

cos ( - + x 2

)

cosx =-cotx. = -sinx

(c) We have

tan (1t-x) =

sin (1t - x) sin x = - - =-tanx. cos(1t - x) -cos x

(d) We have

tan(1t+x) =

sin(1t + x) -sinx = - - =tanx. cos (1t + x) -cosx

Table 2 gives the sine, cosine tangent of some angles less than 90°

Table 2 9

0

15°

18°

22.so

sin

0

.J6-h

.Js-.Ji

4

4

.J6+h

~IO+ 2.Js

4

4

cos

I

tan

0

~25 -10/5

2-J3

5

~~~~~~~I SOLVED Example 1: Given that sin (A + B) =

67.5°

~2-h

36° ~IO -2.Js

~h +1

4

4

J2h

~h+ I ~2h

/5 +1 4

h-I

~5-2.Js

EXAMPLES

Jj and cos (A - B) 2

=

~2-h 2 h+1

~I~~~~~~ Jj ,find A and B, where A and 2

B are positive acute angles. Solution: Here, the given equations are

and

sin (A + B) =

J3

cos (A -B) =

J3

... (1)

2

... (2)

2

From (1), we have

A+B=60°

...(3)

A -B = 30°

...(4)

Also, from (2), we have Adding (3) and (4), we get 2A = 90° => Put this value in (4), finally, we getB= 15°. Example 2: Show that

A =45°.

cos 9 + sin(-9) _ tan(90 + 9) = 3. sin(90 + 9) sin(J80 + 9) cot 9

Trigonometry

283

Solution: The L.H.S. of the given equation is cos9 -sin -cot9 cose + sine-8) tan(90 + 9) =--+----cos 9 -sin 9 cot 9 sin(90 + e) sip(l80 + 9) cot 9 = 1 + 1 +=3 =R.H.S. 1 Example 3: Show that cos 24° + cos 55° + cos 125° + cos 204° + cos 300° = -. 2 Solution: L.H.S. = cos 24° + cos 55° + cos 125° + cos 204° + cos 300° = cos24°+cos 55° +cos (180°-55°)+ cos (180 0+24°)+cos(3600-600) = cos 24° + cos 55° - cos 55° - cos 24° + cos 60°

= cos 60° =

.!. = R.H.S. 2

Example 4: Calculate (I) cos 15° (ii) cos 75° Solution: (i) We know that


cos 15° = cos(45°-300) cos 45° cos 30° + sin 45° sin 30° 1 Jj 1 1 J2+1 -x-+-x-=--

J2 2 J2 2 2J2'

cos 75° = cos(45°+300) cos 45° cos 30° - sin 45° sin 30°

(ii) We have

Jj

1

1

1

Jj-l

J2 x T- J2 x "2= 2J2 . Example 5: Find the value o/cot75°. Solution: cot75° = cot(45°+300) cot 45° cot 30° - 1 cot 45° + cot 30° l.Jj -1 I+Jj Example 6: Show that

Jj-l = Jj+l'

47t r.. 7t 18 - cos 9" = ,,3 Sin 9

. 57t Sin



Solution: We have sin

57t

47t

18 - cos 9

. 57t

= Sin

47t) 18 -Sin. (7t2"-9

-~]

. 57t . 7t (57t +.2:] (57t = Sin - - Sin - = 2 cos 18 18 sin 18 18 18 18 2 2 7t . 7t =2 cos - SIn-

6

=

9

(2X ~) sin %= Jj sin %.

284


Example 7: . Find the vlaue ofsec(-15000) x sin 390° [UPTU B. Pharma2008] Solution: sec (-1500°) x sin 390° = sec(-4 x 360-60) x sin(360 + 30) = sec(-60°) x sin 30°

= sec 60° x sin 30° 1

=2x-=1. 2 7t 97t 37t 57t Example 8: Prove that 2 cos - cos _ + cos -3 + cos - = 0 [UPTU B. Pharma 2007] 13 13 1 13 7t 97t 37t 57t Solution: We have 2 cos - cos - + cos - + cos13 13 13 13 cos (

7t 97t) ('It 97t) 37t 57t 13 + 13 + 13 -13 + cos 13 + cos 13

107t 87t 37t 57t cos + cos - + cos - + cos13 13 13 13 . cos (7t _ 37t) + cos 87t + cos 37t + cos (7t _ 87t) 13 13 13 13 =

~os

37t 87t 37t 87t - + cos - + cos - - cos 13 13 13 13

O. Example 9: If tan A +tan B = a and cot A + cot B = b, prove that 1 1 cot (A + B) = a b . =

Solution: We have

cot (A + B) =

tan(A + B) I-tanAtanB tanA+tanB

tanA+tanB tanA+tanB

tan A + tanB

tan A tanB

a

1 b'

tanA+tanB

tan A tanB tanA+tanB

tan A + tanB tan AtanB tan A tan A tan B

tan B tan A tan B

----+---1 1 tanB tanA 1 cot A + cotB

--+--

Trigonometry

285

Example 10: Obtain the Values of (i) sin 15° [UPTU B. Pharma 2004) [UPTUB. Pharma 2006] (ii) tan 15° sin 15° = sin (45° - 30°) Solution: (i) We have sin 45° cos30° - cos 45° sin 30° 1.J3

1 _ .J3-1

1

J2T- J2'2 - 2J2 . tan 15°

(ii) We have

=

tan(45°-300) tan 45° - tan 30° 1 + tan 45° tan 30°

1 __ 1

=

.J3=.J3- 1 1+_1 .J3+1·

.J3 cos 8° + sin 8° Example 11: Prove that 80 . 80 = cot 37° cos -Sin Solution: We have 1 + tan 8 cos 8° + sin 8° 1- tan 8° cos 8° - sin 8°


0

[Diving Nr and Dr by cos BO]

tan 45° + tan 8° 1- tan 45°tan8° tan (45° + 8°)

= tan 53° = tan(90-37) = cot 37° Example 12:

4

5

Ifsin A = 5 and cos B = 13' Find the values ofsin (A - B) and cos (A + B)

where OLA L7fl2 and 0 LB L7fl2. Solution: We have

sin A

=

cos A =

[UPTUB.Pharma2004[

4

-

5

1t

and 0 < A < - .

2

)1- sin

2

A

~1-(~r =)1- ~~ =~ =~ Again

5 1t cos B = - andO
=

1 + tan A 4 + 4 tan2A = 2 2 tan A - 5 tan A + 2 = (tanA-2)(2tanA-l) =

4 5

4 5

=> 10 tan A => 0 => 0 => tan A = 2 or 112. But 0 SA S 1t/4, so rejecting value 2. Hence, tan A = 112


3Q4


1. Show that (i) tan (45° + 8) + tan (45° -8) = 2 sec 28 (ii) sin 2 A + sin 2 (A - B) - 2 sin A cos B sin (A - B) = sin 2 B (iii)

sin 5A - sin 3A cos 3A + cos 5A

(iv)

cos 4A + cos 3A + cos 2A . 4 . 3A . 2A = cot 3A. sm A+sm +sm

=

tan A

2. Show that (i) (sin 3A + sinA) sin A + (cos 3A - cosA) cos A OO)

C11

2 cos-cos-+cos-+cos1t 91t 31t 51t 13 13 13 13

=

=

0

0

(iii) (cos A - cos B)2 + (sin A - sin Bi = 4 sin2 A - B 2 . A 3A 9A . . 5A (IV) cos 2 A . cos - - cos = sm 5 A sm - . 2 2 2

3. Show that (i) cos 20° + cos 100° + cos 140° = 0 (n) cos 20° cos 40° cos 60° cos 80° = 1/16

(in) cos 10° cos 50° cos 60° cos 70° =

Jj 16

(iv) cos 52° + cos 68° + cos 172° = O. 4. Show that

sin A + sin 3A + sin 5A + sin 7 A cos A + cos 3A + cos 5A + cos 7 A

tan

=

4A

.

5. Show that 2 cos 2A + 1 (i) 2 cos 2A -1

=

tan (60° + A) tan (60° - A)

(ii) cos A + cos (A + 221t) + cos( A _ 2;) =0. 6. Show that cot 48 (sin 58 + sin 38) + cot 8 (sin 58 - sin 38). 7. Show that

sin 8 + sin 8 + sin 48 + sin 58 cos8 + cos 28 + cos48 + cos 58

=

tan 38.

8. Show that 4 cos A cos B cos C = cos (A + B + C) + cos (B + C - A), + cos (C + A - B) + cos (A + B -C). 9. If S9S (A +~B) sin ( C - D) = cos (A - B) sin (C + D), than show that tan A tan B tan C+tanD=O

Trigonometry

305

10. Ifsinx + siny = a and cos x + cosy= 6 Show that

.

x+y

a

(I) tan - - = 2 b

ADDITIONAL SOLVED EXAMPLES (Based on '2A' and '3A' Formulae) Example 1: Show that sin J8° = Solution: Let ~ ~ ~

~ ~

:::>


4

8 58 28+38 28 sin 28 2 sin 8 cos 8 2 sin 8 2 sin 8

~

~

J5 -I

-= = = = = = =

=

18° 90° 90° 90°-38 sin (90° - 38) = cos 38 4 cos 3 8 - 3 cos 8 4 cos2 8 - 3 4 (I -sin2 8)-3

4 sin2 8 + 2 sin 8 - I = 0

-2±)4+I6 -2±2J5 =--8 8 Since, 18° lies in the first quadrant, therefore sin 18° is positive. .

sm8 =

Rejecting the value

-I-J5 4

-:±J5 4

< 0, we get

sin 18° =

J5 -1 4

Remark • Similarly we canfind cos 18° =

J5 +1 ~IO-2J5 , cos 36° = - - ,sin 36° = . 4 4 4

)10+25

Example 2: Show that J + sin 28 - cos 28 J + sin 28 + cos 29

------ =

tan 8.

306


Solution:

(1- cos 29) + sin 29 Consider L.H.S. = ( ) 1+ cos 29 + sin 29 2 sin 2 9 + 2 sin 9 cos 9 2 cos 2 9 + 2 sin 9 cos 9 2 sin 9 (sin 9 + cos 9) sin 9 . -----:,..----:...= - - = tan 9 = R.H.S. 2 cos 9 ( cos 9 + sin 9) cos 9

Example 3: Show that

sin2A (I) - - - - = tan A 1 + cos2A (ii) cosec 2A + cot 2A = cot A (iii) sin 3A + sin 2A - sin A = 4 sin A cos A cos 3A.

2

Solution: (i) (ii) Consider

(iii) Consider

2

2 sin A cos A cos A A R S sin 2A L.H.S. = 1 + cos2A = 2 cos 2 A = sin A =tan = .H .. 1 cos 2A L.H.S. = cosec 2 A + cot 2 A = - - + - - SID 2A sin 2A

1 + cos 2A 2 cos 2 A cos A = = - - = cot A = R.H.S. sin 2A 2 sin A cosA sin A L.H.S. = sin 3A + sin 2A - sin A . A = 2' SA A . A A . 3A +sm . 2A) -sm = ( sm sm-cos--2sm-cos2 2 2 2 5A . A] = 2 cos A [. smT-sm2 2

2cos A2 [2 cos 2 ¥+ 1sin2 ¥-1] A 2 cos-

2

[2 sm-sm .2

3A . A] = 4smAcos-cos. A 3A =RHS

2

2

...

Example 4: lftan 9 = !!....,jind the value ofa cos 29 + b sin 29.

a

Solution: Consider a cos 29 + b sin 29 2

=a

1-2_ a ---...rl. + b _ _ a_ b ] [ b [

C+tan e) (l+tan e)- 1+ ~ 2

1 - tan 9 + b 2

2 tan 9 2

1+ :: 1

Trigonometry Example 5: Find the value a/tan 2r30 ~ Solution: We know that

1- cos A 1 + cosA

A tan2

Put A =45°, we get

H _

1 - cos 45° _ .fi - 1 1 + cos 45° - 1 + _1_ - -./2 + 1 .

tan 22.!.0 2

.fi Example 6: Show that 4 tan 8 (1 - tan 28) tan 4 B = ---'------'J - 6 tan2 8 + tan4 8

L.H.S. = tan 48° = tan C2 x 28)

Solution:

2 tan 28 1- tan 2 28

2x 2 tan 8 1- tan 2 8

1-(

2tan8)2 1- tan 2 8

4tan8 (1- tan 2 8) 2

4

1 - 6 tan 8 + tan 8

(1- tan 2 8)2 - 4 tan 2 8

(1- tan 2

8t

=R.H.S.

1. Show that Ci)

tan 58 + tan 38 58 8 = 4 cos 28 cos 48 tan - tan 3

(ii)

sin 8 + sin 28 = tan 8. 1 + cos 8 + cos 28

2. Show that Ci) sin 4A = 4 sin A cos 3 A - 4 cos A sin 3 A Cii) cos 4A = 1 - 8 sin 2 A cos 2 A CiiI) cos 5A = 16 cos5 A - 20 cos3A + 5 cos A (iv) cos 6A = 32 cos 6 A -48 cos4 A + 18 cos2 A-I.

3. Show that Ci) sin2 72° _ sin2 600 =

..) CII

J5 - 1 8

...) C111

. 7t . 27t . 37t . 47t

5

Sln-Sln-Sln-Sln-=-

5

5

5

5

16

.

7t

. 137t

I

Sln-+Sln-=--

10

(iv) 2 cos 8 =

10

2

~2 + -./2 + 2 cos 48.

307

308


5. Show thattan 15° + cot 15° = 4.

6. Prove that (i) sec 28 - tan 28 =

cos 8 - sin8 8 . 8 cos + sm

(ii) cosec 8 - 2 sin 8 = 2 cot 28 cos 8.

7. Iftan2 8 = 2 tan2 ~ + 1, then show that cos 28 + sin 2 ~ = \

1 3 1 8. If x + - = 2 cos 8, show that x + 3" = 2 cos 38. x x 9. Ifcos8=

Jfg+

cos ~ - e 8 e ~ ,j..' show that tan -=± --.tan-. 1 - e cos,!, 2 1- e 2

10. If2 tan a. = 3 tan 13, show that tan (a. -13) = 5

sin 213 2A -cos I-'

11. Show that (i) cosec 2A + cos 2A = cot A (ii) sec 2A + tan 2A =

(iii)

cos A + sin A (7t . = tan - + A cosA -smA

J

4

A).

I-sinA =tan(.!:.1 + sin A 4 2

1( + -a1) ,show that 2 cos 48 =

12. If cos 8 = -

2

a

1

a 4 + 4"" .

a

I ANSWERS I 4.

~4-~+Ji )4+~+Ji .-(Ji+l)+~4+2.J2. 2 2

2 2

•

CONDITIONAL IDENTITIES Type 1. Identities which involve sines and cosines. Type 2. Identities which involve squares of sines and cosines. Type 3. Identities which involve tangents and cotangents.

Based on Type 1:

IfA + B + C = 7t,prove that (Meerut B. Sc. Biotech 2005\ Example 1: sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C. Solution: We have, sin 2A sin 2B sin 2C = 2 sin (A B) cos (A - B) sin 2C 2 sin C cos (A - B) + sin 2C [given A + B + C:= 7t C~7t-(A + 'B) sin C = sin [7t - (A +'-8)]]

+

+

+

+

Trigonometry

309

2 sin C cos (A - B) + 2 sin C cos C 2 sin C [cos (A - B) + cos C] 2 sin C [cos (A - B) - cos (A + B) [cos C = cos [1t - (A + B)] = - cos (A + B)] . 2 sin C [2 sinA sin B] 4 sin A sin B sin C. Example 2: sin (B + C - A) + sin (C + A - B) + sin (A + B - C) = 4 sin A sin B sin C. Solution: We have sin (B + C - A) + sin (C + A - B) + sin (A + B - C) = sin (1t - 2A) + sin (1t - 2B) + sin (1t - 2C) [since B + C = 1t - A, C + A = 1t - B, A + B = 1t - C] = sin 2A + sin 2B + sin 2 C = 4 sin A sin B sin C [from Ex. no. 1] Example 3: cos 2A + cos 2B - cos 2C = 1 - 4 cos A cos B cos C. Solution: We have, cos 2A + cos 2B - cos 2C = 2 cos (A + B) cos (A - B) - cos 2C = 2 cos (1t - C) cos (A - B) - 2 cos 2 C + 1 = - 2 cos C cos (A - B) - 2 cos 2 C + 1 1 - 2 cos C [cos (A - B) + cos C] 1 - 2 cos C [cos (A - B) + cos {1t - (A + B)} ] 1-2cosC[cos(A-B)-cos(A + B)]

1 - 2 cos C [2 cos A cos B] = 1 - 4 cos A cos B cos C. . B -sm . C = 4 sm-sm-cos-. '. A ,B C . A+ sm E xamp Ie 4: sm 2 2 2 Solution: We have . A +sm . B -sm . C sm

=

=

=

2' A+B A-B - 2' C C sm--cos-sm-cos2 2 2 2

(1t

A-B 2' C C . - - C) - cos--sm-cos2 sm 2 2, 2 2 2 C A-B . C C 2 cos-cos---2sm-cos2 2 2 ·2,

C[

CJ.

=

A-B . 2 cos2' cos-sm2'

=

C [ cos2 cos2'

=

2 cos C [cos! - B _ cos A + BJ =2 cos C [2 sin A sin BJ 2 2 2 2 2 2

=

C . A . B 4 sm-sm-cCls-. 2 2 2

2

A- B-sm. {1t'2--2A+ B}] 2

310


Example 5: cos A + cos B-cos C = 4 cos.icos B sin C -I. 222 Solution: We have eosA+eosB-cosC = 2 eosA+Beos 2 2

A

2

B

_(1_2Sin 2C ) 2

eos{~ _ C}eos A 2

2

2

B -1 + 2 sin2 C 2

C A-B 2' . C = - 2 sm-eos--+ sm 2 - - 1 2 2 2

A B)}} -eosA-2-B1 -1

. C [{ sm . {1t += 2 sm 2"- (- 2

2

. C [ cosA +--eosB . A -- B] -I = 2 sm 2 2 2

A B]_I

= 2 sin C [2eos cos 222

A B. C I = 4 eos-eos-sm-- . 2 2 2 Example 6: sin (B + 2C) + sin (C + 2A) + sin (A + 2B) = 4 sin B - C cos C - A sin A - B. 2 2 2 Solution: We have, sin (B + 2C) + sin (C + 2A) + sin (A + 2B) sin (B + 2C) = sin [(A + B + C) - (A - C)] sin [1t - (A - C)] = sin (A - C). sin(C+2A) = sin(B-A) Similarly. sin (A + 2B) = sin (C - B). Therefore, L.H.S. can be written as sin (A -C) + sin (B-A) + sin (C-B).

'~

= 2 sini (B - C) eosi (2A --B - C)-2 sini (B- C) eosi (B-C) = 2 sin.!. (B - C) [eos.!.(2A - B - C) - eos.!.(B - C)] 2 2 2 = 2 sin.!. (B - C) [2 sin .!.(.A( - C)sin .!.(B - A)] 222 . B-C. C-A . A-B =4sm--sm--sm-222

Example 7: IfA + B + C =

1t

- , prove that

2

cos 2A + cos2 B + cos2 C = 2 + 2 sin A sin B sin C.

[UPTU B. Pharma 2004, 2006)

Trigonometry

311

Solution: We know that cos 2A = 2 cos2 A-I cos 2B = 2 cos2 B-1 cos

2

A

=

L.H.S. =

(1 + cos 2A) 2'

cos

2

B

=

(1 + cos 2B) . 2

.!. (1 + cos 2A) + .!. (1 + cos 2 B) + cos2 C 2

2

1

= 1 + - (cos 2A + cos 2B) + cos2 C) 2 =

1+ ~ [2 cos (2A ; 2B

).COS

e A

; 2B)] + cos 2 C

= 1 + cos (A + B). cos (A - B) + cos2 C = 1 + cos (~ - C ) cos (A - B) + cos2 C as A + B + C =

~

= 1 + sin C . cos (A - B) + (1 - sin2 C) as cos ( ~ - C ) = sin C = 2 + sin C [cos (A - B) - sin C] = 2+sinC [COS(A-B)-Sin( %-(A+B»)] = 2 + sin C [cos (A - B) - cos (A + B)], as sin (

%- e) = cos e

. (A - B + A + B) sm . (A + B - A - B )-] . C [2 sm = 2 +sm

2

.

2

=2 +2 sin A sinB sinC=R.H.S. Example 8:

cosA cosB cosC 2 + + = . sinB sinC sinC sinA sinA sinB

Solution: Taking L.H.S. = sin A cos A + sin B cos B + sin C cos C sin A sin B sin C 2 sin A cos A + 2 sin B cos B + 2 sin C 2sin A sinB sinC

CQS C

sin 2A + sin 2B + sin 2C 2 sin A sin B sinC 4 sin A sin B sin C 2 sin A sinB sinC =2=R.H.S. Example 9: cos 4A + cos 4B + cos 4C = - 1 + 4 cos 2A cos 2B cos 2C.

312

R-emedial Mathematics

Solution: cos 4A + cos 4B + cos 4C = 2 cos (2A + 2B) cos (2A - 2B) + 2 cos2 2C - 1 2 a:s (;at - 2C) cos (2A - 2B) + 2 cos2 2C - 1 2 cos 2C cos (2A - 2B) + 2 cos 2 2C - 1 2 cos2 C [cos (2A -2B) + cos {27t -(2A + 2B)}] - 1 2 cos 2 C [cos (2A - 2B) + cos (2A + 2B)] - 1 = 2 cos 2 C [2 cos 2A cos 2B - 1 = - 1 + 4 cos 2A cos 2B cos 2C.

Based on Type 2 ExamplelO:Showthatsin2i +sin2B +sin2C =I-2sinisinB sinC, 2 2 2 2 2 2 A+B+C = 7t Given that Solution: We have '2 -B + sm '2 -C - + sm 1 - cos2A

sin2 A + sin2 B + sin2 C 222

222

A +-Bc o sA--- B] +sm ,2 [7t A + B] 1- [ cos2'--22 2 :, (cos 2 A - sin2 B = cos(A + B) cos (A -B» =

A+B) 1 - cos ( 2 - cos (A-B) - 2 - - cos 2(A+B) -2-

=

A2 + B) A2 - B) A2 + B)] 1- cos ( - [ cos ( - - cos ( -

=

. Asm, '2B] 1-cos(-A2+ B)- [2 sm'2

= 1 - cos

[~- C] [2 sin Asin B] 2 2 2

_2

,A,B,C I - 2 sm-sm-sm-, 222 Example ll: IfA + B + C = p, Prove that =

2B 2 C 22,A,B,C A cos 2 - + cos - + cos - = + sm - sm - sm 2 2 2 2 2 2 Solution: We have, 2A 2B 2C cos -+cos -+cos 2 2 2

=

=

=

. 2A 2A 2C I-sm -+cos -+cos 2 2 2 A C .2b ,2 2 1+sm --sm -+cos 2 2 2 1 + [cos 2 A + B cos B - A] + cos2 C 2 2 2

Trigonometry

313

.C[cos (A-B) .C] 2 +sm -2- -sm

2

2

A- B) .

A B)}]

. C [ cos ( -2- -sm {1t +2+sm "2- (- 2

2

2+sin

~ [COS( A;B)-COs( A;B)]

A-B) . (A B)]

+ . C [ cos ( -2- -sm -22 +sm

2

. A . B] . C [2 sm-sm2 +sm2 2 2

=

2 + 2 sm-sm-sm-. . A . B . C 2 2 2

Based on Type 3 Example 12: If A + B + C = 1t,prove that tan A + tan B + tan C = tan A tan B tan C [UPTU B. Pharma 2004, 071

Solution: Give that A+B+C

=1t

Example 13: IfA + B + C =

=>

A+B=1t-C tan (A + B) = tan I1-C tanA+tanB - - - - - = - tan C I-tanA tanB

=> =>

tan A + tan B =-tan C+ tan A tan B tan C tan A + tan B + tan C= tan A tan B tan C.

1t, show that

cot B + cot C cot C + cot A cot A + cot B =] + + tan B + tan C tan C + tan A tan A + tan B .

Solution: We have

cotB+cotC tanB+tanC

-----+

cotC +cotA cot A +cotB +----tanC+tanA tanA+tanB ]

1

1

1

--+--

--+--

--+--

tanB+tanC

tanC+tanA

tanA+tanB

~B

1

~C+~C

1

~A+~A

~B

1

-----+ +----tan B tan C tan C tan A tan A tan B = cot B cot C + cot A cot C + cot A cot B.

...(1)

314


Again,

A +B+C

=n

~

A +B

= n - C ~ cot (A + B) = cot (n - C)

cot A cosB-l cot A cotB

----- = -

~ ~

H ence,

cot C

cot A cot B-1 = - cot C cot A - cot B cot C cot A cot B + cot B cot C cot C + cot A = 1.

cot A + cot B cot B + cot C cot C + cot A + +----tan A + tan B tan B + tan C tan C + tan A = cot A cot B cot C + cot C cot A = 1.

Example l4:.if A + B + C =n, tanA tanB tanC tanB tanC tanA --+--+--+--+--+-tanB lanC lanA lanA lanB lanC =~A~B~C+~B~C~A+~C~C~B

Solution: We have, tanA + tanc)+(tanB + tanA)+(tanC + tan B) ( tanB tanB tanC tanC tanA tan A

sin A sinC --+-~A

sinB sin A --+--

~C+~B

sinC sinB --+--

~A+~C

~B

sinB

sinC sin A cosB cosC cos A sin (A + C) cos B sin (A + B) cosC sin (B + C) cos A -~-~--+ +-~-~-cosAcosCsinB cosAcosBsinC cosBcosCsinA -~-~--+

sin (n - B) cos B cos A cosC sinB

sin (n - C) cosC sin (n - A) cos A +-~-~-cosA cosB sinC cosB cos C sin A

sin B cos B cos A cosC sin B

sin C cos C sin A cos A +-----cos A cos B sin C cos B cos C sin A

------+ =

cos B sec A sec C + cos C sec A sec B + cos A sec B sec C.

If A + B + C = n, prove that ·A + SID · BSID · C=4 cos-cos-cosABC l • SID 222 2. cos 2A + cos 2B + cos 2C = - 1 - 4 cos A cos B cos C.

3. cos A + cos B + cos C = 1 + 4 sin A sin B sin C . 2 2 2 sin2A +sin2B+sinC . A . B . C 4. - - - - - - - - = 8 SID-SID-SID-. sin A + sin B + sin C 2 2 2

Trigonometry

. 3A + SIn . 3B + SIn . 3C = 4 cos-cos-cos-. 3A 3B 3C 5. SIn 222 ABC

1t-A

1t-B

1t-C

6. cos- + cos- + cos- =4 cos--cos--cos--. 22222 2 . "2 A + SIn . "2 B + Sin . "2 C = 1 + 4 SIn . (1t-A) . (1t-B) . (1t-C) 7. Sin -2- Slll -2- SIn -2- . .2 A ·2 B .2 C 1 2 A B. C 8. Sin -+SIn --SIn - = - cOS-COS-SIn-.

2

2

2 2 2 2 2A 2B 2C A B. C 9. cos -+cos --cos - =2cos-cos-sm-. 2 2 2 2 2 2 2A

.2B

A

B

. A . B . C

.2C

10. cos --SIn --Sill -. =2 Sill-Slll-Sill-. 2 2 2 2 2 2 11. sin2 A + sin2 B - sin2 C = 2 sin A sin B cos C. 12. cot B cot C + cot C cot A + cot A sin cot B =1. 13. tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C. B

C

C

A

14. tan-tan-+tan-tan-+tan-tan-= 1. 2 2 2 2 2 2

OBJECTIVE EVALUATION MULTIPLE CHOICE QUESTIONS Choose the most appropriate one: 1. In Correct statement is (a)

sin8=-1I5

(b) cos 8 = 1

(c)

sec 8 = 112

(d) tan e

= 20.

2. The value of sin2 20° + sin2 70° is (a)

(b) 0

1

(c) -1

3.

If sin e

(d) 112

= -( 112) and cos 8 =

(.J3 /2») , then e lies in the quad rant

001 (c) III

~ll

(d) N

4. If tan e + cot e = 2, then sin e is

J2 J2

(a)

1

(b) ±

(c)

J2

(d) ±1I

5. Values ofe which satisfy the equation sin e + cos e = 2 sin e cos e, are

( a)

0 21t 31t , 3 ' 2

(b) ~ ~ 3' 2

(c)

21t 41t 21t

(d) None of these

3' 2'

315

316 Remedial Mathematics I f 6 . Th evaueo

cos 54°

tan 20°

tan 36°

cot 70°

--+-(b) 1 (d) 2

(a) 3 (c) 0 7. The value of sin 4620° is

(a)

112

(b) J3/2

(c)

-J3/2

(d) -112

1 1 8. If tan A = - and tan B = -, then (A + B) equal to.

2

3

rrJ6 (c) rrJ4

(b) 0 (d) 7t

(a)

9. If sin A = ~ , sin B = ~ then sin (A + B) is 5 13 (a)

± 35 or ± 45 65 65

(b) ±35 or± 63 65 65 45 33 (d) ±-or±-. 65 65

(c)

± 63 or± 33 65 65 10. Which of the following in not true?

(b) 1 + tan 2 9 = sec2 9 (a) sin2 9 = 1 - cos2 9 2 2 (c) cot 9 - cosec 9 = 1 (d) sin 2 9 + cos2 9 = 1 11. If sin 3A = cos (A - 26°) where 3A is an acute angle, then value of A is (a) A = 29° (c) A = 64°

(b) A = 39° (d) None of these 12. If sin (9 + 36°) = cos 9, (9 + 36°) is an acute angle. Then 9 is equal to (a) 25° (b) 54° (c) 27° (d) 29°

13. If 5 tan 9 - 4 = 0 , then the value of (a)

5sin9-4cos9 is 5sin9 + 4cps9

5

3

(b)

5 6

-

(d) 6

(c) 0

14. Value of tan 5° tan 25° tan 30° tan 65° tan 85° is (a)

J3

(c) 15. If 8 tan x = 15 then sin x - cos x is equal to (a)

(c)

8 17 1

17

(b) _1

J3

(d) O. 17 7 7 (d) 17

(b)

Trigonometry

16. If tan 8

= -3 , then the value of 4

I-cos8 1 + cos8

317

is

5

2

(a)

9

(b)

(c)

I 5

(d) I 9

5

2tan30

17. --""'"2- is equal to

IHan 30 (a) sin 60° (c) tan 60° 18. If 8 tan x = 15 then sin 2 x - cos2 x equal to

(b) cos 60° (d) sin 30°

8

(a)

(b)

17

17

30 17

(c)

(d) None of these.

l+sin8. - 12 th 19. 1ftan 8 - en - - - IS 5' l-sin8 5 (a) 12

(b) 25 13

13 25

(c)

(d) 25.

20. x =cotA+cosAandy=cotA-cosA,then (a) (c)

2..

(;:;J

0 I'

+(x;2f is

(b) - I (d) None of these.

PILL IN THE BLANKS 1. /f2 sin ~ 2

=

1 , then x is equal to

-----

2. /f8 is a positive acute angle such that sec 8 = cosec 60° , then 2 cos28 - I is ----3. (1 + tan 0 + sec 8) (1 + cot 8 - cosec 8) is equal to _ _ __ 4. In a ~ ABC if L B = 90° and sin A =

~ , tan A is equal to _____

5. /f x = a cos 8 and y = b sin 8, b2x 2 + a21 is _____. 6. /f6 + A = 3. Then cosec A equal to _____.

TRUE I FALSE STATEMENT 1. /f8

=

30°, then tan 29 =

2tan8

(TIF)

2.

I-tan 8

2. A,B,C are interior angles of ~ ABC ,then cos B + C = sin.:i..

2

2

(TJF)

318

Remedial Mathematics 3. An equation involving trigonometric ratios of an angle is called a trigonometric identity if it is true for all values of the angle. (T/F)

4. I - sec2 e = tan 2 e. 5.

(T/F)

If A.B, C are interior of A ABC, then tan ( C ;

A)

=

tan

f.

(T/F)

6. The measure of an angle is the amount of rotation from the initial side to the terminal side. (T/F) 7. tan (A + B) = tan A + tan B. (T/F) 8. sec 5 A = cosec (A - 36°), where 5A is an acute angle then A is 21°. (T/F)

9. cosec .JI- cos 2 e

=

1.

(T/F)

ANSWERS MULTIPLE CHOICE QUESTIONS 1. (c)

5. (c)

2. (a) 6. (d)

9. (c)

10. (c)

13. (c) 17. (a)

14. (b) 18. (d)

3. (d) 7. (c) 11. (d) 15. (d) 19. (d)

4. 8. 12 16. 20.

(d) (c) (c) (d) (a).

FILL IN THE BLANK 1. 60° 2 2

5. a b

2. 1/2 6.

3. 2

4. 3/J7

3. True 7. False

8. True

JIO.

TRUE/FALSE 1. False 5. True 9. True

2. True 6. True

4. False

REFRESHER Can We Do?

(Frequently Asked Questions)

1. Find the length of an arc ofa circle of 5 em subtending a central angle 15° [UPTU B. Pharma 2007) 2. Find in degrees the angle subtended at the centre of a circle of radius 10ft by an are oflength 20 ft. [UPTU B. Pharma 2005] 3. Prove the following: (l) t::de + co~e + 2 = sec 2e + cosec 2e [UPTU B. Pharma 2001) (ii) sin 6e + cos 6e = 1 - 3 sin2e cos2e . [UPTU B. Pharma 2007) 1+sine ...) tane+sece-l ( III [UPTU B. Pharma 2004, RGPV B. Pharma 2005] tan e - sec e + 1 cose

Trigonometry

319

4. Prove that the expression, 2 (sin 68 + cos 6) - 3 (sin48 + cos48) is independent ofthe angle 8. [UPTU B. Pharma 20051

5. If cos 8 + sin 8 =

.J2 cos 8, show that cos 8 -

sin 8 =

.J2 sin 8 [RGPV B. Pharma 20031

6. Prove that sin (A + B) sin (A - B)

= sin2A - sin2B

7. Obtain the values of (I) sin 15° (il) (iii'

[UPTU B. Pharma 20031 [RGPV B. Pharma 20011 IUPTU B. Pharma 2004)

C~! 75


1 1°

rUPTU B. Pharma 20061

cos8°+sin8° = cot 37° [UPTU B. Pharma 2002J cos8°-sin8° 9. Prove that tan 138 - tan 98 - tan4 8 = tan 138 tan 98 tan 48. [UPTU B. Pharma 2005) 8 . rwvt:th at

10. If sin A =

i

and cos B =

~

,find the values of sin (A - B) and cos (A + B) 13 where 0 to the base a if d = y. [Meerut B. Sc. Biotech 2005, 20061

..

d =y x = logo y.

...(1)

For Ex;ample: (A) (I) 24 = 16 log2 16 = 4 (i/) 102 = 100 loglO 100 = 2 (iii) 8° = 1 logg 1 = 0

1 (iv) (64)1/6 = 2 log64 2 = -. 6 log2 128 = A real number x such that 2 x = 128 => x = 7 1

log4 2 = A real number x such that 4x = 2 => x = -

2

[.: 4 1/2 = 2]

.

Remarks • Logarithm of a number satisfying the condition (1) is unique. For, if a, ~ are two distinct loganthms of the number y to a base a, then by definition, we have aU = y and a~ = y, when aU = aP.

... (2)

But by properties of powers with positive base different from 1, we conclude from (2) that a = ~. This, if the number y has a logarithm to base a, this logarithm is unique. We denote it by the definition

x = logo y if d = y. • 'log' is the abbreviation of "logarithm". • The logarithm of a number to a given positive real number ("* 1) as base is th!;l index or the power to which the base must be raised in order to make it equal to the given number.

Logarithms

•

323

PROPERTIES OF LOGARITHMS

Here, we assume a > 0, a :1:- 1 ,m > 0 ,n > 0

= Y then x = loga Y' Here, L.H.S. is called exponential form, whereas R.H.S. is corresponding logarithmic form. 2. a l = a, b l = b etc., therefore, logo a = 10gb b = 1. 3. aO = 1, bO = 1 :::) log0 1 =' 0 logb 1 = 0 •

1. if

1

4.

5. Base Change Formula 10gb a = loge a . 10gb c loge a or 10gb a = - logc b 6. The log of the product of two numbers is equal to the sum of their logs.

Remark

• IfXI' x2, ... , xn are positive rational numbers then 7. 8. 9.

10. 11.

12. 13. 14. 15. 16. 17. 18. 19. 20.

•

log (XI' x 2' ... , xn) = logo xI + logo x 2 + ... + logo x n· The log of the ratio of two numbers is equal to the difference of their logs. logo mn = n logo m. p logoq rf = -logo n. q a logan = n. If a > 1, then 0 < ex < ~ :::) logo ex < logo ~. If 0 < a < 1, then 0 < ex < ~ :::) logo ex > logo ~ If a > 1, ex > 1, then logo ex > O. If 0 < a < 1, 0 < ex < 1, then logo ex > O. If 0 < a < 1, ex> 1, then logo ex < O. If a> 1,0 < ex < 1, then log dJ. < O. J fa> 1, ex > I and ex < a, then 0 < log aU < 1. If a> 1, ex> 1 and ex > a, then logoex> 1. If 0 < a < 1, 0 < ex < 1 and ex > a, then 0 < logoex < If 0 < a < 1, 0 < ex < 1 and ex < a, then loga ex > 1.

SYSTEM OF LOGARITHMS

(a) Common Logarithm. In this system we take the base 10. This is also known as Bring's system. For Example. loglo 10 = I, loglo 100 = loglO IO~ = 2, loglo 1000 = 3.

Remark

• If no base is mentioned, the base is always taken as 10.

324


(b) Natural Logarithm. In this system, we take the base e, where e" is an irrational number lying between 2 and 3 and is given by 1 1 1 e =1 + -+-+-+ ..... l! 2! 3!

III

STANDARD FORM OF DECIMAL

To calculate the logarithm of any positive number in decimal form, we always express the given positive number in decimal form ao the product of an integral power of 10 and a number between 1 and 10 c = m x 10k i.e., where k an integer and 1 ~ m ~ 10. For Example: (i) 1234.56 can be written as 1.23456 x 1000 = 1.23456 x 103 (ii) 0.0023 = (0.0023 x 1000) x 10- 3 = 23 x 10-3

•

CHARACTERISTIC AND MANTISSA

The integral part of a logarithm is called the characteristic and the decimal part is called the mantissa. Logarithm to the base 10 are called common logarithms. The characteristics of common logarithms can be written by inspection, using the following rule. "The characteristic of the logarithm (base 10) of a number greater than 1 is less by one than the number of digits in the integral part, and is positive. The characteristic of the logarithm of a positive decimal fraction less than 1, is greater by unity than the number of consecutive zeroes immediately after the decimal point and is negative." On the other hand, to find the mantissa, we used the table of logarithms of numbers. The position of the decimal point in a number is immaterial for finding the mantissa. To find the mantissa of a number, we consider first four digits from the left most side of the number. If the number in the decimal form is less than one and has four or more consecutive zeroes to the right of the decimal point, then mantissa is calculated with the help of number formed by digits, starting with the first non-zero digits. Significant Digit. The digit which are used to find the mantissa of a given number are known as significant di[jits.

~~~~~~I

SOLVED EXAMPLES

I~~~~~~~

Example 1: Express each of the following in exponential form (i) log2 64 = 6 (ij) /oglo 0.01 = - 2. Solution: (i) log264 = 6 ~ 26 = 64. (i/) (oglO 0.01 = - 2 ~ 10- 2 = 0.01. Example 2: Find the values of each of the following form : (i) log9 81

(ii) log 12.4-

Logarithms Solution:

(1) Let lo~ 81 = x. Then x = log9 81 =:> =:> x = 2.

<JX"

= 81 =:>

= 92

<JX"

(ii) Let log h 4 = x.

Then, logh4 = x=:> ( hy = 4 =:> [(2)1/2y =:> 2x/2

= 4 = 22

= 22

~ = 2 =:> x = 4.

=:>

2 Example 3: Rewrite the following equations in the logarithm form : (i) 42/ 3

8

=

(iii) (2 h)

(ii) 50 = J

-2/3

= 1 2

Solution:

(I) 4

3/2

(ii) 50

8 can be written as log4 8

=

= I, can be written as logs I

=i.2

=

o.

~,

can be written as log h ~ = 2 2v2 2 3 Example 4: Rewrite the following equations in the exponential form : (iii) (2 h ) -2/3 =

2

1

(ii) log3 = - 5 243

(i) log2 32 = 5 (iii) log

~

5,,5

5

=

2 3

(iv) logioo (0.1) =

-

1 2

Solution: Logarithmic Form (i)

Exponential Form 32·= 2 5

log2 32 = 5

(il)

log3C:3)= - 5

(iii)

log5JS (5) =

(iv)

loglOO (0.1) =-2

I =r5 243

-

2

"3

5 = (5$)213 I

0.1 = 100- 1/2

Example 5: If logto x = a, find the value of lOa Solution: Here, we have loglo x = a =:> x = 100. Now, lOa - I = 100 x 10-1

Example 6:

If log5

x

=

100

x

10

10

a and loGY2 y

=

I

in terms of x.

a. Find 10020 - I in terms of x and y.

325


Solution: Here, we have logs x =a and log2 y = a, x = 50 and y = 20 20 100 - 1 = (52 x 22i o - 1 = (5 2)20 - 1 x 240 - 2 = 540 - 2 x 240 - 2 540 240

Therefore, Now,

=-x-

52

(50

22

t

(2 0 )4

=--x--

52

22

= x4 x y4 = x4y4

52

22

100

Example 7: Evaluate each of the following: (I) log 5

+ log 2

(il) log 500 - log 5

(iii) 4 log 5 + 2 log 4 (iv) log 6 + 2 log 5 + log 4 - log 3 - log 2 1 (v) -log 36 + log 5 - log 30 2 (VI) log 5 + 2 log 0.5 + 3 log 2. Solution: (i) log 5 + log 2 = log (5 x 2) = log 10 = 1. [By using log (mn) = log m + log n ] (ii) log 500 - log 5 = log( 5~0) = log 100 = 2.

(iiI) 4 log 5 + 2 log 4 = log 54 + log 4 2 = log 625 + log 16 = log (625 x 16) = log 10000 = 4. [': log 10000 = log 104 = 4] (iv) log 6 + 2 log 5 + log 4 - log 3 - log 2 = log 6 + log 52 + log 4 - log 3 -log 2 = log 6 + log 25 + log 4 - (log 3 + log 2) = log (6 x 25 x 4) - log (3 x 2) X = IOg(6X25 3x2 = log'100 = 2.

4)

(v) .!.log 36 + log 5 - log 30 2 = log 6 + log 5 - log 30 = log (6 x 5) - log 30 =

log 30 - log 30

=

O.

=

log (36)112 + log 5 - log 30

Logarithms

327

(vi) log 5 + 2 log 0.5 + 3 log 2 = log 5 + log (0.5)2 + log 23

log 5 + log (0.25) + log 8 = log (5 x 0.25 x 8)

=

= log 10 = I. Example 8:

If log

(m + n) = log m + log n .

Show that

m = _n_ n-1 Solution: Gives log (m + n) = log m + log n => log (m + n) = log mn m+n=mn n =mn - m n = men - 1) n m =-n-l Example 9: If log (mn) = log m - log n Show that n = J Solution: Gives log (mn) = log m - log n => log m + log n = log m -- log n => 2log n = 0 => log n = 0 => log n = log I => n = I Example 10: Show that

[RGPV B. Phrama 20041

=>

[RGPV B. Phrama 2005)

16 25 81 (i) 7 log- + 5 log- + 3 log- = log 2 15 24 80

70 22 7 (ii) log- + log- - log- = 3 log 2 - 2 log 3. 33 /35 18

log~ + 5 log 25 +

3 log!!. 15 24 80 = 7 (log 16 - log 15) + 5 (log 25 - log 24) + 3 (log 81 -log 80) = 7 [log 24 - log (3 x 5)] + 5 [log 52 - log (2 3 x 3)] + 3 [log 34 - log (2 4 x 5)] 3 = 7 [log24 - (log 3 + log 5 )] + 5 [log 52 - (log 2 + log 3)] + 3 [log 34 - (log 24 + log 5)]

Solution: (i) 7

7 [4 log 2 - log 3 - log 5] + 5 (2log 5 - 3 log 2 - log 3) + 3 (4 log 3 - 4 log 2 - log 5) = 28 log 2 - 7 log 3 - 7 log 5 + 10 log 5 - 15 log 2 - 5 log 3 + 12 log 3 - 12 log 2 - 3 log 5

=

=

28 log 2 - 15 log 2 - 12 log 2 - 7 log 3 - 5 log 3 + 12 log 3 - 7 log 5 + 10 log 5 - 3 log 5

=

log 2.

328


70 22 7 = log (70 7 (ii) log-+log--log- x22 -) -Iog33

l35

18

=

33

135

18

log[~x~J log(70xE....x~) =

~ 18

33

l35 18

= IOg(%) = log 8 -log 9 = log 2 3 - log 32 = 3 log 2 - 2 log 3. Example 11: Find the values of x in each of the following: (i)

(ii) log 125 = x log 25

log 144 = log x log 12

(iii) logx 4 + logx 16 + logx 64 Solution: (i) Here. we have

=

12.

log 144 - - - = log x log 12 log 122 - - - =Iogx log 12 2log 12 = log x log 12 log x = 2 i.e.,

logtO x = 2 x = 102 = 100. (ii) Here, we have log 125 log 25

- - - =x log 53 log 52 = x i.e.,

3 log 5 = x 2log5 3 2

- =x i.e.,

3 x =-. 2

(iii) logx 4 + logx 16 + logx 64 = 12 ::::::>

lo~ 22 + logx24 + logx 26 = 12

Logarithms

329

12 logx 2 = 12 10gx 2 = 1 xl =2

=:> =:>

i.e.,

x =2.

If

Example 12:

log x b-e

= log y = log z e-a

, then show that x b + C- a yC + a -

b~ + b - C = I.

a-b

Solution: Let us suppose logx = logy b-e e-a

=

logz a-b

k

=

.

which gives log x = k (b - c), log y = k (e - a), log z = k (a - b) =

+ e - a) log x + (e + a - b) log y + (a + b - c) log z k (b + e - a) (b - c) + k (e + a -- b) (e - a) + (a + b - c) k (a - b)

=

k {(b 2 - e 2) - a(b - c)} + k {(e 2 - ~) - b(e - a) + k{(~ - b 2) - e(a - b)}

(b

=:>

= k {b

2

- e2

+ e2 - ~ + ~ - b 2 }

- k{a (b - c)

= k . 0 - k . 0 = O. Therefore,log x b + c - a + logyc + a - b + logza + b -

log (x b + c - a. y + a -

=:>

x b + c - a.

Hence,

Example 13:

log a If -=

b-e

y

Then

c =

0

= 0 = log

1.

+ a-b. ~ + b - c = 1.

10gb loge b = - - = prove that ~ . b . eC = 1. e-a a-b

--

. log a 10gb loge Solution: Let - - = - - = - b-e

b.~ + b - C)

+ b (e - a) + e (a - b)}

e-a

= k

a-b

log a= k (b - c) log b = k (e - a)

and

log e = k (a - b).

Adding all those, after multiplication by a, b, e, respectively, we get a log a

Therefore,

+ b log b + clog e = ak(b - c) + bk(e - a) + ek(a - b) = O. log ~bbec = 0 = log 1 ~bbec = 1.

Example 14: Solve the equation

clx

=

bX

- C

~ + 5, a, b, e, > '0 but

* 1.

[Meerut B. Sc. Biotech. 20051 Solution: Take logarithm to both sides, we get log? = log bX - C + log ~ + =:>

2x log a

=

(x - c) log b

5

+ (x + 5) log e

=:> x[2 log a - log b - log e] = 5 log e - clog b

330


109(fJ 109(::

r

x

Example 15: Solve the system of equations (i) 5(logy x + logx y) = 26 (ii) xy = 64. 1 Solution: We know that logx y = - - , therefore logy x (i)

IMeerut B. Sc. Biotech 2005)

~5(lOgyx+_l_)=26. logy x

Putting

logy x =p, it gives 5 (p 5p2 - 26p + 5

or

=

+~ J= 26

0

5p2 - 25p - P + 5 = 0 5p (p - 5) - 1 (p - 5) = 0 (p - 5)(5p - 1) =0 P =5,1/5. p = 5, when

~ ~

~

When

i.

(ii)

logy x = 5 ~ x = x . y = y5 . Y = 64 or = 64 ~ y3 = ± 8 1=+ 8 = 23 or y3 = - 8 = (- 2)3

~

i

y =2 or y = -2.

Then (ii)

~

64 64 64 x = - = - =32 orx= -=-32. y

2

-2

But y and x both are used as base in equation (i), so x, y cannot be negative ~ one solution is x = 32, y = 2. When :. (ii)

p =

logy x =

51

~ x =

yl/5

or x 5 =

"*

y

x, y = x 5 = 64 or x 6 = 64 or x 3 8 x =2 or x = - 2 [reject negative value]

~

~

Again (ii)

1 5' then

~

x

64

64 = 32. 2

=-= -

y

This gives another solution. Thus, we get two solutions of the gives system of equations as follows: x = 32, y = 2 or x = 2, y = 32.

Logarithms

~~~~~~~I EXERCISE

7.1

~I~~~~~~~

1. Write the following into logarithmic form (i)

2 8 = 256

(iii) 7 3 = 243

(ii) 10 3 = 1000

(iv) 4- 4 = _1_ (v) 4 3/ 2 = 8. 256 2. Write the following into exponential form (i) log5 25 = 2 (ii) loglo 0.001 = - 3. 3. Find the value of b which satisfies (i)

logJ8 b =

3~

(iii) log J3 x = 4

(ii) loge 210gb 625 = loglO 16 loge 10

(iv) log4 x = 15

(v) logl25

1

X

= -. 6

4. Find (i) log6 16, if logl2 27 = a (ii) log25 24, if log6 15 = a and logl2 18 = [3 (iii) log30 8, if lof30 3 = a and log30 5 = b. 5. If logl2 18 = a and log24 54 = [3, show that a[3 + 5 (a -(3) = I. 6. Without using the table, show that (i)

_1_+_1_ > 2 cos2 1t log4 1t

(ii) log2 17 logl/5 2 log(l/5) > 2

(iii) 110gb a + loga b I ~ 2, where a and b are positive integer not unity. 7. Compute, without using tables (i)

log3 4 log4 5 log5 6 log6 7 log7 8 logg 9 (ii) log3 2 log4 3 log5 4 .... logl5 14 logl6 15. 8. Show that log2 3 is an irrational number. 9. Prove the following: (i)

log a n = I + log b logab n a

10. Show that: (i)

50 log 2 + 2 log 5 - log 3 - 2 log 7 = log 147

(ii)

IOgC~) + logG!)- lOge!) = 0

1· (iii) -log 25 - 2 log 3 + log 18 = 1 2

(iv) .!.. log2 54 + log2 10 - log2 625 = 1 3 (v) loglo 10 + loglo 100 + loglo 1000 + loglo 10000 = 10 (vi) .!..Iog 9 + 2 log 6 + .!..Iog 81 -log 12

2

4

=

331

3 log 3.

332


11. Evaluate each of the following (i) 210g3 5 - 510g32 (ii) (81)lIlog5 3 + 2i og 936 + 3(41og9 7) (iii) log 15 + 2 log 0.5 + 3 log 2 - log 3 - log 5 (iv) log 21 + log 4 + 2 log 5 - log 3 - log 7. 12. Show that (i)

a loga I + 2 log a 2 + 310g a 3 + ... + n loga n

=

22 . 33 . 44 ... nn

(ii)

a loga I + 2 log a 2 + 2 log a 3 + ... + 210ga n

=

(n!i

(iii) loglo tan 1° 10glO tan 2° loglo tan 3° ... loglo tan 50° (iv) loglo tan 1° + 10glO tan2° + ... +loglO tan 89° = o.

=

0

s

I ANSWERS I I. (i)

log2 256 = 8

(ii) log", 1000 = 3

(iv) log4 (1/256) = - 4 2. (i)

52 = 25 (ii) 10-3 = 0.001.

3. (i)

32

(ii)

4. (i)

4(3 -a) 3+a

(ii)

7. (i)

2

(ii)

11. (i)

•

(J2t

2

5-~ 2a~ + 2a

-

4~ + 2

(iii) 3 (1 - a - b).

1 4

(ii) 890

0 (iv) 2

(iii) log7243 = 3

(v) log4 8 = 3/2.

(iii) log2

METHOD TO DETERMINE THE CHARACfERISTIC AND MANTISSA

The characteristic is determined by using the following rule : (i) The characteristic of the logarithm of any number greater than I is one less than the number of digits to the left of the decimal point in the given number. (ii) The characteristic of the logarithm of any number less than 1 is negative and numerically one more than the number of zeroes to the right of the decimal point. For Example: See the following table: S.NO

Given number

1.

63 389.6 3986 6.36

2.

0.4 0.04 0.004 x 10- 1

Characteristic

n -1 ) -2 -3

Explanation One less then the number of digits to the left of the decimal point.

One more than the number of zeroes to the right immediately after the decimal point.

Logarithms

333

Method to Determine the Mantissa The mantissa is determined by using the following rule : (i) The mantissa is the same for the same significant figures in the same order and does not depend on the position of the decimal point. (ii) The mantissa is always taken as positive.

For Example: Given number

Characteristic

Mantissa

Logarithm

5978

3

0.7766

3.7766

597.8

2

0.7766

2.7766

0.005978

-3

0.7766

- 3 + 0.7766

The given number contain the same significant figures, namely 5, 9, 7, 8 in the same order and so the mantissa of their logarithm is the same, they differ only in the characteristic. In log 0.005978, which is equal to - 3 + 0.07766, the characteristic - 3 is negative and the mantissa 0.7666 is positive. To indicate that the negative sign applies to the characteristic only, the '-' sign is put above the characteristic .Thus logO.005978 = - 3 + 0.766 6 = 3.7766. It is read as "bar 3 point 7766"

Remark • To find the mantissa of the logarithm of a number which contain less or more than four digits, make it afour digit number by having zeroes on its right or by condensing it by the rule of approximation which is given below.' - or more t h an i - .IS takes as I ; andless th an i - .IS negIected" . (I) ,,1 2 2 2 (ii) "5 or more than 5 is taken as 10; and less than 5 is neglected" For Example: (i) 6.76236 = 6.7624, upto four decimal places. (ii) 6.7634 = 6.763, upto four decimal places.

Mantissa of the Logarithm of a Given Number To find the mantissa ofiogarithm of a given number we use the standard table ofiogari'thms. The table of logarithms consist of 90 rows and 20 columns. Every row begins with a two digit number 10, 11, .... , 98 99 and every column is headed by a one digit number 0, 1. 2, ... , 9. On the right of the table, a big column is divided into 9 sub columns headed by the digits 1, 2, 3, ... , 9, known as column of mean differences. To find the mantissa of a number, consider first four digits from the left most side of the number. If the number is the decimal point, then mantissa is calculated with the help of the number formed by digits beginning with the first non-zero digits.

Remark • Tofind the logofa given number x, use theformula logx

=

characteristic + mantissa.

334


~~~~~~~I

SOLVED EXAMPLES

I~~~~~~~

Example 1: Find log 756.8. Solution: By neglecting the decimal point, we obtain 7568, which is a four digit number. See the number 75 in the extreme left hand column in the logarithm table. In the horizontal line of75 and under 6 (next digit in the number) we found the number 8785. In the same horizontal line and under 8 (4th digit of the given number), the number found is 8. Adding 8 to the already obtained number 8785, we get 8793. Therefore, the mantissa is 0.8793 => log 756.8 = 2.8793 [.,' characteristic is 2] [characteristic is 1] Similarly log 75.68 = 1.8793

log 0.07568 = 2.8793 [characteristic - 2] Example 2: Find the logarithm of the following number (i) 5395 (ii) 0.002359 (iii) 25795 (iv) 0.005. Solution: (i) The given number 5395 is a four digit number. See the number 53 in the extreme left column is the logarithm table. In the horizontal line of 53 and under 9 (next digit in the given number) we found the number 73.16. In the same horizontal line and under 5 (4 th digit of the number ), the number found is 4. Add this number 4 to 7316 to get 7320, which is the required mantissa of log 5395. Also the characteristic of 5395 is 3. Therefore, log 5395 = characteristic + mantissa = 3.7320. (ii) Firstly, find the four digit number, by getting the first four digits beginning with the first non-zero digit on the right of the decimal point which is 2359. The mantissa of the given number can be determined by the procedure discussed in (i) and given by mantissa of2359 = 3711 + 17 = 3728. Also, the characteristic of 0.00359 is - 3. Hence, log 0.002359 = 3.3728. (iii) Clearly, the characteristic of the logarithm of25795 is 4. To find the mantissa of the given number 25795, consider the four digit number 2579 and apply the same process. we get mantissa of2579 is 4114. Therefore, the logarithm of the given number 2595 is 4.4114. (iv) The characteristic of 0.005 is - 3. To find the mantissa, consider the number 50. See in the row 50, under the column headed by 0 and get the number 6990. Therefore, log 0.005 = - 3 + 0.6990 = 3.6990. Example 3: Find log 11.648. Solution: The characteristic of 11.648 is 1. Now leaving the decimal point, the given number consists of five digits. Condensing it to a four digital number, by the rule of approximation, we get the number 1165. Now follow the same procedure, as above, the mantissa of 1165 is 0.0664. Hence, log(11.647) = log (1 1.65)= 1.0664.

Logarithms

•

ANTILOGARITHM

If log x

=

335

n, then x is called the antilogarithm of n and is written as

x = antilog (n). For Example: (i) log 10 = 1 antilog (1) = 10. (ii)

log 0.0681 = 2.8331 antilog (2.8331) = 0.0681.

Method of Finding the Antilogarithm To find the antilog of a given number, we make use of the table of antilogarithms. The method is almost the same as that for finding the logarithm of a number. The table is divided into three similar sets of columns and the four digits are to be taken in the same manner starting with digits immediately to the right of the decimal point, not excluding zero. The following points must be kept in mind for convenience : (i) If the given number is negative, first make it positive by adding one to the decimal part and by subracting one from the integral part. (ii) Apply the method, same as that used for finding the logarithm of a number. (iii) If the characteristic of the given number is positive and is equal to n, then insert the decimal point after (n + 1) dIgits in the obtained number. If n > 4, then write zeroes on the right side to get (n + 1) digits. (iv) If the characteristic of a given number is negative and is equal to - n or n then on the right side of the decimal point, write (n - 1) consecutive zeroes and then write the obtained number

~~~~~~~I SOLVED EXAMPLES ~I~~~~~~ Example 1: Find antilog 2.3456. Solution: The mantissa is positive and is equal to 0.3456. Now look into the line starting with 0.34. In the horizontal line of 0.34 and under 5 (the next digit of the mantissa), the number obtained is 2213. In the same horizontal line and under 6 (the fourth digit of the given number) in the mean difference columns, the number found 3. Adding 3 to 2213, we get 2216. Now, since the characteristic is 2, the required number must have 3 digit in the integral part. Therefore, antilog 2.3456 = 221.6 Similarly, we can find antilog 0.3456 = 2.216, antilog 1.3456 = 0.2216 antilog 2.3456 = 0.02216, antilog 5.3456 = 0.00002216.

Example 2. Find antilog 3.0675 Solution: In the horizontal line of 0.06 and under 7 we get 1167. In the same horizontal line and under 5 in the mean difference column, we get I. Adding 1 to 1167 we get 1168. Since the characteristic is 3, the required number must have 2 zeroes immediately to the right of the decimal part. Therefore, antilog (3.0675) = 0.001168.

336


Example 3: Find the antilog, each of the following (i) - 1.2084 (ii) - 0.62. Solution: (i) The given number is - 1.2084. Here, we observe that the mantissa of the given number is negative. First we make it positive by adding 1 in following manner - 1.2084 = - 1 - 0.2084 = - 1 - 1 + 1 - 0.2084 =-2+0.7916

= 2.7916. Now, using the antilogarithm table, we find that the number corresponding to the mantissa 0.7961 is 6189.

Since, the characteristic is 2 i.e., - 2, put one zero just after the decimal point to get the antilogarithm of the given number. Therefore, antilog (- l.2084) = antilog (2.7916.) = 0.06189. (li) Consider the given number - 0.62 = - 1 + 1 - 0.62 = -1 + 028 = 1.028. Now, using the antilogarithm table, the number corresponding to 0.23 is 1905 also, the characteristic is T i.e., - l.

Therefore, antilog (- 0.62) = antilog (1.28) = 0.l905. Example 4: Find the values of the following (1) 2.76x4 (iii) 6.42 + 3

(ii) 6.45 x 98l.4

(iv) 0.0064 x 1.507.

Solution: (i) 2.76x4 = (-2 + 0.76) x 4 = - 8 + 3.04 = - 8 + 3 + 0.04 = - 5 + 0.04 =

5.04.

(li) Let x = 6.45 x 98l.4 log x = log 6.45 + log 981.4

::::;>

Therefore (iii)

= 0.8096+2.9919=3.8015. x = antilog (3.8015) = 6331. 1

6.42+3 = -(-6+0.42) 3 -6 0.42 = - + - = - 2 + 0.l4 = 2.l4. 3 3

Lo~arithms

Let x log x

(iv)

=

0.0064 x 1.507, then log (0.0064 x 1.507)

=

3.8062 + 0.1781

=

3.9843.

=

337

x = antilog (3.9843.)

=>

0.09645. Example 5: Find log {(27i x (0.81lls + (90/14), where log 3 Solution: The required value is =

log {(27)3 x (0.81)4/5 + (90)5/4}

=

0.4771213.

27 4 81 5 3 log-+-log---log 90 10 5 100 4 , 4 5 = 3 (log3 3 - I) + -(log3 4 - 2) --(log3 2 + I) 5 4

=

= (9+

I; -%}Og3-(3+~+%)

97 17 -log3-510 20 = 4.6280766 - 5.85

=

2.7780766. Example 6: Using logarithmic table, evaluate the following =

(i)

41.32x20.18 12.69

(.) L · So Iutton: I etx =

(

41.32x20.18 ) 12.69

(iO 112

Then, we have I og x

=

I (41.32X20.18)1/2 og 12.69

=

..!..lo (41.32X20.18) 2 g 12.69

1 = -(log 41.32 + log 20.18 -log 12.69) 2 =

= =

Therefore,

x

=

=

..!.. (1.6162+ 1.3049-1.1035) 2 1 1 "2(2.9211-1.1035)= "2(1.8176) 0.9088. antilo (0.9088) 8.106

338


(ii) Let Then, we have (45.4 )2 ) Iog x = -1 Iog [ 3 (3.2)2 x (5.6)2 =

3"1 [2 log (45.4) -

2 log (3.2) - 3 log (5.6)]

1 = -(2 x 1.6571-2 x 0.5051-3 x 0.7482) 3 =

=

.!.(3.3142-1.0102-2.2446) 3

.!. (0.0594) = 0.0198. 4

x = antilog (0.0198.)

Therefore,

= 1.047.

~~~~~~~I

EXERCISE

7.21~~~~~~~

(ii) 1270 1. (i) 25.795 (v) 0.0074 (iv) 431.5 (vii) 0.3582. 2. Find the antilogarithm of each of the following : (i)

1.4114

(iv) 2.6350

(iii) 0.005 (vi) 0.002598

(ii) 3.1038

(iii) 3.6990

(v) 3.8692

(vi) 3.4146

(vii) 1.5541.

3. If log x = 1.4914 and log y = 2.4669, find the value of each of the following: (i)

x

(iv)

(ii)

xy

.

3

i

(v)

x

(iii)

y

x2

l

x2

Y 4. Using logarithmic table, find the value of each of the following : (i)

(iv)

76.03 x 9.08 101.2x63.17

8113 x 73 / 4

(ii)

(73.56)3 x (0.0371)2 68.21

(v) (0.OOOOI427)ln

(7.29)1/3 x3.26 (vi)

(6.45)3 x (0.0034)113 x981.4 (9.37)2 x (8.93)114 x(0.0617)

.

(iii)

(25.36)2 x 0.4569 847.5

•

Logarithms

339

APPLICATION OF LOGARITHM IN PHARMACEUTICAL PROBLEMS

(1) Half life Period The half life period T of C 14 gives the value of disintegration Constant k from the equation k

0.693 . eauatlOn . .IS = - . The d·· IsmtegratlOn T I

k

2.303 . =- Iog -a- IS.

a-x

t

l4

Exa mple 1: The amount of C isotope in a piece of wood found to be one sixth of its amount present in a fresh piece of wood. Calculate the age of wood: Solution: Half life of C I4 = 5577 years.

Half life period =>

T = 0.693 k

k = 0.693 T

0.693 5577

·· . equation, . 2.303 a name Iy k = - Iog--. Th e dIsmtegratlOn t a-x 0.693 2.303 I 1 X 5577 = - t - og 1/6

2.303 x 5577 x log 6 t = ---------="--0.693 2.303 x 5577 x 0.7782 0.693 t = 20170 year.

(2) Calculation of Boiling Point or Freezing point If the freezing point or the boiling point of a liquid at one pressure is known, it is possible to calculate it at another pressure by the use of the Chaperone classes equation p, ~Hv [ I I] loglo~ = 4.576 7i - T2 .

Example 2: Calculate the vapour pressure of water 90.0°C if its value at JOO.O°C is 76.0 cm. The mean heat of vaporization of water is the temperature range 90° and JOO° C is 542 calories gm. Solution: l1Hv = 542 x 10 caliper mole

P2 =? PI = 76.0 cm T2 =90 + 273 = 363° Abs TI = 100 + 273 = 373° Abs. According to Chaperone clauses equation P2 log1o-

l't

10

~Hv = - [T2 --4.576

-11 ] ,we 11 T2

have

P2 =542 X I8[363-373] glO PI 4.576 363 x 373


Hence, loglo P2

-

542 x 18[363 -373] loglo 76 = 4.576 363 x 373

P2 = 52.88 cm or 528.8 mm.

(3) Calculation of Equilibrium Constants and Concentrations If the concentration of reactants and products are known at equilibrium is a reaction, the equilibrium constant can be calculated.

Example 3: An equilibrium system for the reaction between hydrogen and iodine to given hydrogen iodine at 670 k in a 5 litre flask contains 0.4 mole of hydrogen, 0.4 mole of iodine and 2.4 moles of hydrogen iodine calculate equilibrium constant. Solution: For the reaction H 2(g) + 12 (g) ~ 2HI(g).

k

=

[Hlf [H 2 ][1 2 ]

molar concentration of various species at equilibrium

[H] = 0.4 [I] = 0.4 = 0.08 mol L- I 2 5' 2 5' [HI] = 2.4 = 0.48 5

k

(0.48i (0.08)(0.08) log k = log (0.48)2 - log (0.08)2 = 2 log 0.48 - 2 log 0.08 =2 log 4.8 x 10- 1_ 2 log 8.0 x 10- 2 =

=2 (-I + .6812) - 2 (- 2 + .9031) = 1.5562 k=36

(4) P-H Equation PH =k + log [salt] . a [acid] Example 4: Some of 0.2 m acetic acid are mixed with 50 ml of 0.2 m sodium acetate solution. What will be the PH the mixture ka = 1.85 x ur 5.

We have

[Acid] =

.Q3..

[Salt] =

.Q3..

100 100

pH =pka - log [Sa~t] = - log 1.85 [Acid] pH = - log 1.85 x 10- 5

pH = 4.7325.

x

10- 5 + log 0.2/100 0.21100

Logarithms

341

Example 5: Calculate the PH. Value (i) 0.001 MHC/ (ii) 0.001 M NaOH Solution: (i) Since HC is a strong Acid, H30+ ion concentration is equal to that of the acid [HP1 = [HCl] = 0.001 m - I x 10- 3 m pH = - log [H30+] pH = _ log [I x 10- 3] = - (-3) log 10 = 3. or (ii) Since NaOH is a strong base, it completely ionizes. :. hydroxyl ion concentration is equal to that of the base [OW I] = [NaOH] = 0.01 m = I x 10- 2 kw = [HP+] [OH-] k IxlO- 14 [H 01 = - - = = I x 10- 12 3 [OW] Ix 10-2 PH = - log [H30+] = - log (I x 10- 12) = 12.

I EXERCISE 7.31 1. Calculate the PH of 0.5 molar solution of sulphuric acid 2. Gives that the half life period of Radium is 1580 years. Calculate the disintegration constant and average life 3. The equilibrium Constant for the reaction N2 (g) + 3H2 (g) = 2 NH3 (g). at 715 k is 6.0 x 10- 2 . Ifin a particular reaction there are 0.25 mol L- 1 ofH2 and 0.06 mol L-I of NH3 present, calculate the concentration of N2 at equilibrium. 4. At what temperature will water boil under a pressure of787 mm ? The latent heat of vaporization is 536 cal per gram. 5. The pH of a soft drink is 4.4 . Calculate [HP+] and [OH-]. HINTS TO THE SELECTED PROBLEMS

1. H2 S04 ~ 2H+ + SO~ In a 0.5 M H2So4 solution = 2 x 0.5 mol L- 1 pH = - log [W] = - I x log I = O.

[H1

·s::

•

2. Ha If Ihe perlO

dT

0.693 k

112 = - -

k = 0.693 TI/2

0.693 1580

342


A=

Average life

..!. = __1_-:k

4.3

x

10-4 .

4. 10glO P2 = ~Hv [~-...!..] PI 4.576 1\ T2 10 10 787 = 536 x l8[T2-373] g 762 4.576 373T2

s.

536 x 18[T2 -373] 4.576 3 73T2

10glO 787 - log 762

=

T2

=

374°k.

=

4.4 4.4

pH = - log [HP+]

- log [H30+] log [H30+]

= -

lew lxlO- 14 [OHl = - - = ----:5 [H3 0 +]

3.98x10-

I ANSWERS 1. 0 4. 3740 k or 101 0 C.

2. 4.3 x 10-4 years-I, 2325 years. 5. [Hp+] = 3.98 x lO-l mol L-I , [OH-]

3. 3.84

= 2.5

x 10-10 mol L-I

OBJECTIVE EVALUATION MULTIPLE CHOICE QUESTIONS Choose the most appropriate one: 1. If t.f = bY = cf and logha = log} . then which one of the following will hold True? (a) y = x - z (b) y = x + z (c) ;;

= yz

(d)

2. The domain of the function

~Iogo.sx

(a) (0.5, 00) (c) (0,1)

3.

If loga 6 (a) (c)

= m.

Y.

=

xz .

is (b) (I, 00) (d) (0.5,1).

and loga 3

= n.

then loga (%}s equal to

1 - loga2

(b) 1 + m + n

1- m - n

(d)

1- m + n.

4. If the logarithm of a number to the base·-.[g· is 6, then the number is (a) 512 (c) 216

(b) 343 (d) 36.

Logarithms

5. The equation log3 (3x - 8) = 2 - x has (a) x = 4 (c) x = 2 6. Evaluate log tan 1° + log tan 2° + ... +

the solution (b) x = 3 (d) x = 1 log tan 89°

0

(b)

1

(c)1/2

(d)

.fi

(a)

343

I 28 I 405 · I'f I 14 7. SImp I y og 15 + og 27 = og 196 (a)

(b) 1092

I

(d) .fi 1/2 8. Solve for x if logl6x + (log I6x)2 + (Iogl6x)3 + ... + to OCJ = 1/3. (a) 1/2 (b) 1/2 (c) 2 (d) 3/5 9. If loglO (x 3; ) = 3a + 2b and 2 loglo (x 2y 3) = 2a + 3b, then the value to x at a = 3 in W 100 ~ I~ (c) loglO2 (d) loglo3. 10. If log9o 2 = a, then loglo 25 is equal to (b) 1 - a (a) (I + a) (d) 2 (l + a). (c) 2 (1 - a) (c)

FILL IN THE BLANKS 1. If log .J8 x = 3

i- The value of x in _ _ _ _'

1. If log 2 = 0.3010 and log 3 = 0.4771 the value log 25 _ _ __ 3. log927 - log279 = _ _ __ 4. log7 logs (~ x + 5 + J;) = a then x ____ ' 5. The value of log 3.fi (5832) is equal to _ _ __ 6. The value of 641og8 5 is _ _ _ _ ' 7. If logIQ5 + loglo (5x + 1) = loglo (x + 5) + 1 then x is equal to _ _ _ _' 8. If log 27 = 1.431 then value of log 9 is _ _ __

I

9. 2 loglo 5 + logIQ8 - '2log10 4 = _ _ __ 10.

~

1 1 I ] .IS egua I to _ _ __ + + [ (log a be + I) (Iogh ea + I) (log,. ab + I)

TRUE/FALSE 1. The equation

~

1. IfIOg(a;b) =

log

(~ + 2x) - log J; + 2 = 0 has the solution x = I.

(T/F)

~

[Ioga+logb]a=b.

(TIP)

344


bq 3. IF log,r- = p and log"x = q. the log(alb) x = q _ p . 4. a

log 2 Y'

a

log 2

Z'a

log 2 a

(T/F) (T/F) (T/F) (T/F) (T/F)

= Z.

5. The Number log27 is a rational Number. 6. The value of 3210g 3 is equal to 49. 7. The value of loglo 50000 - log lO 5 is equal to 10.

I ANSWERS MULTIPLE CHOICE QUESTIONS 2. (c)

3. (d)

4. (a)

5. (c)

6. (c)

7. (b)

8. (c)

9. (b)

10. (c).

4. 4 8. 0.954

1. (d)

FILL IN THE BLANK 1. 32

2. 1.398

3. 516

5. 6

6. 25

7. 3

9. 2

10. I

TRU'E/FALSE 1. True 5. False

2. True 6. False

3. True 7. False

REFRESHER

4. True

~I~~~~~~~

Do you know? (After reading this Chapter you must be able to learn the following concepts) Let there be a Number a > 0 and a Y > 0 to the base a if cI = Y .

cI'= Y

¢:>

loga Y = x.

• • • •

logaa = logbb = 1. loga 1 = 0 ,10gb 1 = O. logba. logca = 1. logba = logca logb c .

•

loga(m x n) = logam

• IOga(:)

* 1. Number x is called the logarithm of another variable

= loga m -

+ logan. logan.

• logamn = n logam.

• The integral part of a logarithm is called the characters and the decimal part is called mantissa.

Logarithms

Can we do?

345

(Frequently Asked Questions)

n 1. If log (m + n) = log m + log n, show that m = - n -1


2. If log (mn) = log m - log n, show that n = 1 [RGPV B. Pharma 2005] X 3. Solve the equation ;a = b - c·cf + 5, a, b, c> 0 but not equal to 1 [Meerut B. Sc. (Biotech) 2005] 4. Solve the following equation

xy=64

[Meerut B.Sc. (Biotech 2005]

DDD

SYSTEMS OF COORDINATES

III

INTRODUCTION

Coordinate Geometry is the branch of mathematics in which two numbers are used to represent the position of a point with respect to two mutually perpendicular number lines called coordinate axes. The french mathematician and philosopher Rene Descaotes first published his book La Geometric in 1637 in which he used algebra in the study of geometry. This he did by representing points in the plane by ordered pairs of real number called cartesi an coordinates and representing lines and curves by algebraic equations.

Coordinate Axes The adjoining figure 8.1 shows two number lines XoX and YoY intersecting each other at their zeros. y

3

2 1

X'

0 -3 -2

2

-1

3

X

-1

-2 -3 Y'

Fig. 8.1

XoX' and YoY' are called coordinate axes out ofwhichXoX is called the x - axis, Yo Y is called y-axis and their point of intersection is called the origin.

System a/Coordinates.

347

Remark • Number lines XoX and YaY are sometimes also called rectangular axes as they are perpendicular to each other.

Convention of Signs: The distance measured along OX and Of are taken as positive and those along OX' and Of' are taken as negative as shown in figure 8.1.

Coordinate of a point in a plane Let P be a point in a plane. Let the distance of P from the y-axis = a units. And, the distance of P from the x-axis = b units. Then we say that the coordinates of P are (a, b) a is called the x - coordinates or abscissa of P and b is called the y - coordinate or ordinate. of P. Y

• (a. b) I

I I

I I I

b

: I I

I I I

X'

',M ---=+-------'---'-'."--X 0t---_ a Y'

Fig. 8.2

Remarks • (x, y) and (y, x) do not represent the same point unless x = y. e.g. (5,4) and (4,5) represent two different points.

• In stating the co-ordinates of a point the abscissa proceeds the ordinate. The two are separated by a comma and enclosed in a bracket. Thus a point, whose abscissa is x and whose ordinate is y designated by the notation (x, y) i.e., (abscissa, ordinate) • Since at origin the value of x-coordinates is 0 and the value of y-coordinate is also 0, Therefore, the coordinates of origin = (0,0). • Since for every point on x axis, its distance from x axis is 0 i.e., ordinate is O. Therefore the Coordinate of a point on x-axis are taken as (x, 0). • In the same way, for every point on y-axis its distance from y-axis is 0 i.e., abscissa is 0, therefore, the coordinate ofa point on y-axis are taken as (O,y).

Y

II

I

(-. +)

(+. +)

x' -----:::+----. X 0

(~,I~)

(1~_)

Y'

Fig. 8.3

348

Remedial Mathematics y

Quadrants LetX' OX and YOY be the Coordinate Axes. These axes decide the plane of the paper into four regions, called Quadrants. The RegionsXOY, YOx' ,X'Or' and y' OY are respectively known as first, second, third and fourth quadrants.

II (-,+)

X'--------~-------- X

0

1I1

Using the convention of signs, we have the signs of the coordinates in various quadrant given below. quadrant

nature ofx and y

signs ofcoordinates

XoY

I

x>O,y>O

(+, +)

YoX

II

xO

(-, +)

XoY

III

x

6k

B (4, -9)

= 1

::::>4k-3=-2k-2 ::::> k = 1/6.

Compairing y-coordinate

kx(-9)+(l)x5 = 3 k+l -9k+5 ::::> = 3 ::::>-9k+5=3k+3 k+l ::::> 12k = 2 ::::> k = 1/6. Hence, the required ratio is 1 : 6.

System of Coordinates

357

Example 3: In what ratio does the y-axis divide the line segment joining the point (-4, 5) and (3, -: 7) ? Solution: k=/~

I

I

A (--4. 5)

P (0. y)

B (3.-7)

Fig. 8.9

The line segment joining the points A(- 4, 5) and (3, - 7) is divided by the y axis at the point P(O,y) in the ratio k: 1. Compairing x-coordinate, we have

kx3+lx-4 0=>3k-4=0=>3k=4=>k=4/3. k+1 Therefore the required ratio is 4 : 3. ---- =

Example 4: Find the ratio in which the line 3x + y - 9 = 0 divides the line segment joining A (1, 3) and B (2, 7). Solution: k B

A (1. 3)

(2.7) Line 3x+y-9=0

Fig. 8.10

The equation ofthe given line is

3x+y-9 = O.

...(1)

meets the line segment joiningA(I, 3) and B(2 , 7) at the point P (x , y) and divides the segment internally in the ratio k: I.

. By sectIOn formula: x

2k+1 k+1

7k+3 k+1

= --, y = - -

. ' (2k+1 I.e., CoordmateofPare= - - ,7k+3) -- . k+1 k+1 The point P lies on the line whose equation is given by (I) Therefore, (2k+I)+(7k+3) =0. k+1 k+1 => 6k+3 +7k+3-9k-9 = 0=>4k-3=0

=>

k = 3/4

Hence, the required ratio is 3 : 4

Example 5: Find the coordinates o/point which divides the line joining point (1, 2) and (-3 ,4) in the ratio 2 : 3 internally.


".

358


Solution: LetA (1,2) andB(- 3.4) be the given points. Point P divides A and B in the ratio 2 : 3 PA :PB = 2:3 . :. Coordmates of Pare (2X-3+3XI ,2X4+3X2) ---2+3 2+3

53, 8; 6) i.e., ( -; , I;).

i.e., ( -6 +

Example 6: IfA (- J, - 3), B(1 ,-J) and C(5, J) are the vertices ofa triangle ABC, find the length of median through A. rUPTU B. Pharma 200S1 Solution: Let D be the middle point of BC. A (-1, 3)

Then AD is the median through A and D is

(I

' + 5 -1 + 1),I,e., (3 ,0) th epomt -2-'-2AD = ~{3 _(_1)}2 +{O-3}2

=~42+(_3)2 =5.

o

B (1, -1)

C (5,1)

Fig. 8.11

Example 7: A quadrilateral has the vertices at the points (- 4 , 2) , (2, 6) ,(8, 5) and (9, - 7). Show that the mid points of the sides of this quadrilateral are the vertices of a parallelogram. [UPTU B. Pharma 2004] Solution: LetABCD be the given quadrilateral with vertices A (-4,2), B(2, 6), C(8, 5) and D (9, - 7). Let E, F, 0 and H be the mid points of the sides AB, BC, CD and DA respectively.

2 6 Then the coordinates of E are ( -4 + , 2; ) i.e., (- 1,4). 2 ,. 6+5).I.e., ( 5'2 11) . The coordmates of Fare (2+8 -2-'-2-

(8;9, 5;7}.e.,C; ,-I} and the coordinates of Hare (-4 +9, 2;7}.e,,(%, ~} 2

The coordinates of 0 are

I; , Il·. .c: ,n (s: t,'; ; %} .•..c: ,n

Now, the coordinates of the mid point of EO are ( -1:

and the coordinate of the mid point of FH are

4;

Thus we see that the diagonals EO and FH of the quadrilateral EFGH bisect each other. Hence EFGH is a parallelogram Example 8: Three consecutive vertices ofa parallelogram are A(1 , 2), B (1, 0) and C(4, 0) find the fourth vertex D.


359

Solution: Let the coordinates of the vertex D be (x ,y) Diagonals AC and BD of the parallelogram ABCD bisect each other at M. i.e., M is mid point of AC as well as of BD.

Midpointof

o (x, y)

AC=C~4,2;0)=(%,I)

Mid point of BD =

(x; I, y;O)

C (4, 0)

~-----..,.

... (1)

. .(2)

(1) and (2) are same coordinates of the point M. A(1.2)

B (1,0)

x + 1 = ~ and l = I Fig. 8.12 2 2 2 x = 4 andy = 2 Therefore, the coordinates of Dare (4,2).

Then

I EXERCISE 8.21 1. Find the coordinates of the point which divides the line segment joining the points A(4, - 3) and B(9, 7) in the ratio 3 : 2 2. Find the coordinates of the mid point of the line segment joining the points A(- 5,4) andB(7,-8) 3. Find the ratio in which the point P(m, 6) divides thejoiningA(-4, 3) and B(2, 8). Also, find the value of m. 4. In what ratio does point P(2, - 5) divide the line segment joining A (-3,5) and (4, -9). S. In what ratio is the line segment joining the point A (6, 3) and B (- 2, - 5) divide by the x-axis. Also, find the coordinates of the point of intersection of AB and the x - axis. 6. Find the ratio in which they-axis divides the line segment joining the points A(-4, 10) andB(7,-I).

7. 8. 9. 10.

Also, find the coordinates of their point of intersection. The coordinates of one end point of diameter AB of a circle are A(4. - I) and the coordinates of the center of the circle are C(I. - 3). Find the coordinates of B The three vertices ofa parallelogramABCD, taken in order are A(l, -2), B(3, 6), and C(5, 10) . Find the coordinates of the fourth vertex D. Find the lengths of the medians of a d ABC whose vertices are A(7. - 3) B(5, 3) , and C(3,-I) Let D(3, - 2), E(- 3, 1) and C(4, - 3) be the mid points of the sides BC, CA and AB respectfully of d ABC. Then, find the coordinates of the vertices A, Band C.

I ANSWERS I 1. (7,3). (3, 5)

5. (3,0) 9. 5,5.

2. (1.-2) 6. 4:7,(0,6)

JI6

-2

3. m=5 7. (2. -5)

10. //(-2.0). B(10. -6), C(-4. 2)

4. 5:2

8. (3.2)

360

•


AREA OF A TRIANGLE

Theorem: The Area of a MBC with vertices A(x/,y/), B(x 2, Y} and C(Xj ,y) is given by area (MBC)

11[X/(Y2 - Y3)+ x2(Yj - Y/)+ Xj(Yl - Y2)]/

=

Proof: LetA(x),y), B(x2 'Y2) and C(x3' x 3) be the vertices of the given MBC. Draw AL, BM and CN perpendiculars to the x-axis. Then

ML

= (x) -X2) , LN= (x 3 -x) and MN = (x 3 -x2)

Y

sf-----+-----------"'!C

~2,y~ :

~~y~

I

x'------~----~----~----------~-------x M L N

Y'

/ig.8.13

Area oft). ABC = area (trap. BMLA + area (trap. ALNC) - area (trap. BMNC)

=

[~(AL+BM)XML ] + [~(AL+CN)XLN ]- [~(BM+CN)XMN J.

=

2 (y) + Y2)(X\ -X2) + 2 (y) + Y3)(X3 -XI) -2 (Y2 + Y3)(x 3 -x2)

=

2I [XI (y\ +Y2- Y \-Y3)+ X2 (Y2 + Y3 -Y\ -Y2) + X3 (y\ +Y3 -Y2- Y 3)]

=

2 [X$Y2-Y3)+X2 (Y3-Y$+X 3 (Y\-Y2)]

I

I

I

I

Since, the area is never negative, we have area

1 (MBC) = -Ix\ (Y2 - Y3)+X2(Y3 - yl)+X3(Y\ - Y2)1· 2

Remarks • The area of a triangle is always taken as positive real quantity. Some times the result from the area formula given negative value in that case we reject the negative sign . • The three points (xI'Yt), (x 2'Y2) and (x3 'Y3) are collinear i.e., in a line ifx t (Y2 - Y3) + x 2 (Y3 - Yt) + x3 (y\ - Y2) = 0 because in this situation the area of the triangle reduces to zero.


361

I

~~~~~~~I SOLVED EXAMPLES ~~~~~~~ Example 1: Find the area ofthe triangle whose vertices are (4,3), (5,4) and (1 1,2). Solution: A (4, 3), B (5,4) and C (11, 2) are the three vertices of the given triangle. xI =4,x2 =5,x3 =11 YI = 3'Y2=4'Y3=2

Area of

t1ABC =

21 [XI (Y2 - Y3) + x 2 (Y3 - Y2) + x3 (Yl- Y2) 1

= - [4(4-2)+5(2-3)+11(3-4)] 2 1

= 2[8-5-11]=-4 Rejecting negative sign, we have the area of the given triangle equal to 4 square units. Example 2: Find the value of k so that the point A (- 2 , 3) , B (3, - I) and C (5, k) be collinear.

Solution: We have

=-2,x=3,x3 =5 YI =3'Y2=-I'Y3=k. Xl

1 Area of t1ABC = 2 [Xl (Y2 - Y) + X2 (Y3 -Yl) + X3 (Yl- Y2)] =

21 [-2(-I-k)+3(k-3)+5(3+1)] 1

1

2

2

= - [-2+2k+3k-9+20= - [5k+9] Now, the three points are collinear if the area of flABC = 0 1 -(5k+9) =0 2 5k+9 = 0

i.e.,

k =-9/5.

Example 3: Find the area of the quadrilateral ABCD whose vertices are respectively A (J,J), B (7, - 3), C (12,2) and D (7,21). Solution: Area of quadrilateral ABCD = IArea of (flABC)1 + IArea of flACDI Now

1

1

Areaoft1ABC = 211x(-3-2)+7(2-1)+12x(l+3)1 = 21-5+7+481

= 25 sq units 1 Areaoft1ACD = 211x(2-21)+12(21-1)+7(l-2)1 = .!.1-19+240-71 = 107 sq. units. 2 :. AreaofquadrilateralABCD=25 + 107 = 132 sq. units.

362


Example 4: For what value o/k the points (k, 2 - 2k) (-k + 1 ,2k) and (- 4 - k, 6 - k) are collinear.

Solution: Let the three points be A(x l , YI)

!II

C(x3'Y3). (-4 -k, 6- 2 k).

(k, 2 - 2 k), B(x2 'Y2)

!E ( -

k + I. 2 k) and

If the given points are collinear, then xI (v2 - Y3) + x2 (v3 - YI) +x3 (vI - Y2) = 0

=> k(2k-6+2k)+(-k+ 1)(6-2k-2+2k)+(-4-k)(2-2k-2k) = O. => kC4k-6)-4Ck-I)+C4+k)(4k-2) = 0 => 4~-6k-4k+4+4~+ 14k-S = 0 => S~+4k-4 = 0 => 2~ + k-l = 0 => (2k-l)(k+ I) = 0 k = 1/2ork=-1. => Hence, the given points are collinear for k = 112 or k = - I. Example 5: If the vertices 0/ a triangle have integral coordinates prove that the triangle cannot be equilateral. Solution: LetA (xl'YI)' B (x2 'Y2) and C(x3 'Y3) be the vertices of triangle ABC, then the area of ~ ABC is given by ~ =

'21 [xI (v2-Y3)+x2 CY3-YI)+x3 (vI-Y2)]

= A rational number IfpossibJe let the triangle ABC be an equilateral triangle, then its area is given by

~= =

Jj Cside)2 = Jj CAB)2 4 4 Jj ""4

x

.. num ber a positive

= an irrational number This is a contradiction to the fact that the area is a rational number. Hence. the triangle cannot be equilateral. Example 6: Prove that the/ollowing points are collinear,' (-3, 0) ,(0, -9) and (-2, -3) [RGPV B. Pharma 20041

Solution: The given points will be collinear if the area of the triangle formed by these points is zero. Now area of the triangle 1

= 2'[x 1 (v2-Y3)+x2 (v3-YI)+x 3 (vI-Y2)]

= .!.[-3C-9+3)+O(3-0)+C-2)(O-C-a)] 2

=

.!. [IS - IS] = 0 2.._ ~_-


363

Example 7: The co-ordinates a/vertices Band C a/triangle are (/, - 2) , (2, 4) lies on the line 2x + Y- 2 = O. The area a/the triangle is 8 units. Thenfind the vertices coordinates 0/A. [RGPV B. Pharma 2001 I Solution: Given points are B( 1,2), C(2, 3) and A(x ,y) line on the line 2x +y - 2 = 0 The coordinate of A are (x, 2 - 2x)

Area of A ABC

1

= '2 [xI (Y2 - Y3) + x2(Y3 - YI) + x3 (YI - Y2)

1 :1:8 = '2[x(-2-3)+ 1(2-(2-2x)+2(2-2x+2)] :1:16 =[-5x+I+2x+8-4x] :1:16 =-7x+9

16-9 = -7x (taking positive sign) 7 = -7x. x = - 1 put x = - 1 iny = 2 - 2x and gety = 2 + 2 = 4 . .'. coordinate of A are (- 1, 4) . Taking negative sign - 16 = - 7x + 9

=>

-16-9 = -7x=>x = 2:

put

x = 25/9iny=2-2x

we get

25 18-50 32 y = 2-2 x - = - - = -

:. coordinate of A are

e: '3:)

9

9

9

or (-1,4)

Example 8: Find the area o/the triangle whose vertices are (0, 5) ,(2, 3) and (4, 5) IUPTU B. Pharma20021

Solution: Here xI =O'YI = 5,x2 =2 'Y2 = 3,x3=4'Y3 = 5. :. area of the triangle =

~[(XtY2+X2Y3+X3 Yt)-(YtX2+Y2x3+Y3Xt)]·

= 1. [(0 x 3 + 2 x 5 + 4 x 5) - (- 5 x 2 + 3 x 4 + 5 x 0)] 2

=

1.[(0+ 10+20)-(10+ 12 +0)] 2

=

1. [30-22] = 1. x 8 =4 sq. units 2

2

Example 9: Prove that the points (a,b + c) , (b, c + a), (c, a + b) are collinear. [UPTU B. Pharma 2001, 071

364


Solution: Here x\ = a, y\ = b + e, Y2 = e + a, x3 = e, Y3 = a + b, x 2 = b Now area of the triangle formed by the given points: 1

2" [(xI Y2 + x 2Y 3 + x3 Y\) - (y\ x 2 + Y2 x3 + YJ xI)] 1

= -[{a(e + a) + b (a + b)+e (b + c)} - {(b + c) b+ (e+a) e +(a + b)a}] 2 = .!..[ae + ~+ab + b 2 +be + e2-b2-be+e2-ae-~-ab] =0. 2 Hence, the given points are collinear. Example 10: Find the area ofa triangle formed by the lines: [UPTU B. Pharma 20011 Y = 2x, y = x and y = 3x + 4. Solution: Let the equations of the sides AB, BC and CA of ~ ABC be y - x = 0, y - 2x = 0 and y - 3x - 4 = 0 respectively. Solving these equations in pairs, the coordinates of A, Band Care (-2, -2), (0, 0) and (-4, -8) respectively.

Area of ~ABC

1

= 2"[x\ (Y2- Y 3) + x 2 (Y3 -y\)+x3 (Y\-Y2)]' =

2"1 [(-2)(0+8)+0.(-8+2)+(-4) (-2-0)] 1

= -[-16+8] 2

= .!.. [- 8] = 4 sq. units, neglecting the negative sign.

2 Example 11: Four points A (6, 3), B(- 3, 5), C(4, - 2) and D(x, 3x) are given is such a way I'lDBC 1 I'l ABC = 2"' find x. [UPTU B. Pharma 20061

Solution:

Area of

.!..[x(5 + 2) -3(-2 - 3x) +4 (3x- 5)] 2

~DBC

1

Area of I'lABC

2[6(5 + 2) -3(-2 -3) +4(3 - 5)] 7x+6+9x+12x-20 42+15-8 28x-14 49

1

2

2

4x-2 ---2 7 7 2

x

=

4x-2 11 8

=-


~~~~~~~I

EXERCISE

365

8.3~1~~~~~~~

1. Find the area otthe triangle whose vertices are A (2, 7) B (3, - 1) and C(-5, 6)

2. Find the value of k for which the area formed by the triangle with vertices A (k, 2k), B (- 2,6) C (3, 1) is 5 square units. 3. Showthatthe points A (- 1,1), B (5, 7) and C(8, 10) are collinear. 4. For what value of K are the points A (1, 5) , B (k, 1) and C (4, 11) collinear. 5. If the vertices of a triangle are A (1, k) , B (4, -3) and C (- 9, 7) and its area is 15 sq.units, these find the value of K.

ANSWERS 1. 28. 5 sq. units.

2. K=2, K=2/3

4. K=-1

21 5. K=-3 or K=13

III

LOCUS AND EQUATION TO ALOCUS

Locus: The curve described by a point which moves under given condition or conditions is called its locus. For example: (i) Suppose C is a point in the plane of the paper and P is a variable point in the plane of the paper such that its distance from C is always equal to resay). Obviously all the positions of the moving point P lie on the circumference of a circle whose radius is r. The circumference of this circle is therefore the Locus of the point 0 when it moves under the condition that its distance from point C is always equal to constant r. Fig. 8.14

Equation of the locus of a point The equation of the locus of a point is the relation which is satisfied by the coordinates of every point on the locus of the point.

STEP KNOWLEDGE Step 1: Assume the coordinates of the point say (n, k) whose locus is to be found. Step 2: Write the given condition in mathematical form involving n, k Step 3: Eliminate the variables, if any Step 4: Replace h by x and k by y in the result obtained in step 3. The equation so obtained is the locus of the point which moves under some stated conditions.

366


~~~~~~I

SOLVED EXAMPLES

~I~~~~~~

Example 1: Find the locus of a points P such that the sum of the squares of abscissa and ordinate is equal to the product of abscissa and ordinate. Solution: Let P(h, k) be any point on the locus. :. h is abscissa and k is ordinate of P.

By the given geometrical condition, we get

rt-+JC2= hk Hence, locus of(h, k) is x2 + I = xy Example 2: Find the equation to the locus ofapointequidistantfrom the points A (1, 3) and B (-2,1). Solution: Let P (h, k) be any point on the locus .Then PA = PB (given) => PA 2 = PB2 => (h-I)2+(k-3f= (h+2f+(k-If => 6h+4k = 5 Hence, locus (h, k) is 6x + 4y - 5. Example 3: Find the equation of the locus of a point which moves so that the sum of its distances from (3, 0) and (-3, 0) is less than 9. Solution: Let P(h, k) be the moving point such that the sum of its distance from (3,0) and (- 3,0) is less than 9. PA+PB< 9. Then

~(h-3)2+(k-0)2 +~(h+3)2 +(k--;O)2 => =>

~(h_3)2+k2 < 9-~(h+3)2 +k2 (h- 3) + JC2 < {9 (h-3f+JC2 < SI

~(h-3)2 +k2

r

+(h+3)2+JC2-1S~(h+3i+k2

=>

-12h-Sl < -IS

=> => => =>

4h +27> 6

~(h+3i +k2

~(h+3)2 +k2

(4h+27f > 36 [(h+3)2+JC2] 2 16h +216h+729 > 36(h2 + 6h+9+JC2) 20h2 + 36JC2 < 405 Hence, locus of(h, k) is 20x2 + 361 < 405. Example 4: A point moves so that the sum of its distances from (ae, 0) and (- ae, 0) is 2a, prove that the equation to its locus is 2

i

x 2+2 a b

=

1 whereb 2 =~ (J -~).

System of Coordillates

367

Solution: Let P (h, k) be the moving such that the sum of its distance from A(ae, 0) and B(-ae. 0) is 2a. Th~

M+n=~

=>

~(h-ae)2 +(k-O)2 +~(h+ae)2 +(k -0)2

=>

~(h-ae)2 +k2 (h- ae) + ~ =

=>

=

2a

= 2a-

~(h+ae)2 +k2

4if + (h + ac)2 + ~-4a ~(h+ae)2 +k2 (squaring both side)

-4aeh-4if = (eh + a) =

-4a~(h+ae)2 +k2

~(h+ae)2 +k2

(eh + a)2 = (h + aei + ~ e 2h2 + if + 2aeh = h 2 + ife2 + 2aeh + ~ h2 (1- e2) + ~ = if (l-e2)

h2 a

-+ 2

k2 2

a (1-i)

= 1

Hence, locus of(h, k) is

h2 a

-+ 2 or

k2 2

a (1-i)

= 1

x2 y2 -+= 1 whereb 2 =a2 (I-e2) ' ~ b2

Example 5: A rod of length I slides with its ends on two perpendicular lines find the locus of its mid - point. Solution: Let the two perpendicular lines be the coordinate Axes. Let AB be a rod of length I. Let the coordinate ofA and B be (a, 0) and (0, b) respectively. As the rod slides, the values of a and b change. So a and b are two variables. y

8(0, b)

--~O+-------------~A----~X

(a. 0)

Fig. 8.15

368

Remedial Mathematics Let P(h, k) be the mid point of the rod AB in one of the infinite position it attains. Then

a+O O+b and k = - 2 2 From Ll OAB , we have AB2 = OA 2 + OB2 =if+ b2 =P (2h)2+ (2ki =P (from (1)) 4h2+4~ =p ~

h

= --

Hence, the locus of (h, k) is 4x2 +

~

h=

a b and k = 2 2

-

... (1)

4; = P.

Example 6: If 0 is the origin and Q is a variable point on ~ = 4y. Find the locus ofthe mid pointofOQ. Solution: Let the coordinates of Q be (a, b) and let P(h, k) be the mid point of OQ. Then a+o a o+b b ... (1) h = - - = - andk= - - = - ~a=2handb=2k. 2 2 2 2 Here a and b are two variables which are to be eliminated. Since (a, b) lies on x 2 = 4y. Therefore, a 2 = 4b ~ (2h)2 = 4 (2k) ~ h 2 = 2k [using (1)] ~ h 2 = 2k. Hence, the locus of(h,k) is x 2 = 2y. Example 7: A point moves so that its distance from (3, 0) is twice the distance from (-3, 0). Find the equation of the locus. [UPTU B. Pharma 20051 Solution: Let A represent the point (3,0) , B the point (-3,0). Further, point ( -3,0) and P(h, k) be the moving point. According to the question: PA = 2PB (PAP = 4 (PBi or [(h - 3)2 + (k- 0)2] = 4[(h + 3i + (k- of] h 2 + 9 - 6 h + ~ = 4h2 + 36 + 24h + 4~ ~ 2 ~ 3h + 3~+30h+29 = 0 Hence, the required locus is 3x2 + 3; + 30x + 27 = O. Example 8: Find the locus of a point such that the line segments having end points (2,0) , and (- 2, 0) subtend a right angle at that point. [UPTU B. Pharma 2006] Solution: Let A(2, 0) and B(- 2, 0) be the given points and P(h, k) be the variable point. According to the question L.APB = 90° P(h, k) :. i.e., MPB is a right angle. AB2 = PA 2 + PB2 [2 -(-2)f + [0-0] = [(2 -hi + (0 _k)2] + [(-2 -hi + (0 _k)2] 16 = (2-hi+~+(-2-hi+~ 16 = 4 + h 2 - 4h + 2~ + 4 + h2 + 4h 16 = 2h2+2~+8 A(2.0)

h2+~ = 4.

Hence, the required locus is ~ + ; = 4.

Fig. 8.16


369

Example 9: Find the equation to the locus of a point which moves so that the sum of its distance from (3, 0) and (- 3,0) is less then 9. IUPTU B. Pharma 2003, 041 Solution: LetA(3, 0) andB(- 3,0) be the two given points and (h, k) be the coordinates of the moving point P whose locus is to be found. According to the question PA+PB line AB is perpendicular to line A C. :. ABC is a triangle right angled at A.

(~l) =-1.

378


Example 5: Find k,

if the points (- 1, 3), (8, k) and (2, 1) are collinear.

Solution: The points (- 1,3), (8, k)s and (2, 1) are collinear then slope of AB= slope ofAC.

k-3

k-3

Slope of AB = - - = - 8-(-1) 9 1-3 Slope of AC = - - 2-(-1)

and

k-3 -9-

-2 3

-2

= -3 ~3(k-3)=-18

3k-9=-I8 k =-3 Example 6: Find angle made by the lines x cos 30 0 + Y sin 30 0 + sin 120 0 = 0

with the positive direction of x - axis.


Solution: The equation of the given line is x cos 30 0 + y sin 30 0 + sin 120 0 = 0

y=

...(1)

-cos300x -sin 1200x sin30°

sin 30°

o .J3 12 y=-cot30 x - - 112

~ Y = tan 120 0 x -

.J3 ,

which is the slope intercept form. Hence, the angle made by the given line with the positive direction of x - axis is 120 0 • Example 7: Reduce 4x -r 3y - 9 = 0 to the Normal form andfind the distance (perpendicular [UPTU B. Pharma 2004) distance p) from origin. Solution: We have 4x + 3y - 9 = 0 or 4x + 3y = 9 Dividing both sides by ~(4)2 +(3)2 = 5, we get

i x +~ y 5

5

=

~, which is the normal form 5

~

A

________-::-c

~--------......

B

Fig. 9.3

Hence, the length of perpendicular from the origin to the line is p

=

~. 5

Example 8: Prove that the points (- 1, 0), (3, 1), (2, 2) and (- 2, 1) are the vertices of a parallelogram.

~he

Straight Lines

379

Solution: LetA(- 1,0), B(3, 1), C(2, 2) and D(- 2,1) be the vertices ofthe parallelogram, ABCD taken in order. The mid points of diagonals AC and BD are

(.!.2' 1)

AC= (-1+2 0+2) = 2 '2

and

BD

= (3-2

~) = (.!.

2 ' 2

2'

D

C

1).

Since, the mid points of AC and BD are same. Also, slope of AB x slope of AD #= - 1 A and slope of AC x slope of BD #= - 1 Hence, ABCD is a parallelogram.

Fig. 9.4

1. Find the slope of a line whose inclination to the positive direction of x - axis in antic10ckwise sense is (i) 60° (ii) 0° (iv) 120°. (ii) 150° 2. Find the slope of the line passing through (2,3) and (1,4) 3. Show that the points (1, 3), (2, 5) and (4,9) are collinear. 4. LetA(6, 4) and B(2, 12) be two given points. Find the slope ofa line perpendicular to 5. 6. 7. 8. 9.

10.

AB. Determine x so that 2 is the slope of the line through (2, 5) and (x ,3) Without using Pythagoras theorem, show that the points (1, 2), (4, 5) and (6, 3) represent the vertices of a right angle triangle. Show that the points P(- 4, -5), Q(- 2,2), R(5, 4) and S(3, - 3) are the vertices ofa rhombus. Show thatthe following points represent a rectangle (0, 0), (0, 5), (6 ,5), (6, 0). Show that the following points represent a square (3, 2), (0,5), (-3, 2), (0 -I). Prove that the lines. (i) x + 3y + 4 = 0 and 2x + 6y-7 = 0 are parallel. (li) 2x + 3y + 3 = 0 and 3x - 2y + 5 = 0 are perpendicular.

HINTS TO THE SELECTED PROBLEMS 6. LetA(I, 2), B(4, 5) and C(6, 3) be the vertices of the given triangle.

5-2 3 SlopeofAB= 4-1 ="3 = 1 =m, (say). 3-5 -2 Slope of BC= 6-4 =""2 =-1 = m2 (say). 3-2 1 SlopeofAC= 6-1 ="5 =m 2 (say).

380


1x-I = - 1 => AB 1.. Be. Hence, MBC is right angled.

mIx m 2 =

7 7. SlopeofPQ=-, Slope of RS =

2 7 2

2 7 PQ II RS and QR II PS. So PQRS is a parallelogram. Slope of QR =

2 7

-, Slope ofPS = -

4+5 -3-2 5+4 3-(-2)

Also (slope of PR) x (slope ofSQ) = - - x - - -

9 -5

= - x - =-1 9 5 :. The diagonals PR and QS are perpendicular .'. The Parallelogram PQRS is rhombus.

ANSWERS i

I I

1. (i)

2. 7

•

J3

(ii) 0

4.

(iii)

1

-J3

(iv)

-J3

5. x= 1

2

EQUATION OF LINES IS STANDARD FORM

Slope or tangent form: Find the equation of a line whose y-intercept 'c' and slope 'm' are given.

y

x

Fig.9.S Letthe given line meet y-axis inA and letP(x, 4) be any point on it. As they-intercept of the line is e. :. coordinates of A are (0, C).

The Straight Lines

381

Draw PB 1.. to x-axis and A C 1.. PB.

PB-BC DB BP-OA tan e = - - OB y-c m=--

Then tan

=>

e = -PC AC

= ---

x

=>

y=mx+c

Which is called the slope intercept form of the equation ofa straight line.

Remarks • If c becomes zero, the equationy = mx + c reduces to y = mx which is the equation of a line through the origin. • If m = 0, c:;t: 0, then equation y = mx + c reduces to y parallel to x - axis at a distance c from it. • If m

= c which is the equation of a line

°

= 0, c = 0, then the equation becomes y = which represents the x - axis.

Point Slope Form To find the equation ofa line passing through the given point (x l' y 1) and having slope m: y (x, y) p

k....L.;;.----iR

M

x

Fig. 9.6

Let the given point (xt,Yt) be represented by Q. Let P(x,y) be any point on the line. Draw PM and QL perpendiculars to x - axis from points P and Q and QR 1.. MP. Then and Then

PR =MP-MR =MP-QL=y-Yt QR =LM=OM-OL=x-x t tane = PR = y- Yt

QR

X-Xt

m= Y-Yt x-xI

Y-Yt =m(x,-x t )· Which is the equation of the line in the point slope form.

382


1\vo point form: To find the equation of the straight line passing through two given points: y

--~~~~-------L------~----~--

__

x

Fig. 9.7

Let the two given points be Q(x l' Y 1) and R(x2, Y2)· Let P (x,y) be any point on the line. Draw RL, QM and PN perpendiculars to x - axis from points R, Q and P respectively. Let RS ..L QM anQ QT ..L PN. Then

RS = LM= ON-OL=x 1 -x2 QS =MQ-MS=MQ-RL =YI-Y2 QT=MN=ON-OM=x-x l PT=NP-NT=NP-MQ =Y-YI In MQS, tan9 = QS = YI - Y2 RS xl-x2

...(1)

= PT = Y - YI

... (2)

In flQTP, tan9

QT

x-xI

from (1) and (2), we get

YI- Y2 = Y- Yl XI -X2 x-xI X-XI = Y-YI X-X2 YI-Y2 x-xI

= Y-YI X2 -XI Y2 -YI Y _ YI = Y2 - YI (x-xI). x2 -XI Which is the required equation of line in two point form.

The Straight Lines

383

Intercept form: To find the equation of the line which cuts off intercepts a and b on x-axis and y-axis respectively. y Let the line meet x-axis at point A and y-axis is at point B. As the respective intercepts are a and b. So OA = a and OB = b. Coordinates of A and B are (a, 0) and (0, b) respectively. b Using two point form, the equation ofline is x-a y-o =-o-a b-o

-x +1 a

o

a

x

= ~

Fig.9.S

b

~+~ = 1. a b Which is the equation of the line in the intercept form.

Normal or Perpendicular form: To find the equation ofa line in terms ofthe perpendicular segment p, from the origin to the lines and the angle a which the perpendicular segment makes with the x-axis. Let I be the given line meeting x-axis and y-axis at the points A and B respectively. Let OC ..LlandLAOC= aOC=p. Y OA

Now

OC

= sec a

OA =seca p OA =pseca OB = cosec a OC

Again

DB

= cosec a

----~~--------~-----.x

Fig. 9.9

P OB =p cosec a

Using the intercept form of the equation of the line, the equation of the given line is

x

y

OA

OB

- + - =1. __x_ + __y'--__ p seca

=

1.

p coseca

or xcosa+ysina=p. which is the required equation of the line

384


Parametric form: To find the equation of a straight line in the parametric form: x - x] = Y - y] = r, where r is the parameter. cosS sine Let the given line passes through the point A(xl' y]) and be inclined at an angle e with the positive direction of x-axis. y

P (x, Y)

~~----------~N

--~L-~~--------~--------------~--_x

B

c

Fig. 9.10

Let P(x, y) be any point and AP = r. Draw AB and PC perpendiculars to x-axis from A and P respectively and AN..l Pc. Now AN=BC=OC -OB=x-x] PN = PC - CN = PC - AB = Y - y] Also AP =r. In right angle triangle t.ANP. cose = AN = x-x]. AP r i.e.,

and

x-x] =r. cose

...(1)

sine = PN = Y - y] AP r y- y] =r. sine

...(2)

y:-

From (1) and (2), we get x-x] = Y] =r. cose sme Which is the equation of the line in the parametric form.

Remarks • Parametric form of equation of a line is also known as symmetrical form of equation • from x-x] = Y- Y] =r we have cosS sine ' x =x] + rcose ,Y = y] + rsine, thus the coordinates of any point at a distance r from (xl'y])r are (x] +rcose,y] +rsine)

The Straight Lines

~~~~~~I

SOLVED EXAMPLES

385

I

Example 1: Find the equation ofa line which cuts offan intercept - 2 on the axis ofy cmd makes an angle of 45° with the positive direction ofx-axis. Solution: Here, c=.,..2, andm =tan45° =L. Substituting these values iny= mx + c, we gety =x-2 which is required equation of the line. Example 2: Find the equation ofa line through (4,3) with slope 2. Solution: Equation of line passing through (xI'Y\) and with slope m is y-y\ =m(x-x\) :. The required equation of the line is y-3 =2(x-4) ::::> 2x-y-5 =0. Example 3: Find the equation ofa line which passes through the point (- 2, 3) and makes an angle of 30° with the positive direction ofx-axis Solution: Here 0 = 30,° ::::>

1

m = tanO = tan 30° = .J3'

The point on the line is (- 2,3) Using point slope form, the equation ofline is 1

y-3 = .J3 (x+2) ::::>

.J3y - 3.J3 =x+2

::::> x-.J3y +(3.J3 +2) =0.

Example 4: Find the ratio in which the line segment joining the points (2, 3) and (4, 5) is divided by the line joining the points (6, 8) and (- 3, 2). Solution: The equation of the line joining the points (6, 8) and (- 3, - 2) is

y-8

x-6

- - = - - (Two point form). -2-8 -3-6

y-8 = x-6 -10 -9 9y-72 = IOx-60 or IOx-9y+ 12 =0 Let this line divide the join of (2, 3) and (4, 5) at the point P in the ratio of k: I.

::::>

. Then the coordmates of Pare (4k+2 - - , 5k+3) --. k+l k+1 N ow, the point P on the line (I) 4k+2 5k+3 Therefore, 10 x - - - 9 x - - + 12 =0 k+l k+1 40k+20-45k-27+12k+ 12 =0

...(1)

386


7k =-5 k =-517. Since, the value of k is negative, the line is divided externally. Hence, the required ratio is 5 : 7 externally.

Example 5: Find the equation ofthe line which passes through the point (3, 4) and the sum of intercept on the axes is 14. IUPTU B. Pharma 2008] Solution: Let the intercept made by the line on x-axis be a. Then intercept on y-axis = 14 - a.

its

:. Equation of the line is given by

~+-y- =1 a 14-a As the point (3, 4) lies on it, we have

...(1)

~+_4_ =1 a 14-a 3 (14-a)+4a => 42-3a+4a => c?-13a+42 => (a-7)(a-6) a Putting these values of a in (1),

14a-c? = 14a-c? =0 =0 =

=7,6. we get equation of the lines

~ +.[ = 1 or x + y = 7 7 7 and

~+.[ 6 8

= 1 or4x+ 3y= 24

Example 6: A line is such that its segment between the axes is bisected at the point (xl' YI) Prove that the equation of line is

~+L =1 2xI 2Yl Solution: Let [be the given line which meetsx-axis atA andy-axis atB. Then segmentAB is bisected at the point P (xl' Y\). y

Letthe equation of the line be ~ +.[ = 1 a b

... (1)

:. the coordinates of A and B are (a, 0) and (0, b) respectively. As P is the mid point ofAB. Its coordinate, therefore are given by

(~,%).

We are given that coordinates of Pare (xI'Y\)

a b - =x and - =y

2

I

2

--~~--------~~-----x

o

\

I

Fig. 9.11

The Straight Lines

387

a b =x and - =y 2 I 2 I or a =2x1 andb=2YI :. substituting the value of a and b in (I), equation of the line I becomes ..

~+L =1. 2xI

2YI

Example 7: Find the equation of the straight line passing through the points (- 3, 4) and (1, - 3). [RGPVB.Pharma2004] Solution: Equation Ollhe straight line passing through (xl'YI) and (x2'Y2) is given by

Y-YI = Y2-YI(x_XI) x2 -XI

...(1)

xI =-3'YI =4 x 2 =I'Y2=-3 Putting these values in equation (1) we have

-3-4 --(x+3) 1+3 4(y -4) =-7x-21 4y -16 =-7x-21 7x +4y+5 =0. Y -4

=>

=

Example 8. Find the equation of a line which passes through (2, 3) and whose m is 3. [UPTU B. Pharma 2005] Solution: Equation of line passing through one point is (y -YI) =m(x-x]) ...(1)

m =3,x l =2'YI=3 Putting the values in (1) , the equation of required line is given by Y -3 =3 (x-2)

=> =>

Y -3 =3x-6 3x -y-3 =0.

Example 9. Find the equation ofa line passing through the point (3, - 2) and perpendicular to the line X - 3y + 5 = 0 [UPTU B. Pharma 2006) Solution: Slope of the given line x- 3y+ 5 = 0 is 1

ml =

3"

As the line is perpendicular to line passing through (3,-2). m] xm 2 =-1.

1 -Xm =-1 3 2 => m2 =-3. and required equation is (y + 2) = -3(x - 3) => y+2 =-3x+9

=>

=>

~+y-7=~

388


Example 10. Find the equation of perpendicular bisector of the line segment joining the points A(2, 3) and B(6, - 5) . Solution:

-5-3 -8 Slope = - - = - =-2. 6-2 4

:. Slope of a line perpendicular to the line AB =

e

..!.. 2

The coordinates of the middle point M of AB are

;6, 3 +~-5)) i.e., (4, -1).

Hence, the equation of the perpendicular bisector of AB, i.e., the equation of the line passing through m and perpendicular to AB is 1 y + 1 = -(x-4) 2 => x-2y =6. Example 11. Find the equation of the straight line which passes through 0,2) and is [UPTU B. Pharma 2001) perpendicular to the line 4x - 3y = 8 Solution: The equation of any straight line perpendicular to the line 4x - 3y - 8 = 0 is 3x+4y+A. = 0 ...(1)

If the line (1) passes through the point (1,2) then 3+8+1.=0 1.=-11. Putting A. = -11 in (1) the required equation of the line is 3x+4y-Il =0 Example 12: Find the equation of the straight line passing through the point (a cos 3 e, a sin 3 B) and perpendicular to the line x sec e + y cosec e = a cos2 e.

IUPTU B. Pharma 20071 Solution: The slope of the given line x sec S + Y cosec S = a is -secS. cosecS

-sin S cosS

---I.e.,--

:. the slope of a line perpendicular to the given line =

c~s S

smS Now, the equation of the straight line which passes through the point (a cos 3S ,a sin 3S) . cosS . an d w hose s1ope IS - - IS sinS . 3 cosS 3 Y - a sm S = - . - (x - a cos S) smS => x COS S - y sin e = a (cos4 S - sin4 S) or x cos S - y sin S = a (cos2 S + sin2 S) (cos 2 S - sin2 S) Hence, x cos S - y sin S = a cos2 S. Example 13: Find the equation of the straight line which makes equal intercepts on the IUPTU B. Pharma 20021 axes and passes through the point (3, -5).

The Straight Lines

389

Solution: Let the equation of the straight line be

-=-+E. =1

... (1)

a

b The line (1) makes equal intercepts on the axes, i.e., a = b.

-=- + E.

= 1 or x + y = a a a If this line passes through the point (3, - 5) , then

3- 5

= a or a = - 2

Hence, the required equation is x + Y = - 2 or x+y+2 =0. Example 14: Find the equation ofthe straight line, the portion ofwhich intercepted between the axes is divided by the point (- 2, 6) in the ratio 3 : 2 [UPTU B. Pharma 2007] Solution: Let the equation of the straight line be

-=-+E. =1

a b The line (1) meet x-axis atthe point A (a, 0) andy-axis at the pointB(O,b). Then the point (- 2, 6) divides the line AB is the ratio 3 : 2 By section formula, we have _

=

( 2, 6) -

=>

_2

(2a+3XO 2XO+3Xb) 2 +3 ' 2 +3

= 29

2

and 6 = 3b 5

or a=-5b=10. Putting the value of a and b jn (1) , the required equation of the line is

x y - + - =1 -5 10 y-2x

or

=

10.

Example 15: A straight line, drawn through the point A(2, 1) makes an angle'::' with 4

positive x - axis and intersects another line x + 2y + 1 = 0 at point B. Find the length AB. [UPTU B. Pharma 2003] Solution: The equation of any line passing through the given point A(2, 1) and making an angle .::. with x-axis is 4

~ = ....c..!... = r (say)

...(1) sin 45° \\ cos 45° Where Il represents the distance of any point B on this line from the given point A(2,1) The coordinates (x, y) of any point B on the line (1) are (2+rcos45°, 1 +rsin45°) i.e.,

(2+r.~'1+r.~}

390


Ifthe point B lies on the line x + 2y + 1 = 0, then

(2+r.~)+2(I+r.~)+1

=0

(5+r·1) =Oorr=

-%J2.

5J2

AB = - - -

Hence, the length

3 Example 16: Find the equation of the line passing through the points (4. 3) and (7. 8). IUPTU B. Pharma 2001 ) Solution: The two points are (xI' YI) = (4,3) and (x2,y 2) = (7,8). Using

y - YI

=

Y2 - YI (x - x I) x2 -xI

8-3 y-3 = -(x-4)

7-4

5 3

y-3 = -(x-4) 5x-3y-11 =0. Example 17: Find the slope and the equation of the straight line joining the points (2 -5) and (4. 1). rUPTUB.Pharma2002] Solution: The slope of the line joining the points (2, - 5) and (4, 1) is =

1-(-5) = ~ =3. 4-2 2

Now, the equation of the straight line joining the points (2, - 5) and (4, 1) and whose slope is 3 is y -(-5) =3(x-2)

=>

y+ 5 =3x-6 3x-y =11.

Example 18: Find the equation of the straight line which divides the line joining the point (5. - 2) and (- 5. 8) in the ratio 3 : 4 and is also perpendicular to it. rUPTU B. Pharm 2006] Solution: The equation of the line joining the points (2, 3) and (- 5,8) is

y-3

=

8-3 -5-2 (x-2)ory-3

=

5 -7 (x-1)

- 7y + 21 = 5x - 10 5x+7y =31

...(1)

The slope ofline (1) is -5 and so the slope of the line perpendicular to it will be 7... 7 5 The coordinates (h, k) of the point dividing line (1) in the ratio 3 : 4. are given by

~~

i.e.•

h = 3x(-5)+4x2 andk= 3x8+4x3 3+4 3+4 36 h = 1 andk = - . 7

The Straight Lines

Hence, the equation of the line passing through (h, k) and having slope

391

2. is 5

7

y-k=S(x-h)

36 7 y- - = - (x-(-I))or49x-35y+229=O. 7 5 Example 19: Find the equatiorr ofa line at a distance of3 units from the origin such that the perpendicular from the origin to the line makes an angle tan- i

(~)

with the positive

lUPTU B. Pharma 2006]

direction ofx-axis.

Solution: We have p = 3 and a = tan-1

i

4

c

3 tana=4

3 cos a = -4 an d. sm a = -. 3 4 Hence, the equation of the line in normal form is x cos a + y sin a = p or

4

3

5

5

3 ex BL----4---...J....~

A

Fig. 9.12

xx-+yx-=3

~ 4x+3y=15 Example 20: Find the equation ofthe line which has length ofperpendicular segment from the origin to the line is 4 and the inclination of perpendicular segment with the positive direction ofx-axis is 30°. Solution: The normal form ofthe equation of a line is x cos a + y sin a = p Now, here, p = 4 and a = 30° :. Equation of line is xcos300+ysin30° =4

XX

~

.fj I -+yx - =4 2

2

.fj x+y =8.

Hence, .fj x +Y - 8 = 0 is the required equation of the line. Example 21: Through the point P(3, -5), a line is drawn inclined at 45° with the positive direction ofx - axis. It meets the line x + y - 6 = a at the point Q. Find the length PQ. Solution: The equation of the line through (3, -5) inclined at an angle of 45° (by using parametric form) is

x-3 = y+5 =r. cos 45° sin 45°

392


x-3 = y+5 =r.

-1-

)

J2

J2

.

~

Now, the point Q (

h h+ 3,

5) lies on the line x + y - 6 = O.

~+3 ~-5-6 =0

J2 'J2

~ =8

~

J2

~

r =

8.fi =4J2 2

Hence, required length of PQ = 4

J2 .

1EXERCISE 9.21 1. The x- intercept ofa line is double to itsy- intercept. Ifit passes through (2,3), find its equation. 2. A line makes equal intercept on the coordinate axes and passes through (1,3) find its equation. 3. Find the equation of the line passing through the I?oints (2,3) and (-1, -4). 4. Iflength and inclination of the perpendicular from the origin on the line is 4 and 135° respectively. Find the equation of the line. S. If A(O, 2), B(4, 1), C(1, 3) are the vertices ofa ~ABC, find the equation of 0) side AB (ii) median CF and (iii) attitude on side BC. 6. Find the equation of the line which passes through the point (-3, 8) and the sum of its intercept on the axes is 7. 7. Find the equation of the line through (2, 3) so that the segment of the line intercepted between the axes is bisected at this point. 8. The length of the perpendicular from the origin to a line is 6 and the line makes an angle of 30° with the positive direction ofy - axis_ Find the equation of the line. 9. Find the equation of the line through the point (2, 3) and making an angle of 45° with the x - axis. Also determine the length of intercept on it between A and the line x+y+ 1 =0_ 10. Ifp be the length of the perpendicular drawn from the origin to the line bx + ay = ab

1 1 I show that -+-= 22 2 a b p

The Straight Lines

393

HINTS TO THE SELECTED PROBLEMS 5. Given A = (0, 2), B(4, I) and C(I, 3) (i) Equation of AB is

y-2 x-o x 2-1 = 0-4 =>y-2= -4 =>-4(y-2)=x. =>-4y+ 8 =x=>x+4y= 8. :. x + 4y = 8 is the eqn. of side AB. (ii) Median CF: F = the mid of AB = F=

(0;

1 4 , 2; )

(2,%) and C= (1, 3)

U sing two point form y - YI = x - xI

YI- Y2

xl-x2

y-3/2 x-2 2y-3 x-2 =--=>--=-3/2-3 2-1 3-6 1 2y-3 =-3 (x-2)=>2y-3 =-3x+6 3x + 2y - 9 = 0 is the required equation of median CF.

=> => => (iii) Allude AD 1.. Be.

y

3-1 2 SlopeofBC= - - = --. 1-4 3 3

A(2,3)

Slope of AD = - (': AD 1.. BC) 2 Equation of AD iSY-YI = m (X-XI) 3

--~~--~------x

y-2 ="2(x-O)

3 y-2 =-x 2

Fig. 9.13

2y-4 =3x=>3x-2y+4=0. The equation of a line through A and making an angle of 45° with the x - axis is x-2 y-3 x-2 y-3 cos45° = sin45°

=> -1- = -1-

Ji

=>

Ji

x-2 =y-3 =>x-y+/l =0. Suppose the line meets the line x + y + 1 = 0 at P such that AP = r. Then the coordinates of P are given by x-2 y-3 - - =--=r.

cos45°

sin 45°

r r x =2+ -Ji,y=3+ Ji

394


Thus the coordinate of P are

Since P lies on x + y + I = 0 2

+~ +3 +~ +I J2 J2

0 => J2r = - 6

=

r =-3J2 length of AP = Irl = Thus the length of intercept =

3J2

3J2

I

ANSWERS

1. x + 2y+4=0

2. x+y +4=0

4. x-y-4J2 =0

5. (i)x+ 4y= 8 (ii) 3x+2y-9= 0 (iii) 3x-2y+4 =0.

6. 4x+3y= 12

7. 3x+2y-12=0

9. x-y+ I = 0,

III

3. 7x-3y+2=0

8. .J3x+y-12=0

3J2

TRANSFORMATION OF GENERAL EQUATION IN DIFFERENT STANDARD FORMS

The general equation of a straight line is ax +by + c = 0 which can be transformed to various standard forms as discussed below:

(1) Transformation of ax + by + c

= 0 in the slope intercept form (y = mx + c):

+by +c =0 => by=-ax-c => y =(-~)x +(-~) This 0 is of the formy =mx +c where m =-!!.. and c =-:.. , b b We have ax

Thus for the straight line ax + by + c = 0 m

coeff of x =slope =--ab =---coeff .of y

. y- mtercept =

and

c b

--=-

constant term coeff .of y

"Remark • To determine the slope of a line by the formula m = coeff of x we must transfer all coeff. of y terms in the equation on one side. Transformation of Ax + By + C = 0 intercept form

l;+i=l}

The Straight Lines

395

Ax + By + C =O:::::>Ax+By=-C

We have

Ax

By

- + - =1 -c

x

-c

y

(-~)+(-~)

=1.

This is of the form ~ + 2::. = 1. Thus for the straight line Ax + By + C = O. a b . -C -const. term = ---Intercept on x-axIs = A coeff. of x.

. -C - costant. term Intercept on y-axis = _ = ----B coeff .of y

(2) Transformation of Ax + By + C We have Let

= 0 in normal form (x cos a + y

Ax + By + C =0

sin a

= p): ... (1) ...(2)

x cos a + y sin a - p = 0

be the Normal form of Ax + By + C = O. Then (1) and (2) represent the same straight line. ABC

cosa

- p

sina

cosa = - Ap and sin a = - Bp

C

cos2 a+ sin2 a

C

...(3)

A2p2 B2p2 -+ -2C2 C

=

2

1 = L(A2+B2)

C2

c

P =± - ; = = =

~A2 +B2

But, p denotes the length ofthe perpendicular from the origin to the line and is always positive.

Putting the value of p in (3) we get

-A

cosa

.

-B

= -.j?A=2=+=B=2= , sma = ~ A2 + B2

So, the equation (2) takes the form

-A ~A2+B2

-,====x -A

-,=o=:===~x

~ A2 + B2

-

B

~ A2 + B2

Y-

C

~ A2 + B2

x

This is the required normal form of the lineAx +By + C = o.

396


~~~~~~I

SOLVED EXAMPLES

~I~~~~~~

Example 1. Reduce 3x - 4y + 5 = 0 to slope/orm andfind its intercept on y-axis. Solution: The given equation 3x - 4y + 5 = 0 can be written as 4y = 3x+5 3 5

Y = '4x+

=>

'4'

. 5 Intercept on y-axIs = '4' Example 2. Reduce the lines 3x - 4y + 4 = 0 and 4x - 3y + 12 hence determine which line is nearer to the origin. Solution: We have 3x - 4y + 4 = 0 => -3x+4y =4

= 0 to the normal/orm and

This is the normal form of 3x - 4y + 4 = 0 and the length of the perpendicular from the origin to it is given by. PI

Now

=>

-4x

~(-4)+32

4 5

4x-3y+ 12 =0 -4x+3y =12 3y

+-r=======

12

J(-4)2 +(3)2

4 3 12 --x+-y - 5 5 5 This is the normal form of 4x - 3y + 12 = 0 and the length of the perpendicular from origin

=>

.. . b 12 to It IS given y P2 = S·

Clearly P2 > PI therefore, line 3x - 4y + 4 = 0 is nearer to the origin. Example 3: Reduce 3x + 5y + 4 = 0 to the intercept/orm andfind the y-intercept. Solution: 3x + 5y + 4 = 0 3x + 5y =-4 3x 5y -4 or -+- =-4 -4 -4

x y (-4/3) + (-4/5) = 1 Which is the required intercept form: . . 4 Hence y-mtercept IS - - . 5

The Straight Lines

•

397

POINT OF INTERSECTION OF TWO LINES

Let the two lines be

Alx+Bly+C I =0 A 2x +B21 + C2 = 0 Let (Xl' Yl) be the point of intersection of these two lines:

... (1)

Then Alx l + BIYI + C I =0

... (3)

and AzXl +B211 + C2 = 0 From (3) and (4), we have

...(4)

... (2)

YI B C -B C

I 2 2 I = --'--"'-----"'-'-

X

AIB2 -A2Bt

I

Y

=

I

C1A2 -C2 A1 AIB2 -A2BI

Hence, the coordinates of the point of intersection of the two lines (1) and (2) are

B1C 2 -B2C1 , C IA2 -C2A1 ). ( AIB2 - A2Bl AIB2 - A2Bl

Remarks • To find the coordinates of the point of intersection of two non parallel lines, we solve the given equations simultaneously and the values of x and Y so obtained determine the coordinates of the point of intersection. • The coordinates of the point of intersection determined above do not exist if A 1B2 -A 2 B I =0.

i.e., if ~=!i*~ A2 B2 C2 • If

~ = .!!.L = ~, then the lines are coincident. A2

B2

C2

• If there is only one point which satisfied both equation the system of equations is called consistent. In that case

•

~ * .!!.L * ~. A2

B2

C2

CONDITION OF CONCURRENCY OF THREE GIVEN LINES

Let the equation of the three lines be

a1x +bly + c I =0 a 2x + b21 + c2 = 0 a3x + b)Y + c3 = 0

... (1) ... (2)

...(3)

398


For given lines to be concurrent, no two ofthese lines can he parallel or coincident i.e.,

E1.

*

*

a2 a3 ~ ~ ~ and the point of intersection of any two lines must lie on the third line. Now, the point of intersection of (1) and (2) can be obtained as below: x y 1 ~c2 -b2c I cla2 -alc2 al~ -a2~

...(4)

~c2 -~cI x = -'-=---=-..!.. alb2 -a2~

y = cla2 -alc2 alb2 -a2q Now, the point

(~C2 - b2cI

,cla2 - alc2 ) lies on (3) because the lines are concurrent alb2 - a2bl alb2 - a2 bl a3

-b2CI) + b ( a~C2 lb2 a2~

3

(Cl a 2 -alc2 ) +c =0 . 3 a)b2 - a2q

a3(b lc 2 - b2 c l ) + b 3(c la 2 - a lc 2) + c 3(a l b 2- a 2 b l ) =0. a l (b 2c 3 - b3c2) + b l (c 2a3 - a 2c3) + c l (a2b3 - a3b2) =0. ... (5) Thus for the given three lines to be concurrent, the condition (4) and (5) must hold.

~~~~~~~I SOLVED EXAMPLES I Example 1: Find the coordinates ofthe point ofintersection ofthe lines 2x - y + 3 = 0 and x + 2y-4 = O. Solution: Solving simultaneously the equation 2x - y + 3 = 0 and x + 2y-4 = 0, we obtain

_x_ =-L=_I_ 4-6 3+8 4+1

x -2

y 11

1 -2 11 -=>x = y=-. 5 5' 5 Hence, (-2/5,1115) is the required point ofintersection. Example 2. Show that lines x - y - 6 = 0, 4x - 3y - 20 = 0 and 6x + 5y + 8 = 0 are

=>

-= -

=

concurrent. Also, find their common point of intersection. Solution: The given lines are x- y- 6=0 4x-3y-20 =0 6x+5y+& =0 Solving (1) and (2) by cross multiplication, we get __ x_= y '20-1& -24+20

1

-3+4

...(1) ...(2) ...(3)

The Straight Lines

399

x =2,y=-4 Thus, the two lines intersect at the point (2, - 4). Putting x = 2, y = - 4 in (3) , we get

6 x 2+5x x (-4)+8=0 so (2, -4) lies on (3). Hence, the given lines are concurrent and their common point of intersection is (2, - 4).

3: Prove that the lines 2x + 3y - 13 = 0 x + 2y - 8 = 0 and 3x - y - 3 concurrent Solution: Solving the equations 2x+3y-13 =Oandx+2y -8=0 x y 1 We have -24+ 26 -13 + 16 4-3 .

-.E~.ample

=

0 are

or x =2,y=3 The lines will be concurrent if the point (2,3) satisfies the equation of third line. Putting the coordinates (2, 3) in 3x - y - 3 = 0, we have 3 (2)-3 -3 =0

o = 0 , which is true. Hence, the lines are concurrent. Example 4: Find the value of k, so that the lines

x-2y+1 =0 2x - 5y + 3 = 0 and 5x - 4y + k = 0 are concurrent. Solution: The equation of the lines are : x-2y+ 1 =0 2x-5y+3 =0 5x-4y+ k =0 Solving (1) and (2) x =~=--6+5 2-3 -5+4 x -1

... (1) ...(2) ...(3)

=l:'...=..!... -1

-1

x = l,y= 1 :. The point ofintersection of(1) and (2) is (1, 1). This pointwiII lie on (3) if5 -4 + k= 0 or k=- 1. Thus for concurrency of(1) and (2) and (3) ,k=-l. Example 5: Find the equation of the line which is perpendicular to the line 3x-2y+4 = 0 and passes through the point of intersection of the lines x + 2y + 1 = 0 and y = x + 7. [UPTU B. Pharma 2008]

Solution: The eqn. of a line perpendicular to 3x - 2y + 4 = 0 is 2x+3y+1 =0

...(1)

400

Remedial Mathematics Point of intersection of x + 2y + 1 = Uand y = x + 7 is x = -5, Y = 2 is (-5, 2). Line (1) passes through this point so 2(-5)+3(2)+1.. =0 -10+6+1.. =0

A =4 puting, we get

2x+3y+4 =0.

•

ANGLE BETWEEN TWO INTERSECTING LINES

Theorem 1: Prove thatthe angle Q between the line y = mIx + C1 andy = m jX + C2 is given m)-m2

by tan () = ---"--=1+m)m2

Proof: Let I) and 12 be two lines y = m) x + c) and y = m2 x + c2 respectively. Let I) intersect 12 at P making an angle e between them. Let I) and 12 meet x - axis at Rand Qrespectively. Let I) and 12 angle a. and ~ respectively, with the positive direction of x - axis. y

Q

J3

o

X

/1

'1

Fig. 9.14

a. = e+ ~

The exterior angle

e e tan e

tan

or

tan

=a.-~ =tan(a.-~)

e=

tan a. - tan~ 1 - tan a. tan ~ m) -m2 1+m)m2

e = tan-)

m) - m2 1+m)m2

The Straight Lines

401

Remark • The value oftan e can be bot4 positive and negative because between two lines there we two distinct angles. If thIs value is +ve, then the angle between the lines is acute and if it is -ve the angle is obtuse. Theorem 2: Prove that the angle Qbetween thelinesalx +bJy + c J = Oandar + b§ + c2 = 0 alb2 is given by tan () = la2bl 1· ala2 + bl b2

r + by; + c2 =0. Then.

Proof: Letm l andm2 be the slopes of the linesalx+bly+c) =Oanda ml

Now

tane

=

-;1

andm

2=

-b:2

=II~:)~J

tan e

tan e = la2b) - alb21 ala2 + b)b2

e = tan-II albl -

alb21· a)a2 + b)b2

Hence,

Condition of perpendicularity: Two lines are perpendicular, ifthe angle between them is a right angle i.e., a = 90° tan a = tan 90° = 00 m)-m2

=00

1 +mlm2

1 + m)m2 =-1 m) m2 =-1 Hence, two lines are perpendicular ifthe product of their slopes is -I.

=> =>

Condition of parallelism: Two lines are parallel, if the angle between them is either 0 or 1t, i.e., a = 0 or 1t ., tan a = tan 0 ortan 1t = 0

=> => =>

m)-m2

=0 1 + mlm2 m l -m2 =0 m l =m2


~~~~~~~I SOLVED EXAMPLES ~I~~~~~~ Example 1: Find the acute angle between the line 9x + 3y - 5 = 0 and 2x + 4y + 3 = O. Solution: We have 9x + 3y - 5 = 0 ...(1) 2x + 4y+ 3 =0 ...(2) 9 m = -- =-3 Slope of(l) =>

3

I

Slope of(2) =>

m2 =

-"42 = -"21

The acute angle between the lines is given by tan 8 =

Iml+ - m21 1

mlm2

~3 + 1/21 tan 8 = 1+3/2

1

=>

tan 8 =

1-551=>tan8= 1 =>8=45°.

Example 2: IfA(- 2, 1), B(2, 3) and C( - 2, - 4) are three points, find the angle between BA and Be. Solution: Let m l and m2 be the slope of BA and BC respectively. Then

m

I

=

3 -1 2-(-2)

2

1

4

2

-=-

-4-3 7 andm = - - = 2 -2-2 4 Let Q be the angle between BA and Be. Then ml-m2 tan 8 = --'.-~ l+mlm2

110/81

7/4-0/2)1 2 11+ 7/4 (1/2) = 15/8 = ± "3

=>

8 = tan- I

=>

(j).

Example 3: Find the angle between the following lines: (z) x cos a l + y sin a l = PI and x cos a 2 + y sin a 2 = P2 X Y an d ---= x Y 1. -+-

••) ( II

a

b

b

a

Solution:

. . cos al I f x cos a l +ysma l =Pllsm l = --.--=-cota l (i) Thesopeo smal cosa2 . . The s Iope 0 f x cos Clz +Y sm Clz = P2 IS m2 = - - . - - = - cot a 2 sma2

The Strqight Lines

Now,

403

1 m tan 8 =lm - 2 1 1+mlm2

cot (XI + cot (X2 1 + cot (X2 cot (X2

= -

I I ---+-tan (XI tan (X2 1+_1_ _1_ tan (XI tan (X2

tan (XI - tan (X2 ( ---'-----=-= tan (X 1 + tan (XI tan

=>

(X2

)

- (X I

2

e =(XI-(X2

x y. (ii) Slope of - + - = 1 IS m a b I

(11 a) lib

x y. -(lIb) Slopeofb"-~ = hsm 2 = (-lIa)

Hence m I m2 = ( - ;).

b a

= - -- = - -

(~) = -

•

a

=-b"

I

=> The lines are at right angles. =>8=90° Example 4: The angle between two lines is 45°. If the slope ofone ofthem is 114. Find the slope of other. Solution: Here, 8 = 45°, m 1 = 114 . ~et the slope of the required line be m 2•

Now tan 450 = / ml - m2 /

l+mlm2

=>

4 1 = /11 - m / => 1 + m = ± l+ml4 "4

(~_ m) 4

. m 5 -3 -3 for + ve SIgn 1 + - = - - m => - m = => m = - . 4 4 4 4 5 . m I 5 3 5 for - ve SIgn 1 + - = - - + m => - = - m => m = 4 4 4 4 3

The possible slope of the lines are

~, -3. 3

5

Example 5: Find the angle between the lines:

x-yJ) -5 =OandJ)x+y-7=0

[UPTU B. Pharma 2001] .

404 Remedial Mathematics Solution: The given two lines are: x-yfj -5 =0. fj x+ y-7 =0

and Here

m

l

m2

...(1)

...(2)

= Slope of the line (1) = __1_ = _1_ -fj J3

- fj r:; = Slope of the line (2) = -1- = - ,,3

m l x m2 =-1. Hence, the two lines are at right angles. Example 6: The line joining (- 5, 7) and (0, - 2) is perpendicular to the line joining (1, 3) and (4, x). Then find x. [UPTUB. Pharma2003] Solution: Here m l = Slope of the line joining the points (- 5,7) and (0, -2).

clearly

-2-7 0-(-5)

-9 5

and m2 = Slope of the Iinejoining the points (1,3) and (4, x).

x-3 x-3 =--=-4-1 3 If the given two lines are perpendicular, then m1m2 =-1

(_~)

(X;3)

=-I~-9(x-3)=-15

~ x = 14/3. Distance of a point from a line: Let ax +by + c = 0 be any equation ofthe line and P(x, y) be any point in space, then the perpendicular distance d of the point p from the line is

d =

lax +bYl +cl l

L

Fig. 9.17

~a2 +b2

Distance between two parallel lines:

+ +

+ +

Let ax by c 1 = 0 and ax by c 2 = 0 be two equation of parallel lines, then the distance between the two lines is given by

Fig. 9.18

The Straight Lines

~~~~~~I

SOLVED EXAMPLES

405

~I~~~~~~

Example 1: Find the length ofthe perpendicular drawn from the point (- 2, 3) on the line 12x - 5y + I = o. Solution: We know that the length of the perpendicular segment from point (x\, YI) on

ax+by+c=Ois

11

+l'IY+cl a 2 +b 2

:.Here a =I2,b=-5,c=1,x l =2'YI=3 :. length of the required perpendicular segment =

112X(-2)+(-5)X3+II ~(l2)2 + (_5)2

= 1-24 -15 + 11 = 1- 38 1= 38.

13 13 13 Example 2: Find the distance between the parallel lines 2x - 3y + 9 = 0 and 4x - 6y + I = O. Solution: As the given lines are parallel, they have same distance between them throught out. So we shall find the distance of any point on the first line from the second line (0, 3) is a point on the line 2x - 3y + 9 = o. Perpendicular distance ofthe point (0,3) from 4x - 6y + 1 = 0 is

4X-6X3+II

~42 + (_6)2 1

=

(-17) ..J52

17

=

2J13

1~ . 2"l3 Example 3: Are the points (2, - 4) and (0, 5) on the same or opposite sides of the line 2x-5y + 6 = O? Solution: Perpendicular distance of (2, - 4) from the given line is _ 2x2-5(-4)+6 PI ~4+25 Hence, the distance between the given lines is

4+20+6

59

...(1)

Perpendicular distance of(O, 5) from the given line is

2xO-5x5+6

.J4 + 25 -25 +6

59 -19

=

59·

...(2)

Since (1) and (2) are of opposite signs, therefore, the point are on opposite sides of the given line.

406


Example 4: Which ofthe lines 2x - y + 3

= 0 and x - 4y - 7 = 0 is farther from the origin?

IRGPV B. Pharma 2001] Solution: The length of perpendicular from (xl'YI) on ax + by + c = 0 is =

laxj;b:~:cl

Length of perpendicular of2x - y + 3 = 0 from origin

= 12XO-0+31=

~4+1

PI

~

15·

and length of perpendicular of x - 4y - 7 = 0 from origin

7 P -10-4XO-71--

~1 +16

2

.Iff"

as PI> P2 ... 2x - y + 3 = 0 is farther from origin. Example 5: Find the distance between the two parallel straight lines y = mx + c and y = mx + d. [R-GPV B. Pharma 2002] Solution: Puttingy = 0 in y = mx + c, we get x = - elm. Thus ( - : ,0 }s a point on the line

y = mx + c. Length of perpendicular from ( - : ,0) to y = mx + d is given by

p

~ mX~+d ~I~I

Example 6: Find the distance between the paral/ellines 3x + 4y

=

12 and 3x + 4y

=

3

[UPTU B. Pharma 2004J Solution: The given lines are 3x+4y =12

3x+4y =3 ...(2) Putting x = 0 we gety = 3. Thus (0, 3) is a point on the line (I). The perpendicular distance between the lines (I) and (2) is = the length of perpendicular from the point (0,3) to the line (2) 3xO+4x3-3

9

~9+ 16

5

1. Find the length of the perpendicular from the origin on the line 4x- 3y = 7. 2. Find the distance ofthe point (3, -2) from the line 7x-5y-29 = O. Determine whether the point lies on the origin side of the line.

3. For what value of kwill the point (3, k) lie on the origin side of the line 2x + 3y+ 6 =

o.

The Straight Lines

407

4. Find the foot of the perpendicular drawn from the point (- 2, - I) on to the line

3x+2y-5=O. 5. Show thatthe point (l, 2) is equidistant from the lines 5x - 2y- 9 = 0 and 5x - 2y + 7 = O. 6. Find the distance between the pair of parallel lines 2x- 3y+ 4 = 0 and 4x-6y-5 = O. 7. If a and b are the intercepts of a line on the x and y axis respectively and P be its perpendicular distance from the origin then show that -;- = -\- + ~. P a b

HINTS TO THE SELECTED PROBLEMS 4. Let P (- 2, - I) = (xl' y\) and M = (h, k) be the foot of the perpendicular On to

3x+2y-5=O. .

h-x\ a

Now (h, k) are gIven by - -

(ax! + by! +c) a2 + b 2

k - y\

= --= b

h+2 = k+1 = -(-6-2-5)

=>

3

2

9+4

h+2 = k+1 =1 3 2 h+2 =3,k+I=2 h=l,k=1

=> => =>

:. The foot of perpendicular (1, 1).

7. ~+l. a b

=

1 =bx + ay =ab =>bx + ay-ab = 0

I

ANSWERS

I

~, origin lie on the opposite side of the line.

1. 7/5

2.

3. kx\ ,*x2 (t) One-one Onto Function: ifa jimction fA ---+ B is both one-one and onto i.e., the different points in A are joined to different points in B and no point in B is left vacant.

A

B

0+-------+0 Fig. 10.9 One-one onto/unction

Remarks • One-one onto mapping is also known as bijective or one-to-one.

A

B

• For a one-one onto function, Rang = Co-domain andx\ ,*x2 =::> j(x\) j(x2 )

'*

• j(x\) =j(x2 ) =::>x\ =x2

(g) Many-one Into Function: Afunctionf: A ---+ B which is both many-one and into function is Fig. 10.10 Many-one into/unction called a many one into function i.e., two or more points in A are joined to. some points in B and there are some points in B which are not joined to any point in A. Therefore, for many-one into function. (i) Rank c Co-domain (ii)x\ ,*x2 =::> j(x\) = j(x2 ).

(h) Many-oneOnto Function: ifjUnctionf: A ---+B is both many-one and onto function, then it is many one onto jUnction i.e., in B, one point is joined to at least one point in A and two or more points in A are joined to some points in B.

A

B

~ Fig.lO.n Many-one onto/unction

416 Remedial Mathematics Therefore, for many-one onto function (i) Range = Co-domain (ii) XI -:F-X2 ~ j(x l ) = f(x2)

Working Procedure [For Checking the injectivity (one-one) ofthe Function] STEP KNOWLEDGE

Let x andy be two arbitrary elements in the domain off Step 1: Takef(x) = fly) Step 2: /fwe get, X = y, after solvingf(x) = fly) Then, f: A ~ B is one-one.

Working Procedure [For Checking the Surjectivity (onto) of a Function] STEP KNOWLEDGE

Step 1: Step 2: Step 3: Step 4: Step 5:

Take an arbitrary element y in the co-domain Putf(x) =y Solve f(x) = y for x and obtain x in terms ofy Get the equation ofthe form x = g(y) Ifx = g(y) belong to domain off,for all values ofy, thenfis onto.

~~~~~~I

SOLVED EXAMPLES

~I~~~~~~

Example 1: Letf: R -+ R be a function defined by

f(x)

=

{

3X -1, when x > 3 x 2 - 2, when - 2 ~ x ~ 3 2x+3, when x 0 { Xlif x_

, ,, ,

I----t

-2

Fig.l0.t8

426


The graph of the modulus function ,f(x) = Ix I is shown in Figure

-4

-3

-2

2

3

4

Fig. 10.19

Here,

/(1) =/(-1)= I /(2) =/(-2) =2

/(3) =/(-3)=3

Remark • Modulus Function is always an even function. (g) Linear Function: Afunctionf A ~ B o/the/orm/(x) = ax + b, where a, b

a linear function. The graph of a linear function is always a straight line For Example: Graph of/(x) = 4x - 2 is shown in Figure 6 5 4

3

2

-4

-3

-2 -1

3 -1

-2

Fig. 10.20

4

E

R is called

Functions and Limits

x f(x)

2

o

2

-2

6

427

(h) Quadratic Function: Afunctionf A ~ B is called a quadraticfunction ~fit is oftheform y = d + bx + c, where a, b, c E R, a ;rOo f(x)

-3 -2

-4

234

-1

Fig. 10.20 (a)

f(x)

The graph of such a function is called parabola. For Example:Graph of f{x) (figure 10.20)

I

f~)

=

-2 2

8

2x2 is shown in

-1

2

2

8

(i) Exponential Function: Afunctionf A -+ B ofthe form f(x) = where a > 0 and x E R is called an exponential function.

cr,

The exponential function d will never be negative for all a> 0 and x E R. Therefore, domainoff(x) isR and range off(x) is the set of positive real numbers.

-3 -2

-1

2

3

Fig. 10.21

For Example: The graph of the exponential function.f{x) = 2x is shown in Fig. 10.21

-2

-1

1/4

112

o

G) Logarithmic Function: For any a > 0, a .: x> 0 is called logarithmic function. By definition oflogarithms, we have

aY=xlogax=y

2 2 ~ 1.

00

4

a function f(x) defined by f(x) = logax,

428


Here it is clear that, x >0 for all Y E R and a> 0, a::;; I. So j(x) is defined for all x> 0. Thus. domain ofj(x) is the set of positive real numbers. The graph of the logarithmic function is shown in fig 10.22. f(x) f(x)

=loga x, a > 1

--~~---*--------------------

o

__ x

Fig. 10.22

III ALGEBRA OF FUNCTIONS The algebraic operations of addition, subtraction, multiplication and division, yield new functions. Let us see the following definitions. Definition: Let f and g be two real valued functions with domain D} and D2 respectively. If D = DI (') D2 ::;; $, then (I) The sum function, denoted by f + g, is defined by if+ g) (x) = j(x) + g(x), with domain D. (ii) The difference function, denoted by f - g is defined by if-g) (x) = j{x)- g(x), with domainD. (iii) The product function, denoted by f g is defined by ifg) (x) = j(x) . g(x), with domain D. (iv) The quotient function, denoted by

f

g

is defined by

f)(x) = f(X» ,with domain D', where (g g(x D' = {x:x

E

D,g(x)::;;O}::;;$

(v) The reciprocal function denoted by

1

f

is defined by

(~ }X)= f~X)' with domain D", where D"= {x:x E Dl'j{x)::;;O} (vi) If c is any real number, then scalar multiple off by c, denoted by cf, is defined as (cf) (x) = cj{x), with domain D} (vii) Iff is a function, then f f is denoted by f2,j2! is denoted by f3 and so on. Also ift(x) = {f{x)}n, withdomainD}.


429

IlI!I COMPOSITION OF FUNCTIONS Let f: A ~ Band g: B ~ C be two real valuedfunctions. Then the composition offand g denoted by g of, such that g of: A ~ C is defined by (go j)(x) =g(f)(x» This is also known as function of a function or resultant of a function. Similarly, (fo g) (x) = f(go ex»

Remarks

rl

• Iff:X ~ Y is a one-one onto mapping, thenfo = Iyandf-l 0 f= Ix • If f:X ~ Y and g:y ~ Z be two one-one onto mappings, then the mapping of go fis also one-one and onto. • go fmay exists whilefo gmay not exists. • If g ofand f o g both exit, they may not be equal. • g ofexists ifand only if the range offis a subset of domain of g. Simiiarly,Jo gexists if range of g is a subset of domain off

General Theorems Theorem 1: The composition offUllctiolls is associative.

Remark • The composition offunctions is not commutative, i.e.f° g:l= g 0 f Theorem 2: The composition ofany function with the identity function is thefunction itself. Theorem 3: Letf: A ~B, g: B~A betwofonctionssuchthatgof=IA . Thenfisan ilyection and g is a Surjection. Theorem 4: Letf A ~ Band g: B ~ C be two functions, then (i) g 0 f: A ~ C is onto => g : B ~ C is onto. (il) g 0 f: A ~ C is one-one => f: A ~ B is one-one. (iii) g 0 f: A ~ C is onto and g : B ~ C is one-one => f: A ~ B is onto. (iv) g 0 f: A ~ C is one-one andf: A ~ B is onto => g : B ~ C is one-one. Theorem 5: The inverse ofbijectivefunction is unique. Theorem 6: Iff A ~ Band g: B ~ Care two bijectivefunctions, then g of: A ~ C is a bijection and

ISOLVED EXAMPLES I Example l(a): IffR~R isdefinedbyj(x) =x2 -3x + 2,findf(f(x». Solution: Since j(x) = x2 - 3x + 2 Therefore, j(j(x» = j(x2 - 3x + 2) =(x2-3x+2i-3(x2-3x +2)+ 2 =x4 +9x2 +4-6x3 -12x+4x2-3x2 + 9x-6 + 2 = x4 - 6x3 + 1Ox2 - 3x

430


Example l(b): LetfN-+R beafunctionsuchthat.f{x) =2x-3 andg: Z -+R beafunCtion

such that g(x)

x-3

Find g of N -+ R. 2 Solution: Consider, (go.f) (x) = gf(x) = g(2x- 3) = --.

= 2x-3-3 = 2(x-3) =x-3

2

2

Example2: LetA = {I, 2, 3, 4, 5}, LetfA -+A andg: A -+A be defined by f(1) = 3./(2) = 5./(3) = 3./(4) = 1./(5) = 2 g(1) = 4, g(2) = 1, g(3) = 1, g(4) = 2, g(5) = 3. Find (10 g) and (g oj). Solution: Here, we have,

«0 gXI) =.f{g(1»=/4) = I g)(2) =.f{g(2» =.f{I) = 3 «og)(3) =.f{g(3»=.f{I)=3 g)(4) =.f{g(4» =.f(2) = 5 g)(5) =.f{g(5» =.f(3) = 3 (g0.f)(1) =g(f(l)=g(3)= I (go/)(2) = g(f(2» = g(5) = 3 (go.f)(3) = g(f(3» = g(3) = I (g0.f)(4) = g(f(4» = g(I»=4 (go.f)(5) = g(f(5» = g(2) = 1. Example 3: Let j. g: R -+ R be two functions defined by

«0 «0 «0

f(x) Find (i) f+ g (iv) g-f (vii) fig _ Solution: Since, we have

=

../x-I andg(x) (ii) g

+f

=

~4-x2

'If x ER. (iii)

(v) fg (viii) glf

(vi) gf

.f{x) = ../x-I andg(x)= ~4_x2 Therefore, the domain off= [1, 00 [ = D J (say) and domain of g = [ -2,2] = D2 (say) D = D J I1D 2 = [1, 00 [11 [-2, 2] = [I, 2] *~ Define Then, we have (I) «+gXx)=.f{x) + g(x)= ../x-I + ~4-x2 ,withdomainD (il) (g +j)(x) = g(x) +.f{x) = ~4 - x 2 + ../x-I, with domain D (iii)

«- g)(x)

= .f{x)- g(x) =

(iv) (g-j)(x) = g(x)-.f{x) =

../x-I- ~4-x2 ,withdomainD

~4-x2 - ../x-l

f- g


(v) (fg)(x)=1(x) . g(x) = ..Jx-l· ~4_X2 =

431

~(x-l)(4-x2), with domain D

~4_x2 . ..Jx-l = ~(4-x2)(x-I) ,withdomainD

(vi) (gf)(x) = g(x) .1(x) =

(vii) The domain of f is obtained by deleting those points x at which g(x) = 0 from D. g

Therefore, the domain of f isD'= {x:x g

E

D,g(x);eO} = [1,2]

f)(x) = f(x) = ~ = (g g(x) ~4-x2

~

(viii) The domain D" of

~(

X-I)

4-x2

is given by

D' = {x: x

E

D,j{x);e O} = ]1, 2]

Th""for~ (; }Xl ~ ;~:~ ~ ~~a ~-:12) Example 4: Iff(x) =x2 and g(x) = 3x. Find the value of(g Of) for x = 1, 2, 3 Solution: Here, we have f(1)=1 2 =1 (gof)(l) =g(f(l»=g(l)=3 xl =3 f(2) =22 =4 (gof)(2) =g(f(2»=g(4)=3 x4= 12 Therefore, 1(3) =3 2 =9 Now, (gof)(3) = g(f(3» = g(9)= 3 x 9=27. This can be illustrated in (figure 10.23)

c

B

A

Fig. 10.23

Example 5: Show that bf: R - (OJ

~ R,

given by f(x)

itself Solution: (i)fis one-one: Letx,y E R- to} such thatf(x) =f(y) Then, f(x) = f(y)

3 x

3

y x=y

3 x is invertible and it is. inverse of

= -

432


j(x) =j{y) x=y

=>

Since, x,y are arbitrary, therefore,j(x) = j{y)

=> =>

x=y'ilx,y f is one-one

{ii)fis onto Let y be an arbitrary element of R - {O}, thenj(x) =y 3

-

=>

x

=y 3

x= y

=> Therefore, for eachy

E

. R - {O}, there eXIsts -3

y

E

R - {O}, suc h that

j(x) = f(~) = _3_ = Y Y

3/y

=>

fis onto. Therefore,f is bijective and hence invertible. To find;l, letj(x) = y, then j(x) =y

=>

3

- =y x 3

=>

x=y

=>

;1 (y) = ~ y

3 ;I(x) = - = j(x). x Hencefis the inverse of itself.

=>

Example 6: Ifthefun.ctionfR ~ R is given by j(x) = x 2 + 2 and g:R ~ R be given by g(x) x = --1' Findfogandg of

x-

Solution: Here, we obse\'Ve that, the range off= domain of g and range of g = domain off fog and g a fboth exist. Therefore,

1 ~)=(_x_)2 =~+2

Consider,

(fog)(x) =j(g(x» =

J~x-I

and

(g oj) (x) = g(/{x»

gC-~ + 2) =

=

x-I

x 2 +2

(x-I)

x 2 +2 =2- (x +2)-1 x +I 2


433

Example7: lffR-+Rbegivenby 2 2 f(x) = sin x + sin ( X +

~ ) + cos x cos ( .x + ~ ). V X E R

andg: R ~ R be such thatg (5/4) = 1 then show that g of: R ~ R is constant function. f(x) = sir?x + si n2 (x + n/3) + cosxcos(x + n/3) Solution: Given that =

~[2sin2 x+2sin2 (x+~ )2cosxcos( x+~)]

=

~ [1- cos 2x + 1- cos ( 2x + 23

=

~ [% - cos 2x - cos ( 2x + 23

=

~ [ %- {cos 2x + cos(2x + 23

=

~[%-cos( 2x+~ )cos~+ cos( 2X+~)]

=

~ [% - cos( 2x + ~ ) + cos( 2x + ~ ) ] = ~ V X

1t )

1t

~

+ cos ( 2x +

)

~) ]

~

1t

Now,

g o.f(x) = g(j(x» = g(5/4) = 1

=>

g oj(x) = 1, V X E R.

Hence, g of: R ~ R is a constant function. Example 8: Iff R -+ R be a jUnction given by f(x) = ax + b, and b such that fO f = lIt Solution: Here, we have fof=IR which implies, foj(x) = fix), V x E R => j(f(x» =x, V x E R(:.fR(x)=x) => j(ax + b) =x, V X E R => a(ax+b)+b=x, V XER => (if-l)x+ab+b=O,VxER => if - 1 = 0 and ab +b = 0 a =± 1 andb(a+ 1)=0 => a = 1, then b = 0 If If Therefore, Hence,

) }

\;j

~

+ cos ( 2x + ) + cos ]

+ cos ( 2x + ) ]

E

R

X E R. Find the value ofa

a=-I,thenb(a+l)=O, V bER

a = -1 and b may take any real value. either a = 1 and b = 0, or a = - I and b can take any real value.

434


Example 9: Which ofthefollowingfunctions are odd or even or neither? (I) j(x) =tanx+3 cosecx+x (ii) j(x)=lxl+ 1 (iii) j(x) = Ix-21 Solution: (i) Here, we have;f(x) = tan x + 3 cosec x + x ~ fe-x) =tan(-x)+3 cosec(-x) + (-x) = -tan x - 3 cosec x - x = -(tanx + 3cosec x + x) = -f(x)

Therefore,J(x) is an odd function. (ii) We have,J(x) = Ix I + 1 ~ fe-x) =I-xl+ 1 = Ixl+ 1 =f(x) j(x) is even function. ~ (iii) We have, j(x) = Ix-21 ~ j(-x) = l-x-21 =1-(x+2) I = Ix+21 ~ j(-x) #=j(x)orft.:-x)#=-j(x) Therefore,J(x) is neither even nor odd function. Example 10: Iff(X)=tOg (I+X) showthatf(x)+fty) I-x

=

f(x+y) l+xy [UPTU B. Pharna 200SJ

Solution: It is given that

~

j(x) =10g(I+X) I-x

...(1)

fiy)=10 g (I+ Y ) l-y

...(2)

Adding (1) and (2) we get

I

f{x)+fiy) = 10g(I+X)+ 109(l+ y I-x l-y) =

Again

f(~) l+xy

log[Cl + x)(1 + y)] = [1 + x + y + XYJ (l-x)(l-y) l-x-y+xy

=[I+2yj l_x+y l+xy

= IOg[l+X+ y++xy] l-x- y+xy Using (3) and (4), we conclude that

j(x)+fiy)

...(3)

=1 x+ y) J~I+xy

...(4}


435

Example 11: Ifis an identity function what isfe j? What isff Solution: Letfbe an identity function on a set X. Thisf: X ~Xis identity if .f(x) = x

'V X E

X

...(i)

fe.f(x) =j{f(x)} =.f(x) from (I) =x from (i) fe.f(x) =x

So

.fI(x) =.f(x).f(x) = x. x

Again

From (I)

=x2 3

Example 12: Iff(x) = x -

:3

. gIven . Solution: It IS t h at j(x)

So

then show thatf(x) = feu = o.

= x

.f(x) =

3

1 -'3

x

f(~)

(x :3) {(~J -C:xJ} 3

=

_

3 1 1 3 x -=--x 3 3 x x

1

x3 = 3

x

x6 = 1

=> =>

Now

x=1

{(f(x)}

x= I

1 =1--=0 1 =13__ 13

1

{f(~)L=1 = OJ _1 =0 3

So

.f(x)

=~~) =0.

Example 13: Letf: A ~ B such that.f(x) = x- 1 and g: B ~ C such that g(y) = y.. Find fe g(y). Then

.f(x) =x-l andg(y)=y feg(x) =j{g(x)} =.f(x2)=x2-1

So

feg(x)

Solutio,n: Given that

=x2-1.

436


Exercise 14: Given thatf(x)

I

x-I

= --, g(x) = - - . Find the value ofg[f(x)].

I-x x I x-I j(x) = - , g ( x ) = I-x x

Solution:

g[f(x)]

So

=g[I~J=[~IJ= I-I~~X) I-x

=x

(I-x)

g[f(x)] =x.

Hence

~~~~~~~I

EXERCISE 10.2\~~~~~~~~

1. If A = {a, b, e, d} andfcorresponding to the Cartesian product {(a, b), (b, d), (e, a), (d, e)}. Showthatfis one-to-one from A ontoA. Find]l. 2. If A is anon-empty setandf, g: A ~A, such thatfo g= go f= IA' Show thatfand g are bijections and g = ]1. 3. LetX= {-2,-I, 0,1,2, 3} and Y= to, 1,2, ... , 10} andf:X ~ Ybe a function such that j(x) =~, V x EX, Findjl (A), where A = {O, 1, 2,4} 4. Find the inverses of the following functions, if exist I (I) j(x)=--x+4 3

(iii)j(x) =

x-I (ii) j(x)= ,x* 1 x+l

~I-x2 ,o~x~ 1

5. IfA = {a, b, e, d} andfcorrespondstothecartesianproduct {(a, b),(b, d),(e, a),(d, e)}. Show thatfis one-one from A onto B. Find]l. 6. Iff R ~ R is a bijection given by j(x) = ~ + 3, Find]1 (x). 7. IffR ~ R is defined byj(x) = 3x-7. Showthatfis invertible and find]l.

8. Let fA

~

B be a function, such that

(I) A = {0,-1,-3, 2}, B= {-9,-3, 0, 6} and.f{x)= 3x. Find]l. 9. Letf R ~ R be given by.f{x) = (x + 1 -1, x ~ -1. Show thatfis invertible. Also find the setS= {x:.f{x) =]1 (x)}. 10. Find]1 (3), if exist, when.f{x) = x 3 +4, wheref: R ~ R. 11. Iff= {(5, 2), (6, 3)}, g= {(2, 5), (3, 6)}. What is the range offandg? Findfo g. 12. If.f{x) = ~ - 1, g(x) = 3x + 1, then describe the following function: (I) g 0 f (ii) f o g (iii) go g (iv) fo!

p

13. If.f{x)= X-I, verify that ifo]l)(x) =x. x+l 14. Iff R ~ Rand g: R ~ R are defined by.f{x) = x + 2 and g(x) = ~ +5. Findfo g and

go!


437

15. Iff and g are two real valued functions such that.f{x) = :x? - 5 and g(x) = 2x + 3, Find fog. 16. Letf: R ~ R be defined by f(x) =

+' x +1

Findf(f(2))

1

17. Iff(x) = - , showthatf(f{f(x)})=x. I-x

18. Iff R~ R, where.f{x) =:x?+ 2 andg: R ~ R, whereg(x) = 1__1_, then find I-x (I) fog (ii) gof 19. LetA = {x E R: ~x ~ 1}, iffA ~A is defined by

°

j(x) =

then show that (f0 f)(x) = x, v

X, {

if x E Q

I-x, if

X E

x~Q

A.

20. LetfZ ~ Z and g: Z ~ Zbe defined by.f{n) = 3n, v n

E

Z and g: Z ~ Zbe defined by

n / 3, if n is a multiple of \:f n E Z g(n) = { 0, if n is not a multiple of 3 \:f n E Z gof=Izandfo g::tlz ·

Show that


1. A = {a, b, c, d} andfA

~A

A x A = {(a, b), (b, d), (c, a), (d, c)}

It is clear that under J, each element of A is mapped to a unique and every element of A so f is one-one and onto

rl = {(b, a), (d, b), (c, a), (d, c)} I 4. (i) .f{x)= -3"x+4 y = -'!'x+4. Solving for x

Let

3 x =3(4-y)= I2-3y rI(x) =12-3x

=> So (il) .f{x)

x-I

= -,

x ;r-I

x+I

Let

x-I y=x+I (x+ l)y =(x-I) xy-x =-(1 +y) x(y-I) =-(1 +y)

Fig. 10.24

I

438


_ -{I + y) 1+ Y =y-l 1- y

x-

rl (x) = l+x ifx¢ 1 I-x

so

f(x)=~I-x2

(iiI)

y

Let Solving for

=

O:s;x:s;1

~1-x2

1 = l_x2 x2 = 1-1

so

x=

~1-i

rl(x) =

~1-x2

,0:S;x:S;1

8. (i) f:A-+B A = {0,-1,-3,2} B= {-9,-3,0,6}

fix) = 3x, show that Herej(x) is one-one onto (bijection) sorl exist rl (x) = {CO, 6), (-3,-1), (-9, -3), (6, 2)} (il) fA -+ B:j(x) = x2

A={1,3,5, 7,9},B={0, 1,9,25,49,81} Herej(x) is not bijective sincej(x) is not onto (0 is the element is B which is not mapped by any element of A underfsor l orfdoes not exist. 15• .f{x)=x2-5,g(x)=2x+3 fo g(x)=j{g(x)} =j(2x+ 3)=(2x+ 3i-5 =4x2+9 + 12x-5 =4x2+ 12x+4

17.

A

B

Fig.lO.2S

1

f(x) = I-x fiflf(x)4} =

f(fC~J=f(I_~l=f(~l I-x

= f (

I-X)

-=;-

1 1 = 1_(I-X) = 1+ (I-x)

-x 19. Wehavef:A -+AwhereA = {x f(x) =

E

I-x

R, 0 :S;x:S; l}

X, if XEQ { I-x, if XIi!OQ

x

x

=x+(I-x)

=x

Fllnctions and Limits

439

JV(x)} = {f(X), if X e Q

f(1-x), if xr£Q X,

if xeQ

l-(1-x), if xr£Q

=

{

=

{X,

if xeQ x, if xr£Q

j(f(x)} = x, V

so

X

eA.

I

ANSWERS

1. j l = {(b, a), (d, b), (a, c), (c, d)}

3. jl(A) = {O, -I, I, -2, 2} 4. (i) j l (x) = 12 - 3x,

(ii) jl(x)

= 1+ x I-x

5. r l = {(b, a), (d, b), (a, c), (c, d)} 6. rl(x)

= (x -

x+7 7. rl(x) = -3-

3)1/3, V X e R

8. (i)r l = {(-9, -3), (-3, -1), (0, 0), (6, 2)},

(ii)

I-I

does not exist

9. S= {O, I}

11. fog(2)=2,fog(3)=3

=>j{range) = (2,3) g(range) = (5, 6) 12. (i) gof=3:x?-2 (iii) 9x + 4

(il) 9:x?+7x

(iv) x4 - 2x2

14. gof=~+8x+ 13,fog=~+7

16.

15. 4:x?+ 12x+4

•

~ 29

CONCEPT OF LIMIT

Consider a function 2

j(x) = x - 9 + 6x + 18 x 2 -6x +9

The value ofj{x) at x = 2 is of the form

Q,

o

which is meaningless or indeterminate.

Therefore, in this case, we cannot divide:x? - 9 by x - 3, because at x = 3, x - 3 is zero. Now suppose x is not exactly equal to 3 but x tends to 3. Then x - 3 is not equal to zero, thus in this case we can divide the numerator:xl- - 9 by denominator x - 3.

440


We therefore obtain 2

fix) = x -9 = (x-3)(x+3) =(x+ 3). x-3 x-3 Now, if x is very close to 3, thusj(x) comes to 6

(::t I

x:~: -61 = 1x2:~;6X 1= I

1

=lx+31· Now Ix + 3 I can be made as small as we please by letting x tends to 3. Hence from the above observations, we observe that when x takes the fixed value 3, the value ofj(x) comes to be a meaningless number but when x tends to 3.j(x) tends to 6. Which in fact gives the limit ofj(x) as x tends to 3. Mathematically, we can write . x 2 _9 Llm-- =6. x-t3 x-3 Limit ofJ{x) at x = a. In many cases we obtain k, = ~. k (say). Then we call kthe limit ofj(x) atx = a. Mathematically, we write

lim f(x) = k.

x-ta

Some Important Results on Limits Ifj(x) and g(x) are two function, then (i) Lim [f(x) ± g(x)] = Lim f(x) ± g(x) . x--+a x-ta

(ii) Lim[f(x)· g(x)] = {Lim f(x)} {Lim g(x)} . x-ta

x--+a

x-ta

Lim I(x) (iii) Lim f(x) = X--+a , with Lim g(x)"* a. x-ta g(x) Lim g(x) x-ta x--+a

(iv) Lim [cj(x)] =c Lim j(x). x-+a

X~(I

Some Standard Limits x" -I 1. Lim-- =n. x--+' x-I

2. Lim

3. Lim(I+.!. J =e.

4.

5. Lim (I + x)'/x =e.

6.

x--+a

X

x-tro

x--+o .

aX-I

x--+o

x

7. Llm-- =Ioga

8.

L. tanS Im--=l S .

10.

9.

8--+0

e

x"-a" x-a

= n·a"-'.

Lim(I+~J =~. X

x--+ro

L. log a (1 + x) 1m = logae.

x --+

x

00

L. sinS Im-- =1 S .

8--+0

L. logx Im-- =0.

x--+oo

X


~~~~~~~I

SOLVED EXAMPLES

I~~~~~~~

Example 1: Evaluate thefollowing limits: (I) lim (x 3 -xt + 1)

(ii)

lim (I + x+.x2 + .... + X 10)

x~I

(iii)

lim

x~4

x~-I

4

. ax 2 +bx+c (iv) hm 2 ' a + b + C *- O.

3

~ x-2

x--->I cx +bx+a

Solution: (i) lim (~-.x2+ 1)= 1 _1 + 1 = 1-1 + 1 = 1. 2

3

x~l

(ii)

lim (I + x + .x2 + ... +x IO ) = 1 + (- 1) + (-1 i + ... + (-I) 10

x~-I

=1-1+1-1+ ... +1=1. (iii)

4 +3 lim(4x+3) lim _x_ _ = .::!.x~-=-4_ __ x~4

x-2

4x4+3

19 2

4-2

lim(x-2) x~4

2 ax +bx+c 2 (iv) x~ cx +bx+a

r

ax(I)2 +b(I)+c

a+b+c

c(l)2 +b(I)+c

a+b+c

---,:....:.,..._...0....:._ =

=I .

Example 2: Evaluate the following limits: (i)

(iii)

. 3x 2 -x-lo hm ---,,-.-2

x~2

. (2x-3«Fx -I) (ii) I1m ~---,,--'-'-----'x--->I

x -4

2x2 +x-3

4

lim x _81 x--->32x 2 -5x-3 2 . 3x -x-IO

Solution: (i) hm

x--->2

2

x -4

=

(x-2)(3x+5)

lim -'-------'--'-----'-

x~2

(x-2)(x+2)

= lim 3x+5 x~2

=3x2+5 =!.!. 2+2

x+2

4

I. (2x-3)(Fx -1) · (2x-3)(Fx -1) (ii) I1m = 1m ~-"":"":-----'x--->I

x~1 (2x+3)(x-I)

2x2 +x-3

=

lim (2x-3)(Fx-I) x--->I (2x+3)(Fx +1)(Fx -I)

lim _....:('--2_x_-~3:"..)_ x--->I (2x +3)(Fx + 1) -1

2xl-3

----,=--- = --- =5x2 10 (2 xl + 3)(.Ji + 1)

(iii)

· I1m

x4 -81

x~32x

2

=

-5x-3

(x-3)(x+3)(x 2 +9)

I.

1m ~-'-...:....--":"":""-----'-

x--->3 =

2

lim (x+3)(x +9) x~3

=

(x-3)(2x+l)

(2x+1)

(3+3)(3 2 +9) 2x3+1

6x18 108 =--=-

7

7

441

442

(iv)


' x4 -4 \1m 2 x~.J2 x +3J2x-S =

=

(x 2 +2)(x 2 -2) 1m ~--;=i-'----;:,=x~.J2 (x+4J2)(x-J2) I'

2

lim (x +2),(x+J2)(x-J2) x~.J2 (x+4J2),(x-J2)

2

= lim

(x +2),«x+J2) x~.J2 x+4J2

_ (2+2)(J2 +J2) _ sJ2 _ s -

(.fi +4.fi)

-

s.fi -S'

Example 3: Find the following limits

J"m - J2a

lim

(it)

x~a

x-a

, ,/3-x-l

(iii) h m - - -

x~2 2-x Solution: (i) Here, the function involving surds, so rationalizing the numerator

, I1m

x~o

.J1;7 -~ 2

=

X

=

1m

[JI;;2 -Pi[JI;;2 +~] r:--:; r:--:;

' \1m

2 ---;==--===-

I'

x~o

x 2 [vfl+x 2 +vfl-x 2 ]

X~o[~ +J17] 2

[~+.J1-O]

=~

= 1

2

'

(ii) Diagonalizing the numerator, we ha"e

\' [../x+a -fi;][../2+a +..!2a] ' ../x+a -fi; \1m = 1m -=-------===,---=:----=x~a x-a x~a (x-a)[../x+a+fi;] =

' (x+a)-2a \1m ~---;==,..---;===x~a(x-aH../x+a +fi;]

' 1 = \1m ---,,==---.==-

x~a[../x+a +fi;] 1

=

1

../a+a +fi; = 2J2a '


(iii) Rationalizing the numerator, we have . .J3-x-I 11m x-+2 2-x

=

=

=

. [.J3-x-I][.J3-x+l] lim !:....:-c---~===--~ x-+2 (2-x)[.J3-x +1] . (3-x)-I hm - - - - ; = = = - x-+2 (2 -x)[.J3 -x + 1] . 2-x I1m - - - - ; = = = - x-+2 (2 - x) [..13 - x + I]

=

. 1 IIm-..,,=~-

x-+2[.J3-x+I] 1

1 -=1+1 2

[..13-2 +1] Example 4: Evaluate x bX (/) lim a x-+o x

. eX_e- x (ii) h m - - x-+o x x bX . aX -I_b x +1 Solution: (i) lim a = 11m - - - - x-+o x x-+o X X . [aX - 1 b - 1] = 11m - - - - x-+o x x

.

aX -1

.

bX_I

= IIm---hm--

x-+o x x-+o X loga -10gb = log(a / b). . e2i -1 . e2x -1 2 hm--= h m--·x-+o xe x x-+o 2x eX =

-x

X

(ii) lim e -e x-+o x

=

2

2

eO

1

= Ix- = Ix- =2. n

· d the posItIve . . .Integer n so that I·1m xn _3Examp Ie: 5 F In x-+3 x-3

=

108 .

n 3n Solution: Since lim ~ = 108 x-+3 x-3 n(3)'1-1 = 108 n.(3)n-1 =3(3)4--1 n =4.

=> =>

Example 6: Evaluate

·log(I+x 3 )

(I) I1m -='---x-+o sinx

(iii)

lim x-+7t/4

sinx-cosx x-1t/4

. I-cosx·.Jcos2x (iz) I1m 2 x-+o x . 1+ cos2x (iv) I1m 2. x-+7t/2(1t-2x)

.

443

444

Remedial MathemaIlCS

0Iog(1+x3 ) 10Iog(1+x3 ) Solution: (I) 11m 3 = 1m 3 x~o sm x x~o 3 sm x 0

0

x -3x

=

lim 2- log (l + x 3 ) ( 3 )1/xl x~Ox3 = lim log l+x

x~o

o (sin x)3 hm - x~o

(1)3

X

= loge =! = 1.

1 (

ii)

r l-cosxJroS2x x~ x2

=

1

r l-COSXJroS2X[1+COSX~] x~ x2 1+cosx.Jcos2x

l-cos2 xocos2x 0 = \1m --:------==~ 2 x~o x (1 + cos x.Jcos 2x ) 0 l-cos2 x(1-2sin 2 x)

- \1m

---,----==~

- x~o

=

x 2 (1 + cos x..Jcos 2x )

0 l-cos2 x+2cos 2 xosin 2 x 11m ----::-.,.-----::==___

x~o

x 2 (I +cosx.Jcos2x)

2 sin2 x+2cos xsin2 x 1Im--,-----==,2 x---M) x (1 + cos x.Jcos 2x) 0

=

10 + cos 2 x ) 0 sm x - olm ( 1 2 = 11m ( -0)2 x~o X x~o 1 + cos x.Jcos 2x =

(iii)

(1)2 x 1+ 2 (1)2 = i 1+1.Ji 2

0

o sm o (1t - + Y ) - cos (1t - + y) lim smx-cosx = lim 4 4 x~x/4 (x-1t/4) y~O y 01t

=lim

1to

1t

o1to

sm-ocosy + cos- °sm y - cos-ocosy +sm- osmy 4 4 4 4

Y

x~o

1

1

0

1

1

F2

F2

0

-cosy +-smy --cosy +-sm y = lim F2

F2 y

y~O

=

lim~(Siny)=~ lim siny

y~oF2

y

J2 y~O

=~Xl=~=F2o F2

Ji

y


1 + cos 2x 2 x~1t/2 (n - 2x)

=

0

11m

(iv)

445

l+cos2x

0

11m

x~1t/2 4 (n--x )2 2

2

= lim

2 cos x x---)1t/2 (n 4 --x 2

=

)2

L

2 lim sin y

2

y---)o

i

= lim L

cos 2(n -- y ) 2

x---)O 2

y2

= .!.( lim Siny)2 = .!.(I)2 ==.!. 2 y~O Y

2

2

Example 7: Evaluate

cosx - cos a 1l m - - - - x---)a x-a 0

(UPTU B. Pharma 2004)

o

Solution: We have lim cosx - cos a x---)a x-a 0 -sinx-O 10 ( 0 ) 0 = 11m == 1m -smx =-sma x---)a 1-0 x---)a Example 8: Evaluate e X+e- x _2

[form -]

o

o

hm---::---


x---)o x2 Solution: We have

eX +e- x -2 e 2x -2ex +1 = hm ----,,---x---)o x2 X---)o x 2eX o

[:0 e-x =

0

hm

1

~]

e

lim[~(eXx-:-1)2] (lim~)o[lim(eX _1)2) x---)O eX x---)O =

x---)o eX

=

X

(T}W2 = I

Example 9: Evaluate the following limits: (I)

lim x~O

(ii)


x

lim.JI+; -I

x---)O (iii)

~ -I

(Meerut B. Pharma 2003)

X

esinx -I lim - - -

.Y~O

(Meerut oB. Pharma 2006)

x

Solution: (t) We know that d X

=

e 10ge a Using this, we get

=

ixlogea)_1 hm---x---)o X

I

lim~ x---)o X

0

0

446


=

(x loge a) (x loge a)2 ] + ... -1 [1+ l! + · 2! IIm=-----~------~~------~ x--+O

X

_ . x[

- hm - loge a +

x(loge a)2 + ...] 2!

x--+Ox

_ . [

- hm loge a+ x--+O

X(IO ge a)2]_ + ... -logea. 2!

(ii) Expanding ~ = (l + x)ll2 by Binomial Theorem, we get

lim

.Jl+x

x--+O

-

1

.O+x)1I2_1

LIm -'------'---

=

X

x--+a

= lim

X

[1+Lx+ m(~-I) h .. ]-1 2 2. x

= (iii) Expand

!~;[~-ix+ . .J=~.

e inx by exponential series 2

.

esinx_I

hm---x--+o

=

x

sinx sin x 1] [1 +--+---+ ... lim 1! 2! x--+O

x

3

_- I'1m (Sinx) sinx sin x ... ] - - . [1 +--+---+ x--+o x 2! 3!

(Sinx)

3

x ]

· - - . I'1m [I +sinx - - +sin - - + ... I1m x--+o X x--+o 2! 3! =lxl=1. =

Example 10: Evaluate lim

x 2 +4

2 .

(x - 2) Solution: Divide the Nr. and Dr. by x 2 , we get X--+OO

1+0 =1. 1-0+0



447

III ONE SIDE LIMITS (I) Right hand limit: Afunctionfis said to approach I as x approaches afrom right if

corresponding to an arbitrary positive number 8> 0 such that

8,

there exists a positive number

Ij(x)-ll <E, whenevera<x 0 such that

Ij(x)-/I <E, whenevera-o·<x

lim j(x) = limf(a+h) x~a+O

h~O

(il) To find tile limit on left, put a - h for x inf(x) and then take limits as h ~ 0

=>

lim j(x)

=

x~a-O

limj(a-h). h~O

II1II LIMIT AT INFINITY AND INFINITE LIMITS (A) Limits at infinity: (I) Afunctionj(x) is said to tends to a limit I as x ~ -00 iffor given e> 0, however small, there exists a positive number 0, such that lj(x)-/I<e'V x~o => I-E <j(x) < 1+ E 'V X ~ 0 and we write

lim j(x) X~-o:J

= I.

448


(il) A function f(x) is said to tends to a limit I as x ~ there exists a positive number 0> 0 such that

00 iffor

given I: > 0, however small,

Ij(x)-/I 0, there exists a positive number 2 such that

°

x EA,O 0, :3 0> 0 such that Jj(x) -j(a) J < E whenever 0 < J x - a J < 0.

~~~~~~I

SOLVED EXAMPLES

I~~~~~~

2

Example 1: Show thatj(x) = x -J is continuous/or all values o/x except x = 1. x-I

Solution: If =>

x"* 1, thenj(x)=(x+ 1)=Apolynomial j(x) is continuous for all values of x"* 1.

If x = 1,j(x) is of the form

Q, which is not defined and so the functionj(x) is discontinu-

o

ous atx= 1. Example 2: Show that the/unctionf(x) is defined by 2

x ,x"* I lex) = { 2,x= 1 is discontinuous at x = 1. Solution: Here the value of/ex) atx = I is 2 => /(l) =2. Now,

RHL=j(1+0)= Iim/(l+h)= lim (l+hi=1 x-+o

also

x-+o

lliL=j(1-0)= Iimj(l-h)= lim (I-hi =1 x-+o

x-+o

450


Therefore, we have .1(1 +0) =.1(1-0)*.1(1) => j{x) is not continuous at x = 1. Example 3: Examine whether or not the function .l(x) = {Si:2X, when x

*0

2, whenx=O is continuous at x = O.

Solution: Given that

.l(x) =2, whenx=O .1(0) =2

=>

NowRHL=f(O+O)= lim.l(O +h)= lim [sin2(O+h)] x--+o x--+o (O+h) =2J:. 1im sinx =IJ

i

lRL

and

x

x--+o

=.I(~)=

lim .l(O-h) = lim [sin2(0-h)] =2. h--+O (0 - h)

h--+O

Therefore, we have .1(0+0) =.1(0-0)=.1(0)=2. Hence,j{x) is continuous at x = 2. Example 4: Afunction.l(x) is defined asfollows j(x)

=

1+XifX:=>2 { 5-xifx~2

check the continuity of.l(x) at x

=

2.

Solution: Here, we have .1(2) = 1 + 2 or 5 - 2 = 3

...(1)

RHL =.1(2+0)= lim f(2+h)

Now,

x--+o

= lim [5-(2+h)]= lim [3-h)]=3 ... (ii) h--+O

h--+O

lRL =.1(2 -0) = lim .l(2-h) = lim [1 +(2-h)] = 3 .... (iii)

and

h--+O

h--+O

Now, from (I), (ii) and (iii), we have .1(2+0) =.1(2)=.1(2-0)=3. Hence, the functionj{x) is continuous atx = 2. Example 5: Test thefollowingfunctionfor continuity at x = 0: (I) .l(x) = xsin..!.. ,x* O,.I(x) = 0 atx= O. x (il) .l(x)=

1

l-e

-lIx

,x*O,j{x)Oatx=O.

[Meerut B. Sc. Biotech 20061


451

Solution: (i) Here, we have

lliL =j(O-O) = lim j(O-h)= lim./(-h) h~O

h~O

= lim (-h)sin(_1 )

-h

h~O

· h· 1 = I1m SInh~O

h

= 0 x a finite quantity lying between I and - 1 = 0 RHL =./(O + 0) = lim itO -h) = lim j(h)

and

h~O

h~O

· h· 1 =0. = I1m SInh~O h .1.0) = 0 given

Also

./(0 + 0) =./(O - 0) =./(0).

=>

Hence, the function./(x) is continuous at x = O.

(ii) Here we have

lliL =./(0-0)= lim j(O-h) h~O

=

limj(-h) =

h~O

lim~ =0

h~O

l-e

RHL =./(0-0)= lim j(O+h)

and

h~O

l_lfh = 1. l-e Also, .10) =0 => ./(0 + 0) "#./(O - 0) =./(0). Hence,./(x) is discontinuous at x = 0 and this discontinuity is of first kind. Example 6: Examine the continuity at x = 1 ojthe jitnction j(x) = 5x-4, when 0 50x 50 1 = 4x3 - 3x, when 1 < x < 2. [Meerut M. Sc. Chemistry 2004] Solution: At x = 1,./(1) = 5.1-4 = 1. = lim j(h) = lim h~O

Now

h~O

R.H.L. = Limj(x) x~l+

= Lim./(l

+h)

h~O

= Lim [4(1 + h)3 -3 (l + h)], using lower rule h~O

=[4xl-3 x l]=1.

L.H.L. = Limj(x) = Limj(l-h)

Also

x~l-

h~O

= Lim[5(I-h)-4] =5 x 1-4= 1 h~O

R.HL = L. H. L = 1 => lim j(x) x~l

Obviously

lim j(x) =./(I) x~l

=> ./(x) is continuous at x

=

I.

=

I

452


~~~~~~~I

~I~~~~~~

EXERCISE 10.3

1. Ifj{x) =:? - 3x + 5, then findf(O). 2. Ifj{x)= 3:? +4x+3, then fmdj(O),j(1) andf(-I).

3. Ifj{x) =:? - 3x + 6, then find the value of f(3 + h) - f(3) h 4.

Ifj{x) = loge

(~::) , then prove thatj{x) + j{y) = f( T)

1 __ 1 x2 2

5. Ifj{x) =

1t

I 1+-x2 1t

6. Ifj{x) =

, then find

f(~) . 4

2

~ , then prove that: x =

f(x) x-I f(x)-I 7. Which of the following functions are even or odd? (i)

- x3

(ii) x 50

(iii) x +!

I-x l+x

9. Ifj{x) = - - , then prove that:

1+ f(x)f(y) f(x)- f(x 2 ) 1+ [f(x)f

x-y l+xy

=

8. Ifj{x)= x-I ,then prove that: f(x)-f(y) x +1

(iv)

x

-

-I

2

10. Evaluate the following limits: (I)

lim 6:?-4x+ I

(ii) lim x~1

x~o

2

C~ r1m 2x -3x+6 III x~2 3x 2 -6x+8

x 2 -a2 x~a x-a . x 2 -3x (vii) hm2 x~a x -9 3x2 -4x+ I (ix) lim 2 x~1 x -4x+3 11. Evaluate the following limits: (v)

lim

3x 2 + 2x+3 lim 2 x~O 5x + 2x + 4 x 2 -4x+5 (iiz) lim 2 x~oo x - 7 x + 12 (I)

(iv)

(VI)

(viii)

(il)

(iv)

x 2 +2x-1 2x+3

2 r1m-x-I

x~1

lim

x-I x 2 -x-6

x~3x

3

-2x-3

2x2 -7x+6 lim x~25x2 -llx+2

lim

5-x2

x~oo3x+5x

lim x~oo

2.

~7x4 _5x 3 +4x2 +3 3x 2


453

12. Evaluate the following limits: lim (I + px

(l)

x-+o

X

(iii) lim a x-+o x

i

(il)

Ix

bX

eX _e- x

.

hm--x-+o x

(iv) lim (I+xt-I x-+o

x

x

lim aX +b -2 x 13. The functionf(x) is defined as follows:

(v)

x-+o

5x-4 f(x)= { 4x 3 -3x '

O<x~1

l<x 1

j(x)=

[RGPVB. Pharma 20041

Does lim j(x) exists. x-+I

ANSWERS 1.

5

5.

-

2. j(O)=3,j(I)= 1O,j(-1)=2

3. 3 +h

15

17

7. (I) odd 10. (I)

(ii) even

2 5 5 (vi) 4 (il)

(v) 2a

-

(iii) odd

(iv) odd.

(iii)

(iv) 2

1 2

(vii) -

(viii) -

9

(Lx) - 1.

3 5

H. (I) -

(il)

12. (I) eP

(il) 2

5

(iv)

(iii)

a

(iii) 10gb

17 3

(iv) n

(v) logab. 13. At x = 4, function has infinite discontinuity and is continuous at all other points in

R. 14. does not exist

454


OBJECTIVE EVALUATION MULTIPLE CHOICE QUESTIONS Choose the most appropriate one: 1.

IffR -4 R is defined by j{x) = x 2 - 3x + 4, for all x E R. then]1 (2) is (a) {I, 2} (b) (1,2) (c) [1,2] (d) {- 2}

2.

Let A be a set containing 10 distinct elements. Then the total number of distinct functions from A toA is (b) 1010 (d) 2 10 - 1 (c) 2 10 (a) 10!

3.

The number of bijective functions from the set A to itself, if A contains 108 elements is (a) 108 (b) (108)! (c) (108)2 (d) 2 108

4.

Iff: N x N -4 N is such thatj{m, n) = m + n for all n numbers, then which of the following is true. (a) fis one-one but not onto

E

N, where N is the set of all natural

(b) fis neither one-one nor onto

(c) fis one-one and onto (d)fis onto but not one-one

5.

6.

If R denotes the set of all real numbers then the functionf: R -4 R defined by j{x) = Ix I is (a) one-one only (b) onto only (c) both one-one and onto (d) neither one-one nor onto ThemappingfR+ -4R defined byj{x) = 10gloX, (whereR+ isthe set of all positive Number) is (a) (c)

7.

only one-one mapping both one-one and onto

(b) only onto mapping (d) None of these

On the set Z of all integers define f Z - {O}

I n

j{n) =

-4 Z

as follows:

.

2' n IS even 0, n is odd

Then.! is (a) onto but not one-one (b) one-one but not onto (c) one-one and onto (d) into

8.

9.

10.

11.

R be defined by j{x) = 2x + sinx, for x

R. Then fis (b) one-one but not onto (a) one-one and onto (c) onto but not one-one (d)neither one-one nor onto Iff: A -4 B is surjective then (a) n(A) ~ nCB) (b) n(A) = nCB) (c) n(A) ~ nCB) (d) none of these If A = {2, 3, 4, 5}, then which of the following relation is a function from A to itself. (a) fi = {(x,y):y=x+ I} (b)fi = «x,y):x+y>6} (c) /3= «x,y):x>y} (d) f4={(x,y):x+y=7} Let the functionf: R

· The vaIue 0 f I1m x--+2

(a)

I 5..J3

-4

~1+J2+x

-..J3 .

x-2 (b)

E

6~

IS

1

1

(c)

7..J3

(d)

8..J3


12.

x+sinx is equal to x-eosx (a) 0 (b) -I

455

lim

x4'"

13.

lim

e= -ell

xJI+Y

= -

x'-(l +y)

=y(l +x)or(x2 -y)+(x'-y-yx)=0

On squaring, we get

y.JI+x.

484 Remedial MatMmatics :::) :::) :::)

(~-y)+xy(x-y) =0 (x +y)(x-y)+xy(x-y) =0 x+y+xy=Oory(1 +x)+x =0

x

y=-1+ x Now, differentiating both sides, with respect to x we get

d

d

dy = (1 + x) d; (-x) - (-x)d;(l + x) = _ (1 + x) + x dx

Cl+x)2

O+x)2 (1 + x)2 .


Example 3: Find : ' when y= sin C. Splution: We have

y = sin c1

-ely = cos ex . -d (eX) = cos e...x . e-" dx

dx

dy dx =

c .cos C.

Example 4: Find : ' when y = cos ~ log sin x.


Solution: We have y = cos ~ . log sin x

r d d c dx = cos"X· dx (logsinx)+logsinx· dx (cos"x).

dy

=

cos~ ._._l-.~(sinx) smx dx

+ log sin x

dx

=

cos~ 1" r .1- 1-_·cosxogsmx·sm"x sinx 2 Fx

=

r log sin x x sin~ cos"X· cot x . 2~

Example S: Find : ' when y = tan (.; ). Solution: We have y = tan

·(-sin~) .~~


(~)

(.!.)

-dy = ~ tan dx dx x =

sec

2

(~)- ~ ~) = sec (~ 2

-(

)- ( -

x12 )

Differentiation

Example 6: Find : ' when y = ~x2

. dy SolutIOn: dx

d

= dx (e

3x 2

(RGPV B. Pharma 2001 ]

)

=~x2 .6x=6xo~x2

0

(RGPV B. Pharma 2003]

Example 7: Find : ' when y = log cos x Solution: We have y = log cos x

dy did - = -(log cos x) = -_o_ocosx dx -dx . cosx dx = _1_0 (-sinx) =-tanxo cos x

dy xX· Example 8: Find dx ofthefunctiony = x Solution: Here,y= x

<Xl

xX'"

:::> y=xY:::>logy=ylogx Now, differentiating with respect to x, we get

! dy = ydx :::>

(! -IOgX) y

dy dx

Example 9: Find :

2: x

y~(logx) + dy log x dx

dy =0

or! dy = y!.. + dy log x ydx x dx

dx

:::> -

dx

y2 =

(1- y logx)x

ofthe function log xy =

~ +y

0

Solution: Here, log(xy) =~+yorlogx+ logy=~+yo Now, differentiating both sides with respect to x, we get !..+! dy = 2x + 2ydy or (!-2Y)dy = 2x-!.. x ydx dx y dx x or

(2x 2 - I)

(1- 2y2) dy

y

Therefore, Example 10: ljx3 +

485

dx dy dx

x =

(2x 2 -I)y (I _2y2)x

0

y = 3axy;find dy dx

Solution: We have,x3 + y = 3axy Differentiating both sides with respect to x, we get

.!!-. ~ +.!!-.l = ~ (3axy) dx

dx

dx

486


=>

dy (3y-3ax)- = 3ay-3x2 dx

=>

dy 3(V-ax) - =3(ay-~) dx dy dx

Hence,

ay_x 2 y2_ax

=

Example 11: if ~I- x 2 + JI- y2 dy dx

Solution: We have putting

~I -

a(x-y), then prove that

=

P

~I- x 2

=

x 2 + ~I -

i

= a(x - y) x = sin 9 and y = sin 4> we get

~I- sin 2 9 + ~I- sin 2 4> 2cos(9;4>}os(9;4»

=>

cot (

9

; 4»

=

a{2cos(9;4>)sin(9;4>)}

=

a

9 - 4> = coc l a

=>

2

=>

sin-I x - sin-I y Differentiating w.r.t. x, we get 1

~I - x Hence,

e- sin 4»

cos 9 + cos 4> = a(sin 9 - sin 4»

=> =>

= a(sin

_ 2

1

JI - i

=

2 coC! a

dy =0

dx dy = dx

JI-

y2 1- x 2

III WGARITHMIC DIFFERENTIATION To fmd the derivative of a function, which is of the form of the product of functions or quotient of function or a function ofthe form (j(x--,<x). In this case, we take the logarithms on both sides and then differentiate. This process is known as Logarithmic differentiation.

Differentiation

~~~~~~~I Example 1: Find :

SOLVED EXAMPLES

~I~~~~~~

ofthefunctiony = xxx

XX Solution: Here,y= x Taking logarithms on both sides we get XX logy = log (x ) orlogy =xX logx. Now, differentiating w.r.to x, we get I dy d ) + -ex d X) logx - = x x -(logx ydx dx dx

..!. dy

ydx

Hence, Example 2: Find :

= xx.!. + XX (1 + logx)logx [.: ~exX) = XX (I + IOgx)] x dx

:

=y~-l +xXlogx(l + log x)].

:

= xxx [x x- 1 + XX 10gx(1 t log x)] .

ofthefollowingfunctions:

(i) y = ax-3 + bx1 + cx9/ 2 sinx

(ii)y=xX (iv) Y = (ax + bY (iii) y = (1 + xY (v) y = (cos x/og x. Solution: (i) We have, y = ax-3 + bx3 + cx9/2 sinx So,

dy =

dx

~eax-3 + bx3 + cx9/ 2 sin x) dx

= a..!!.-ex-3 ) + b..!!.-ex3 ) + c..!!.-ex9/ 2 sin x) dx dx dx =-3ax-4 + 3bx2 + c(x9/ 2

!

sin x +

!

ex9/2)Sinx)

=-3ax-4+3bx2 + c(x9/2cosx+~x7/2SinX)

en)

Here, y = xX. Taking logarithms on both sides, we get logy =xlogx. Now, differentiating w.r.to x, we get .Idy

-- =

ydx

d d x-(logx)+-(x)logx dx dx

487

488


or

1 dy 1 - - = x- + 1 logx ydx x

or

dy = y(1 + logx) dx

Hence,

dy dx =x'{1 + log x)

y = (1 + xy. Taking logarithms on both sides, we get logy =xlog(l + x). Now, differentiating both sides with respect to x, we get

(iii) Here,

~ dy

ydx

=

x~[log(l + x)] + ~(x)log(l + x) dx

dx

~ dy = x_I_+ I.log(l + x) I+x

ydx

: =y[l:X + log(l + X)]. Hence,

:

=(1

+XyC:

x + log(l + x)

J-

y = (ax + by Taking logarithms on both sides, we get = x log(ax + b)

(iv) Here,

logy

Now, differentiating both sides w.r.to x, we get

~ dy

=

or

~ dy

= x

1 d (ax + b) + I.log(ax + b) (ax+b)dx

or

~ dy

=x

I a + log( ax + b) (ax+b)

or

:

=

Hence,

:

=(ax+br[ax':b + log(ax + b)] .

ydx

(v) Here,

yd..

ydx

x~log(ax + b) + ~(x) ·Iog(ax + b) dx

dx

Y[ ax': b + log(ax + b)]

y = (cosx)logx

Taking logarithms on both sides, we get logy = logx log (cos x).

Differentiation

489


~ dy

ydx

=

logx~log(cosx) + ~logx ·log(cosx) dx

dx

or

~ dy = logx_l_~(cosx) + !log(cos x)

or

1 dy sinx 1 - - = -logx-- + -log(cosx) y dx cosx x

or

~ dy

or

:

cosx dx

y dx

ydx

x

= !log(cosx) - tanxlogx x

= (cosxl ogx

[~IOg(cOSX) - tanxlogx]

Example 3: Differentiate (sin xf with respect to x.


Solution: Lety=(sinxt

Taking log on both sides logy = log (sin log y = x log sin x. differentiating w.r.t. x

xt

1 .dy · IOgSlllX'-'X . d - = xd- lOgSlllX+ Y dx dx dx . =

1 d. Iogsmxx . 1 x·--·-smx+ sinx dx

cosx . x-.-+ 10gSlllx smx =x cot x + log sinx. =

:

= y(x cot x + log sin x) = (sinxt [x cot x + log sin x]

Example 4: Find : Solution: We have

Let Then

when y = xX + (sin x)logx


y = xX + (sin x)logx u =xX and v= (sinx)logx.

y=u+v dy du dv -=-+dx ~dx dx Consider u = xX . Taking log on both sides, log u = log xX log u = X )ogx.

... (1)

490


differentiating both sides w.r.to x 1 du

1 = x·-+logx. u dx x 1 du _ . - = 1 + log x u dx du dx = u(1 + log x)

du - =.r(1 + logx) dx , v = (sin x)logx.

Consider

...(2)

Taking log on both sides log v = log x . log(sinx). differentiating w.r.t. x Idv I I 1,·1 = ogx· -.-. cosx + ogsmx· -. v dx smx x = logx· cot x +! logsinx. x

dv = v(logx. cot x + !Iogsinx) dx x dv = (sin X)logx (log x . cot x + ! log sin x) dx x Using (2) and (3) in (1), we get :

= xX(l + log x) + (sinx)logx(logx. cot x +

...(3).

~logSinx).

ExampleS: Ify= (tan x)cot x + (cotx)tanx, find : y = (tanx)cotx + (cot x)tan x u = (tanx)Cotx and v =,(cotx)tan x

Solution: We have Let then

y=u+v

dy du dy -=-+dx dx dx u = (tanx)Cotx

Now taking log, we get

log u = log (tan x)cotx log u = cot x log (tan x) diff. w.r.t. x, we get 1 ! dy = cotx.__.sec 2 x + log(tanx)·(-cosec 2 x) u dx tan x = cosec 2x - cosec2x log(tan x) = cosec2 x(1-log(tan x)

...(1)

Differentiation

dy dx = (tan x)cotx . cosec 2x(1 -log(tan x»

Now

491

... (2)

v = (cotx)tanx

taking log, we get log v = log (cotx)tanx = tan x log (cot x) diffw.r.t. x, we get

.! dy

v dx

= =

tanx._I_.(-cosec 2x) + log(cotx).sec 2 x cot x -sec2 x + log (cot x)· sec 2x

dv dx = (cotx)tanx· sec2 x(log(cotx)-I)

... (3).

Using (2) and (3) in (1)~ we get :

=(tanx)cotx·cosec2x(I-log(tanx» + (cot x)tanx. sec 2x(log(cotx)- 1)

Example 6: Findthe :

ofthefollowingfonctions:

1 + cosx (il) - - sinx

(i) t? log sin 2x Solution: Here,y = t? log sin 2x

dy = ~(eX logsin2x) = eX ~(logsin2x) + ~(eX)logsin2x dx dx dx dx =

eX -1-~sin2x + eX logsin2x

sin2x dx

_._1_ cos 2x ~ (2x) + eX log sin 2x sm2x dx = 2t? cot 2x + t? log sin 2x. 1 + cosx y= sinx = eX

Oi) Here,

So,

dy d dx = dx

(1

+ cosx sinx =

)

sinx~(l + cos x) dx

(1 + cosx)~sinx dx

(sinx)2

sinx(- sin x) - (1 + cosx)(cosx) (sin x)2 · x-cosx-cos 2 2x -sm -(l+cosx)

=

(sinxp - (1 + cos x) (1 + cosx)(l- cos x)

= 1- cos2 x -1 1- cosx

492


Example 7: Find the derivative of the following function (t) ax + by + c = 0 (ii) Y = 4ax (iii) :?+y=if (iv) :?+Y=4d. Solution: (i) The given equation is ax + by + c = o. Differentiating w.r.to x, takingy as a function of x, we get d dy dy a-(x)+b- =o=> a+b- =0 dx dx . dx

dy=-!!. b· dx (ii) The given equation is y = 4ax. Differentiating w.r.to x, takingy as a function of x, we get

=>

dy dy 4a dy 2a 2y- = 4a => - = - => - = - . dx dx 2y dx Y

y

(iii) The given equation is:? + = if. Differentiating both sides w.r.t. to x, we get

dy dy dy 2x+2y dx =O=>2x= -2y dx => dx =

x -y'.

(iv) The given equation is:? + y = 4ax2 . Differentiating both sides with respect to x, we get

2x + 2Y dy = 8ax dx 2Y :

=8ax-2x=2x(4a-l)

dy = x(4a-l) dx y

ExampleS: Find : (t) x 3 +

ofthe following functions:

I = 3d

..

xm

(II) -

am

ym

+- m = 1 b

(iii) x = aCt - sin t), y = a(I - cos t). Solution: (i) Here, the given equation is x 3 + = 3d. Differentiating both sides, w.r. to x, we get

I

2 2dy 2dy ., 3x +3y - =6axor 3y - =6ax-3.r=3x(2a-x) dx dx

or

dy = x(2a - x) dx y2

.:\ H h· . . xm ym 1 (llJ ere, t e gIven equatIOn IS - + - m = . am b

Differentiation

Differentiating, both sides w.r to x, we get

1 d (m d x ) +1m-d- (m) y =-(1) am dx b dx dx 1 dy mx m-J m-J --+-m y =0 am bm dx

or

bmmxm- 1 + ammym-l dy =0 dx

or

or (iii) Here, x = a (I - sin I), y = a (1 - cos I)

SO, :

=a(l-cos/)and

~

=a[-(-sint)]

or

dy . dl =as1O/.

So,

dy = dy 1dt = a sin t dx dx 1dt a(l- c.os I) 2sint 12cosl 12

2sin2112

sint 1- cost

cost 12 s1Ol12

= -.-- =

t

cot-. 2

dy 1 So, dx = cot2". Example 9: Ify = Ax2? -x + 8 andfor x = 1, :

= 2, then obtain the value ofA.

Solution: Here,y=Ax2?-x+ 8, and for x = I, :

=2

So,

dy = A!!...(x2 e X )-!!...(x)+!!...(8) dx dx dx

dx

l

or Now, put

=>

x=1 dy =2 'dx 2 =A{e+2e} -I, 311 A=-=-=>A=-. 3e e e

493

494


~~~~I MISCELLENEOUNS SOLVED EXAMPLES ~I~~~ +x Example 1: Find the differential coefficient at x = 0, ify = aX + - I-x

~

Solution: We have y = aX + ~I + x

1- x

Differentiating w.r.t. x, we get

1(I+X)-1/2{10-X)-0+X)(-l)} -dy = a xloga+--dx 2 1- x 0- x)2

IJ§-X IJ§-X{ +

= a Xl oga+- - - {I-X+I+X} 2 1+ x (1- x)2 =

a X loga+- - 2

x

1

2 } (1- x)2

x=o

Putting

(:)X=o = aOlOga+~~~:~ {(1_20)2} 1 2 2 12

= loga+-·- =loga+ 1

Example 2: Differentiate sin- J

u

(1-

x) w.r.t. .j;.

I+x

x) ,v= .j;

Solution: Let = sin-I (1l+x Differentiating w.r.t. x

du dx

-;=1_=(=I=_=x=)~2

- (1 + x) - (1- x) dv I (1 - x)2 ' dx = 2.j;

I+x

l+x

=

~O + x)2

- (1- x)2

I dv .j;(1 + x)' dx Therefore,

=

(-2)

dv (1 + x)2 'dx

1 2.j;

du 1 / I -2 dv = ,- .j;(1 + x) 2.j; = 1 + x

=

1 2.j;

Differentiation

~+

Example 3: Ify=

l.

495

Then show that

dy = x-a dx 2x...r;;;

SOluti~n: We have

y =

~+

l

Differentiating w.r.t. x, we get

dy dx

= _1 _ I

..ra 2.[;

+..ra( __ 1 )

I

2x3/2

= ~(x-a) =

2x

x-a

,...-.

2xvax

X

Example 4: y = e

x+ex+e +

dy dx

then show that

=----L.


I_y· X

Solution: We have y = e

x+ex+e +

y

... (i)

= £f+Y

Differentiating w.r.t. x we get

!

= eX + Y ( 1 +

!)

= y ( 1+

!)

from (i)

!(1-y) =y

dy dx

=-L l-y

1. Find the derivatives (or differential coefficients) ofthe following functions: . (x + 1)(x + 2) (ii) y = log sin ~ (I) y= (x + 3)(x + 4) x4 _ 5x 2

(iii) y= 5x6 + 7x

2. Find

!

0) y=

1 + tan x

(iv) y= 1

of the following functions:

JC::)

(iii) y= log[.J(x + 1) - .J(x -1)]

(ii) y=

-tanx

.

rn

496


3. Find the derivatives of the following function~ : (i) y=(x-2)(x+2)(x-3)(x+3) (ii) y=(x+ l)(aJ-21) (iii) y = sin-I (tan x) (iv) y = (tan xy. 4. Differentiate the following functions: (i) y=(sinx)COSX

(iv) y = tf tan4 x.

(iii) y = sin x log x

5. Find :


(i) y = log sin x + cos-I(t!) + x4 sec x (ii) y = (logxY (iv) y=sec(~-2x+ I).

(iii) log (;-) = x + y

6. Find the derivatives of the following functions: (i) £?x cos 3x (ii) log (sin-I x 4) (iii) x 3 - ; - 3axy = O. Find :

for the following implicit functions.

7.xy=x+y 8. (~+IP=XY

9. sin(xy) + x =~-y y

2

10. ytf + 2x sin x = cos Y 11. sin (x + y) = log (x + y) 12. e

X

Y -

=

IOg(;-)

13. x· 2Y + 2x = Y 14. tan-I (~+ =9 15. 3 sin (xy) + 4 cos (xy) = 5 16. ycosx=x-y

I)

17. If

fY

~

V-; + fy

dy x-17y = 6, show that dx = 17x _ y

dy 18. Ifsiny=xcos (a +y) prove that dx

19. If

=

cos 2 (a + y) cosa

~I _ x' + ~I _ y' ~a(x'-l)prov.that dy ~ dx

I 2 2 -I Y dy x +Y 20. If logvx + y = tan - prove that - = - x dx x-y

X'%

l

l_x 6

Differentiation

(v)

~sin(a tan-1 x)= dx

481

cos(a tan- 1 x) ~(a tan- 1 x) dx

= ~cos(atan-l x). a+x

1. Differentiate sin x w.r.t. x.

2. Ify= 3 x4 tr + 5, then find :

.

3. Differentiate (.x2 + 7x + 2) (£f - sin x) 4. Differentiate (x - 2) (x - 3) w.r.t. x. using product rule. Differentiate the same after expanding as a polyomial. Verify that the two answers are the same. x 2 +ex 5. Differentiate y = I 20 w.r.t. x. ogx + sinx+cosx 6. Differentiate . w.r.t. x. smx-cosx

d (. 7. Evaluate dx sm x 2) 8~ Differentiate log sin x w.r.t. x. 9. Differentiate e COS X w.r.t. x.

10. Find:, if x = a cos G,y = b sin G. 11. Find : ' if x = a(G + sin G), y = a(l - cos G). 12. Find : ' ifx= a(t- sin t),y= a(l-cot t)

rRGPV B. Pharma 2001 )

13. Find : ' x = log t + sin t, y = l + cos t

[RGPV B. Pharma 2002),


2.

:

="

=

!

4 x (3x e + 5)

~(3x4eX)+~(5) dx

dx

= 3 ~(x4ex) = 3 [x4 ~(eX) + eX ~(X4)] dx dx 4 = 3[x £f+£f4~] =3~ £f(x+4).

dx

482


dy d . 5. dx = dx [(x2+7x+2)(eX-smx)]

=(x2+7x+2)· =(x2+7x+2)

!

(eX-sinx) + (eX-sinx)

!

(x2+7x+2).

[~(eX) - ~(SinX)] + (eX - SinX)[~(x2) +~(7x) + ~(2)]

dx dx dx dx = (x2+ 7x + 2) (eX -cosx] + (eX - sin x) (2x + 7). = (eX -cosx)x2 + (9 eX-7 cosx- 2 sinx)x + 9 eX -2 cosx-7 sinx.

dx

6. dy = ~(sinx+cosx). dx dx sin x - cosx

(sin x - cos x) . ~(sinx + cos x) - (sin x + cos x) . ~(sinx - cosx) dx dx (sinx - cosx)2 _ (sin x - cos x) . (cos x - sinx) - (sin x + cos x) (cos x + sin x) (sin x - cos x)2 (sin2 x + cos 2 X

-

2sinx cos x) - (sin 2 x + cos 2 x + 2sin x cos x) (sin x - cos x)2

2 --------;:(sin x - cosx)2 .

7. Lety = sin x2 = sin twhere t=x2 dy dt Therefore dt = cos t and dx = 2x. Hence, by chain rule, we have dy dy dt -=_0-

dx

dt dx = cosx2· 2x= 2xcosx2. 8. Lety = log sin x = log t, where t = sin x. dy 1 dt Therefore, dt = t . and dx = cosx.

By chain rule. we get dy

dt 1 = -.-cosx =cotx. dt dx smx

dy

-= _.-

dx

11. Wehave :

=a(l +cosO),

de =asin9

dy dy dx asin9 9 -=-= =tan-. dx d9 a(l + cos 9) 2

Differentiation

483

ANSWERS 1. xcosx+sinx 2. 3x3~(x+4) 3. (~-cosx)x2 + (9 ~ -7 cosx-2 sinx)x+ 9 ~-2 cosx-7 sinx 2 (log x + 20)(2x + eX) + (x +

4. 2x+ 1 6.

5. -2

10. -cote

(log x + 20)2

7. 2x cosx2

(sin x - cosx)2

eX)(~ + 10) 9. _ sinx eCosx

8. cot x

12.

11. tan e/2

cosec 2t (I-cost)

13.

t(i - sint) (I + tcot!)

III DIFFERENTIATION OF IMPLICIT FUNCTIONS A function which can be expressed in terms of independent variable x is known as explicit function. On the other hand, A function which is not explicit, is known as implicit function, or we can say A function which cannot be expressed directly in terms of independent variable x is implicit function. For example,:xY + ;f = a is an implicit function, because this function cannot be expressed in terms ofx. To find the :

of implicit function, we differentiate each

term with respect to x treating y as a function of x and then separating : . A method by which we find the :

of implicit function is known as implicit differentiation.

~~~~~~I Example 1: Find :

SOLVED EXAMPLES

l~~~~~~~

ofthe implicitfunction ~ + 2hxy + by + 2gx + 2fy + c = 0

Solution: Differentiating the given equation with respect to x, we get

2ax+2h(xdY +Y)+2bydY +2g+2fdy =0 dx dx dx

dy (hx + by + f) dx dy dx Example 2: Find : Solution: Here, ~

afthefunction

x~l+ y+ y~l+x

x..JI+Y + y~ =0 x..JI+Y = -

On squaring, we get

x 2(I +y)

=-

(ax + hy + g) ax+hy+g hx+by+ f·

=0.

y~ .

=;(1 +x)or(~-;)+(~y-;x)=O

484 Remedial Mathematics ~ ~ ~

(x2-y)+xy(x-y) =0 (x+y)(x-y)+xy(x-y) =0 x+y+xy=Oory(l +x)+x =0 x

y=--l+x Now, differentiating both sides, with respect to x we get d d dy = (l + x) d; (-x) - (-x)d;(l + x) = _ (l + x) + x ~

(l+x)2

dx

(l+x)2

----~

(l + x)2 .


Example 3: Find : ' when y= sin l? S.olution: We have

y = sin l? J J -dy, = cosex . d- ( eX) =COSe·e

dx dy dx

dx

= l?

. cos l?

Example 4: Find : ' when y = cos £ Solution: We have y = cos £

:


log sin x.

. log sin x = cos£·

~(lOgSinX)+lOgSinX. ~(COS£).

= cos£ ._._l-.~(sinx) + smx dx

logsinx.(-sin£).~£ dx

cos £ 1 . . I .1_1= -_·cosxogsmx·smvx sinx 2£

r

= cOSvX ·cotxExample S: Find : ' when y Solution: We havey =

=

log sin x x sin £

2£

. [RGPV B. Pharma 2001]

tan (.; ).

tan(~) -dy = ~ tan('!') dx dx x

=

sec2(~} ~ -(~) = sec2(~}( - :2)

= __ 1 sec2 x2

(.!.)x

Differentiation

Example 6: Find :

' when y = t?~

485


. dy d 3 x2 Solution: dx = dx (e )

=e3:;?- . 6x=6x· t?:;?-. [RGPV B. Pharma 20031

Example 7: Find : ' wheny = log cos x Solution: We have y = log cos x

dy dx

=

dId -dx (log cos x) =. cosx . dx . cosx

= _1_. (-sinx) =-tanx. cos x

dy Example 8: Find dx ofthe function y Solution: Here,y = x

=

x

xx. '"

XX '"

==> y =xY ==> logy= y logx Now, differentiating with respect to x, we get

~ dy

ydx

==>

=

y~(logx) + dy log x dx

(~log x) dy - ~ = 0 y dx x

Example 9: Find :

or

dx

==> dy = dx

i

~ dy

ydx

(1- ylogx)x

= y.!.. + dy log x x dx

.

ofthe function log xy = x? + ;- .

Solution: Here, log(xy) =x? +;- orIogx+ logy=x? +;-. Now, differentiating both sides with respect to x, we get

.!.. + J. dy x

or

(1- 2i) dy y dx

+ 2y dy or dx

(~ y

2Y ) dy = 2x -

(2x 2 -I)

x

dy _ (2x 2 - l)y dx - (1 _2y2)x .

Therefore, Example 10: If xl +

= 2x

ydx

?

dy = 3axy;find dx

?

Solution: We have x 3 + = 3axy Differentiating both sides with respect to x, we get

!!...-~ +!!...-l = ~(3axy) dx

dx

dx

dx

.!.. x

486


=>

(31-3ax) dy

=>

3(V-ax) dy =3(ay-x2)

dx

3ay-3x2

=

dx

dy = ay

Hence,

+ ~1 dy dx

~1 -

Solution: We have

2

i-ax

dx

Example 11: if ~1 - x 2

_X

y2 = a(x - y), then prove that

n

~1- x 2

=

x 2 + ~1- y2 = a(x - y) x = sin 8 and y = sin 4> we get

putting

~1- sin 2 8 + ~1- sin 2 4>

a(sin 8 - sin 4» cos 8 + cos 4> = a(sin 8 - sin 4»

=>

=

=> => 8 - 4> = coC! a 2 => sin-! x - sin-! y = 2 coC! a Differentiating w.r.t. x, we get

=>

1

~l Hence,

III

_

x

2

dy =0

1

~l -

i

dx

dy = / _ y2 dx 1- x 2

LOGARITHMIC DIFFERENTIATION

To find the derivative of a function, which is of the form of the product of functions or quotient offunction or a function of the form (f(x--,<x). In this case, we take the logarithms on both sides and then differentiate. This process is known as Logarithmic differentiation.

Differentiation

~~~~~~~I Example 1: Find :

SOLVED EXAMPLES

I

ofthefunctiony = xxx.

XX Solution: Here,y= x . Taking logarithms on both sides we get XX logy=log(x ) orlogy=~logx. Now, differentiating w.r.to x, we get 1~ d X) logx - = x x -d (I ogx) + -(x ydx dx dx

-.!.. ~ = xx..!.. + xX(l + 10gx)logx [.: !!.-(XX) = XX (1 + log X)] ydx

Hence, Example 2: Find :

x

dx

:

= y[!!-I + ~ logx(l + log x)].

:

=

XXX [xx-I +xxlogx(ltlog x )].

ofthe followingfunctions:

(i) y = ax-3 + b~ + cJJI2 sinx (iii) y = (1

(ii) Y =~

+ xy

(iv) Y

=

(ax + by

ogx

(v) y = (cos x/ . Solution: (i) We have, y = ax-3 + bx3 + cx9/2 sinx

~

S

dx

0,

=

!!.-(ax-3 +bx3 +cx9/ 2 sinx) dx

=

a!!'-(x-3 ) + b!!'-(x3 ) + c!!.-(x9/ 2 sin x) dx dx dx

= -3ax-4 + 3bx2 + c( x 9/ 2

!

sin x +

!

(X9/2)Sinx)

=-3ax-4+ 3bx2 + c( x 9/ 2 cosx +~x7/2 sin x ) y =~. Taking logarithms on both sides, we get logy =xlogx. Now, differentiating w.r.to x, we get

(n) Here,

. -.!.. ~ = x!!.-(logx) + !!.-(x) log x ydx

dx

dx

487

488


1 dy ydx

1 x

or

- - = x- + 1 logx

or

dy = yO + logx) dx

dy = r(l + logx) dx (iii) Here, y =(1 +xt· Taking logarithms on both sides, we get logy =x logO + x). Now, differentiating both sides with respect to x, we get Hence,

-

..!.. dy =

x..:{..[log(l + x)] + ..:{..(x)log(1 + x} dx dx

..!.. dy =

x _1_ + I . 10g(1 + x) l+x

ydx

ydx

Hence,

y[l:X +IOg(l+X)].

:

=

:

=(1 +xtC: x + log(l +

X))-

y = (ax + bt Taking logarithms on both sides, we get logy = x log(ax + b) Now, differentiating both sides w.r.to x, we get

(iv) Here,

1dy d d - - = x-10g(ax + b) + -(x) ·Iog(ax + b) ydx dx dx or

-- =

1 dy yd'C

x

1 d (ax+b)+l·log(ax+b) (ax+b)dx

"or

-- =

1 dy ydx

x

1 a+ 10g(ax+b) (ax+b)

b+ loge ax + b)]

or

:

= y [ ax:

Hence,

:

= (ax + bt[ ax: b + log(ax + b)] .

(v) Here,

y

= (cos x)logx

Taking logarithms on both sides, we get logy = logx log (cos x).

Differentiation

489


~ dy

ydx

=

logx~log(cosx) + ~logx ·!og(cosx) dx

dx

or

-- =

ldy ydx

1 d 1 logx---(cosx) + -log(cosx) cosxdx x

or

-- =

1 dy y dx

sinx 1 -logx--+-log(cosx) cosx x

or

~ dy

=

~ logecos x) -

or

:

=

(cosx)logx

ydx

x

tan x log x

[~IOg(COSX)-tanxIOgx]

Example 3: Differentiate (sin xl with respect to x. Solution: Lety=(sinxt Taking log on both sides logy = log (sin xt logy = x log sin x. differentiating w.r.t. x 1 dy y' dx

=

lRGPV B. Pharma 2002J

d · I . d x dx logsmx+ ogsmx· dx ·x.

I d. Iogsmxx . 1 = x·--·-smx+ sinx dx cosx . + logsmx smx = x cot x + log sin x. = X-.-

:

=

y(x cot x + log sin x)

=

(sinxt [x cotx + log sin x]

Example 4: Find: wheny =r + (sinx)logx y =r + (sinx)logx u =randv=(sinx)logx. y=u+v dy du dv -=-+dx "dx dx Consider u =r . Taking log on both sides, log u =logr log u = X logx.

rRGPV B. Pharma 20011

Solution: We have Let Then

... (1)

490


differentiating both sides w.r.to x 1 du

udx

=

1

X ._+ logx. x

1 du =l+logx u dx du dx = u(1 + log x) du dx = ~(1 + logx) Consider

...(2)

v = (sinx)logx.

Taking log on both sides log v = logx· log(sinx). differentiating w.r.t. x Idv = Iogx· -.-' 1 cosx + I ' 1 ogsmx· -. v dx smx x = logx.cotx+..!.logsinx.

x

dv = v(logx. cot x + ..!.logsinx) dx x dv = (sin X)logx (log x . cot x + ..!. log sin x) dx x Using (2) and (3) in (1), we get = XX (1 + log x) + (sin X)logx (log x . cot x +

:

...(3).

~IOgSinX).

ExampleS: Ify= (tan x)cot x + (cotx)tanx, find :

y = (tanx)cotx + (cot x)tan x u = (tan x)cotx and v =1 (cot x)tan x

Solution: We have Let

y=u+v

then

dy du dy -=-+dx dx dx u = (tanx)Cotx

Now taking log, we get

log u = log(tanx)Cotx log u = cot x log(tan x) diff. w.r.t. x, we get

1 dy u dx

- -

= =

1 cotx·--·sec 2 x + log(tanx)·(-cosec 2x) tan x cosec 2x - cosec2x log(tan x)

= cosec 2 x(1-log(tan x)

...(1)

Differentiation

= (tanx)CotX. cosec2x(l-log(tanx))

: Now

491

...(2)

v = (cotx)tanx

taking log, we get log v = log (cotx)tanx =tanxlog(cotx) diff w.r. t. x, we get Idy 1 2 2 - - = tanx·--·(-cosec x) + log(cotx)·sec x v aX cot x = -sec2 x + log (cot x)· sec2x

dv aX = (cot x)tanx. sec 2 x(log (cotx)-I)

...(3).

Using (2) and (3) in (l), we get :

= (tanxt otx . cosec2x(l-10g (tan x)) + (cotx)tanx. sec2x(log(cotx)-l)

Example 6: Find the :

(i)

~ log sin

ofthe followingfimctions :

1 + cosx (ii) - - sinx

2x

Solution: Here,y = ~ log sin 2x dy = .!!...-(e X10gsin2x) = eX .!!...-(logsin2x) + .!!...-(eX) log sin 2x

aX

aX

= eX

aX

aX

-1-.!!...-sin2x + eX logsin2x sin2x aX

= eX _._1_ cos 2x.!!...- (2x) + eX log sin 2x sm2x aX = 2~ cot 2x + ~ log sin 2x. (ii) Here,

So,

y= dy

1 + cosx sinx d

aX = aX

(I + )

cosx sinx =

sinx.!!...-(1 + cos x) - (1 + cos x).!!...-sin x aX aX (sinx)2

sin x(.... sin x) - (1 + cos x)(cos x) (sinx)2 . 2

2

_ -sm x-cosx-cos x _ -(1 + cos x) (sinx)2 - 1- cos 2 X -(1+cosx) (1 + cos x)(1 - cos x)

-I 1 - cos x

492


Example 7: Find the derivative of the following function (i) ax + by+ c = 0 (it); =4ax (iii) :? = if (iv) :? = 4d. Solution: (i) The given equation is ax + by + c = O. Differentiating w.r. to x, taking y as a function of x, we get

+;

+;

d dy dy a-(x)+b- =o=> a+b- =0 dx dx . dx dy a => dx (ii) The given equation is; = 4ax. Differentiating w.r.to x, takingy as a function of x, we get dy dy 2y- =4a=> dx dx

4a dy => 2y dx

= -

2a Y

=-.

(iii) The given equation is:? + ; = if. Differentiating both sides w.r.t. to x, we get

dy dy dy 2x+2y dx =0=>2x= -2y dx => dx =

x -y'.

(iv) The given equation is:? + ; = 4ax 2 . Differentiating both sides with respect to x, we get

2x + 2 y dy dx 2Y : dy dx Example 8: Find :

=

8ax

=

8ax-2x =2x(4a-l)

=

x(4a-l) y

ofthefollowingfunctions:

xm ym (ii) - + = I am bm (iii) X = a(t-sin t),y = a(1- cos t). Solution: (i) Here, the given equation is ~ + I = 3d. Differentiating both sides, w.r. to x, we get (i) x 3 +1=3d

2 2dy 2dy , 3x +3y dx =6axor 3y dx =6ax-3x-=3x(2a-x)

or

dy dx

=

x(2a-x) y2

" . . . xm ym (11) Here, the gIven equatIOn IS - + - m am b

=

1.

Differentiation

Differentiating, both sides w.r to x, we get 1d 1d d - - ( x m)+--(ym) = -(1) m am dx b dx dx 1 dy mx m-I m-l --+-my - =0 m am b dx

or

bmmxm- 1 + am mym-I dy = 0 dx

or

or (iii) Here, x=a(t-sint),y=a(l-cost)

dx

So, dt or

=

dy. a (1 - cos t) and dt = a[- (- SIn t)]

dy . di =asm t.

sint

dy = dy I dt = a sin t dx I dt a(1 - C:.os t)

So, dx

1- cost

2sintl2costl2 costl2 t = - - - = cot2sin2 tl2 sintl2 2. dy

So, dx Example 9: Ify

=

=

t cot"2'

A~t? -x + 8 andfor x =

1, :

Solution: Here,y=A~t?-x+ 8, and for x = 1, :

So,

=

2. then obtain the value ofA.

=2

dy = A!!.-(x2 eX )-!!.-(x) + !!.-(8) dx dx' dx dx

or

Now, put

x=l dy =2 'dx 2 =A{e+2e} -1,

=>

A=-=-=>A=-. 3e e e

311

493

494


I MISCELLENEOUNS SOLVED EXAMPLES I Example 1: Find the differential coefficient at x

~ +'x

= 0, ify = aX + - I-x

• Solution: We have y = a X + J§+x --

I-x

, Differentiating w.r.t. x, we get

ely dx

-

Putting

= a

xloga+--I (I+X)-1I2 {IO-X)-(1+X)(-I)} 2 1- x (1- x)2

=

a Xl oga+-I~-X{I-X+I+X} -2 I+x (1- x)2

=

2 } a X loga+- - - { 2 I + x (1- x)2

I~-X

x =0

(:1=0

= =

aOIOga+~J~:~ {(l_20)2} I 2 log a + - . - = log a + 1 2

12

Example 2: Differentiate Sin-/(I-X) w.r.t. I+x Solution: Letu= Sin-1(I-X),v= I +x Differentiating w.r.t. x

J;.

J; - (I + x) - (1- x) dv

du dx

r==(=1==)'="2

(1- x)2

1

' dx =

2J;

1- 1: :

=

l+x (-2) ~(1 + x)2 - (1- x)2 (1 + x)2 ' dv

J;(1 + x)' dx Therefore,

du

I

dv = - J;(1+x)

/

1

=

2J;

1

2J;

-2 =

I+x

dv

dx

=

1 2J;

Differentiation

Example 3: Ify=

~+

to

495

Then show that

dy x-a dx = 2x";-;;;

Soluti~n: We have

y =

~+~

Differentiating worot. x, we get

dy dx

=

_I _ I +

Fa 2£

Fa( __ 1 ) 2x3 / 2

I x-a = --(x-a) = - 2x3/2 2x";-;;;

Example 4: y = e

x+ex+eX +

dy dx

0

then show that

=-L

(UPTU B. Pharma 2003J

I-yO X

Solution: We have y = e

x+ex+e +

y = ?+y Differentiating wor.t. x we get :

... (i)

1

= eX + y ( + : ) = y

(I + : )

from (i)

dy(1-y) =y dx dy y -=-dx I-y

1. Find the derivatives (or differential coefficients) of the following functions: (x+l)(x+2) (ii) Y = log sin ~ (1) y= (x + 3)(x + 4) o

(iii) y =

2. Find :

(0 y=

x4 _ 5x 2 5x6 + 7x

(iv) y=

1+ tan x 1- tan x

0

ofthe following functions:

~C::)

(iiO y= log[~(x + 1) - ~(x -I)]

(ii) y =

JlGf}

(iv) y=(a2+5x+7r2

o

496


3. Find the derivatives of the following functions: (i) y=(x-2)(x+2)(x-3)(x+3) (ii) y=(x+ l)(aJ-21) (Ui) y=sin- I (tan x) (iv) y=(tanxt. 4. Differentiate the following functions: (ii) y = sin(e X ) +.!.. + log(x2 + x)

(i) y= (sinx)COSX

X

(iv) y = t? tan4 x.

(iii) y = sin x log x

5. Find :


(i) y = log sin x + cos-I(t?) + x4 sec x (ii) y= (logxt

IOg(~) =x+y

(Ui)

(iv) y = sec (x 2 -2x + 1).

6. Find the derivatives of the following functions: (i) cos 3x (ii) log (sin- I x 4) (Ui) x 3 3axy = O.

ea

-1-

Find :

for the following implicit functions.

7. xy=x+ y 8. (~+y)2=xy

9. sin(xy)+~ =~-y y 10. yt? + 2x sin x = cos Y 11. sin (x + y) = log (x + y) 12. e

X -

Y

=

IOg(~)

13. x·2Y+2x =y 14. tan- I (X2+y) = 9

15. 3 sin (xy) + 4 cos (xy) = 5 16. y cos x = x - y

HJ¥

dy x-17y 17. If -+ - =6, show that - = - - ' x y dx 17x-y dy cos 2 (a + y) 18. Ifsiny=xcos(a+y)provethat dx = cosa

19. If

JI-x' +Jl- y' ~a(x'-l)provethat dy ~ x'§ dx

I 2

20. If log" x + y

2

= tan

-I

Y dy X +Y - prove that - = - x dx x-y

i

l-x6

Differentiation

497

21. If EI + eV = EI + Y, prove that ely = _ eX (eY -I) dx eY(eX -I) 22.

If~=EI-Y,provethat

dy logx - = . dx (1 + logx)2

IRGPV B. Pharma 2001)

. dy x+tanx 23. Fmd dx' wheny = ---=2,.---x -I 24. Differentiate w.r.t. x

IRGPV B. Pharma 2004]

2

ax +h (ii) - - - sinx + cosx

[RGPV B. Pharma 2001, 20021

25. Differentiate with respect to x

t! +e-x t! _e-x

[~GPV

B. Pharma 2001

r

26. Differentiate with respect to x eX + tan x


cot x - x 2 5. Find :

' when --

(i) y = sin2(lo&,e sec x).

[RGPV B. Pharma 2002] [RGPV B. Pharma 2005] [RGPV B. Pharma 2005]

(ii) y=sin- I (.r+2x) (iii) y = sec(3x + 2) + (3x + I) \13


1.

. (x + I)(x + 2) (I) y= (x + 3)(x + 4) Taking log on both sides, we get logy= log (x+ I) + log (x+ 2)-log(x+ 3)-log (x+4),

Idy

Diff. w.r.t.x,

dy dx

~--y

I

1

y dx = x + 1 + (x + 2) {

2x + 3 (x + I)(x + 2)

I I (x + 3) - (x + 4)

2x + 7 } (x + 3)(x + 4)

2 2 2 = y{2X(X +7x+12-x2 -3x-2)+3(x +7x+12)-7 (x +3X+2») (x 2 + 3x + 2)(x 2 + 7x + 12) 2

2

(x + I)(x + 2) { 4x + 20x + 22 ) 4x + 20x + 22 = (x+3)(x+4) (x+3x+2)(x 2 +7x+12) = (x 2 +7x+12)2

498 Remedial Mathematics (iJ) y = log sin :?-

dy Diff. w.r.t. x, we have dx

=

1

2

,

-.--2 . (cos x ) 2x =2xcotxSIOX

1 + tan x (iii) y= I-tanx Diff. both sides by Quotient rule dy sec 2 x(l- tan x) + sec 2 x(1 + tan x) -=

dx

2. (i)y=

.

(1- tan x)2

2sec 2 x =---7" (1- tan x)2

J(I-

x) = (1- x)1/2 1+ x (1 + x)1I2

-

Diff. both sides by Quotient Rule dy = dx

_!_I_~(l+x)-!~ 1 2~ 2 (~ (1 + x)

_

1 {(l + x) + (1- x)} _

- 1

(1 + x)~l- x 2

(l + x)~l- x 2

- - 2'

-

(ii) Do as Question No.2 (i) (iii) y = 10g[.Ji+l- ~] , diff. w.r.t. x both sides we have

dy dx

1

=

I)

-

2~

2(..jx+i _~)~x2 -I

2~X2

1

-(..jx+i-~)

Y = (2:?- + 5x +

(iv)

(1

~x+l-~ 2~x+1

-I

7r2, diff. both sides w.r.t. x

dy = - 2 (4x + 5) = - 2(4x + 5) dx (2x + 5x + 3 (2x + 5x + 7)3 3. (i)y= (x-2)(x+ 2)(x-3)(x+ 3) = (x 2-4)(:?--9) Difference w.r.t. x by product rule

7t

dy dx =2x(:?--9)+2x(:?--4)=4~-26x. (ii) Do

as in above Question

(iii) y = sin-1 (tan x), diffw.r.t. x

dy

1

dx

~l- tan 2 x

-=

2 sec x

=

cos x

~cos2 x - sin 2 x

2 sec x

=

secx .Jcos2x

=

~

secxvsec2x

Differentiation

(iv) y = (tan x't, taking log on both the sides

logy=x log (tan x), 1~ x 2 Diffw.r.t. x - - = log(tan x) + --sec x y dx tan x ~ dx 4. (i)

=

y {log tan x + x sec x cosec x}

,= (tanx't {log tanx+x sec x cosec x} y = (sin x)COSX taking log on both the sides logy = cos x log sin x, diffw.r.t. x 1~ y dx

-- =

:

=

. I' cosx -smx ogsmx+cosx-sinx

y (cos x cot x- sinx log sin x)

= (sin x)COSX (cos x cot x

y= sin(eX ) +.!.. + log(x 2 + x),diffw.r.t.x x

(ii)

~

-

(iii)

(iv)

=

x x 1 1+ 2x cos(e )e - - + - -

x2

x2 + X

dx Y= dy = dx y=

sinx (log x), diffw.r.t. x byproduct rule sinx - - + cosx(logx) x eX tan 4x, diffw.r.t. x by product rule

~

eX tan4 x + 4 tan3 x sec2 x eX

dx

= =

5. (i)

- sin x log sin x)

y= ~

dx

eX tanX (tan3 x + 4 sec 2 x) log sin x + cos -I (eX) + x4 sec x, diff, w.r.t. x

cos x = -.- smx

eX

r.---;; + 4x

'11- e2x

3

4

sec x + x sec x tan x

eX 3 = cotx- ~ +x secx(xtanx+4) 2x 1- e (ii) y={logx't Taking log on both sides logy =x log (log x) Diff. w.r.t. x both sides

.!. ~

=

_x_(.!..) logx x

ddxO'

=

Y

y dx

+ log(log

x)

{_1_ X)} xl {_1_ + + log(log

logx

= (log

logx

IOg(lOgX)}

499

500


IOg(~) =x + Y

(in)

logx-Iogy =x + y, . . 1 Idy dy Dlffw.r.t. x both sIdes ;- dx = 1 + dx

y

dy dx

(1 +~) 1+.!.-x y =

dy = y(I - x) dx x(I + y)

y = sec (.x2 - 2x + 1), diff. w.r. t. x both sides

(iv)

dy =2(x-I)sec(.x2-2x+ 1) tan (.x2-2x+ 1) .fix = 2(x-I) sec (x- Ii tan (x- 1)2 y = ;.x cos 3x, diff. w.r.t. x

6. (i)

= 2;.x cos 3x + (- 3 sin 3x) e2x

:

=;.x (2 cos 3x - 3 sin 3x) y = log (sin -I x 4) diff. both w.r.t. x

(n) Let

dy

-

3x

I

sin- x4

dx (in) ~

1

=

4x3

4x3

~I- x 8

sin-I x4 ~I- x 8

x-===

-y- 3axy = 0, diff. w.r.t. x 2

-

3i

:

- 3a( y + x : ) = 0

dye y 2 +ax+x ) 2 -ay=o -dx dy x 2 -ay dx = y2 + ax .

7.

xy=x+y~y=

x

1

- - = 1+-x-I x-I

. dy-I Dlff. w.r.t.x, - = 2 dx (x-I)

8. (x 2 +li=xy 2

Diff.w.r.t.x, 2(x +i){2X+2Y

!}

= x: +y

Differentiation

=:)

{4y(x 2 +

501

i) - x} dy = y-4x(x2 + y) dx i) + i) - x

dy = y-4x(x 2 +

dx

4y(x

2

9. sin (xy)+ ~ =x2-y

... (1)

i

. Xdy) 1 2x dy = 2x- dx dy Dlff.w.r.t.x, cos(xy) ( y + - +2'-3"dx y y dx =:)

(

=:) (

=:)

xcos(xy) - -2X)dy +1 y3 dx

=

2x - - 1 - ycos(xy)

i

xy2 cos(xy) - 2; + y2 ) :

{xy2 cos(xy) - 2(xy2 - i

=

~-I-l cos (xy)(Multiplying byy)

- ysin(xy) +

l) dy

=

dx

2xy-1

-l cos (xy)

Putting the value of'::' from (1) y dy

2xy2

-l cos(xy) -I

=:)-=-----"---:=---~~--=----=-

dx

2ysin(xy) + xi cos(xy) - 2xi + 3i

10. yEf + 2x sin x =

cos Y

Diff. w.r.t. x, ye X + eX dy + 2 x log 2 sin x + 2 x cos x dx

= -

dy = - [ye X + 2 x cosx + 2 x sinx log2] (eX + sin y)

dx

11. sin (x + y) = log (x + y) Diff.w.r.t.x, COS(x+Y){I+ : }

12.

=:)

{COS(X + y) -

=:)

dy dx =-1.

Ef-Y=

_I_}

dy x+y dx

= _

=

X~y {I+ : }

I_}

{COS(X + y) _ _ x+y

x log- = logx-Iogy Y

Diff. w.r.t. x, we have eX - Y

(I -:)

= .; -

~:

sin y dy dx

50l Remedial Mathematics

:(;_eX-Y) ely =

~'_eX-Y

=

y(1-xe~-Y)

dx x(l 13. x.2Y+2x=y

ye

y,)

X -

Diff. w.r.t x, we have 2Y + x2Y log 2 ~- + 2 log 2 = :

x

2Y + 2x log 2 ely =

dx 14. tan -1

ely (1 - x . 2Y log 2)

=

dx

[2 + 2x IOg2] Y

1- x2Y log2 (.x2 +;l) = 9 =>.x2 +;l = tan 9

Diff. w.r.t. x, we have 2x + 2Y :

=

0

ely x =>dx=-y'

15. 3 sin (xy) + 4 cos (xy) = 5 DifEw.r.t.x, 3COS(XY){Y+X:}-4Sin(xy){y+x:} =0 ely {3xcos(xy) - 4xsinx(xy)} =-y{3 cos (xy)-4 sin (xy)}

dx

:

.

=

-(~) ... (1)

16. ycosx=x-y Diff. w.r.t.x, - ysinx + (cos x) : ely (1 + cos x)

dx

=

=

1- :

1 + Y sin x

ely - 1 + ysinx - ( 1 +XSinx) - - - - - -1- dx l+cosx l+cos+x (1 + cos x)

17.

Jf+H=6 =>y+x = 6Fxy =>.x2+;l-34xy= 0 Diff. w.r.t.x, 2x + 2Y: - 34( x: + y) =0

1 + xsinx + cos x (1 + cosx)2

Differentiation

503

x+ dy (y-17x)-17y =0 dx dy = x-17y dx 17x-y 18. siny = x cos (a + y)

!

Diff. w.r.t.x,

... (1)

(cosy) = cos(a +

y) -

xsin(a +

y{!)

dy {cosy + xsin(a + y)} = cos (a + y) dx dy {COSy + siny sin(a + y)} = cos (a + y) dx cos(a + y)

From(l) dy {cos(a+ y)cosy+sinysin(a+ y)} =cos2 (a+y) dx dy = cos 2(a + y) cosa

dx

1 2 19. -log(x + y 2) 2

=

tan -1 -Y x

xdy 2x + 2y dy _ 1 dx -y] ) 2 2(x + y2) dx - 1+ ~~ [ x 2

1

(

x+ydy=xdy-y dx dx x+ y= (x- y)dy dx dy = x+ y dx x-y 20. tr +e'=tr+ Y,

D·ff..w.r.t.x, e 1

x +eY dy dx -- e x + Y

~_~+Y= dy(-e>'+ex+ Y ) dx

dy = eX{l-e>') dx e>'(e X -I)

{I + dy} dx

504


ANSWERS 1. (i)

(iii)

2. (i) _

4x 2 + 20x + 22 (x 2 + 7x + 12)2 x 2(-10x 7 + 100x 5 + 21x2 - 35) (5x 6 + 7x)2 1

(iv)

1

2~x2

(iv)

(ii)

~(1- x 2 )

(iii) _

(li)

-1

3. (i) 4~ - 26x

2x cot x 2 2sec 2 x

(1- tanx)2 -1 2x.Jx(x + 1) - 2(4x + 5) (2x 2 + 5x + 7)3

(ii) 8~ + 6x2 -21

(iii) sec x~r-(s-ec-2-x-)

(iv) (tan xY (x sec x cosec x + log tan x)

4. (i)(sin x)COS x (cos x cot x - sin x log sin x) (ii)

~+

eX cos(eX) _

x .. ,

1

sinx

(m) - - + cos x og x x

5. (i) cot x -

h

1-e h

2: + 1 x +x

(iv) eX tan 3 x (4 sec2 x + tan x)

+ x 3sec x(x tan x + 4) (n) (logx)X [_1_ + log (log X)]

(1- x)y (iii) (1 + y)x

~gx

(iv) 2(x-1)sec(x2-2x+ l)tan{x2-2x +

1) 6. (i) ~ (2 cos 3x- 3 sin 3x) 2

oo') x -ay (111 2 ax-y 1- Y -1 7. - - or ---:X -1 (x _1)2

9.

2xy2

-l cosexy)-l

2ysinxy + xy2 cosxy _ 2xy2 + 3y2

8. 10.

y-4x(x 2 +i) 4y(x 2 + y2) _ (x) _[ye X +2x cosx+2xsinxlog2] eX + siny X

11.-1

12. L[xe x x ye

14. (-x/y)

Y

- Y

-1] -1

Differentiatioll

15. -(y/x)

16.

27. (i) 2 tan x sin (loge sec x)

(ii)

(iii) 3 sec(3x +2) tan(3x +2) +

_

50S

1+ysinx 1 + cosx 1

~1- (x 2 + 2x)2

(2x + 2)

1 2/3 (3x + 1)

SECOND ORDER DERIVATIVES

It is known that derivative of y w.r.t. x (if exists) is denoted by :

and is called the first

derivative ofy, Further, derivative Of: w.r.t. x (ifit exists) is denoted by

~:;

and is called the second

derivative ofy,

(dy) = second derivative ofy w.r.t. x, 3 2 Similarly, d ;, ~(d2:;) derivative of d :; w.r.t. x, dx dx dx dx 2

d2 y Thus dx

= -d -

dx dx =

=

In general dny denotes the nth derivative ofy w.r.t. x, dx n

Other Symbols 1. :

2.

is also denoted by y 1 or y'

d 2y dx is also denoted by Y2 or y"

3. :

is also denoted by Dy, where D is the operator

!.

d 2y -2

d2 is also denoted by d y, where D2 is the operator - 2 ' dx dx 4. Letj{x) be a differentiable function, then!' (x) denotes the first derivative of/ex) w.r.t. d x. Thus!' (x) = dx {{(x)} . . d Sllmlarly/" (x) = dx

if' (x)} =

d 2 {f(x)}

dx 2

= second derivative of/ex) w.r.t. x.

S86


STEP KNOWLEDGE To Find the Higher Ordered Derivatives Step 1. Let the given function be y. Step 2.

(i) Differentiate the given function w.r.t. x to get :

(ii) If both base and power in the given function are variables then first take logarithm and then differentiate to get : . 2

Step 3 . Step 4.

AT diffi . dy t d y )Vow I erentlate dx w.r.t. x to ge dx 2

·

If a particular expression is to be obtained, simplify the expression involved after obtaining first derivative making use of the given relation between x andy and if required also use the expression for first derivative obtained After simplification find the higher derivatives.

I SOLVED

EXAMPLES I

Example 1: Find the second derivatives ofthe followingfunctions (i) y = x 3 log x (ii) e6x cos 3x 3 Solution: (i) Let y = x log x Differentiating w.r.t. x, we get dy

I + 3x2 . log x x Again differentiating w.r.t.x, we get -

dx

= X

d 2y --2 =

dx

3

•-

2 I 2x + 3x . - + 6x . log x = 5x + 6x log x x

(ii) Lety = e6x cos 3x Differentiating w.r.t. (1) w.r.t. x, we get

..

:

=e6x·6cos3x+e6x(-sin3x)·3 = 6e6x cos 3x - 3e6x sin 3x ... (2)

Again differentiating (2) w.r.t. x, we get 2

d

dx

;, =

6[e6x. 6 cos 3x+ e6x (-3 sin 3x)] - 3 [e6x . 6 sin 3x+ e6x . 3 cos 3x] = e6x [36 cos 3x-18 sin 3x-18 sin 3x-9 cos 3x]

e6x (27 cos 3x - 54 sin 3x) = 27 e6x (cos 3x- 2 sin 3x) tanx Example 2: Ify = e , prove that =

cos 2 x

d

2

dx

dy ;' - (1 + sin2x)- =0

dx

... (1)

Differentiation

S07

Solution: Given,y=etanx .. logy=tanx ... (l)

~ dy

ydx

= sec2 x or dy = y sec2 x

... (2)

dx

2 dy cos x - =y

or

...(3)

dx

Differentiating again W.T.t. x, we get 2

cos 2 x d y _ 2cosxsinx dy = dy

dx 2

dx

dx

2

2 d y . dy or cos x dx 2 -(1+sm2x) dx =0.

Example 3: Ify =

~

d 2y

+ sinx,jind2 dx


Solution: We have y = ~ + sin x -dy = -d (X) e +d- (. smx) dx dx dx =~+cosx.

2 d y dx2

=

d (dy) d x dx dx = dx (e + cosx)

d x + -(cosx d ) = -(e) dx dx = ~-sinx.

d 2y dy Example 4: Ify = a cos (log x) + b sin (log x ), show that x 2 - 2 + x - + Y = 0 dx dx [RGPV B. Pharma 20031 Solution: Giveny = a cos (log x) + b sin (log x) ...(1) Differentiating (1) W.T. t. x, we get ..

dy = -asin(logx) . .!. + bcos(logx) . .!. dx x x

x . dy = - a sin (log x) + b cos (log x) dx Again differentiating again w.r.t. x, we get

or

d 2y dy 1. 1 x - - + 1· - = - acos(logx)· - - b sm(logx)·~ dx x x or

2 2d y dy . x dx 2 +x dx =-[acos(logx)+bsm(logx)]=-y [From(l)]

2 dy 2d or x dx; +x dx +y =0.

508


d 2y

dy

Example5:lfx=(sin- 1 xi,provethat(i-x2) dx 2 = x dx +2 Solution: Given. y = (sin -I x)2

... ( 1)

Differentiating (1) w.r. t. x, we get dy

dx =

squaring, we get

o-x

2

{:f

2

.

. Sin

-I X·

~l

1 r:-21 2 dy 2.-1 _ x 2 or" 1 - x- dx = Sin X

=4(sin-1 xi=4y[From(l)]

Again differentiating both sides w.r.t. x, we get 2 dy d y (dy)2 dy (l-x 2 )2-·-+(-2x) = 4dx dx 2 dx dx

Dividing both sides by 2 : ' we have 2 2 d y dy (l - x ) =x·-+2 dx 2 dx'

d 2y Example 6: Wheny = a sinx + b cosxfind dx 2 .


Solution: As y = a sin x + b cos x. on differentating w.r.t. x, we have dy

dx

=

a cos x -

b .

Sin X.

Again differentiating w.r.t. x, we get d 2y

.

- 2 =-asInx-bcosx dx

Example 7: lfx =

d 2y d 2y - 2 =-y=> - + y =0. dx dx 2 a(cos t + t sin t)

Y

=

a (sin t - t cos t) Find

d2 y

-2 .

dx

Solution: We have, x = a( cos t + t sin t) y = a(sin t - t cos t)

dx dt

=

. ) a( -. Sin t + t cos t + Sin t = at cos t

dy . t - cos t) . dt = a( cos t + t Sin = at Sin t

Differentiation

509

dy dy/dt atsint -=--=--=tant dx dx/ dt atcost 2

d y dx 2

=

!!....(dy) dx dx

=

~(dy). dt dt dx

=

dx

d 121 -(tant)·-- = sec t · _ dt at cos t at cos t

sec 3 t

at Example 8: Ify = x + tan x, show that 2

2 d y cos x·---2y+2x =0 dx 2 Solution: We have y = x + tan x

dy = 1 + sec 2 x dx d 2y

dx 2

= 2 sec x . sec x tan x = 2 sec2 x tan x

Now consider LHS d 2y 2 cos x . -2- - 2y + 2x = cos 2 x(2 sec 2 x tan x) - 2(x + tan x) + 2x dx = 2 tanx-2x-2 tan x + 2x =O=RHS

~~~~~~~I

EXERCISE 11.4

~I~~~~~~~

1. Find the second order derivative of the following functions: (i) log x lRGPV B. Pharma 20051 (ii) x2 + 3x + 2 (iii) x cos x (iv) If sin 5x

(v) sin (log x) dy

d 2y

2. Ifx=a(9-sin9),y=a(1-cos9),find dx,AIsofind dx 2 . 1- t 2 2t . 3. Ifcosx= - - 2 andsmy= - - 2 ,0$t$1. l+t l+t 2

Show that d ;' is independent of t. dx


d 2y 1t 4. Ifx= 3 sin t- sin 3t,y= 3 cos t-cos 3t, find - 2 att= -. dx

3

2

d y 5. Ify=~+tanx,showthat-2 =6x+2sec2 xtanx dx

d 2y dy 6. Ify= A t? + Bet/x, show that dx 2 -(p + q) dx + pqy= 0 [RGPV B. Pharma 2001]

ANSWERS (iii) - x cos x - 2 sin x

(it) 2

(iv)

2~(5

cos 5x -12 sin 5x)

(v)

-sin(logx) + cos(logx) 2 X

dy 8 d 2y 1 48 2 - = cot- - = --cosec• dx 2' dx 2 4a 2

4.

OBJECTIVE EVALUATION MULTIPLE CHOICE QUESTIONS (Choose the most appropriate one) 1. lfy = 2.x4 + 3~ + 2.x + 5 , then : (a)

8~ + 9x2 + 2

2. lfy = (

(a) d dx

3. -

(a)

Fx +

1 1 + 2"

(b) 8x4 + 9x3 + 2

then :

(d) None of these.

is equal to

(b) 1 - -

x

1

2x

1 (c) 1 - 2

x

1

(d) 1+-

2x

l-sin2x 1+ sin2x

sec2(~+x )

4. Ify = xx

(a)

l),

=

oo

x ---

,

(b)

then dy dx

y x(l+ ylogx)

sec2(~-x)

(c)

-sec2(~-x)

i

(c)

Y x(l-ylogx)

(d) None of these.

=

(d)

x(l+ ylogx)

(d)

i x(l+ ylogx)

s. ~ (x2 . e" sin x) = dx

(a) xe 2 (2sinx+x sinx+ cos x) (c)

xr:'(2sinx+xsinx+cosx)

(b) xe'" (2sinx + x sinx - cos x)

(d) None of these.

Differentiation

511

6. If y = x", then dy is equal to dx (a)

(b) x' + logx

1 + logx

~logx+~logx+Jlogx+--oo then

7. Ify= (a)

(b)

1

x(2y-l)

1 x(2y+ I)

(c) x'logx

:

1 (c)

8. Ify=logu,x+ 10gx10+ 10g1010+ 10gxX, then :

(a) (c)

xloge 10

(b)

x(logex)2

loge 10 x loge 10 -x(log e x)2

=? 2y -I

(d)

2y+ 1

(d)

x,[X

is equal to

x loge 10

-X10glO e

(d) None of these.

9. If y = x,[X ,then dy is equal to dx

2+logx

2$

(a)

(b)

2-logx 2$

,[X

[2

+ 10gx] I 2"x

[2 -2$

10gx]

(c)

x

(b)

tan X.lOg cosx+ cotx.logsinx (Iogcosx)2

10. Ify = logco sinx. then dYequals sx dx (a)

cot x.log cos x + tan x.log sin x (Iogsinx)2

(c)

cot X.lOg cosx + tan x.logsin x (Iogcosx)2

11. IfxJ' = (a)

~ -Y then

x+ y x(I+logy)

(d) None of these

dy equals: dx (b)

Y x(l+ logx)

12. Ify =eux cos bx then dy equals: dx (a) eOx (a cos bx + b sin bx) (c) eox (b cos bx + a sin bx)

(c)

x-y

x-y x(l+ logx)

(d)

x(l + logx)2

(b) eox (a cos bx - b sin bx) (d) eax (b cos bx - a sin bx)

x

I+e dy 13. Ify = - - x ' then - is equals. I-e dx

14. Ify = (I - x l/4) (I - xll2) (I + xl/4), then dy equals. dx (a) -I

(b) I

I-cosx . 15. IfJ{x) = - - . - , thenf'(7tl2) IS equal to. I-smx (a) 0 (b) (c)

(c)

$

00

(d)

(d) -x.

does not exist

512


FILL IN THE BLANKS 1. Iff(x) = Ix - 21 and g(x) = Ix I, then

g'(x) = - - ,

2x+I

2. Iff(x) = log tan -4-' thenf'(O) = _ _ ' 2X-I) ,2 dy 3. Ify=f ( - andf'(x)=smx ,then = _ _, x2 +1 dx 4. Iff(x) = ~ - al and g (x) = f[[(f{x»] then g '(x) = _ _ ' d(I+X2+X4) 6. If = ax + b then a = b= , dx l+x+x2 ' --'--

7. Ify=log(x+ JI+x2 ),thenY2(0)= _ _ ,

TRUE/FALSE Write 'T' for true and' F' for false statement dn

n

1. If u = ax + b, then ~ [[(ax + b)] is equal to - n [[(x)], dx n du

(T/F)

dy= - 5x 2. Ify=tan- I -4x- + tan-I (2+3X) - - then -, 1+ 5x2 3 - 2x dx 1+ 25x 2 dx d 2y dy d 2x d 2y 3 Ifx=f(t) andy =g(t) and i f =0 then - , = -,' dt dt 2 dt dt 2 . , dx 2

(T/F)

(T/F)

4. If xy = C2 , where C is a constant and if u is any function of x, then du du x-+y-=O dx dy

(T/F)

ANSWERS MULTIPLE CHOICE QUESTIONS 1. (a) 5. (c)

9.

(c)

13. (d)

2. 6. 10. 14.

3. (c) 7. (a)

(c)

(d) (c)

11.

(a)

15. (d)

(c)

4. (d) 8. (c) 12. (b)

FILL IN BLANKS

G)

, (2X-I)2{2+2X-2X2} 3. SI\1 -22' 2 X +I (x + I)

1. 1.

2. cosec

4. 1.

5. e

6. a = 2, b = - I

2. F

~

1

7. 0,

TRUE/FALSE 1. T

T

~

T

Differentiation

513

REFERSHER Do you know? After reading the chapter, you should be able to know the following concepts: • Derivative of a functionf(x) is the limiting value of 8x as 8x exist finitely and is denoted by :

ox - 0, provided the limit

.

• The derivative of the sum of two function is equal to the sum oftheir derivatives. d d d dx [ =>

_'!'fdt 6

SinFx Fx dx

f

Fx

Solution: Put

=

=t

.!.x-1/2 dx = dt 2

dx =2dt

Fx

t

= -.!.loo-(1-3x 2 )+C. 6!:>

IRGPV B. Pharma 20041

528


-- fsinx2dt=2 fsin+dt =-2cost+C fsinFxdx Fx

= - 2 cos

Fx + c.

cotx Example 7: Evaluate f I (. ) dx . og smx

IRGPV B. Pharma 20021

'. - f cotx dx SolutIOn. Let / - I (. ) og smx put log(sin x) = t or cot xdx = dt /=

dt

ft

=Iogt+ C

= log [log sin x] + C. Example 8: Evaluate

Solution: Let / =

sinxcosxdx

facos

2

sinxcosxdx

facos

2


. 2

x+bsm x dx

. 2

x+bsm x

a cos 2 x + b sin 2 x = t => (-2a cos x sin x + 2b sin x cos x) dx = dt => (2b - 2a) sinx cos x dx = dt dt sin x cos xdx = 2b-2a dt 1 /= I -- = logt t 2(b - a) (2b-2a)

put

f

I

2(b-a) sinx

Example 9: Evaluate

fsm. (x-a) dx

Solution: Put So that

(x-a)=t dx = dt f

log [a cos 2 x + b sin 2 x] + C.

sinx dx sin(x-a)

=

)UPTU B. Pharma 2007)

Fin(t+a) dt sin!

= fsintcosa~ cost sin a dt smt =

cos a fdt

+ sin a feot! dt

= cos a. t + sin a log I sin tl + C = (x-a) cos a + sin a log Isin (x-a)1 + c.

Integration

529

Example 10: Evaluate (i)

e

x

.

-SIUX

feX +cosx

dx

IUPTU B. Pharma 20011

rUPTU B. Pharma 20011 IUPTU B. Pharma 20021 (iv)

Jlogsecx tan x dx

fUPTU B. Pharma 20031

Solution: (i) Put t? + cos x = t So that (~- sin x) dx = dt ex - sin x

fe

X

•

dx =

+SIUX

Jdt - = log t + C = log (~ + cos x) + C. t x2

(ii) We have

/=

J1+ x

6

dx

putting x 3 = t => 3x2 dx = dt /=

.!.J~dx = '!'f-t- dt 6 2 3 1+ x

3 1+ t

f

1 ro = -1- -2t- dt = -logO + t2) + ~ 3 x 2 1+ t2 6

(iii) / = JO + log X )2 dx x Let

1 + log x

1 dx x

= t => -

/ =

= dt

fl2 dt = -t

3

3

+C =

(1 + log x)3

(iv) Put log sec x = t So that

1 - - sec x tan x dx = dt secx tan x dx = at tanx dx

flogsecx

= =

fl-t dt

= log t + C

log (log sec x) + C.

3

+ C.

530


Example 11: Find 2

l+x dx 1+x4

f

/=

Solution: Consider

(1+~)

=

Let (x -

~)

f( x--,;1)2

dx

+2

= t. Then (1 + x~ ) dx = dt

/-- ft 2 +(.J2)2 dl

/ =

= _1_tan- 1 _1_ + C

12

.J2

~ tan-1 ( ~l) + C.

Example 12: Evaluate the following integral:

(I)

f OgX x

(il)

dx

Solution: (i) Here, the given integral is f

logx dx. x

1 log x = t, so - dx = dt x

Put

Iogxdx - f d - t 2 c- (Iogx) 2 C - t 1- -+ +

f

So,

2

x

(ii) Here, put 1 + sin 2 x Put 2 sin x. cos x = dt Therefore,

III

rinxcosx dx' 1+sin 2 x

2

(on putting t = log x)

= 1

1 f 1 . dx = - tdt = -log t fsinxcosx l+sm x 2 2 2

1 log (1 + sin 2 x) + C. 2

=-

INTEGRATION BY PARTS

The method of 'integration by parts' is very powerful method of finding the integral. If the integrand is the product of two different functions, then the method of "integration by parts" can be used. If fix) and g(x) are two functions ofx , then

ff(x)g(x)dx =j(x) fg(x)dx- f[!f(x){fg(X)dx}]dx.

llltegration

531

The method of 'integration by parts, can be written into words as follows:

The integral of the product of two function = First function x integral of the second function - integral of (the derivative offirstfunction x integral of the secondfunction) STEP KNOWLEDGE

• There is no general rule to choose the first and second function, but remember the following points: (I) If the second function is not given the unity may be taken as the second function. (ii) The integral of second function must be known. (iii) If necessary, then the above formula can be applied more than once.

(iv) If the integral is of the form fxnf(x) dx where n is positive, then ~ must be taken as the first function. (v) Trick using here is ILATE which stands for I ~ Inverse Trigonometric, L ~ Logarithmic function A ~ Algebraic function, T ~ Trigonometric function E ~ Exponential function.

~~~~~~~I SOLVED EXAMPLES ~I~~~~~~ Example 1: Evaluate the following integrals: (i)

fx 2 sin x dx

Solution:

(ii)

f x 2e3x dx.

CO Here, the given integral is

fx 2 sin x dx .

Now, integrating by parts, taking x 2 as first function, we get 2

fx sinxdx

=x2

fsinxdx-

f{~x2 fSinxdx}dx

=

_x2 cos X - f{2x.(-cosx)}dx

=

-x2 cosx + 2

=-x2 cosx + 2

fx.cosxdx [x. fcosxdx- f{! (x). fcosx dx}dx ]

=-x2cosx+2 [xsinx-

fsinx~]

= - x2 cos x + 2x sin x + 2 cos x + C. where C is the constant of integration. (ii) Here, the given integral is fx 2e3X dx. Now, integrating by parts, taking x 2 as the first function, we get 2 3X

fx e

dx

=x2 =

fe

3X

x 2e3x

2 3X dx- f{! x fe dx}dx

-3-- f2X

e3x

J

dx =

x 2 e3x

2

-3--3 fx.e 3X dx.

532


Example 2: Evaluate (I)

x x dx Je----,,(1 +X)2


(il)

JeX(-sinx+2cosx)dx.


Solution: (i) We have

Jex-x-dx (1+x)2 =

=

fex (x+l)-l dx J (x+l)2

Jex[-I _ _ I x+l (x+l)2

Jdx.

Jex[f(x) + f'(x)]dx wherej(x) = _1_ x+l x x = I(x) Je dx+ fe f'(x)dx =

x x = ~ j(x) - Je f'(x) dx + Je f'(x)dx

applying integration by parts to the first integral taking 1

~ j(x)+C =~-

x+l

~

as the second function.

+c.

(iz) We have jeX(-sinx+2cosx)dx s =

j(_e 2X sinx + 2e 2x cosx)dx _e 2x 2e 2x -2--1 (2 sinx-cosx) +-2--1 (2 cosx+ sinx) + C 2 +1 2 +1

[

~

I

e .: jeoxsinbxdx= 2 2 (asinbx-bcosbx+C a +b eOX and fe~ cosbxdx = 2 2 (acosbx+bcosbx)+C. a +b e 2x = [-2 sinx+ cosx+4 cosx+ 2 sin x] + C 5 = e2x cos x + C.

Integration

Example 3: Evaluate Solution: We have

f~ log x dx

533

(UPTU B. Pharma 20081

x

f~logX dx x II

I

=> Integrating by parts, we get = 10gx{ -;) - f;{;) dx

x

f-I dx x

1

1

= -!...Iogx +

2

= --Iogx--+C. x x Example 4: Evaluate the/ollowing integrals. (I)

f(a2_~2)3/2dx

f(loge x )2 dx

(ii)

Solution: 0) Here the given integral is Put

=>

x = a sin

f(a

2

(iv)

n fx logxdx .

1 2 3/2 dx . -x )

e

dx = a cos e d8

So,

~ fsec 2 8d8 a

12 I { =-tan8+C=2

a

a

x

~a2 _x 2

} +c.

(ii) Here, given integral is f(lOge x)2 dx. Here, the second function is not given so, we take second function as unity. Therefore, f(loge x)2.1dx = (loge xi

fl dx -

2 f{! (loge x ) fl.dx }dX

=x(logex)2- f 210ge x !....xdx X

=x(logexi-2 flogex.1 dx =x(loge x i-2 [f lOge x

fl dx- f{! loge x fl dx}dx ]

= X (loge x)2 - 2x logex + 2 J-!-x dx X

=x (logex)2 -2x loge x + 2x + C.

534


2

X

(iiI) Here, the given integral is Jx e dx.

Integrating by parts, taking x2 as the first function, we get

22 2 2 2 2 Jx e dx = x Je dx- J{!x Je dx}dx

=x2~-2 Jx.exdx

=x2~-2[X JeXctx- J{!xJex ctx}ctx] x2 ~ - 2x~ - 2~ + C = ~ (x2 - 2x + 2) + C

=

(iv) Here, given integral is Jx n logxdx. Now, integrating by parts, taking logx as the first function, we get n n n Jx logxdx = log x Jx ctx - J{! logx Jx dx}dx s

JI

x n+1 x n+1 ;n+1 xn =logx·--- ---dx =logx·--- J--dx n+1 x n+1 n+l n+1 xn+1 I x n+1 = log x - - - - - - - - + C n+l n+1 (n+l) x n +1 = --(logx-n+I)+C n+1

[RGPV B. Pharma 2004].

Example 5: Integrate log x. Solution: Jlogx dx = J(logx).l dx =

(log x.) . x-

f!'.x dx x

Integrating by parts taking I as second function

= x log x - J1dx = x log x - x = x (log x - I) Example6:Evaluate Jxtan- I xdx.

[RGPVB.Pharma2001]

Solution: Jxtan- I x = tan-I x JXdx-

{!

tan-I x Jxdx

Jdx.

I x2 xx 2 =tan- x -2'J + 2 dx 2 I x I

1JI + x 2 -I dx

2

=tan-I x xx - - 2 2 2

= tan-I x. -I

=tan

XX

2

x 2

l+x2

-![ 2

Jldx-

J-I-dx] l+x2

x .1 I -I xx-- -x+ - tan x+ C. 2 2 2

Integration

2

Example 7: Evaluate flog (1 + x ) dx.

535


2

Solution: We have flog (1 + x )dx = log (1

+~) fl.dx

-

{!

log(1 + x 2 ) fl.dx

Jdx.

I

= log (1

+~)x- f~.2x.xdx I+x

J--dx 1+ x X2

=xlog(l+x2)-2

2

=x log (1 + x 2)-2

=xlog(l

+X

2

I+x 2 -1

J I+x 2 dx )-2[ •fldx- J~dx] I+x

=x log (I + x 2) -2 [x-tan- 1 x] + C =x log (1 + ~)-2x+ 2 tan- 1 x + C.

ExampleS:

dx . f+ x -I

[RGPV B. Pharm 2004]

Solution: Putx2 - I = t

=>

2xdx = dt dt xdx = - .

2

J-x -x-Id x 2

I I = -I JI-dt =-logt = -log(~-I)+C. 2 t 2 2

Example 9: Evaluate fxe ox dt. Solution: fxe

ox

dt =x

(RGPV B. Pharama 2003]

(~eax ) - (~.eax) dx

Integrating by parts taking x as find function. = !... eax _ a

~

reoxdx = !... eax __l_eax = ~(x-~) e ax + C. aJ' a a2 a a


Example 10: Evaluate Jxlogxdx. x2 2

Solution: Jxlogxdx = (log x). -

2

1 x J-dx x 2

Integrating by parts taking x as second function. =

~~ logx- ~ fxdx 2

2

1 2

1 2

2

4

= -x logx- -x +C.

{

536


Example 11: Integrate tan -Ix

IRGPVB. Pharma 20011

Solution: ftan- I xdx= ftan- I x.ldx Integrating by part x taking x as first function.

f

=xtan- I x- -1 - 2x -dx 2 1+x2

1 =xtan- I x- -Iog(l +~)+C. 2

Two Important Formulae eOX

ax cos bx dx = -2--2 (a cos bx + b sin bx) + C a +b ax ax (ii) Je sin bx dx = ~ (a sin bx - b cos bx) + C a +b Example 12: Evaluate the following integrals: (i) fe

(i)

fx 2 cosx dx

(ii)

3 fx cos x dx

(iii)

fx 2 sinx dx

(iv)

fex.cosx dx

2 Solution: (i) Here, the given integral is fx cos x dx . Integrating by parts, taking x 2 as first function, we get 2 2 fx cosxdx =x fcosxdx-

{~(x2)·fcoSXdxJdx

=x2 sinx-2 fxsinxdx

=~ sinx-2[ fx(-cosx)- f{~ (x) fSinxdx}dx] = ~ sin x - 2 [- x cos x - f cos x dx ] = ~ sin x + 2x cos x - 2 sin x + C. 3 (ii) Here, the given integral is fx cosx dx .

Integrating by parts, taking x 3 as the first function, we get 3 3 fx cosxdx = x fcosxdx-

{!

(x)3 fcosxdx Jdx

=x3 sinx-3 fx 2 sinxdx

= x 3 sinx-3 [x 2 fsinxdx- f{2x·fsinxdx}dx] = x 3 sinx+3x2 cosx-6[x.sinx- fsinxdx] =x3 sinx + 3x2 cosx- 6x sinx- 6 cosx + C.

where C is the constant of integration. = (x 3 - 6x) sin x + (3x 2 - 6) cos x + C

Integration

537

(iii) Here, the given integral is fx 2 sinx dx .

Integrating by parts, taking ~ as the first function, we get 2

2

fx sinx dx = x fsinxdx- f(2x fsinxdx)dx

=-~ cosx + 2 fxcosxdx s =-~cosx+2 [xsinx- fsinxdx] = - x 2 cos x + 2x sin x + 2 cos x + C (iv)Let

f = fex.cosxdx.

Integrating by parts, taking cos x as the first function, we get fexcosxdx =cosx.~+ fsinx.exdx.

Again integrating by parts, we get fex cosx dx = ex cos x + sinx~- fcosx.exdx

+

I = ~ cos x ~ sins x - I 2I = ~ cos x + ~ sin x

1=

III

!(~ cosx+~sinx)= 2

eX (cosx+sinx)+C

2

INTEGRATION BY PARTIAL FRACTIONS

If the given function is of the type I(x) ,wherej{x) andg(x) both are polynomials, then we g(x) may assume that the numeratorj{x) and denominator g(x) have no common polynomial factor and that the degree ofj{x) is less than the degree of g(x). If the degree ofnumeratorj{x) is greater than or equal to the degree of denominator g(x), then we have to dividej{x), by g(x) so that I(x) = hex) + rex) , where hex) is the quotient (a polynomial) and r (x) is the g(x) g(x) ; remainder whose degree is less than the degree of g(x}. The method in which we change the I(x) into partial fractions, depends on factors of the denominator. We know that every g(x) polynomial can be expressed as a product oflinear and irreducible quadratic factors with real coefficient. So, we have the following cases: (I) If all the factor of denominator are linear and non repeated, then for each linear non-

repeated factor (ax + b), there corresponds a fraction of the form _A_, where A is ax+b a constant, is to be obtained. (ii) If all factors in the denominator are linear but some of them are repeated, then for each linear factor (ax + b repeating n times, there correspond the sum of r partial fractions.

t

538


(iii) Ifthe factor in the denominator are linear and irreducible but quadratic factor and are non repeated, then for each irreducible quadratic factor a:? + bx +c, which occurs only one in the denominator, there correspond a partial fractions of the form

A 2

ax +bx+c

~~~~~~~I

SOLVED EXAMPLES

Example 1: Evaluate the integral

x+4

J3+2x-x

x+4

So,

dx

---- =

Putting x = 3, we get A =

7.. and putting x = -I, we get B = ~ 4

x+4

J3+2x-x 2

4

7 3 + 4(3 -x) 4(1 +x)

x+4 So, we have ----::3+2x-x 2

dx = 7.. J-I-dx~ J-I_ dx; Integrating 4 3-x 4 I+x =-


7 3 -log(x-3)+ -Iog(l +x)+C

4

3x+2 (x-2)(x+1)2

or

(3x+2)

4

3x+2

J(x-2)(x+ 1)

Solution: Here, the given integral is Consider

2

x+4 A B --+--. (3-x)(1+x) 3-x I+x x+4=(l+x)A+(3-x)B

Consider

Hence,

dx .

J3+2x-x

Solution: Here, the given integral is

x+4

2

~I~~~~~~

2

dx .

3x+2

J(x-2)(x+ I)

2

dx .

ABC (x-2) (x+l) (x+I)2

=

--+--+---

=

A (x + 1)2 + B(x- 2)(x + 1) + C (x- 2)

. 8 d . I On puttmg x = 2, we get A = '9' an putting x = - I, we get C = "3 . Now, on comparing the coefficient of x2 , we get A + B = 0

8 B=-A=--

i.e., So, we have

9

3x+2 (x-2)(x+ I)

8

2

8

---+--9(x-2) 9(x+l) 3(x+I)2

Integration

539

Hence I 3x+2 dx =! f_l_ dx _! f-I-dx+~ f 1 dx , (x-2)(x+I)2 9 x-2 9 x+1 3 (x+I)2 8 8 1 1 = - log (x - 2) - - log (x + 1) - - - - + C 9 9 3 (x + 1) =

!IOg(X-2)_ 1 +C 9 x+1 3(x+l) ,

Example 3: Evaluate (1) (ii) (iii)

fX:X2 fX-X dx

[UPTUB. Pharma 20051 [UPTU B. Pharma 2004]

3 3

f(x -l)(x - 2)(x - 3) dx X

(') · 1 nr vve have SoIUhOR:

- -12

x+x

=

[UPTU B. Pharma 2006) 1 = -I - -1- , reso I" , I tifaCtIons, ' vmg mto partla so x(1 + x) x I + x

fx~x2 = R~-l~Jdx I~dx- I-I dx l+x x =

= log lxi-log II + xl + c.

=IOgl~I+c. I+x 1 1 I ABC (ii) Here, - = = = - + - - + - - (say), 3 2 x-x x(1-x ) x(1 +x)(1-x) X I-x (1 +x) To find A suppress x in the given fraction and put x = 0 in the remaining fraction.

A=

So

Then

I = I similarly B = _1_ = ~ and C=-1/2 (1 + 0)(1 - 0) 1(1 + 1) 2 .

I I 1 -=-+--x-x 3 X 2(1-x) I - dx f-x-x 3

2(1+x)

= f'!"dx+'!"f-I_dx_'!"f-I_ dx

x

2 I-x

(l+x)

2

I I = log x - -log (1 - x) - - log (1 + x) + Cs

2

= log x -

=

~ 10g(1 - x 2 ) 2

~IOg[~)+c. 2 (1-x 2)

2

+C

540


(iiI) Here degree ofthe numerator is not less or degree lower than the denominator. We divide the numerator by the denominator till the remainder is of lessor degree than the denominator

~ ABC ------- = 1+--+--+-~-Q~-~~-~

~-Q

~-~

~-~

3

3

We have

1 1 2 A = (1-2)(1-3) = "2,B= (2-1)(2-3) =-8

and

3 27 C = (3-1)(3-2) = 2"

3

..

x3 (x-Q(x-2)(x-3

= 1+

I 2(x-l)

8 (x-2)

27 2(x-3)

------+---

3

Hence

I(x-l)(x-2)(x-3) x dx - II dx + I dx I(x-2) 8 dx I 27 dx z(x-l) + 2(x-3) =x +


27 "21 log 1x-11-8Iog Ix-21 + 2" log 1x-31 + c.

7X 2 +3x+l x(x + 1)

f


7x 2 +3x+l x(x + 1)

f

To change the integrand into partial fraction, first we divide (7~ + 3x + 1) by (~+ x), because the degree of numerator must be less than the degree of denominator. So we get 7x 2 +3x+l 1-4x - - - - =7+-x(x+l) x2+x So,

f f

2 l 4x 7x- +3X+l - = 7 d x +-- x(x+l) x 2 +1

f

l - 4X =7x+ -2--dx x +x

f

... (1)

l - 4x Now, to finds -2--dx. x +x

f

1-4x A B - =-+-2 x +x x x+l => 1-4x =A (x-l)+ Bx putting x = - 1 => B = - 5 and putting x = 0 => A = 1

Let

Thus,

1-4x 5 x2+x =~- x+l

... (2)

Integration

541

I-4X = II--5 I- 1d x =Iogx-5 log (x + 1) I-2--dx x +x x x+I

So,

Hence, by equation (1), we get 7x2 +3x+I x(x+I) dx=7x+logx-5log(x+I)+C.

J

Ix2 +xx _ 6 dx Solution: /= I dx = J( 2;( 3) dx. x +x-6 xx+ Example 4:

2

Let

[RGPV B. Pharma 2002J

X

x A B =--+-(x-2)(x+3) (x-2) (x+3)· x =A(x+3)+B(x-2)

Putting x =-3 in eq. (1), =>-3 =-B =>B= Putting x = 2 in eq. (1), => 2 = 5A => A = x Ix 2 +x-6

=

...(1)

~. 5

~. 5

~f_I-dx+~f-I-dx 5 x-2

5 x+3

2 3 = -log(x-2)+ -log(x+3)+C. 5 5

Example 5: Evaluate Solution: Let! =

f(2x + I)(x x

J(2x+I)(x x

2

2

+1)

+ I)

dx

dx

. x A Bx+C puttmg (2x + 1)(x 2 + 1) = (2x + 1) + (x 2 + I) x

A(x 2 +1)+(Bx+C)(2x+l)

(2x+1)(x 2 +1) (2x+I)(x2+1) x =A (x2+ 1)+(Bx + C)(2x+ 1) putting x = -1/2 and 0

when

1

x=-2

-.!. = ~A=-~ when =>

2 4 5 x =0 O=A+C

C=~

5

[RGPV B. Pharma 2004J

542


equating the coefficient of x 2 is both side of equation o =A +2B

=~.

B = A

2

5

x dx=-'3.f dr: +~JX+2dx 2 + I) f(2x + l)(x 5 (2x + I) 5 x 2 + 1 . 2 J dx 1 Jxdx 2 J 1 ="5 (2x + 1) + 5" x 2 + 1 +"5 dx x 2 + 1 dx . =

Example 6: Evaluate

f(x

2

putting

5

10

-

f(x

2

5


dx + l)(x-l) ]-

Solution: Let

-~log(2x+I)+~log(x2 +1) + '3. tan-I x+C

dx +l)(x-l)

A

Bx+C x 2 +1

----;;;----- = - - + - -

(x 2 +I)(x-I)

(x-I)

I = A (x 2 + 1) + (Bx + C) (x - 1). x=landO.

::::>

putting When x = I

2A

::::>

=

I

A =112

::::>

when x = 0 ::::> A - C = 1 ::::> C =112-1 =-112 equating the coeff ~ = 0 in both sides of the equation. A +B=O

B=-~. 2

Now f

2

1

(x +I)(x-I)

dx =.!. f_l_dx_'!'fx+I dx 2 x-I 2 x2 +1

=

~ 2

f-I- dx - .!. f xdx -.!. f-I- dx x -I 2 x2 +1 2 x2 +1 .

= .!.log(x-I)- ~log(~+ 1)- ~tan-IC+ C. 2 4 2 2x+ 1 dx Example 7. Evaluate f (2x + 3) (3x _ 4) .

'.

- f(2x+3)(3x-4) (2x + I) dx

SolutIOn. LetI-


Integration 2x+l (2x + 3)(3x - 4)

Let

A (2x + 3)

B (3x - 4)

---+---

(2x+ 1) = A (3x-4) + B(2x + 3) x = 4/3 in eq (1)

=> putting

543

=>

~+I=B(~+3)

=>

17 11 11 -B=-=>B=3 3 17

...(1)

3 x= -2

When

-3 + 1 A ( - ~ - 4)

=>

=

-2= -17 A 2

A=~.

=>

17

f(2x+3)(3x-4) 2x + 1 dx

~

=

1

17 f(2X+3)

dx + .!.!. dx 17 f(3X-4)·

2

11

17

51

.

= -log(2x+3)+ -log(3x-4)+C.

2x dx Example 8: Evaluate (x -I)(x + 3) .

f

Solution: Let (

=>

(RGPV B. Pharm 20021

2x A B 1)( 3) = - - + - - . x- x+ (x-I) (x+3) 2x (x-I)(x+3)

A(x+3)+B(x-I) (x-I)(x+3)

2x = A (x + 3) + B (x - 1) => putting x = - 3 in eq (1) => - 6 = - 4 B

(1)

B=~=~.

=>

4 2 put in x = 1 in eq (1) => 2 = 4A => A = 112

2x dx f(x-I)(x+3)

1

=

2"

f(x-I) dx 3 f dx +2" x+3

1 3 = - log(x-I)+ -log(x+3)+C.

2

Example 9: Evaluate

x2 2

f;x -x

- 2x

2

dx

(RGPV B. Pharma 2001)

544


Solution:l=

fx

x2 3

2

-x -2x

dx =

f x(x

=I

2 2X

-x-2)

dx

x dx=I x dx (x 2 -x-2) (x-2)(x+ 1) .

x A B =--+--. (x-2)(x+I) (x-2) (x+l)

Let

x (x-2)(x+I)

A(x+I)+B(x-2) (x-I)(x+l)

x =A (x + 1)+ B(x-2) B = 1/3.

putting x =-1 in(J)

~.

Putting x = 2 in eq (l) , => 2 = 3A => A =

fx

3

x2

_x 2 -2x

dx =

3

~ I-1-dx+! I 3 x-2

dx

3 x+I

2

1

= -log (x - 2) + - log (x + 1) + C. 3 3 Example 10: Evaluate the integral

I( x- I)~2x+ 1) dx.•


I(x -1)~2X + 1) dx .

.

ConsIder

=>

x (x-I)(2x+I)

A

B

(x-I)

(2x+l)

= - -+---

x=(2x+I)A+(x-I)B.

. cr x = 1 we get A · cr x = - -I we get B = -I = -3'land puttll1 N ow, puttm b ' b 2' 3

Thus,

x (x-I)(2x+I)

Hence, f

x dx = !f_l_dx+!f-I- dx (x-I)(2x+l) 3 x-I 3 2x+I

---+--3(x-l) 3(2x+l)

=

1

"3 log (x -

1) +

1

"6 log (2x + 1) + C

2


f(x-I)x (x-2) dx.


f(x-I).~ (x-2) dx.

3

2

(1)

integration

x2 ABC D = --+--+--+-(x-I)\x-2) x-I (x_I)2 (x-I)3 x-2

Let

=>x2=(x-li(x-2)A +(x-l)(x-2)B+(x-2)C+ (x-I)3 D put x = I => C = - I put x = 2 => D = 4 Now, we want to find the coefficients A and B. On comparing the coefficient ofx3 in (1). we get A +D=O. Now comparing the constants terms in (1), we get -2A + 2B-2C-D =0 Putting the values of C and D, we get A =-4,B =-3

545

... (1) ... (2)

2

x 4 3 I 4 ---::---- = - - - - - - - - + - (x-I)\x-2) x-I (x_I)2 (x-I)3 x-2'

Thus, Hence

,

x2 I 3 f- -I d x - f--dx+4 1 f--dx I dx=-4 f-dx+f(x-I)3(x-2) x-I 2 (x_I)2 (x_I)3 x-2 3 I =log(x-I)+ --+ 2 +4Iog(x-2)+C x-I 2(x-l)

(X-2)

3 I =4log - - + + +C. x-I x-I 2(x-I)2

Evaluate the following Integrals 1.

Fx dx JSinFx

3.

J(4x+2)~x2 +x+l dx

sec 2 (2 tan- 1 x) dx 5. J 1+x2 7. J cosec: dx log tan2 9.

11.

J4x3~5-x2 dx

J.

dx x x sm-+tan2 2

4.

fcosecx.log(cosecx-cotx)dx

6.

J1- sin 2x dx x+cos 2 x

8. JI-cotx dx 1+ cot x

546 Remedial Mathematics HINTS TO THE SELECTED PROBLEMS

fit

1. /=

dx

Fx

Put

=t

_l_dx =dt

2.Jx

~

=2dt

Fx

J =-2 cos

I dx - =t=> - 2 =-dt x x

Put 2 J=- cos tdt

f

= -

=

.Jx +C

i

l+cos 2X) 2 ( ':cos x = -2 - -

rrl+cos 2t) 2 dt

=- J~

I(1 + cos2t)dt, Integrating

-i (1+ Si~2t)+C

= __ 1 -!sin~+C,

2x 3. J

=

4

x

f(4x+2)~x2+X+ldx

P~

~+x+l=t

=>

(2x+I)dx=dt J =2 It ll2 dt, t 3/2 = 2-+C 3/2

=

4 _t 3/2 +C 3

=.i (~+x+ 1)3/2+C 3

4. /= fcosecx,log(cosecx-cotx)dx log (cosec x - cot x) = t

Put I

- - - - - ( - cosec x cot x + cosec2 x) dx = dt cosecx-cotx

J= ftdt

=

t2 1 -+C= -{loglcosecx-cotxl}2+C

2

2

lntegration

tan-I x = 1

Put

_l_dx =dt l+x2 2

1= fsec (2t) dt, Integrating

tan2t

.

= ~+C

puttIng the value of t, we have

= -1 tan (2 tan_I x) + C 2

6. 1=

f I - sin 2x dx 2

x+cos x Put

x + cos 2 x = t (l - 2 sin x cos x) dx = dt (l - sin 2x) dx = dt

1= Jdt, Integrating t I = log t + C = log Ix + cos 2 xl + C

7. 1=

x

cosecx

flogtan~ dx; put log tan -2

= t, differentiating

2

x)

I We get -I- ( sec 2 - -dx =dt x 2 2 tan2

=> =>

dx =dt 2 sin x / 2 cos x / 2 cosec x dx = dt t =

dt

.

f-,Integratmg t

=

log t + C

=

log log tan

putting the value of t

I

8. 1= JI-cotx dx = Js~nx -cosx dx l+cotx smx+cosx Put cos x + sin x = t => - (sin x - cos x) dx = dt

~I + C

547 \

548


T= -

dl

ft

= -log 1 + C,

Putting the value of 1 / =

9. /=

-log I cos x + sin xl + C

f4x3~5-x2 dx

= f4x3~5-x2dx = f4x2.~5-x2 ·xdx 2 dt Put5-x-=I=>-xdx= -

2

/=4 f(5-t)v't( -

~)

=-2 f(5v't -p/2)dt, Integrating t 3/2 2 512 } =-2 { 5 3/2 -Sf +C

20 13/2 +~t512 +C , putting the value of t

= -

3

5

=_ 20 (5_x2)3/2+ ~(5-x2)512+C 3

5

X

10. /- f(x+l)e

-

cos 2 (xe x )

dx

Put x eX = t => (eX + x eX) dx = dt => eX (x + 1) dx = dt / =

Jcosd~ t =

fsec 2 t dt ; Integrating

= tan t + C, putting value of t =

tan (xeX) + C

ANSWERS I 1. -2 cosJ;

3. ~ (x2 + x + 1)3/2 + C

2x 4 x 1 4. - {log Icosec x - cot xl}2 + C 2

5. '!'tan (2 tan- 1 x) + C

6. log ~+ cos2 xl + C

7. log Ilog tan ~I + C

8. -log Icos x + sin xl + C

3

.; .~

1 1. 2 C 2. ----sm-+ s

2

Integration

549

10. tan (xeX) + C 11.

J21 log Icosec (1t/4 -x/2) - cot (1t/4 + x/2) + C

~~~~~I

MORE SOLVED EXAMPLES

Exampel:Evaluate

f~dx.

1

1

eX

l+~

I+e

Solution: I

I~~~~~~

l+e

= f--dx = f--dx = f - dx x x l+e-

eX

Putting 1 + eX = t~ eX dx=dt

1= fdt = logltl + C = logll + eXl + C t

sin2x

fa sm x+b cos x dx sin2x Solution: We have 1= f a sm x+b cos x Example 2: Evaluate

2. 2

2

2. 2

2

2

2

Let a2 sin 2 x + b 2 cos 2 x = t ~ (2a2 sin x cos x - 2b 2 sin 2x cos x) dx = dt ~ (a2 sin 2x- b2 sin 2x) dx = dt

. dt sm 2x dx = -2--2

a -b

Putting in (1), we get /=

=

ft(a

dt 2

2

-b)

dt = -2--2 log It1/ + C

a-b

~ log 1a2 sin2 x + b2 cos2 xl + c. a -b

dt

Example 3: Find

fx{l +x )

Solution: Let! =

fx(1+xn) =

n

dt

f nxn-1dxn X

(1+x ) (multiplyby~-l into Nr. and Dr.)

put

(I+~=t~xn=t-l

~ - 1 dx

=

dt ~ ~ - 1 dx = dt n

550


f

1 dt 1= -;; (t-1)t

Therefore,

=

Integrating

I

Solution: Let/=

1 __ !)dt t

n

= ~ {log It n

=

Example 4: Evaluate

n_ J\J-1

~

2

11- log ItI +

c = ~ log It -11 t

n

~IOg/~I+c n xn +1

(2x+5)

fvxI

(by partial fraction) +C

( ... putting the value of t)

. dx

+3x+ 1

f (2x +5) = f (2x2+3) +2 dx Jx +3x+l Jx +3x+1 _J (2x+3) dx+ J 2

~x2 +3x+ 1

I

Suppose

J (2x+3)

=

~x2 +3x+1

1

2

~x2 +3x+ 1

dx

~+3x+2 = t

put

(2x+ 3) dx = dt

f~

Now

11 =

12 =

f-Jx2+3x+l = 2 (23991)

=

2dx

dx •

1/2

x + x+---+

4 4

=210 g

!( x+%)+~x2 +3X+l!+C2

1=11 + 12

Now

Example 5: Evaluate

J cotx

dx.

log sin x

Solution: Let/=

J cotx

dx

logsinx Putting log sin x = t:::::;> _._1_ cos x dx = dt smx cot x dt = dt Then

I = fdt = log ItI + c

t

= log I log sinxl + C

dx '" (1)

Integration

Example 6: Evaluate Solution: Let/=

fFxx(1x +1 dx

1

fFxx(

x +1

dx

Putting Fx + I = t, differentiating w.r.t. x

_1_dx =dt 2Fx

~

=2dt

Fx

Then

j

2dt r = = 2 log It + C = 2 log I + II + C. t '

"X

f

Example 7: Evaluate Jsec3 xtanxdx .

j= Jsec 3 xtanxdx

Solution: Let

j= Jsec 2 x(secxtanx)dx

•

Putting, sec x = t, differentiating w.r.t. x, we get

secxtanxdx=dt 3

/=

Then

t rt 2 dt = -+C J' 3'

Now, putting the value of t, we have 3 _ sec x C j --+.

3

2

logx Example 8: Evaluate f--dx. x Solution: Let

j =

=

flogxx

P

2

dx

log x dx

x

Putting log x = t, and differentiating We get

1 - dx x

=

dt

2t2 /=2 ftdt = T+C=F+C

Then

After putting the value of t, we have I = (log x)2 + C. Example 9: Evaluate

fx

2

tan-I x 3 l+x

6

dx

551

552


Solution: Let Let us put tan-I x 3 = t, differentiating

1 l+(x )

2_

---:-3-::-2 . 3x

dx - dt

2

x dt --dx =1+x6 3 Then

1=

t2

1

3' Jt dt = 6" + C

Putting the value of t, we gets (tan-I x 3 )2 j=

In

6

+C.

n1sin- 1 x

Example 10: Evaluate •

Solution: Let! =

l_x2

dx .

IenlSin-l x ~ dx

,,1- xputting sin- x = t, differentiating w.r.t. x,s we get 1

1

,-:--;; dx = dt

"1-x2 enlSin7"1 x

(After putting the value of t)

---+C. m

1. Evaluate the following integrals: (I) (iiI)

(2+3:0 x )dx g

I(8-6xi 1 dx

2

(ii)

f3sec (3x+9)dx

(iv)

f

(ii)

Jsec xtanxdx

(iv)

fxsinx 2 dx

2;':3

2. Evaluate the following integrals: (I) (iil)

f sin2x dx cos 2 2x

fxlogx 1 dx

7

Integration

3. Evaluate the following integrals: (l)

f

a

b+ce x

eX

dx

(ii)

2 pec (1ogx) dx x 4. Evaluate the following integrals: (iii)

(l)

fcosecxdx

(iii)

f~dx 1+x8 .

f-d x I +e 2x 2 sec x dx 1+ tanx .

(iv)

f

(ii)

f

cotx dx logsinx

3

S. Evaluate the following integrals:

f2. IOg logxdx x (iii) NCa 2 -x 2 )dx (/)

(ii)

fxJ']+; 1 dx

(iv)

fxsin xcosxdx.

(ii)

fxcos 2xsinxdx

(ii)

fJ(l J2dx dx + cos2x) .

(ii)

fJl +sinxdx

(iv)

flog(l;X2) dx. x

(ii)

f xll2 log x dx

(iv)

seX -[x(1ogx)2 +2Iogx]dx.

(ii)

f-(xlogx+l)dx x

(iv)

fcosx.c~s 2x.cos 3xdx.

6. Evaluate the following integrals: (I)

fxcos 2xdx

(iii)

f

1

x2~(1 +x2)

dx

7. Evaluate the following integrals: Ci)

flog sin-I x dx

~(l-x2)

8. Evaluate the following integrals: (i)

fXdx +x

.JI

(iii) fcos 4 xdx

9. Evaluate the following integrals: (/)

fxe

3X

dx

(iii) f l+x dx (2+x)2 10. Evaluate the following integrals: (l) (iii)

f

tanx dx secx+cosx

f

sin2x+l

~(x+sin2 x)

dx

x

eX

SS3

554


11. Evaluate the following integrals: (/)

f.}1 + cosx dx

(ii)

r~S2x dx.

(ii)

J

(iv)

J(sinx -cosx)2 dx.

smx

12. Evaluate the following integrals: (i)

(iii)

Jcotxdx dx logsinx

f cosx-sinx

dx

(cosx+sinx)2

x+3

~(1-x2)

dx


fx{I+(logx)2} I dx

(ii)

J

6x-8 3x2 -8x+5 dx

(iii) r o t dx .


xex dx f(X+I)2

cos2x dx (cos x + sinx) 15. Evaluate the following integrals: (iiI) J

(ii)

4x 3 J -4 d x 5x +7

(iv)

Fin~x-a) dx. smx

(ii)

f

(iv)

J

(il)

feX +e1 2x dx

(tv)

x 2 +8 fdx x 3 +4x .

(ii)

f(x -1)(2x x dx + 1)

3

(l)

(iii)

J(x-a)(x-b)(x-c) x dx f

3xdx (x_l)2(x+3)

16. Evaluate the following integrals: 3x 2 (l)

f(X+l)(X+2)(X+3) dx

f

2x+ 1 dx (x+2)(x_l)2 17. Evaluate the following integrals: (iii)

(i)

(iiI)

f

x2+4x+l dx x 3 +2x2 -xx2

f(x 2 + l)(x 2x dx 2 + 3) .

x-l (x + 1)(x 2 + 1) dx

x2 dx. (x-a)(x-b)

Integration

555

18. Evaluate the following integrals: (I)

fx.sin-I xdx

(ii)

fxsin-3 xdx

(iii) fsinxlog(cosx)dx

(iv)

fsinxlog(secx+tanx)dx.

19. Evaluate the following integrals:

r

X 7 + dx 2x+3 20. Evaluate the following integrals:

(i)

fsec 4 xtanxdx.

(1)

HINTS TO THE SELECTED PROBLEMS 6

7

2. (ii)

fsec xtanxdx = fsec x secxtanxdx s, Put sec x = t and then integrate

fxlogx 1 dx

(iii)

1 Put log x = t => - dx = dt x

Put ~ = t => x dx = dt 2

. 2dx (iv) t = 'f )xsmx / =

f..!..sintdt =- ..!..cos t + C=-..!.. cosx2 + C 2 2 2

adx 3. (l) /= - b+cex =

f -x fb~-x ae dx +c

[divide Nr and Dr by eX]

Put b e-x + c = t => b e-x dx = dt / =

ba fdt t

= -

ba log t + C

=-::b log (c + be-X) + C eX

f-l+e- d x

(ii) /=

2x

dt • Put eX = t => eX dx = dt, / = f--2' i~tegrating I+t = tan-I t + C = tan-I eX + C 2

(iii) /

=

sec (log x) dx f

x I

1

Putlogx=t=> -dx=dt/= fsec 2 tdt, integrating x = tan t + C = tan (log x) + C

556

Remedial Mathematics 2

. ) / = Jsec x dx dx puttmg . 1 + tan x = t ( IV l+tanx 4. (I) /= fcosecxdx = J-.-1-dx = smx

~ x 2sin-cos-

J

2

2

2

= ~ fsec x / 2 dx [dividing Nr and Dr by cos2 x!2] 2

Put tan ~ 2

tanx!2

=

t 1 2 x - sec -dx =dt

2

2

dt. . / = Jt" mtegratmg

= log t + C = log ( tan ~) + C 2 2 5. (iii) f ~ a - x dx = f ~ .1 dx, integrating by parts =

~a2_x2

.x- fi(a2-x2rI/2(-2x)Xdx

=X~a2_x2 =

x

a 2 _x 2 _a 2 - f

12 va -x 2

-

~ a -x

dx

f~a2-x2 +a 2 f h 2 2 a -x

=x~a 2 -x 2 -/+~ sin- 1 ~+C a 2I = x I

va2 -x 2

/ =

~ I 2

+ ~ sin- 1 ~ + C a 2

va

2

+ ~ sin-I ~ + C -x 2 a 2

(iv) /= fxsinxcosxdx =

i Ix(2sinxcosx)dx

=

i fxsin2xdx (integrating by parts)

=

i fx( -C~2X) -

R-C~2X)dx

=

xcos2x sin2x C +--+ 4 8

=

~ (sin 2x-2x cos 2x) + C 8

Integration

6. (il) /= Ix cos2x sin x dx =

(iii)

~{x(2SinX cos2x) dx

=

~ fx(sin 3x -

=

~ fxsin3x dx -

=

~[x( -C~S3X) + Si~3X _ x(-cosx) + sin x] + C

=

~(sin3x 18

sin x) dx fxsinx dx, integrating by part

3xcos3x) + .!.(xcosx - sinx) + C

fx 2Q 'put x I +x2

2

=

tan 9 and integrate

log (sin- i x) . ~ = dx, put sm- i x l-x 2 Taking I as II function

7. (1) /=

(il) /=

f

J .,fidx

f

..II + cos2x

.,fidx

~2 cos 2 x

=

= t and integrate by part.

Isecxdx = log (sec x +tanx)+C

8. (t) /= f xdx = f{(x+I)-I} dx= f{.jl+x--I-}dx .jl+x =

(it)

.jl+x

.jl+x

[~(l+X)3/2 _2CI+x)I12 ]

NI +sin x dx Use: ..JI + sin x =

X 'x x cos 2 -+ 2·2X sm -+ 2sm-cos-

2

222

x . x)2 ( x . x) = cos'2+ sm '2 ( cos'2+ sm '2

t 4 (iii) lcos x)dx Use: cos4 x = (cos2 x)2 = . Slog(1+x (IV) /= 2

2

)

(

I + c0s 2x 2

)2

dx

X

= =-

=-

flog(l+x)2.~dx, integrating by parts x

f x(l+x) 1 2\ dx .!. log (1 + xl) + I 2 dx l+x .!. log (I + xl) x

X

= -

2

.!. log (1 + xl) + 2tan- i x + C x

557

558 Remedial Mathematics 9. (I) 1= fxe 3x dx, integrating by parts taking x as first function. (il) I = Jx ll 2 log x dx, integrating by parts taking x as first function.

+x )2 dx = S(21+x

(iii) 1=

f2 + x - I { 1 1} 2 dx = J - - 2 dx. integrating (2+x) 2+x (2+x) 1

=log(2+x)+ - - + C (2+x) x

(iv) 1= feX[x(logx)2 + logx]dx = Je [(lOgX)2

+~logX Jdx

2ex = fe x (log x)2dx + J-logx dx x integrating by parts, I integral 2 2ex = Je X(log x)2 - f-(logx) eX dx+ f-(logx)dx x x

f

S

tan x dx = sinx dx = J sinx dx secx+cosx cosx(secx+cosx) l+cos 2 x put cos x = t =:> -sin x dx = dt -dt = --2 integrating = coC1 +C=coC1(cosx)+C

/=

f1+t sin2x+1 I . dx f,\}x+sm x

(iii) t =

2

Put x + sin2 x + t =:> (1 + 2 sin x cos x) dx = dt =:> (1 + sin x) dx = dt 1=

f~

(iv) /= fcosxcos2xcos3x dx =

=

i

=

±

;integrating = 2~x+sin2+x +C

i

f(2cosxcoS2x)cos3x dx

J(cos3x + cosx)cos3x dx =

~ f(2cos 2 3x+ 2cosxcos3x)dx

JW+cos6x)+cos4x+cos2x} dx . . 1( sin6x sin4x sin2X) mtegratmg = - X + - - + - - + - - +C 4

= 11. (i) 1=

Use

6

4

2

~(l2x+2 sin 6x+ 3 sin4x+ 6 sin2x) + C 48

fJ1 + cosx dx I+cosx =2cos 2 ~ 2

Integration

r~S 2x dx

(ii) ] =

smx

Use:

12. ] =

cos 2x = 1 - 2 sin2 x

cotx

. dx ; put log sin x = t => cot x dx = dt fIogsmx t =

dt. . mtegratmg

Jt'

= log t + C = log x sin x + C

.. f rzdx x+3 (ll)]=

vI - x~

=

f rzdx+ x f rzdx 3 vI - x~

vi - x~

=

f rz+3sm xdx . -I x

vi - x~

Put 1 - x 2 = t in I integrating x dx = _ dt 2

=-~f~

i

+3sin- x+C

lJi

=----+3 sin-I x + C 2 t(112) =-

(iii) / = 13. (i)

]=

(ii)

. dx put cos x + sin x = t and then integrating f(cosx+smx) dx 1· fx{1 + (logx)2} putlogx=t => -dx=dt x 2

dl.

.

f-1+1- 2 ' mtegratmg

= =

cosx-sinx

~1-x2 +3 sin- i x+ C

tan- i 1 + C = tan- i (log x) + C

f

6x-8 dx 3x2 -8x+5 Put 3x2 - 8x + 5 = 1 and integrate

]=

(iii) fCosFx dx

Fx Put Fx = 1 and integral

14. (l) /-

f(xxe+ x1)2 dx --

X

f(x+l-1)e (x + 1)2

Now do same as in Question 9 (iv). (iii) ] =

fcosx+smx cos 2~ dx

Use: cos 2x = cos 2 X

-

sin 2 x

1 1)

f x( = e x + 1- (x + 1)2 dx

559

560

Remedial Mathematics (iv) [= rlll~X-a) dx = pinxcosx.-sinacosx = dx

smx

smx

= f(cosa-sinacotx)dx =

x cos a - sin a log sin x + C 3

15. (I)

x J(x-a)(x-b)(x-c) dx

~

~

~

= J(a-b)(a-c)(x-a) + (b-a)(b-c)(x-b) + (c-a)(c-b)(x-c)

After breaking into partial fraction =x+

~ ~-~~-~

(il) [=

~

log(x-a)+

0-~0-~

log(x-b) +

2 ~-~~-~

log(x-c)+C

J(x+l)(x x-1 dx = n __I_+_2_1_)dx By partial fraction 2 +1) Jl x+l x +1

Now integrating we have I -log (x + I) + -log (x 2 + I) + C 2

/ =

2

(iii)

j=

2

2

x dx J(x-a)(x-b)

=

{a (a-b)(x-a)

f

-

} dx b (a-b)(x-b)

b2

a2

=x+ --log (x -a)- --log(x-b)+C a-b a-b 3

16

.

j=

3x dx J(X+l)(x+2)(X+3)

=

=

{ 3 f 2(x+l)

. . -12- + 27}dx, mtegratmg (x+2) 2(x+3)

3 27 -log(x+l)-12Iog(x+2)+ -log(x+3)+C

2

2

JeX (ldx+ eX) = ( -eX1 - -1+l)dx e-- dx, mtegratmg = (x e -eX eX + 1 X

(il)

f

dx = eX + e2x

=_ 2

(iii) /=

).

~ log (eeX

+ 8 dx = J x Jxx +4x x(x 3

1

2 2

•

~ + IOg(e + 1)+ C eX eX X

X

+ 1) + C=-

+8 +4x)

=2Iogx- -log(x2+4)+C 2

dx=

( 2 - -x-) dx x x 2 +4 (By breaking into partial fraction)

Integration

17. (iii) /=

561

2x

f(x 2 + 1)(x2 + 3) dx

Putx2=t:::::>2xdx=dt

f(t+I~:t+3) = (2(/+1) 2(t~3»)dt

/=

1 1 = -log(t+ 1)- -log(t+3)+C

2

2

1

,2

1

,,

1

(X2 + 1)

= -log(.x-+ 1)- - log(x-+3)+C= -log - +C. 2 2 2 x 2 +3

fx sin-I x dx, integrating by parts, we get 1 1 x dx ,= -sm x . -I dx = -sm x- f r:--? x+- f -x r:--?

18. (1) / =

X

2

•

2

-I

2

2

2

-I

2

2\fI-x2

2

\f1-x2

2 2 = _x sin- 1 x+.!.{f l-x dx- J--=dx=} 2 2 ~I-x2 ~I-x2 =

2 x . -I x+1 -sm 2 2

{J

r:--?I 2 dx \fl-x-

J~1-x2 dx

}

x . -I I x 2 I . -I . -I = -sm x+- [{ -~ +-sm x} -sm ] x +C 2 2 .... 2 2 2 2

x . -I x+-x\fl-x1 r:--?I 2 I. -I x+ C = -sm --sm 2 4 4 (ii)

fxsin 3 xdx Use: sin 3x = 3 sin x - 4 sin3 x and then integrating by parts.

(iv) /= fsinxlog(secx+tanx)dx, integrating by parts. II

I

= -cos x log (sec x + tan x) - f-cosxsecxdx = - cos x log (sec x + tan x) + x + C. 20. (I)

3

Jsec 4 xtanxdx = fsec x (sec x tan x) dx Put secx= t:::::> sec x tan x dx = dt 3

t

4

sec 4 x

/ = Jt dt = 4+ C = - 4 - +c.

562


ANSWERS 1. (i) (2 + 3logx)2 6

(ii) tan (3x + 9)

(iii)

1 6(8-6x) ,

. 1 (IV) -log(2x+3).

2 1 2. (i) -sec x 2

(ii) .!. sec 7 x

(iii) log log x

(il) tan-I €f

(iii) tan log x

7

I (iv) - -cos x 2 . 2

3. (i) - ~ log (c + be--,) b (iv) log (1+ tan x).

. x 4. (I) log tan "2

(ii) log (log sin x)

5. (i) log x (log log x-I)

(ii) -2 log (cosec 8 + cot 8) where x

122

x

a

2 .

(iii) "2"a--x-+T sm

4

-;;'

(ii)

~ (sin 3x 18

3x cos 3x) + .!. (x cos x - sin x)

2

..j1;;2

(i ii)

x 7. (i) sin-I x [log sin- 1 x-I] 8. (i)

tan2 8,

I (iv) g(sin2x-2xcos2x).

-I(X)

6. (i) .!. (2x sin 2x + cos 2x)

=

(ii) log (sec x + tan x),

[~(l+X)3/2_2(\+X)I/2J

(ii) 2 ( s in

~-

cos

(. .) 41(34 + sm. -4x8 + sm. 2)x + C /II

-

- X

(iv) -.!.log (1+~) + 2 tan- 1 x.

x

e: (x-~) x

9. (i)

1

(iii) - - + log (2 + x)

2+x 10. (i) coe 1 cos x (iii)

2~(x+sin2 x),

11. (i) 2.fisin!",

2

(ii)

~x3/2Iogx_ix3/2 3

9

(iv) (log xi €f.

(ii) €f log x,

Civ) J...(2sin6x+3sin4x+6sin2x+ 12x).

48

(ii) log tan!" + 2 cos x. 2

~) + C

Integration

12. (i) log log sin x

(iv) 3 sin-I x-

~(1-x2),

(iv) x+ !cos2x 2

(iii) - . - - SInX+cosx

13. (1) tan-I log x 14. (i)

(ii) log (3x2 - 8x + 5)

_e_

(iii) 2 sin

Fx .

x

(ii) ! log (5x4 + 7) l+x 5 (iv) x cos a - sin alog sin x.

15. (i) x+

a3 ~-~~-~

log(x-a)+

+

b3 ~-~~-~

(iii) sin x + cos x

log(x-b)

c3

1 log (x-c) (ii) -log(x2+ I)-log(x+ I), (c-a)(c-b) 2

~IOg(X-l)_i_l_

(iii)

x+3

10

4x-1

2

a b2 (iv) --log(x-a)---log(x-b)+C a-b a-b 16. (i)

563

i

2

log (x + 1)-12Iog (x + 2) + 27 log (x + 3),

2

(ii) log

(l+e:X)-e

lx ,

(iii) ..!..IOg(X-I) __I_,

3

x+2

x-I

(iv) 2Iogx-! log (x2 + 4). 2 17. (i) log (x2_1) x+2

(ii) ..!.. log (x - I) + ! log (2x + I), 3 2

1 x2 + I (iii) -log-2- +c. 2 x +3 2

.) x . -I x ~I 2 I._I 18 • (1 -Sin x+-\ji-x- --Sin 244 I (ij) 36 [3x (cos 3x- 9 cos x) - sin 3x + 27 sinx],

(iii) cos x (I -log cos x), (iv) x - cos x log (sec x + tan x). 19. (i) 2x + ..!..log (2x + 3)

2

20. (i) ! sec 4 x 4

(ii)

..!.. log (1 + x 3 ). 3

~~

564


III

DEFINITE INTEGRAL

Ifj(x) is a continuous and non-negative function over a closed interval [a, b] then

J:

f(x)dx

is called the definite integral ofj(x) between the limits a and b (b > a). If Jf(x)dx = F(x) + c, then

J:

f(x)dx = [F(x) + c]~

F(b)- F(a) is a defmite value.

=

Here, a is called the lower limit and b is called the upper limit and the interval called the range of integration.

ra, b] is

Remarks

· J:

J:

• If F(b) - F(a) in not a definite value, then the integral

III 1.

= a and a = b.

f(x) dx represents the area bounded by the lines x

f(x)dx is indefinite.

PROPERTIES OF DEFINITE INTEGRALS

J: f(x)dx =0.

2. TIre value of definite integral is independent of tIre variable of integration. Le.,

J:

f(x)dx

=

then

J:f(x)dx

=

f(u)du.

[F(x)]: =F(b)-F(a)= J:f(u)du

3.

J:

4.

J;f(x)dx+ J:f(x)dx= J:f(x),wherea

21 = 1t J;logsinxdx = 21t

7t/2

J0

= 21t(

,

logsmxdx

CJ:

a

f(x)dx

=2

J: f(x)dxif f(2a-x) = f(Xf) a

-~log2)

=1t2 log! 2

)2

9 1t/2 'lC/2 3. /= r 1t/2(-,d9 = r 92 cosec2 9d9 = (_92 cot 9)Q'2- r -29cot9d9 Jo sm9 Jo Jo

= (-0 + 0)-2 J;/2 -9cot9d9 = +2 [(9lQg Sin 9)~/2 - J;/2logSin 9d9 ] =+2 [(0-0)- J;l2l0gSin9d9] =+2

(~IOg2)

[Since J;12 log sin 9 = -21tlOg2]

=1tlog2 5. /=

7t/4

J 0

log (1 + tan 9)d9

= J;/4 10g {1

+tan(~+9 )}d9

= r1t/4l0g{1+ tan1t/4-tan9 '}dx Jo l+tan1t/4tan9 = r1t/410g{1+ I-tan9}d9 = r1t/410g{ 2 }dx Jo, l+tan9 Jo 1+tan9

r = Jo

7t/4

log2dx-

J1t/4 0 log (1 +tan 9)d9

Integration

2I=log2[x]

1tI4=

o

579

~log2 4

7t

/= glog2 7. / = J1t12

.Jtii.TU

dx

... (1)

..Jtanx +..Jcotx

o

... (2) _ 7t _ f1t'2 21dx - -

Adding (l) and (2)

o

2

7t

/=4 r1t12

8 • /= JI0

cos 2 xdx sinx+cosx

... (1)

=

r/2 o

r

Adding (1) and (2)

.

cos

2(7t

'2- x

)

sin(~-x )+cos(~-x)

2 sin x dx = 0 cosx+sinx 2 2 21= f1t'2 cos x+sin x dx o cosx+sinx /2

=

r/2 cosx+sinx dx

=

J

0

1t12

0

dx

x x I-2sm -+2sm-cos2 2 2 .2X.

Dividing Nr and Dr by cos2 ~ 2 r1t12

21- JI o

Lettan ~ = t => sec2 ~ dx = 2 dt

2

dx

2

sec 2 x/2dx

1+2tan~-tan2~ 2 2

... (2)

580


Also when x = 0 ~ t = 0 x = ~ => t = 1 2 2/-

rl

-2

2dt

- Jo 21+I-t 2 -

rl

dt

Jo (J2)2 _(t_I)2

~2' 2~[IOglt:::n = _1 [lOg J2 _IOg(J2-1)] 2J2 J2 J2+1 = _1_10g{J2-1}=_I_IOg J2+1 s J2 J2+1 J2 J2-1

=_110 {(J2+1)(J2+1)} J2 g (J2 -1)(J2 +1) =

So

/=

~10g(J2+1)2= ~10g(J2+1)

~10g(J2+1)

9. /= f:sinm ncos 2m +1 xdx =f(x),say Here,f(x) = sinmx cos 2m + 1 x f(1t-x) So

sinm(1t -x) CQs(2m+ I) (1t-x) = - sinmx cos2m + 1 X =.f(-x)

=

f:

/ =0 Since

f(x)=Oiff(2a-x)=-f(x).

OBJECTIVE EVALUATION MULTIPLE CHOICE QUESTIONS Choose the most appropriate one: 1.

2.

3\ogx

fe (a)

4

\

+ It dx is equal to log (x4 + 1) + C (c) .!..log(x4 +I)+C

J.1

(x

4

(b)

_log(x4+ I)+C.

(d) -.!..Iog (x4 + I) + C.

4

tan4 ..rx sec2 ..rx dx is equal to.

(a)

~ tan 5 ..rx + C

(b) '!"tanS..rx+

(c)

2 tan 5 ..rx + C

(d) None of these.

s

S

c.

Integration

3.

rot x. tan x dx 2 is equal to sec x-I (a) cotx+x+ C (e) -cotx +x+ C 3

2

4.

5.

) dx is equal to: ( I+x+-+-+x x 2! 3! (a) e'+C (e) e"x+ C

fFx +-Ix-2 dx (a)

(a) (e)

7.

. (x - 2)3/2] + C

!

x3 - +5x 3 x3 - +5x+2 3

(e)

x o +1 --+if +C a+1 x u+ 1 aX --+--+C a+I loga

(c)

1

- [log (sec x + tan x)] + C 2

(d)

!

3

2

[x 3/2 _ (x - 2)3/2] + C .

x3 (b) - -5x 3 x3 (d) - + 5x + I. 3

x o+ 1 (b) --+logx+C. a+1

(d) None of these.

(b) [log (sec x + tan x)] + C.

(d) None of these.

cose# . fe# Fx dx equal to IS

(a) sin e# + C (b) cos e

10.

.!. [x3/2 + (x + 2)3/2] + C

fsecxlog(secx+ tan x) dx is equal to: (a) sec 2 + se~ tan x + C

9.

(b)

f(eolOgx + exlogU)dx is equal to (a)

8.

.!. [x3/2 -

'3

(b)-e'+C (d) None of these.

=?

[x 312 + (x - 2)3/2] + C 2 f(x 2 + 5) dx =

(e)

6.

(b) cotx-x+ C (d)-cotx-x + c.

Fx +C

f-ll-sin2xdx

(b) 2 sin e# + C

(d) 2 cos e# + C.

=

(a) sin x + cos x + C (e) secx+tanx + C

(b)

sinx-cosx + C

(d) secx-tanx + C.

II. Ifj{x) = f(a + b -x), then f: x f(x)dx is equal to

(a) (a+b) lXf(x)dx

(b) i(a+b) l Xf (x)dx

(e)

(d) i(b-a) lXf(X)dx.

(b-a) lXf(x)dx

581

582 12.


r • 1

x+2 dx Jx 2 +2x-3

(a)

2J3 1 ----log3 3 2

2J3 1 (b) -+-log3 3 2

(b)

2J3 I J33 +2) ---Iog( 3 2

2J3 1 J33+2) (d) ---Iog(

.

(e)

re/3 -log tan 3re/2 3 (re/2-log sin re/12)

14. The value of

r

(b) 2 (2re/3 -log tan 5rt/2)

(d) None of these.

e,/xdx is (b) 2e2 (d) 3e2.

e2 (e) 4e2

(a)

r

15. The value of the integral (a) (e)

16.

2

r/2 xsinx . - - d x IS -1t/3 cosx

13. The value of the mtegral (a)

3

/2

1

logtanxdx is (b) rt/4 (d) re

rt/4 0

2 I1t12 . sin x dx is equal to 1 smx+cosx

(b) J2log( J2 + 1)

(a) rt!2 (e)

~ log ( J2

(d) None of these.

+ 1)

r

17. The value of the integral

(a)

12 3/2 dx O(l+x)

(b)

112

!J2 2

(d) J2

(b) 1 1t

2 18. IfI= I: sin xdx, then. 2

(d) /= 4 I1t/2 0 sin 2 xdx

(a) /=2 S:1t sin xdx (e)

/= S:1tcos 2 xdx

19. The value ofthe integral (a) (e)

0 log 112

20. The value of (a) 1 (e) 0

r

. 2 (d) /= 11t/4 0 sm xdx.

III (x+ ~(x2 + l)dx is (b) log2 (d) None of these.

/2 sin[log(x+N +1)]dx is -1t/2 (b) -1 (d) None of these.

Integration

583

FILL IN THE BLANKS l. If

f 4ege

X +6e-X x

-4e

dx =Ax+

-x

2.

f x(x3I + I) dx --

3.

fe

4.

eX f .J1+e-

5.

2x

Blog (ge

2x

-4) + CthenA =

,B _ _ _ and C-= _ __

1 -2x dx =A tan- I 2x+ C then A = +e 2x

dx-

f sinx-cosx 1 dx = f sm x+ 1dx = cos x f l .IS= . 3

6.

7. 8. 9.

2

2sinx+sin2x

fJl:X3 dx = e -I dx is = The value of f eX +1 2x -2--

1

1t/2

f 4 cos 2 x+ 9'sm 2 x dxis= _ __ The value of f sec xcos 2xdxis = _ __

10. The value of 0 11.

2

f1t/2

2

3

12. The value of Jo

tan xdx is= _ __

~ dx is = _ __

13. The value of

f4

14. The value of

"" f xe-

o l+v2x+1 x 212

-0

dxis= _ __

f1t/2 2

15. The value of Jo

TRUE/FALSE

x cosxdx is= _ __

I:

1.

fa f(x)dx

2.

IIo .J(1 + x)(1 + x3) dx in less then or equal to ~8

3. 4.

I:

I:

=

[f(x) + f(-x)]dx

I f(x) Idx = 0 => f(sinx) dx = 2

I:

1:

12

(f(x))2 dx

(T/F) (T/F)

=0

(TIF)

f(sinx)dx

5. Ifj{x) is an odd function then

(TIF)

Io f(cosx)dx = 2 Jo

(1t/2

1t

f(cosx)dx

(TIF)

584


ANSWERS MULTIPLE CHOICE QUESTIONS 2. (a) 1. (c) 5. 9. 13. 17.

3. (d) 7. (c)

4. (a)

(c)

6. (d)

(b) (b) (b)

10. (a)

11. (b)

8. (d) 12. (b)

14. (b) 18. (a), (b)

15. (c) 19. (a)

20. (c)

16. (c)

FILL IN BLANKS 3 35 1. A= -- B= - CER.

2'

4. sin- 1 (If) + C 6. - log leos xl + 8.

2.

36'

5.

2

C

3.

2

J2 log tan (x12 -n/8) + C 1 1 7. - log (1- eos x) + - log (1 + eos x) + CA

2"1 eos2 x + tan x + c..

6

6

9. log (If +

3 10. tanx+ sin 2x-2x + C. 1

x +1

1

~ log (x 3/2 + ~1+x2 ) + C

12. -log(e/2)

.!.IOgl4-1 + 3

11.

e-~

+ C.

n/12

n

14. 0

13. 2-log2

2

15. --2. 4

TRUE/FALSE 1. True 5. False

2. False

3. True

4. True

Do you know? After reading this chapter you should be able to know the following concepts:

• Letf(x) be a function and c be a constant. d d dc dx [f(x)+C] = dx [f(x)] + dx =F(x). Hence ff(x)dx =F(x)+C, where the symbol

f is an integral sign and Cis constant

of integration. The function F(x) is called the indefinite integral of integrand fix) • The integral of the product ofa function with a constant is equal to the product of the constant and integral of that function. • The integral of the sum or difference of two function is equal to the sum or difference of their integrals. fLfi(x)±h(x)dx

=

f.li(x)dx± fh(x)dx

Integration

585

• Ifj{x) andg(x) are two functions of x , then

ff(x)g(x)dx

=

f(x) fg(x)dx- f! f(x){ fg(x)dx}dx .

• Ifj{x) is a continuous and non-negative function over a closed internal [a, b] then

! f(x)dx

t

is called the defmite integral ofj(x) between the limits a and b. (b > a), then

f(x)dx = [f(x) +

CJ: =f(b)-f(a) is a definite value. Here, a is called the lower and

b is called the upper limit and the interval [a, b] is called constant of integration. Can we do? (Frequently Asked Questions) Evaluate the following integrals: [UPTU B. Pharma 2001]

3.

f cosx dx 2 sin x x . f e -smx dx eX +cosx x 2dx

fl +x6


4.

f tanx dx


5.

fO + logx)2 dx x

(UPTU B. Pharma 2002]

6.

f5+~OSX


f

(UPTU B. Pharma 2007]

1. 2.

logsecx

8.

sinx dx sin(x-a) fx 3sinx 2dx

9.

x 2 tan- 1 x f 1+x2 dx

7.


(UPTU B. Pharma 2004] (UPTU B. Pharma 2006]

. -I

10.

fxsm x dx ~1_x2

(UPTU B. Pharma 2006J

11.

feX(l+x)2 x dx

[UPTU B. Pharma 2007J

12.

fe 2x (-sin x + 2cosx)dx


13.

fa 2

14.

~x2

fx:x2

IUPTU B. Pharma 2003] [UPTU B. Pharma 2005]

586

15.


JX~X3 J(x-1)(x-2)(x-3) x ch

[UPTUB. Pharma 2004)

3

16.


x2

1 + ch x4 +1

17.

J

18.

J1t/4 1t/2 cosa cosec ada

19. Sh owthat 20. Evaluate

[UPTUB. Pharma 20(5) 2

r

o

/2

.Jsinx .Jsinx+.Jcosx

J-;-log x ch x

[UPTU B. Pharma 2002) ch

= -7t


4 [UPTU B. Pharma 2008)

DOD

Appendix-I "MENSURATION AND ITS PHARMACEUTICAL APPLICALITONS" Some important results: 1. Volume of cuboid = length x breadth x height. 2. Curved surface area of cuboid = 2 x height (length + breadth). 3. Total surface area = 2 [length x breadth + breadth x height + length x height]. 4. Volume of a cube = (edge)3.

i.

5. Curved surface area of cube = 4 (edge 6. Total surface area ofa cube = 6 (edge)2. 7. Cylinder: (Right circular cylinder): Let r be the radius of the base and h be the height of the cylinder. Then, (I) Volume = 7t,2h. (ii) Curved surface area = 27trh.

(iii) Total surface area = 27t (h +r). 8. Cone: (Right circular cone) Let r be the radius, h the height and 1the slant height ofthe cone. The

Volume= .!.7tr 2h. 3 (il) Curved surface area = 7tr (/+ r).

(I)

(iii) Total surface area 1= ~h2 + r2 9. Sphere: Let r be radius of the hemisphere. Then, (I)

Volume=

~7tr3 . 3

(iz) Surface area = 47t,2.

10. Hemi sphere: Let r be the radius of the hemi-sphere. (I)

Volume=

3. 7tr3 . 3

(iz) Curved surface area = 27t,2. (iii) Total surface area = 37t,-2.

588


~~~~~~~ SOLVED EXAMPLES

Example 1: The metal cylinder ofradues 18 cms and height 80 cms is melted to prepare spheres ofdiameter 12 cms. Find the number ofspheres prepared [RGPV B. Pharma 20021 Solution: Volume of cylinder = 1t?h =1t x (l8i x 80 = 25920 1t cubic cm.

4

Volume of sphere = -1tr 3

3

4

=-1t x (6)3 3 = 2881t cubic cm.

Number of spheres prepared =

259201t 2 = 90. 881t

Example 2: Find the volume ofa capsule having a cylinder ofheight 5 mm and radius 4 mm on both sides of cylinder, there are hemispheres of radius 4 mm. [RGPV B. Pharma 2005] Solution: Height ofthe cylinder, h = 5 mm.

---..-1)

(-.J....-1

Shape

r = radius of the cylinder = radius of hemisphere = 4 mm. Volume of capsule = volume of cylinder + 2 x volume of hemisphere

=1t?h+2(~1tr3 ) =

22 2 4 22 3 -x(4) x5+-x-x(4) 7 3 7

= 1760 + 5632 = 10912 =519.619cubicmm. 7 21 21 Example 3: Write the volume ofcapsule in the shape ofcylinder bounded by hemisphere (as shown in fig given below on the ends. length = 8 mm. diameter = 4 mm . [RGPV B. Pharma 2003]

.,--;

Appendix

589

Solution: Volume of capsule = volume of cylinder + 2 x volume of hemisphere

=1tYlh + 2 x

~ 1t?

=1t(2i x 8+

± x 3

321t = 321t + -

= --

3

=

3

1t

x (2)3

1281t

3

.

cubic mm.

134.01 cubie mm.

Example 4: Find the volume ofa right circular cone ofradius 3 cm and height 7 cm and also find the surface area of the whole one. [RGPV B. Pharma 2001] Solution: Volume of cone = .!.1tr2h 3 1 22 -x-x3x3x7 3 7 = 66 cubic cm. =

Surface area = 1tr (I + r) = 22 x 3

7 =

(.J58 + 3)

22 -x3(7.483+3) 7

66 x 10.483 = 98.8489 cm. 7 Example 5. Write the volume oftablet in the shape ofcylinder top bounded by cone (as shown is fig) on one end height of cone = 5 mm. height of cylinder =5 mm. =

[RGPV B. Pharma 2007] radius of cylinder = 4 mm. Solution: Let hI = height of cone = 5 mm. h2 = height of cylinder = 5 mm. r = radius of cone = radius of cylinder = 4 mm. Volume of tablet = Volume of cone + volume of cylinder 1

= -1tYlh l + 1tYlh2· 3

=.!.1t(4i x 5+1t(4i x 5 3

=

(j 1) +

80 1t = ~ x 80 1t

= 320 x 22 = 335.24 cubic mm. 3 7


EXERCISE 1. The base radii of two right circular cones ofthe same height are in the ratio 3 : 5. Find 2. 3. 4. 5. 6.

the ratio of their volumes. The circumference of the base of a 9 m high wooden solid cone in 44 m. Find the volume ofthe cone .(use 11 = 2217.) Circumference of the edge of hemispherical bowl in 132 em. Find the capacity of the bowl. (use 11 = 2217). How many spherical lead shots each 4.21 em in diameter can be obtained from a rectangular solid lead with dimensions 66 em, 42 em and 21 em? A solid cylinder has a total surface area 462 sy. em. Its {;urved surface area is one third of the total surface area. Find the volume of the cylinder. A rectangular sheet of paper 44 em x 18 em is rolled along its length and a cylinder is formed. Find the volume ofthe cylinder [use 11 = 2717].

I 1. 9: 25.

5. 53gem

3

.

ANSWERS I

2. 462m3 . 6. 2772em3 .

4. 1500.

tiC 1t9U•IfIJ

TABLES RELATED TO MATHEMATICS

COMMON LOGARITHMS ·x

1

0

!

2

3

I

4

5

10 I .0000 0043 0086 012810170 0212 0212· I I II 1 0414 1,0453 0492 053 1 10569 0607 , I 0607 i 121.079210828 0864 089910934 0969 0969 13 ."391"73 1206 1239 1271 1303 1303 14 .1461 1492 1523 1553 1584 1614 15 .1761 1790 1818 1847 1875 1903 16 .2041 2068 2095 2122 2148 2175 17 .2304 2330 2355 2380 2405 2430 18 .2553 2577 2601 2625 2648 2672 19 .2788 2810 2833 2856 2878 2900 20 .3010 3032 3054 3075 3096 3118 21 3222 3243 3263 3284 3304 3324 22 .3424 3444 3464 3483 3502 3522 23 .3617 3636 3655 3674 3692 3711 24 .3802 3820 3838 3856 3874 3892 25 .3979 3997 4014 4031 4048 4065 26 .4150 4166 4183 4200 4216 4232 27 .4314 4330 4346 4362 4378 4393 28 .4472 4487 4502 4518 4533 4548 29 .4624 4639 4654 4669 4683 4698 30 .4771 4786 4800 4814 4829 4843 31 .4914 4928 4942 4955 4969 4983 32 .5051 5065 5079 5092 5105 5119 33 .5185 5198 5211 5224 5237 5250 34 .5315 5328 5340 5353 5366 5378 35 .5441 5453 5465 5478 5490 5502 36 .5563 5573 5587 5599 5611 5623 37 .5682 5694 5705 5717 5729 5740 38 5798 5809 5821 5832 5843 5855 39 .5911 5922 5933 5944 5955 5966 40 .6021 6031 6042 6053 6064 6075 41 .6128 6138 6149 6160 6170 6180 42 .6232 6243 6253 6263 6274 6284 43 .6335 6345 6355 6365 6375 6385 44 .6435 6444 6454 6464 6474 6484 45 .6532 6542 6551 6561 6571 6580 46 6628 6637 6646 6656 6665 6675 47 6721 6730 6739 6749 6758 6767 48\.6812 6821 6830 6839 6848 6857 49 .6902 6911 6920 6928 6937 6946 No. n=3.14159 e = 271828

7

8

9

0253 0294 0334 0374 0645 0682 0719 0755 1004 1038 1072 1106 1335 1367 1399 1430 1644 1673 1703 1732 1931 1959 1987 2014 2201 2227 2253 2279 2455 2480 2504 2529 2695 2718 2742 2765 2923 2945 2967 2989 3139 3160 3181 3201 334513365 3385 3404 3541 3560 3579 3598 372913747 3766 3784 3909 3927 3945 3962 408214099 4116 4133 4249 4265 4281 4298 4409 4425 4440 4456 4564 4579 4594 4609 4713 4728 4742 4757 485714871 4886 4900 4997 5011 5024 5038 5132 5145 5159 5172 5263 5276 5289 5302 5391 5403 5416 5428 5514 5527 5539 5551 5635 5647 5668 5670 5752 5763 5775 5786 5866 5877 5888 5899 5977 5988 5999 6010 608:1 6096 6107 6117 6191 6201 6212 6222 6294 6304 6314 6325 6395 6405 6415 6425 6493 6503 6513 6522 6590 6599 6609 6618 6684 6693 6702 6712 6776 6785 6794 6803 6866 6875 6884 6893 6955 6964 6972 6981

6.." + 42 40 39 37 35 34 33 32 30 28 26 25 24 22 21

log r!' log e-P

1.5657

4

1 2 3

5

6

7

8

9

29 28 27 26 25 24 23 22 21 20 16 18 15 17 14 17 13 15 13! 15

34 32 31 30 28 27 26 26 24 22 21 20 19 18

38 36 35 33 32 31 30 29 27 25 23 22 22 20 19

ADD

4 4 4 4 4 3 3 3 3 3 3 2 2 2 2

8 8 8 7 7 7 7 6 6 6 5 5 5 4 4

20 124 19 2 4 1812 4 18 2 4 17 2 3 16 2 3 16 2 3 15 2 3 15 I 3 14 1 3 14 1 3 13 I 3 13 I 3 13 I 3 12 1 2 12 I 2 12 I 2 11 1 2 11 1 2 11 1 2 10 1 2 10 1 2 10 1 2 10 1 2 10 1 2 9 12 9 12 9 I 2 9 I 2

log

0.49715 0.43429

1 04343

P

6

13 17 12 16 12 16 11 15 11 14 10 14 10 13 10 13 9 12 8 11 8 10 7 10 7 10 71 9 6 18 6 8 6 8 5 7 5 7 5 7

21 20 19 19 18 17 16 16 15 14

5 6 5 6 5 6 4 6 4 6 4 6 4 5 4 5 4 5 415 415 4 5 314 3 4 3 4

8 8 8 7 7 7 7 6 6 6 6 6 6 6 8 8 8 8 8 8 8 8 4 4

I

4 4 4 4 4 4 3 4 3 4 3 4

3 3 .3 3 3 3

13 12 12 11

11

25 24 23 22 21 20 20 19 18 17

10 9 9 9

11 13 111 13 11 113 10 112 10 10 II 91 11 9 10 8 10

1"

2 0.8686 1 1314

3 1.3029 2.6971

4 1.7372 22628

15 14 14 14 I3

17 16 16 15 14 14 14 13 13

13 12 12 II 8 10 II 13 8 9 10 12 8 9 10 12 8 9 10 12 7 8 10 11 7 8 10 11 7 8 10 11 7 8 9 10 7 8 9 10 7 8 9 10 6 7 8 9 617 8 9 7 8 9 661 7 8 9 7 8 9 61 7 8 7 8 5 6 7 8 5 6 7 8

;I:

No. In x = loge x = (lIM) log 10 X log" = log 10 X = M loge X

]7

JO 12114 16 18

log

o ~6222

(lIM) = 2.30259 M= 0.43429

1.63778

5 2.1715

6 26058

7

8

J 0401

34754

9 39087

10 4.3429

38285

33942

49599

4.5256

4.0913

56571

I

COMMON LOGARITHMS 1

.,01123456789

I

+ 6998

7007

7016

7024

7033

7042

7050

7059

7067

9

ADD 1 2 3 !4 4 5 6 7 8

51 .7076\7084 521.7160 7168 531.724317251 541.7324 7332 55 .7404 7412

7093 7177 7259 7340 7419

7101 7185 7267 7348 7427

7110 7193 7275 7356 7433

7118 7202 7284 7364 7443

7126 7210

7292 7372 7451

7135 7218 7300 7380 7459

7143 7226 7308 7388 7466

7152 7235 7316 7396 7474

8 8 8 8 8

I 2 2 1 2 2 I 2 2 1 2 2 I 2 2

56 57 58 59

.7482 7490 .7559 7566 .7634 7642 .7709 1 7716 60 .778217789

7497 7574 7649 7723 7796

7505 7582 7657 7731 7803

7513 7589 7664 7738 7810

7520 7597 7672 7745 7818

7528 7604 7679 7752 7825

7536 7612 7686 7760 7832

7543 7619 7694 7767 7839

7551 7627 7701 7774 7846

8 8 8 7 7

I 1 I 1

61 62 63 64 65

7860 7931 8000 8069 8136

7868 7938 8007 8075 8142

7875 7882 7945 7952 8014 8021 8082 8089 814918156

7889 7959 8028 8096 8J62

7896 7966 8035 8101 8169

7903 7973 8041 8109 8176

7910 7980 8048 81\6 8182

7917 7 1 1 2 7987 1 7 I I 2 8055 1 711 1 2 8182 I 7 I 1 2 8189 '\7 I I 2

66 8195 67 .8261 68 .8325 698338 70 .8451

8202 8267 8331 8395 8457

8209 8274 8338 8401 8463

8215 8280 8344 8407 8470

8222 8287 8351 8414 8476

8228 8293 8357 8420 8482

8235 8299 8363 8426 8488

8241 8306 8370 8432 8494

8248 8312 8376 8439 8500

8254 8319 8382 8445,1 8506 I

7 11 I 2 3 3 6 I 1 2 2 3 6 I I 2 2 3 6 1 1 2 2 3 6 I I 2 2 3

4 4 4 4

14 14

71

72 8573 73 .8633 74 8692 75 .8751

8519 8579 8639 R698 8756

8525 8585 8645 870.+ 8762

8531 8691 8651 8710 9768

8537 8597 8657 8716 8774

8543 8603 8663 8722 8779

8549 8609 8669 8727 8785

8555 8615 8675 8733 8791

8561 8621 8681 8739 8797

8567 8627 8686 8745 8802

6 11 1 2 234 6 I I 2 2 3 4 6 I 1 2 2 3 4 6 I I 2 2 3 4 6 I I 2 2 3 4

4 4 4 4 4

76 77 78 79 80

88m! .8865 .8921 .8976 .9031

8814 8871 8927 8982 9036

8820 8876 8932 8987 9042

8825 8882 8938 8993 9047

8831 8887 8943 8998 9053

8837 8893 8949 9004 9058

8842 8899 8954 9009 9063

8843 8904 8960 9015 9069

8854 8910 8965 9020 9074

8859 8915 8971 9025 9079

6 6 6 6 5

81 82 83 84 85

.9085 .9138 .9191 .9243 .9294

9090 9143 9196 9248 9299

9096 9149 9201 9253 9304

9101 9154 9206 9258 9309

9106 9159 9212 9263 9315

9112 9165 9217 9269 9320

9117 ,9122 9170 19175 9222 9227 9274 9279

9133 9186 9238 9289 9340

5 112233445 5 1112233445 5 5 5 1112233445

86 87 88 89 90

.9345 .9395 .9445 .9494 .9542

9350 9400 9450 9499 9547

9355 9305 9455 950.+ 9552

,;>360 9410 9460 9509 9557

9365 9415 9465 9513 9~62

9370 9420 9469 9518 9566

9325 '19330 9375 9380 9425 9430 9474 , 9479 9523 \9528 9571 9576

9128 9180 9232 9284 9335 9385 9435 9484 9533 9581

9390 9440 9489 9538 9586

5 112233445 5 10 11223344 5 011223344 5 011223344 5 011223344

91 9590 92 9638 93 968;5 94 .9731 95 9777

9595 9643 9689 9736 9782

9600 9647 9694 9741 '1786

9605 9652 9699 9745 9791

9609 9657 9703 9750 9795

9614 9661 9708 9754 9800

9619 9666 971 J 9759 9805

9624 9671 9717 9763 9809

9628 9675 9722 9768 9814

9633 9680 9727 9773 9818

5 5

96 9823 97 .0368 98, 9912 99: .9956 !

9827

9832 9377

9836 9841 9881 9886 992619930 9969 9974

9845 9890 9934 9978

'1850 9894 9939 9983

9854 9899 9943 9987

9859 9903 9948 9991

Y863 9908 9952

4011222334 4011222334 4011222)34 401 122233·1

50 .6990

I

6m 123)4561789

.7853 .7924, .7993 f .8062 .8129

8513

9Sn

99i7 9

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