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)
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Remedial Mathematics S.K. Pundir
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[email protected] 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 ..................................................................................,.......... 160 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 ................................................................. 6189 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 ........................................................................................................
~1126
3.1 3.2 3.3
Introduction .................................................................................................................. 90 Determinant of Order Two ...................................................... ...................................... 90 Determinant of Order Three ...................................................... .................................... 91
3.4 3.5
Cofactors and Minors of an Element ...................................................... ................ 92 Properties of Determinants ........................................................................................ 9S
3.6
Cramer's Rule ............................................................................................................. 117 Refresher ............................................. ............................. ... ......... ............. ......... ......... 12S
(vi) 4.
Contents
MATRICES ............................................................................................................. ;•. 127210 4.1 Introduction ............................................................................................... ,................ 127 4.2 Types of Matrices .................................................................................................... 127 4.3 Determinant of Square Matrix .................................................................................. 129 4.4 SingularandNonSingularMatrix ...........................................................~ .................... 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 SkewSymmetric 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 Etransformation .......................................................... 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 Nonhomogeneous Equations .................................................................................. 193 4.28 Conditions for Consistency .................................................................................... 193 4.29 Conditions for a System of Nequations in Nunknowns to have a ......................... . Unique Solution ........................................................................................................ 194 Objective Evaluation ................................................................................................. 206 Refresher ................................................................................................................... 209 ¥
5.
••••••••••••••••••••••••
MEASURE OF CENTRAL TENDENCY ............................................................. 211254 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 ..................................................................................................... 255321 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 ........................................................................................................... 322345 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 ............................................................................. 346374 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 ........................................................................................... 375411 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 .................................................................................... 412458 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 ................................................................................................... 459514 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 ........................................................................................................... 515586 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 .................................................................................................................... 587594
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+bb = Ob => 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
03 2x =3
2x+33
=>
=
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(111::._ Y ) = lOy
IO
5
...(1 )
QJladratic Equations
110_y_1O(y1O) = lOy 5 11Oy2y+20 = lOy => 1303y = lOy => Now, adding 3y on both sides of (2), we get 1303y+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+66 = 56 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 = 2x8+8 x+IO =2x => 10 = 2xx => 10 = x or x= 10 =>
Example 4: Solve
x +3 2
x6 8
=
H ere, th · . .IS x + 3 =x6 · e gIven equatIOn S o IutIOn: 2 8
=>
8(~+ 3)
=x6
8x +8x3 =x6 2 => 4x+24 = x6 Now, adding 24 on both sides, of the above equation, we get 4x+2424 = x624 4x = x30
=>
4
Remedial Mathematics
4xx = 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(a2p2)+b(ap) = 0 => a(ap)(a+p)+b(ap) = 0 => (ap)[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(x3)1 (x3) = => (x3)(x1) = x =
0 0 0 0
1,3
Quadratic Equations
5
E~a mple 2. Solve x2  2x  3 = 0
Solution: Here, we havex22x3 = 0 ~ x23x+x3 = 0 ~ x(x3)+1(x3)=0 ~ (x3)(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, wehave4x23x1 =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 ~ a27x =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
Remedial Mathematics
=>
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 C1S)±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(3i4xlxS
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 {32' +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±J9001100
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 + 2xi / 3
4 x+4
= 
(x + 2) + 2 (x + 1)
4
(x + I)(x + 2)
x+4
=
5
5
i}
8
Remedial Mathematics
(3x+4)(x+4) 3.x2+ 16x+ 16 x2 4x8 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 215y+2 = 0 (y2)(2yl) = 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 25125y6 = 0 (5y+ 6)(5yl) = 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) x2x12=0 (iii) x2+1=0 (iv) x24x+3=0 2. Solve the following equations by quadratic method. (l) x27x+12=0 (ii) x24x+7=0 2 (iii) 2x 3x+ 1 =0 (iv) 27.x210x+ I =0 (v) 3x2x1O=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.x2lOx=3x15
7 5. Solve the following equations:
(ii)
4x + 9
2X2  3x  5 3
=
5
6. Solve the following equations:
(/)
3x 7 16 =
5
x5
7. Solve the equation: pqx2  (p2 + 8. Solve:
~{2(x2 
x + I)}
=
l) x + pq = 0
3x4
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) Flx) ~ ) vl;x) ="6 +
(1/1)
13
HINTS TO THE SELECTED PROBLEMS 2. (I) .x27x+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) x24x+7=0.
x= l,b=4,c=7. b ± ~r:b2:_4ac x =
2a 4 ± ~( 4)2  4 x I x 7 2xl
9
10 Remedial Mathematics 7±.J4948
7±1
2
2
7+1  
71
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 8xx 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
~+ 14x3x42 = 0
=7
=>
=7
=> x(x+ 14)3 (x+ 14) = 0
=7
=>
(x+ 14)(x3)
=
0
= 30
=>
x x
=
14 +3
or
5.
~ll)
2X2  3x  5
4x+9 5
3
=> => =>
10~15x25 = 12x+27 1O~27x52 = 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)
=>
3x7 5

= 6
16 x5 (3x7)(x 5)
=
80
27 ±53 20
=
Quadratic Equations
3~15x7x+35 = 80
3~22x45 = 0
3~27x+ 5x45 = 0 3x(x9)+5(x9) = 0 (3x+ 5)(x9) = 0 3x+5 = 0
x
or x9=O
5
= 
3
or x=9.
7. pq~_(p2+q2)x+ pq= 0 => pqx2 _ p 2x  q2x + pq = 0 => px (qxp)q (qxp) = 0 => (qxp) (pxq) = 0 either qx  p = 0 or px  q = 0 p q x=or x=q P 2 8. )2(x  x + 1) = 3x4 Squaring both sides. 2 (~x + 1) = (3x4)2
=> =>
Y2x+2 = 9~+ 1624x. 7~22x+ 14 = 0 x= =
~x7xI4
+22±) (22)2
2x7
22±~484392 14 22 ±
J92
22 ± 2
14 let
=>
53
14
11 ±
53
7
5X = y 1126y+3 = 0
y= =
126 ± )(126)2  4 x 1 x 3 2xl ±126
ANSWERU (ii) 4,3
(iii) (i,  i)
11
12
Remedial Mathematics
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. i9yl0 = 0 => ;10y+yl0 = 0 => y(y1J)+ 1(yl0) = 0 => (ylO) (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  xI
(
2x+I )2 4 10 ( + 9 = O.
J
xI
=> => => =>
xI
ylOy+9
=>
= 0
= 0
y9yy+ 9 = y(y9)1 (y9) = (y1)(y9) = y =
0 0 0
1 or y=9
2x + 1)2 y _ 1 => ( 
Now
rUPTU B. Pharma 20051
xI
. 2x + 1 2x + I SolutIOn: We have ( )4 10 ( J2 + 9
xI
2x + I
2x + 1 _ I = 1=>± xI
2x + I   =1 xI 2x+l =xI or 2x+ 1 =x+ I x = 2 or x=O.
=>
xI
=> =>
= I
or
2
2x + I 2x + 1 =9 => =±3. ) xI xI
y = 9 =>  
Again
(
2x + I =3 xI
=>
or
2x+ I = 3x3
=>
x = 4
or
2x+ I =3 xI
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+2y3 =0=>y+3yy3=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 xl =0 x=
1 ± .ji+4
2
1±
.J5
=
2
y =3
Also, 2
x =3 x +1 x 2 = 3x3 ~+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: 4x5.2x+4 = 0 which can be written as (2 Xi  5 (2X) + 4 = 0 Put '? = y, we get y5y+4 = 0 which is quardrat in y. 'So1ving (l) for y, we get (y4)(yl) = 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 + y2 = 0, which is quadratic iny,
=>
Quadratic Equations
15
Solving for y, we get (y+2)(yl) = 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 xI
SOLVED EXAMPLES
+JXl x
Put
~
x
xI
13, (.'(*I,x*O) 6
=
· H ' . .IS ere, t I 1e gIven equatIOn So I utlOD:
~I~~~~~~
~   + Fxll   = 13 xI
x
6
= y, then the above equation reduces to
I 13 7 Y +  =  => 6y  13y + 6 = 0
6 Solving for y, we get y
(2y3)(3y2) = 0
3 2
y =
2'3
Now,
y
12 => ~ xIx = 12 => _x_ = (1)2 => x = 2. 2 5
also,
y=~=>~ x =~3 3 x I
=
xI
16
Remedial Mathematics
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 163y8 8164y+y8 8y(y8)+ 1 (y8) (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
7149yy+7 = 0 7y(y7)1 (y7) = 0 (y7) (7yl) = 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 215y+2 = 0 214yy+2 = 0 2y(y2) I(y2) = 0 (y2)(2yl) = 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.x29=0
4. 5. (.x23x+3i(xl)(x2)=7 7. 3x2 + 7x 1 +5=0
(~)2 5(~)+6=0 x+a x+a
6. (.x25xi30(.x25x)216=0 8. (.x25x+7)2_(x2)(x3)= 1 10. 3x + 1 + ~ = ~ (x E R) x + 1 3x+ 1 2
x l+x 13 11.   +   = l+x x 6
18
Remedial Mathematics
13. 5x+ 1 +5 2 x = 53 + 1
14. 3x+rX2 =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 =>;8y9=0 => ;9y+y9 = 0 =>y(y9) + 1 (y9)=0 => (y9)(y+ I) = 0 =>y9=0 or y+ 1 =0 => x 2 = 9 or x 2 =  1 => y = 9 or y =  1 => x = ± 3 or x = ± i 3. substitute xI = y. to obtain 3; + 7y + 5 = O.
Let
a)
4. substitute (x = y. to obtain;  5y + 6 = O. x+a 5. (~3x+3i(xI)(x2)=7 => (x 2 3x+3)2_(x23x+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 xI
Then
=> => => => =>
=
y.
;12=y ;+y12=0 ;+4y3y12=0 (y+4)(y3) = 0 y = 3 or y=4 xI = 3 or xI =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+24120 = 0 => y+ 10y96 = O. which is quadratic iny, solving fory, we get y (y + 16)  6 (y + 16) = 0 => (y6) ()It 16) = 0 Y =6,16 => Now, Y = 6 x 2 +5x = 6 => ~+5x6 = 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 (2x7)(x3)(x+ 3)(2x+ 5) = 91 => [(2x7)(x + 3)][(x3)(2x+ 5)] =91 => [2.x2x  21] [2x 2 x  15] = 91 Put 2x2 x = y, we get (y21) (y15) = 91 ;36y+31591 = 0 => ;36y+224 = 0 => (y8)(y28) = 0 => Y = 8,28 => Now, Y = 8 2x2 _x = 8 => 2x2 x8 = 0 =>
1±J65
=>
x =
also,
y = 28
4
2xlx = 28 2 2x x28 = 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+49y61 = 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 xl5x+6 = 0 (x2) (x3) = 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+2x56 = 0 (3x+ 14)(x4) = 0 (3x + 14) or (x4) = 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
Remedial Mathematics
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 + 17x84 = 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+8JY7 =\
=~ 1  6x + 8 
~5x2
 6x  7 = 1
Quadratic Equations
23
Squaring both the sides, we get
(y + S) + (y  7)  2,Jr(y+S"')(y7) = 1
~y2 + Y _ 56
y =
=> Again squaring, we get
y = y+y56 Y = 56 5x2 6x = 56 5~6x56 = 0 (5x+ 14)(x4) = 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)(xl)(x2) = 30 1. (i)x(x+ l)(x+3)(x+4) = ISO (iii) (x5)(x7)(x+4)(x + 6) = 504 (iv) x(2x+ l)(x2)(2x3)=63 (v) (~3x1O)(~5x6)= 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)(xl)(x2)=30 ~ [(2x + 3)(xl)[(2x+ 5)(x2)] = 30 ~ (:zx2+x3)(2~+xlO) = 30 ~Let 2x2+x = y. ~
(y3)(ylO)=30
~
Y13y+30 =30
~
y13y y(y13)
~
= 0 =
0
y = 0 ot Y  13 ~ 0
~
:zx2+x = 0 or 2xl+x13=0
~
1±)(1)2 4x2x(13) x (2x + 1) = 0 or x = '2x2
~
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 3x36 =  .J(2x + 8) (x + 5) squaring again, we get
(3x36i =4 (2x+ 8) (x+ 5) ~ 9 (x12)2 =4 [:zx2 + 10x+ 8x+ 40] ~ ·9 (xl + 14424x) =4[:zx2+18x+40] ~ 9xl + 144 x 9  24 x 9x = 8xl + 160 + 72x. ~ xl144x+ 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 
Solution: Here, the given equation is
EXAMPLES
~I~~~~~~
5x + 4 = x  4
~x 2 
=> ~(x  4)(x + 4)  ~(x I) (x  4)
16 =
~x 2 
(x4)
5x + 4 = x  4
26 Remedial Mathematics
:::::> .J(x4) [.Jx+4 ..r;=I.Jx4] =0 Now,either .Jx4 =0:::::>x4=0:::::>x=4 or[.Jx +4 ~  .Jx 4]
=
0
.Jx+4..r;=I = ~x4
:::::>
On squaring, we get. 4(~+3x4) = (x+7)2=~+ 14x+49
4~+ 12x16 = ~+ 14x+49 3~12x65 = 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 + ~
Solution: Here, the given equation is
= .J5 (x 1)
~x 2 + 2x  3 + ~(x2
 x) =
:::::> ~(x 1)(x + 3) + Jx (x 1) = J5 (x 1) :::::>
JX1[.Jx+3+FxJ5] = 0
~ = 0 :::::>xl =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) =22x
Again squaring 4x (x + 3) = (2  2x)2 :::::> 4~+ 12x = 48x+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 + (5119) =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 (xl)(x3) = O~x= 1,3
Now, ~ ~
Also
y
=
8 ~ ~rx24x+}9 = 8
~
~4x+ 19 = 64
~
~4x45 = 0
~
~+5x9x+45 = 0 x (x + 5)  9 (x + 5) = 0 ~ (x+ 5)(x9) = 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
Solution: Here, the given equation is
~5x2
 6x +·8 
~5x2
 6x  7 = 1
...(1)
28
Remedial Mathematics
Let
~5x2
 6x + 8 = A and J5x 2  6x  7 = B, then given equation reduces to
Also
=> =>
A B = 1 A2_B2 = (5x 26x+8)(5x2 6x7)= 15 (AB)(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
=> =>
5~5x56 = 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 = (2x23x5)(x23x+4)=~9 (A  B)(A + B) = x2  9 (x+3)(A+B) = (x3)(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 ~3x5,= x 2 ~3x5 = 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.
~x25x+6+~x29 = ~2x2
2.
~x2 +2x3 +~
=
llx+15
~5(x1)
3. ~2x2  5x  2  ~2x2  5x  9 = 1 4.
~4x2 7x15 _~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]
~x3[~x,2+~x+3J
=
=
=
~2x2  6x  5x + 15
~2x (x  3)  5 (x  3)
~(x3)(2x5)
=>~x3[~x2+~x+3~2x5J =0 either
~ x  3 = 0 => x = 3.
or ~ x  2 + ~ x + 3  ~2x  5 squaring both sides.
x2+x+3+2~x2 ~x+3 =2x5 1+5 = 2~x2 ~x+3
6= 2F=2F+3 =>
3 =  ~(x 2)(x + 3)
squaring again, we get 9 = (x2)(x+ 3) 9 = ~x6
29
30 Remedial Mathematics x2x 15
=
x= x= 4.
~4x2 :::)
7x15
_~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) :::) ~(x3)(4x+5) ~x(x3)
=>
~(x  3)(x + 3)
~(x3)(x+3)
~x3)[~4x+5)J;~x+3J =0 ~x3
either or
=
=
~(x  3)(x + 3)
=
0,x=3
~4x+5J;~x+3. =0
:::)
~4x+5 J; = ~x+3
squaring both sides.
4x+5+x2J; ~4x+5
=
2+3.
4x+53 = 2J; ~4x + 5 :::)
:::) :::) :::) :::) :::)
4x+2 = 2J; ~4x + 5
2x+l=J;~4x+5 2 4x + 1 +4x = x(4x15) 4x2 + 1 + 4x = 4x2 + 5x 5x4x = 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~
=12in(l),weget
12y+2 = 0 ly=o y(yI) = 0 y = 0,1 1 Y = o=>x+ =0 x
32
Remedial Mathematics
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 ~ = 3x2 2x(x2) + 1 (x2) = 0 (x2)(2x+ 1) = 0 x2=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
Solution: Here, the given equation is
(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 l2y15 = 0
=> Now, if
=>
(y5)(y+3) =0 y = 3,5 1
y=5=>x=5 x .x25xl = 0 x= x
orif
.x2 + 3xl 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.x2x+2=0
{5±m 3±v'i3} 2'
2
2. ~ 3.x23x+2 =0
r x;
(x + ;
3. 4x4 4~ 7.x24x+4 = 0
4.
5. 2x4_~+ 14.x29x+2 =0 7. x 44x3 3.x24x+ 1 =0
6. x4+13(~+x)=2.x2 8. x 5  4x4 + ~ + .x2  4x + 1 = 0
9. (x+;f %(x;)4 =0 11. x 6  x 5 + x4 .x2  xI
=
0
.
6(
+ 1)  5
=
0
34 Remedial Mathematics HINTS TO THE SELECTED PROBLEMS 1. 2x4x3_11~x+2=0 divide by~, we get
2x2  xII
_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+~) 21yI5 216y+5y15 2y(y3)+5(y3) (y3)(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
~ y6y7 = x4+ 13 (x 3 +x)=al x4+ lal3x3 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+
y3y4 = 0 x2 4x + 1 = 0
putting x = 1 ~ (1)54 (_1)4 + ( 1)3 + (_1)1 4 x (I) + 1
=141+1+4+1=0 so (x + 1) is a factor of given equation. (x+ l)(x45x3 +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 25(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 y5y+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. 112 ' 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,xb>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). (xa)(xb) =  ve CaseU It is possible if one factor is positive and the other is negative. Let xa = +ve>O,xb=ve a, x < b or a < x < b i. e. x lies between a and b (a < b) or xa = 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)(x5) = [x(3)] (x5)
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, thenxa > 0 andx ~ < 0 so that (xa) (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 (xa) 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+~b24ac 2a
and ~ =
b~b24ac ...!.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(xa)(3(xa)=O (xa)(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
Remedial Mathematics
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
po,+~.
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)
rtc+p+ 1 = lc
0,2 + 20, + 1
p2 + 2P + 1
0,2 + 20, + C
p2 + 2P + c
... (2)
::+'::'
(0,+1)2
(P+l)2
(o'+l)20c)
(p+l)20c)
'::': +
[using (1)]
a +1 P+1 (a + 1)  (P + 1) = += =1 O,p pO, 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 +2aI) (4a +2aI)a
2
2
1 1 =a·OOa·
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 (x9)(xl)=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+~)22a~= .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
Remedial Mathematics
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) = 273.%(3)3.%[92.%]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. + ~)24ex.~ = 144 ~ p24.45 = 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 218x+45 = 0 => (x3)(x15) = 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 ).
Solution: Here, the given equation is
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  bcca) = 0 Given that, the sum of its roots is equal to zero. _ (a + b  2c) = 0 1 2cab = 0
or
a+b c=  2 Product of roots =
=
... (1)
ab bc ca 1
=abc(a+b)
ab_(a~b») ~+x2+ 1 +2x = 61 => ~+2x60 = 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 => ~+ 16x384 = 0 => x2+2x48 = 0
=> => => => =>
~+8x6x48=O
x (x + 8)  6(x + 8) = 0 (x+8)(x6) = 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"
50 Remedial Mathematics
1 8x
8 15
8x+x x (8  x)
8 15
1
+X
8
8
~8x  x 2
15 64x  8x2 0
•
120 = 8x264x+ 120 = x28x+ 15 = 0 (x5)(x3) = 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 =
.. => => => => => => =>
=
9x227x162 = 0 x23x18 = 0 x2  6x + 3x  18 = 0 x (x  6) + 3 (x  6) = 0 (x6)(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(x4) = 0 x i.e., x = 0 or x=4 ;, Since with x = 0 the triangle is not possible, hence x = 4 x1 xI, 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
52 Remedial Mathematics
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+ 25x x (x + 25)
500 x (x + 25) = 12500 => ~+25x12500 = 0 => (x+ 125)(x100) = 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 = (2xI) m and third side = (x + l)m. Applying Pythagoras Theorem, we get (2xli =x2 +(x+li => 4~4x+ 1 = x 2 +x2 +2x+ 1 x 2x2 6x = 0 i.e., x 2 3x = 0 => x(x3) = 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 31,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+lOx x (x +10)
120
x(x+ 10) = 1200 => ~+ IOx1200 = 0 => .x2+40x30x1200 = 0 => x(x+40)30(x+40) = 0 => (~+40)(x30) = 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+250x750000 = 0 On factorising we get (x + 1000) (x750) = 0 x = 10000rx750
=>
Rejecting the negative values of x, we get x = 750 The usual speed of the plane = 750km/hr
"21 hrs
54 Remedial Mathematics
~~~~~~~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(x3) =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 =15x 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 8Y45y+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(3xl) =60 2
3x2 x24 =0
8 x=  or3 3 since, sides of the triangle can not be negative x = 3 and sides =5xand3xl = 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 8x20 = 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 = (xl0) Kmlhr.
56 Remedial Mathematics
Time taken by fast train to cover 600 Km = 600 hrs
x
Time taken by slow train to cover 600 Km = 600 hrs xl0 600 _ 600 = 3 xl0 x 600x  600(x 10) x(xl0) =3
=> =>
x'l10x2000 x Neglecting x Speed to fast train speed of slow train
0 = 50 or x=40 = 40, we getx = 50 andxlO = 5010 =40 = 50 Km/hr = 40 KmIhr.
=
ANSWERS 1. 5. 9. 11. 13.
(24 and21)or (21 and24) 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 + xI = 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 xI = 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) lOor6 (c) lOor6 (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)
x24.J3x9 =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) x29x20=0 (b) x29x+200
..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
Remedial Mathematics
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 +2x1O =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. xI 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. xI
xI
[UPTUB.Pharma2005j
60
Remedial Mathematics
[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+ 5y7 =0 (ii) 5x8y= 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 8y2 = 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,x2y=1. Solution: The given equations are and putting
2x+3y = 16
...(1)
x2y =1
... (2)
x = 5 andy = 2 ip(l) we get IRS =2x5+3x2=16=RHS.
62
Remedial Mathematics
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
Remedial Mathematics
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 (33)=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 = ;; (24x). '
Simultaneous Linear Equatioll 65
Now
1
x =1
y=
3"
x =2
y=
3"
1
[24(1)]=2. (24x2)]=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. 3xy =2 Now
y=3x2.
x =1
y=3 x l2=1.
x =2
y=3 x22=4.
66
Remedial Mathematics
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
9x3y = 6
~
y= (9x6) 3
x =0
~
Y =  (9 x 0  6) =2 3
x =3
~
y= 3(9 x 36)=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+If++++++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.
Simultaneous Linear Equation
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 =62x x Y
4
I
4
2
2
2
x2y+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
Remedial Mathematics
Solution: Graphof2x3y= 5. 2x3y = 5
'3I (2x5).
=>
y =
Now
x = I 1
'3 (2x 15)=1.
Y =
x = 4 y
'I3 (2 x 4  5) =  1.
=
Thus, we have the following table rr~~
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. Graphof6y4x=3 6y4x = 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 of6y4x = 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
~
5xy7 = 0 xy+ 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+2y4 =0 ... ( 1) 3x+2y+ 1 =0 ...(2)
~ =~=I·!!L=~=I·.:L=44 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 x2y =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
Remedial Mathematics
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 +yl =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 = (2kl), b2 = (kl), c2 =(2k+ 1).
where
for no solution, we must have
~
=
a2
!!L *.:L . b2
c2
1 1 3 =* kl (2k+l)' 2k 1
3 I 1 1 =   and   *  2k  1 k 1 k  1 2k + 1 (3k3 =2kl)and(2k+l*kl) 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=3,2x3y=7. 3. 2x+ 3y=4, 3xy=5. 4. 4x+3y=5,2yx=7. 5. 3x+y+ 1 =0,2x3y*8=0. 6. Solve the following system of linear equations graphically. 2x3y17=0,4x+y13 =0. Shade the region between the lines and the xaxis. 7. Solve the following system of linear equations grapJ,;r l l 2x+y5 =0,x+y3 =0. fmd the points where th~ graph meet the xaxis. 8. Solve the following system of linear equations graphically 2xy4=0;x+y+ 1 =0,.
Simultaneous Lilleal" Equation
71
9. Solve the following system of linear equations graphically. xy+ I =0, 3x+2y12=0. Calculate the area bounded by these lines and the xaxis. 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) 2x3y= 5, 6x9y= 15. (iii) 6xlOy=3,3x5y=7. ( iv)
~+.,[ =3 x2y=2.
3 2 ' 11. For what value of k, the system of equations kx+2y= 5, 3x4y= 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,(2kI)x+(kl)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 noninteger 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 crossmultiplication method.
72
Remedial Mathematics
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
2xy3 =0 4xy 5 =0
... (1) ... (2)
y =2x3 from(l): Substitute y from (3) in (2), we get 4x(2x3)5 =0
... (3)
4x  2x  3  5 = 0 2x=2=:.x=1 Substituting x = I in (3), we get
y =23 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 signthen 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 )
Simultaneous Linear Equation
6x5y=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 30x25y =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+ 3y9 =0 4x+6y18 =0 3y =92x from(l).
92x y=
... (1) ... (2)
... (3)
3
substitutingy from (3) in (2), we get.
=>
4x + 6 (
9
~2x ) 
18 = 0
4x+ 2 (92x)18 = 0
=>
4x4x + 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
=
92x 3' 92x
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
Remedial Mathematics
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
702y455y =360 265 247y =95 Substitutingy =
135
95 y 247
5
5
13 =>y13
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
=>
u2v=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
62v =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
Remedial Mathematics
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+3y17=O ... (1) 3x2y6 = 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
=>
1834
.::....y=51+12 49·
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+y2 = 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+y2 =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 ==
ab
and
(ab) a
ab
a 2b 2 (ab)
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 auab *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+y3 =0
3. 2x+y=35 3x+4y=65 5. 4x7y+28=0 5y7x+9=0. 2 x
3
5 x
4
6.
Solve  +  = 13 and =2, wherex*O,y *0.
7.
ax+by=(ab)
y
y
8. ~= l.
~ + l. = (0 + b) a
b
a
b
1
x
1 y
2 x
y
ax + by= if + b2 .
bxay= (a + b) 9.
°
4x+3y47=0. 4. 3x5y+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:
4x7y+28 = 0 7x+5y+9=0
... (1 )
... (2)
By cross multiplication, we have x 7 x 95 x 28
=>
y
28 x (7)  9 x 4
x
y
63 140
19636
x
=>
4 x 5  (  7)( 7)
=
2049·
=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+3v13 = 0 5u4v+2 = 0
We get
y
...(1) ... (2)
8. The given equations are bxay = 0 ax+by(el+b 2) = o. 9. The given equations are bx+ ayab (a+ b) = 0
b2 X + ely  2el b2 = O. 10. Put
1
x
= u and  =v. y
EXERCISE 2.3 ~~~~~~~~~ Solve 1. xy=3
3x2y= 10 3. 2x+y=7 4x3y=1 5. 4x3y=8 29 6xy= 
3 8 7.2x+5y=3 5 3x2y== . 6
2. 2x+ 3y=0 3x+4y=5.
4. l1x+ 15y+23=0 7x2y20=0 6. 0.4x1.5y=6.5 O.3x + O.2y = 0.9
8. 31x+23y=39 23x+31y= 15.
80
Remedial Mathematics
9 4 10.    = 8.
9. ax+ by=a b
x Y 13 7  +  = 101, (x;toO,y;t:O) x y
bxay= 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
xy 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
3x2y

2
  +   =2 where (2x+ 3y);toO, (3x2y);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=(72x) substitutingy = (7  2x) in (2), we get 4x3(7 2x) = 1 4x21 +6x =1 => lOx = 20 => x=2 => from (1): 2 x 2+y=7 y=3. 4. The given equations are l1x + 15 y =  23 7x2y = 20 Multiplying (1) by 2 and (2) by 15 we get 22x+30y = 46 105x30y = 300
... (1 )
... (2)
...(3) ...(4)
Simultaneous Linear Equation
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 = (1395315)= 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 = (x5) years. The son's age 5 years ago = (y5) years. ... (1) .. (x5) = 7 (y5) => x7y=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
Remedial Mathematics
x3y = 10
=:>
...(2)
on subtracting (1) from (2)
4y =40
=:>
y= 10.
x3 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. (x4) = 4 (y4) x4y+ 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
2x5y18 =0 x =4y12. from (1) Substituting x from (3) in (2), we get 2(4y12)15y18 =0 or 8y245y18 =0 or 3y =42 or
y
=
... (2) ... (3)
14
substitutingy = 14 in (2) we get
x=4 x 1412=:>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. 4x5y =4
...(1)
Simultaneous Linear Equation
Number formed on reversing the digits = (lOy + x). .. IOx+y18 = lOy+x => 9 (xy) =18 => xy=2. Multiplying (2) by 5 and subtracting (1) from the result. We get x = 6. Substituting x = 6 in (2), we get y = (62)=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 (2xy) = 0 => 2xy= 0 => y=2x ... (1) => Thus, unit digit = 2 x tens digit (unit digit) > (tens digit) so y > x yx =3 ...(2) using (1) is (2), we get (2xx) =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
Remedial Mathematics
3x yl
=> => and
3 2 6x3y =3
=>
6x=3y3
2x3y =1 x + 12 3y
1 = 3
...(1)
=>
3x+36=3y
3x3y = 36 xy=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: 2xy1 = 0
xy+4 = o. Y = 2xl
...(3)
from (1) substitutingy from (3) is (2), we get
x(2xI)+4 =0
... (1)
... (2)
=>
x=5.
substituting x = 5 in (3), we get
y =101=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.
Simultaneous Linear Equation
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 (8x19y). and B's monthly saving = Rs (7x  16y).
..
8x19y =2500 7x16y =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 x3y =2 x+4y =5 Adding (1) and (2) we gety = 7 substituting y = 7 in (i) we get x3 x 7 =2 ~ x=23. the larger number = 23 and the smaller number = 7.
... (1) ...(2)
...(1) ...(2)
86
Remedial Mathematics
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 19x8y =12
... (2)
10. Let the required fraction be .::.. The according to the given condition, we have
y
x +1 y +1
4
x5 y5
1
  =  and   = . 5
2
5 (x + 1) = 4 (y + 1) and 2 (x  5) = (y  5). 5x4y =1 2xy =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 3y6 = 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) kci' 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
J8x2 fjy=5
J8x2 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  x3y+5=0 3
3 3 3 x+ 3y+ 1 = 0 (d) xy+)=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
Remedial Mathematics
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.
Simultaneous Linear Equation
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 4x5y=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 2x3y = 5, 6y4x= 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)
=356=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::~ :::Ia
12
=a
I: : :::I+ I: :
::~I
a13
ll (a22a33a32a23)a 12 (a2Ia33a31a23)+ a 13 (a21a32a3Ia22)
2
IA I
For example,
3
= I
5
2 3 2
4
1 31_ 31 31+411 21 2 I 4 I 42 2
=21
=2(2 +6)3(112)+4(28) = 16+3924=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
Remedial Mathematics
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.
__
COFACTORS 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 cofactor of the element a , .. Hence, in a determinant of order n 1J n
laijl
=ai1AI1+ai2A,2+····+a,~in= 'L,aijAij jl
Now, let ~J be the (nl) x (n 1) submatrix 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 = 152= 13
M l2 =
Ml3 =
I~ ~I = 1  6 =  5
M21 =
M22 =
I~ ~I = 20 2 = 18
M23 =
M31 =
I~ ~I =63 =3
M32 =
Mil
M33
I~ ~I =54= 1 I~ ~I = 15 1 = 14 I~ ~I =46=2 I~ 211 =81=7 '
=I~ ~1=123=9.
The cofactors 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
Remedial Mathematics
Example 3: Find the minor and cofactors ofelements ofthe following determinant
2 3
5
6
0
4
1
5 7
Solution: We have
0
M1·1 
41 = 020 =20
5 1
7
7
M12
=
I~
Ml3
=
I~ ~I =300=30
M2l
3 51 =2125 =4 = 15 7
M22
=
I~
M 23
=
I~
M31
3 = 10 :1 =120=12
M32
=
M33
=I~
=20
Al3
=30
A2l
=4
41 =424=46
51 =145 =19
7
31 5 =10+3=13
I~
All
A22 =19
A 23 =13
A3l
=12
:1 = 830= 22
A32
= 22
31 0=0+18=18
A33 =18.
Example 4: Write the cofactors 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 =abbc A22 = Cl + 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
Remedial Mathematics
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 byl 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~ ~I21~ ~I + 31~ ~I
= 1. (61)2. (43) + 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~ ~Iol~ ~I + ll~ ~I =4(10)0+ I (04) =44=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~ ~Iol; ~Iol; ~I =xy=R.H.S. Example 6: Without expanding, show that
be ea ab ea ab be ab be ea
=
0
Solution: We have
be ea ab ea ab be ab be ea Example 7: Without expanding, show that
b 2 e2
0 ea
ab 0 ab bc 0 be ca
(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+c2b=0 Operating Rl ~ Rl + R3  2R2 , we get x+ 1 x+2
x+a
0
0
a+c2b
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
Remedial Mathematics
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 ca
= abc [(b  a) (c 2 ~)  (b 2  a2) (c  a)] = abc [(b a)(ca) {(c + a)  (b + a)}] = abc (ba) (ca) (c+ a ba) = abc (a b)(b  c)(ca) 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
abe 2b 2e
2a bea 2e
2a 2b =(a+b+el eab
Solution: Operating R] ~ R] + R2 + R3, we get
a b e 2a 2a 2b bea 2b 2e 2e eab
a + b +e a + b + e a + b + e 2b bea 2b 2e 2e eab [Take (a + b + c) out from Rd
=(a+b+e) 2b bea 2b 2e 2e eab (Operate C2 ~ C2  C] and C3 ~ C3  C])
o
0
=(a+b+e) 2b bea o (expand by R]) 2e 0 abe =(a+b+e) 1 (abe)(abe)=(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
Remedial Mathematics
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
ba b 2 _a 2
0 ea
e 2 _a 2
ba b 2 _a 2 ea
on expanding the determinant along C 1 =
(ba)(ea) 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)(ca) [c + a (b + a)] = (b a)(ca) (c b)
= (a b}(b c)(ca). 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 cb ac
o
ab bc
Applying R2 ~ R2  R\, R3 ~ R3  R\
=(a+b+c) ICb acl ab bc = (a+ b + c) {(bd(ab)(ac)} = (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) addingA 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: Wehave3A4B =3 [2
Example 3:
[~ ~] + [~ _~] [~=~ ~:~] [~
/
1 3 1]_4[1 2 3 ] 1 5 01
[~ ~ l~][~ 48 4] 12
1 7]
64 = [ 00
98 3(4)] = [2 3(4) 1512 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~] 21 C =
~[_~
0(1)
11
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 1I] [ 3
I 2
I 2
2
8 7
Example 5: Solve the/ollowing equations/or A and B, 2AB =
. Solution: GIven 2A  B = [33
3 3
[~ 33 20] and2B +A = [J4
J
4
~]
~l
Multiplying both sides by 2, we get
3 3
4A2B = 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] 61 6+4 44 A
~J
=.5!.fO5
5 10
~] =[~
!] =[105 1 2
~]
5 10
~]
136
Remedial Mathematics
Now substituting the value of A in equation (2), we get 2B
= =
Example 6:.if A = [ 1/3
sin
2
[42 1(1) 51] 11 42 40
=
[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 etan e sin e+cos e] = [1 11] 1/3+2/3 cosec 2ecot 2 e 1 Example 7: .if A
=
[~
;
:]
and B =
456
Solution: We have 3A  4B = 3
[~ ~ ~] thenfind 3A  4B . 001
[~ ~ ~14 [~ ~ ~l 456
=
001
[~ ! 1~1[~ ~ ~l 12 15 18
=
•
04 30 60 94 [ 120 150
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 prefactor 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
Remedial Mathematics
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+21 0+02
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) 1.11.2  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
Remedial Mathematics
945
88+0
88+0] = 88+0 945 88+0 [ 88+0 88+0 945
~ [~ ~ ~] 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 =
[41 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
Remedial Mathematics
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 12.2
... (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 12k
=[O+2k).34k.1 k.3+(l2k).1 =
1+2k
][3 4] I
k.
4k] then 12k
I
(1+2k).4+ C4k)CI)] = [3+2k 4Ck1)] k.(4) + (l2k).(I) k+1 12k
[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 [~ ~]
[UPTU B. Pharma 2002)
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 + BD = 0
Solution: We have A + B  D = 0 or D = A+B
1 2] [3 2] [13 22] [ 2 0] [P q] = [2:  ~0] . = [: ~
D = 3 4 +
or
5 6
D
Hence
= 3 + 1 45
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
Remedial Mathematics
(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 ~ a2b 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] [12 1+1] [1 0] =
1
=
22 2+1
01
a +b aI] ~I] = [abb b+ I 2
and
A2+B2
Also
A
=[1°
2 0]+[a +b aI] I ab  b b + I
B 
1+a
0]
[I+a
2 x 2+b
(1+a)2 40] = [ (2+b)(al) Now given that (A + B/ = A2 + 8 2. Hence from (I) and (2) , we get 2 [a +bl (1+a)2 [ (2+b)(al) 4 = abb
0]
a I
Hence
abI]
+ G=;]+[: ~I] G:: =;:;]=G::
(A+ Bi = [ 2+b
or
=[a 2ab+bl 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]. Find2A3B. 3
[~ ~ ~]andB=[~ ~ ~].Find3A2B. 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 ~lFindA25A =
1
[~ ~2 =; o~]
5. IfA = 2 1 3, B = 2 1 andC, = 4 3 1 12 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
Remedial Mathematics
~~~~~~~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 = Si Ai Proof: Let A i and SI exist. Since A and B are non  singular, therefore, (AB) (B 1 AI) = A (BSI) AI, (By associative law) (By inverse property)
= AlAI =AA I =1
Similarly, (SI AI) (AB) = I (SI AI) (AB) = (AB)(SI AI)=I i.e., SI AI is tIl1e inverse of AB or (AB)I = SI AI and as such AB is also non  singular Theorem 4: IfA is a non  singular matrix, then (Air i =A. Proof: Let AI be the given matrix instead of A, then (AI)A =A(AI)=l.
This show that A is the unique inverse of AI i.e., A =(Airl Theorem 5: The inverse of the transpose ofa matrix A is the transpose ofthe inverse ofA, i.e., (AT i = (Ai) '. Proof: We have AAi = Ai A .. (AA I ), =f=(AIA),or(AI),A'=I=A'(A I ), Hence, (A"I)' is the inverse of A' (~.4')1 = (AI),
•
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 AI 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),A1 =A1 (A 1),=1. Hence AI 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 =
[56 15+15
=(11)
Also
(adj A)A =
2+2] = [II 65 0
0] 11
[~ ~] = IAI I
[=~ ~2] [~ ~5]
= [56
3+3 = [
10+10] 65
~ 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
=112 313 =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
Remedial Mathematics
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 + 15 244+6 12615] 000 25103 000
= [
=
[
Example 4: Find the inverse ofA
8410
368+0
20 0
28 12
9] 0
6
28
14
=
[~
156+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(2425)2(1215)+3(1012) =1+66 =1. Now
AI
~ 01A {~ ~
J [~ ~: ~I]
ISS
156
Remedial Mathematics
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(232)7(10 24) +3(40+6)
682 =70. as IAI :#: 0 so AI 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
..
AI =
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 AI exists.
= 1(4435)2(33+10)+3(218) =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 AI =
A31 =2 A32=4 A33=2
4
=
133 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
[UPTU B. Pharma 2001)
IAI = 1(61)2(43)+3(29) = 5 221 =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
AI
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) = 42+25 =27
Since IAI:t; 0, therefore AI 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
AI
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(122)3(161)+4(83) = 2045 +20 =5.
IAI * 0, Therefore AI 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 AI
[UPTU B. Pharma 200032008) Solution: We have
[~1
22
2~]
Matrices
159
! :]4[~ 2~]5[~ ~ ~]
2 A 4A5/= [: 889221001
9 4 5 880
880] = 8=8=0 945 880 [ 880 880 945
n
~ [~ ~
Now we have to find AI Multiplying both sides of A2  4A  51 = 0 by AI, we get AI (A 2 4A 51) = O. 2 AI A 4A IA5A I 1=0 or
5A
1
~A41~ [~
~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 AI.
[UPTU B. pharma 20041
2I 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
Remedial Mathematics
Now A 3 6A 2 +9A41 =
[
22 21 21 22
21; 65 [ 6 21
21
22
21
5
0 0 0] = 0 0 0 =0. [ 000 Now we have to find AI Multiplying both sides of A 3_ 6A 2 + 9A  41 = 0 by AI, we get AI (A 3 6A 2 +9A,4J) = 0 £1 A 36A I A 2 +9A I A4AI 1 = 0 A 2 6A +914A = 0 4A I = A26A +91.
[~ ~]f' ~}9[~ ~] 5
1
0
6
2 1
0
5
1
[: AI 
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) AI. Solution:We have
(i)
IAI
=
[~ ~ ~] I
I
2
1 (43)2(63)+ I (32) =16+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 AI
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
Remedial Mathematics
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
cososinosinocosoj
[ sin Ocos Ocos 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
.
II
.
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 li    9
2
Hence, the matrix A is unitary .
2
2
2
163
164 Remedial Mathematics
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]
[RGPV B. Pharma 20011
Matrices
165
I ANSWERS
[~2 ~2] 4
1.
4.
9.
11.
2.
5 2
[~I1
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 nonsingular matrix, then its inverse exists. Hence, we have AI (AX) = £1 B
X=A1B This matrix equation provides solution for the variables x, y and z .
166 Remedial Mathematics
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 AI 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 = [
AI =
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 (10)1(01)= 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
AI
~ 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(11)1(12)+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 II
I
11
2
=2,
1
=
= 0.
1=3.
A 22 =3,
A 23 = I,
A31
=2,
A33 =2.
[~ ~3 ~ 1= [~ ~3 ~ 1 2
AI
IAI . Then we have
= (_1)1 + 1 11 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: 2xy+3z=l, x+2yz=2, 5y5z=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)=105 + 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 2xy= 13k, 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
AI=~(adjA)=.!.[2 IAI
We have
=>
I]
I 2
5 1
_I 1 [2 X=A B="5 1
I]
2
1][13k] 2 2+k
1 1 x =  [45kl, y=  [3+5kl,
5
5
1 [26k+2+k] 1+3k+4+2k
="5
I [45k] 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 :. AI B.
Where
Now
IAI
~ [~ : ~] ~
100)2(20)+
1(01)~
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 ="42 [ 1
3
1
2o 1] 2
x=l,y=l, z=l.
2 3
2
3
14\
~4
170
Remedial Mathematics
Example 6: Using matrices solve the equations 2xy + 3z =9 x+y+z=6 xy+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 (11)+3
(11)~46~2.
1
The cofactors ofthe elements of IAI are
2 AI2 = 0 A\3 =2
All =
A21
A31
=1 A 23 = 1
AI
0 1
A33 =3
2) [2
1 = 0 3 2
~ ~~IA ~ ~IP2
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. 2xy=2 7x+3y=5 3x+4y=3 3. 4x3y=3 4. 2x+5y= 1 3x  5y = 7 3x + 2y = 7
Matrices
5. 2x+y+z= 1 x2yz= 3/2 5y5z=9 7. 2x+3y+5z=5 x2y+z=4 3xy2z=3
171
6. xy+z=4
2x+y3z=O x+y+z=O 8. xy+2z=7 3x+4y5z=5 2xy+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.
[RGPV B. Pharma 2001]
By finding inverse of coefficient matrix.
[RGPV B. Pharma 20011
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.
AI = 
/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 nonzero 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 nonsingular.
Remarks • If the rank of a matrix is zero, then matrix is a null matrix • The rank of every nonzero 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 nonzero entry. (ii) The number of zeros before the first nonzero 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 nonzero 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 nonzero rows since rank ofA is equal to the number of nonzero 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 rrowed square matrix whose determinant is nonzero, 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 nonzero 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 nonzero number. (iii) Addition of any row to K times the other row, where K is any nonzero number.
Remarks • lfthe elementary transformation (or E transformation) is performed on rows, then it is called rowtransformation. • If the Etransformation is performed on column, it is called columntransformation .
•
ELEMENTARY MATRICES
Definition: A matrix which is obtained by a single Etransformation is called an elementary matrix. For example
[:
A~ [~
0 1 o , 0 0I :] cto '][' 0 o 0 0 10 8 4
~']
2
174
Remedial Mathematics
Here first E matrix is obtained from 13 by interchanging C 1 and C3 columns and the second Ematrix is obtained by RI~R2 + 2R2.
Remarks • All the elementary matrices ar~ non  singular. • Each elementary matrix possesses its inverse.
III
INVARIANCE OF RANK UNDERETRANSFORMATION
Casel. Elementary transformation (E transformation) do not change the rank ofa matrix. Case ll. Multiplication of the elements of a row by a nonzero 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 Etransformation. • The rank ofa matrix does not change by a columntransformation.
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 ofEtransformations, 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 ofEtransformations.
Corollary 2. IfA is a matrix oforder m x n ofrank r, then there exist nonsingular 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 ofEtransformations, 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., AA. (il) Symmetry. Let A and B be two matrices of order m x n and AB. This implies if B is obtained from A by a finite number of Etransformation, then A can also be obtained from B by a finite Etransformations. 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 Etransformation 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 Etransformations. 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 nonsingular matrix P such that PA =
[~]
... (i)
where G is a matrix of order r1 x n ofrankrl and 0 is a zero matrix oforder(mr 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 nonzero rows, therefore the matrix.
[~] B Cannot have more than r 1 nonzero 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 nonsingular 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 (3636) +2 (1818) + 3 (12  12)= 0
detA 2 = 1 (4848)+2(2424)+4(1212)=0 detA 3 = 1(7272)+3(2424)+4(1818)=0 det A4 = 2(7272)+3 (4848)+4(3636)=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 nonzero matrix. Hence, the rank of the given matrix is 1.
Matrices
(il)
A
=
177
[~ ! ~l 456
det A = 1(2425)+2 (2018) + 3 (15 16)
=1+43=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 = 46 =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 nonzero rows. Therefore, rank of A is equal to the number of nonzero 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 nonzero rows. Hence the rank of A2=2.
Example 3: Use Etransformation 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 nonzero rows. Hence, the rank of A = 3.
178
Remedial Mathematics
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 nonsingular 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 Etransformation on the matrix A on the L.H.S. of(1) until A reduced to the normal form. In this process we apply Erow transformation to prefactor 13 ofR.H.S. of(1) and Ecolumn transformation to postfactor 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
Remedial Mathematics
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 Etransformation, 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 nonzero 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~C3Cl
0
ba b3
_a 3
(ba)(ca)
.. (ba)(ca)
c3
C,+C,C,
0
a
b3
ca , 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 +cab 2 ab
1
I
= (Ja)(ca)[(c 2 +cab 2 ab)O] = (b  a) (c a) [(c 2  b2 ) + a (c b)] = (b  a)(ca) [(c b)(c + b + a)].
IAI = (ab)(bc)(ca)(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 nonzero minors of order 1. Hence rank (A) = 1.
182
Remedial Mathematics
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 =ca:#=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 =ba:#=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 nonzero 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
Remedial Mathematics
21. Change the following matrix A into normal form and find its rank
WA
~ U_~ ! ~l
(i,) A
~ [~ ~ ~l :
HINTS TO THE SELECTED PROBLEMS
1. A =
[~ ~ ~l 012
detA = 1 (20)2(40)+3(20) =28+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 (2425)2 (1820)+3(1516) =1+43 =0 Rank(A) < 3.
The square submatrices of A are Al = and
[~ !~], A2 =[!
det(A,) = 46=2*0 Rank(A) = 2
!]
etc.
Matrices
4. LetA =
185
1 25 70 85] [o 0 08 0
Since the matrix A is of Echelon form and contains three nonzero 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 nonzero rows. Hence Rank of A=3.
6. LetA =
[~ ~ ~ =~] 1
o
0 1
1 I 11
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
Remedial Mathematics
[~ ; i =~] Then A is converted to Echelon form having two nonzero rows. Hence Rank of A = 2.
7. A=
2 3 1 4I] [6 3 07 1 1 2
3
1
32
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 Nonzero 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 (612)1 (4 + 9) =1 + 65 =0
=> Rank of (A) < 3 The square submatrices of A are
A = t
and Hence Now
[12 31] [13 I]4 etc. ,A 2 =
detA t = 3 2 =5 *0 => Rank (A) =2. RankA = 2 B
1 2 Ij [
= 6
12 6 5105
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Remedial Mathematics
Performing R2 ~ R2 + 6R], R3 ~ R3 + 5R]
_[~1 ~ ~1] :. B is reduced to Echelon form, which having one nonzero 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 (8064)2 (6472) =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 Etransformation applied on 'A' and viseversa and further since we know that Etransformation 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
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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) NonHomogeneous 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 +···+amnxnO
.
... (1)
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Remedial Mathematics
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 nontrivial 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 nontrivial 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 (nr) 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
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Remedial Mathematics
Example 2: Find the solution o/the/ollowing system o/linear homogeneous equations: x+y+z=O 2xy3z = 0 3x5y+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
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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 nonzero 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 .
°
•
NONHOMOGENEOUS EQUATIONS
Let us consider a system of equations which are nonhomogeneous 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.
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Remedial Mathematics
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 NEQUATIONSIN NUNKNOWNS TO HAVE A UNIQUE SOLUTION
if A be an nrowed nonsingular 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 nrowed nonsingular,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 AI, we have AI AX=A l BorIX=A 1 B
17.
or
X = AI 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
AI AX = AI 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 xy+z=2 2x+yz=l. Let
and assume that the... exists a matrix X"
Then
[~ ~1
iJ[l[H
[~l
such that AX" B.
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Remedial Mathematics
Performing R2 ~ R2  R\ and R3 ~ R3  2R\
we get
[~~2 ~] [~] o 1 3
= [
z
:4] 11
Performing R \ ~ R2 + R3
~ [~;~ ~2][~] ~ [~l:J => => => =>
x2z =5,2y=4,y3z= 11 x2z = 5,y=2, 2 + 3z= 11 x =2z5,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 + yz =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 y3z =18 1,y= 3,z= 5.
Matrices
197
Example 4: Solve the following equations by matrix method x2y + 3z =6 3x+y4z =7 5x3y+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 nonzero rows and rank A = rank [A I B] = 2. Thus, the equation are consistent.
i.e., Let
x2y+3z = 6 7y13z = 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
y5z =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 + 2yz = 3, 3xy+2z= 1, 2x2y+ 3z=2,xy+ 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 nonzero rows. Thus rank A = rank of [A IB] = 3. Therefore the equations are consistent
and
x2yz = 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 + 2y2z =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 69
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
22 ] 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 nonzero 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 + y2z =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~R23Rl,R3 ~R32Rl'
[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 nonzero rows therefore the rank [A I B] = 3 and we see that
A
[~ ~l :
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Remedial Mathematics
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 AI

[~
1 1 1 2 1 A3
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 2xy + 3z= 9,x +y +z = 6,xy+ 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 ~ x3y8z+ 10=0, 3x+ y4z =0, 2x+ 5y+ 6z 13 = 0.
Matrices
203
4. 5x+3y+7z=4,3x+26y2z=9, 7x+2y+ lOz=5. 5. 5x6y+4z= 15, 7x+4y3z= 19, 2x+y+ 6z=46. 6. 2xy+3z= 8,x+2y+z=4, 3x+y4z=O. 7. Show that the following equations are not consistent.
2xy+ z=4, 3xy+ z= 6, 4xy+ 2z= 7 x+yz= 9. 8. Prove that the following system of equations have a unique solution
5x+3y+ 14z=4,y+2z= l,xy+2z=O. 9. Use matrix method to equation to solve 3x+y+2z=3, 2x3yz=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~R22RI,R3 ~R3RI
[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 NonZero 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
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Remedial Mathematics
Now R2 ~R23R2,R3 ~R32RI 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 '
~R3R2
We get

I~]
The Rank of [A IB] = Number of NonZero 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
x3y8z =10 y+2z = 3 Letz = c, theny = 3 2c andx= 2c1.
which gives
Hence the solution of the given system of equation isx= 2cl,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.
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Remedial Mathematics
ANSWERS 1. x= 1,y=2,z=3
2. x=3,y=0,z=1
3. Consistent:x=2cl,y=32c,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) Prefactor
(b) Postfactor
(c) Cofactor
(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
Remedial Mathematics
[' , '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 (AI) =   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 nonnull 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 = AI Ir I. If A is a nonsingular matrix, then (ATrl = (AIl. The inverse of a nonsingular 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
Remedial Mathematics
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"
[~ ~ ~]
~ !] ~
IUPTU B. Pharma 2001]
IUPTU B. Pharma 2002]
:]
usingA'4A  51" 0 and hence find AI. [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 Ai.
IUPTU B. Pharma 2006]
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

[UPTU B. Pharma 2001]
~]
find Ai. 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
Remedial Mathematics
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. Shortcut method 3. Stepdeviation 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. Shortcut 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=xA. 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
Remedial Mathematics
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)
(21)(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
Measure of Central Tendency
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. Shortcut 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
Remedial Mathematics
Step 4. Sum all the deviation to obtain I:.fd.
x=
A + I:.fd n 3. StepDeviation 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(xA)
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). 110 3
Class Interval Frequency
1120 11
2130 7
3140
4150 15
4
5160
6170 7
o
7180 3
Solution: Class Interval
Mid"aluex
Frequencyf
ft
110
5.5
3
16.5
1120
155
11
170.5
2130
25.5
7
178.5
3140
4
142.5
4150
35.5 45.5
15
682.5
5160
55.5
0
0
6170
65.5
7
458.5
7180
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
Oll
1122
2233
3344
4455
5566
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
3344
38.5
4455
49.5
5566
60.5
011 1122 2233
Total
"iJ= 103
27 34 28 0 15 16
Lfd' = 58
218
Remedial Mathematics
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) =38.56.194=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(nI) (n2) n(nI) 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 [n1Co +
(nI)C + 1
... + (nl)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'
2025 2530 3035 3540 4045 4550 5055 5560
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
Measure of Central Tendency
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
Remedial Mathematics
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.
Measure of Central Tendency
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=zz
then
fd = f(zz) '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 = zz 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
Remedial Mathematics
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 xA ~= n 'Lfd' M=Then Mean 'Lf Put the values of d', we get 'Lf(x  A) h 'Lf
=! h
'Lf (XA) 'Lf
=!
'Lfx _ A'Lf h' 'Lf 'Lf
M
=
i(MA)
Thus, we can say that the arithmetic mean is not independent of the change of origin and scale.
Measure of Central Tendency
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: 48 812 1216 1620 2024 2428 2832 3236 3640 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
Remedial Mathematics
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 2025 2530 3035 3540 4045 4550 44 Frequency 18 102 57 19 12. Compute the arithmetic average of the following data: 58 810 1012 1215 1517 1720 2025 05 2530 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: 1520 Class 05 510 1015 2025 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
07
717
1421
2128
2835
3542
4249
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
Remedial Mathematics
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
Remedial Mathematics
~~~~~~~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
Remedial Mathematics
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,
Measure of Central Tendency
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.Gnp = (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
Remedial Mathematics
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 Glogx = 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 010 1020 2030 3040 4050 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 010 1020 2030 3040 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
Remedial Mathematics
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.
Measure o/Central Tendency
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 010 1020 2030 3040 4050 Frequency 4 5 11 6 4 Solution: Class
Mid value x
Frequency
J/x
j7x
010 1020 2030 3040 4050
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 010 1020 2030 3040 Frequency 5 8 3 4 Solution: We have Class
Mid value x
Frequency(f)
J/x
j7x
010 1020 2030 3040
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
Remedial Mathematics
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
Measure of Central Tendency
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)
Measure a/Central Tendency
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 AKA
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
Remedial Mathematics
{, )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
4050 19
504> 25
6070
36
7080 72
8090 51
90100 43
6.Calculate the harmonic mean of the following frequency distribution Class Frequency
04 4
48 12
812
20
1216 9
1620 5
7.Compute the harmonic mean of the following frequency distribution Class Frequency
4050 12
504>
10
6070 15
7080 17
8090 8
90100 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
Remedial Mathematics
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(lr)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
INF 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
Remedial Mathematics
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
Measure a/Central Tendency
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)
Cumulative frequency
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
Remedial Mathematics
~)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 2030 3040 4050 5060 6070
/ Solution:
8
30
'2fj
16
(j)
Cumulative frequency
8 26 30 20 16 N= 100
8 34
(x)
2030 3040 4050 5060 6070
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 4050. So, 4050 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.
~NF Then, median is given by = I + 2 /
xi
(~x 10034) = 40+ 2
= 40+
30
xl0
(5034) 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
140150 5
150160 10
160170 20
170180 9
180190 6
190200 2
Measure of Central Tendency
247
Solution: Calculation of median Cost o/living index
No.'o/weeks
Cf
140150 150160 160170 170180 180190 190200
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 160170. Thus, 160170 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
Cumulative 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
Remedial Mathematics
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
57 7
810 12
1113 19
1416 10
1719 2
5. Compute the median for the following frequency distribution Variable 4550 SO55 5560 6065 6570 7075 7580 8085 8590 9095 Frequency 2 3 5 7 9 11 7 2 3 6. Calculate the median for the following frequency distribution Variable 05 510 1015 1520 2025 Frequency 4 16 15 2 2 7. Calculate the median for the following frequency distribution 010 1020 2030 3040 4050 5060 6070 7080 Variable Frequency
2
18
30
45
35
20
6
8. Compute the median ofthe following frequency distribution: Age 1519 2024 2529 3034 3539 Frequency
4
20
38
24
10
3 4044 4
9. Compute the median ofthe following frequency distribution: Class Frequency
04 4
46 6
68 8
812 12
1218 7
1820 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.
Measure a/Central Tendency
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  fl xi 2f  f1 It
I = Lower limit of the modal class,
where,
f= Frequency of the modal class, 11 = Frequency preceding the modal class,
fi
= Frequency succeeding the modal class,
i = Class width.
If11 andfj are both (or one) is greater than fthen we use the following formula Mode = 1+
It fi + f1
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
12.517.5
2
20
17.522.5
22
25
22.527.5
19
30
27.532.5
14
35 40 45
32.537.5 37.542.5 42.547.5 47.552.5 52.557.5
3 4 6 1
50 55
1
Here, by inspection, we see that the maximum frequency is given by = 22. So the class 17.522.5 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 4421
a
f  f1 xi 2f  f 1 +fi 222 x5 2x22219
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
020 2040
14
4060 6080 80100
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 4060. So for this
1=40,1=27'£1 =x,fi =y, i=20
Measure of Central Tendency
251
Then mode, M
= 1+
o
111 21  11  fi
48 = 40+
8
=
xi
27x x20 54x y
27x x20 54x y
8 (54xy) = 20 (27x) 2 (54xy) = 5 (27x) 1082x2y = 1355x 3x2y =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
100200
200300
300400
400500
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.200300 is the modal class in which 1=200,J= 25,L 1 = 12,J= 15,;= 100. Using the formula Mode = 1+
111 21  11  fi
= 200+
xi
2512 xl00 501215
= 200 + 1300 = 200 r 56;52 23 =256.52
Example 7: Compute the mode ofthefollowing distribution Class Frequency
1020 22
2030 35
3040 40
4050 62
5060 50
6070 45
7080 40
Solution: Here, the frequencies are regular and highest frequency 62 belongs to the class 4050. Thus 4050 is ~emodalclass in which 1=40,J=62,11 =40,fi =50, i= 10.
252
Remedial Mathematics
Using the formula Mode = 1+ =
f  f1 2f  f1 
40+
fi
xi
6240 xlO 2x624050
= 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 010 1020 2030 3040 4050 5060 No. of articles 14 17 22 26 23 18 [MEERUT B.Se. BIOTECH2005]
Solution: Table for Calculation of Mean Weight in 9 ms No ofartic.:lesf
010 1020 2030 3040 4050 5060
Mid value x
d=x35
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
Cumulative frequency
010 1020 2030 3040 4050 5060
14 17 22 26 23 18
14 31 53 79 102 120
C~O
r
item = 60 th item.
It belongs to 3040 class. Thus, using the formula Median
=
1+
~ (~  F )
= 30+~(6053) = 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
Measure o/Central Tendency
253
Mode Since highest frequency is 26 which belongs to 3040 class. Thus, using the formula Mode = 1+
I  II
21  II  Ii
= 30+
X
h
26 22 xlO 2x 26 22 23
= 30+~ =35,71. 5245
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 4060 608080100 100120120140 140160 160180180200 No. 0/students 6 9 11 14 20 15 10 8 7 5. Find the mode ofthe following frequency distribution:
Height Frequency
120124 125129 130134 135139 140144 145149 150154 2 8 15 20 10 5 5
6. Find the mode of the following frequency distribution: Marks Frequency
1025 6
2540 20
5570 26
4055 44
7085 3
85100
7. Find the mode for the following frequency distribution: 2025 Age No.o/persons 50
2530 70
3035 80
3540 180
4045 150
4550 120
5055 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
010 7
1020 8
2030 10
3040
4050 40
5060 35
6070 10
7080 7
10. Find the missing frequency for the given frequency distribution:
Classes Frequency
010 14
1020
2030 27
3040
4050 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~'rx
+ 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
Remedial Mathematics
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 xaxis, cosec e and cot e are not defined. When the terminal ray coincides with yaxis, 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 xaxis. 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
 =
Icos8
cosec 8 + cot 8.
Solution: Here, we have
L.H.S.
=
1 + cos 8 1 cos8 1 + cos 8 1 + cos 8 Icos 1 + cos8
/x
1 +cose sine
1 sine
cos 8 sine
=+
= cosec 8 + cot 8 = R.H.S. Example 2: Show that
tan8cot8 . 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
[RGPV B. Pharma 20021
(sec A  tan A)2
sin A)2 1 = ( cosA  cosA
=
(Isin A)2 cosA
(1 sin A)2 = (1 sin A)2 = cos
2
A
(lsin 2
A)
(1 sin A)2 (lsinA)(l +sinA)
= IsinA =R.H.S. 1+ sin A Example 4: Sh ow t hat
tanS + seeS 1 1 + sinS eosS = =  tanseeS + 1 easS IsinS
[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
(IsinS)
(1 
sin 8)
cos 2 S
=    cos8(1sin S)
268
Remedial Mathematics
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)33ab (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 Icos 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 = Icos 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)(sec8tan8)
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+sin8cos8)21cos8 [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
Remedial Mathematics
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 mn
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+nyl4(mn)2 = [(mn)(m+n)f Therefore, 4(m 2 + n2 + 2mn)4 (m 2 + n22mn) = [(mn)(m + n)f => 4m 2 + 4n 2 + 8mn  4m 2 4n2 + 8mn = (m 2 n2i => 16mn=(m2n 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 = (ab)(~ + 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~~216X2 = (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 4x21) (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
Remedial Mathematics
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  sin8 = sin8  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. Icot Itan (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 sec8tan8=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  sine = sine  cosec8  cot 8
or
+cosec e + cot e cosec e  cot e
2 sin e
274
Remedial Mathematics
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 secacos9 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 isx 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] Coordinates of point Q are (a,  b). X' t:=*"'r+ X A xeoordinates = cos ( x) a = cos (x) => cos(x) = a= cosx [By()] => = sin (x) :. ycoordinate Y' =>b=sin(x) Fig. 6.10 => sinex) = 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 = xeoordinate of P = a > 0 sin x = ycoordinate 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
coordinates 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 coordinates of point S are (a,  b) cos x = a>O sinx = b 0; cot x For both angle x and angle 21t + x, the coordinates 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~="...1JI_
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 Tratios . • 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 Tratios. 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
Remedial Mathematics
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(1tx)=cosx
(i +
x ) = cos x.
(b) cos(1t+x)=cosx
(d) sin(1t+x)=sinx. (c) sin (1tx) =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 (xy) = cos x cosy + sin x siny. 8. For all real x, (a)sin2x =2sinxcosx (b) cos 2x = cos2 X  sin2 x = 2 cos 2 xI = 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 (xy) (b) 2 sin x cosy = sin (x + y) + sin (xy) (c) 2 sin x siny = cos (xy) cos (x + y) (d) 2 cos x siny = sin (x + y) sin (xy). 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) sinxsiny = 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 (xy) = coi xsin 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
Remedial Mathematics
(b) We have tan
sin(~+ x) 2
(% + X)
1t
cos (  + x 2
)
cosx =cotx. = sinx
(c) We have
tan (1tx) =
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
.J6h
.Js.Ji
4
4
.J6+h
~IO+ 2.Js
4
4
cos
I
tan
0
~25 10/5
2J3
5
~~~~~~~I SOLVED Example 1: Given that sin (A + B) =
67.5°
~2h
36° ~IO 2.Js
~h +1
4
4
J2h
~h+ I ~2h
/5 +1 4
hI
~52.Js
EXAMPLES
Jj and cos (A  B) 2
=
~2h 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 + sine8) 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(3600600) = 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
[UPTU B. Pharma 2004)
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
Jjl
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
Jjl = Jj+l'
47t r.. 7t 18  cos 9" = ,,3 Sin 9
. 57t Sin
[UPTU B. Pharma 2002)
[UPTU B. Pharma 2003)
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
Remedial Mathematics
Example 7: . Find the vlaue ofsec(15000) x sin 390° [UPTU B. Pharma2008] Solution: sec (1500°) x sin 390° = sec(4 x 36060) 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) ItanAtanB 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 _ .J31
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°
[UPTU B. Pharma 2002]
0
[Diving Nr and Dr by cos BO]
tan 45° + tan 8° 1 tan 45°tan8° tan (45° + 8°)
= tan 53° = tan(9037) = 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 = (tanA2)(2tanAl) =
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
[UPTU B. Pharma 1006)
3Q4
Remedial Mathematics
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 coscos+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 ~ ~ ~
~ ~
:::>
[UPTU B. Pharma 2001)
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
IJ5 4
:±J5 4
< 0, we get
sin 18° =
J5 1 4
Remark • Similarly we canfind cos 18° =
J5 +1 ~IO2J5 , 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
Remedial Mathematics
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 smcos2smcos2 2 2 2 5A . A] = 2 cos A [. smTsm2 2
2cos A2 [2 cos 2 ¥+ 1sin2 ¥1] A 2 cos
2
[2 smsm .2
3A . A] = 4smAcoscos. 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
12_ 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 AI.
3. Show that Ci) sin2 72° _ sin2 600 =
..) CII
J5  1 8
...) C111
. 7t . 27t . 37t . 47t
5
SlnSlnSlnSln=
5
5
5
5
16
.
7t
. 137t
I
Sln+Sln=
10
(iv) 2 cos 8 =
10
2
~2 + ./2 + 2 cos 48.
307
308
Remedial Mathematics
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).
IsinA =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)} ] 12cosC[cos(AB)cos(A + B)]
1  2 cos C [2 cos A cos B] = 1  4 cos A cos B cos C. . B sm . C = 4 smsmcos. '. 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 AB  2' C C smcossmcos2 2 2 2
(1t
AB 2' C C .   C)  cossmcos2 sm 2 2, 2 2 2 C AB . C C 2 coscos2smcos2 2 2 ·2,
C[
CJ.
=
AB . 2 cos2' cossm2'
=
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 smsmcCls. 2 2 2
2
A Bsm. {1t'22A+ B}] 2
310
Remedial Mathematics
Example 5: cos A + cos Bcos C = 4 cos.icos B sin C I. 222 Solution: We have eosA+eosBcosC = 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 AB 2' . C =  2 smeos+ sm 2   1 2 2 2
A B)}} eosA2B1 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 eoseossm . 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(BA) Similarly. sin (A + 2B) = sin (C  B). Therefore, L.H.S. can be written as sin (A C) + sin (BA) + sin (CB).
'~
= 2 sini (B  C) eosi (2A B  C)2 sini (B C) eosi (BC) = 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 . BC. CA . AB =4smsmsm222
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 AI cos 2B = 2 cos2 B1 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(AB)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
Remedial 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 =I2sinisinB 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 (AB)  2   cos 2(A+B) 2
=
A2 + B) A2  B) A2 + B)] 1 cos (  [ cos (   cos ( 
=
. Asm, '2B] 1cos(A2+ B) [2 sm'2
= 1  cos
[~ C] [2 sin Asin B] 2 2 2
_2
,A,B,C I  2 smsmsm, 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 Ism +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 (AB) .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)]
AB) . (A B)]
+ . C [ cos ( 2 sm 22 +sm
2
. A . B] . C [2 smsm2 +sm2 2 2
=
2 + 2 smsmsm. . 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=1tC tan (A + B) = tan I1C tanA+tanB      =  tan C ItanA 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
Remedial Mathematics
Again,
A +B+C
=n
~
A +B
= n  C ~ cot (A + B) = cot (n  C)
cot A cosBl cot A cotB
 = 
~ ~
H ence,
cot C
cot A cot B1 =  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 coscoscosABC 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 SIDSIDSID. sin A + sin B + sin C 2 2 2
Trigonometry
. 3A + SIn . 3B + SIn . 3C = 4 coscoscos. 3A 3B 3C 5. SIn 222 ABC
1tA
1tB
1tC
6. cos + cos + cos =4 coscoscos. 22222 2 . "2 A + SIn . "2 B + Sin . "2 C = 1 + 4 SIn . (1tA) . (1tB) . (1tC) 7. Sin 2 Slll 2 SIn 2 . .2 A ·2 B .2 C 1 2 A B. C 8. Sin +SIn SIn  =  cOSCOSSIn.
2
2
2 2 2 2 2A 2B 2C A B. C 9. cos +cos cos  =2coscossm. 2 2 2 2 2 2 2A
.2B
A
B
. A . B . C
.2C
10. cos SIn Sill . =2 SillSlllSill. 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. tantan+tantan+tantan= 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)
5sin94cos9 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
Icos8 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' lsin8 5 (a) 12
(b) 25 13
13 25
(c)
(d) 25.
20. x =cotA+cosAandy=cotAcosA,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.
Itan 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+secel ( 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
[UPTU B. Pharma 2002)
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
Remedial Mathematics
(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 103
•
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 nonzero 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 101
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
326 Remedial Mathematics
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_ n1 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 =nl 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
Remedial Mathematics
70 22 7 = log (70 7 (ii) log+loglog 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 be
= log y = log z ea
, then show that x b + C a yC + a 
b~ + b  C = I.
ab
Solution: Let us suppose logx = logy be ea
=
logz ab
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 =
be
y
Then
c =
0
= 0 = log
1.
+ ab. ~ + b  c = 1.
10gb loge b =   = prove that ~ . b . eC = 1. ea ab

. log a 10gb loge Solution: Let   =   =  be
b.~ + b  C)
+ b (e  a) + e (a  b)}
ea
= k
ab
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
Remedial Mathematics
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
Remedial Mathematics
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) 103 = 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 nonzero digits.
Remark • Tofind the logofa given number x, use theformula logx
=
characteristic + mantissa.
334
Remedial Mathematics
~~~~~~~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 nonzero 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
Remedial Mathematics
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+loglog 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 log3510 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+ 1.30491.1035) 2 1 1 "2(2.92111.1035)= "2(1.8176) 0.9088. antilo (0.9088) 8.106
338
Remedial Mathematics
(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 1.65712 x 0.50513 x 0.7482) 3 =
=
.!.(3.31421.01022.2446) 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.
ax
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 ax 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[363373] glO PI 4.576 363 x 373
340 Remedial Mathematics
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) PH 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 102 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 LI 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
Remedial Mathematics
A=
Average life
..!. = __1_:k
4.3
x
104 .
4. 10glO P2 = ~Hv [~...!..] PI 4.576 1\ T2 10 10 787 = 536 x l8[T2373] 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 104 yearsI, 2325 years. 5. [Hp+] = 3.98 x lOl mol LI , [OH]
3. 3.84
= 2.5
x 1010 mol LI
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
Remedial Mathematics
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
[RGPV B. Pharma 2004]
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 yaxis 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 yaxis = a units. And, the distance of P from the xaxis = 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 coordinates 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 xcoordinates is 0 and the value of ycoordinate 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 xaxis are taken as (x, 0). • In the same way, for every point on yaxis its distance from yaxis is 0 i.e., abscissa is 0, therefore, the coordinate ofa point on yaxis 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
::::>4k3=2k2 ::::> k = 1/6.
Compairing ycoordinate
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 yaxis 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 xcoordinate, we have
kx3+lx4 0=>3k4=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+y9=0
Fig. 8.10
The equation ofthe given line is
3x+y9 = 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+39k9 = 0=>4k3=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.
[RGPV B. Pharma 20051
".
358
Remedial Mathematics
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 (2X3+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 +{O3}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.
System of Coordinates
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 xaxis. Also, find the coordinates of the point of intersection of AB and the x  axis. 6. Find the ratio in which theyaxis 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
•
Remedial Mathematics
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 xaxis. 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\(Y2Y3)+X2 (Y3Y\)+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.
System of Coordinates
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(42)+5(23)+11(34)] 2 1
= 2[8511]=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(Ik)+3(k3)+5(3+1)] 1
1
2
2
=  [2+2k+3k9+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(32)+7(21)+12x(l+3)1 = 215+7+481
= 25 sq units 1 Areaoft1ACD = 211x(221)+12(211)+7(l2)1 = .!.119+24071 = 107 sq. units. 2 :. AreaofquadrilateralABCD=25 + 107 = 132 sq. units.
362
Remedial Mathematics
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(2k6+2k)+(k+ 1)(62k2+2k)+(4k)(22k2k) = O. => kC4k6)4CkI)+C4+k)(4k2) = 0 => 4~6k4k+4+4~+ 14kS = 0 => S~+4k4 = 0 => 2~ + kl = 0 => (2kl)(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 (v2Y3)+x2 CY3YI)+x3 (vIY2)]
= 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 (v2Y3)+x2 (v3YI)+x 3 (vIY2)]
= .!.[3C9+3)+O(30)+C2)(OCa)] 2
=
.!. [IS  IS] = 0 2.._ ~_
System of Coordinates
363
Example 7: The coordinates 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(23)+ 1(2(22x)+2(22x+2)] :1:16 =[5x+I+2x+84x] :1:16 =7x+9
169 = 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
=>
169 = 7x=>x = 2:
put
x = 25/9iny=22x
we get
25 1850 32 y = 22 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. [3022] = 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
Remedial Mathematics
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 + e2b2be+e2ae~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) (20)] 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+12x20 42+158 28x14 49
1
2
2
4x2 2 7 7 2
x
=
4x2 11 8
=
System of Coordinates
~~~~~~~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
Remedial Mathematics
~~~~~~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 => (hI)2+(k3f= (h+2f+(kIf => 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
~(h3)2+(k0)2 +~(h+3)2 +(k;O)2 => =>
~(h_3)2+k2 < 9~(h+3)2 +k2 (h 3) + JC2 < {9 (h3f+JC2 < SI
~(h3)2 +k2
r
+(h+3)2+JC21S~(h+3i+k2
=>
12hSl < 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=~
=>
~(hae)2 +(kO)2 +~(h+ae)2 +(k 0)2
=>
~(hae)2 +k2 (h ae) + ~ =
=>
=
2a
= 2a
~(h+ae)2 +k2
4if + (h + ac)2 + ~4a ~(h+ae)2 +k2 (squaring both side)
4aeh4if = (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 (le2)
h2 a
+ 2
k2 2
a (1i)
= 1
Hence, locus of(h, k) is
h2 a
+ 2 or
k2 2
a (1i)
= 1
x2 y2 += 1 whereb 2 =a2 (Ie2) ' ~ 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 + [00] = [(2 hi + (0 _k)2] + [(2 hi + (0 _k)2] 16 = (2hi+~+(2hi+~ 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
System of Coordinates
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
Remedial Mathematics
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.
k3
k3
Slope of AB =   =  8(1) 9 13 Slope of AC =   2(1)
and
k3 9
2 3
2
= 3 ~3(k3)=18
3k9=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.
[UPTU B. Pharma 2007)
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
= (32
~) = (.!.
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 + 6y7 = 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.
52 3 SlopeofAB= 41 ="3 = 1 =m, (say). 35 2 Slope of BC= 64 =""2 =1 = m2 (say). 32 1 SlopeofAC= 61 ="5 =m 2 (say).
380
Remedial Mathematics
1xI =  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 32 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 yintercept 'c' and slope 'm' are given.
y
x
Fig.9.S Letthe given line meet yaxis inA and letP(x, 4) be any point on it. As theyintercept of the line is e. :. coordinates of A are (0, C).
The Straight Lines
381
Draw PB 1.. to xaxis and A C 1.. PB.
PBBC DB BPOA tan e =   OB yc 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 =MPMR =MPQL=yYt QR =LM=OMOL=xx t tane = PR = y Yt
QR
XXt
m= YYt xxI
YYt =m(x,x t )· Which is the equation of the line in the point slope form.
382
Remedial Mathematics
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= ONOL=x 1 x2 QS =MQMS=MQRL =YIY2 QT=MN=ONOM=xx l PT=NPNT=NPMQ =YYI In MQS, tan9 = QS = YI  Y2 RS xlx2
...(1)
= PT = Y  YI
... (2)
In flQTP, tan9
QT
xxI
from (1) and (2), we get
YI Y2 = Y Yl XI X2 xxI XXI = YYI XX2 YIY2 xxI
= YYI X2 XI Y2 YI Y _ YI = Y2  YI (xxI). 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 xaxis and yaxis respectively. y Let the line meet xaxis at point A and yaxis 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 xa yo =oa bo
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 xaxis. Let I be the given line meeting xaxis and yaxis 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
Remedial Mathematics
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 xaxis. 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 xaxis from A and P respectively and AN..l Pc. Now AN=BC=OC OB=xx] PN = PC  CN = PC  AB = Y  y] Also AP =r. In right angle triangle t.ANP. cose = AN = xx]. AP r i.e.,
and
xx] =r. cose
...(1)
sine = PN = Y  y] AP r y y] =r. sine
...(2)
y:
From (1) and (2), we get xx] = 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 xx] = 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 ofxaxis. Solution: Here, c=.,..2, andm =tan45° =L. Substituting these values iny= mx + c, we gety =x2 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 yy\ =m(xx\) :. The required equation of the line is y3 =2(x4) ::::> 2xy5 =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 ofxaxis 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
y3 = .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
y8
x6
  =   (Two point form). 28 36
y8 = x6 10 9 9y72 = IOx60 or IOx9y+ 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+2045k27+12k+ 12 =0
...(1)
386
Remedial Mathematics
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 xaxis be a. Then intercept on yaxis = 14  a.
its
:. Equation of the line is given by
~+y =1 a 14a As the point (3, 4) lies on it, we have
...(1)
~+_4_ =1 a 14a 3 (14a)+4a => 423a+4a => c?13a+42 => (a7)(a6) a Putting these values of a in (1),
14ac? = 14ac? =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 meetsxaxis atA andyaxis 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
YYI = Y2YI(x_XI) x2 XI
...(1)
xI =3'YI =4 x 2 =I'Y2=3 Putting these values in equation (1) we have
34 (x+3) 1+3 4(y 4) =7x21 4y 16 =7x21 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(xx]) ...(1)
m =3,x l =2'YI=3 Putting the values in (1) , the equation of required line is given by Y 3 =3 (x2)
=> =>
Y 3 =3x6 3x y3 =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
=>
=>
~+y7=~
388
Remedial Mathematics
Example 10. Find the equation of perpendicular bisector of the line segment joining the points A(2, 3) and B(6,  5) . Solution:
53 8 Slope =   =  =2. 62 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 = (x4) 2 => x2y =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+4yIl =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 xaxis atthe point A (a, 0) andyaxis 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 y2x
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 xaxis 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
Remedial Mathematics
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
83 y3 = (x4)
74
5 3
y3 = (x4) 5x3y11 =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. 42 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(x2)
=>
y+ 5 =3x6 3xy =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
y3
=
83 52 (x2)ory3
=
5 7 (x1)
 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
yk=S(xh)
36 7 y  =  (x(I))or49x35y+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 ofxaxis.
Solution: We have p = 3 and a = tan1
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 BL4...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 ofxaxis 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
x3 = y+5 =r. cos 45° sin 45°
392
Remedial Mathematics
x3 = y+5 =r.
1
)
J2
J2
.
~
Now, the point Q (
h h+ 3,
5) lies on the line x + y  6 = O.
~+3 ~56 =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
y2 xo x 21 = 04 =>y2= 4 =>4(y2)=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
xlx2
y3/2 x2 2y3 x2 ==>=3/23 21 36 1 2y3 =3 (x2)=>2y3 =3x+6 3x + 2y  9 = 0 is the required equation of median CF.
=> => => (iii) Allude AD 1.. Be.
y
31 2 SlopeofBC=   = . 14 3 3
A(2,3)
Slope of AD =  (': AD 1.. BC) 2 Equation of AD iSYYI = m (XXI) 3
~~~x
y2 ="2(xO)
3 y2 =x 2
Fig. 9.13
2y4 =3x=>3x2y+4=0. The equation of a line through A and making an angle of 45° with the x  axis is x2 y3 x2 y3 cos45° = sin45°
=> 1 = 1
Ji
=>
Ji
x2 =y3 =>xy+/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 x2 y3   ==r.
cos45°
sin 45°
r r x =2+ Ji,y=3+ Ji
394
Remedial Mathematics
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. xy4J2 =0
5. (i)x+ 4y= 8 (ii) 3x+2y9= 0 (iii) 3x2y+4 =0.
6. 4x+3y= 12
7. 3x+2y12=0
9. xy+ I = 0,
III
3. 7x3y+2=0
8. .J3x+y12=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=axc => 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 xaxIs = A coeff. of x.
. C  costant. term Intercept on yaxis = _ = 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
Remedial Mathematics
~~~~~~I
SOLVED EXAMPLES
~I~~~~~~
Example 1. Reduce 3x  4y + 5 = 0 to slope/orm andfind its intercept on yaxis. Solution: The given equation 3x  4y + 5 = 0 can be written as 4y = 3x+5 3 5
Y = '4x+
=>
'4'
. 5 Intercept on yaxIs = '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
4x3y+ 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 yintercept. 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 ymtercept 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
Remedial Mathematics
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 + 2y4 = O. Solution: Solving simultaneously the equation 2x  y + 3 = 0 and x + 2y4 = 0, we obtain
_x_ =L=_I_ 46 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 4x3y20 =0 6x+5y+& =0 Solving (1) and (2) by cross multiplication, we get __ x_= y '201& 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+3y13 =Oandx+2y 8=0 x y 1 We have 24+ 26 13 + 16 43 .
.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
x2y+1 =0 2x  5y + 3 = 0 and 5x  4y + k = 0 are concurrent. Solution: The equation of the lines are : x2y+ 1 =0 2x5y+3 =0 5x4y+ k =0 Solving (1) and (2) x =~=6+5 23 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 3x2y+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 = tanII 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
402 Remedial Mathematics
~~~~~~~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 =
1551=>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
=
43 7 andm =   = 2 22 4 Let Q be the angle between BA and Be. Then mlm2 tan 8 = '.~ l+mlm2
110/81
7/40/2)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:
xyJ) 5 =OandJ)x+y7=0
[UPTU B. Pharma 2001] .
404 Remedial Mathematics Solution: The given two lines are: xyfj 5 =0. fj x+ y7 =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
27 0(5)
9 5
and m2 = Slope of the Iinejoining the points (1,3) and (4, x).
x3 x3 ==41 3 If the given two lines are perpendicular, then m1m2 =1
(_~)
(X;3)
=I~9(x3)=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
= 124 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
4X6X3+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 2x5y + 6 = O? Solution: Perpendicular distance of (2,  4) from the given line is _ 2x25(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
2xO5x5+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
Remedial Mathematics
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
= 12XO0+31=
~4+1
PI
~
15·
and length of perpendicular of x  4y  7 = 0 from origin
7 P 104XO71
~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. [RGPV 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+4x33
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 7x5y29 = 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+2y5=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 4x6y5 = 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+2y5=O. .
hx\ a
Now (h, k) are gIven by  
(ax! + by! +c) a2 + b 2
k  y\
= = b
h+2 = k+1 = (625)
=>
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 + ayab = 0
I
ANSWERS
I
~, origin lie on the opposite side of the line.
1. 7/5
2.
3. kx\ ,*x2 (t) Oneone Onto Function: ifa jimction fA + B is both oneone 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 Oneone onto/unction
Remarks • Oneone onto mapping is also known as bijective or onetoone.
A
B
• For a oneone onto function, Rang = Codomain andx\ ,*x2 =::> j(x\) j(x2 )
'*
• j(x\) =j(x2 ) =::>x\ =x2
(g) Manyone Into Function: Afunctionf: A + B which is both manyone and into function is Fig. 10.10 Manyone 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 manyone into function. (i) Rank c Codomain (ii)x\ ,*x2 =::> j(x\) = j(x2 ).
(h) ManyoneOnto Function: ifjUnctionf: A +B is both manyone 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 Manyone onto/unction
416 Remedial Mathematics Therefore, for manyone onto function (i) Range = Codomain (ii) XI :FX2 ~ j(x l ) = f(x2)
Working Procedure [For Checking the injectivity (oneone) 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 oneone.
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 codomain 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_
, ,, ,
It
2
Fig.l0.t8
426
Remedial Mathematics
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
Remedial Mathematics
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 ::;; $, then (I) The sum function, denoted by f + g, is defined by if+ g) (x) = j(x) + g(x), with domain D. (ii) The difference function, denoted by f  g is defined by ifg) (x) = j{x) g(x), with domainD. (iii) The product function, denoted by f g is defined by ifg) (x) = j(x) . g(x), with domain D. (iv) The quotient function, denoted by
f
g
is defined by
f)(x) = f(X» ,with domain D', where (g g(x D' = {x:x
E
D,g(x)::;;O}::;;$
(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}.
Functions and Limits
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 oneone onto mapping, thenfo = Iyandfl 0 f= Ix • If f:X ~ Y and g:y ~ Z be two oneone onto mappings, then the mapping of go fis also oneone 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 oneone => f: A ~ B is oneone. (iii) g 0 f: A ~ C is onto and g : B ~ C is oneone => f: A ~ B is onto. (iv) g 0 f: A ~ C is oneone andf: A ~ B is onto => g : B ~ C is oneone. 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) =(x23x+2i3(x23x +2)+ 2 =x4 +9x2 +46x3 12x+4x23x2 + 9x6 + 2 = x4  6x3 + 1Ox2  3x
430
Remedial Mathematics
Example l(b): LetfN+R beafunctionsuchthat.f{x) =2x3 andg: Z +R beafunCtion
such that g(x)
x3
Find g of N + R. 2 Solution: Consider, (go.f) (x) = gf(x) = g(2x 3) = .
= 2x33 = 2(x3) =x3
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) gf (vii) fig _ Solution: Since, we have
=
../xI andg(x) (ii) g
+f
=
~4x2
'If x ER. (iii)
(v) fg (viii) glf
(vi) gf
.f{x) = ../xI 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)= ../xI + ~4x2 ,withdomainD (il) (g +j)(x) = g(x) +.f{x) = ~4  x 2 + ../xI, with domain D (iii)
« g)(x)
= .f{x) g(x) =
(iv) (gj)(x) = g(x).f{x) =
../xI ~4x2 ,withdomainD
~4x2  ../xl
f g
Functions and Limits
(v) (fg)(x)=1(x) . g(x) = ..Jxl· ~4_X2 =
431
~(xl)(4x2), with domain D
~4_x2 . ..Jxl = ~(4x2)(xI) ,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) ~4x2
~
(viii) The domain D" of
~(
XI)
4x2
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 oneone: 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
Remedial Mathematics
j(x) =j{y) x=y
=>
Since, x,y are arbitrary, therefore,j(x) = j{y)
=> =>
x=y'ilx,y f is oneone
{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~xI
and
(g oj) (x) = g(/{x»
gC~ + 2) =
=
xI
x 2 +2
(xI)
x 2 +2 =2 (x +2)1 x +I 2
Functions and Limits
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 => (ifl)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
Remedial Mathematics
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) = Ix21 Solution: (i) Here, we have;f(x) = tan x + 3 cosec x + x ~ fex) =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 ~ fex) =Ixl+ 1 = Ixl+ 1 =f(x) j(x) is even function. ~ (iii) We have, j(x) = Ix21 ~ j(x) = lx21 =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) Ix
=
f(x+y) l+xy [UPTU B. Pharna 200SJ
Solution: It is given that
~
j(x) =10g(I+X) Ix
...(1)
fiy)=10 g (I+ Y ) ly
...(2)
Adding (1) and (2) we get
I
f{x)+fiy) = 10g(I+X)+ 109(l+ y Ix ly) =
Again
f(~) l+xy
log[Cl + x)(1 + y)] = [1 + x + y + XYJ (lx)(ly) lxy+xy
=[I+2yj l_x+y l+xy
= IOg[l+X+ y++xy] lx y+xy Using (3) and (4), we conclude that
j(x)+fiy)
...(3)
=1 x+ y) J~I+xy
...(4}
Functions and Limits
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) =xl andg(y)=y feg(x) =j{g(x)} =.f(x2)=x21
So
feg(x)
Solutio,n: Given that
=x21.
436
Remedial Mathematics
Exercise 14: Given thatf(x)
I
xI
= , g(x) =   . Find the value ofg[f(x)].
Ix x I xI j(x) =  , g ( x ) = Ix x
Solution:
g[f(x)]
So
=g[I~J=[~IJ= II~~X) Ix
=x
(Ix)
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 onetoone from A ontoA. Find]l. 2. If A is anonempty 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) =
xI (ii) j(x)= ,x* 1 x+l
~Ix2 ,o~x~ 1
5. IfA = {a, b, e, d} andfcorrespondstothecartesianproduct {(a, b),(b, d),(e, a),(d, e)}. Show thatfis oneone 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) = 3x7. 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)= XI, 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!
Functions and Limits
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. Ix
18. Iff R~ R, where.f{x) =:x?+ 2 andg: R ~ R, whereg(x) = 1__1_, then find Ix (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
Ix, 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
HINTS TO THE SELECTED PROBLEMS
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 oneone 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(4y)= I23y rI(x) =123x
=> So (il) .f{x)
xI
= ,
x ;rI
x+I
Let
xI y=x+I (x+ l)y =(xI) xyx =(1 +y) x(yI) =(1 +y)
Fig. 10.24
I
438
Remedial Mathematics
_ {I + y) 1+ Y =yl 1 y
x
rl (x) = l+x ifx¢ 1 Ix
so
f(x)=~Ix2
(iiI)
y
Let Solving for
=
O:s;x:s;1
~1x2
1 = l_x2 x2 = 11
so
x=
~1i
rl(x) =
~1x2
,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 oneone 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)=x25,g(x)=2x+3 fo g(x)=j{g(x)} =j(2x+ 3)=(2x+ 3i5 =4x2+9 + 12x5 =4x2+ 12x+4
17.
A
B
Fig.lO.2S
1
f(x) = Ix fiflf(x)4} =
f(fC~J=f(I_~l=f(~l Ix
= f (
IX)
=;
1 1 = 1_(IX) = 1+ (Ix)
x 19. Wehavef:A +AwhereA = {x f(x) =
E
Ix
R, 0 :S;x:S; l}
X, if XEQ { Ix, if XIi!OQ
x
x
=x+(Ix)
=x
Fllnctions and Limits
439
JV(x)} = {f(X), if X e Q
f(1x), if xr£Q X,
if xeQ
l(1x), 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 Ix
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)
II
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
Remedial Mathematics
We therefore obtain 2
fix) = x 9 = (x3)(x+3) =(x+ 3). x3 x3 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. xt3 x3 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.
xta
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 xta
(ii) Lim[f(x)· g(x)] = {Lim f(x)} {Lim g(x)} . xta
x+a
xta
Lim I(x) (iii) Lim f(x) = X+a , with Lim g(x)"* a. xta g(x) Lim g(x) xta x+a
(iv) Lim [cj(x)] =c Lim j(x). x+a
X~(I
Some Standard Limits x" I 1. Lim =n. x+' xI
2. Lim
3. Lim(I+.!. J =e.
4.
5. Lim (I + x)'/x =e.
6.
x+a
X
xtro
x+o .
aXI
x+o
x
7. Llm =Ioga
8.
L. tanS Im=l S .
10.
9.
8+0
e
x"a" xa
= 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
Functions and Limits
~~~~~~~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
~ x2
x>I cx +bx+a
Solution: (i) lim (~.x2+ 1)= 1 _1 + 1 = 11 + 1 = 1. 2
3
x~l
(ii)
lim (I + x + .x2 + ... +x IO ) = 1 + ( 1) + (1 i + ... + (I) 10
x~I
=11+11+ ... +1=1. (iii)
4 +3 lim(4x+3) lim _x_ _ = .::!.x~=4_ __ x~4
x2
4x4+3
19 2
42
lim(x2) 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 xlo hm ,,.2
x~2
. (2x3«Fx I) (ii) I1m ~,,'''x>I
x 4
2x2 +x3
4
lim x _81 x>32x 2 5x3 2 . 3x xIO
Solution: (i) hm
x>2
2
x 4
=
(x2)(3x+5)
lim ''''
x~2
(x2)(x+2)
= lim 3x+5 x~2
=3x2+5 =!.!. 2+2
x+2
4
I. (2x3)(Fx 1) · (2x3)(Fx 1) (ii) I1m = 1m ~"":"":'x>I
x~1 (2x+3)(xI)
2x2 +x3
=
lim (2x3)(FxI) x>I (2x+3)(Fx +1)(Fx I)
lim _....:('2_x_~3:"..)_ x>I (2x +3)(Fx + 1) 1
2xl3
,= =  =5x2 10 (2 xl + 3)(.Ji + 1)
(iii)
· I1m
x4 81
x~32x
2
=
5x3
(x3)(x+3)(x 2 +9)
I.
1m ~'...:....":"":""'
x>3 =
2
lim (x+3)(x +9) x~3
=
(x3)(2x+l)
(2x+1)
(3+3)(3 2 +9) 2x3+1
6x18 108 ==
7
7
441
442
(iv)
Remedial Mathematics
' x4 4 \1m 2 x~.J2 x +3J2xS =
=
(x 2 +2)(x 2 2) 1m ~;=i';:,=x~.J2 (x+4J2)(xJ2) I'
2
lim (x +2),(x+J2)(xJ2) x~.J2 (x+4J2),(xJ2)
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
xa
, ,/3xl
(iii) h m   
x~2 2x 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 +vflx 2 ]
X~o[~ +J17] 2
[~+.J1O]
=~
= 1
2
'
(ii) Diagonalizing the numerator, we ha"e
\' [../x+a fi;][../2+a +..!2a] ' ../x+a fi; \1m = 1m ====,=:=x~a xa x~a (xa)[../x+a+fi;] =
' (x+a)2a \1m ~;==,..;===x~a(xaH../x+a +fi;]
' 1 = \1m ,,==.==
x~a[../x+a +fi;] 1
=
1
../a+a +fi; = 2J2a '
Functions and Limits
(iii) Rationalizing the numerator, we have . .J3xI 11m x+2 2x
=
=
=
. [.J3xI][.J3x+l] lim !:....:c~===~ x+2 (2x)[.J3x +1] . (3x)I hm     ; = = =  x+2 (2 x)[.J3 x + 1] . 2x I1m     ; = = =  x+2 (2  x) [..13  x + I]
=
. 1 IIm..,,=~
x+2[.J3x+I] 1
1 =1+1 2
[..132 +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
= IImhm
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 x3
=
108 .
n 3n Solution: Since lim ~ = 108 x+3 x3 n(3)'11 = 108 n.(3)n1 =3(3)41 n =4.
=> =>
Example 6: Evaluate
·log(I+x 3 )
(I) I1m ='x+o sinx
(iii)
lim x+7t/4
sinxcosx x1t/4
. Icosx·.Jcos2x (iz) I1m 2 x+o x . 1+ cos2x (iv) I1m 2. x+7t/2(1t2x)
.
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 lcosxJroS2x x~ x2
=
1
r lCOSXJroS2X[1+COSX~] x~ x2 1+cosx.Jcos2x
lcos2 xocos2x 0 = \1m :==~ 2 x~o x (1 + cos x.Jcos 2x ) 0 lcos2 x(12sin 2 x)
 \1m
,==~
 x~o
=
x 2 (1 + cos x..Jcos 2x )
0 lcos2 x+2cos 2 xosin 2 x 11m ::.,.::==___
x~o
x 2 (I +cosx.Jcos2x)
2 sin2 x+2cos xsin2 x 1Im,==,2 xM) 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 smxcosx = lim 4 4 x~x/4 (x1t/4) y~O y 01t
=lim
1to
1t
o1to
smocosy + cos °sm y  cosocosy +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
Functions and Limits
1 + cos 2x 2 x~1t/2 (n  2x)
=
0
11m
(iv)
445
l+cos2x
0
11m
x~1t/2 4 (nx )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 xa 0
(UPTU B. Pharma 2004)
o
Solution: We have lim cosx  cos a x)a xa 0 sinxO 10 ( 0 ) 0 = 11m == 1m smx =sma x)a 10 x)a Example 8: Evaluate e X+e x _2
[form ]
o
o
hm::
(UPTU B. Pharma 2007)
x)o x2 Solution: We have
eX +e x 2 e 2x 2ex +1 = hm ,,x)o x2 X)o x 2eX o
[:0 ex =
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)
(UPTU B. Pharma 2006)
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 hmx)o X
I
lim~ x)o X
0
0
446
Remedial Mathematics
=
(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
hmx+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. 10+0
[UPTU B. Pharma 1995)
Functions and Limits
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, wheneverao·<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~aO
limj(ah). 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 => IE <j(x) < 1+ E 'V X ~ 0 and we write
lim j(x) X~o:J
= I.
448
Remedial Mathematics
(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. xI
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(10)= Iimj(lh)= lim (Ihi =1 x+o
x+o
450
Remedial Mathematics
Therefore, we have .1(1 +0) =.1(10)*.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(Oh) = lim [sin2(0h)] =2. h+O (0  h)
h+O
Therefore, we have .1(0+0) =.1(00)=.1(0)=2. Hence,j{x) is continuous at x = 2. Example 4: Afunction.l(x) is defined asfollows j(x)
=
1+XifX:=>2 { 5xifx~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 [3h)]=3 ... (ii) h+O
h+O
lRL =.1(2 0) = lim .l(2h) = lim [1 +(2h)] = 3 .... (iii)
and
h+O
h+O
Now, from (I), (ii) and (iii), we have .1(2+0) =.1(2)=.1(20)=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
le
lIx
,x*O,j{x)Oatx=O.
[Meerut B. Sc. Biotech 20061
Functions and Limits
451
Solution: (i) Here, we have
lliL =j(OO) = lim j(Oh)= 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 =./(00)= lim j(Oh) h~O
=
limj(h) =
h~O
lim~ =0
h~O
le
RHL =./(00)= lim j(O+h)
and
h~O
l_lfh = 1. le 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) = 5x4, when 0 50x 50 1 = 4x3  3x, when 1 < x < 2. [Meerut M. Sc. Chemistry 2004] Solution: At x = 1,./(1) = 5.14 = 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
=[4xl3 x l]=1.
L.H.L. = Limj(x) = Limj(lh)
Also
x~l
h~O
= Lim[5(Ih)4] =5 x 14= 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
Remedial Mathematics
~~~~~~~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) xI f(x)I 7. Which of the following functions are even or odd? (i)
 x3
(ii) x 50
(iii) x +!
Ix l+x
9. Ifj{x) =   , then prove that:
1+ f(x)f(y) f(x) f(x 2 ) 1+ [f(x)f
xy l+xy
=
8. Ifj{x)= xI ,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 xa . 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 +2x1 2x+3
2 r1mxI
x~1
lim
xI x 2 x6
x~3x
3
2x3
2x2 7x+6 lim x~25x2 llx+2
lim
5x2
x~oo3x+5x
lim x~oo
2.
~7x4 _5x 3 +4x2 +3 3x 2
Functions and Limits
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
.
hmx+o x
(iv) lim (I+xtI x+o
x
x
lim aX +b 2 x 13. The functionf(x) is defined as follows:
(v)
x+o
5x4 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
Remedial Mathematics
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 oneone but not onto
E
N, where N is the set of all natural
(b) fis neither oneone nor onto
(c) fis oneone and onto (d)fis onto but not oneone
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) oneone only (b) onto only (c) both oneone and onto (d) neither oneone nor onto ThemappingfR+ 4R defined byj{x) = 10gloX, (whereR+ isthe set of all positive Number) is (a) (c)
7.
only oneone mapping both oneone 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 oneone (b) oneone but not onto (c) oneone and onto (d) into
8.
9.
10.
11.
R be defined by j{x) = 2x + sinx, for x
R. Then fis (b) oneone but not onto (a) oneone and onto (c) onto but not oneone (d)neither oneone 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 .
x2 (b)
E
6~
IS
1
1
(c)
7..J3
(d)
8..J3
Functions and Limits
12.
x+sinx is equal to xeosx (a) 0 (b) I
455
lim
x4'"
13.
lim
e= ell
xJI+Y
= 
x'(l +y)
=y(l +x)or(x2 y)+(x'yyx)=0
On squaring, we get
y.JI+x.
484 Remedial MatMmatics :::) :::) :::)
(~y)+xy(xy) =0 (x +y)(xy)+xy(xy) =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 .
[RGPV B. Pharma 2004)
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.
[RGPV B. Pharma 2001]
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~) .~~
[RGPV B. Pharma 2001]
(~)
(.!.)
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
Remedial Mathematics
=>
dy (3y3ax) = 3ay3x2 dx
=>
dy 3(Vax)  =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(xy), 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
=>
sinI x  sinI 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 = ax3 + 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 = ax3 + bx3 + cx9/2 sinx So,
dy =
dx
~eax3 + bx3 + cx9/ 2 sin x) dx
= a..!!.ex3 ) + b..!!.ex3 ) + c..!!.ex9/ 2 sin x) dx dx dx =3ax4 + 3bx2 + c(x9/ 2
!
sin x +
!
ex9/2)Sinx)
=3ax4+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
Remedial Mathematics
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
Now, differentiating both sides w.r.to x, we get
~ 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.
[RGPV B. Pharma 2002]
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
[RGPV B. Pharma 20011
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
Remedial Mathematics
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(1log(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(Ilog(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 · xcosxcos 2 2x sm (l+cosx)
=
(sinxp  (1 + cos x) (1 + cosx)(l cos x)
= 1 cos2 x 1 1 cosx
492
Remedial Mathematics
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 :
=8ax2x=2x(4al)
dy = x(4al) 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  =6ax3.r=3x(2ax) 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 ) +1md (m) y =(1) am dx b dx dx 1 dy mx mJ mJ +m y =0 am bm dx
or
bmmxm 1 + ammyml dy =0 dx
or
or (iii) Here, x = a (I  sin I), y = a (1  cos I)
SO, :
=a(lcos/)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
Remedial Mathematics
~~~~I MISCELLENEOUNS SOLVED EXAMPLES ~I~~~ +x Example 1: Find the differential coefficient at x = 0, ify = aX +  Ix
~
Solution: We have y = aX + ~I + x
1 x
Differentiating w.r.t. x, we get
1(I+X)1/2{10X)0+X)(l)} dy = a xloga+dx 2 1 x 0 x)2
IJ§X IJ§X{ +
= a Xl oga+   {IX+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 = sinI (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 = xa 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
= ~(xa) =
2x
xa
,....
2xvax
X
Example 4: y = e
x+ex+e +
dy dx
then show that
=L.
[UPTU B. Pharma 2003]
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)
!(1y) =y
dy dx
=L ly
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
Remedial Mathematics
3. Find the derivatives of the following function~ : (i) y=(x2)(x+2)(x3)(x+3) (ii) y=(x+ l)(aJ21) (iii) y = sinI (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 :
of the following functions:
(i) y = log sin x + cosI(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 (sinI 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. tanI (~+ =9 15. 3 sin (xy) + 4 cos (xy) = 5 16. ycosx=xy
I)
17. If
fY
~
V; + fy
dy x17y = 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 xy
X'%
l
l_x 6
Differentiation
(v)
~sin(a tan1 x)= dx
481
cos(a tan 1 x) ~(a tan 1 x) dx
= ~cos(atanl 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. smxcosx
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(lcot t)
rRGPV B. Pharma 2001 )
13. Find : ' x = log t + sin t, y = l + cos t
[RGPV B. Pharma 2002),
HINTS TO THE SELECTED PROBLEMS
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
Remedial Mathematics
dy d . 5. dx = dx [(x2+7x+2)(eXsmx)]
=(x2+7x+2)· =(x2+7x+2)
!
(eXsinx) + (eXsinx)
!
(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 eX7 cosx 2 sinx)x + 9 eX 2 cosx7 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 cosx2 sinx)x+ 9 ~2 cosx7 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 (Icost)
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 ~ ~ ~
(x2y)+xy(xy) =0 (x+y)(xy)+xy(xy) =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 .
[RGPV B. Pharma 2004)
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 £
:
(RGPV B. Pharma 2001]
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
[RGPV B. Pharma 20011
. 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
Remedial Mathematics
=>
(313ax) dy
=>
3(Vax) dy =3(ayx2)
dx
3ay3x2
=
dx
dy = ay
Hence,
+ ~1 dy dx
~1 
Solution: We have
2
iax
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 [xxI +xxlogx(ltlog x )].
ofthe followingfunctions:
(i) y = ax3 + b~ + cJJI2 sinx (iii) y = (1
(ii) Y =~
+ xy
(iv) Y
=
(ax + by
ogx
(v) y = (cos x/ . Solution: (i) We have, y = ax3 + bx3 + cx9/2 sinx
~
S
dx
0,
=
!!.(ax3 +bx3 +cx9/ 2 sinx) dx
=
a!!'(x3 ) + b!!'(x3 ) + c!!.(x9/ 2 sin x) dx dx dx
= 3ax4 + 3bx2 + c( x 9/ 2
!
sin x +
!
(X9/2)Sinx)
=3ax4+ 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
Remedial Mathematics
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   = x10g(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
Now, differentiating both sides w.r.to x, we get
~ 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
Remedial Mathematics
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(1log(tan x)
...(1)
Differentiation
= (tanx)CotX. cosec2x(llog(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(l10g (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 xcosxcos x _ (1 + cos x) (sinx)2  1 cos 2 X (1+cosx) (1 + cos x)(1  cos x)
I 1  cos x
492
Remedial Mathematics
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
=
8ax2x =2x(4al)
=
x(4al) y
ofthefollowingfunctions:
xm ym (ii)  + = I am bm (iii) X = a(tsin 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 =6ax3x=3x(2ax)
or
dy dx
=
x(2ax) 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 mI ml +my  =0 m am b dx
or
bmmxm 1 + am mymI dy = 0 dx
or
or (iii) Here, x=a(tsint),y=a(lcost)
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
Remedial Mathematics
I MISCELLENEOUNS SOLVED EXAMPLES I Example 1: Find the differential coefficient at x
~ +'x
= 0, ify = aX +  Ix
• Solution: We have y = a X + J§+x 
Ix
, Differentiating w.r.t. x, we get
ely dx

Putting
= a
xloga+I (I+X)1I2 {IOX)(1+X)(I)} 2 1 x (1 x)2
=
a Xl oga+I~X{IX+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/(IX) w.r.t. I+x Solution: Letu= Sin1(IX),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 xa dx = 2x";;;;
Soluti~n: We have
y =
~+~
Differentiating worot. x, we get
dy dx
=
_I _ I +
Fa 2£
Fa( __ 1 ) 2x3 / 2
I xa = (xa) =  2x3/2 2x";;;;
Example 4: y = e
x+ex+eX +
dy dx
0
then show that
=L
(UPTU B. Pharma 2003J
IyO 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(1y) =y dx dy y =dx Iy
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
Remedial Mathematics
3. Find the derivatives of the following functions: (i) y=(x2)(x+2)(x3)(x+3) (ii) y=(x+ l)(aJ21) (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 :
of the following functions:
(i) y = log sin x + cosI(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 x17y 17. If +  =6, show that  =   ' x y dx 17xy dy cos 2 (a + y) 18. Ifsiny=xcos(a+y)provethat dx = cosa
19. If
JIx' +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 xy
i
lx6
Differentiation
497
21. If EI + eV = EI + Y, prove that ely = _ eX (eY I) dx eY(eX I) 22.
If~=EIY,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! +ex t! _ex
[~GPV
B. Pharma 2001
r
26. Differentiate with respect to x eX + tan x
[RGPV B. Pharma 2002]
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
HINTS TO THE SELECTED PROBLEMS
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+12x2 3x2)+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= Itanx 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= (x2)(x+ 2)(x3)(x+ 3) = (x 24)(:?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 = sin1 (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+cosxsinx
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
Remedial Mathematics
IOg(~) =x + Y
(in)
logxIogy =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(xI)sec(.x22x+ 1) tan (.x22x+ 1) .fix = 2(xI) 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
sinI 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+xI xI
. dyI Dlff. w.r.t.x,  = 2 dx (xI)
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 = y4x(x2 + y) dx i) + i)  x
dy = y4x(x 2 +
dx
4y(x
2
9. sin (xy)+ ~ =x2y
... (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
=
~Il cos (xy)(Multiplying byy)
 ysin(xy) +
l) dy
=
dx
2xy1
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.
EfY=
_I_}
dy x+y dx
= _
=
X~y {I+ : }
I_}
{COS(X + y) _ _ x+y
x log = logxIogy Y
Diff. w.r.t. x, we have eX  Y
(I :)
= .; 
~:
sin y dy dx
50l Remedial Mathematics
:(;_eXY) ely =
~'_eXY
=
y(1xe~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=xy 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+;l34xy= 0 Diff. w.r.t.x, 2x + 2Y:  34( x: + y) =0
1 + xsinx + cos x (1 + cosx)2
Differentiation
503
x+ dy (y17x)17y =0 dx dy = x17y dx 17xy 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=xdyy dx dx x+ y= (x y)dy dx dy = x+ y dx xy 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{le>') dx e>'(e X I)
{I + dy} dx
504
Remedial Mathematics
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(sec2x)
(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
1e 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(x1)sec(x22x+ l)tan{x22x +
1) 6. (i) ~ (2 cos 3x 3 sin 3x) 2
oo') x ay (111 2 axy 1 Y 1 7.   or :X 1 (x _1)2
9.
2xy2
l cosexy)l
2ysinxy + xy2 cosxy _ 2xy2 + 3y2
8. 10.
y4x(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
Remedial Mathematics
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 3x18 sin 3x18 sin 3x9 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
(UPTU B. Pharma 2006)
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
Remedial Mathematics
d 2y
dy
Example5:lfx=(sin 1 xi,provethat(ix2) 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
ox
2
{:f
2
.
. Sin
I X·
~l
1 r:21 2 dy 2.1 _ x 2 or" 1  x dx = Sin X
=4(sin1 xi=4y[From(l)]
Again differentiating both sides w.r.t. x, we get 2 dy d y (dy)2 dy (lx 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 .
[RGPV B. Pharma 2004]
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 =asInxbcosx 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 tanx2x2 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(9sin9),y=a(1cos9),find dx,AIsofind dx 2 . 1 t 2 2t . 3. Ifcosx=   2 andsmy=   2 ,0$t$1. l+t l+t 2
Show that d ;' is independent of t. dx
510 Remedial Mathematics
d 2y 1t 4. Ifx= 3 sin t sin 3t,y= 3 cos tcos 3t, find  2 att= . dx
3
2
d y 5. Ify=~+tanx,showthat2 =6x+2sec2 xtanx dx
d 2y dy 6. Ify= A t? + Bet/x, show that dx 2 (p + q) dx + pqy= 0 [RGPV B. Pharma 2001]
ANSWERS (iii)  x cos x  2 sin x
(it) 2
(iv)
2~(5
cos 5x 12 sin 5x)
(v)
sin(logx) + cos(logx) 2 X
dy 8 d 2y 1 48 2  = cot  = cosec• dx 2' dx 2 4a 2
4.
OBJECTIVE EVALUATION MULTIPLE CHOICE QUESTIONS (Choose the most appropriate one) 1. lfy = 2.x4 + 3~ + 2.x + 5 , then : (a)
8~ + 9x2 + 2
2. lfy = (
(a) d dx
3. 
(a)
Fx +
1 1 + 2"
(b) 8x4 + 9x3 + 2
then :
(d) None of these.
is equal to
(b) 1  
x
1
2x
1 (c) 1  2
x
1
(d) 1+
2x
lsin2x 1+ sin2x
sec2(~+x )
4. Ify = xx
(a)
l),
=
oo
x 
,
(b)
then dy dx
y x(l+ ylogx)
sec2(~x)
(c)
sec2(~x)
i
(c)
Y x(lylogx)
(d) None of these.
=
(d)
x(l+ ylogx)
(d)
i x(l+ ylogx)
s. ~ (x2 . e" sin x) = dx
(a) xe 2 (2sinx+x sinx+ cos x) (c)
xr:'(2sinx+xsinx+cosx)
(b) xe'" (2sinx + x sinx  cos x)
(d) None of these.
Differentiation
511
6. If y = x", then dy is equal to dx (a)
(b) x' + logx
1 + logx
~logx+~logx+Jlogx+oo then
7. Ify= (a)
(b)
1
x(2yl)
1 x(2y+ I)
(c) x'logx
:
1 (c)
8. Ify=logu,x+ 10gx10+ 10g1010+ 10gxX, then :
(a) (c)
xloge 10
(b)
x(logex)2
loge 10 x loge 10 x(log e x)2
=? 2y I
(d)
2y+ 1
(d)
x,[X
is equal to
x loge 10
X10glO e
(d) None of these.
9. If y = x,[X ,then dy is equal to dx
2+logx
2$
(a)
(b)
2logx 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)
xy
xy 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. Ie dx
14. Ify = (I  x l/4) (I  xll2) (I + xl/4), then dy equals. dx (a) I
(b) I
Icosx . 15. IfJ{x) =   .  , thenf'(7tl2) IS equal to. Ismx (a) 0 (b) (c)
(c)
$
00
(d)
(d) x.
does not exist
512
Remedial Mathematics
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) = _ _ ' 2XI) ,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 + tanI (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)
, (2XI)2{2+2X2X2} 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
.!.x1/2 dx = dt 2
dx =2dt
Fx
t
= .!.loo(13x 2 )+C. 6!:>
IRGPV B. Pharma 20041
528
Remedial Mathematics
 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
[RGPV B. Pharma 2002)
. 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 = 2b2a dt 1 /= I  = logt t 2(b  a) (2b2a)
put
f
I
2(ba) sinx
Example 9: Evaluate
fsm. (xa) dx
Solution: Put So that
(xa)=t dx = dt f
log [a cos 2 x + b sin 2 x] + C.
sinx dx sin(xa)
=
)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 = (xa) cos a + sin a log Isin (xa)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 = '!'ft 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
= =
flt dt
= log t + C
log (log sec x) + C.
3
+ C.
530
Remedial Mathematics
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
~ tan1 ( ~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
33 fx.e 3X dx.
532
Remedial Mathematics
Example 2: Evaluate (I)
x x dx Je,,(1 +X)2
(UPTU B. Pharma 2007)
(il)
JeX(sinx+2cosx)dx.
[UPTU B. Pharma 2004)
Solution: (i) We have
Jexxdx (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 21 (2 sinxcosx) +21 (2 cosx+ sinx) + C 2 +1 2 +1
[
~
I
e .: jeoxsinbxdx= 2 2 (asinbxbcosbx+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
fI 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(logexi2 flogex.1 dx =x(loge x i2 [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
Remedial Mathematics
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· Jdx 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 = (logxn+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 = tanI x JXdx
{!
tanI x Jxdx
Jdx.
I x2 xx 2 =tan x 2'J + 2 dx 2 I x I
1JI + x 2 I dx
2
=tanI x xx   2 2 2
= tanI x. I
=tan
XX
2
x 2
l+x2
![ 2
Jldx
JIdx] l+x2
x .1 I I xx x+  tan x+ C. 2 2 2
Integration
2
Example 7: Evaluate flog (1 + x ) dx.
535
[RGPV B. Pharma 2001]
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
Jdx 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 [xtan 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
Jx xId x 2
I I = I JIdt =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
(RGPV B. Pharma 2002]
Example 10: Evaluate Jxlogxdx. x2 2
Solution: Jxlogxdx = (log x). 
2
1 x Jdx 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
Remedial Mathematics
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 = 22 (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 sinx2 fxsinxdx
=~ sinx2[ 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 sinx3 fx 2 sinxdx
= x 3 sinx3 [x 2 fsinxdx f{2x·fsinxdx}dx] = x 3 sinx+3x2 cosx6[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
Remedial Mathematics
(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+2xx
x+4
So,
dx
 =
Putting x = 3, we get A =
7.. and putting x = I, we get B = ~ 4
x+4
J3+2xx 2
4
7 3 + 4(3 x) 4(1 +x)
x+4 So, we have ::3+2xx 2
dx = 7.. JIdx~ JI_ dx; Integrating 4 3x 4 I+x =
Example 2: Evaluate the integral
7 3 log(x3)+ Iog(l +x)+C
4
3x+2 (x2)(x+1)2
or
(3x+2)
4
3x+2
J(x2)(x+ 1)
Solution: Here, the given integral is Consider
2
x+4 A B +. (3x)(1+x) 3x I+x x+4=(l+x)A+(3x)B
Consider
Hence,
dx .
J3+2xx
Solution: Here, the given integral is
x+4
2
~I~~~~~~
2
dx .
3x+2
J(x2)(x+ I)
2
dx .
ABC (x2) (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 (x2)(x+ I)
8
2
8
+9(x2) 9(x+l) 3(x+I)2
Integration
539
Hence I 3x+2 dx =! f_l_ dx _! fIdx+~ f 1 dx , (x2)(x+I)2 9 x2 9 x+1 3 (x+I)2 8 8 1 1 =  log (x  2)   log (x + 1)     + C 9 9 3 (x + 1) =
!IOg(X2)_ 1 +C 9 x+1 3(x+l) ,
Example 3: Evaluate (1) (ii) (iii)
fX:X2 fXX 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 II dx l+x x =
= log lxilog II + xl + c.
=IOgl~I+c. I+x 1 1 I ABC (ii) Here,  = = =  +   +   (say), 3 2 xx x(1x ) x(1 +x)(1x) X Ix (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 =+xx 3 X 2(1x) I  dx fxx 3
2(1+x)
= f'!"dx+'!"fI_dx_'!"fI_ dx
x
2 Ix
(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 (1x 2)
2
+C
540
Remedial Mathematics
(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 = (12)(13) = "2,B= (21)(23) =8
and
3 27 C = (31)(32) = 2"
3
..
x3 (xQ(x2)(x3
= 1+
I 2(xl)
8 (x2)
27 2(x3)
+
3
Hence
I(xl)(x2)(x3) x dx  II dx + I dx I(x2) 8 dx I 27 dx z(xl) + 2(x3) =x +
Example 3: Evaluate the integral
27 "21 log 1x118Iog Ix21 + 2" log 1x31 + c.
7X 2 +3x+l x(x + 1)
f
Solution: Here, the given integral is
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 14x     =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+ 2dx x +x
f
... (1)
l  4x Now, to finds 2dx. x +x
f
14x A B  =+2 x +x x x+l => 14x =A (xl)+ Bx putting x =  1 => B =  5 and putting x = 0 => A = 1
Let
Thus,
14x 5 x2+x =~ x+l
... (2)
Integration
541
I4X = II5 I 1d x =Iogx5 log (x + 1) I2dx x +x x x+I
So,
Hence, by equation (1), we get 7x2 +3x+I x(x+I) dx=7x+logx5log(x+I)+C.
J
Ix2 +xx _ 6 dx Solution: /= I dx = J( 2;( 3) dx. x +x6 xx+ Example 4:
2
Let
[RGPV B. Pharma 2002J
X
x A B =+(x2)(x+3) (x2) (x+3)· x =A(x+3)+B(x2)
Putting x =3 in eq. (1), =>3 =B =>B= Putting x = 2 in eq. (1), => 2 = 5A => A = x Ix 2 +x6
=
...(1)
~. 5
~. 5
~f_Idx+~fIdx 5 x2
5 x+3
2 3 = log(x2)+ 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
Remedial Mathematics
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
[RGPV B. Pharma 2003]
dx + l)(xl) ]
Solution: Let
~log(2x+I)+~log(x2 +1) + '3. tanI x+C
dx +l)(xl)
A
Bx+C x 2 +1
;;; =   +  
(x 2 +I)(xI)
(xI)
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 =1121 =112 equating the coeff ~ = 0 in both sides of the equation. A +B=O
B=~. 2
Now f
2
1
(x +I)(xI)
dx =.!. f_l_dx_'!'fx+I dx 2 xI 2 x2 +1
=
~ 2
fI dx  .!. f xdx .!. fI dx x I 2 x2 +1 2 x2 +1 .
= .!.log(xI) ~log(~+ 1) ~tanIC+ C. 2 4 2 2x+ 1 dx Example 7. Evaluate f (2x + 3) (3x _ 4) .
'.
 f(2x+3)(3x4) (2x + I) dx
SolutIOn. LetI
[RGPV B. Pharma 2001)
Integration 2x+l (2x + 3)(3x  4)
Let
A (2x + 3)
B (3x  4)
+
(2x+ 1) = A (3x4) + 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)(3x4) 2x + 1 dx
~
=
1
17 f(2X+3)
dx + .!.!. dx 17 f(3X4)·
2
11
17
51
.
= log(2x+3)+ log(3x4)+C.
2x dx Example 8: Evaluate (x I)(x + 3) .
f
Solution: Let (
=>
(RGPV B. Pharm 20021
2x A B 1)( 3) =   +   . x x+ (xI) (x+3) 2x (xI)(x+3)
A(x+3)+B(xI) (xI)(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(xI)(x+3)
1
=
2"
f(xI) dx 3 f dx +2" x+3
1 3 =  log(xI)+ log(x+3)+C.
2
Example 9: Evaluate
x2 2
f;x x
 2x
2
dx
(RGPV B. Pharma 2001)
544
Remedial Mathematics
Solution:l=
fx
x2 3
2
x 2x
dx =
f x(x
=I
2 2X
x2)
dx
x dx=I x dx (x 2 x2) (x2)(x+ 1) .
x A B =+. (x2)(x+I) (x2) (x+l)
Let
x (x2)(x+I)
A(x+I)+B(x2) (xI)(x+l)
x =A (x + 1)+ B(x2) 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
~ I1dx+! I 3 x2
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.•
Solution: Here, the given integral is
I(x 1)~2X + 1) dx .
.
ConsIder
=>
x (xI)(2x+I)
A
B
(xI)
(2x+l)
=  +
x=(2x+I)A+(xI)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 (xI)(2x+I)
Hence, f
x dx = !f_l_dx+!fI dx (xI)(2x+l) 3 xI 3 2x+I
+3(xl) 3(2x+l)
=
1
"3 log (x 
1) +
1
"6 log (2x + 1) + C
2
Example 11: Evaluate the integral
f(xI)x (x2) dx.
Solution: Here, the given integral is
f(xI).~ (x2) dx.
3
2
(1)
integration
x2 ABC D = +++(xI)\x2) xI (x_I)2 (xI)3 x2
Let
=>x2=(xli(x2)A +(xl)(x2)B+(x2)C+ (xI)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 + 2B2CD =0 Putting the values of C and D, we get A =4,B =3
545
... (1) ... (2)
2
x 4 3 I 4 :: =         +  (xI)\x2) xI (x_I)2 (xI)3 x2'
Thus, Hence
,
x2 I 3 f I d x  fdx+4 1 fdx I dx=4 fdx+f(xI)3(x2) xI 2 (x_I)2 (x_I)3 x2 3 I =log(xI)+ + 2 +4Iog(x2)+C xI 2(xl)
(X2)
3 I =4log   + + +C. xI xI 2(xI)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.
J4x3~5x2 dx
J.
dx x x sm+tan2 2
4.
fcosecx.log(cosecxcotx)dx
6.
J1 sin 2x dx x+cos 2 x
8. JIcotx 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(cosecxcotx)dx log (cosec x  cot x) = t
Put I
     (  cosec x cot x + cosec2 x) dx = dt cosecxcotx
J= ftdt
=
t2 1 +C= {loglcosecxcotxl}2+C
2
2
lntegration
tanI 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= JIcotx 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
Remedial Mathematics
T= 
dl
ft
= log 1 + C,
Putting the value of 1 / =
9. /=
log I cos x + sin xl + C
f4x3~5x2 dx
= f4x3~5x2dx = f4x2.~5x2 ·xdx 2 dt Put5x=I=>xdx= 
2
/=4 f(5t)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+ ~(5x2)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
= fdx = fdx = 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 = 22
a b
Putting in (1), we get /=
=
ft(a
dt 2
2
b)
dt = 22 log It1/ + C
ab
~ 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 nxn1dxn X
(1+x ) (multiplyby~l into Nr. and Dr.)
put
(I+~=t~xn=tl
~  1 dx
=
dt ~ ~  1 dx = dt n
550
Remedial Mathematics
f
1 dt 1= ;; (t1)t
Therefore,
=
Integrating
I
Solution: Let/=
1 __ !)dt t
n
= ~ {log It n
=
Example 4: Evaluate
n_ J\J1
~
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 =
fJx2+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 fdx. 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
tanI x 3 l+x
6
dx
551
552
Remedial Mathematics
Solution: Let Let us put tanI 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 (tanI x 3 )2 j=
In
6
+C.
n1sin 1 x
Example 10: Evaluate •
Solution: Let! =
l_x2
dx .
IenlSinl x ~ dx
,,1 xputting sin x = t, differentiating w.r.t. x,s we get 1
1
,:;; dx = dt
"1x2 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(86xi 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 .
fd 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 sinI x dx
~(lx2)
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
Remedial Mathematics
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 cosxsinx
dx
(cosx+sinx)2
x+3
~(1x2)
dx
13. Evaluate the following integrals: (l)
fx{I+(logx)2} I dx
(ii)
J
6x8 3x2 8x+5 dx
(iii) r o t dx .
14. Evaluate the following integrals: (l)
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~xa) 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(xa)(xb)(xc) 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) .
xl (x + 1)(x 2 + 1) dx
x2 dx. (xa)(xb)
Integration
555
18. Evaluate the following integrals: (I)
fx.sinI xdx
(ii)
fxsin3 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 ex + c = t => b ex dx = dt / =
ba fdt t
= 
ba log t + C
=::b log (c + beX) + C eX
fl+e d x
(ii) /=
2x
dt • Put eX = t => eX dx = dt, / = f2' i~tegrating I+t = tanI t + C = tanI 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.1dx = smx
~ x 2sincos
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(a2x2rI/2(2x)Xdx
=X~a2_x2 =
x
a 2 _x 2 _a 2  f
12 va x 2

~ a x
dx
f~a2x2 +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
+ ~ sinI ~ + C x 2 a 2
(iv) /= fxsinxcosxdx =
i Ix(2sinxcosx)dx
=
i fxsin2xdx (integrating by parts)
=
i fx( C~2X) 
RC~2X)dx
=
xcos2x sin2x C ++ 4 8
=
~ (sin 2x2x 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 lx 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+xI}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 + 2smcos
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 + Jlogx 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 sinI 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.
.
f1+1 2 ' mtegratmg
= =
cosxsinx
~1x2 +3 sin i x+ C
tan i 1 + C = tan i (log x) + C
f
6x8 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+l1)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~Xa) dx = pinxcosx.sinacosx = dx
smx
smx
= f(cosasinacotx)dx =
x cos a  sin a log sin x + C 3
15. (I)
x J(xa)(xb)(xc) dx
~
~
~
= J(ab)(ac)(xa) + (ba)(bc)(xb) + (ca)(cb)(xc)
After breaking into partial fraction =x+
~ ~~~~
(il) [=
~
log(xa)+
0~0~
log(xb) +
2 ~~~~
log(xc)+C
J(x+l)(x x1 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(xa)(xb)
=
{a (ab)(xa)
f

} dx b (ab)(xb)
b2
a2
=x+ log (x a) log(xb)+C ab ab 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 sinI 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\fIx2
2
\f1x2
2 2 = _x sin 1 x+.!.{f lx dx J=dx=} 2 2 ~Ix2 ~Ix2 =
2 x . I x+1 sm 2 2
{J
r:?I 2 dx \flx
J~1x2 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\flx1 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)  fcosxsecxdx =  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
Remedial Mathematics
ANSWERS 1. (i) (2 + 3logx)2 6
(ii) tan (3x + 9)
(iii)
1 6(86x) ,
. 1 (IV) log(2x+3).
2 1 2. (i) sec x 2
(ii) .!. sec 7 x
(iii) log log x
(il) tanI €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 xI)
(ii) 2 log (cosec 8 + cot 8) where x
122
x
a
2 .
(iii) "2"ax+T sm
4
;;'
(ii)
~ (sin 3x 18
3x cos 3x) + .!. (x cos x  sin x)
2
..j1;;2
(i ii)
x 7. (i) sinI x [log sin 1 xI] 8. (i)
tan2 8,
I (iv) g(sin2x2xcos2x).
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 sinI x
~(1x2),
(iv) x+ !cos2x 2
(iii)  .   SInX+cosx
13. (1) tanI 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(xa)+
+
b3 ~~~~
(iii) sin x + cos x
log(xb)
c3
1 log (xc) (ii) log(x2+ I)log(x+ I), (ca)(cb) 2
~IOg(Xl)_i_l_
(iii)
x+3
10
4x1
2
a b2 (iv) log(xa)log(xb)+C ab ab 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(XI) __I_,
3
x+2
xI
(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) log2 +c. 2 x +3 2
.) x . I x ~I 2 I._I 18 • (1 Sin x+\jix 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
Remedial Mathematics
III
DEFINITE INTEGRAL
Ifj(x) is a continuous and nonnegative 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(2ax) = 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 [(00) 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/4tan9 '}dx Jo l+tan1t/4tan9 = r1t/410g{1+ Itan9}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 I2sm +2smcos2 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
Remedial Mathematics
Also when x = 0 ~ t = 0 x = ~ => t = 1 2 2/
rl
2
2dt
 Jo 21+It 2 
rl
dt
Jo (J2)2 _(t_I)2
~2' 2~[IOglt:::n = _1 [lOg J2 _IOg(J21)] 2J2 J2 J2+1 = _1_10g{J21}=_I_IOg J2+1 s J2 J2+1 J2 J21
=_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(1tx) So
sinm(1t x) CQs(2m+ I) (1tx) =  sinmx cos2m + 1 X =.f(x)
=
f:
/ =0 Since
f(x)=Oiff(2ax)=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 xI (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 +Ix2 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) cotxx+ C (d)cotxx + c.
Fx +C
fllsin2xdx
(b) 2 sin e# + C
(d) 2 cos e# + C.
=
(a) sin x + cos x + C (e) secx+tanx + C
(b)
sinxcosx + C
(d) secxtanx + 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(ba) lXf(X)dx.
(ba) lXf(x)dx
581
582 12.
Remedial Mathematics
r • 1
x+2 dx Jx 2 +2x3
(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/2log 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 +6eX 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 sinxcosx 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
Remedial Mathematics
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 2x2x + C. 1
x +1
1
~ log (x 3/2 + ~1+x2 ) + C
12. log(e/2)
.!.IOgl41 + 3
11.
e~
+ C.
n/12
n
14. 0
13. 2log2
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 nonnegative 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
[UPTU B. Pharma 2001]
4.
f tanx dx
[UPTU B. Pharma 2001]
5.
fO + logx)2 dx x
(UPTU B. Pharma 2002]
6.
f5+~OSX
[UPTU B. Pharma 2001]
f
(UPTU B. Pharma 2007]
1. 2.
logsecx
8.
sinx dx sin(xa) fx 3sinx 2dx
9.
x 2 tan 1 x f 1+x2 dx
7.
[UPTU B. Pharma 2001]
(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
[UPTU B. Pharma 2004]
13.
fa 2
14.
~x2
fx:x2
IUPTU B. Pharma 2003] [UPTU B. Pharma 2005]
586
15.
Remedial Mathematics
JX~X3 J(x1)(x2)(x3) x ch
[UPTUB. Pharma 2004)
3
16.
[UPTU B. Pharma 2006)
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
[UPTU B. Pharma 2005]
4 [UPTU B. Pharma 2008)
DOD
AppendixI "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 hemisphere. (I)
Volume=
3. 7tr3 . 3
(iz) Curved surface area = 27t,2. (iii) Total surface area = 37t,2.
588
Remedial Mathematics
~~~~~~~ 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+xx(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 xx3x3x7 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
590 Remedial Mathematics
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 eP
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