Comprehensive Basic Mathematics Vol 2

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Copyright © 2005, New Age International (P) Ltd., Publishers Published by New Age International (P) Ltd., Publishers All rights reserved. No part of this ebook may be reproduced in any form, by photostat, microfilm, xerography, or any other means, or incorporated into any information retrieval system, electronic or mechanical, without the written permission of the publisher. All inquiries should be emailed to [email protected]

ISBN (13) : 978-81-224-2535-2

PUBLISHING FOR ONE WORLD

NEW AGE INTERNATIONAL (P) LIMITED, PUBLISHERS 4835/24, Ansari Road, Daryaganj, New Delhi - 110002 Visit us at www.newagepublishers.com

Dedicated to all my teachers


Foreword Mathematics is considered as the mother of Science as well as the mother of Philosophy. In fact, it is the basic knowledge on which any human endeavour in the field of intellectual pursuit is based. But somehow, in our academic field Mathematics is considered as a subject tough to comprehend and to master. The otherwise enjoyable subject becomes sometimes to the students a frightening nightmare and in the course of my long career in the field of education, I have found many students even becoming nervous at the mention of Mathematics. That is, in my experience, mostly because of the way it is taught by some teachers and the way some of the books are written. My esteemed colleague Smt. Veena, G.R. is not only a good teacher who would make the students to enjoy Mathematics, but also a good “Narrator” of Mathematics. Extremely resourceful and completely committed, Veena has brought into this book her direct experience in the class room as a teacher. In lucid style, she has written this book making it students-friendly, teacher-friendly and teachinglearning focused. While congratulating her for this good work, I wish that both the students and the teachers would welcome this book and make use of it.

PROF. K.E. RADHAKRISHN A ADHAKRISHNA Principal, Surana College, Bangalore-560004.


Preface This book is written as per the Syllabus of Basic Mathematics for II Year Pre-University Course of Karnataka. Each topic has been discussed exhaustively as per the requirements of the latest syllabus. The book has been written in a very simple style, to enable students to understand the subject effectively. More focus is given for a systematic approach to enhance the grasping of the subject so that all types of questions could be answered well in each chapter. I hope this book will satisfy all the requirements of the students for learning the subject successfully and getting through the examination with flying colours. My sincere thanks to Prof. K.E. Radhakrishna, Principal, Surana College, an eminent academician and educationist for his foreword. I also thank Ms. Sudha S., Department of Mathematics, Surana College for reading the manuscript and identifying the unforeseen computational errors. I am also thankful to Mr. R.K. Gupta, Chairman, Mr. Saumya Gupta, Managing Director, New Age International Pvt. Ltd., New Delhi and Mr. Vincent D’souza, Branch Manager, Mr. Babu V.R., Marketing Manager, Bangalore branch for accepting to publish the book. I sincerely welcome criticism, views and suggestions from readers.

G.R. VEEN A EENA [email protected]


Contents Foreword Preface

1.

Mathematical Logic

1

1.1 1.2 1.3

1 1

1.4 1.5 1.6

2.

3.

vii ix

Introduction Propositions Logical Connectives and Compound Propositions Conjunction, Disjunction, Conditional, Biconditional, Negation Tautology and Contradiction Logical Equivalence Converse, inverse and Contrapositive of a Conditional

2 9 12 14

Per muta tion and Combina tion erm utation Combination

21

2.1 2.2 2.3 2.4 2.5

Introduction Fundamental Principle Permutation and Combination Factorial of a Positive Integer Permutation 2.5.1 Linear Permutation 2.5.2 Value of npr 2.5.3 Value of npn 2.5.4 Value of 0 2.6 Permutation of Things of which Some are Alike 2.7 Circular Permutation 2.8 Combination 2.8.1 Value of ncr 2.8.2 Complementary Combinations

21 21 22 22 22 23 23 23 24 28 31 33 33 34

Pr oba bility Proba obability

46

3.1 3.2 3.3 3.4 3.5

46 46 47 52 56

Introduction Terminology Definition of Probability Addition Rule of Probability Conditional Probability

xii

Contents

3.6

4.

5.

6.

7.

8.

9.

Multiplication Rule

56

Binomial Theor em heorem

65

4.1 4.2

65 65

Introduction Statement of Binomial Theorem

Par tial Fractions artial

88

5.1 5.2

88 89

Definitions Partial Fractions

Matrices & Determinants

105

6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 6.10 6.11 6.12

105 105 105 107 108 117 120 136 144 146 153 164

Introduction Matrix Types of Matrices Algebra of Matrices Transpose of a Matrix Determinants Properties of Determinants Minor, Co-factor, Adjoint and Inverse of a Square Matrix Characteristic Equation of a Square Matrix Cayley Hamilton Theorem Solution of Linear System of Equations Application of Matrices in Business Problems

Ra tio and Pr opor tions, Var ia tions Ratio Propor oportions, aria iations

182

7.1 7.2 7.3 7.4 7.5 7.6 7.7

182 182 190 192 202 209 213

Introduction Ratio Proportion Direct Proportion or Direct Variation Problems on Time and Work Problem on Time and Distance Problems on Mixture

Aver ages era

221

8.1 8.2 8.3

221 221 222

Introduction Arithmetic Average or Mean Combined Average

Bill Discounting

234

9.1 9.2 9.3

234 234 235

Introduction Terminology Formulae

Contents

10. Stoc ks and Shar es Stocks Shares 10.1 10.2 10.3 10.4

Stock Shares Distinction between Stock and Shares Terminology

11. Lear ning Cur ve Learning Curv 11.1 11.2 11.3 11.4 11.5

Introduction Learning Curve The Learning Curve Ratio Graphical Representation of Learning Curve Learning Curve Equation

12. Linear Pr ogramming Pro 12.1 12.2 12.3

Introduction Linear Programming Solution to Linear Programming Problem

13. Cir Circcles 13.1 13.2 13.3 13.4 13.5 13.6 13.7

Definitions Equation of a Circle Point of Intersection of a Line and a Circle Equation of Tangent to the Circle x2 + y2 + 2gx + 2fy + c = 0 at the Point (x1, y1) on it Length of the Tangent from the Point (x1, y1) to the Circle x2 + y2 + 2gx + 2fy + c = 0. Condition for the Line y = mx + c to be a Tangent to the Circle x2 + y2 = a2 and point of contact Condition for the Line lx + my + n = 0 to be a Tangent to the Circle x2 + y2 + 2gx + 2fy + c = 0.

xiii

246 246 246 246 247

255 255 255 255 255 257

263 263 263 264

278 278 278 292 298 299 300 301

14. Par abola ara

317

14.1 14.2 14.3 14.4 14.5

317 317 318 320 321

Introduction Parabola Equation of the Parabola in the Standard Form Different Forms of Parabola with Vertex (0, 0) Different Forms of Parabola with Vertex (h, k)

15. Limits and Continuity 15.1 15.2 15.3

Introduction Constants and Variables Function

343 343 343 343

xiv

Contents

15.4 15.5 15.6

Limits Standard Limits Continuous Functions

344 345 356

16. Dif ential Calculus Difffer erential 16.1 16.2 16.3 16.4 16.5 16.6 16.7 16.8 16.9

366

Introduction Derivative of a Function Derivative of Some Standard Functions from First Principles Rules of Differentiation Differentiation of Composite Functions Differentiation of Implicit Functions Differentiation of Parametric Functions Logarithmic Differentiation Successive Differentiation

17. Applica tion of Der ves pplication Deriivati tiv 17.1 17.2

429

Derivative as a Rate Measure Maxima and Minima

18. Integration 18.1 18.2 18.3 18.4 18.5 18.6

Standard Integrals Algebra of Integrals Substitution Method Integration by Partial Fraction Method Integration by Parts Integrals of the Type e x f x + f ′ x dx

z

af af

Introduction Properties of Definite Integrals Application of Definite Integrals to Find Area

20. Applica tion of Calculus in Business pplication 20.1

429 438

455

19. Def inite Inte grals Definite nteg 19.1 19.2 19.3

366 366 367 371 386 390 393 397 401

455 456 461 466 474 477

484 484 490 494

504

Terminology

504

Examina tion Cor ner Examination Corner

515

• • • • • •

Blue Print Model Question Paper 1 Model Question Paper 2 Model Question Paper 3 Chapterwise Arranged Question Bank Gist and Formulae

516 517 520 522 526 546

1 Mathematical Logic 1.1 INTRODUCTION: Logic is the science dealing with principles of reasoning. We can find all the different ways of solving a problem by logical reasoning. The English Mathematician George Boole is the founder of mathematical logic. To express the principles of reasoning, a symbolic language has been developed. This symbolic language is called mathematical logic or symbolic logic. Mathematical logic finds application in switching circuits, digital computers and other digital devices.

1.2 PROPOSITIONS: A proposition is a statement which in the given context is either true or false but not both. The propositions are denoted by small letters p, q, r... Examples: 1. Sum of two even integers is even integer. 2.

3 is a rational number.

3. Earth is flat. 4. Delhi is the capital of Karnataka. 5. 7 is a prime number. 6. 5 − 7 = −2. Note: The statements involving opinions, question marks, exclamatory mark, command, wish are not propositions. Examples: 1. Logic is interesting. 2. What a beautiful weather! 3. Where are you going? 4. Please sit down. 5. May God bless you.

2

Basic Mathematics

TR UTH VALUE: The truthness or falsity of a proposition is called its truth value. If a proposition is TRUTH true it is denoted by ‘T’ and if it is false it is denoted by ‘F’. Example: The truth value of 1. 5 + 6 = 11 is ‘T’. 2. ‘Asia is in India’ is F. 3. ‘Today is Sunday’ is either ‘T’ or ‘F’ in the given context i.e., on a particular day it is only one of ‘T’ or ‘F’.

1.3 LOGICAL CONNECTIVES AND COMPOUND PROPOSITIONS Two or more simple propositions are connected by using the words ‘and’, ‘or’, ‘if ... then’, ‘if and only if’. These words or phrases are called logical connectives. Any proposition containing one or more connectives is called a compound proposition. The simple propositions occurring in a compound proposition are called its components. TRUTH ABLE:: The truth values of the compound proposition for all possible truth values of its TR UTH TABLE components is expressed in the form of a table called truth table. For a compound proposition with only one proposition, Truth table consists of 2 possibilities (either T or F). For a compound proposition with two propositions truth table consists of 22 = 4 possibilities. For a compound proposition with 3 propositions truth table consists of 23 = 8 possibilities. CONJUNCTION:: If p and q are 2 simple propositions. Then the proposition ‘p and q’ is called the CONJUNCTION conjunction of p and q. It is denoted by p ∧ q. Example: If p : 7 is a prime number. q : 6 is an even number, then p∧q : 7 is a prime number and 6 is an even number. The truth value of the compound proposition p ∧ q depends on the truth values of p and q. Note that the conjunction of p and q is true only when both p and q are true, otherwise it is false. Truth Table

p T T F F

q T F T F

p∧q T F F F

DISJUNCTION DISJUNCTION:: If p and q are 2 simple propositions, then proposition ‘p or q’ is called the disjunction of p and q. It is denoted by p ∨ q. Example: If p : 2 is rational number.

Mathematical Logic

3

q : 2 is odd number, then p ∨ q = 2 is rational or 2 is odd number. The truth value of p ∨ q depends on the truth values of p and q. Note that the disjunction of p and q is false only when both p and q are false. Otherwise it is true.

Truth Table

p

q

p∨q

T T F

T F T

T T T

F

F

F

CONDITION AL (IMPLICA TION) CONDITIONAL (IMPLICATION) TION):: If p and q are two simple proposition, then the proposition if ‘p ... then q’ is known as conditional or implication. It is denoted by p → q or p ⇒ q. Example: If p : 6 is an even number. q : 6 is divisible by 2, then p → q : If 6 is an even number then 6 is divisible by 2. The truth value of p → q depends on the truth values of p and q. Note that p → q is false only when p is true and q is false. Truth Table

p T T F F

q T F T F

p→q T F T T

BICONDITION AL (DOUBLE IMPLICA TION OR EQ UIV ALENCE): If p and q are simple propoBICONDITIONAL IMPLICATION EQUIV UIVALENCE): sitions, then the proposition ‘p if and only if q’ is called biconditional or double implication. It is denoted by p ↔ q. Example: If p : k is odd number. q : k2 is odd number, then p ↔ q : k is odd number if and only if (iff) k2 is odd number. Note that p ↔ q involves both the conditionals p → q and q → p.

4

Basic Mathematics

a

f a

f

p ↔ q is p → q ∧ q → p

∴

The biconditional p → q is true if p and q are both true or both false i.e., if p and q have same truth values. Otherwise it is false. Truth ta ble tab

a

f a

p

q

p→q

q→ p

T T

T F

T F

T T

p ↔ q i. e., p → q ∧ q → p T F

F F

T F

T T

F T

F T

f

NEGA TION: If ‘p’ is a proposition then the proposition ‘not p’ is called negation of p. It is denoted NEGATION: by ~p. Example: If p : 6 is odd number then ~p : 6 is not an odd number. If p is true then ~p is false and if p is false then ~p is true. Truth ta ble tab

p T

~p F

F

T

WORKED EXAMPLES I. Write the following propositions in symbols: 1. An integer is even if and only if it is divisible by 2. Solution: Let p : An integer is even.

q : It is divisible by 2. ∴ The given proposition is p ↔ q. 2. If 6 + 3 = 7, then 7 − 3 = 6 Solution: Let p : 6 + 3 = 7 q:7−3=6 Then the given proposition is p → q 3. I play chess or I study at home. Solution: Let p : I play chess q : I study at home.

Mathematical Logic

The given proposition: p ∨ q. 4. It is raining and the ground is wet. Solution: Let p : It is raining. q : The ground is wet. The given proposition : p ∧ q. 5. If it rains today then government declares a holiday and we are happy. Solution: Let p : It rains today q : Government declares a holiday r : we are happy. Given proposition: p → (q ∧ r). 6. If a number is not real then it is complex. Solution: Let p : A number is real. q : It is complex. Given proposition in symbols: ~p → q. 7. If Rama is intelligent or hardworking then logic is easy. Solution: Let p : Rama is intelligent q : Rama is hardworking r : Logic is easy. Given proposition: (p ∨ q) → r. 8. If 3 is not odd and 2 is not even then 7 is not odd or 8 is not even. Solution: Let p : 3 is odd q : 2 is even r : 7 is odd s : 8 is even. Given proposition: (~p ∧ ~q) → (~r ∨ ~s) II. Express each of the following compound propositions in sentences if p : The question paper is difficult. q : I get good marks. r : I can go abroad. 1. p ∧ ~q Solution: The question paper is difficult and I do not get good marks. 2. q → r If I get good marks, then I can go abroad. 3. (~p ∧ q) → r Solution: If the question paper is not difficult and I get good marks then I can go abroad. 4. r ↔ q Solution: I can go abroad iff I get good marks. 5. (p ∧ ~q) → ~r

5

6

Basic Mathematics

Solution: If the question paper is difficult and I do not get good marks then I can not go abroad. III. If p, q, r are propositions with truth values T, F, T respectively then find the truth value of the following propositions: 1. p ∧ ~q p T

q F

p∧ ~ q T

~q T

Given Given

So Truth value of p ∧ ~q is T when p is T and q is F. 2. p → (q ∧ r) q

r

q∧r

p→ q∧r

T

F

T

F

F

q F

~p F

3. (~p ∧ q) ↔ r

p T

a f

p

−p ∧ q F

a ~ p ∧ qf ↔ r

r T

F

4. (p ∨ ~q) ↔ (q ∧ ~r) (1)

p T

q F

p∨ ~ q T

~q T

Given Given

(1) ↔ (2)

(2)

r T

q∧ ~ r F

~r F

a p∨ ~ qf ↔ aq ∧ ~ r f F

Given

5. p → (q → ~r)

a

p

q

r

~r

q →~ r

p → q →~ r

T

F

T

F

T

T

IV. Write the truth values of the following propositions if p : 2 is even prime number. q : 2 is rational number. r : 7 is a composite number. 1. p ∧ (q ∨ r) Given

p : 2 is even number – True proposition. So Truth value of p : T. Similarly Truth value of q : F Truth value of r : F

f

Mathematical Logic

7

a f

p

q

r

q∨r

p∧ q∨r

T

F

F

F

F

∴ Truth value of p ∧ (q ∨ r) is F. 2. p → (q ∧ ~ r)

∴

a

p

q

r

~r

q∧ ~ r

p → q∧ ~ r

T

F

F

T

F

F

f

Truth value of p → (q ∧ ~ r) is F. 3. p ↔ (~ q ∧ ~ r)

a

p

q

~q

r

~r

~ q∧ ~ r

p ↔ ~ q∧ ~ r

T

F

T

F

T

T

T

f

∴ Truth value of p ↔ (~ q ∧ ~ r) is T. V. 1. A certain compound proposition (p ∧ q) → r is known to be false. Find the truth values of p, q and r.

a p ∧ qf → r is F

Given: This implies ⇒

p ∧ q is T and r is F

[3 T → F is F]

p is T, q is T and r is F

3 T ∧ T is T

2. A certain compound proposition (p ∧ q) → (r ∨ ~s) is known to be false. Find the truth values of p, q, r and s. Given:

a p ∧ qf → ar∨ ~ sf is F

⇒

p ∧ q is T and r ∨ ~ s is F

⇒

p is T , q is T and r is F, ~ s is F

⇒

p is T , q is T , r is F and s is T .

a

f a f

3 T → F is F 3 T ∧ T is T and F ∨ F is F 3 ~ F is T .

3. A certain compound proposition ~ p∧ ~ q ∧ r ∧ s is given to be True. Find the truth values of

p, q, r and s. Given:

a~ p∧ ~ qf ∧ ar ∧ sf is T

⇒

~ p∧ ~ q is T and r ∧ s is T

⇒

~ p is T , ~ q is T and r is T , s is T

⇒

p is F, q is F; r is T ; s is T .

3 T ∧ T is T 3 ~ T is F .

8

Basic Mathematics

VI. Construct the truth table for the following compound propositions: 1. p → ~p

p T F

p →~ p

~p F T

F T

Explana tion: A compound proposition with one component will have 2 possibilities either T or F. Explanation: Write ~p finally p → ~p. 2. ~p → ~q.

p T T F

q T F T

~p F F T

~q F T F

~ p →~ q T T F

F

F

T

T

T

Explana tion: A compound proposition with 2 components will have 4 possibilities. Write 2 T and 2F Explanation: under p, alternatively T, F under q to get all possible combinations. Now ~p is T when p is F and vice versa. ~q is T when q is F and vice versa. ~p → ~q is F only when ~p is T and ~q is F. Otherwise it is T. 3. p ↔ (p ∧ q)

a

p∧q T F F F

p T T F F

q T F T F

q T F T F

~ p∨q T F T T

p↔ p∧q T F T T

f

4. p ∨ ~(~p ∨ q)

p T T F F

~p F F T T

a

~ ~ p∨q F T F F

f

a

p∨ ~ ~ p ∨ q T T F F

f

Mathematical Logic

9

5. p ∧ (q ∨ r)

a f

p

q

r

q∨r

p∧ q∨r

T T

T T

T F

T T

T T

T T F F F

F F T T F

T F T F T

T F T T T

T F F F F

F

F

F

F

F

Explana tion: A compound proposition having 3 components will have eight possibilities. Write 4T and Explanation: 4F under p, 2T, 2F, 2T, 2F under q and alternately T and F under r to get all the possible combinations of truth values. Now q ∨ r is F only when both q and r are F. Otherwise it is T. p ∧ (q ∨ r) is T only when p is T and (q ∨ r) is T otherwise it is F. 6. (p → ~q) ↔ (q ∧ ~ r)

a p →~ q f ↔ aq ∧ ~ r f

p T

q T

r T

~q F

p →~ q F

~r F

q∧ ~ r F

T T

T F

F T

F T

F T

T F

T F

F F

T F F F F

F T T F F

F T F T F

T F F T T

T T T T T

T F T F T

F F T F F

F F T F F

T

1.4 TAUTOLOGY AND CONTRADICTION A tautology is a compound proposition which is always true irrespective of the truth values of its components. A contradiction is a compound proposition which is always false irrespective of the truth values of its components. Note: To determine whether a given proposition is a tautology or a contradiction, construct its truth table. If its truth value for all possibility is True, then it is tautology. If its truth value for all possibility is False then it is contradiction. Otherwise it is neither tautology nor contradiction.

10

Basic Mathematics

WORKED EXAMPLES 1. Prove that p ∧ ~p is a contradiction.

p T F

p∧ ~ p

~p F T

F F

From the last column p ∧ ~p is a contradiction. 2. Prove that p ∨ ~p is a tautology.

p T

~p F

p∨ ~ p T

F

T

T

From the last column p ∨ ~p is a tautology. 3. Find whether p → ~p is tautology or contradiction or neither

p T

~p F

p →~ p F

F

T

T

From the last column, clearly p → ~p is neither tautology nor contradiction. 4. Prove that (p ∧ ~q) ↔ q is neither tautology nor contradiction.

p T T F F

q T F T F

~q F T F T

p∧ ~ q F T F F

a p∧ ~ qf ↔ q F F F T

From the last column it is clear that (p ∧ ~q) ↔ q is neither tautology nor contradiction. 5. Prove that (p → q) ↔ (~q → ~p) is a tautology.

a xf

p T T F F

q T F T F

~q F T F T

~p F F T T

p→q T F T T

a yf

~ q →~ p T F T T

X↔Y T T T T

From the last column it is clear that (p → q) ↔ (~q → ~p) is a tautology.

Mathematical Logic 11

6. Prove that (p ∨ q) ∧ (~p ∧ ~q) is a contradiction.

~q F T

p∨q

T F

~p F F

T T

~ p∧ ~ q F F

T F

T T

F T

T F

F T

p

q

T T F F

a p ∨ q f ∧ a ~ p∧ ~ qf F F F F

From the last column, the given proposition is contradiction. 7. Examine whether (p → q) ∧ (q → r) is a tautology or contradiction or neither

a p → qf ∧ a q → r f

p T

q T

r T

p→ q T

q→r T

T T

T F

F T

T F

F T

F F

T F F

F T T

F T F

F T T

T T F

F T F

F

F

T

T

T

T

F

F

F

T

T

T

T

From the last column it is clear that the given proposition is neither tautology nor contradiction. 8. Prove that

p T T T T F F F F

q T T F F T T F F

r T F T F T F T F

a p → qf ∧ a q → r f → p→ q T T F F T T T T

q→r T F T T T F T T

p → r is a tautology

a p → qf ∧ a q → r f T F F F T F T T

p→ r T F T F T T T T

From the last column clearly the given proposition is tautology.

a p → qf ∧ a q → r f → T T T T T T T T

p+r

12

Basic Mathematics

1.5 LOGICAL EQUIVALENCE: Two propositions X and Y are said to be logically equivalent if and only if they have same truth values and we write X ≡ Y. Note that X ≡ Y if and only if X ↔ Y is always true or X ↔ Y is tautology.

SOME STANDARD RESULTS ON LOGICAL EQUIVALENCE: 1. Prove that p ≡ ~(~p).

1

2

3

p

~p

~ ~p

T

F

T

F

T

F

a f

from column (1) and (3) p ≡ ~(~p). 2. Prove that ~(p ∧ q) ≡ ~p ∨ ~ q

1

2

3

a

4

f

5

6

7

~p

~q

~ p∨ ~ q

p

q

p∧q

~ p∧q

T

T

T

F

F

F

F

T

F

F

T

F

T

T

F

T

F

T

T

F

T

F

F

F

T

T

T

T

From columns 4 and 7, ~ (p ∧ q) ≡ ~p ∨ ~q. 3. Prove that ~(p ∨ q) ≡ ~p ∧ ~ q

1

2

3

4

5

p

q

~p

~q

p∨q

~ p∨q

T

T

F

F

T

F

F

T

F

F

T

T

F

F

F

T

T

F

T

F

F

F

F

T

T

F

T

T

From column 6 and 7, ~ (p ∨ q) ≡ ~p ∧ ~ q.

a

6

f

7 ~ p∧ ~ q


4. Prove that ~ (p → q) ≡ p ∧ ~ q

1

2

3

4

p

q

p→q

T T

T F

T F

~ p→ q F T

F F

T F

T T

F F

a

f

5

6

~q F T

p∧ ~ q

F T

F F

F T

From columns 4 and 6, the given propositions are logically equivalent. 5. Prove that ~ (p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p)

1 p T T F F

2 q T F T F

3 p↔q T F F T

a

4

~ p↔q F T T T

a

f

5 ~p F F T T

f a

6

7

~q F T F T

8

p∧ ~ q F T F F

f a

q∧ ~ p F F T T

f

From columns 4 and 9, ~ p ↔ q ≡ p∧ ~ q ∨ q∧ ~ p Note: The results

a f ~ a p ∨ q f ≡ ~ p∧ ~ q ~ a p → qf ≡ p∧ ~ q ~ a p ↔ qf ≡ a p∧ ~ qf ∨ a~ p ∧ qf ~ p ∧ q ≡ ~ p∨ ~ q

are used to find the negation of compound propositions.

WORKED EXAMPLES: I. Negate the following: 1. 6 is an odd number or 3 is an even number. Let p : 6 is an odd number. q : 3 is an even number. Given proposition in symbol is p∨q

9

a p∧ ~ qf ∨ aq∧ ~ pf F T T T

14

Basic Mathematics

a

f

~ p ∨ q ≡ ~ p∧ ~ q

We know

∴ Negation: 6 is not an odd number and 3 is not an even number. 2. He is rich and He is not happy Let p : He is rich. q : He is happy. Given: p∧~q

a

f

a f

~ p∧ ~ q ≡ ~ p∨ ~ ~ q

We know

≡~ p ∨ q ∴ Negation: He is not rich or He is happy. 3. If the cow is big, then it is healthy. Let p : Cow is big. q : It is healthy. Given proposition in symbols: p → q

a

f

~ p → q ≡ p∧ ~ q

We know

∴ Negation: Cow is big and it is not healthy. 4. If the triangles are not equiangular then the sides are not proportional. Let p : The triangles are not equiangular. q : The sides are not proportional. Given proposition: p → q Its negation: ~ ( p → q)

a

f

~ p → q ≡ p∧ ~ q

But

Negation: The triangles are not equiangular and the sides are proportional. 5. 6 is even if and only if it is divisible by 2. Let p : 6 is even. q : 6 is divisible by 2. Given: p ↔ q. Its negation is ~(p ↔ q) But

a

f a

f a

~ p ↔ q ≡ p∧ ~ q ∨ ~ p ∧ q

f

∴ Negation: 6 is even and it is not divisible by 2 or 6 is not even and it is divisible by 2.

1.6 CONVERSE, INVERSE AND CONTRAPOSITIVE OF A CONDITIONAL Let p → q be the given conditional then the conditional q → p is called converse of p → q. The conditional ~p → ~q is called inverse of p → q. The conditional ~q → ~p is called contrapositive of p → q. Note that contrapositive is converse of inverse.


WORKED EXAMPLES: Wr ite the con ver se ver se and contr apositi ve of the ffollo ollo wing conditionals: Write conv erse se,, in inv erse contra positiv ollowing 2 1. If x = 3, then x = 9. Solution: Let p : x = 3, q : x2 = 9. Given: p → q Converse: q → p If x2 = 9, Then x = 3 Inverse: ~p → ~q If x ≠ 3. Then x2 ≠ 9. Converse: ~q → ~p If x2 ≠ 9. Then x ≠ 3. 2. If two triangles are congruent then they are similar. Let p : 2 triangles are congruent. q : 2 triangles are similar. Given: p→q Converse: q→p If 2 triangles are similar then they are congruent. Inverse: ~p → ~q If 2 triangles are not congruent then they are not similar. Converse: ~q → ~p If 2 triangles are not similar then they are not congruent. 3. If cows can fly then birds cannot fly. Let p : cows can fly. q : Birds can fly. Given: p → ~q Converse of p → q is q → p Converse of p → ~q is ~q → p i.e. If birds cannot fly then cows can fly. Inverse of p → q is ~p → ~q So inverse of p → ~q is ~p → ~(~q) ~p → q i.e. If cows cannot fly then birds can fly. Contrapositive of p → q is ~q → ~p ∴ Contrapositive of p → ~q is ~(~q) → ~p i.e., q → ~p. If birds can fly then cows cannot fly.

16

Basic Mathematics

4. If I work hard then I can score 90% and I can go for engineering. Solution: Let p : I work hard. q : I can score 90% r : I can go for engineering. Given p → (q ∧ r) Converse: q∧r→p i.e. If I can score 90% and I can go for engineering then I work hard. Inverse is ~p → ~(q ∧ r) i.e., ~p → ~ q ∨ ~r If I do not work hard then I can not score 90% or I cannot go for engineering. Contrapositive is ~(q ∧ r) → ~p i.e., (~q ∨ ~r) → ~p If I cannot score 90% or I cannot go for engineering then I do not work hard. 5. If e is not irrational and π is rational then 6 is not even or 2 is odd. Let X : e is not irrational and π is rational Y : 6 is not even or 2 is odd. Given X→Y Converse: Y→X If 6 is not even or 2 is odd then e is not irrational and π is rational. Inverse: ~X → ~Y If e is irrational or π is not rational Then 6 is even and 2 is not odd. Contrapositive: ~Y → ~X If 6 is even and 2 is not odd then e is irrational or π is not rational.

REMEMBER • p ∧ q is T only when both p and q are true otherwise it is false i.e., T ∧ T is T otherwise it is F. • p ∨ q is F only when both p and q are false. Otherwise it is True i.e., F ∨ F is F otherwise it is T. • p → q is F only when p is true and q is false i.e., T → F is F otherwise it is T. • p ↔ q is T only when both p and q are together True or False, i.e., T ↔ T and F ↔ F is T otherwise it is F. • ~p is T when p is F and vice-versa. • Tautology is a compound proposition which is always ‘true’ for all possible combinations of the truth values of its components. • Contradiction is a compound proposition which is always ‘False’ for all possible combinations of the truth values of its components. • 2 propositions X and Y are logically equivalent if and only if they have identical truth values. It is denoted by X ≡ Y. • ~ (~p) ≡ p


• • • • • • •

~ (p ∧ q ) ≡ ~ p ∨ ~ q ~ (p ∨ q) ≡ ~p ∧ ~q ~ (p → q ) ≡ p ∧ ~ q ~ (p ↔ q) ≡ (p ∧ ~ q) ∨ (~p ∨ q) Converse of the conditional p → q is q → p Inverse of the conditional p → q is ~p → ~q Contrapositive of conditional p → q is ~q → ~p.

EXERCISE ite the ffollo ollo wing compound pr opositions in symbols. I. Wr Write ollowing propositions 1. If a triangle is equilateral then all the sides of the triangle are equal. 2. If I don’t go to picnic then I will study at home. 3. Sun rises in the east and earth is not flat. 4.

2 is irrational or 5 is real.

5. A number is prime if and only if it is not composite. 6. If 2 + 2 ≠ 4 and 6 + 6 ≠ 12, then 4 + 7 = 6 or 5 + 3 = 9. 7. ABC is a right angled triangle if and only if one of the angle = 90° and square on the hypotenuse = sum of the squares on other 2 sides. 8. a + ib = x + iy iff a = x and b = y. II. If p, q and r ar aree 3 pr propositions truth espectiv ely truth opositions with tr uth vvalues alues T, F and T respecti vel y then ffind ind the tr uth values of the ffollo ollo wing: ollowing: 1. 5.

a p ∨ qf → ~ r p → a p ∨ qf

a f a p ↔ rf∨ ~ q

2. ~ p → q ∧ r 6.

3.

a p ∧ qf ↔ ~ a q ∨ r f

uct the tr uth ta ble ffor or the ffollo ollo wing pr opositions: III. Constr Construct truth tab ollowing propositions: 1. ~ p∧ ~ q 5.

2. p∧ ~ q

a p ∧ qf ↔ a q∧ ~ r f

a

6. p∨ ~ p ∧ q

3. ~ p → ~ q

f

a f

4.

a p ∧ qf ∨ r ↔ p a

4. ~ p →~ q

f

7. ~ p → q ∧ r

ify w hether the ffollo ollo wing compound pr opositions ar gies or contr adictions or IV. Ver erify whether ollowing propositions aree tautolo tautolog contradictions neither:

a

1. p → p ∨ q

a f

f

10.

3. p∧ ~ p

6.

7.

a ~ p ∧ q f ∧ aq → p f a p → qf ∧ a q → ~ r f → a p → r f p ∨ a q ∧ r f ↔ a p ∨ qf ∧ a p ∨ r f

5. ~ ~ p ↔ p 8.

2. p ∨ ~ p

9.

a p ∧ qf ∨ p ~ p ∧ a p∨ ~ q f ∧ q

a

4. p ∧ ~ p ∨ q

f

18

Basic Mathematics

hether the ffollo ollo wing compound pr opositions ar gicall y equi valent: V. Find w whether ollowing propositions aree lo log ically equiv 1. 3. 5.

a p ∧ qf → a p ∨ qf ; a p ∨ qf → a p ∧ qf p → aq ∧ rf ; a p → q f ∧ a p → rf p ∧ a q ∨ r f ; a p ∧ qf ∨ a p ∧ r f

2.

a p ∨ q f ∨ r ; p ∨ aq ∨ r f

4. p → q ; ~ p → ~ q

a

f

a

6. ~ p → q ∧ ~ r ; p ∧ ~ q ∨ r

f

gate the ffollo ollo wing compound pr opositions: VI. Ne Neg ollowing propositions: 1. 5 is odd and 6 is even. 2. Cow is not big or it is black. 3. If 2 lines are parallel then they do not intersect. 4. I will pass the examination iff the questions are easy. 5. p∧ ~ q 6. p →~ q

a

7. p ∧ q → r

f

8. p ↔~ q ver se con ver se and contr apositi ve of the ffollo ollo wing: VII. Find the in inv erse conv erse contra positiv ollowing: 2 1. If x is even then x is even. 2. If a2 + b2 = c2 and a2 = c2 then b2 = 0. 3. If a number is real then it is rational or it is irrational. 4. If Smitha gets a first class then she is either intelligent or hard working. 5. If 3 is not prime and 7 is not an odd number then 37 is not an even number or 73 is an odd number. 6. ~p → ~q 7. 8. 9. 10.

a p ∧ qf → a p ∨ qf p → aq → r f p → a~ q →~ rf a p ∨ qf → r.

ANSWERS I. 1. p → q

2. ~ p → q

3. p∧ ~ q

4. p ∨ q

5. p ↔~ q

6.

a f

a

7. p ↔ q ∧ r

8. p ↔ q ↔ r

f

a ~ p∧ ~ qf → a r ∨ sf

II. 1. F

2. T

3. F

4. F

5. F

6. T


III.

1.

~ p∧ ~ q F F F

p∧ ~ q F T 2.

T

F F

3.

a

~ p →~ q T T F

4.

T

5.

a

6.

IV. 1. 5. 9. V. 1. 3. 5. VI. 1. 2. 3. 4. 5. 6. 7. 8.

Tautology Tautology Contradiction

p∨ ~ p ∧ q T T T

f 7.

p→q∧r T T T T T F

T

T T

2. Tautology 6. Contradiction 10. Tautology

No, not logically equivalent. Logically equivalent. Logically equivalent.

f

F

a p ∧ qf ↔ a q∧ ~ r f F T T T T F

~ p →~ q T F T

3. Contradiction 7. Neither

F F

4. Neither 8. Neither

2. Logically equivalent. 4. No, not logically equivalent. 6. Logically equivalent.

It is not odd or 6 is not even Cow is big and it is not black. 2 lines are parallel and they intersect I will pass the examination and the questions are not easy or I will not pass the examination and the questions are easy. ~p ∨ q p∧q ~p ∨ (q ∧ ~r) (p ∧ q) ∨ (~p ∧ ~q)

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Basic Mathematics

VII. 1. Inverse: If x is not even then x2 is not even

Converse: If x2 is even then x is even. Contrapositive: If x2 is not even then x is not even. 2. Inverse: If a2 + b2 ≠ c2 or a2 ≠ c2 Then b2 ≠ 0.

Converse: If b2 = 0 then a2 + b2 = c2 and a2 = c2. Contrapositive: If b2 ≠ 0 then a2 + b2 ≠ c2 or a2 ≠ c2. 3. Inverse: If a number is not real then it is not rational and it is not irrational.

Converse: If a number is rational or it is not irrational then it is real. Contrapositive: If a number is not rational and it is not irrational then it is not real. 4. Contrapositive: If Smitha is neither intelligent nor hardworking then she doesn’t get a first class.

Converse: If Smitha is either intelligent or hardworking then she gets a first class. Inverse: If Smitha does not get first class then she is neither intelligent nor hardworking. 5. Contrapositive: If 37 is even and 73 is not odd then 3 is prime or 7 is odd.

Converse: If 37 is not even or 73 is odd then 3 is not prime and 7 is not odd. Inverse: If 3 is prime or 7 is odd then 37 is even and 73 is not odd. 6. Inverse: p → q Converse: ~q → ~p Contrapositive: q → p

a

f a f Converse: a p ∨ qf → a p ∧ qf Contrapositive: a ~ p∧ ~ qf → a ~ p∨ ~ qf 8. Inverse: ~ p → a q∧ ~ r f Converse: a q → r f → p Contrapositive: a q ∧ ~ r f → ~ p 9. Inverse: ~ p → a ~ q ∧ r f Converse: a ~ q →~ r f → p Contrapositive: a ~ q ∧ r f →~ p 7. Inverse: ~ p∨ ~ q → ~ p∧ ~ q

10. ~ p ∧ ~ q → ~ r r → p∨q ~ r → ~ p∧ ~ q

2 Permutation and Combination 2.1 INTRODUCTION: In our daily life we come across situations where we have to select or arrange certain things out of a given number of things. This selection or arrangement involves a principle known as fundamental principle which is illustrated by the following example. Suppose that in a auditorium there are 4 different entrance doors (say I1, I2, I3 and I4) and there are 5 different exit doors (say O1, O2, O3, O4 and O5). In how many ways can a person enter and leave the auditorium? If a person enters the auditorium through the door I1, he can go out by any one of the exit doors O1 O2 O3 O4 O5. So there are 5 ways of leaving the auditorium if the person enters it through door I1. Similarly corresponding to the entrance door I2 there are 5 ways of leaving the auditorium. Altogether 5 + 5 + 5 + 5 = 20 different ways. In general if there are m different entrance doors and n different exit doors, a person can enter and leave the auditorium in mn ways. This is fundamental principle.

2.2 FUNDAMENTAL PRINCIPLE:

Fig. 2.1

If one event can be done in m different ways and after it has been done in one of these ways, a second event (which is independent of the first) can be done in ‘n’ different ways then the two events together can occur in mn ways. The extension of this principle (also called the mnp ... principle) to the case of more than 2 events is obvious. Example: 1. A boy and a girl have to be selected from a group of 5 boys and 6 girls. In how may ways can the selection be made?

22

Basic Mathematics

Solution: Here First operation is selecting a boy from a group of 5 boys. This can be done in 5 ways. After this is done, the second operation is selecting a girl from 6 girls. This can be done in 6 ways. By fundamental principle, the total number of selections = 5 × 6 = 30 ways. 2. There are 4 candidates for the post of manager. 3 candidates for the post of officer and 5 for the post of clerk. In how many ways can these posts be filled? Solution: A manager may be selected in 4 ways. An officer may be selected in 3 ways and A clerk may be selected in 5 ways. By fundamental principle, the 3 posts together can be filled in 4 × 3 × 5 = 60 ways.

2.3 PERMUTATION AND COMBINATION: Suppose 3 members (say a, b, c) went to a cinema where they get only 2 tickets. So it is required to select 2 members out of 3 members (a, b, c). The following selections are possible: ab, bc, ca. So there are 3 ways of selecting 2 members out of 3 members. In symbols this can be written as 3c2. It is called number of combination of 3 members taken 2 at a time. Now consider 3 symbols α, β and γ. If we wish to arrange these letters taken 2 at a time, we get the following arrangements αβ, βγ, γα, βα, γβ, αγ. These arrangements are called number of permutations of 3 symbols taken 2 at a time. In symbols this can be written as 3p2. Note that in arrangement order is important. Therefore αβ is different from βα, whereas in selection order is not important. Hence ab is same as ba and so it is regarded as only one selection.

2.4 FACTORIAL OF A POSITIVE INTEGER: The product of first n natural numbers i.e. 1. 2. 3. ... (n − 1). n is called factorial n. It is represented by the symbol n or n!. ∴

a f

a f

n ! = 1 ⋅ 2 ⋅ 3 ... n − 1 ⋅ n = n n − 1 ... 3 ⋅ 2 ⋅ 1

1=1

So

2 = 2 ×1

3 = 3× 2 ×1 4 = 4 × 3 × 2 × 1 = 4 3 = 4 × 3 × 2 and so on.

In general

a f

n = n n − 1 = n n − 1 ⋅ n − 2 and so on.

2.5 PERMUTATION: An arrangement of all or part of a set of objects in some order is called permutation. If the objects are arranged along a straight line it is called a linear permutation. If the objects are arranged around a circle then it is called circular permutation.

Permutation and Combination 23

2.5.1 Linear Permutation: Each of the different arrangements in a straight line that can be made by taking some or all of a number of things at a time is called linear permutation. The number of permutation of n distinct things taken r at a time is denoted by npr. 2.5.2 Value of npr: The number of permutation of n distinct things taken r at a time i.e. npr will be same as the number of ways in which r blank places can be filled up with n given objects. As the first place can be filled in by any one of the n objects, there are n ways of filling the first place. After having filled in the first place by any one of the n things, there are (n − 1) objects left. Hence 2nd place can be filled in (n − 1) ways. Similarly 3rd place can be filled in (n − 2) ways and so on. Hence rth place can be filled in (n − (r − 1)) ways i.e., (n − r + 1) ways.

1st n

Position of the object Number of ways

2 nd n −1

a f a

3 rd n−2

f

...... ......

r th

an − r + 1f

By fundamental principle, r places can be filled by any r of n objects in n (n−1) (n−2) ... (n−r +1) ways. n

∴

a fa fa f a

f

pr = n n − 1 n − 2 n − 3 ... n − r + 1

Multiplying and Dividing by (n − r) (n − r − 1) ... 2 ⋅ 1 in RHS we get n

pr =

a fa

fa f a a fa

fa fa f

f

n n − 1 n − 2 n − 3 ... n − r + 1 n − r n − r − 1 ... 2 ⋅ 1 n − r n − r − 1 ... 2 ⋅ 1 n

pr =

n n−r

Hence value of

n

pr =

n . n−r

2.5.3 Value of npn: The number of permutation of n distinct objects taken all at a time is npn. It is same as the number of ways in which n blank spaces can be filled up with n given objects.

Position of the object Number of ways

1st n

2 nd n −1

3 rd n−2

...... ......

n − 1th 2

n th 1

As the first place can be filled in by any one of the n objects there are n ways of filling the first place. After having filled in the first place, 2nd place can be filled in (n − 1) ways and so on. ∴ By fundamental principle, n

a fa f

pn = n n − 1 n − 2 ... 2 ⋅ 1

24

Basic Mathematics n

∴ ∴

pn = n n

Value of

pn = n .

2.5.4 Value of 0 : n

We have

pr =

n n−r

r = n.

Put n

pn = n

But ∴

n n−n

pn = n n=

n 0

0=

n n

Cross-multiplying,

0 =1

⇒

WORKED EXAMPLES: 1. Evaluate: ( a)

5

(b)

p2

We have

n

pr =

6

n

7

p7

n n−r

2. If npn = 720 find n. We have

(c)

p3

a

5 = 5 × 4 = 20. 5−2

5

p2 =

6

p3 = 6 × 5 × 4 = 120.

7

p7 = 7 = 5040.

pn = n = 720 given

f


2 3 4 5 6

n = 6 × 5 × 4 × 3 × 2 ×1

n= 6 ⇒

n = 6.

3. If np2 = 72, find n. We have

720 360 120 30 6 1

np 2

= n (n − 1) = 72. n (n − 1) = 9 × 8 [By inspection: 72 = 9 × 8]. ⇒ n = 9. 4. In how many of the permutations of 7 things taken 4 at a time will (a) One thing always occur (b) One thing never occur? Solution: (a) Keeping aside the particular thing which will occur, the number of permutation of 6 things taken 3 at a time = 6p3 = 6 × 5 × 4 = 120. Now this particular thing can take up any one of the four places and so the total number of ways = 120 × 4 = 480 ways. (b) Leaving aside the particular thing which has never to occur, the number of permutation of 6 things taken 4 at a time = 6p4 = 6 × 5 × 4 × 3 = 360 ways. 5. How many 4 digit numbers can be formed with the digits 2, 4, 5, 7, 9. (Repetitions not being allowed). How many of these are even? Number of 4 digit numbers that can be formed with the digits 2, 4, 5, 7, 9 (without repetitions) = 5p4 = 5 × 4 × 3 × 2 = 120. Th H T U Since we require an even number, we must have 2 or 4 in the unit’s place. After filling the unit’s place by 2 or 4, the remaining 3 places (Ten’s, Hundred’s and Thousand’s) can be filled by remaining digits 5, 7, 9, 2 or 4 in 4p3 ways 4 × 3 × 2 = 24 ways. ∴ Number of even numbers that can be formed 2 or 4 4 = 2× 4 p3 = 2 × 24 = 48

p3

6. How many numbers can be formed by using any number of digits 3, 1, 0, 5; no digit being repeated in any number: Solution: The number of single digit numbers = 3p1 (excluding zero) = 3. The permutation of 4 digits taking 2 at a time are 4p2 but 3p1 of these have zero in ten’s place so reduce to single digit number. ∴ Number of 2 digit numbers = 4p2 − 3p1. Similarly number of 3 digit numbers = 4p3 − 3p2. Number of 4 digit numbers = 4p4 − 3p3.

26

Basic Mathematics

Total number of numbers

d p − p i+d p − p i+d p − p i = 3 + a 4 × 3 − 3f + a 4 × 3 × 2 − 3 × 2 f + a 4 × 3 × 2 × 1 − 3 × 2 × 1f = 3 + a12 − 3f + a24 − 6f + a24 − 6f = 3+

4

2

3

1

4

3

3

2

4

4

3

3

= 3 + 9 + 18 + 18 = 48. 7. There are 4 Kannada books, 3 Hindi books and 5 English books. In how many ways can these be placed on a shelf if the books of the same language are to be together? Solution: Since the books of the same language are to be together. Let us consider the 4 Kannada books as 1 unit, 3 Hindi books as another unit and 5 English books as a different unit. Then we have to arrange 3 3 different units. This can be done in p3 = 3 = 6 ways. 4 But the 4 Kannada books remaining together can be arranged in p4 =

4 ways. Similarly, 3 Hindi

books can be arranged in 3 ways and 5 English books can be arranged among themselves in 5 ways. ∴

Number of permutations = 3 × 4 × 3 × 5

=3×2×4×3×2×1×3×2×5×4×3×2×1 = 6 × 24 × 6 × 120 = 103680 ways. 8. In how many different ways can 5 examination papers be arranged in a row so that the best and the worst papers may never come together. Solution: Without any restrictions the 5 exam papers can be arranged among themselves in 5p5 = 5 = 120 ways. Considering now, the best and the worst papers as a single paper, we have only 3 + 1 = 4 papers. Now these 4 papers can be arranged taking all at a time in 4 p3 = 4 × 3 × 2 = 24 ways. But in each of the 24 ways, the best and worst papers can be arranged among themselves in 2 = 2 ways. ∴ Total number of ways, where the best and the worst paper always come together = 24 × 2 = 48. Hence required number of arrangements where best and the worst paper never come together = 120 − 48 = 72 ways. 9. In how many ways can 5 boys and 3 girls be seated in a row so that (a) each girl in between 2 boys (b) no two girls sit together (c) all the girls are together. Solution: (a) First we arrange 5 boys in a row. This can be done in 5p5 = 5 = 120 ways. . Consider one such arrangement:

B1 ↓ B2 ↓ B3 ↓ B4 ↓ B5


There are 4 places available for 3 girls, so that each girl is between 2 boys. The 4 places can be filled by 3 girls in 4p3 ways = 4 × 3 × 2 = 24 ways. For one arrangement of boys, there are 24 ways of arranging girls. ∴ For 120 arrangement of boys = 120 × 24 = 2880 ways. (b) Again we arrange 5 boys in a row. This can be done in 5 = 120 ways. Considering 1 such arrangement *B 1 *B 2 * B 3 *B 4 * B 5 * There are 6 places available for the girls so that no two girls are together. The 6 places can be filled by 3 girls in 6p3 ways = 6 × 5 × 4 = 120 ways. ∴Required number permutations = 120 × 120 = 14400 ways. (c) We require all the 3 girls to be together. So we consider 3 girls as one unit. Now 5 boys and 1 unit (of 3 girls) can be arranged in 5 + 1 = 6 = 720 ways. Again 3 girls, wherever they are together can be arranged in 3 = 6 ways. ∴ Number of permutations = 720 × 6 = 4320 ways. 10. How many different words can be formed with the letters of the word ‘ORDINATE’. (a) Beginning with O (b) beginning with O and ending with E. (C) So that vowels occupy odd places. (a) Keeping ‘O’ in the first place. We can arrange the remaining 7 letters in 7 places in 7

p7 = 7 ways = 5040 ways.

O

∴

7

p7

Number of words that can be formed with the letters of word ‘ORDINATE’ that begins with 0 = 5040. (b) Keeping O in the first place and E in the last place, the remaining 6 letters can be arranged in 6 place in 6p6 = 6 = 720 ways. O

E

6

p6

(c) In the word ‘ORDINATE’ O, I, A, E are vowels. There are 4 odd places and 4 vowels. The 4 vowels can be arranged in 4 odd places in 4p4 = 4 ways = 4 × 3 × 2 × 1 = 24 ways.

28

Basic Mathematics 1

st

3

rd

5

th

7

th

Remaining 4 even places can be occupied by remaining 4 letters in 4p4 ways = 4 ways = 24 ways.

∴Total number of permutations = 24 × 24 = 576. 11. If the letters of the word MAKE be permuted and the words so formed be arranged as in dictionary what is the rank of the word? Solution: The alphabetical order of the letters is AEKM. If A is fixed in 1st place, the other 3 letters can be permuted in 3 = 3 × 2 = 6 ways. . The number of words begin with A = 6. Similarly the number of words which begin with E and K = 3 = 6 each. The words beginning with M are MAEK and MAKE. ∴ Rank of the word = 6 + 6 + 6 + 1 + 1 = 20. 12. Prove that the number of ways in which n books can be placed on a shelf when 2 particular books

a

f

are never together is n − 2 ⋅ n − 1. Solution: Regarding the 2 particular books as one book, there are (n − 1) books now, which can be arranged in

n −1

pn −1 = n − 1 ways. Now these two books can be arranged in 2 ways. Therefore the

number of permutations in which 2 particular books are always together = 2 . n − 1 . The number of permutations of n books without any restriction = n . ∴ The number of permutations in which 2 particular books never occur together

= n − 2. n −1 = n ⋅ n − 1 − 2. n − 1

a = an − 2 f

= n −1 n − 2

f

n −1

Hence proved.

2.6 PERMUTATION OF THINGS OF WHICH SOME ARE ALIKE Theorem Theorem: If x is the number of permutations of n things taken all at a time of which one set p are alike, another set q are alike and so on. Then prove that x=

n p ⋅ q ...


Proof: Replacing p like objects by ‘p’ unlike objects, q like objects by ‘q’ unlike objects and so on, we arrive at a stage where all n objects are distinct. By permuting these unlike objects amongst themselves each of the x permutations would give rise to p ⋅ q ... permutations. Hence x permutations give rise to x ⋅ p ⋅ q ... . But the number of permutations of n distinct objects taken all at a time = n . x ⋅ p ⋅ q ... = n

∴ ⇒

x=

n . p ⋅ q ...

WORKED EXAMPLES: 1. In how many ways can letters of the word ‘INDIA’ be arranged? Solution: There are 5 letters in the word ‘INDIA’ of which I occur 2 times. ∴

5 2

Number of words possible =

= 5 × 4 × 3 = 60. 2. In how many ways can the letters of the word ‘PERMANENT’ be arranged so that 2E’s are always together. Solution: There are 9 letters in the word ‘PERMANENT’ of which E occurs 2 times, N occurs 2 times. If 2 E’s are together, taking them as one letter we have to arrange 8 letters in which N occurs two times. ∴

Number of arrangements =

8 2

= 8 × 7 × 6 × 5 × 4 × 3 = 20,160 3. In how many ways can be letters of the word ‘HOLLOW’ be arranged so that the 2L’s do not come together. Solution: There are 6 letters in the word HOLLOW in which L occurs 2 times, O occurs 2 times. ∴

Number of arrangement =

=

6 2⋅ 2 6× 5× 4×3×2 2 ×1× 2 ×1

= 180. If 2 L’s are together, taking them as one letter, we have to arrange 5 letters in which O occurs 2 times.

5 = 5× 4×3 2

∴

Number of arrangements in which 2 L’s are together =

∴

= 60. Number of arrangements in which 2L’s do not come together = 180 − 60 = 120.

30

Basic Mathematics

4. Find the number of permutations of the word ‘EXCELLENCE’. How many of these permutations (i) begin with E (ii) begin with E and end with C (iii) begin with E and end with E (iv) do not begin with E. Solution: The word ‘EXCELLENCE’ contains 10 letters of which E occurs 4 times C occurs 2 times, L occurs 2 times. Number of permutations =

10 4⋅ 2⋅ 2

10 × 9 × 8 × 7 × 6 × 5 = 37800. 2 ×1× 2 ×1

=

( i) Put E in the first place and arrange the rest. Now there are 9 letters of which E occurs 3 times, L occurs twice and C occurs twice. ∴

9 3⋅ 2 ⋅ 2

Number of permutations =

9×8×7×6×5× 4 2 ×1× 2 ×1

=

= 15,120. (ii) Put E in the first place and C in the last place, and arrange the rest. There are 8 letters of which E occurs 3 times, L occurs twice. ∴


=

8 3⋅ 2 8× 7×6×5×4 = 3360. 2

( iii) Put one E in the first place and another E in the last place and arrange the rest. There are 8 letters of which E occurs 2 times, L occurs twice and C occurs twice. ∴


=

8 2⋅ 2⋅ 2 8×7×6×5× 4×3 2 ×1× 2 ×1

= 5040. (iv) Number of permutations that do not begin with E = Total number of permutations. — Number of permutations that begin with E. = 37800 − 15120 = 22680. 5. How many numbers less than 3 millions can be formed using the digits of the number 2123343?


Solution: Million

HTh

TTh

Th

H

T

U

1 or 2

Since the number should be less than 3 millions, we can have either 1 or 2 in the million’s place. Keeping 1 in the million’s place and arranging the rest. There are 6 numbers 2, 2, 3, 3, 3, 4 of which 2 repeats twice, 3 occurs thrice. ∴

Number of numbers =

6 6×5× 4 = 2⋅ 3 2

= 60. Now keeping 2 in the million’s place and arranging the rest. There are 6 numbers 1, 2, 3, 3, 3, 4 in which 3 occurs thrice. ∴

Number of numbers =

6 =6×5×4 3

= 120. Total number of numbers less than 3 millions. = 60 + 120 = 180.

2.7 CIRCULAR PERMUTATION: If we have to arrange 4 letters A, B, C, D in a row. Then 2 of the arrangements would be ABCD, BCDA and we treat these two arrangements as different. But if we arrange along the circumference of the circle, the two arrangements

and

are one and the same.

So we conclude that circular permutations are different only when the relative order of the objects is changed; otherwise they are same. In circular permutation of n different things one thing is kept fixed and the balance (n − 1) things are arranged relative to it in n − 1 ways. If the clockwise and anticlockwise orders are distinguished, the required number of permutations = n −1 . ∴ Number of ways in which n persons can occupy the chairs in a round table = n − 1 . If the clockwise and anticlockwise orders are not distinguished then required number of permutations =

n −1 2

.

32

Basic Mathematics

∴ Number of ways in which n flowers or n beads are strung to form garland or necklace =

n −1 2

.

WORKED EXAMPLES: 1. In how many ways 5 people sit around a table? Solution: Fixing the position of one person, the remaining 4 persons can sit around a table in 4 ways = 4 × 3 × 2 × 1 = 24. ∴ 5 people can sit round a table in 24 ways. 2. In how many ways can 7 different jewels be strung into a necklace? Solution: Keeping one jewel fixed, remain 6 jewels can be arranged in 6 = 720 ways. Since clockwise and anticlockwise arrangements are same, Required number of permutations.

1 × 720 = 360 ways. 2 3. In how many ways can 7 people be arranged at a round table so that 2 particular persons always sit together. Solution: First, the two particular persons can be arranged in 2 = 2 ways. . Considering them as one fixed person, the remaining 5 persons can be arranged in 5 = 120 ways. ∴ The required number of permutation = 120 × 2 = 240. 4. A round table conference is to be held between delegates of 9 countries. In how many ways can they be seated if 2 particular delegates must not sit next to each other? Solution: The number of ways in which 2 particular delegates must not sit next to each other = Total number of permutations — Number of permutations in which 2 particular delegates sit next to each other. Now Total number of circular permutations of 9 delegates = 9 − 1 = 8 Considering 2 delegates as one fixed person, the remaining 7 delegates can be arranged in 7 ways. 2 delegates again can be arranged in 2 ways. ∴ Number of permutations in which 2 particular delegates sit next to each other = 7 × 2 ∴ Required number of permutation

a f

= 8 − 7 × 2 = 8 7 − 7 × 2 = 7 8 − 2 = 6 7 = 30240

5. In how many ways can 6 persons sit around a table so that all shall not have the same neighbours in any 2 arrangements? Solution: 6 persons can sit round a table in 5 = 120 ways. But each person will have the same neighbours in clockwise and anticlockwise arrangements.


∴

Required number of ways =

5 120 = = 60. 2 2

6. In how many ways can 4 gentlemen and 4 ladies sit down together at a round table so that no two ladies may come together. Solution: Let the gentlemen first take up their seats. They can sit in 3 = 6 ways. When they have been seated, there remain 4 places for the ladies each between 2 gentlemen. Therefore the 4 ladies can sit in 4 places in 4 = 24 ways. Required number of ways = 3 × 4

∴

= 6 × 24 = 144.

2.8 COMBINATION Each of the different groups or selection which can be made by taking some or all of a number of things at a time (irrespective of the order) is called a combination. The number of ways of selection of n different things taken r at a time is called the number of combination of n different things taken r at a time. It is written as ncr. 2.8.1 Value of ncr: The number of combinations of n different things taken r at a time can be arranged in r ways. nc

∴

r

combinations will produce n cr × r permutations.

Now n cr × r = Number of permutations of n different things taken r at a time = npr. n

∴

cr × r = n pr

⇒

n

But

n

∴

n

cr =

pr r

pr =

n n−r

n

cr =

n . n−r⋅ r

Note: (1) When r = n. 1. When r = n. n

∴

n

cn = 1

cn =

n n = =1 n − n⋅ n 0⋅ n

3 0 =1

34

Basic Mathematics

2. When r = 0 n

cr = n c0 = n

∴

n =1 n−0⋅ 0

c0 = 1

3. When r = 1. n

n n n −1 = =n n − 1⋅ 1 n −1

c1 =

nc 1

∴

= n.

2.8.2 Complementary Combinations: 1. Prove that ncr = ncn−r Proof: 1. By using formula for ncr: n

LHS: n

RHS:

n

n n−r⋅ r

cr =

cn − r = cn − r =

...(1)

n n−n+r⋅ n−r

n

=

r⋅ n−r

n n−r⋅ r

...(2)

From (1) and (2) n

cr = n cn − r

Proof: By analytic method: nc is selecting r things from n things. If we select r things from n things then ( n − r) things are left. r ∴ For every combination of (n − r) things, there corresponds a combination of r things. nc = nc ∴ r n − r. Hence proved. n n n + 1 cr 2. Prove that cr + cr − 1 = Proof: By using formula for ncr nc + nc LHS: r n−r

= = =

n n + n−r⋅ r n − r −1 ⋅ r −1

a f

n n − r ⋅r ⋅ r −1 n n − r ⋅r ⋅ r −1

+ +

n n − r + 1⋅ r − 1 n

an − r + 1f ⋅ n − r ⋅ r − 1


=

LM OP N Q LM n − r + 1 + r OP n n − r ⋅ r − 1 N r an − r + 1f Q n ⋅ an + 1f n +1 =

=

n + 1c r

=

=

n 1 1 + n − r ⋅ r −1 r n − r +1

n − r + 1⋅ r

n +1− r ⋅ r

= RHS.

n + 1c r

Proof by analytic method: is the total number of combination of n + 1 things taken r at a time which is nothing but the combination that contain a particular thing (ncr) plus the combination that do not contain a particular thing (ncr − 1). n +1

cr = n cr + n cr −1

i.e.,

WORKED EXAMPLES: 1. Find the value of (i) 7c3 (ii) 5c2 + 5c1 ( i) We have

∴

= 5

(ii) Now

6

c2 =

n

cr =

n n−r⋅ r

7

c3 =

7 7 − 3⋅ 3

7×6×5× 4 = 35. 4 × 3 × 2 ×1

c2 + 5c1 = 5+1c2 = 6 c2 6 6×5× 4 = = 15. 6 − 2⋅ 2 4⋅2 OR

5

c2 + 5 c1 = =

5 5 + 5 − 2⋅ 2 5 − 1⋅ 1 5× 4× 3 5× 4 + 3×2 4 ⋅1

= 10 + 5 = 15.

3 n cr + n cr −1 = n +1cr

36

Basic Mathematics

2. Find n if nc20 = nc6 Solution: In the formula ncr = ncn − r We observe r + n − r = n n

∴

c20 = n c6

⇒ 20 + 6 = n ⇒ n = 26. 3. If nc7 = nc23, find nc29. n

d3

c7 = n c23

⇒ n = 7 + 23 n = 30 n

Now,

c29 = 30 c29 =

30 30 − 29 ⋅ 29

= 30. 4. If

20c r+2

=

20c 2r − 6,

then find r. 20

Given

cr + 2 = 20 c2 r − 6

⇒ r + 2 = 2r − 6 2 + 6 = 2r − r

⇒ r=8

OR 20

⇒ r + 2 + 2 r − 6 = 20

n

cr = n cn − r ⇒ r + n − r = n 3r − 4 = 20 3r = 24 r = 8.

⇒ 5. If nc2 = 36, find n. Solution:

cr + 2 = 20 c2 r −6

n

c2 =

n = 36 n−2⋅ 2

a f

n n −1 ⋅ n − 2 = 36 n − 2⋅2

⇒

a f

n n −1 = 36 2

n

cr = n cn − r ⇒ r + n − r = n

i


⇒ 6. If nc5 = 24 nc4, find n.

n (n − 1) = 72 n (n − 1) = 9 × 8 n = 9. n

(By inspection)

c5 = 24 ⋅ n c4

n n = 24 ⋅ n − 5⋅ 5 n − 4⋅ 4 n− 4⋅ 4 = 24 n − 5⋅ 5

⇒

an − 4f ⋅ n − 5 ⋅ 4 = 24 n − 5⋅5⋅ 4

n − 4 = 120 n = 120 + 4 n = 124. n n 7. If pr = 60 and cr = 10, then find n and r. Solution: We have

n

n

cr =

pr r

∴

10 =

⇒

r=

60 r

60 =6 10

r= 3

⇒ Given:

r = 3. np np

r

3

= 60 = 60

n = 60 n−3

a fa

f

n n −1 n − 2 n − 3 = 60 n−3

⇒

n (n − 1) (n − 2) = 60 n (n − 1) (n − 2) = 5 × 4 × 3. n = 5.

2 2 3 5

60 30 15 5

1 (By inspection)

38

Basic Mathematics

8. If

18c

r

Given

=18cr + 2, then find rc5. 18c

r

=

18c r+2

LM c = c OP N ⇒ r + n − r = nQ n

3

r ≠r+2 r + r + 2 = 18

r

n

n− r

2 r = 16

⇒

r = 8. r

c5 = 8 c5 =

=

8 8 − 5⋅ 5

8×7×6× 5 = 56. 3 × 2 ×1× 5

9. In how many ways can 4 persons be selected from amongst 9 persons? How many times will a particular person be always selected? Solution: The number of ways in which 4 persons can be selected from amongst 9 persons

= 9c4 =

9 9×8× 7×6 = = 126. 9 − 4⋅ 4 4 × 3 × 2 ×1

Let a particular person is selected always. Then we have to select 3 persons from the remaining 8 persons. This can be done in 8c3 ways =

8 = 56. 8 − 3⋅ 3

10. A student has to answer 7 out of 10 questions in an examination. How many choices has he, if he must answer the first three questions. Solution: There are 10 questions of which a student must answer first 3 questions. Remaining 4 questions (3 he has to answer 7 questions) can be selected among 10 − 3 = 7 questions in 7c4 ways. ∴

Number of combinations = 7 c 4 =

=

7 7− 4⋅ 4

7×6×5× 4 3 × 2 ×1× 4

= 35 ways. 11. In how many ways 5 red and 4 green balls can be drawn from a bag containing 7 red and 8 green balls. 7 Solution: Number of ways of drawing 5 red balls from 7 red balls = c5 =

=

7 7 − 5⋅ 5

7×6× 5 = 21. 2× 5

Permutation and Combination 39 8 Number of ways of drawing 4 green balls from 8 green balls = c4 =

= ∴

8 8− 4⋅ 4

8× 7×6×5 = 70. 4 × 3× 2 ×1

Total number of ways (By fundamental principle) = 70 × 21 = 1470.

12. Find the number of (a) Straight lines (b) triangles that can be drawn from 20 points of which 4 are collinear. Solution: Two points are needed for a straight line. If none of the 20 points are collinear then we would get 20c2 straight lines. But 4 points are given to be collinear. So we would not get 4c2 lines, instead we get only one straight line containing all the 4 points. ∴

Number of straight lines =

= =

20

c2 − 4 c2 + 1

20 4 − +1 20 − 2 ⋅ 2 4 − 2 ⋅ 2 20 × 19 4 × 3 − +1 2 2

= 190 − 6 + 1 = 185. (b) We need 3 non-collinear points for a straight line. If none of the 20 points are collinear then we would get 20c3 triangles. Since 4 points are given to be collinear, we would not get 4c3 triangles from these points. ∴ Number of triangles = 20c3 − 4c3

=

20 4 − 20 − 3 ⋅ 3 4 − 3 ⋅ 3

= 1140 − 4 = 1136. 13. A committee of 10 members is to be chosen from 9 teachers and 6 students. In how many ways this can be done if ( i) The committee contains exactly 4 students. (ii) There is to be a majority of teachers. ( iii) There are atleast 4 students. ( iv) There are at most 7 teachers. Solution: ( i) The committee contains exactly 4 students and 10 − 4 = 6 teachers. 4 students can be selected out of 6 students in 6c4 ways and 6 teachers out of 9 teachers can be selected in 9c6 ways.

40

Basic Mathematics

∴ The number of selections =6c4 × 9c6

= =

(ii) As ( a) ( b) (c ) ( d) ∴

9

c6 × 6 c 4 = 1260

( b)

9

c7 × 6 c3 = 720

(c )

9

c8 × 6 c2 = 135

( d)

9

c9 × 6 c1 = 6

( iii) As ( a) ( b) (c )

6 9 × 6 − 4⋅ 4 9 − 6⋅ 6

6×5× 4 9×8×7× 6 × 2 ×1× 4 6 × 3 × 2 ×1

= 15 × 84 = 1260. there is to be majority of teachers, the committee may consist of 6 teachers 4 students 7 teachers 3 students 8 teachers 2 students 9 teachers 1 student. Number of selection =

( a)

∴

Total number of selections = 1260 + 720 + 135 + 6 = 2121. there is to be at least 4 students, the committee may consist of 4 students 6 teachers 5 students 5 teachers 6 students 4 teachers. Number of selections =

( a)

6

c4 × 9 c6 = 1260

( b)

6

c5 × 9 c5 = 756

(c )

6

c6 × 9 c 4 = 126

So Total number of selections = (iv) As ( a) ( b) (c ) ( d)

(By fundamental principle)

1260 + 756 + 126 = 2142. there is to be atmost 7 teachers, the committee may consist of 7 teachers and 3 students 6 teachers and 4 students 5 teachers and 5 students 4 teachers and 6 students.


∴

Number of selections = (a )

9

c7 × 6 c3 = 720

(b )

9

c6 × 6 c 4 = 1260

(c )

9

c5 × 6 c5 = 756

(d )

9

c4 × 6 c6 = 126

∴

Total number of selections = 720 + 1260 + 756 + 126 = 2862.

14. A team of eleven is to be chosen out of 16 cricketers of whom 4 are bowlers and 2 others are wicket keepers. In how many ways can the team be chosen so that there are at least 3 bowlers and at least one wicket keeper. Solution: 4 Bowlers

2 Wicket keepers

10 others

3 4 3 4

1 1 2 2

7 6 6 5

(a) (b) (c ) (d) Number of ways

∴

(a)

4

c3 × 2 c1 × 10 c7 = 960

(b)

4

c4 × 2 c1 × 10 c6 = 420

(c )

4

c3 × 2 c2 × 10 c6 = 840

(d)

4

c4 × 2 c2 × 10 c5 = 252

Total number of ways = 960 + 420 + 840 + 252 = 2472. 15. Arun has 7 friends, 4 of them are boys and 3 are girls. His sister, Aalekya has 7 friends, 4 of them are girls and 3 of them are boys. In how many ways can they invite for a party of 3 girls and 3 boys. So that there are 3 of Arun’s friends and 3 of Aalekya’s friends.

42

Basic Mathematics

Solution: Arun's friends

Aalekya's friends

4 boys

3 girls

3 boys

4 girls

3 2 1 –

– 1 2 3

– 1 2 3

3 2 1 –

(a) (b) (c ) (d)

∴ Number of selections: ( a)

4

c3 × 3c0 × 3 c0 × 4 c3 = 16

( b)

4

c2 × 3 c1 × 3 c1 × 4 c2 = 324

(c )

4

c1 × 3 c2 × 3c2 × 4 c1 = 144

( d)

4

c0 × 3c3 × 3 c3 × 4 c0 = 1

∴ Total number of selections = 16 + 324 + 144 + 1 = 485. 16. How many diagonals are there in a octagon? Solution: Number of diagonals in octagon = 8 c2 − 8

[3 octagon has 8 sides]

= 20.

REMEMBER: • •

n

n

n n−r

pr =

p0 = 1,

•

n

pn = n

•

n

p1 = n

• Permutation of n objects of which p objects are of one kind, q are of another kind and so on is n . p ⋅ q ...

• Number of circular arrangement of n persons round a table = n − 1 • Number of circular arrangement of n beads or n flowers to form a necklace or garland =

•

n

cr =

n n−r⋅ r

n −1 . 2


•

n

cn = 1

•

n

c1 = n

•

n

c0 = 1

•

n

cr = n cn − r

•

n

cr + n cr −1 = n +1cr

• Number of straight lines that can be drawn from n points of which p points are collinear = nc − pc + 1. 2 2 • Number of triangles that can be drawn from n points of which p points are collinear = nc3 − pc3. • Number of diagonals in a polygon of n sides = nc2 − n.

EXERCISE I. Find the value of: 1. 10p3 2. 12p3 3. 15c8 4. 8c5 5. 14c10 II. Find n if 1. np2 = 90 2. nc3 = 20 III. Find r if 1. 6pr = 360 2. 13pr = 156 IV. Find n and r if 1. npr = 240 and ncr = 120 2. npr = 336 and ncr = 56 V. 1. If np4 = 12, np2 = 120, find n. 2. If np4 = 56, np2 = 120, find n. VI. 1. How many 3 digit numbers can be formed by using the digits 9, 7, 6, 5, 3, 2 (repetitions not allowed)? (a) How many of these are less than 400? (b) How many of these are multiples of 5? (c) How many of these are multiples of 2? 2. In how many ways can the letters of the word ‘STRANGE’ be arranged so that (a) The vowels never come together. (b) The vowels are never separated. 3. A shelf contains 6 Hindi books, 5 Kannada books and 8 English books. In how many ways can they be arranged so that (a) Hindi books are together? (b) Hindi books are together and Kannada books are together.

44

4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21.

Basic Mathematics

(c) Books of the same languages are together. ( d) No two English books are together. How many arrangements of the letters of the word SUNDAY can be made if the vowels are to appear only in the odd places. How many numbers of four different digits can be formed using the digits 0, 1, 2, 3, 4, 5? How many of them are even? If the letters of the word GATE be permuted and the words so formed be arranged as in a dictionary what will be the rank of the word? Find the number of permutations of the letters of the word INSTITUTION, when all the letters are taken at a time. How many of them (i) Have 3T’s together (ii) begin with 2N’s. Find the number of permutations of the letters of the word ASSASSINATION. How many of them (i) have 3 A’s together (ii) begin with 2 N’s. Find the number of permutation of letters of the word TOMORROW. How many of them have (i) 3O’s together. (ii) End with 2R’s. Find the number of ways in which 6 different beads can be arranged to form a necklace. In how many ways can 5 persons sit around a table. In how many ways can 4 boys and 4 girls be seated round a table so that no two boys are adjacent. In how many ways can 7 persons sit around a table so that all shall not have the same neighbours in any two arrangements? A round table conference is to be held between delegates of 20 countries. In how many ways can they be seated if 2 participants may wish to sit together always. (i) If nc10 = nc6, find n. (ii) If 43cr − 6 = 43c3r + 1, then find r. From 8 lecturers and 4 students a committee of 6 is to be formed. In how many ways can this be done so that the committee contains (i) exactly 2 students (ii) atleast 2 students. How many (i) straight lines (ii) triangles are determined by 12 points, no three of which lie on the same straight line. How many (i) straight lines (ii) triangles are determined by joining 20 points in a plane of which 6 are collinear. In how many ways a student can choose 8 questions from a set of 12 questions if the questions 1 and 10 are compulsory. Find the total number of diagonals of a hexagon. Out of 3 books on maths, 4 books on Physics and 5 books on Chemistry, how many collections can be made, if each collection consists of ( i) exactly one book on each subject (ii) at least one book on each subject.

ANSWERS I. 1. 720 II. 1. n = 10

2. 132 2. n = 6.

3. 6435

4. 56

5. 1001


III. 1. IV. 1. V. 1. VI. 1. 2.

r=4 r=2 n =16, r = 2 n=4 120 (a) 40 (a) 3600 (b) 1440

2. n = 8, r = 3 2. n = 8 (b) 20 (c) 40

( b) 10 ⋅ 6 ⋅ 5

3. (a) 13 6

(c) 3 ⋅ 6 ⋅ 5 ⋅ 8

4. 144 5. 300, 156 6. 14 7.

11 3⋅ 3⋅ 2

8.

12 144

9. 10. 11. 12. 13.

3360 60 24 144 360

( i) 30240 ( i)

10 24

( i) 360

14. 2 × 18 15. 16. 17. 18. 19. 20. 21.

(i) 16 (i) 420 (i) 12c2 (i) 176 210 9 (i) 60

(ii) 12 (ii) 672 (ii) 12c3 (ii) 1120

(ii) 3255.

(ii) 10080 (ii)

10 72

(ii) 120

12 (d) 11 ⋅ p8

46

Basic Mathematics

3 Probability 3.1 INTRODUCTION: The term probability refers to the chance of happening or not happening of an event. The theory of probability provides a numerical measure of the elements of uncertainity. It enables us to take decision under conditions of uncertainity with a calculated risk. The theory of probability has its origin in the games of chance, related to gambling for instance throwing a dice or tossing a coin. Generally speaking, the probability of an event denotes the likelihood of its happening. The value of probability ranges between zero and one. If an event is certain to happen its probability would be 1 and if it is certain that the event wouldn’t take place, then the probability of its happening is zero. Ordinarily in social sciences probability of the happening of an event is rarely 1 or 0. The reason is that in social sciences we deal with situation where there is always an element of uncertainity about the happening or not happening of an event.

3.2 TERMINOLOGY: Before we give definition of probability, it is necessary that we familiarise ourselves with certain terms that are used in this context. (i) Random eexper xperiment: xper iment: It is an experiment which if conducted repeatedly under homogeneous conditions doesn’t give the same result. The result may be any one of the various possible outcomes. For example: If a die is thrown it wouldn’t always fall with number 3 up. It would fall in any one of six ways which are possible. (ii) Trial and eev vent: The performance of a random experiment is called a trial and the outcome – an event. Event could be either simple or compound (or composite). An event is called simple if it corresponds to a single possible outcome. Thus in tossing a die, the chance of getting 3 is a simple event (Q 3 occurs in a die only once). However the chance of getting an odd number is compound (Q odd numbers are more than one — 1, 3 and 5). (iii) Exhausti Exhaustiv ve cases: All possible outcomes of an event are known as exhaustive cases. In the throw of a single die the exhaustive cases are six, as the die has only 6 faces each marked with different

Probability

(iv)

(v)

(vi)

(vii)

47

numbers. Similarly the number of exhaustive cases in tossing 2 coins would be 4: HH, HT ,TH and TT (H-Head, T-tail). oura Favour able cases: The number of outcomes which result in the happening of a desired event are called favourable cases. Thus in a single throw of a die the number of favourable cases of getting an odd number are 3 (i.e. 1, 3 and 5). Mutually xclusi lusiv Mutuall y eexc xc lusi ve cases: Two or more cases are said to be mutually exclusive if the happening of any one of them excludes the happening of all others in a single experiment. Thus in a throw of a single die, the events 5, 4 and 3 are mutually exclusive. Equally likel ely Equall y lik el y cases: Two or more events are said to be equally likely if the chance of their happening is equal, i.e., there is no preference of any one event over the other. Thus in the throw of a die, the coming up of 1, 2, 3, 4, 5 or 6 is equally likely. Independent dependent Inde pendent and de pendent eev vents: An event is said to be independent if its happening is not affected by the happening of the other events. So in the throw of a die repeatedly coming up of 5 on the first-throw is independent of coming up of 5 again in the second throw. However we are successively drawing cards from a pack without replacement, the event would be dependent.

3.3 DEFINITION OF PROBABILITY: We can define probability in 3 ways. (i) Mathematical or classical definition. (ii) Statistical or Empirical definition. (iii) Subjective approach or set theoretic approach definition. i Mathematical or classical definition: If there are ‘n’ mutually exclusive, exhaustive and equally likely simple events in a trial, ‘m’ of them are favourable to the occurrence of an event A. Then probability or chance of occurrence of A equal to

af

P A =

Number of favourable cases Total number of all possible equally likely cases

af

P A =

m . n

Note: 1. Since 0 ≤ m ≤ n we have 0 ≤ or

m ≤ 1. n

af

0 ≤ P A ≤ 1.

2. When m = n, P (A) = 1 and when m = 0, P (A) = 0, i.e. when m = n, the event A is certain and when m = 0, the occurrence of event A is an absolute impossibility. 3. The probability of non-occurrence of A is denoted by

48

Basic Mathematics

di

P A =

Number of unfavourable cases . Total number of all possible cases

di

P A = =

n−m n

n m − n n

af P d A i = 1 − P a Af P a Af + P d A i = 1 .

=1− ∴ ⇒

m =1− P A n

4. The main disadvantage of mathematical method is that it fails when there are infinite number of possible outcomes and it cannot be applied to trials where the outcomes are not equally likely. ii Statistical or Empirical definition of probability: If a random experiment is repeated for an indefinitely large number of times under identical conditions, then the limiting value of the ratio of the number of times an event occur to the total number of trials is said to be the probability of occurrence of the event, provided the limit is a definite finite number. If T is the number of trials and event A occurs f times in ‘T’ trials, then the probability of occurrence

af

of event A is given by probability = P A = lim

T →∞

F f I. HTK

We use this method when the elementary events are not equally likely and the exhaustive number of cases in a trial is infinite. The limitation of this method is that in practice an identical experimental condition doesn’t exist while repeating a random experiment for a large number of times. Moreover the relative frequency i.e.

f may not attain a unique limiting value when T → ∞. T

iii Subjective probability or set theoretic approach: A set of points representing all possible elementary outcomes of a random experiment is called the sample space (S). The number of all possible sample points in the sample space S is represented by n (S). The definition of probability is based on the following assumptions. (i) Total number of elementary events in the sample space (S) is finite say N (ii) N elementary events of the experiment are equally likely.

af

P A =

Number of elementary events favourable to event A Total number of equally elementary events in S.

Probability

i.e.,

49

Number of sample points in A. a f nnaaASff = Total number of sample points in S.

P A =

Note: A pack of cards contain 52 cards, 26 red cards and 26 black cards. Among 26 black cards, 13 are calavar and 13 are spade. Among 26 red cards, 13 are Diamond and 13 are hearts. Each symbol contain A, Q, K, J, 2, 3, 4, 5, 6, 7, 8, 9, 10. (13 cards) So there are 4 A’s, 4 Q’s, ... 4-10’s. A is also known as Ace. Face cards are A, Q, K and J.

WORKED EXAMPLES: 1. If one card is drawn at random from a well shuffled pack of 52 cards. Then find the probability of each of the following. (a) Drawing an ace card, (b) Drawing a face card, (c) Drawing a diamond card, (d) Drawing either spade or hearts, (e) Not drawing an ace of hearts. Solution: (a) One card can be drawn out of 52 cards in 52c1 = 52 ways = n (S). One ace card can be drawn out of 4 ace cards in 4c1 = 4 ways = n (A). ∴Probability of drawing an ace card =

af af

n A Number of favourable cases = Total number of all possible equally likely cases n S

=

4 1 = . 52 13

(b) A face card can be drawn out of 12 face cards in 12c1 = 12 ways. ∴Number of favourable cases = 12 Total number of all possible equally likely cases = 52c1 = 52. ∴Probability of drawing a face card =

12 3 = . 52 13

(c) A diamond card can be drawn out of 13 diamond cards in ∴Probability of drawing a diamond card

=

13c 1

= 13 ways.

13 1 = . 52 4

(d) There are 13 spade and 13 hearts cards in a pack of cards. Either a spade or a heart can be drawn in 26c1 = 26 ways.

50

Basic Mathematics

∴Probability of drawing either a spade or a hearts card =

26 1 = . 52 2

(e) There is one ace of hearts.

1 . 52

∴

Probability of drawing an ace of hearts =

∴

Probability of not drawing an ace of hearts = 1 −

1 51 . = 52 52

2. Three balls are drawn at random from a bag containing 6 blue and 4 red balls. What is the probability that two balls are blue and one is red? Solution: The bag contains (6 + 4) = 10 balls. 3 balls can be drawn out of 10 balls in 10c3 =

10 = 120 ways. 10 − 3 ⋅ 3

∴Total number of cases = 120. Now 2 blue balls can be drawn out of 6 in 6c2 =

6 = 15 ways. 6−2⋅ 2

1 red ball can be drawn out of 4 in 4c1 = 4 ways. ∴2 blue balls and 1 red ball can be drawn in 15 × 4 = 60 ways. ∴Number of favourable cases for the event = 60. ∴Probability of drawing 2 blue balls and 1 red ball =

60 1 = .. 120 2

3. Three unbiased coins are tossed. What is the probability of obtaining (a) all heads (b) two heads (c) one head (d) atleast one head (e) atleast two heads (f) All tails. Solution: There are 23 = 8 mutually exclusive exhaustive and equally likely cases HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. (a) Probability of all heads =

(b) Probability of 2 heads =

(c) Probability of 1 head =

1 8

[There is only one HHH amongst 8 possiblities]

LM MN

HHT 3 → HTH 8 THH

LM MN

HTT 3 → THT 8 TTH

(d) Probability of atleast one head =

7 8

OP PQ

OP PQ

Probability

LM MM N

HHH 4 1 HHT (e) Probability of atleast 2 heads = = → HTH 8 2 THH (f) Probability of all tails =

51

OP PP Q

1 . 8

4. The marks obtained by 100 students are given below. Marks 0 − 10 11 − 20 21 − 30 31 − 40 41 − 50 51 − 60 61 − 70 71 − 80 81 − 90 91 − 100 No. of students 5 10 13 14 17 7 11 8 9 6

If a student is selected at random from the entire group of 100 students, find the probability that his marks (i) is under 40. (ii) above 50 (iii) either between 31 to 40 or 41-50. (i) Total number of students = 100. Number of students obtaining marks less than 40 = 5 + 10 + 13 + 14 = 42. ∴

Required probability =

42 21 = . 100 50

(ii) Number of students scoring above 50 = 7 + 11 + 8 + 9 + 6 = 41 ∴


41 100

(iii) Number of students obtaining marks between 31 to 40 = 14 and number of students obtaining marks between 41 to 50 = 17. ∴ Total number of students whose score is either between 31 to 40 or 41 to 50 = 14 + 17 = 31. ∴


31 100

5. If a pair of dice is thrown, find the probability that the sum of digits is neither 7 nor 11. Solution: A pair of dice is thrown. ∴ n (S) = 6 × 6 = 36. Let A be an event of getting the sum 7 and B be an event of getting the sum 11. Then ∴

A = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} n (A) = 6 B = {(5, 6) (6, 5)} n (B) = 2. Probability of getting 7 =

af af

6 n A = . 36 n S

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Basic Mathematics

Probability of getting 11 =

af af

2 n B = . 36 n S

Probability of getting either 7 or 11 = ∴

6 2 8 + = . 36 36 36

Probability of getting neither 7 nor 11 = 1 −

8 28 7 = = . 36 36 9

3.4 ADDITION RULE OF PROBABILITY: Sta tement: If A and B are 2 events, then probability that at least one of them occurs is given by Statement:

a

f af af a

f

P A∪ B = P A + P B − P A∩ B

Pr oof: Consider the Venn diagram. The shaded portion denotes A∪B, i.e., set of all outcomes where Proof: some of the outcomes are common to both A as well as B. So P (A∪B) is the probability of happening of atleast one of the events A and B. U A

B A ∩ B

Fig. 3.1

P (A) + P (B) is the sum of all the probabilities in A and all the probabilities in B. So the probability in A ∩ B has been added twice in P (A) as well as in P (B). So we must subtract P (A ∩ B) once from P (A) + P (B) to obtain probabilities in A∪B. ∴

a

f af af a

f

a

f af af a

f

P A∪ B = P A + P B − P A∩ B

Cor ollar y: If A and B are mutually exclusive, then P (A∪B) = P (A) + P (B) Corollar ollary: Proof: Pr oof: From addition rule we have P A∪ B = P A + P B − P A∩ B

Since A and B are mutually exclusive, A and B are disjoint sets. ∴ ⇒ ∴ ⇒

A∩B = φ

a f P a A ∪ Bf = P a Af + P a Bf − 0 P a A ∪ Bf = P a Af + P a Bf P A∩ B = 0

Probability

53

Note: 1. P (A∪B) means P (A or B) i.e., probability of happening of atleast one of the events A and B. P (A ∩ B) means P (A and B) i.e., probability of happening of both the events A and B. 2. P (A∪B∪C) is probability of happening of atleast one of the events A, B and C. It is given by addition rule as

a

f af af af a f − P a B ∩ C f − P aC ∩ Af + P a A ∩ B ∩ C f

P A∪ B∪C = P A + P B + P C − P A∩ B

This can be proved by writing the Venn diagram. U B

A

C

Fig. 3.2

WORKED EXAMPLES: 1. A ticket is drawn from a bag containing 25 tickets bearing number 1, 2, 3, ..., 24, 25. Find the probability of its bearing a number which is either even or a multiple of 3. Solution: The events ‘even number’ and ‘a multiple of 3’ are not mutually exclusive as there are some numbers which are even as well as multiples of 3. Ex: 6, 12, 24. ∴ P (an even number or a multiple of 3) = P (an even number) + P (a multiple of 3) − P (an even number and a multiple of 3)

ck

ph ma

fr ma

P 2, 4, 6, 8, ..., 24 + P 3, 6, 9, ..., 24 − P 6, 12, 24 =

fr

12 8 3 17 . + − = 25 25 25 25

[Since there are 12 even numbers from 1 to 25, 8 multiples of 3 and 3 numbers which are even as well as multiples of 3 from 1 to 25]. 2. What is the probability of getting either total of 7 or 11 when a pair of dice is tossed? Solution: Total outcomes when a pair of dice is tossed = 6 × 6 =36. The events ‘a total of 7’ and ‘a total of 11’ are mutually exclusive events. ∴ P (a total of 7 or 11) = P (a total of 7) + P (a total of 11).

54

Basic Mathematics

= P {(6, 1), (5, 2), (4, 3) (3, 4) (2, 5) (1, 6)} + P {(6, 5) (5, 6)}

=

6 2 8 2 + = = . 36 36 36 9

3. The probability that a contractor will get a plumbing contract is 2/3 and the probability that he will not get an electric contract is 5/9. If the probability of getting atleast one contract is 4/5, what is the probability that he will get both the contracts. Solution: Let A be an event that a contractor gets plumbing contract. B be an event that a contractor gets electrical contract. Then Given:

af

P A =

2 3

di

5 P B = . 9

af

di

P B = 1− P B = 1−

5 4 = . 9 9

a f 45 P a A ∩ Bf = ?

P A∪ B =

From addition rule,

a

f af af a 4 2 4 = + − P a A ∩ Bf 5 3 9

f

P A∪ B = P A + P B − P A∩ B

⇒

a

f

P A∩ B = =

a

2 4 4 + − 3 9 5

30 + 20 − 36 45

f

P A∩ B =

14 . 45

∴ Probability that a contractor will get both the contracts =

14 . 45

4. One card is drawn from a pack of 52 cards, what is the probability that the card drawn is neither red nor king. Solution: The event ‘card drawn is red’ and ‘card drawn is king’ is not mutually exclusive because there are two cards in the pack which are red as well as king. ∴ P (card drawn is red or king)

Probability

55

= P (card drawn is red) + P (card drawn is king) − P (card drawn is red and king) Since there are 26 red cards, 4 king cards and 2 cards which are red as well as king, P (Cards drawn is red or king),

= ∴

26 4 2 28 . + − = 52 52 52 52

P (card drawn is neither red not king) = 1 − P (card drawn is red or king)

=1−

28 52 − 28 24 = = . 52 52 52 =

6 . 13

5. A card is drawn at random from a well shuffled pack of 52 cards. What is the probability that it is a heart or a queen or black card. Solution: Let A be an event that ‘the card drawn is heart’. B be an event that ‘the card drawn is queen’ and C be an event that ‘the card drawn is black’. A, B and C are not mutually exclusive events. So

a

f af af af a f − P a B ∩ C f − P a C ∩ Af + P a A ∩ B ∩ C f .

P A∪ B∪C = P A + P B + P C − P A∩ B

There are 13 heart cards, 4 queen cards, 26 black cards, 1 heart queen card 2 queen black cards and No heart black card. No card which is heart, queen and black. ∴

a

f

P A∪ B∪C =

13 4 26 1 2 + + − − −0+0 52 52 52 52 52

a

f

P A∪ B∪C = = ∴

13 + 4 + 26 − 1 − 2 52

40 10 = . 52 13

P (Card drawn is heart or queen or black) =

10 . 13

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Basic Mathematics

This can be illustrated by the following Venn diagram.

12 other cards

Queen 12 1 Hearts 1 2 24 Black

Fig. 3.3

3.5 CONDITIONAL PROBABILITY: Let us consider the following example. A fair dice is thrown and the number that appeared is even. What is the probability that the number 4 has appeared? Since it is given that ‘the number appeared is even’. Possible outcomes are no longer {1, 2, 3, 4, 5, 6} but only {2, 4, 6}. Out of these 3 possible outcomes there is 1 outcome in favour of appearance of 4. ∴ Probability of appearance of 4, given that an even number has appeared =

1 . 3

Formally, if E is the event ‘The number that appeared is 4’ and F the event ‘Number that appeared is even’ then P (E/F) denotes the probability of E given that F has happened.

a f

1 P E F = . 3 The probability that an event B occurs subject to the condition that A has already occurred is called the conditional probability of occurrence of the event B. It is denoted by P (B/A).

∴

3.6 MULTIPLICATION RULE: Sta tement: The probability of the simultaneous occurrence of 2 events is the product of the probability Statement: that one of the events will occur and the conditional probability that the other event will occur given that the first event has occurred. If A and B are two events then P (A ∩ B) is the probability of their simultaneous occurrence and it is given by

a f af a f P a A ∩ Bf = P a Bf ⋅ P a A Bf P A∩ B = P A ⋅P B A

or

Pr oof: Let A be any event with sample points n (A), i.e., P (A) > 0. If n (S) is the total number of sample Proof: points in S and B = another event such that A and B are not disjoint sets. (i.e., A∩B ≠ φ). Let the sample

a f nnaaASff

points in A∩B be n (A∩B). Then from definition P A =

Probability

57

f n anAa∩SfBf

a

P A∩ B =

S

A

B

A ∩ B

Fig. 3.4

Now Let P (B/A) = Conditional probability of event B given A has occurred = =

a

f a

f af a f af af af af P a A ∩ Bf P a B Af = P a Af P a A ∩ Bf = P a Af ⋅ P a B Af

n A∩ B n A∩ B n S P A∩ B = = n A n A n S P A

i.e., ⇒ Similarly we can establish

a

f af a f

P A∩ B = P B ⋅P A B Alternative Proof:

If A and B are 2 mutually dependent events then conditional probability of B when A has occurred is proportional to P (A∩B).

a f a

f

P B A ∝ P A∩ B

i.e,

a f

a

f

P B A = K ⋅ P A ∩ B where K is the constant of proportionality To find K, Conditional probability of A when A has occurred is 1 ∴

a f

a

P A A = KP A ∩ A

af

1 = KP A

⇒

K=

1 P A

af

f

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Basic Mathematics

K=

Substituting

a f

a

1 in P B A = KP A ∩ B P A

af

f

We get

a f P a1Af ⋅ P a A ∩ Bf P a A ∩ Bf = P a Af ⋅ P a B Af

P BA = ⇒

Cor ollar y: If A and B are independent events then P (A ∩ B) = P (A) ⋅ P (B). Corollar ollary: oof: If A and B are independent events then P (B/A) = P (B) and P (A/B) = P (A). Proof: Pr ∴ By multiplication theorem,

a

f af a f

P A∩ B = P A ⋅P B A Substituting P (B/A) = P (B) we get

a

f af af

P A∩ B = P A ⋅P B

Hence proved. Note: 1. The multiplication theorem can be extended. For 3 dependent events A, B and C

a

f af a f a

f

P A∩ B∩C = P A ⋅P B A ⋅P C A∩ B 2. If there are n independent events A1, A2, .... An,

b

g b g b g b g b g

P A1 ∩ A2 ... ∩ An = P A1 ⋅ P A2 ⋅ P A3 ... P An .

WORKED EXAMPLES: 1. A pair of dice is thrown and sum of the numbers on the two dice comes to be 7. What is the probability that the number 4 has come on one of the dice? Solution: Let the events A and B be such that Event B: Sum of numbers on the two dice is 7. Event A: The number 4 has come. Total outcomes when a pair of dice is thrown = 36 = n (S)

a f ma f a f a f a f a f a fr 6 P a Bf = . 36

P B = P 6, 1 , 5, 2 , 4, 3 , 3, 4 , 2, 5 , 1, 6

To get P (A∩B) select the outcomes favourable to A from the outcomes that are favourable to B. ∴

a

f ma f a fr

P A ∩ B = 4, 3 , 3, 4 = From Multiplication theorem,

a

2 . 36

f af a f

P A∩ B = P A ⋅P A B

Probability

59

a f P aPAa∩AfBf

P A B =

⇒

2 2 1 P A B = 36 = = . 6 6 3 36

a f

∴ Probability that the number 4 has come on one dice given that sum of numbers on 2 dice is 7 =

1 . 3

2. Two cards are drawn from a pack of 52 cards with replacement (i.e., the second card is drawn after replacing the first card in the pack). Find the probability that (a) Both are ace, (b) First card is jack and second card is king, (c) One is king and other is queen. Solution: (a) In a pack of 52 cards, there are 4 ace cards. ∴ P (both cards are ace)

=

4 4 1 . × = 52 52 169

(b) P (First card is jack and second card is king)

=

4 4 1 . × = 52 52 169

[Q There are 4 jack and 4 king cards in a pack of 52 cards] (c) P [One card is king and other is queen] = P [First is king and 2nd is queen] ∪ P [First is queen and 2nd is king]

=

4 4 4 4 × + × [Using addition theorem P (A∪B) = P (A) + P (B)] 52 52 52 52 =

1 1 2 . + = 169 169 169

3. Two cards are drawn without replacement from a pack of 52 cards. What is the probability that (i) both are queen. (ii) both are diamond cards. (iii) one is king and the other is ace. Solution: We are taking 2 cards from 52 cards. This can be done in 52c2 ways. There are 4 queen cards from which we require 2 queen cards. This can be done in 4c2 ways. 4

∴

Required Probability =

=

c2 4×3 = c2 52 × 51

52

1 . 221

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Basic Mathematics

OR There are 4 queen cards in a pack of 52 cards. ∴ Probability of drawing first queen card =

4 52

Since the card drawn is not replaced, we are left with 51 cards and 3 queen cards. ∴ Probability of drawing 2nd queen card = ∴ Probability of both queen cards = (ii) Required probability =

13 52

3 51

4 3 1 . × = 52 51 221

c2 c2 13 × 12 1 = 2 ×1 = . 52 × 51 17 2 ×1

(Since there are 13 diamond cards).

(iii) There are 4 favourable choices to take out a king and 4 favourable choices to take out an ace. ∴ Number of favourable cases = 4c1 × 4c1 In total, there are 52 cards out of which any 2 cards can be taken. ∴ Required probability =

4

c1 × 4 c1 52 c2 =

4×4×2 8 . = 52 × 51 663

4. A lot contains 10 items of which 3 are defective. 3 items are chosen from the lot at random one after another without replacement. Find the probability that all the 3 are defective. Solution: Let A, B and C be the events of drawing defective items in the first, second and third drawing respectively. Hence Probability of all the three items being defective is given by

a

f af a f a

f

P A∩ B∩C = P A ⋅P B A ⋅P C A∩ B =

3 2 1 × × 10 9 8

=

1 . 120 OR

3 defective items can be picked from 3 defective items in 3c3 ways.

Probability

3 items can be picked from 10 items in

10c

3

ways. 3

∴

61


c3 c3

10

1 6 = 10 × 9 × 8 10 × 9 × 8 3 × 2 ×1 =

1 . 120

5. Anil and Bharath appear in an interview for 2 vacancies. The probability of their selection being

1 1 and respectively. 7 5 Find the probability that (i) both will be selected (ii) only one is selected (iii) none will be selected (iv) atleast one of them will be selected. Solution: Let A: Anil be selected. B: Bharath be selected.

af

P A =

Given

P (both will be selected)

a

af

1 1 and P B = 7 5

f a = P a Af ⋅ P a Bf

= P A and B = P A ∩ B

=

f

1 1 1 × = . 7 5 35

(ii) P (Only one will be selected)

= P A and B or A and B

d i d i = P a Af ⋅ P d B i + P d A i ⋅ P a Bf 1 1 1 1 = ⋅ F 1 − I + F1 − I ⋅ 7 H 5K H 7K 5 = P A∩ B ∪ A ∩ B

=

1 4 6 1 ⋅ + ⋅ 7 5 7 5

=

4 6 10 2 + = = . 35 35 35 7

[Q A and B are independent events]

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Basic Mathematics

(iii) P (none will be selected)

d i = P d A ∩ B i = P dA i ⋅ PdB i 1 1 = F1 − I F1 − I H 7 K H 5K = P A and B

=

6 4 24 × = . 7 5 35

(iv) P (at least one of them will be selected) = 1 − P (none of them will be selected)

1−

24 11 = . 35 35

REMEMBER:

a f nnaaASff P d A i = 1 − P a Af

af

• P A = •

0 ≤ P A ≤ 1.

• P (A∪B) means P (A or B) and P (A∩B) means P (A and B) • Addition rule: P (A∪B) = P (A) + P (B) − P (A∩B) • For mutually exclusive cases P (A∪B) = P(A) + P(B)

a f af af af a f a f a f a f • P a A ∩ Bf = P a Af ⋅ P a B Af • P a A ∩ Bf = P a Bf ⋅ P a A Bf • P a A ∩ Bf = P a Af ⋅ P a Bf if A and B are independent events. • P a A ∩ B ∩ C f = P a Af ⋅ P a B Af ⋅ P aC A ∩ Bf • If A , A , A , ... A are independent events P b A ∩ A ∩ ... ∩ A g = P b A g P b A g ... P b A g • P A∪ B∪C = P A + P B + P C − P A∩ B − P B∩C − P C∩ A + P A∩ B∩C

1

2

3

n

1

2

n

1

2

n

EXERCISE 1. A bag contains 100 tickets each bearing a distinct number from 1 to 100. A ticket is drawn from the bag. Find the probability that

Probability

63

(a) the ticket bears an odd number. (b) the ticket bears a number divisible by 3. (c) the ticket bears a number divisible by 3 or 5. 2. A pair of dice is thrown. Find the probability that the sum will be (a) equal to 4 (b) less than 4 (c) greater than 4. 3. A bag contains 4 white, 5 red and 6 green balls. 3 balls are drawn at random. What is the probability that (a) All are green (b) All are white. 4. A card is drawn from a pack of playing cards. What is the probability of drawing (a) black card (b) king card (c) diamond card. 5. A box has 3 silver and 2 gold coins. 2 coins are drawn at random. Find the probability that (a) both the coins are silver ones. (b) both are gold coins (c) one of them is a silver coin. 6. A problem in statistics is given to 3 students A, B and C. Their probabilities in solving it are

1 1 1 , and respectively. What is the probability that the problem would be solved? 8 6 4 7. The following table gives a distribution of wages of 1000 workers.

a f

Wages Rs.

No. of workers

120 − 140 141 − 160 161 − 180 181 − 200 201 − 220 221 − 240 241 − 260 9

118

478

200

142

35

18

An individual is selected at random from the above group. What is the probability that his wages are (a) under Rs. 160 (b) above Rs. 200 (c) between Rs. 160 and 200. 8. One card is drawn from a pack of 52 cards. What is the probability that the card drawn is (a) either red or king (b) either king or queen 9. A ticket is drawn from a bag containing tickets bearing numbers 1 to 25. Find the probability that the number is either even or a multiple of 3. 10. A coin and a die are thrown. What is the probability of getting a head or an even number. 11. A box contains 50 bolts and 150 nuts. Half of the bolts and half of the nuts are rusted. If one item is chosen at random, what is the probability that it is rusted or a bolt? 12. A die is rolled. If the outcome is an odd number, what is the probability that it is prime. 13. A pair of dice is rolled. If the sum on the 2 dice is 9. Find the probability that one of the dice showed 3.

af

14. If A and B are 2 events in a sample space S such that P A =

d

i

d i

a

f

1 5 3 , P B = , P A∪ B = . 2 8 4

Find (a) P (A∩B) (b) P A ∩ B . 15. 3 children are randomly selected from a class. What is the probability that (a) all 3 were born on Monday (b) 2 were born or Friday and the other on Tuesday (c) none were born on Wednesday.

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Basic Mathematics

ANSWERS 1. (a)

1 2

(b)

33 100

(c)

47 100

2. (a)

3 36

(b)

3 36

(c)

30 36

3. (a)

4 91

(b)

4 455

4. (a)

1 2

(b)

1 13

(c)

1 4

5. (a)

3 10

(b)

1 10

(c)

3 5

7. (a)

127 1000

(b)

195 1000

(c)

678 1000

8. (a)

28 52

(b)

2 13 5 8

6.

290 64

9.

17 25

10.

3 4

11.

12.

2 3

13.

1 2

14. (a)

1 8

(b)

1 4

4 Binomial Theorem 4.1 INTRODUCTION: We know

a x + af a x + af a x + af

1

= x+a

2

= x 2 + 2ax + a 2

3

= x 3 + a 3 + 3ax x + a

a

f

= x 3 + 3 x 2 a + 3xa 2 + a 3 . The general expansion for (x + a)n where n is positive integer is given by Sir Isaac Newton and is known as Binomial Theorem.

4.2 STATEMENT OF BINOMIAL THEOREM: If x and a are real numbers and n is any positive integer then

a x + af

n

= x n + n c1 x n −1a + n c2 x n − 2 a 2 + n c3 x n −3 ⋅ a 3 + ... n cn x n − n a n . = x n + nx n −1 ⋅ a +

a f

n

Since,

n

n

c1 =

c2 =

a fa

f

n n − 1 n −2 2 n n − 1 n − 2 n −3 3 x ⋅ a + ... + a n . ⋅x ⋅a + 2! 3! cr =

n n − r⋅ r

n n − 1⋅ 1

a f

=

n⋅ n −1 n −1

a fa

= n.

n n −1 n n n −1 n − 2 , c3 = 2! 3!

f

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Basic Mathematics

and so on ... ncn = 1. By taking 1 = C0 , nc1 = C1, n c2 = C2 and so on ... n cn = 1 = Cn . We can write,

a x + af

n

= C0 x n + C1 x n −1a 2 + C2 x n −3 a 3 + ... Cn a n .

Note: (1)

a x + af a x + af a x + af

1

= x + a : Power : 1, Number of terms : 2

2

= x 2 + 2ax + a 2 : Power : 2, Number of terms 3.

3

= x 3 + 3 x 2 a + 3 xa 2 + a 3 . Power : 3, No. of terms : 4.

The number of terms in the expansion of (x + a)n is n + 1. (2) In the expansion of (x + a)n, from left to right the power of x decreases by 1 from one term to the next term and the power of a increases by 1. (3) In any term, the power index of x plus the power index of a = n. (4) The co-efficients C0, C1, C2 ... Cn are called binomial co-efficients. (5) The general term in the expansion is

n

− cr x n −r ⋅ a r which is (r + 1)th term. Hence Tr +1 = n cr x n r ⋅ a r .

cr = n cn − r , the binomial co-efficients of first and last term are equal: Second and penultimate term are equal and so on. (7) If n is even then the number of terms in (x + a)n is (n + 1) which is odd. So there will be only one

(6) Since

n

middle term and it is TF n

I . If n is odd then the number of terms in (x + a) H2 K

n

+1

even. So there will be two middle terms. They are TF n +1 I and TF n +1

H2K

H

2

is (n + 1) which is

I K

+1

WORKED EXAMPLES:

F 2I 1. Expand: G x + J H yK FG x + 2 IJ H yK 5

2

2

5

= x + c1 ⋅ x 5

5

4

F2I ⋅G J Hy K 2

F 2I + c ⋅x G J Hy K F 2I + c x G J + Hy K 2

1

5

3

2

2

3

5

2

3

2

5

c4 x 1

FG 2 IJ Hy K 2

4

+ 5 c5 x 0

FG 2 IJ Hy K 2

5

Binomial Theorem

67

FG IJ H K

2 5× 4 3 4 5× 4×3 2 8 5× 4×3×2 16 32 + ⋅x ⋅ 4 + ⋅x ⋅ 6 + ⋅ x ⋅ 8 + 10 2 3! 4! y2 y y y y

= x 5 + 5x 4 ⋅

= x5 +

F a bI 2. Expand H + K . b a F a + bI = F aI H b aK H bK

10 x 4 x 3 80 x 2 80 x 32 40 + + 6 + 8 + 10 . y2 y4 y y y

4

F aI F bI + 4 × 3 F aI F bI + 4 × 3 × 2 F aI F bI + F bI H b K H a K 2 H b K H a K 3! H b K H a K H a K a a b b = F I + 4F I + 6 + 4 ⋅F I + F I . H bK H bK H aK H aK F 1I F F 1II 3. Expand G x − J = G x + G − J J H yK H H yKK F 1 I 4 ⋅ 3 d x i FG − 1 IJ + 4 ⋅ 3 ⋅ 2 d x i FG − 1 IJ + FG − 1 IJ = dx i + 4 d x i G − J + H y K 2 H y K 3! H y K H y K 4

4

3

+4

2

4

2

2

3

4

4

4

4

2

2

2 4

2 3

2 2

= x8 −

d

2

4. Simplify: 3 + 2

i + d3 − 2 i 5

2

3

4

2

4 x6 6x 4 4 x2 1 + 2 − 3 + 4. y y y y

5

Solution: From Binomial theorem we have

a x + af

5

= x 5 + 5x 4 ⋅ a +

a x + af

5⋅4 3 2 5 × 4 × 3 2 3 5 × 4 × 3× 2 4 xa + a 5 . ⋅x a + ⋅x ⋅a + 2 3! 4!

5

= x 5 + 5 x 4 a + 10 x 3 a 2 + 10 x 2 a 3 + 5 xa 4 + a 5 .

...(1)

5

= x 5 − 5 x 4 a + 10 x 3 a 2 − 10 x 2 a 3 + 5xa 4 − a 5 .

...(2)

Changing a to −a we get

a x − af

Adding (1) and (2) we get

a x + af + a x − af 5

5

= 2 x 5 + 20 x 3 a 2 + 10 xa 4 .

= 2 x x 4 + 10 x 2 a 2 + 5a 4

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Basic Mathematics

Putting x = 3 and a = 2 we get

d3 + 2 i + d3 − 2 i 5

5

LM N

d 2i = 6 81 + 180 + 5 a 4f = 2 ⋅ 3 3 4 + 10 ⋅ 32

2

+5

d 2 i OPQ 4

= 6 261 + 20 = 1566 + 120 = 1686.

d

5. Simplify: 1 + 2

i − d1 − 2 i 4

4

We have from Binomial theorem

a x + af

4

= x 4 + 4 x 3a +

a x + af Replacing a by −a.

4

a x − af

Subtracting (2) from (1)

4

4×3 2 2 4×3×2 3 x a + x a + a4 2 3!

= x 4 + 4 x 3a + 6 x 2 a 2 + 4 x a3 + a 4

...(1)

= x 4 − 4 x 3a + 6 x 2 a2 − 4 x a3 + a 4

...(2)

a x + af − a x − af 4

Putting x = 1 and a =

4

= 8 x 3 a + 8 xa 3

2 we get

d1 + 2 i − d1 − 2 i 4

4

= 8 ⋅ 13 ⋅ 2 + 8 ⋅ 1 ⋅

d 2i

3

= 8 2 + 8⋅2 2

= 8 2 + 16 2 = 24 2.

Fa I 6. Find the middle term in the expansion of H + bxK . x 8

Solution: The expansion has 8 + 1 = 9 terms. So T8 2

+1

is the middle terms i.e., T5 is the middle term.

We have

Tr +1 = n cr ⋅ x n − r ⋅ a r . Comparing

F a + bxI Hx K

8

with (x + a)n,

Binomial Theorem

x=

We get

a x

a = bx n=8

r +1= 5 r=4 Substituting in

Tr +1 = n cr ⋅ x n − r ⋅ a r

F a I ⋅ abx f H xK a c F I ⋅ a bx f H xK 8− 4

T5 = 8 c4 =8

4

5

4

= 8 c4

a4 5 5 ⋅b ⋅ x x4

= 8 c4

a4b5 x5 x4

=

5

8 8−4⋅ 4

a 4b5 ⋅ x

=

8 × 7 × 6 × 5 × 4/ 4 5 ⋅a b x 4 4/

=

8×7×6×5 × a4b5 x 4 × 3× 2 ×1

= 2 × 35a 4 b 5 x

= 70a 4 b 5 x.

F H

7. Find the middle term in the expansion of 2 x +

y x

I K

11

.

Solution. The expansion has 11 + 1 = 12 terms. So there are 2 middle terms. Tn +1 and Tn +1 2

2

+1

are middle terms i.e., T12 and T12 2

2

+1

= T6 and T7 are middle terms.

69

70

Basic Mathematics

F H

Comparing 2 x + For

y x

I K

11

a

with x + a

f

n

we get x = 2x, a =

y and n = 11. x

T6 , r + 1 = 6 ⇒ r = 5.

Tn +1 = n cr x n − r ⋅ a r T6 =

T6 =

y c ⋅ a2 x f ⋅F I H xK y c ⋅ a2 x f ⋅ F I H xK 11− 5

11

5

5

= 11

=

formula

5

6

5

11 y5 ⋅ 26 ⋅ x6 ⋅ 5 11 − 5 ⋅ 5 x 11 × 10 × 9 × 8 × 7 6 ⋅ 2 ⋅ xy 5 5 × 4 × 3 × 2 ×1

T6 = 29568 ⋅ xy 5 . For,

T7 , r + 1 = 7 ⇒ r = 6

Tr +1 = n cr x n − r ⋅ a r .

a f

T7 = 11c6 ⋅ 2 x

⋅

F yI H xK

6

11 y6 ⋅ 25 ⋅ x 5 ⋅ 6 11 − 6 ⋅ 6 x

T7 = T7 =

11− 6

11 × 10 × 9 × 8 × 7 5 y 6 ⋅2 ⋅ . 5 × 4 × 3 × 2 ×1 x

T7 = 14784 ⋅

y6 . x

Fx − 1 I . H2 x K 9

8. Find the middle term in the expansion of

2

The expansion has 9 + 1 = 10 term. So there are 2 middle terms Tn +1 and Tn +1 2

2

+1

are middle terms.

Binomial Theorem

T10 and T10 2

2

+1

are middle terms.

T5 and T6 are middle terms.

Fx − 1 I Comparing H2 x K

9

with (x + a)n we get

2

x=

x 1 , a = − 2 , n = 9. 2 x

T5, r + 1 = 5 ⇒ r = 4.

For

Tr +1 = n cr ⋅ x n − r ⋅ a r . T5 = 9 c 4 ⋅

F xI ⋅F− 1 I H 2K H x K 9− 4

4

.

2

T5 =

9 × 8 × 7 × 6 x5 1 ⋅ ⋅ 4 × 3 × 2 × 1 25 x8

T5 =

126 x 5 32 × x 8

T5 =

63 . 16 x 3

For T6, r + 1 = 6 ⇒ r = 5.

T6 = 9 c5 ⋅

F xI ⋅ F− 1 I H 2K H x K

= −126 ⋅ =−

9 −5

2

x4 1 ⋅ 2 4 x 10

126 −63 . = x 6 8x 6

F H

2 9. Find the middle term in the expansion of 3 x −

The expansion has 10 + 1 = 11 terms. So T10 2

+1

= T6 is the middle term.

F H

Comparing 3 x 2 −

y 3

I K

5

10


y 3

I K

10

.

71

72

Basic Mathematics

x = 3x 2 a=−

y 3

n = 10 r +1= 6 r=5

Tr +1 = n cr x n − r ⋅ a r .

d i

T6 = 10 c5 ⋅ 3x 2

10 − 5

F yI H 3K

⋅ −

5

T6 =

− 10 5 10 y 5 ⋅3 ⋅ x ⋅ 5 5⋅ 5 3

T6 =

−10 × 9 × 8 × 7 × 6 10 5 ⋅x ⋅y 5 × 4 × 3× 2 ×1

T6 = −252 ⋅ x 10 ⋅ y 5 . 10. Find the middle term in the expansion of

F H

x−

2 x

I K

10

.

The expansion has 10 + 1 = 11 terms. So T10 2

+1

is the middle term

T6 is the middle term. Comparing

LM N

x−

2 x

OP Q

10


n = 10, x=

x

a=−

2 x

r+1=6 r=5

aformulaf 2 c d xi ⋅F− I H xK

Tr +1 = n cr ⋅ x n − r ⋅ a r T6 = 10

10 − 5

5

5

Binomial Theorem

T6 = c5 10

⋅

a−2f

5

x5

10

=−

=−

5 ⋅x2

10 − 5 ⋅ 5

5

⋅x2

−5

⋅ 25

10 × 9 × 8 × 7 × 6 −5 2 5 ⋅x ⋅2 5 × 4 × 3× 2 ×1

= −252 x − 5 2 ⋅ 2 5 = −8064 x − 5 2

F H

11. Find the co-efficient of x2 in the expansion of 3 x +

F H

Comparing 3 x +

1 2x

I K

8

1 2x

with (x + a)n we get x = 3x, a =

I. K 8

1 and n = 8. 2x

Tr +1 = n cr x n − r ⋅ a r formula

a f ⋅ FH 21x IK 1 c ⋅ a3 x f ⋅ F I H 2x K

Tr +1 = 8 cr 3 x Tr +1 = 8

8− r

r

8−r

r

r

= 8 cr ⋅ 38− r ⋅ x 8− r ⋅

1 2 ⋅ xr r

= 8 cr ⋅ 38 −r ⋅ 2 − r ⋅ x 8 − r − r Tr +1 = 8cr ⋅ 38 − r ⋅ 2 − r ⋅ x 8 −2 r . To get the co-efficient of x2, equating power index of x to 2. 8 − 2r = 2 8 − 2 = 2r

6 = 2r r = 3. Substituting r = 3 in (1),

T3+1 = 8 c3 ⋅ 38−3 ⋅ 2 −3 ⋅ x 8 −2 a 3f

73

74

Basic Mathematics

T4 =

8 ⋅ 35 ⋅ 2 − 3 ⋅ x 2 8 − 3⋅ 3

T4 =

8 × 7 × 6 35 2 ⋅ ⋅x 3 × 2 23

T4 = 56 × ∴

35 × x2. 3 2

Co-efficient of x2 is 35 × 7.

Fx H

12. Find the co-efficient of x25 in the expansion of

F H

4 Solution. Comparing x −

1 x3

I K

4

1 − 3 x

I K

15

.

15

with (x + a)n we get x = x4, a = −

1 and n = 15. x3

Tr +1 = n cr x n − r ⋅ a r formula

d i

Tr +1 = 15 cr ⋅ x 4

15 − r

F H

⋅ −

1 x3

a f ⋅ a −1f ⋅ a −1f

I K

r

Tr +1 = 15 cr ⋅ x 60 − 4 r ⋅ −1 2 ⋅ x −3r = 15cr ⋅ x 60 − 4 r −3r Tr +1 = 15 cr ⋅ x 60 − 7r

r

r

To get the coefficient of x25, equating power index of x to 25. 60 − 7r = 25 60 − 25 = 7r 35 = 7r

⇒

r = 5.

Substituting r = 5 in (1)

a f

T5+1 = 15 c5 ⋅ x 25 ⋅ −1 5

a f

T6 = −15 c5 ⋅ x 25 3 −1 5 = −1 ∴

Co-efficient of x25 = −15c5.

F H

13. Find the co-efficient of x60 in 2 x 2 −

3 x3

I K

25

.

...(1)

Binomial Theorem

F H

Comparing 2 x 2 −

3 x3

I K

25

75

3 and n = 25. x3

with (x + a)n we get x = 2x2, a = −

Tr +1 = n cr ⋅ x n −r ⋅ a r formula

d i

Tr +1 = 25 cr ⋅ 2 x 2

25− r

F H

I K ⋅ a −3f ⋅ x ⋅ a −3f ⋅ a −3f

⋅ −

= 25 cr ⋅ 2 25− r ⋅ x 50 −2 r

r

3 x3

r

= 25 cr ⋅ 2 25− r ⋅ x 50 −2 r −3r = 25 cr ⋅ 2 25− r ⋅ x 50 −5r

−3 r

r

r

To get the co-efficient of x60, equating power index of x to 60. i.e.

50 − 5r = 60

⇒

5r = 50 − 60 5r = −10 r = −2.

F H

Since r is negative, there is no x60 term in the expansion of 2 x 2 −

F 3x 14. Find the constant term in the expansion of G H2 F 3x − 1 I with a x + af . Comparing G H 2 3x JK 2

We get x =

2

1 − 3x

3 x3

I K

25

.

I. JK 9

9

n

3x 2 1 , a = − , n = 9. 2 3x Tr +1 = n cr x n − r ⋅ a r formula Tr +1 = 9 cr

= 9 cr

F 3x I ⋅ F − 1 I GH 2 JK H 3x K 3 ⋅x a−1f ⋅ 2

9−r

9 −r

r

18 − 2 r

r

29−r

3r ⋅ x r

a f

= 9 cr ⋅ 39 −r − r ⋅ x 18 −2 r − r ⋅ −1 r ⋅ 2 −9 + r

...(1)

76

Basic Mathematics

To get the constant term, equating power index of x to zero ∴

18 − 2 r − r = 0 18 − 3r = 0

⇒

3r = 18 r = 6.

Substituting r = 6 in (1)

a f ⋅ a −1f

T6 +1 = 9 c6 39 − 6 −6 ⋅ x 0 ⋅ −1 6 ⋅ 2 −9 + 6 9 ⋅ 3 −3 9−6⋅ 6

=

6

⋅ 2 −3

9×8×7 1 1 7 ⋅ ⋅ = . 3 × 2 × 1 27 8 18

F 15. Find the constant term in the expansion of G 2 H F 1 I Comparing G 2 x − J with a x + af . x xK H

x−

1 x

IJ xK

23

.

23

n

We get

x = 2 x, a = −

1 x x

and n = 23.

Tr +1 = n cr x n − r ⋅ a r Tr +1 =

23

F 1 IJ c ⋅ d2 x i ⋅G− H x xK 23 − r

r

r

= 23 cr ⋅ 2 23− r ⋅ x = 23 cr ⋅ 2 23− r ⋅ x = cr ⋅ 2 23

23− r

23 − r 2

⋅ x −r ⋅ x

23− r r −r − 2 2

23 − r − 2 r − r 2 ⋅x

= 23 cr ⋅ 2 23− r ⋅ x

23 − 4 r 2

−

a−1f

a f

⋅ −1 r

a f

⋅ −1 r

a f

⋅ −1 r .

To get the constant term equating power index of x to zero. 23 − 4 r = 0

r 2

r

Binomial Theorem

⇒

77

4r = 23

r=

23 . 4

Since r is a fraction there is no term independent of x or there is no constant term.

F x − 2 IJ 16. Prove that the constant term in the expansion of G H2 xK FG x − 2 IJ with a x + af 2 x K Pr oof: Comparing H Proof:

10

is

2

45 . 64

10

n

2

x=

We get

x 2 , a = − 2 and n = 10. 2 x

Tr +1 = n cr x n − r ⋅ a r formula Tr +1 =

10

F x IJ c ⋅G H2K

10 − r

⋅

r

= cr ⋅ x 10

Tr +1 = 10 cr ⋅ x

5−

r 2

F −2 I Hx K

r

2

a f

⋅ 2 −10 + r ⋅ −2 r ⋅ x −2 r

r 5− − 2 r 2

a f

⋅ 2 −10 + r ⋅ −2

To get the constant term equating power index of x to zero.

5−

r − 2r = 0 2

10 − r − 4r =0 2 10 − 5r = 0

⇒

5r = 10 ⇒ r = 2.

Substituting r = 2 in (1) we get

a f

T2 +1 = 10 c2 ⋅ x 0 ⋅ 2 −10 + 2 ⋅ −1 2 ⋅ 2 2 =

10 × 9 −8 ⋅ 2 ⋅1 ⋅ 4 2 ×1

=

10 × 9 × 4 45 = . 64 2 × 28

r

...(1)

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Basic Mathematics

F x − 2 IJ Hence the constant term in the expansion of G H2 xK

10

is

2

45 . 64

17. Find the value of (0.99)5 correct to 4 decimal places. Solution: We know 0.99 = 1 − 0.01

a0.99f = a1 − 0.01f

∴

5

5

We have from Binomial theorem,

a x + af

5

= x 5 + 5x 4 ⋅ a +

5×4 3 2 5× 4×3 2 3 5× 4×3×2 x a + ⋅x ⋅a + ⋅ xa 4 + a 5 2 3! 4!

Replacing a by −a.

a x − af

= x 5 − 5x 4 a + 10 x 3 a 2 − 10 x 2 a 3 + 5 xa 4 − a 5 .

5

Taking x = 1 and a = 0.01, we get

a1 − 0.01f = 1 − 5a1f a0.01f + 10 a1f a0.01f −10 a1f a0.01f + 5 a1fa 0.01f − d0.01 i. 5

4

5

2

3

3

4

2

5

= 1 − 0.05 + 0.001 − 0.00001 + 5 × 10 −8 − 1 × 10 −10

a

≈ 0.9414801.

f

≈ 0.9415 Correct to 4 decimal places . 18. Prove that sum of Binomial co-efficients of order n = 2n. Also prove the sum of odd binomial coefficients = sum of even Binomial co-efficients = 2n − 1. Proof: Pr oof: We have from Binomial theorem.

a x + af

n

= C0 x n + C1 x n −1 ⋅ a + C2 x n − 2 a 2 + ... + Cn a n

...(1)

To get the sum of binomial co-efficient, Taking x = a = 1 we get

a1 + 1f

n

= C0 ⋅ 1n + C11n −1 ⋅ 1 + C2 ⋅ 1n − 2 ⋅ 1 + ... + Cn . 2 n = C0 + C1 + C2 + ... + Cn . C0 + C1 + C2 + ... + Cn = 2 n .

i.e.,

Now, Taking x = 1 and a = − 1 in (1)

a1 − 1f

n

a f

a f

= C0 1n + C11n −1 . −1 + C2 1n −2 −1 2 + ... + C n

...(2)

Binomial Theorem

0 = C0 − C1 + C2 − C3 − ... + Cn

a−1f

3

1

a f

79

...(3)

a f

= −1, −1 2 = 1, −1 3 = −1 & so on.

Adding (2) and (3) we get

2 n = 2C0 + 2C2 + 2C4 +... 2 n = 2 C0 + C2 +... 2n = C0 + C2 + C4 +... 2 C0 + C2 + C4 ... = 2 n −1

⇒

Hence sum of even binomial co-efficients = 2n − 1. Subtracting (3) from (2) we get

2 n = 2C1 + 2C3 + 2C5 + ... 2n = C1 + C3 + C5 +... 2 C1 + C3 + C5 + ... = 2 n −1 .

⇒

Hence sum of odd binomial co-efficients = 2n − 1. 19. The 1st, 3rd and 5th term in the expansion of (a + b)n are respectively 32, 240 and 90. Find a, b and n. Given: In the expansion of (a + b)n, T1 = 32 , T3 = 240 and T5 = 90.

We know

a

Comparing a + b

f

n

a

Tr +1 = n cr x n − r ⋅ a r

with x + a

f

n

formula

we get x = a, a = b and n = n.

For 1st term r + 1 = 1 ⇒ r = 0. For 3rd term r + 1 = 3 ⇒ r = 2. For 5th term r + 1 = 5 ⇒ r = 4.

T1 = n c0 a n − 0 ⋅ b 0 = 32. 1 ⋅ a n = 32 a n = 32

T3 = n c2 ⋅ a n − 2 ⋅ b 2 = 240

...(1)

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Basic Mathematics

a f

n n − 1 n− 2 2 ⋅a ⋅ b = 240 2

a f

n n − 1 ⋅ a n −2 ⋅ b 2 = 480

...(2)

T5 = n c4 ⋅ a n − 4 ⋅ b 4 = 90

a fa

fa f

n n −1 n − 2 n − 3 ⋅ a n − 4 ⋅ b 4 = 90 4!

a fa fa f

n n − 1 n − 2 n − 3 ⋅ a n − 4 ⋅ b 4 = 2160 a n = 32.

From (1) Since n is a +ve integer. By inspection, ∴ Substituting

...(3)

25 = 32. a = 2 and n = 5. a = 2 and n = 5 in (2).

a f

5 5 − 1 ⋅ 2 5− 2 ⋅ b 2 = 480 5 × 4 × 2 3 ⋅ b 2 = 480 b2 =

480 20 × 8

b2 = 3 ⇒

b = 3.

Verification: By

a fa fa f

n n − 1 n − 2 n − 3 a n − 4 ⋅ b 4 = 2160 5 × 4 × 3 × 2 × 2 5− 4 ⋅

d 3i

4

= 2160

120 × 2 × 9 = 2160 2160 = 2160.

Hence a = 2, n = 5 and b =

3.

20. Using Binomial theorem prove that 6n − 5n always leaves the remainder 1 when divided by 25. We know, 6n = (1 + 5)n We have from Binomial theorem.

Binomial Theorem

a x + af a1 + 5f

n

n

= x n + nx n −1 ⋅ a +

=1+ n⋅5 +

6 n = 1 + 5n + 6 n − 5n = 1 +

a f

n n −1 ⋅ x n − 2 ⋅ a 2 + ... + a n . 2!

a f

a fa

f

n n −1 2 n n −1 n − 2 3 ⋅5 + 5 + ... + 5 n . 2 3!

a f

n n −1 2 ⋅ 5 + ... + 5n . 2

a f

a fa

f

n n −1 2 n n −1 n − 2 ⋅ 5 3 + ... + 5 n . 5 + 2 3!

Now right hand side contains all terms containing 5n except 1st term 1. ∴ when RHS is divided by 25, it leaves the remainder 1. 21. Find the greatest term in the expansion of (x − y)20 when x = 12 and y = 4

a x − yf = LMNx FH1 − xy IK OPQ

20

20

Consider

LM N

= x 20 1 −

y x

OP Q

20

F I H K 21 − r F y I = r H xK

Tr +1 20 − r + 1 y = Tr r x Tr +1 Tr When y = 4 and x = 12

F I H K

Tr +1 21 − r 4 21 − r = = 12 3r Tr r If Tr +1 ≥ Tr . Then 21 − r ≥ 3r 21 ≥ 3r + r

21 ≥ 4r ⇒

4r ≤ 21 Greatest value of r is 5

∴ Greatest term in the expansion is Tr +1 = n cr ⋅ x n − r ⋅ a r with

r +1= 5 r=4

n = 20, x = 12 and a = −4.

a f

T5 = 20 c 4 ⋅ 12 20 − 4 ⋅ −4

4

81

82

Basic Mathematics

T5 = 20 c4 ⋅ 1216 ⋅ 4 4 . 22. Find the greatest term in the expansion of (2p + 2q)17 when p = 12 and q = 14.

a2 p + 3qf = a2 pf ⋅ LMN1 + 23qp OPQ 17 − r + 1 3 F q I T = ⋅ G J 2 H pK T r 18 − r 3 F 14 I = r 2 H 12 K T 18 − r L 7 O = T r MN 4 PQ

17

17

17

r +1 r

r +1 r

If Tr +1 ≥ Tr

a18 − rf 7 ≥ 4r 126 − 7r ≥ 4 r 126 ≥ 11r 11r ≤ 126.

∴

The greatest value of r is 11. Greatest term is 11th term.

Tr +1 = n cr ⋅ x n − r ⋅ a r

a f ⋅ a3qf ⋅ a24f ⋅ a 42f .

T11 = 17 c10 ⋅ 2 p

= 17 c10

17 −10

7

10

10

23. In the expansion of (1 + x)43, the co-efficients of (2m + 1)th and (m + 2)nd terms are equal, find m. Solution. Given T2m + 1 = Tm + 2 We have Here

Tr +1 = n cr ⋅ x n − r ⋅ a r . formula n = 43, x = 1 and a = x.

T2 m +1 = 43c2 m ⋅ 143− 2 m ⋅ x 2 m . T2 m +1 = 43 c2 m ⋅ x 2 m .

...(1)

Now Tm + 2 = 43 cm +1 ⋅ 143− m +1 ⋅ x m +1

3 143− 2 m = 1

Binomial Theorem

Tm +2 = 43 cm +1 ⋅ x m +1

83

...(2)

Given (1) = (2) ∴

43

c2 m ⋅ x 2 m = 43 c m +1 ⋅ x m +1

c2 m ⋅ x 2 m = 1. 43 c m +1 ⋅ x m +1 43

43 ⋅ x 2m 43 − 2 m ⋅ 2 m =1 43 ⋅ x m +1 43 − m + 1 ⋅ m + 1 43 − m + 1 ⋅ m + 1 2 m −a m +1f ⋅x = 1. 43 − 2 m ⋅ 2 m 43 − m − 1 ⋅ m + 1 2 m − m −1 ⋅x = 1. 43 − 2 m ⋅ 2 m 42 − m ⋅ m + 1 m −1 ⋅x = 1 = x0. 43 − 2 m ⋅ 2 m

⇒

m−1=0

⇒

m =1 .

[By equation power index of x]

Verification: By Substituting m = 1 we get

42 − 1 ⋅ 1 + 1 ⋅ x1−1 = 1 43 − 2 1 ⋅ 2 ⋅ 1

af

41 ⋅ 2 0 ⋅x =1 41 ⋅ 2 1 = 1. 24. The 21st and 22nd terms in the expansion of (1 + x)44 are equal. Find x. Solution: Given T21 = T22 We know

Tr +1 = n cr ⋅ x n − r ⋅ a r . Formula

Comparing (1 + x)44 with (x + a)n we get

x = 1, a = x and n = 44. T21 = 44 c20 ⋅ 144 −20 ⋅ x 20

84

Basic Mathematics

44 ⋅ x 20 44 − 20 ⋅ 20

=

T22 = 44 c21 ⋅ 144 − 21 ⋅ x 21 44 ⋅ x 21 44 − 21 ⋅ 21

=

T21 = T22

Given ∴

44 44 ⋅ x 20 = ⋅ x 21 44 − 20 ⋅ 20 44 − 21 ⋅ 21

⇒

44 − 21 ⋅ 21 x 21 = 44 − 20 ⋅ 20 x 20 23 ⋅ 21 =x 24 ⋅ 20 23 ⋅ 21 ⋅ 20 =x 24 ⋅ 23 ⋅ 20 21 24

⇒

x=

⇒

7 x= . 8

REMEMBER: •

a x + af

n

= x n + nx n −1 ⋅ a +

a f

a fa

f

n n − 1 n− 2 2 n n − 1 n − 2 n−3 3 ⋅a + ⋅ a + ... + a n . x x 2! 3!

• There are (n + 1) terms in the expansion of (x + a)n. The power indices of ‘x’ go on decreasing by 1 and those of ‘a’ go on increasing by 1 at each stage so that the sum of power indices is n. − • The general term or (r + 1)th term is given by Tr +1 = n cr x n r ⋅ a r .

• If n is even the number of terms in the expansion of (x + a)n is (n + 1) which is odd. ∴ There will be only one middle term i.e. Tn . 2

+1

• If n is odd, the number of terms in the expansion of (x + a)n is (n + 1) which is even. ∴ there will be 2 middle terms: Tn +1 and Tn +1 . 2

2

+1

Binomial Theorem

85

• To find the term containing xm in the expansion of (x + a)n i.e., to find the co-efficient of xm, write Tr + 1. Simplify and equate power index of x to m. Get r and substitute the value of r in Tr + 1. • For getting the term independent of x or constant term, equate the power index of x to zero after writing Tr + 1 simplify then get the value of r. If r is a positive integer greater than 0, substitute in Tr + 1. • If r is negative or a fraction then conclude that there is no term independent of x in the expansion.

EXERCISES I. Expand b y using Binomial theor em: by theorem:

F x + 1I H xK FG 2 p − q IJ H q 2pK F y+ 1 I GH y JK

a

4

1.

3.

5.

2. 1 + xy 6

4.

f

7

F 2a − b I H 3K

6

5

II. Find the indicated term in the expansions: 1. 4th term in (2 +

F 1I 2. 12th term in G y + J H yK F x 3I 4. 8th term in G − J H 2 yK

a)7

F 1 IJ 3. 3rd term in G x + H xK F bI 5. 10th term in H 2 a + K a

10

4

2

12

.

III. Find the mid dle ter m(s) in the ffollo ollo wing: middle term(s) ollowing: 1.

FG x + 1 IJ H 2 xK F a − bI H b aK

20

5.

Fa + 3 I H2 a K 2

Fx − 1 I H2 x K F a − abI Hb K F 3x − x I GH 2 JK

10

2.

14

3.

13

2

12

4.

2

19

6.

9

86

7.

Basic Mathematics

FG 2 x H

2

1 − x

IJ K

11

8.

Fx + 3 I H2 x K

19

2

IV m inde pendent of x in the ffollo ollo wing eexpansions: xpansions: IV.. Find the ter term independent ollowing

F 4a − 3 I GH 3 2a JK FG x − 2 IJ H2 xK F 2x + 1 I H xK 2

1.

2.

Fx − 3 I H xK

4.

FG 2 x − 1 IJ H xK

6.

Fx − 1 I H xK

9

10

3.

2

10

5.

2

10

4

15

21

2

V. Find the coef coeffficient of:

d

1. x 23 in x 2 − x

i

F H

20

F H

4 4 2. x in x +

I K 1 in F 2 x + I H xK

1 1 4 3. 17 in x − 3 x x

F H

15

2

3

F H

6.

1 1 in 2 − 4 x x

2.

d

I K b 3

6 3 4. a b in 2 a −

5

5 5. x

1 x3

I K

15

I K

9

8

VI. Find the vvalue alue of: 1. 3.

d2 + 3 i + d2 − 3 i d1 + 5 i + d1 − 5 i 5

5

5

i d 6

2 +1 −

i

2 −1

6

5

VII.

F 1. Prove that in the expansion of G ay H

2

b + y

IJ K

25

. There is no term independent of y.

2. The second, 3rd and 4th term of expansion of (x + y)n are 108, 54 and 12 respectively. Find x, y and n. 3. Find the value of (1.01)5 correct to four decimal places. 4. Prove that the sum of odd binomial co-efficients of order n = 2n − 1. 5. Prove that the sum of binomial co-efficient of order n = 2n.

Binomial Theorem

ANSWERS 4 2 I. 1. x + 4 x + 6 +

4 1 + 4 2 x x

2 2 3 3 4 4 5 5 6 6 7 7 2. 1 + 7 xy + 21x y + 35 x y + 35x y + 21x y + 7 x y + x y

3. 64

FG p IJ H qK

6

− 96

FG p IJ H qK

6 6 4. 64a − 64a b +

4

+ 60

FG p IJ H qK

2

− 20 +

LM N

II. 1. 560 a 3 −3

3 −7

⋅3 ⋅ x y

4.

c7 2

III. 1.

20

c10 2 −10

7

13

7.

19

5. 1760

c5 ⋅ 64 ⋅ x 19 2 and 11c5 ⋅ 32 ⋅ x 7 .

IV. 1. 2268 3.

FG IJ H K

3 q 4 p

4

+

q6 . 64 p 6

b9 . a6 63 8x5 c6 312 ⋅ b 6

4.

12

6.

15309 13 5103 14 x and x . 8 16

8.

19

4.

c9 ⋅ 39 ⋅ 2 −10 x −9 and 19 c9 ⋅ 310 2 −9 ⋅ x −11 .

V. 1. −1140

26 . 27

15

c10 ⋅ 2 5

6. − 21 c7

5. No term independent of x.

c3 ⋅

−

2. 405

45 64

9

4

3. 6 x 3

2. −

c9 39 ⋅ 2 −10 ⋅ x −8 and 19 c10 310 ⋅ 2 −9 ⋅ x −10

−11

FG IJ H K

3 q 4 p

c2 y −9

3. −14 c7 5.

−

OP Q 2.

10

2

b6 80 4 2 160 3 3 20 4 a b − a b + + a 2 b 2 − ab 5 + . 3 27 27 81 729

1 5 1 y 3 + 5 y 2 + 10 y + 10 + + 2 y y y

5.

FG IJ H K

15 q 4 p

2.

15c 8

3. −1365

5. 80

6. 2 4 ⋅ 8c 4 .

VI. 1. 724

2. 140 2

3. 352

VII. 2. x = 3, y = 1 and n = 4.

3. 1.0510.

4.

87

5 Partial Fractions 5.1 DEFINITIONS: 1. Pol ynomial: An expression of the form olynomial:

a0 x n + a1 x n −1 + a2 x n −2 + ... + an where a0 , a1 , a2 ... an are constants, x is a variable is called a polynomial in x of degree n (provided a0 ≠ 0). 2. Pr oper and impr oper fr actions: If f (x) and g (x) are two polynomials in x, then Proper improper fractions:

af af

f x is called g x

a rational fraction. If the degree of f (x) is less than degree of g (x) then it is called proper fraction. Otherwise it is called an improper fraction. By division, an improper fraction can always be reduced to the sum of a polynomial and a proper fraction. Examples: 1.

2.

4x −1 is a proper fraction since degree of numerator (=1) is less than degree of denomi2 x 2 + 8x − 1 nator (=2). x2 −1 is an improper fraction since degree of numerator = Degree of denominator = 2. x +1 2

a f

x2 − 1 x2 −1 By division = Note that x + 1 2 x2 + 2x + 1

a f

1 x + 2x + 1 x − 1 2

2

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

2x + 2 x2 − 1 −2 x − 2 =1− 2 =1+ 2 2 x + 2x + 1 x + 2x + 1 x +1

a f

We get

which is polynomial and a proper fraction.

5.2 PARTIAL FRACTIONS:

a f f

1 2 x − 4 + 2 x +1 + = x +1 x − 4 x +1 x − 4

a fa

Consider

=

x − 4 + 2x + 2 3x − 2 = 2 x +1 x − 4 x + x − 4x − 4

a fa

f

=

3x − 2 . x − 3x − 4 2

Here we have expressed the sum of 2 proper fractions as a single proper fraction. The reverse process of expressing a single proper fraction as the sum of the two or more proper fractions is called as ‘Resolving into partial fractions’. The following rules are used to resolve a proper fraction into partial fractions: 1. To each linear factor (ax + b) which occurs only once as a factor of the denominator, there

A where A is constant. ax + b

corresponds a partial fraction of the form

x2 − 1 A B . = + 2x + 3 x + 1 2x + 3 x +1

a fa

Example:

f a f a

f

2. To each linear factor (ax + b) which occurs r times as a factor of the denominator, there corresponds r partial fractions of the form, A1 A2 + ax + b ax + b

a

Example:

A3

f aax + bf 2

+

x2 x 2x + 3

a

f

3

=

3

+ ... +

Ar

aax + bf

r

where A1 , A2 , Ar are constants.

A B C + + 2x + 3 x 2x + 3

a

f a

D

f a2 x + 3f 2

+

4

.

Note, here corresponding to linear factor x which occur only once in denominator, we have taken only one constant A and corresponding to linear factor 2x + 3 which occur 3 times, we have taken 3 constants B, C and D. 3. To each non factorisable quadratic factor ax2 + bx + c which occur only once as a factor of denominator, there corresponds a partial fraction of the form constants.

3x − 7

Example:

d x + 1ia x − 1f 2

=

Ax + B C + 2 x +1 x −1

Ax + B where A and B are ax + bx + c 2

90

Basic Mathematics

Observe here, corresponding to x2 + 1, which is quadratic and non-factorisable we have taken Ax + B and corresponding to x − 1 which is linear we have taken one constant C. 4. To each repeated non-factorisable quadratic factor (ax2 + bx + c) which occurs r times as a factor of the denominator, there corresponds r partial fractions of the form.

A1 x + B1 A2 x + B2 + 2 ax + bx + c ax 2 + bx + c

d

i

2

+ ...

dax

Ar x + Br 2

+ bx + c

i

r

where A1, A2, ..., Ar and B1 B2, ... Br are constants.

x2 − 1

d2 x + 1i

Example:

2

3

=

Ax + B Cx + D Ex + F . + + 2 2 3 2 2 x + 1 2x + 1 2x 2 + 1

d

i d

i

WORKED EXAMPLES: I. Resolve into partial fractions: 1.

x −1 x+3 x−4

a fa

f

This is a proper fraction. Resolve into partial fractions:

x −1

A

B

a x + 3fa x − 4f = x + 3 + x − 4 Multiplying both sides by (x + 3) (x − 4)

a

f a

f

x −1 = A x − 4 + B x + 3

Put ⇒

x−4=0 x = 4 in (2)

af a f 3 3 = B a 7f ⇒ B = . 7

4 −1 = A 0 + B 4 + 3

Put

x+3=0 x = −3 in (2)

a

f af

−3 − 1 = A −3 − 4 + B 0

−4 = −7 A ⇒ A =

Substituting

A=

−4 4 ⇒A= −7 7

af

4 3 and B = in 1 7 7

...(1)

...(2)

Partial Fractions

91

4 3 x −1 = 7 + 7 x+3 x−4 x+3 x−4

a fa

f

=

2.

4 3 + 7 x +3 7 x −4

a f a

f

4x − 1 x2 − 1 4x − 1 4x −1 = 2 x −1 x +1 x −1

a fa f

This is a proper fraction. Resolve into partial fractions.

A B 4x − 1 = + x −1 x +1 x −1 x +1

a fa f Multiplying by (x − 1) (x + 1)

a f a f

4x − 1 = A x + 1 + B x − 1

Put

x+1=0

a f

x = −1 in (2)

af a f 5 −5 = B a −2f ⇒ B = 2

4 −1 − 1 = A 0 + B − 1 − 1

x−1=0

Put i.e.

af

x = 1 in (2)

a f af 3 3 = A a2 f ⇒ A = 2

4 1 −1 = A 1+1 + B 0

Substituting A =

3 5 and B = in (1) 2 2 3 5 4x − 1 4x − 1 = 2 + 2 = x −1 x +1 x −1 x +1 x2 − 1

a fa f

=

3 5 . + 2 x −1 2 x +1

a f a f

...(1)

...(2)

92

3.

Basic Mathematics

x x − 5x + 6 2

x x x = 2 = x − 5x + 6 x − 3x − 2 x + 6 x x − 3 − 2 x − 3

a f a f

2

x

a x − 3fa x − 2f .

=


x

a x − 3fa x − 2f Multiplying by (x − 3) (x − 2)

a

=

A B + x−3 x−2

f a

...(1)

f

x= A x−2 +B x−3

x−2=0 x = 2 in (2)

Put

af a f 2 = B a −1f ⇒ B = −2

...(2)

2= A 0 +B 2−3

.

x−3=0 x = 3 in (2)

Put

a f af

3= A 3−2 + B 0 3= A⇒ A=3 .

Substituting A = 3 and B = −2 in (1)

x

a x − 3fa x − 2f 4.

=

3 −2 3 2 = − . + x −3 x −2 x −3 x −2

2x −1 x + 1 x + 2 2x + 1

a fa

fa

f


A B C 2x − 1 = + + x + 1 x + 2 2x + 1 x + 1 x + 2 2x + 1

a fa

fa

f

Multiplying by (x + 1) (x + 2) (2x + 1)

a

fa

f a fa

f a fa

2x − 1 = A x + 2 2x + 1 + B x + 1 2x + 1 + C x + 1 x + 2

Put

a f

x + 1 = 0 ⇒ x = −1 in (2)

a

fb a f g a f a f

2 −1 − 1 = A −1 + 2 2 −1 + 1 + B 0 + C 0

...(1)

f

...(2)

Partial Fractions

93

a fa f

−3 = A +1 −1 ⇒ A = +3 .

x + 2 = 0 ⇒ x = −2 in (2)

Put

a f

af a

f ba f g af

2 −2 − 1 = A 0 + B −2 + 1 + 2 −2 + 1 + C 0

a fa f

−5 = B −1 −3 ⇒ B =

−5 . 3

af 1 1 1 2 F − I − 1 = A a 0f + B a0f + C F − + 1I F − + 2I H 2K H 2 KH 2 K 1 3 −1 − 1 = C F I F I H 2 K H 2K 3 8 −2 = C F I ⇒ C = − H 4K 3 5 −8 A = 3, B = − and C = in a1f 3 3 2x + 1 = 0 ⇒ x = −

Put

Substituting

1 in 2 2

−8 3 −5 3 2x − 1 3 + = + . x + 1 x + 2 2x + 1 x + 1 x + 2 2x + 1

a fa 5.

a

2x − 1 x+2 x−3

fa f

fa

f

2

This is a proper fraction. Resolve into partial fraction.

a

2x − 1 x+2 x−3

fa f

2

Multiplying by (x + 2) (x − 3)2

a f

2x − 1 = A x − 3

2

=

A B C + + x +2 x −3 x −3

a f

a

fa f a

+B x+2 x−3 +C x+2 x−3=0

⇒ x = 3 (Put x = 3 in Eq. (2))

af

af af a 5 = C a 5f ⇒ C = 1

2 3 −1 = A 0 + B 0 + C 3 + 2

x+2=0

f

...(1)

2

f

...(2)

94

Basic Mathematics

∴ x = −2 (Put x = −2 in Eq. (2))

a f

a f af af 5 1 −5 = A a −5f ⇒ A = − =− 25 5

2 −2 − 1 = A −2 − 3 2 + B 0 + C 0 2

Put

x = 0, A = −

1 and C = 1 in (2) 5

a f

+ B 2 −3 + 1 2

af

2 0 −1= − −1 = − −1 − 2 + −3 +

1 −3 5

2

a fa f a f

9 − 6B + 2 5

9 = −6B 5 1 9 −6 = = −6 B ⇒ B = 5 × −6 5 5

a f

af

1 1 Substituting A = − , B = and C = 1 in 1 5 5

a

2x − 1 x+2 x−3

fa f

=

2

15 1 −1 5 + + x +2 x −3 x −3

a f

=−

6.

x2 +1

a x − 2f d x

2

−4

2

1 1 1 . + + 5 x +2 5 x −3 x−3 2

a

f a f a f

i x2 + 1

a x − 2f d x

2

− 22

i a =

x2 + 1 x2 + 1 = x−2 x−2 x+2 x+2 x−2

fa

fa

f a

fa

f

2


a

x2 + 1 x+2 x−2

fa

f

2

Multiplying with (x + 2) (x − 2)2

a

x2 + 1 = A x − 2

f

=

2

A B C + + x+2 x−2 x−2

a

a

fa

f a

f a

f

...(1)

2

+ B x+2 x−2 +C x+2

f

...(2)

Partial Fractions

x − 2 = 0 in (2)

Put

⇒ x=2

af af a f 5 5 = C a 4f ⇒ C = . 4

22 + 1 = A 0 + B 0 + C 2 + 2

x +2 = 0 in (2) ⇒ x = −2

Put

a−2f

2

a

+ 1 = A −2 − 2

f

2

a f

5 = A 16 ⇒ A = x = 0, A =

Put

1=

1=

5 . 16

af

5 5 and C = in 2 16 4

a f

5 −2 16

af af

+B 0 +C 0

2

a fa f 54 a2f

+ B 2 −2 +

a f

5 5 × 4/ + −4 B + 16 2 4 1−

5 5 − = −4 B 4 2

11 11 4 − 5 − 10 = −4 B ⇒ B = − = 4 −4 16 4

a f

B= A=

Substituting,

af

5 11 5 , B= and C = in 1 16 16 4

x2 + 1

a x + 2fax − 2f =

11 . 16

2

5 11 5 16 16 4 = + + x+2 x−2 x−2

a

5 11 5 + + 16 x + 2 16 x − 2 4 x−2

a

f

a

f a

f

2

.

f

2

95

96

7.

Basic Mathematics

dx

2x + 3 2

ia

f

+9 x−3


dx

2x + 3 2

=

i a x − 3f

+9

Multiplying by (x2 + 9) (x − 3)

a

Ax + B C + 2 x +9 x−3

fa f d

2 x + 3 = Ax + B x − 3 + C x 2 + 9

...(1)

i

2 x + 3 = Ax 2 + Bx − 3 Ax − 3 B + Cx 2 + 9C.

2 x + 3 = Ax 2 + Cx 2 + Bx − 3 Ax − 3B + 9C. Equating the co-efficients of like powers.

x 2 : 0 = A + C ⇒ A = −C

...(2)

x : 2 = B − 3A

...(3)

1 : 3 = −3B + 9C ⇒ 1 = − B + 3C

...(4)

Substituting Eq. (2) i.e., A = −C in (3)

a f

B − 3 −C = 2 B + 3C = 2 Equation (4):

− B + 3C = 1 6C = 3 ⇒ C =

A = −C ⇒

A=−

1 2

− B + 3C = 1

−B + 3 ⇒B= ∴

F 1I = 1 H 2K

3 1 −1 = 2 2 B=

1 2

1 2

Partial Fractions

A=

Substituting

d

af

1 1 −1 , B = and C = in 1 2 2 2

1 1 1 − x+ 2 2 = + 2 2 2 x−3 x +9 x +9 x −3 2x + 3

ia f

=

8.

97

−x + 1

d

2 x +9 2

1 . 2 x−3

i a +

f

8x 2 + 1 x3 − 1 8x 2 + 1 8x 2 + 1 8x 2 + 1 = = x 3 − 1 x 3 − 13 x − 1 x2 + x + 1

a fd

i

Resolve into partial fractions:

8x 2 + 1

a x − 1fd x

2

i

+ x +1

=

A Bx + C + 2 x −1 x + x +1

...(1)

a fd i + 1 = A d x + x + 1i + a Bx + C fa x − 1f

Multiplying by x − 1 x 2 + x + 1

8x 2

2

8 x 2 + 1 = Ax 2 + Ax + A + Bx 2 − Bx + Cx − C

Equating the co-efficients of like powers:

x2 : 8 = A + B

...(2)

x : 0= A− B+C

...(3)

: 1= A−C

...(4)

1 Solving (3) and (4)

A− B+C = 0 A−C =1 2A − B = 1 Solving (2) and (5)

A+ B=8 2A − B = 1 3A

= 9⇒

A=3

...(5)

98

Basic Mathematics

Substituting A = 3 in (2) 3+ B = 8 B=8−3

B=5 Substituting A = 3 in (4)

A−C =1 3−C =1

3 −1 = C ⇒ C = 2 Substituting A = 3, B = 5 and C = 2 in (1)

8x 2 + 1

a x − 1fd x 9.

2

i

+ x +1

=

3 5x + 2 + x − 1 x2 + x + 1

2x + 1 2x + 1 = 3 x + x x x2 + 1

d

i


2x + 1

d

i

x x +1 2

=

A Bx + C + 2 x x +1

...(1)

Multiplying by x (x2 + 1)

i a

d

fa f

2 x + 1 = A x 2 + 1 + Bx + C x 2 x + 1 = Ax 2 + A + Bx 2 + Cx

Equating the co-efficients of like powers:

x2 : 0 = A + B

Now Substituting

x

: 2=C

1

: 1= A

A + B = 0 ⇒ 1 + B = 0 ⇒ B = −1 A = 1, B = −1 and C = 2 in (1) 2x + 1

d

i

x x +1 2

=

1 −x + 2 + . x x2 + 1

Partial Fractions

10.

99

x2 − 2 x 2 + x − 12

This is an improper fraction. Since degree of numerator = degree of denominator = 2.

1 x + x − 12 x − 2 2

2

x 2 + x − 12 (− ) (− ) ( + ) − x + 10 x2 − 2 − x + 10 =1+ 2 2 x + x − 12 x + x − 12

∴

...(1)

Consider

− x + 10 − x + 10 − x + 10 = = 2 x + x − 12 x + 4 x − 3 x − 12 x x + 4 − 3 x + 4

a

2

=

f a

f

− x + 10

ax + 4f a x − 3f .


a

A B − x + 10 = + x +4 x −3 x+4 x−3

fa f

Multiplying by (x + 4) (x − 3) we get

a

f a

− x + 10 = A x − 3 + B x + 4

f

x−3=0⇒ x=3

Put

af a f 7 = B a7f ⇒ B = 1.

−3 + 10 = A 0 + B 3 + 4

x+4=0

Put

x = −4

a f

a

f af

− −4 + 10 = A −4 − 3 + B 0

a f

14 = A −7 ⇒ Substituting A = −2 and B = 1 in

a

A = −2.

(2)

1 −2 − x + 10 + = x +4 x −3 x +4 x −3

fa f

...(2)

100

Basic Mathematics

Substituting this in equation (1)

1 x2 − 2 −2 = 1+ + 2 x +4 x −3 x + x − 12 =1−

11.

2 1 . + x+4 x−3

8x 2 + x − 2 4 x 2 + 5x + 1 This is an improper fraction. Since degree of numerator = Degree of denominator = 2.

2 4 x + 5x + 1

8x +

2

2

x−2

8 x 2 + 10 x + 2 ( − ) ( −) ( −) − 9x − 4 −9 x − 4 8x 2 + x − 2 =2+ 2 2 4 x + 5x + 1 4 x + 5x + 1

∴

9x + 4 8x 2 + x − 2 =2− 2 4 x + 5x + 1 4 x 2 + 5x + 1

...(1)

9x + 4 9x + 4 = 4 x 2 + 5 x + 1 4 x 2 + 4 x + 1x + 1

Consider

=

9x + 4 9x + 4 = x + 1 4x + 1 4x x + 1 + 1 x + 1

a f a f a fa

f


A B 9x + 4 = + x + 1 4x + 1 x + 1 4x + 1

a fa Multiplying by (x + 1) (4x + 1)

f

a

f a f

9x + 4 = A 4x + 1 + B x + 1 4x + 1 = 0

Put

4 x = −1 ⇒ x = −

1 4

F 1 I + 4 = A a0f + B F − 1 + 1I H 4K H 4 K

9 −

...(2)

Partial Fractions

−

101

F I H K

9 3 +4= B + 4 4

F I H K

7 3 ⇒ =B 4 4 Put

a f

B=

7 3

.

x + 1 = 0 ⇒ x = −1

ba f g af 5 −5 = A a −3f ⇒ A = . 3

9 −1 + 4 = A 4 −1 + 1 + B 0

Substituting A =

5 7 and B = in (2) 3 3 5 7 9x + 4 3 = + 3 x + 1 4x + 1 x + 1 4x + 1

a fa

f


RS T

53 73 8x 2 + x − 1 =2− + 2 x + 1 4x + 1 4 x + 5x + 1 =2−

12.

UV W

5 7 . − 3 x + 1 3 4x +1

a f a

f

x3 − 1 . x3 + x This is an improper fraction since degree of numerator = Degree of denominator = 3.

1 x + x x −1 3

3

x3 + x ( − ) ( −) − x −1 x3 − 1 x +1 −x −1 =1− 3 = 1+ 3 3 x +x x +x x +x

∴ Consider

x +1 x +1 This is a proper fraction. = 3 x + x x x2 + 1

d

i

...(1)

102

Basic Mathematics

Resolve into partial fractions

d

x +1

i

x x +1 2

A Bx + C + 2 x x +1

=

Multiplying by x (x2 + 1)

i a

d

...(2)

fa f

x + 1 = A x 2 + 1 + Bx + C x x + 1 = A x 2 + A + Bx 2 + Cx Equating the co-efficients of like powers

x2 : 0 = A + B x : 1= C

1

:1= A

A + B = 0 ⇒ 1 + B = 0 ⇒ B = −1

Substituting A = 1, B = −1 and C = 1 in (2)

d

x +1

i

x x +1 2

=

1 −x +1 + x x2 + 1


RS T

1 −1x + 1 x3 =1− + 2 3 x x +x x +1

=1−

UV W

1 −x + 1 − . x x2 + 1

REMEMBER: Proper fraction •

ba x + b gb 1

1

Partial fraction

Nr a2 x + b2 a3 x + b3 ...

gb

g

Nr

•

aax + bf acx + d f

•

dax

A B + ax + b ax + b

a

2

Nr 2

ia

+ bx + c dx + e

A B C + + + ... a1 x + b1 a2 x + b2 a3 x + b3

f

dax

Ax + B 2

+ bx + c

i

+

f a 2

+

C cx + d

f

C dx + e

Provided ax2 + bx + c is non-factorisable. If ax2 + bx + c is factorisable as (a1x + b1) (a2x + b2), then replace ax2 + bx + c by (a1x + b1) (a2x + b2). Then resolve,

Partial Fractions

b

Nr A B C as + + a1 x + b1 a2 x + b2 dx + e a1 x + b1 a2 x + b2 dx + e

gb

f b

ga

g b

g a

103

f

If an improper fraction is given to resolve, first by dividing the numerator by denominator write the given fraction as the sum of the polynomial and the proper fraction. Then the proper fraction is resolved into partial fractions.

EXERCISE Resolve into partial fractions: 1.

4.

7.

10.

8x − 1

a x − 2fa x − 3f

2.

fa f

9 x +1 x + 2

a fa

f

1 x −1 1+ x

a fa f

5.

2

8.

2

11.

3x − 1

13.

a x + 2fd1 − x + x i

14.

x2 − 2 x −4 x +3

a

19.

2 x 2 − 3x − 4 x2 − x − 6

a x − 1fd x dx

2

− 5x + 6

i

6.

2

ia f

+ 2x + 1 x − 1

3x + 2 x3 + x

9.

12.

a x − 2f d2 x + 3i 2

2 x 2 + 3x + 2 17. x2 − x − 2

fa f

20.

2x + 3 x − 3x + 2 2

a

3x + 5 x+2 2 x−3

fa f

x2 + 1

4 x 2 + 3x − 1

a x − 2f d x

2

−4

5x 2 + 1 x3 − 1

15.

a x + 1fd x + 1i 2

x 3 + 7 x 2 + 17 x + 11 18. x 2 + 5x + 6

x 4 − 3 x 3 − 3x 2 + 10 x +1 2 x − 3

a fa f

ANSWERS 1.

3 5 + x −2 x +3

1 1 1 4. 2 x + 1 − x + 2 + 2 x + 3

a f

a f

1 −14 6. 25 x + 2 + 5 x + 2

a

f a

f

2

+

2.

5 1 − x −1 x +1

5.

2 3 4 + − x −1 x − 2 x − 3

14 25 x − 3

a f

i

x −1

x2 + 2x + 4

2

16.

3.

x 2 − 10 x + 13

1 x +1 x + 2 x + 3

a fa

4x + 6 x2 − 1

3.

7 5 − x − 2 x −1

104

Basic Mathematics

9 9 9 7. x + 1 − x + 2 − x + 2

a

f

5 0 3 8. − 2 x + 1 + x + 1 2 + 2 x − 1

a f a f

2

5 11 5 9. 16 x + 2 + 16 x − 2 + 4 x − 2

a

f

a

f a

1 1 1 10. 4 x − 1 + 4 1 + x + 2 1 + x

a f a f a f

11.

2 −2 x + 3 + 2 x x +1

12 13x + 4 − 14. 11 x − 2 11 2 x 2 + 3

a

d

16 1 − 3 x − 2 3 x −1

i

a f a f a x − 2f + 16 a17x + 1f − 4 a x11+ 1f

17. 2 +

20.

f

f

a f

2

2

12.

2 3x + 1 + x −1 x2 + x + 1

13.

15.

x −1 + x + 1 x2 + 1

16. 1 −

1 2 + x+3 x−4

18.

a x + 2f + x 1+ 2 + x 2+ 3

19. 2 −

2 1 + x+2 x−3

2

−

17 . 16 x − 3

a f

x −1 + x + 2 1 − x + x2

6 Matrices & Determinants 6.1 INTRODUCTION: The theory of matrices was developed in 1857 by the French mathematician Cayley. It was not well advanced till 20th century. But now a days matrices are powerful tool in modern mathematics having wide applications.

6.2 MATRIX: A matrix is an arrangement of numbers in rows (horizontal lines) and columns (vertical lines). The arrangement is usually enclosed between square brackets [ ] or curved brackets ( ) or pairs of vertical lines || ||. The matrix is usually denoted by a capital letter. Order of a matrix = Number of rows × Number of columns. If a matrix has m rows and n columns, then Order = m × n (read as m by n)

Example: Matrix A =

LM1 N4

2 5

OP Q

3 has the order 2 × 3. 6

6.3 TYPES OF MATRICES: 1. Rectangular ma tr ix: If the number of rows is not equal to number of columns in a matrix, then matr trix: that matrix is called rectangular matrix.

LMa A= b MNc

1

Example:

1 1

a2 b2 c2

a3 b3 c3

a4 b4 c4

a5 b5 c5

OP PQ

3× 5

w ma tr ix or Ro w vvector: ector: If a matrix has only one row, then it is called row matrix. (a) Ro Row matr trix Row

Example:

X= 1

2

3 1×3

tr ix or column vvector: ector: If a matrix has only one column then it is called column (b) Column ma matr trix matrix.

106

Basic Mathematics

LM1OP Y= 2 MN3PQ

Example:

3 ×1

2. Squar tr ix: If the number of rows is equal to number of columns in a matrix, then it is called Squaree ma matr trix: square matrix.

LM MN

a1 A = b1 c1

Example:

a2 b2 c2

a3 b3 c3

OP PQ

3× 3

In a square matrix, the entries from the left top corner to the right bottom corner are called principal diagonal elements. So in the above example a1 b2 c3 are Principal diagonal elements. (a) Diagonal matrix: If in a square matrix all the non-diagonal elements are zero, then it is called diagonal matrix.

Example:

A=

LM1 0OP , B = LM20 N0 2Q MN0

0 −1 0

OP PQ

0 0 3

(b) Scalar matrix: If in a diagonal matrix, all the diagonal elements are equal, then it is called scalar matrix.

Example:

A=

LM5 0OP , B = LM06 N0 5Q MN0

0 6 0

0 0 6

OP PQ

(c) Identity matrix or unit matrix: If in a scalar matrix all the diagonal elements are equal to 1, then it is called unit matrix or identity matrix.

Example:

I2 =

LM1 0OP , I N0 1 Q

3

LM MN

1 = 0 0

0 1 0

0 0 1

OP PQ

iangular ma tr ix: A square matrix is called upper triangular if all the non-diagonal (d) Upper tr triangular matr trix: elements below the principal diagonal are zeroes.

LMa MN 00

1

Example:

a2 b1 0

a3 b2 c1

OP PQ

wer tr iangular ma tr ix: A square matrix is called lower triangular if all the non-diagonal (e) Lo Low triangular matr trix: elements above the principal diagonal are zeroes.

LMa MN bc

1

Example:

1

1

0 b2 c2

0 0 c3

OP PQ


107

Null matrix: If each element of a matrix is zero then it is called null matrix or zero matrix.

A=

Example:

LM0 N0

OP Q

0 0

LM N

0 0 , B= 0 0

0 0

OP Q

6.4 ALGEBRA OF MATRICES: tr ices: Two matrices of same order are said to be equal iff the corresponding 1. Equality of ma matr trices: elements are equal,

LM N

a1 i.e., b 1

OP LM Q N

OP Q

a2 3 = b2 −1

4 iff a1 = 3, b1 = −1, a2 = 4, b2 = 0. 0

dition and subtr action of ma tr ices: Two or more matrices of the same order can be added or 2. Ad Addition subtraction matr trices: subtracted. It is the matrix obtained by adding or subtracting the corresponding elements i.e., If

Then

Illustration: If

A=

LM a Na

OP Q

4

LM N

b a3 and B = 1 b4 a6

a2 a5

1

b2 b5

b3 b6

LM a + b a + b a + b OP Na + b a + b a + b Q a −b a −b a −b O A − B = LM a b a b a − − − b PQ N 1 2 3 1 A=L MN0 −1OPQ and B = LMN4 6OPQ , then 1+ 3 2 +1 4 3 A+ B= L MN0 + 4 −1 + 6OPQ = LMN4 5OPQ 1− 3 2 −1 1 −2 A− B= L MN0 − 4 −1 − 6OPQ = LMN−4 −7OPQ A+ B=

1

1

2

2

3

3

4

4

5

5

6

6

1

1

2

2

3

3

4

4

5

5

6

6

OP Q

Scalar m ultiplica tion: If A is any matrix and k is any scalar or constant, then kA is a matrix obtained multiplica ultiplication: by multiplying every element of A by k.

LM a a a OP , then Na a a Q ka ka ka O kA = LM PQ ka ka ka N 1 2O L 1 2O L 3 6O Illustration: If A = LM N3 4QP . Then 3A = 3 MN3 4QP = MN9 12QP . i.e.,

if A =

1

2

3

4

5

6

1

2

3

4

5

6

Ma tr ix m ultiplica tion: The product of 2 matrices exists only when the number of columns in 1st Matr trix multiplica ultiplication: matrix is equal to the number of rows in the 2nd matrix.

108

Basic Mathematics

If A is a matrix of order m × n and B is a matrix of order n × p. Then AB exists and is of order m × p. To get the elements of AB, the elements of 1st row of A multiplied by the corresponding elements of the first column of B and the products are added. The sum is the element in the first row, first columns of AB. Similarly other elements are obtained.

i.e., If

A = a1

a2

a3

LM MN

b1 & B = b2 b3

1× 3

OP PQ

, then AB exists and is of order 1 × 1. 3×1

AB = a1b1 + a2 b2 + a3 b3 A=

2. If

LM a Na

21

d

OP PQ

3 × 2 and is of order 2 × 2.

LM a b Na b

+ a12 b21 + a13 b31 + 11 a22 b21 + a23 b31

11 11

21

b12 b22 b33

i

Then AB exists 3 2 × 3

AB =

LM MN

OP Q

b11 a13 & B = b21 a23 b31

a12 a22

11

OP Q

a11b12 + a12 b22 + a13 b33 . a21b12 + a22 b22 + a23 b33

Illustration:

If

A=

LM1 N4

2 5

AB exists and is of order 2 × 2.

3 6

OP Q

2×3

LM 1 + 4 + 9 N4 + 10 + 18 14 32 =L MN32 77OPQ .

AB =

LM MN

1 & B= 2 3

4 5 6

OP PQ

, then 3× 2

OP Q

4 + 10 + 18 16 + 25 + 36

Note: Note:(1) A + B = B + A i.e., commutative law with respect to addition holds good. (2) AB ≠ BA in general i.e., commutative law with respect to multiplication doesn’t hold good in general.

6.5 TRANSPOSE OF A MATRIX: If A is any matrix then the matrix obtained by interchanging rows and columns of A is called transpose of A. It is denoted by A′ or AT. For example: If

A=

LM1 2OP N3 4Q

2× 2

then A T =

LM 1 3OP N2 4 Q

2× 2


1 A = LM 4 N

If

2 5

OP Q

3 6

then A

T

2 ×3

LM 1 = 2 MN 3

4 5 6

OP PQ

3× 2

If A is of order m × n then AT is of order n × m.

WORKED EXAMPLES:

LM 1 0OP and B = LM 2 3OP Then find (i) A + B (ii) 2A − 3B (iii) A + 2A′ N−4 3Q N−1 −6Q 1+ 2 0+3 1 0 2 3 A+ B= L (i) MN−4 3OPQ + LMN−1 −6OPQ = LMN−4 + a−1f 3 + a−6fOPQ 3 3 =L MN−5 −3OPQ 2 3O 1 0O − 3L 2 A − 3B = 2 LM (ii) P M N−4 3Q N−1 −6PQ 2−6 0−9 2 0 6 9 =L MN−8 6OPQ − LMN−3 −18OPQ = LMN−8 − a−3f 6 − a−18fOPQ −4 −9 =L MN−5 24OPQ 1 −4 1 0 2 −8 1 0 A + 2 A′ = L (iii) MN−4 3OPQ + 2 LMN0 3OPQ = LMN−4 3OPQ + LMN0 6OPQ L 1 + 2 0 + a−8fOP = L 3 −8O . =M N−4 + 0 3 + 6Q MN−4 9PQ L 1 2 −1OP L 1 3 2O 2. If A = M 4 0 −3 and B = M 1 −1 5P MN 1 −1 5PQ MN6 2 0PQ 1. If A =

Then find (i) (A + B)′ (ii) 2A′ − 3B.

(i)

LM MN

1 A+ B= 4 1

LM 1 + 1 A + B = 4 +1 MN1 + 6

2 0 −1 2+3 0 + −1 −1 + 2

a f

OP PQ

LM MN

1 −1 −3 + 1 6 5

OP PQ

OP PQ

3 −1 2

2 5 0

LM MN

5 −1 1

2 −1 + 2 −3 + 5 = 5 7 5+0

OP PQ

1 2 5

109

110

Basic Mathematics

a

⇒

(ii)

L f MM N

2 ′ A+ B = 5 1

2 A′ − 3B = 2

LM 1 MN−21

LM 2 8 MN−42 −06 LM 2 − 3 2 A′ − 3B = MN−24−−183 2

7 1 5 4 0 −3

OP PQ

1 1 −1 − 3 1 5 6

OP PQ

3 −1 2

LM MN

LM MN

9 −3 6

6 15 0

→

LM 4 × 1 MN−1 0× ×a11f

1

2

−3

B

1× 3

3×1

a a a

fOP L fP MM fQ N LM 4OP BA = 1 2 −3 MM−01PP B N Q BA = 1 × 4 + 2 × 0 + a −3fa −1f 4×2 0×2 −1 × 2

4 × −3 4 0 × −3 = 0 −1 × −3 −1

 →

BA = 4 + 0 + 3 = 7 .

∴

AB ≠ BA.

4. If

−1 3 −12

−3 , then find AB and BA. Is AB = BA?

LM 4 OP AB = 0 MM−1PP N Q AB =

OP PQ

2 5 0

OP PQ 8−9 2 − 6O L −1 0 − a −3f −2 − 15P = M 1 −6 − 6 10 − 0PQ MN −20 3 2 −2 − 3 18 10

=

LM 4OP 3. If A = M 0P and B = 1 NM−1PQ

OP PQ

5 −1 2

LM MN

1 A= 1 1

1 1 1

OP PQ

1 1 , then find A2 . 1

8 0 −2

OP PQ

−12 0 +3

−4 −17 10

OP PQ


A2 = A ⋅ A

LM1 1 1OP LM1 1 A = 1 1 1 1 1 MN1 1 1PQ MN1 1 L1 + 1 + 1 1 + 1 + 1 = M1 + 1 + 1 1 + 1 + 1 MN1 + 1 + 1 1 + 1 + 1 L3 3 3O L1 1 = M3 3 3P = 3 M1 1 MN3 3 3PQ MN1 1 2

A2

5. Find x and y if Given

LM x 3OP + LM2 N4 5Q N3

OP LM Q N

y 5 = 7 −1

7 4

OP PQ

1 1 = 3 A. 1

OP Q

⇒

x + 2 = 5, 3 + y = 7

⇒

x = 5 − 2, y = 7 − 3

⇒

x = 3 and y = 4.

⇒

OP PQ

1+1+1 1+1+1 1+1+1

Hence A2 = 3A.

LM x 3OP + LM2 yOP = LM5 N4 5Q N3 −1Q N7 x+2 3 + yO L 5 = LM 4 3 5 + + a−1fPQ = MN7 N

6. Find x and y if

OP PQ

1 1 1

OP Q 7O 4PQ

7 4

LM 2 1OP LM xOP = LM7OP N−1 0Q N yQ N2Q LM 2 1OP LMx OP = LM7OP N−1 0Q N yQ N2Q LM 2 x + y OP = LM7OP ⇒ N−1a xf + 0 a yfQ N2Q − x = 2 ⇒ x = −2.

2x + y = 7

a f

2 −2 + y = 7

2x + y = 7 . −x + 0 = 2

111

112

Basic Mathematics

y = 7+4 y = 11.

So, x = −2 and y = 11 7. Find x and y given that

LM 1 2OP LMx OP = LM2OP N−3 7Q N yQ N1Q LM 1 2OP LMx OP B = LM2OP N−3 7Q N yQ N1Q LM x + 2 yOP = LM2OP N−3x + 7yQ N1Q   →

Given

⇒ ⇒

x + 2y = 2

...(1)

−3x + 7 y = 1

...(2)

(1) × 3 + (2)

3x + 6 y = 6 −3x + 7 y = 1 13 y = 7 y=

⇒

y=

Substituting

x+2

7 in (1) 13

F 7I =2 H 13K x=2− x=

14 13

26 − 14 13

x=

8. Find A and B if A + B =

7 13

12 . 13

LM7 0OP and A − B = LM3 0OP . N2 5 Q N8 3Q


LM7 0OP N 2 5Q L3 0OP A− B= M N8 3Q L7 0OP + LM3 0OP A+ B+ A− B= M N2 5Q N8 3Q L10 0OP 2A = M N10 8Q 1 L10 0 O L5 0 O = A= M 2 N10 8 PQ MN5 4 PQ L7 0OP A+ B =M N 2 5Q LM5 0OP + B = LM7 0OP N 5 4 Q N 2 5Q L7 0OP − LM5 0OP B=M N2 5 Q N5 4 Q L 2 0OP B= M N−3 1Q A+ B=

Adding

⇒

Given

⇒

Ver if ica tion: erif ifica ication:

A+ B= 9. If A =

Given:

LM4 N1

OP Q

LM5 0OP + LM 2 0OP = LM7 0OP N5 4Q N−3 1Q N2 5Q

0 , then prove that A2 − 2A − 8I = 0 where 0 is the null matrix. −2 A=

LM4 N1

0 −2

OP Q

L4 0OP LM4 0OP B A = A⋅ A = M N 1 −2Q N 1 −2Q L 4 × 4 + 0 × 1 4 × 0 + 0 a −2fOP =M N1a4f + a−2fa1f 1a0f + a−2fa −2fQ  →

2

113

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Basic Mathematics

LM16 + 0 0 + 0OP = LM16 0OP N 4 − 2 0 + 4Q N 2 4Q L4 0OP = LM 8 0OP 2A = 2 M N 1 −2 Q N 2 −4 Q L 1 0OP = LM 8 0OP 8I = 8 M N0 1Q N0 8Q

=

Consider LHS, A 2 − 2 A − 8I =

= Hence proved. 10. If A =

LM1 N3

OP Q

LM16 − 8 − 8 N 2−2−0 LM N

2 1 ,B = −1 2

OP LM 0OP − LM8 0OP Q N −4Q N0 8Q 0 − 0 − 0 O L 0 0O = = R.H.S. 4 − a −4f − 8PQ MN 0 0PQ LM16 N2

0 8 − 4 2

OP Q

LM N

0 2 and C = 1 −1

OP Q

1 , 1

then prove that A (B + C) = AB + AC.

B+C =

Consider

LM 1 0OP + LM 2 1OP = LM3 1OP N2 1Q N−1 1Q N1 2Q

L1 2OP LM3 1OP B = LM3 + 2 A B+C = M N3 −1Q N1 2Q N 9 − 1 L5 5OP A a B + Cf = M N8 1Q  →

1+ 4 3−2

OP Q ...(1)

Now

L1 AB = M N3

 →

L1 AC = M N3 L5 AB + AC = M N1

OP LM 1 0OP B = LM1 + 4 Q N2 1Q N3 − 2

2 −1

OP LM 2 Q N −1 2 O L0 + −1PQ MN 7

→

2 −1

OP B = LM2 − 2 Q N6 +1 3O L5 5O = 2PQ MN8 1PQ 1 1

From (1) and (2) A (B + C) = AB + AC

OP LM Q N

OP Q

0+2 5 = 0 −1 1

2 −1

OP LM Q N

3 2

1+ 2 0 = 3 −1 7

OP Q ...(2)


L1 11. If A = M N1

LM OP MM PP NQ

OP Q

1 2 , B = 6 and C = 2 2 2

3 0

115

1 , then prove that A (BC) = (AB) C

LM1OP BC = 6 MM2PP NQ

LM1 × 2 1 × 1OP Consider 2 1 B = M6 × 2 6 × 1P NM2 × 2 2 × 1PQ LM 2 1OP BC = 12 6 MM 4 2PP N Q L 2 1OP L1 × 2 + 3 × 12 + 2 × 4 1 3 2O M L A a BCf = M N1 0 2PQ MMN124 26PQP B = MN1 × 2 + 0 × 12 + 2 × 4 L2 + 36 + 8 1 + 18 + 4OP = LM46 23OP A a BC f = M N 2 + 0 + 8 1 + 0 + 4 Q N10 5 Q →

  →

1×1+ 3 × 6 + 2 × 2 1×1+ 0 × 6 + 2 × 2

OP Q ...(1)

Now

L1 AB = M N1

3 0

2 2

OP LM16OP B = LM1 × 1 + 3 × 6 + 2 × 2 OP = LM1 + 18 + 4OP = LM1 + 18 + 4OP = LM23OP Q MMN2PPQ N1 × 1 + 0 × 6 + 2 × 2Q N 1 + 0 + 4 Q N 1 + 0 + 4 Q N 5 Q

2

1

B = LMN235 ××22

  →

a ABf C = LMN235OPQ  →

OP LM Q N

OP Q

23 × 1 46 = 5 ×1 10

23 5

...(2)

From (1) and (2) A (BC) = (AB) C. Hence proved. 12. If A =

LM3 4OP and B = LM 1 N0 5 Q N2

Consider

OP Q

a

f

−1 , then prove that A + B ′ = A′ + B′ 1

LM3 N0 a A + Bf′ = LMN43 A+ B=

OP LM Q N 2O 6 PQ

4 1 + 5 2

OP LM Q N

4 −1 = 1 2

OP Q

3 6

...(1)

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LM 3 N4 L3 A′ + B′ = M N4 A′ =

OP Q 0O L 1 + 5PQ MN −1

LM N

0 1 and B ′ = −1 5

OP Q

2 1

OP LM Q N

2 4 = 1 3

2 6

OP Q

...(2)

From (1) and (2)

a A + Bf′ = A′ + B′ Hence proved.

LM 1 2OP , then find X such that AX = I where I is identity matrix of order 2 × 2. N −1 7 Q L 1 2OP, To find X such that AX = I A=M Given N−1 7Q LM3 A is of order 2 × 2 OP a bO L 2 × 2, So X=M Let MM IXisisofoforder P N c d PQ order 2 × 2. PQ N

13. If A =

Consider

AX = I

LM 1 2OP LMa N −1 7Q N c  →

⇒

b d

OP B = LM a + 2c Q N − a + 7c

a + 2c = 1 − a + 7c = 0 9c = 1 c = 1 9.

⇒

; ;

OP LM Q N

1 b + 2d = 0 −b + 7d

OP Q

0 1

b + 2d = 0 − b + 7d = 1 9d = 1 d = 1 9.

Substituting c = 1/9

Substituting d = 1/9

a+2

b+2

F 1I = 1 H 9K

a =1− 7 a= . 9

2 9

F 1I = 0 H 9K

b=−

2 9


LM N

a Hence X = c

14. If A =

L OP MM Q M N

7 b = 9 1 d 9

LM3 4OP , B = LM1 N0 5 Q N2

−

OP Q

OP PP LMN Q

OP Q

2 9 =1 7 1 9 1 9

−2 1

−1 , then find X such that A + 2X = B. 4

Solution. Since A is of order 2 × 2. B is of order 2 × 2. X is of order 2 × 2. Now A + 2X = B 2X = B − A

X= X=

LM N

1 B− A 2

OP LM Q N

3 −1 − 4 0

1 1 2 2

LM N

X=

1 −2 2 2

X=

LM−1 N1

4 5

OP Q − 5 2O . − 1 2 PQ

OP Q

−5 −1

6.6 DETERMINANTS: Every square matrix is associated with a unique real number called its determinant value. Determinant of a Square Matrix of Order 2: If A =

LM a Na

1 3

OP Q

a2 , then determinant A is denoted by |A| or det A and its value is a1a 4 − a2 a3 . a4

Example: If A =

LM1 2OP , then A = 1 × 7 − 2 × 3 = 7 − 6 = 1. N3 7Q

Determinant of a Square Matrix of Order 3:

If

LM MM N

a1 A = a4 a7

a2 a5 a8

OP PP Q

a3 a6 , then a9

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A = a1

a5 a8

a6 a4 − a2 a9 a7

b

g

a6 a4 + a3 a9 a7

b

a6 a9

g b

A = a1 a5 a9 − a6 a8 − a2 a4 a9 − a7 a6 + a3 a4 a9 − a6 a7

g

For example,

LM MN

1 A= 2 3

If

A =1

1 1

2 1 1

OP PQ

3 −1 , then 2

−1 2 −2 2 3

−1 2 +3 2 3

1 1

a f a f a f 1 a3f − 2 a7f + 3 a −1f

= 1 2 +1 − 2 4 + 3 + 3 2 − 3

3 − 14 − 3 = −14.

Singular and non-singular ma tr ices: If A = 0 for a matrix A then A is said to singular matrix. If matr trices: A ≠ 0 then A is called non-singular matrix.

WORKED EXAMPLES: 1. If A =

LM 1 N4

OP Q

2 , Then find A . −1

Solution: Given A =

LM 1 N4

OP Q

2 −1

a f af

A = 1 −1 − 2 4 −1 − 8 = − 9.

2. If A =

LM1 2OP and B = LM 2 1OP , then find AB . N0 1 Q N−1 1Q L 1 2OP LM 2 AB = M N0 1Q N−1 L1a2f + 2 a−1f =M N0 a2f + 1a−1f  →

Solution:

OP B Q 1a1f + 2 a1fO P 0 a1f + 1 a1fQ

1 1


LM2 − 2 1 + 2OP = LM 0 3OP N 0 − 1 0 + 1Q N−1 1Q AB = 0 − a −3f = 3.

AB =

OR AB = A ⋅ B =

1 0

2 2 ⋅ 1 −1

1 1

af a f a f a1fa2 + 1f = 1a3f = 3. 1O 1 P is singular P 6 PQ = 1 1 − 2 0 ⋅ 2 − −1

LM1 2 3. Find x if the matrix 0 MM1 2x N LM1 2 1OP Given 0 MM1 2x 61PP is singular. N Q

1 0 1

∴

1

x 2

a

2 x 2

1 0 −2 6 1

1 1 = 0. 6 1 0 +1 6 1

x =0 2

f a f a f 6 x − 2 − 2 a −1f + 1 a − x f = 0

1 6x − 2 − 2 0 − 1 + 1 0 − x = 0

6 x − 2/ + 2/ − x = 0 5x = 0 ⇒

1 4. If 1 0

2 3 6

2 x = 2, then find x. 2

x=0 .

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1 1 0

Given

1

3 6

2 3 6

1 x −2 2 0

a

f a

2 x =2 2 1 x +2 2 0

f af

3 =2 6

1 6 − 6x − 2 2 − 0 + 2 6 = 2

af

6 − 6 x − 2 2 + 12 = 2 6 − 6 x − 4 + 12 = 2 −6 x + 14 = 2 −6 x = 2 − 14 ⇒

x 5. Solve for x : 2 −1

2 5 2

x=2 .

−1 x = 0. x

a

f a f a f x a3 x f − 2 a 3 x f − a 9 f = 0

x 5x − 2 x − 2 2 x + x − 1 4 + 5 = 0

3x 2 − 6 x − 9 = 0

3x 2 − 9 x + 3x − 9 = 0

a f a f a3x + 3fa x − 3f = 0

3x x − 3 + 3 x − 3 = 0

3x + 3 = 0

or

x −3= 0

or

x = +3.

3 x = −3

x = −1 Hence

x = −1 or

< +3 x −9 x

27 x 2

−6 x

x = 3.

6.7 PROPERTIES OF DETERMINANTS: 1. Prove that the value of a determinant remains same when its rows and columns are interchanged i.e., prove that A = A′ .

LMa A= b MMc N

1

Pr oof: Let Proof:

1

1

a2 b2 c2

a3 b3 c3

OP PP Q


LM MM N

a1 A ′ = a2 a3

Then

b1 b2 b3

c1 c2 c3

121

OP PP Q

A = A′ .

To prove:

a1 A = b1 c1

Consider

= a1

b

b2 c2

a2 b2 c2

b3 b − a2 1 c3 c1

g

a3 b3 c3 b3 b + a3 1 c3 c1

b

g b

b2 c2

a1 b2 c3 − b3c2 − a2 b1c3 − b3 c1 + a3 b1c2 − b2 c1

g

a1b2 c3 − a1b3c2 − a2 b1c3 + a2 b3c1 + a3 b1c2 − a3 b2 c1

...(1)

Now

a1 A ′ = a2 a3

b1 b2 b3

c1 c2 = a1 b2 c3 − b3c2 − b1 a2 c3 − a3 c2 + c1 a2 b3 − a3 b2 c3

a1 A ′ = a2 a3

b1 b2 b3

c1 c2 = a1b2 c3 − a1b3c2 − a2 b1c3 + a3 b1c2 + a2 b3 c1 − a3 b2 c1 c3

b

g b

g b

g

From (1) and (2) |A| = |A′|. Hence proved. 2. Prove that a determinant changes its sign when 2 of its rows are interchanged.

LMa A= b MMc N

1

Pr oof: Let Proof:

1

1

LM MM N

b1 B = a1 c1

b2 a2 c2

b3 a3 c3

To prove: A = − B . Now

a2 b2 c2

a3 b3 c3

OP PP Q

OP PP is the matrix obtained by interchanging 1st and 2nd rows. Q

...(2)

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Basic Mathematics

a1 A = b1 c1

b

g

a2 b2 c2

a3 b3 c3

b

g b

= a1 b2 c3 − b3 c2 − a2 b1c3 − b3c1 + a3 b1c2 − b2 c1

g

= a1b2 c3 − a1b3c2 − a2 b1c3 + a2 b3c1 + a3 b1c2 − a3 b2 c1

b1 B = a1 c1

b2 a2 c2

...(1)

b3 a3 = b1 a2 c3 − a3 c2 − b2 a1c3 − a3 c1 + b3 a1c2 − a2 c1 c3

b

g b

g b

g

= a2 b1c3 − a3 b1c2 − a1b2 c3 + a3 b2 c1 + a1b3c2 − a2 b3c1

= − − a2 b1c3 + a3 b1c2 + a1b2 c3 − a3 b2 c1 − a1b3 c2 + a2 b3 c1

...(2)

From (1) and (2) B =−A

⇒

A =−B .

Hence proved. 3. Prove that the value of a determinant is zero if any two of its rows are identical.

a1 Pr oof: Let A = a1 Proof: b1

a2 a2 b2

a3 a3 be the determinant whose 2 rows, (1st and 2nd) are identical. b3

To prove A = 0

Now

a1 A = a1 b1

a2 a2 b2

a3 a3 = a1 a2 b3 − a3 b2 − a2 a1b3 − a3b1 + a3 a1b2 − a2 b1 b3

b

g

b

g

b

g

= a1a2 b3 − a1a3 b2 − a1a2 b3 + a2 a3 b1 + a1a3 b2 − a2 a3 b1

= 0. Hence proved. 4. If every element of any row of a determinant is multiplied by a non zero constant k then prove that the whole determinant is multiplied by k.

LMa A= b MMc N

1

Pr oof: Let Proof:

1 1

a2 b2 c2

OP PP Q

LM MM N

a3 a1 b3 and B = b1 c3 kc1

a2 b2 kc2

a3 b3 kc3

OP PP Q


123

which is obtained by multiplying 3rd row of A by k. To prove B = kA.

a1 Now A = b1 c1

a2 b2 c2

a1 B = b1 kc1

a3 b3 = a1 b2 c3 − b3c2 − a2 b1c3 − b3c1 + a3 b1c2 − b2 c1 c3

b

a2 b2 kc2

b

g

b

g b

g

...(1)

a3 b3 = a1 kb2 c3 − kb3c2 − a2 kb1c3 − kb3 c1 + a3 kb1c2 − kb2 c1 kc3

g

b

g

b

b

g b

g b

= k a1 b2 c3 − b3 c2 − a2 b1c3 − b3 c1 + a3 b1c2 − b2 c1

af

g

g

B = k . A from 1

Hence proved. 5. If each element of any row of a determinant is the sum of two terms, then prove that the determinants can be expressed as the sum of two determinants i.e.,

a1 + x b1 Prove that c1

a2 + y b2 c2

a3 + y a1 b3 = b1 c3 c1

Pr oof: L.H.S.: Proof:

b

= a1 + x

g bc

2

2

= a1

b2 c2

b

a2 b2 c2

a3 x b3 + b1 c3 c1

a1 + x b1

a2 + y b2

a3 + z b3

c1

c2

c3

g

b3 b − a2 + y 1 c3 c1

b

g

b3 b + a3 + z 1 c3 c1

RS T

b3 b2 +x c3 c2

b3 b1 − a2 c3 c1

b3 b1 − a2 c3 c1

b3 b1 + a3 c3 c1

b3 b1 +y c3 c1

y b2 c2

z b3 c3

b2 c2

UV W

b3 b1 + a3 c3 c1

b2 b1 +z c2 c1

c3 b1 −y c3 c1

b3 b1 +z c3 c1

b2 c2

Rearranging,

= a1 a1 = b1 c1

b2 c2 a2 b2 c2

Hence proved.

a3 x b3 + b1 c3 c1

y b2 c2

b2 b2 +x c2 c2

z b3 = RHS. c3

b2 c2

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Basic Mathematics

6. Prove that the value of a determinant is not altered if to the elements of any row the same multiples of the corresponding elements of any other row are added.

a1 i.e., prove that b1 c1

a3 a1 + kb1 b3 = b1 c3 c1

a2 b2 c2

a1 + kb1 b1 c1

Pr oof: R.H.S.: Proof:

a1 = b1 c1

a2 b2 c2

a1 = b1 c1 a1 = b1 c1

a2 + kb2 b2 c2

a3 kb1 b3 + b1 c3 c1

a2 b2 c2

a3 b1 b3 + k b1 c3 c1 a2 b2 c2

a2 + kb2 b2 c2

a3 + kb3 here R1′ = R1 + kR2 b3 c3

b

g

a3 + kb3 b3 c3

kb2 b2 c2

kb3 b3 using property 5 . c3

b2 b2 c2

b3 b3 using property 4 . c3

a

f

a

f

a3 b3 + k 0 3 2 rows are identical . c3

a fa

a1 = b1 c1

a2 b2 c2

f

a3 b3 = L.H.S. c3

Hence proved. 7. Prove that if all the elements in a row of a determinant is zero then the determinant value is zero. a1 Pr oof: Let A = b1 Proof: 0

a2 b2 0

a3 b3 whose 3rd row is zero row. 0

A = a1

b2 0

af

b3 b − a2 1 0 0

af

af

b3 b + a3 1 0 0

b2 0

= a1 0 − a2 0 + a3 0 = 0. Hence proved.


125

8. In a determinant if all the elements on one side of the principal diagonal are zeros, then prove that the value of the determinant is equal to the product of the elements in the principal diagonal.

a1 0 A = Pr oof: Let Proof: 0

a2 b1 0

a3 b2 be the determinant in which all the elements to the left of princ1

cipal diagonal are zeros. A = a1b1c1

To prove:

Consider

a1 0 0

a2 b1 0

a3 b1 b2 = a1 0 c1

b

b2 c1

− a2

0 0

b2 c1

+ a3

0 0

b1 0

af af

g

= a1 b1c1 − 0 − a2 0 + a3 0 = a1b1c1 . Hence proved.

Note: The above properties which are proved for the rows also holds good for columns because |A| = |A′| from property 1. So • The determinant changes its sign when 2 of its columns are interchanged i.e.,

a1 b1 c1

a2 b2 c2

a3 a2 b3 = − b2 c3 c2

a1 b1 c1

a3 b3 c3

a1 • The value of the determinant is zero when 2 of its columns are identical i.e., b1 c1

a1 b1 c1

a3 b3 = 0 . c3

• If every element of any column of a determinant is multiplied by constant k. Then the whole

ka1 determinant is multiplied by k i.e., kb1 kc1

a2 b2 c2

a3 a1 b3 = k b1 c3 c1

a2 b2 c2

a3 b3 c3

• If each element of any column of a determinant is sum of two terms then the determinant can be expressed as the sum of 2 determinants.

i.e.,

a1 + x b1 + y c1 + z

a2 b2 c2

a3 a1 b3 = b1 c3 c1

a2 b2 c2

a3 x b3 + y c3 z

a2 b2 c2

a3 b3 c3

• The value of a determinant is not altered if to the elements of any column the same multiples of the corresponding element of any other column are added.

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i.e.,

a1 b1

a2 b2

c1

c2

a2 + ka3 b2 + kb3 c2 + kc3

a3 a1 b3 = b1 c3 c1

a3 b3 c3

c2′ = c2 + kc3

• If all the elements in a column of a determinant are zero then the determinant is zero.

a1 b1 c1

i.e.,

0 0 0

a2 b2 = 0 c2

WORKED EXAMPLES: 3860 1. Find the value of: 3862

3860 3862

3861 3863

3861 R′ = R2 − R1 Using property 6 with k = −1 3863 2 =

3860 2

3861 3860 =2 2 1

3861 1

a f

= 2 3860 − 3861 = 2 −1 = −2 81 2. Evaluate: 84 87

82 85 88

83 86 89

Solution: 81 84 87

81 = 3 3

82 3 3

82 85 88

83 86 89

R2′ = R2 − R1 R3′ = R3 − R2

83 3 = 0 3 2 rows are identical 3

a

x−y 3. Without expanding prove that y − z z−x

y−z z−x x−y

z−x x−y =0 y−z

f


x−y y−z z−x

LHS:

0 y−z z−x

99 4. Evaluate: 100 101

101 102 103

y−z z−x x−y 0 z−x x−y

z−x x − y R1′ = R1 + R2 + R3 y− z 0 x−y=0 y−z

3 1st row is zero.

104 105 106 R2′ = R2 − R1

R3′ = R3 − R1 99 = 1 2

101 1 2

104 99 1 =2 1 2 1

101 1 1

104 1 = 0 Since 2 rows are equal. 1

5. Solve for x:

x+2 2 2

3 x +3 3

4 4 =0 x+4

Taking C1′ = C1 + C2 + C3 x+9 x+9 x+9

3 x+3 3

4 4 =0 x+4

Taking x + 9 common

a

1 x+9 1 1

f

3 x +3 3

x+9=0 ⇒ or x = −9

1 1 1

4 4 =0 x+4

3 x +3 3

R2′ = R2 − R1

4 4 =0 x+4

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Basic Mathematics

and R3′ = R3 − R1

1 0 0

3 x

4 0 =0 x

0

⇒ 1⋅ x ⋅ x = 0 x2 = 0

⇒ x=0 Hence x = −9 or x = 0. 1+ x 6. Prove that x x

y 1+ y y

z z =1+ x + y + z 1+ z

1+ x x x

LHS:

y 1+ y y

z z 1+ z

1+ x + y + z = 1+ x + y + z 1+ x + y + z

1 1+ x + y + z 1 1

a

f

C1′ = C1 + C2 + C3

y 1+ y y

y 1+ y y

1 1+ x + y + z 0 0

a

f

z z 1+ z y 1 0

z z 1+ z

R2′ = R2 − R1 R3′ = R3 − R1 z 0 1

a1 + x + y + zfa1fa1fa1f = 1 + x + y + z = R.H.S. 1 7. Prove that 1 1

a b c

a2 b2 = a − b b − c c − a c2

a

fa fa f


1 1 1

LHS:

a b−c c−a

1 =0 0

a2 b2 c2

a b c

R2′ = R2 − R3 R3′ = R3 − R1

a2 1 2 b −c = 0 c2 − a2 0

a b−c c−a

2

a2 b−c b+c c−a c+a

a fa f a fa f

Taking (b − c) and (c − a) common from R2 and R3

1 b−c c−a 0 0

a fa f

a 1 1

a2 b+c c+a

R3′ = R3 − R2

a2 b+c a−b

a 1 0

1 b−c c−a 0 0

a fa f Taking (a − b) common from R3

a

a 1 0

1 a−b b−c c−a 0 0

fa fa f

a2 b+c 1

aa − bfab − cfac − af a1f a1f a1f = aa − b fa b − cfac − a f = RHS. Hence proved. x 8. Prove that p p

p x q

q q = x − p x −q x + p+q x

a

fa

x p p

LHS: x+ p+q = x+ p+q x+ p+q

p x q

fa

p x q

f

q q C1′ = C1 + C2 + C3 x

1 q q = x+ p+q 1 1 x

a

f

p x q

q q x

R2′ = R2 − R1 R3′ = R3 − R2

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Basic Mathematics

1 = x+ p+q 0 0

p x−p q−x

1 = x + p+q 0 0

p x− p − x−q

a

f

a

f

q 0 x−q q 0 x−q

a

f a

f

q 0 1

R3′ = R3 + R2

Taking (x − p) common from R2 and (x − q) from R3

a

1 x− p x−q x+ p+q 0 0

fa

fa

p 1 −1

f

1 = x− p x−q x+ p+q 0 0

a

fa

fa

f

p 1 0

q 0 1

a fa fa f = a x − pfa x − qfa x + p + qf = RHS. = x − p x − q x + p + q 1⋅ 1⋅1

9. Prove that

1 x

1 y

x3

y3

1 z = x−y y−z z−x x+y+z

a

fa fa fa

1 x

1 y

1 z

x3

y3

z3

z3

LHS:

0 = x−y x 3 − y3

0 y−z y3 − z3

0 = x−y x − y x 2 + xy + y 2

a

fd

a

fa f

= x−y y−z

f

C1′ = C1 − C2 C2′ = C2 − C3

1 z z3

i a

0 1 2 x + xy + y 2

0 y−z x − y y 2 + yz + z 2

fd

0 1 2 y + yz + z 2

i 1 z z3

1 z z3


Expanding

a x − yfay − zf 0 − 0 + 1d y + yz + z i − d x + xy + y i a x − yfa y − zf y + yz + z − x − xy − y a x − yfa y − zf az − xfa z + xf + y a z − xf a x − yfa y − zf a z − x fa x + y + zf = a x − yfa y − zfa z − x fa x + y + z f . 2

2

2

2

2

2

2

2

Hence proved.

10. Prove that

a + b + 2c c c

LHS:

a b + c + 2a a

a + b + 2c c c

b b =2 a+b+c c + a + 2b

a

a b + c + 2a a

f

3

b b C1′ = C1 + C2 + C3 c + a + 2b

2a + 2b + 2c = 2a + 2b + 2c 2a + 2b + 2c

a b + c + 2a a

b b c + a + 2b

1 =2 a+b+c 1 1

a b + c + 2a a

b b c + a + 2b

a

f

1 =2 a+b+c 0 0

a

f

a

fa fa

a a+b+c 0

b 0 a+b+c

fa

=2 a+b+c 1 a+b+c a+b+c

= 2 (a + b +

− a2 11. Prove that ab ac

LHS:

ab −b2 bc

c)3.

Hence proved.

ac bc = 4a 2 b 2 c 2 −c 2 −a 2 ab ac

ab −b2 bc

ac bc −c2

f

R2′ = R2 − R1 R3′ = R3 − R1

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Taking a, b, c common from C1, C2 and C3 respectively we get

−a abc b c

a −b c

a b −c

Taking a, b, c common from R1, R2 and R3 respectively

−1 1 1

aabcfaabcf ⋅

1 −1 1

1 R′ = R2 + R1 1 2 R ′ = R3 + R1 −1 3

−1 a b c ⋅ 0 0

1 0 2

2 2 2

Expanding

1 2 0

a f af af a4f = 4a b c = RHS.

a 2 b 2 c 2 −1 0 − 4 − 1 0 + 1 0 a2b 2c2

12. Prove that

LHS:

1+ a 1 1

1 1+ b 1

2 2 2

F H

1 1 1 1 1 = abc 1 + + + a b c 1+ c 1+ a 1 1

1 1+ b 1

I K

1 1 1+ c

Taking a, b, c common from R1, R2 and R3 respectively we get

1+ a a 1 abc ⋅ b 1 c

1 a 1+ b b 1 c

1 a 1 b 1+ c c

1 a

1 a 1 1+ b 1 c

1 a 1 b 1 1+ c

1+ abc ⋅

1 b 1 c


1+ abc ⋅

1 1 1 + + a b c 1 b 1 c

1+

1 1 1 + + a b c 1 1+ b 1 c

aabcf FH1 + 1a + b1 + 1c IK

1+

1

1

1 b

1+

1 c

1 c

1 1 1 + + a b c 1 b 1 1+ c

133

R1′ = R1 + R2 + R3

1 1 b

aabcf FH1 + a1 + b1 + 1c IK

C2′ = C2 − C1 C3′ = C3 − C1

1 b 1+

1 c

1 1 b 1 c

0

0

1

0

0

1

aabcf FH1 + a1 + b1 + 1c IK ⋅1 ⋅ 1 ⋅1 1 1 1 = a abcf F 1 + + + I Hence proved. H a b cK a 2 13. Prove that a bc

b b2 ca

c c 2 = ab + bc + ca b − c c − a a − b . ab

a

fa fa fa

a a2 bc

LHS:

b b2 ca

c c2 ab

a−b a2 − b2 bc − ca

a

a−b a−b a+b −c a − b

a

fa

f

f

C1′ = C1 − C2 C2′ = C2 − C3

b−c b2 − c2 ca − ab

c c2 ab

b−c b−c b+c −a b − c

f a fa f a f

c c2 ab

Taking (a − b) common from C1, (b − c) common from C2

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1 = a−b b−c a+b −c

1 b+c −a

c c 2 C1′ = C1 − C2 ab

0 = a−b b−c a−c a−c

1 b+c −a

c c2 ab

a

a

fa f

fa f

Taking (c − a) common from C1

c c 2 R2′ = R2 − R3 ab

0 = a − b b − c c − a −1 −1

1 b+c −a

0 = a−b b−c c−a 0 −1

1 a+b+c −a

a a

fa fa f

fa fa f

c c 2 − ab ab

Expanding

a fa fa f d i a = aa − bfa b − cfac − af − c/ + ab + ac + bc + c/ aa − bfab − cfac − afaab + bc + caf. Hence proved.

= a − b b − c c − a 0 − 1 0 + c 2 − ab + c 0 + a + b + c 2

1 2 14. Prove that x x

LHS:

x 1 x2

2

x2 2 x = 1 − x3 . 1

d

i

x 1 x2

1 x2 x

x2 x C1′ = C1 + C2 + C3 1

1 + x + x2 = 1 + x + x2 1 + x + x2

d

= 1+ x + x

2

i

x 1 x2 1 1 1

x 1 x2

x2 x 1 x2 x 1

f


i

1

x

x2

= 1 + x + x2 0 0

1− x x2 − x

x − x2

i

1 0 0

x 1− x −x 1 − x

d

d

= 1 + x + x2

135

R2′ = R2 − R1 R3′ = R3 − R1

1 − x2 x2 x 1− x 1 − x2

a f a f d i Taking (1 − x) common from R2, (1 − x) from R3

a fa f

3 1 − x 2 = 12 − x 2 = 1 − x 1 + x

d

= 1 + x + x2

x2 x 1+ x

x 1 −x

1 1− x ⋅ 1− x 0 0

ia f a f

expanding, we get

d i a f d1 + x + x i = d1 + x + x i a1 − x f ⋅ d1 + x + x i d1 + x + x i a1 − xf = d1 − x i = RHS. = 1 + x + x2 1 − x 2

2

2

2

2

2 2

2

a fd

i

3 1 − x 1 + x + x 2 = 13 − x 3 = 1 − x 3

3 2

Hence proved.

Given

⇒

x y z

1 + x3 1 + y 3 = 0, then prove that 1 +xyz = 0. 1 + z3

x2 y2 z2

x 15. If x, y, z are all different and y z x y z

x2 y2 z2

x2 y2 z2

1 x 1+ y 1 z

1 + x3 1 + y 3 = 0. 1 + z3 x2 y2 z2

x3 y3 = 0 z3

(using property of determinants)

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x2 y2 z2

x y z

⇒

x2 y2 z2

x y z

⇒

1 1 1 + xyz 1 1 1 x2 y2 z2

1 x 1 + xyz y 1 z

1 1=0 1

x 1 + xyz y z

a

⇒

f

x2 y2 = 0 z2

x y z

LMBy interchanging R and R OP N Then R and R Q 1

2

x2 y2 z2

3

1 1=0 1

x 1 + xyz = 0. 3 x, y and z are all different, y z

⇒

2

x2 y2 z2

1 1≠0. 1

Hence proved.

6.8 MINOR, CO-FACTOR, ADJOINT AND INVERSE OF A SQUARE MATRIX: Let [aij] be a square matrix of order n × n. Then minor of an element aij is the determinant obtained by deleting the row and the column containing it (i.e. ith row & jth column). If minors are multiplied by (−1)i + j i.e., with proper signs + or − we get co-factors. Adjoint of a matrix is the transpose of co-factor matrix. Illustration: 1. If A =

LM1 N3

minor minor minor minor

of of of of

OP Q

2 , then 4 1 2 3 4

= = = =

4 3 2 1

Co-factor matrix =

Adjoint of A = columns.

LM 4 N −3

LM 4 N−2

OP Q

LM N

+ −3 Multiplying by 1 −

OP Q

− +

OP Q

−2 Taking transpose of co-factor matrix i.e., interchanging rows and 1


LM MM N

1 2. If A = 2 3

OP PP Q

2 3 1

3 1 , then 2

3 minor of 1 = 1

minor of 2 =

2 3

2 minor of 3 = 3

LM MM N L1 1 = 4 − 3 = 1. M2 MM3 2 N LM1 3 = 2 − 9 = −7. 2 MM3 1 N

1 1 = 6 − 1 = 5. 2 2 3

Similarly minor of 2 :

2 1

OP PP Q 3O 1P P 2PQ

2 3 1

3 1 2

2 3 1 2 3 1

3 1 2

LM MN

OP PP Q

1 3 = 4 − 3 = 1. 2 2 3

minor of 3 :

1 3

3 = 2 − 9 = −7. 2

minor of 1 =

1 3

2 = 1 − 6 = −5. 1

minor of 3 =

2 3

1 3 = 2 − 9 = −7. 2 1 3

minor of 1 =

1 2

3 = 1 − 6 = −5. 1

minor of 2 =

1 2

2 = 3 − 4 = −1. 3

LM 5 Hence matrix of minors = MM−71 N

LM MM N

2 3 1

1 −7 −5

−7 −5 −1

2 3 1

3 1 2

OP PP Q

OP PP Q

3 1 2

OP PQ

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LM+ Multiplying with − MM+ N

− + −

OP PP Q

+ − the corresponding elements +

LM MN

5 We get co-factor matrix = −1 −7

OP PQ

−1 −7 +5

−7 +5 −1

Taking transpose, we get Adjoint matrix

LM 5 ∴ Adjoint of A = −1 MM−7 N

−1 −7 5

OP PP Q

−7 5 −1

INVERSE OF A SQUARE MATRIX: If A is a non-singular square matrix of order n × n then there exists a square matrix B of order n × n such that AB = BA = I where I is the identity matrix of order n × n. Here B is called inverse of A. It is denoted by A−1. ∴

A ⋅ A −1 = A − 1 ⋅ A = I Inverse of A can be found by using the formula

i.e.,

A −1 =

adjoint of A . A

A −1 =

adj A . A

Illustration: 1. To find inverse of a matrix of order 2:

Let

LM 1 −1OP N2 2 Q L 1 −1OP A=M N2 2Q A = 1 a2 f − a −1fa 2f = 2 + 2 = 4 ≠ 0.

∴

A is non-singular. Hence A−1 exists. A −1 =

adj A formula A


LM 1 N2

Now minor of 1 = 2 minor of −1 = 2 minor of 2 = −1 minor of 2 = 1 ∴ Co-factor matrix =

LM2 N1

OP Q

−1 2

OP Q

LM N

−2 + . By multiplying minors with 1 −

Taking transpose we get

139

− +

OP Q

LM 2 1OP N−2 1Q L 1 1 OP 1O M 2 4 . = 1PQ M 1 MN− 2 14 PPQ

Adjoint of A =

A −1 =

∴

LM N

2 1 ⋅ 4 −2

Note: We can write adjoint of a square matrix of order 2 × 2 directly by interchanging the principal diagonal elements and changing the signs of secondary diagonal elements. Examples:

LM 1 7OP , then N2 3Q L 3 −7OP interchange 1 and 3, change the signs of 2 and 7. Adj A = M N−2 1Q L 1 −7OP, then adj A = LM 6 7OP (ii) If A = M N2 6 Q N−2 1Q L −1 5OP, then adj A = LM7 −5OP (iii) If A = M N −2 7Q N2 −1Q L 0 −5OP , then adj A = LM−2 5OP (iv) If A = M N−6 −2Q N 6 0Q L −1 −6OP , then adj A = LM−7 6OP (v) If A = M N −8 −7Q N 8 −1Q (i) If A =

2. To find inverse of a square matrix of order 3:

Let

LM1 A= 1 MM2 N

3 2 0

2 2 0

OP PP Q

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1 A=1 2

3 2 0

2 2 2 =1 0 0

2 1 −3 0 2

1 2 +2 0 2

af a f a f = 1 a0 f − 3 a0 − 4 f + 2 a 0 − 4 f =1 0 −3 0− 4 +2 0−4

0 + 12 − 8 = 4 ≠ 0

Hence A−1 exists.

A −1 =

LM1 A= 1 MM2 N

Now

minor of 1 =

2 0

2 =0 0

1 minor of 3 = 2

2 = 0 − 4 = −4 0

1 minor of 2 = 2

2 = 0 − 4 = −4 0

3 Similarly minor of 1 = 0

2 =0 0

1 2

2 = 0 − 4 = −4 0

1 minor of 2 = 2

3 = 0 − 6 = −6 0

minor of 2 =

Similarly minor of 2 =

3 2

2 =6−4=2 2

minor of 0 =

1 1

2 =2−2=0 2

minor of 0 =

1 1

3 = 2 − 3 = −1 2

adj A . A 3 2 0

2 2 0

OP PP Q

2 0


LM0 ∴Matrix of minors = 0 MM2 N We get

−4 −4 0

LM MM N

− + −

4 −4 0

−4 6 −1

OP PP Q

+ −4 −6 Multiplying with − + −1

LM0 Co-factors matrix = 0 MM2 N

+ − +

141

OP PP Q

OP PP Q

Taking transpose we get

LM 0 Adj A = 4 MM−4 N A −1 =

∴

A

OP PP Q

0 −4 6

2 0 −1

Adj A 1 = 4 A

LM 0 MM−44 N

0 −4 6

LM 0 = 1 MM−1 N

0 −1 32

12 0 −1 4

−1

OP PP Q

2 0 −1

OP PP Q

Note: We can write the adjoint of a square matrix directly by writing the minors multiplied by proper signs (+ or −) column wise.

LM 1 If A = M 1 MN 2

i.e.

Then

LM MM Adj A = M MM MN

2 0 1 − 2 1 2

2 0 2 0 2 0

LM 0 Adj A = 4 MM−4 N

3 2 0

3 0 1 2 1 − 2 −

0 −4 6

OP PP Q

2 2 0

2 0 2 0 3 0

3 2 1 − 1 1 1

OP PP Q

2 0 −1

2 2 2 2 3 2

OP PP PP PP Q

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WORKED EXAMPLES: 1. Find A−1 if A =

LM 1 N2

7 −6

OP Q A −1 =

Solution:

a

adj A formula A

L −6 adj A = M N −2

Now

A=

1 2

A −1

LM 6 = M 20 MN 202

OP Q

−7 1

F By interchanging principal diagonalI GG elements and changing signs of JJ H secondary diagonal elements. K

7 = −6 − 14 = −20 −6

A −1 =

∴

f

LM −7OP 1Q N 7O L3 7O M P 20 = 10 20 P 1P M1 1 P. − P M − P 20 Q N10 20 Q

1 −6 −20 −2

LM1 2OP N3 −1Q L1 2OP A=M N3 −1Q L−1 −2OP adj A = M N−3 1Q

2. Verify A. adj A = |A|.I for the matrix A =

Given:

Hence

L1 2OP = LM−1 −2OP B A ⋅ adj A = M N3 −1Q N−3 1Q L 1a−1f + 2 a−3f 1a−2f + 2 a1fOP =M N3a−1f + a−1fa−3f 3a−2f + a−1fa1fQ L−1 − 6 −2 + 2OP = LM−7 0OP =M N−3 + 3 −6 − 1Q N 0 −7Q  →

Consider


LM 1 0OP = −7I MN0 1PQ 2 = 1a−1f − 3 a2 f = −1 − 6 = −7. −1

A ⋅ adj A = −7

A=

But

1 3

∴

3. Find

A.adjA = |A|.I. A−1

Solution:

Now

LM 1 if A = 0 MM 1 N

2 −1 1

OP PP Q

3 1 2

A −1 =

adj A A

1

2 −1 1

A= 0 1

3 1 2

= 1 (−2 −1) − 2 (0 − 1) + 3 (0 + 1) = −3 + 2 + 3 = 2 ≠ 0. Hence A−1 exists.

Now

LM 1 A= 0 MM 1 N

2 −1 1

OP PP Q

3 1 2

LM −1 1 − 2 3 2 MM 01 21 11 23 −11 Adj A = − MM 1 2 1 2 − 0 MM 01 −11 − 11 21 01 N LM−2 − 1 −a4 − 3f 2 + 3OP = −a−1f MM 0 + 1 −a12−−23f −−11PP N Q LM−3 −1 5OP Adj A = 1 −1 −1 MM 1 1 −1PP N Q

3 1 3 1 2 −1

OP PP PP PP Q

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A

∴

adj A 1 = = 2 A

−1

4. Verify AA −1 = A −1 A = I if A =

LM 1 2OP . N−1 3Q A=

Solution: Given A=

1 −1

A −1 =

∴

LM−3 MN 11

OP PQ

−1 −1 1

5 −1 −1

LM 1 2OP N−1 3Q 2 = 3 − a−2 f = 5. 3 LM N

adj A 1 3 = 5 1 A

OP Q

−2 1

1 2O 1 L3 −2 O 1 3+2 = LM = LM B P M P 1 3 1 1 − N Q 5 N Q 5 N−3 + 3 1 L 5 0 O L 1 0O = =I AA = M 5PQ MN 0 1PQ 5 N0  →

Now

AA

−1

OP Q

−2 + 2 2+3

−1

Similarly,

A

−1

LM N

  →

1 3 ⋅A= 5 1

OP LM QN

1 −2 ⋅ 1 −1

LM a f 6 − 6OP N a f 2 + 3Q 1 L 5 0 O L 1 0O = M = =I 5PQ MN0 1PQ 5 N0 =

OP B Q

2 3

1 3 − −2 5 1 + −1

AA−1 = A−1A = I

Hence

6.9 CHARACTERISTIC EQUATION OF A SQUARE MATRIX: If A is any square matrix of order n × n and I is identity matrix of the same order then |A − λI| = 0 where λ is a constant is called characteristic equation of a square matrix A. The roots of the equation |A − λI| = 0 (i.e., value of λ which satisfies this equation) are called characteristic roots or eigen values. Examples: 1. If A =

LM1 N2

OP Q

4 then 3


A − λI =

=

OP LM 0OP Q N 1Q 0 O L1 − λ 4 PλQ = MN 2 3 − λOPQ

LM1 N2

4 1 −λ 3 0

LM1 4OP − LMλ N2 3 Q N 0

A − λI =

1− λ 2

4 3− λ

Characteristic equation A − λI = 0

1− λ 2

∴

4 =0 3−λ

a1 − λf a3 − λf − 8 = 0 3 − 3λ − λ + λ2 − 8 = 0

λ2 − 4λ − 5 = 0. This is the characteristic equation.

a f a f aλ − 5faλ + 1f = 0

λ λ − 5 +1 λ − 5 = 0

λ = 5 or λ = −1.

Hence the characteristic roots (or eigen values) are 5 and −1.

LM 4 A = −2 MM−2 N

2. If

Then

4−λ A − λI = 0 ⇒ −2 −2

a4 − λf 1 −0 λ

0 1 0

OP PP Q

1 0 1

0 1− λ 0

0 −2 − 0 +1 1− λ −2

1 0 = 0. 1− λ 1− λ =0 0

a4 − λfa1 − λfa1 − λf + 0 + 2 a1 − λf = 0 a1 − λf a4 − λfa1 − λf + 2 = 0 1 − λ = 0 or a4 − λ fa1 − λ f + 2 = 0

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λ = 1 or 4 − λ − 4λ + λ2 + 2 = 0 λ2 − 5λ + 6 = 0 λ2 − 3λ − 2 λ + 6 = 0

a f a f aλ − 3faλ − 2f = 0

λ λ −3 −2 λ−3 = 0

λ = 1 or λ = 3 or λ = 2.

∴ Characteristic roots or eigen values are 1, 2 and 3.

6.10 CAYLEY HAMILTON THEOREM: Statement: Every square matrix A satisfies its characteristic equation, A − λI = 0.

Illustration: If A =

LM1 N2

2 3

OP Q

Then characteristic equation is A − λI = 0.

1− λ 2

i.e.,

2 =0 3−λ

a1 − λfa3 − λf − 4 = 0 3 − 3λ − λ + λ2 − 4 = 0 λ2 − 4 λ − 1 = 0.

This is characteristic equation. 2 By Cayley Hamilton theorem ‘A’ satisfies this equation. So A − 4 A − I = 0 where I is identity matrix of same order as that of A. Now to verify Cayley Hamilton theorem, We have to verify A2 − 4A − I = 0.

OP LM1 2OP B Q N2 3 Q L1 + 4 2 + 6OP = LM5 8 OP A =M N2 + 6 4 + 9Q N8 13Q L1 2OP = LM4 8 OP 4A = 4 M N2 3Q N8 12Q L1 =M N2

→

Now

A

2

2

2 3

...(i)


LM1 0OP N0 1 Q L5 8 OP − LM4 8 OP − LM1 0OP − 4A − I = M N8 13Q N8 12Q N0 1Q L5 − 4 − 1 8 − 8 − 0 OP = LM0 0OP . =M N8 − 8 − 0 13 − 12 − 1Q N0 0Q I=

A2

Hence verified. Note: From (1)

A2 − 4 A − I = 0.

Operating by A−1 A −1 ⋅ A 2 − 4 A −1 ⋅ A − A −1 I = 0.

A − 4 I − A −1 = 0 ⇒

A −1 = A − 4 I .

OP LM 0OP Q N 1Q 2O L4 0 O − 3 PQ MN 0 4 PQ 2O −1PQ

LM1 N2 L1 =M N2 L−3 =M N2

2 1 −4 3 0

A −1 = A −1

A −1

Hence we can find inverse of a square matrix by using Cayley Hamilton theorem.

WORKED EXAMPLES: 1. Find the eigen values of the matrix

LM1 4OP . N3 2 Q

Solution: Characteristic equation is

A − λI = 0 1− λ 3

4 =0 2−λ

a1 − λfa2 − λf − 4 a3f = 0 2 − 2 λ − λ + λ2 − 12 = 0

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λ2 − 3λ − 10 = 0 λ2 − 5λ + 2 λ − 10 = 0

a

f a

f

λ λ −5 +2 λ−5 = 0 λ = 5 or λ = −2.

Hence the eigen values are λ = 5 and λ = −2. 2. Find the characteristic roots of the matrix

LM1 N3

OP Q

1 . −1

Solution: Characteristic equation is A − λI = 0.

1− λ 3

1 = 0. −1 − λ

a1 − λfa−1 − λf − 3 = 0 −a1 − λ fa1 + λ f − 3 = 0 −d1 − λ i − 3 = 0 2

2

−1 + λ2 − 3 = 0

λ2 − 4 = 0 ⇒ λ2 = 4 ⇒ λ = ±2. Hence characteristic roots are +2 and −2. 3. Verify Cayley Hamilton theorem for the matrix

LM 1 N2

OP Q

−1 . 6

Solution: Characteristic equation is A − λI = 0

1− λ 2

−1 = 0. 6−λ

a1 − λfa6 − λf − 2 a−1f = 0 6 − 6λ − λ + λ2 + 2 = 0 λ2 − 7λ + 8 = 0 .

This is characteristic equation. By Cayley Hamilton theorem every square matrix obeys its characteristic equation. Hence it is required to verify A2 − 7A + 8I = 0


OP LM 1 −1OP B Q N2 6 Q LM 1 − 2 −1 − 6 OP = LM−1 −7OP N2 + 12 −2 + 36Q N14 34Q 1 −1O L 7 −7O 7 A = 7 LM N2 6QP = NM14 42QP L 1 0OP = LM8 0OP 8I = 8 M N0 1Q N0 8Q L1 = A⋅ A = M N2

→

Now

A

2

A2 − 7 A + 8I

Now LHS: =

=

LM−1 N14

−1 6

OP LM −7OP + LM8 Q N 42Q N0 −7 − a−7f + 0O L0 = 34 − 42 + 8 PQ MN0

7 −7 − 34 14

LM −1 − 7 + 8 N14 − 14 + 0

0 8

OP Q

OP Q

0 = R.H.S. 0

Hence verified.

4.

LM 1 Verify Cayley Hamilton theorem for the matrix A = 2 MM 1 N

−1 1 2

OP PP Q

0 0 −1

Solution: Characteristic equation is A − λI = 0 1− λ 2 1

a1 − λf 1 −2 λ

−1 1− λ 2

0 0 =0 −1 − λ

a f

2 0 − −1 1 −1 − λ

0 +0=0 −1 − λ

a1 − λf a1 − λfa−1 − λf − 0 + 1c2 a−1 − λf − 0h + 0 = 0 −a1 − λ f 1 − λ − 2 a1 + λ f = 0 − a1 − λ f d1 − λ i − 2 − 2λ = 0 −d1 − λ − λ + λ i − 2 − 2λ = 0 2

2

2

2

3

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− λ3 + λ2 − λ − 3 = 0. This is characteristic equation. To verify Cayley Hamilton theorem, we have to verify − A 3 + A 2 − A − 3 I = 0

LM 1 = A⋅ A = 2 MM 1 N LM 1 − 2 + 0 = 2+2+0 MM 1 + 4 − 1 N

 →

A

Now

2

A

A =A 3

2

2

−1 1 2

LM−1 = 4 MM 4 N

LM−1 − 4 + 0 = 4−2+0 MM 4 − 2 + 1 N A

A=

Now

LM−5 MM 23 N

−1 −5 −3

OP PP Q

−1 1 2

−1 − 1 + 0 −2 + 1 + 0 −1 + 2 − 2

LM−1 ⋅A= 4 MM 4 N

3

OP LM 1 PP MM21 QN

0 0 −1

 →

−2 −1 −1

0 0 1

OP LM 1 PP MM21 QN

−1 1 2

0 0 1

1− 2 + 0 −4 − 1 + 0 −4 − 1 + 2

LM−5 = 2 MM 3 N

OP PP Q

0+0+0 0+0+0 0 + 0 +1

OP PP Q

−2 −1 −1

OP PP B Q

0 0 −1

OP PP Q

0+0+0 0+0+0 0 + 0 −1

OP PP Q

−1 −5 −3

0 0 −1

LM MM N

0 1 0

0 1 0 and 3I = 3 0 0 −1

OP PP B Q

0 0 −1

OP PP Q

LM MM N

OP PP Q

0 3 0 = 0 1 0

0 3 0

0 0 3

OP PP Q

LM MM N

0 3 0

Consider LHS, − A3 + A2 − A − 3I

LM−5 =− 2 MM 3 N

−1 −5 −3

OP PP Q

LM MM N

0 −1 0 + 4 4 −1

−2 −1 −1

OP PP Q

LM MM N

3 0 0 − 2 1 1

−1 1 2

3 0 0 − 0 0 −1

0 0 3

OP PP Q


LM MM N

5 −1−1− 3 = −2 + 4 − 2 − 0 −3 + 4 − 1 + 0

OP PP Q

1− 2 +1− 0 5 −1−1− 3 3 −1− 2 + 0

LM MM N

0 = 0 0

0 0 0

0 0 0

0+0−0−0 0+0−0−0 1+1+ 1− 3

OP PP Q

Hence Cayley Hamilton theorem is verified. 5. By using Cayley Hamilton theorem find A−1 if A =

LM 1 7OP N −6 5Q

Characteristic equation = A − λI = 0

1− λ −6

7 =0 5−λ

a1 − λfa5 − λf − a−6fa7f = 0 5 − 5λ − λ + λ2 + 42 = 0

λ2 − 6λ + 47 = 0. By Cayley Hamilton theorem, A2 − 6A + 47I = 0 Operating by A−1, A −1 A 2 − 6 A −1 ⋅ A + 47 A −1 ⋅ I = 0

A − 6 I + 47 A −1 = 0 ⇒

47 A −1 = 6 I − A

A −1 =

Now

∴

1 6I − A 47

LM1 0OP = LM6 0OP N0 1 Q N0 6 Q L6 0OP − LM 1 7OP = LM5 6I − A = M N0 6 Q N −6 5Q N6 1 L 5 −7O . A = 1PQ 47 MN6 6I = 6

−1

OP Q

−7 1

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6. Find

A−1

LM3 if A = 4 MM 1 N

2 −1 3

OP PP Q

1 0 by using Cayley Hamilton theorem. 1

Solution: Characteristic equation: A − λI = 0

3−λ 4

2 −1 − λ 3

1

1 0

=0

1− λ

a3 − λf −13− λ 1 −0 λ − 2 41 1 −0 λ + 1 41 −13− λ = 0 a3 − λf a−1 − λfa1 − λf − 0 − 2 4 a1 − λf − 0 + 1 12 − a−1 − λf = 0 −a3 − λ f d1 − λ i − 8 a1 − λ f + 12 + 1 + λ = 0 2

−3 + 3λ2 + λ − λ3 − 8 + 8λ + 13 + λ = 0 − λ3 + 3λ2 + 10λ + 2 = 0

According to Cayley Hamilton theorem,

− A3 + 3 A 2 + 10 A + 2 I = 0 Operating by A−1

− A3 ⋅ A −1 + 3 A2 ⋅ A −1 + 10 AA−1 + 2 A −1 ⋅ I = 0 − A 2 + 3 A + 10 I + 2 A −1 = 0

A −1 =

⇒

1 2 A − 3 A − 10 I . 2

Now,

LM 3 A = A⋅ A = 4 MM 1 N LM 9 + 8 + 1 = 12 − 4 + 0 MM 3 + 12 + 1 N

OP LM 3 PP MM41 QN

 →

2

A2 =

2 −1 3

1 0 1

6−2+3 8 +1+ 0 2−3+ 3

LM18 MM168 N

7 9 2

2 −1 3

OP PP Q

3+ 0 +1 4+0+0 1+ 0 +1 4 4 2

OP PP Q

OP PP B Q

1 0 1


A

2

LM 3 3A = 3 4 MM 1 N LM 1 10 I = 10 0 MM0 N LM18 7 − 3 A − 10 I = 8 9 MM16 2 N

OP LM PP MM Q N 0 0 O L10 1 0P = M 0 P M 0 1PQ MN 0 4O L 9 6 P4P − MM12 −3 2 PQ MN 3 9

2 −1 3

LM MM N

−1 = −4 13

A

∴

−1

1 9 0 = 12 1 3

1 2 −7

OP PP Q 0 0O 10 0P P 0 10 PQ 3O L10 0P − M 0 P M 3PQ MN 0 6 −3 9

OP PP Q

153

3 0 3

0 10 0

0 0 10

OP PP Q

1 4 −11

LM MM N

−1 1 2 1 = A − 3 A − 10 I = −4 2 2 13

1 2 −7

OP PP Q

1 4 −11

6.11 SOLUTION OF LINEAR SYSTEM OF EQUATIONS: Consider 3 equations in 3 variables:

a1x + b1y + c1z = d1 a2x + b2y + c2z = d2 a3x + b3y + c3z = d3. The solution is the value of x, y and z which simultaneously satisfy the above equations. The solution can be obtained by various methods. Cramer’s rule and matrix method are 2 such methods of solving the system of equations. 1. Cramers rule:

Let

a1 ∆ = a2 a3

b1 b2 b3

c1 c2 c3

Multiplying both sides by x

a1 x∆ = x a2 a3

b1 b2 b3

c1 c2 c3

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Basic Mathematics

a1 x x∆ = a2 x a3 x

b1 b2 b3

c1 c2 C1′ = C1 + C2 y + C3 z c3

a1 x + b1 y + c1 z x∆ = a2 x + b2 y + c2 z a3 x + b3 y + c3 z d1 x∆ = d2 d3

b1 b2 b3

a f

x∆ = ∆ x

⇒

x=

Similarly,

c1 c2 c3

c1 c2 = ∆ x say c3

∴

a1 ∆ = a2 a3

b1 b2 b3

∆x . ∆ b1 b2 b3

c1 c2 c3

Multiplying by y

a1 y∆ = a2 a3

b1 y b2 y b3 y

a1 y∆ = a2 a3 a1 y ∆ = a2 a3 ⇒

c1 c2 c3

C2′ = C2 + xC1 + zC3

b1 y + a1 x + c1 z b2 y + a2 x + c2 z b3 y + a3 x + c3 z d1 d2 d3

c1 c2 = ∆ y say c3

a f

y∆ = ∆ y y=

c1 c2 c3

∆y ∆


z=

Similarly

∆z . ∆

Note: If there are 2 equations in 2 variables, Say a1x + b1y = d1 a2 x + b2 y = d2 ∆=

Then assume

a1 a2

b1 b2

∆x =

d1 d2

b1 b2 which is obtained by replacing the 1st column by d1 and d2

∆y =

a1 a2

d1 nd d2 which is obtained by replacing 2 column by d1 and d2.

By Cramer’s rule,

x=

∆y ∆x and y = . ∆ ∆

2. Matrix method: To solve

a1x + b1y + c1z = d1 a2x + b2y + c2z = d2 a3x + b3y + c3z = d3

LM a A= a MMa N

1

Assume

2 3

b1 b2 b3

OP PP Q

LM OP MM PP NQ

LM MM N

c1 x d1 c2 , X = y and D = d2 c3 z d3

OP PP Q

Then matrix equation is AX = D Operating by A−1 [3 A is non-singular, A−1 exists]

A −1 ⋅ AX = A −1 D IX = A −1 D X = A −1 D

i.e.,

LM xOP MM yzPP = A NQ

−1

D.

155

156

Basic Mathematics

Here A−1 can be calculated either by using Cayley Hamilton theorem or by using the formula

A −1 =

adj A . A

WORKED EXAMPLES: I. Solve the following by Cramers rule: 1. 3x + 4y = 11 2 x + 3y = 8 ∆=

Let

3 2

af af

4 = 3 3 − 2 4 = 9 − 8 = 1. 3

af af

∆x =

11 8

4 = 11 3 − 4 8 = 33 − 32 = 1 3

∆y =

3 2

11 = 3 8 − 11 2 = 24 − 22 = 2 8

af af

x=

∆y ∆x and y = . ∆ ∆

x=

1 2 and y = 1 1

x = 1 and y = 2. 2. 2x + 4y = 7 x − 7y = 6. Let

∆=

2 1

a f af

4 = 2 −7 − 4 1 = −14 − 4 = −18. −7

a f af

∆x =

7 6

4 = 7 −7 − 4 6 = −49 − 24 = −73. −7

∆y =

2 1

7 = 2 6 − 7 1 = 12 − 7 = 5. 6

a f af

x=

∆y ∆x and y = ∆ ∆

x=

5 −73 and y = −18 −18

x=

73 5 and y = − . 18 18


3. 2x −4y + 3z = 3 3x + 3y + 2z = 15 5x − 2y + 2z = 7

2 ∆= 3 5

Let

=2

a f

3 −2

a

−4 3 −2

2 3 − −4 2 5

3 2 2 2 3 +3 2 5

3 −2

f a f a f 2 a10f + 4 a−4 f + 3 a−21f

2 6 + 4 + 4 6 − 10 + 3 −6 − 15

20 − 16 − 63 ∆ = 20 − 79 = −59.

−4 3 −2

3 ∆ x = 15 7 =3

a f

3 −2

2 15 − −4 2 7

3 2 2 2 15 +3 2 7

3 −2

= 3 (6 + 4) + 4 (30 − 14) + 3 (−30 − 21) 3 (10) + 4 (16) + 3 (−51) 30 + 64 − 153 ∆x = 94 − 153 = −59. 2 ∆y = 3 5 =2

15 7

2 3 −3 2 5

3 15 7

3 2 2

2 3 +3 2 5

15 7

= 2 (30 − 14) − 3 (6 − 10) + 3 (21 − 75) = 2 (16) − 3(−4) + 3(−54) = 32 + 12 − 162 ∆y = 44 − 162 = −118.

157

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Basic Mathematics

−4 3 −2

2 ∆z = 3 5

a f

3 −2

=2

3 15 7 15 3 − −4 7 5

15 3 +3 7 5

3 −2

= 2 (21 + 30) + 4 (21 − 75) + 3 (−6 − 15) = 2 (51) + 4 (−54) + 3 (−21) 102 − 216 − 63 ∆z = 102 − 279 = −177.

x= y= z=

∆ x −59 = =1 ∆ −59 ∆y ∆

=

−118 =2 −59

∆ z −177 = =3 ∆ −59

x = 1, y = 2 and z = 3.

∴ 4. x − y − 2z = 3 2x + y + z = 5 4x − y − 2z = 11

1 ∆= 2 4

Let

=1

1 −1

a f

1 2 − −1 4 −2

−1 1 −1

−2 1 −2

a f

1 2 + −2 4 −2

1 −1

= 1 (−2 − (−1)) + 1(−4 − 4) − 2 (−2 − 4) = 1 (−2 + 1) + 1(−8) − 2 (−6) = −1 −8 + 12 = +3. 3 ∆x = 5 11

=3

1 −1

a f

1 5 − −1 11 −2

−1 1 −1

−2 1 −2

a f

1 5 + −2 11 −2

1 −1


a fg a f a f 3 a−2 + 1f + 1 a−21f − 2 a−16 f 3 a−1f + 1a −21f − 2 a−16f

b

3 −2 − −1 + 1 −10 − 11 − 2 −5 − 11

−3 − 21 + 32 −24 + 32 = 8. 1 ∆y = 2 4

3 5 11

−2 1 −2

2 5 a f 4 −2 11 −2 4 11 = 1a−10 − 11f − 3 a−4 − 4 f − 2 a22 − 20 f = −21 − 3 a −8f − 2 a2 f =1

5

1

−3

2

1

+ −2

= −21 + 24 − 4 = −25 + 24 = −1.

1 ∆z = 2 4

−1 1 −1

3 1 5 =1 −1 11

a f

5 2 − −1 11 4

b a fg a

f a 11 + 5 + 1 a2 f + 3 a−6 f

5 2 +3 11 4

1 11 − −1 5 + 1 22 − 20 + 3 −2 − 4

f

16 + 2 − 18 = 0.

x=

∴

∆y ∆x ∆ ,y= and z = z ∆ ∆ ∆

x=

8 0 −1 ,y= and z = 3 3 3

8 −1 x= , y= and z = 0. 3 3 II. Using matrix method solve the following system of equations: 1. 2x − 3y = 4 3 x + 2y = 5

1 −1

159

160

Basic Mathematics

A=

Let

LM2 N3

OP Q

LM OP NQ

AX = D X = A−1D.

Matrix equation

adj A A

A −1 =

Now

LM 2 3OP and A = 2 3 N−3 2Q 3O adj A 1 L 2 = = PQ M 3 2 − 13 N A

a f

−3 = 4 − −9 = 4 + 9 = 13 2

Adj A =

∴

A −1

Now

OP LM4OP B Q N5Q 1 L 8 + 15O X= 13 MN−12 + 10PQ 1 L 23O X= M P 13 N −2 Q LM 23 OP x LM OP = M 13 P N yQ M− 2 P N 13 Q LM N

→

1 2 X= A D= 13 −3 −1

x=

⇒ 2. 2x − 3y = 4 4x − 5y = 10. Let

LM OP NQ

−3 4 x ,X= and D = 2 5 y

A=

LM2 N4

3 2

23 −2 ; y= . 13 13

OP Q

LM OP NQ

LM OP N Q

−3 4 x ,X= and D = −5 10 y

Matrix equation: ⇒

AX = D X = A−1D.

Now

A −1 =

adj A A


Adj A =

LM−5 3OP and A = 2 a−5f − a−3f 4 = −10 + 12 = 2. N−4 2Q adj A 1 L −5 3O = M A = 2 N −4 2 PQ A −1

OP LM 4 OP B Q N10Q 1 L −20 + 30 O = M 2 N −16 + 20 PQ 1 L10 O L 5O X= M P=M P 2 N 4 Q N2 Q LM x OP = LM5OP N y Q N2 Q LM N

→

1 −5 X = A −1 D = 2 −4

⇒

x = 5 and y = 2.

3. x + y = 1 y+z=7 z+x=2 Given equations:

Let Matrix equation: ⇒ Now

Consider

3 2

x + y + 0z = 1 0x + y + z = 7 x + 0y + z = 2

LM1 A= 0 MM1 N

1 1 0

LM OP MM PP NQ

OP PP Q

LM OP MM PP NQ

x 0 1 1 , X = y and D = 7 . z 1 2

AX = D X = A−1D A −1 =

LM MM N

1 A= 0 1

adj A A 1 1 0

af a f

0 1 1

OP PP Q

A = 1 1 − 1 −1 = 1 + 1 = 2.

161

162

Basic Mathematics

LM 1 MM 00 Adj A = − MM 1 MM 01 N

1 1 1 1 1 0

1 0 1 1 1 − 1

0 1 0 1 1 0

−

1 1 1 − 0 1 0

0 1 0 1 1 1

OP PP PP PP Q

LM 1 −1 1OP L 1 −1 1O adj A = −a−1f MM −1 −a−11f −11PP = MMM−11 11 −11PPP Q N Q N L 1 −1 1OP adj A 1 M 1 1 −1 = A = 2 M A MN−1 1 1PPQ L 1 −1 1OP LM1OP 1 LM 1 − 7 + 2 OP 1M 1 1 −1 7 = 1+ 7 − 2 X = A D= P M P 2M 2M MN−1 1 1PQ MN2PQ MN−1 + 7 + 2PPQ L−4O L−2O 1M P M P X= 6 = 3 2M P M P MN 8PQ MN 4PQ LM x OP LM−2OP MM yz PP = MM 43PP ⇒ x = −2, y = 3 and z = 4. NQ N Q −1

−1

x = −2, y = 3 and z = 4. 4. x + y + z = 3 x + 2y + 3z = 4 x + 4y + 9z = 6.

Let Matrix equation: ⇒ Now

LM MM N

1 A= 1 1

1 2 4

OP PP Q

LM OP MM PP NQ

LM OP MM PP NQ

x 1 3 3 , X = y and D = 4 z 9 6

AX = D X = A−1D. A −1 =

adj A A


LM1 1 1OP A= 1 2 3 MM1 4 9PP N Q A = 1 a18 − 12f − 1 a9 − 3f + 1 a 4 − 2 f

Consider

A = 6 − 6 + 2 = 2.

LM+ 2 3 − 1 1 + 1 1 OP MM 41 93 41 91 21 31 PP adj A = − MM 1 9 + 1 9 − 1 3 PP MM + 11 24 − 11 41 + 11 21 PP N Q LM 18 − 12 −a9 − 4f a3 − 2f OP adj A = −a9 − 3f MM 4 − 2 −a94−−11f −a23−−11fPP N Q LM 6 −5 1 OP adj A = −6 MM 2 −83 −+21PP N Q L 6 −5 1OP adj A 1 M 8 −2 A = = −6 2 M A MN 2 −3 +1PPQ −1

Now

X=A

−1

L6 1M D = −6 2M MN 2

 →

LM MM N LM x OP LM2OP X= y = 1 . MM z PP MM0PP NQ NQ

−5 8 −3

OP PP Q

LM OP MM PP NQ

18 − 20 + 6 4 1 1 X = −18 + 32 − 12 = 2 2 2 6 − 12 + 6 0

⇒

x = 2; y = 1; z = 0.

OP LM3OP PP MM46PP B QN Q

1 −2 +1

163

164

Basic Mathematics

6.12 APPLICATION OF MATRICES IN BUSINESS PROBLEMS: 1. A company sold 22 scooters, 15 bikes and 10 mopeds in January and 20 scooters, 20 bikes and 12 mopeds respectively in June. Represent the data in matrix form. Scooters Bikes Mopeds January 22 15 10 June 20 20 12 ∴

LM N

22 Corresponding matrix = 20

15 20

10 12

OP Q

2. Suppose the matrices A and B represent the number of items of different kinds produced by 2 manufacturing units in one day.

LM OP MM PP NQ

LM OP MM PP NQ

2 1 A = 5 and B = 3 Compute 2A + 5B what does 2A + 5B represent? 7 6 Solution: 2A + 5B

LM OP LM OP MM PP MM PP NQ NQ LM 4OP LM 5OP LM 4 + 5OP = 10 + 15 = 10 + 15 MM14PP MM30PP MM14 + 30PP N Q N Q N Q L 9O = M 25P . MM44PP N Q 2 1 =2 5 +5 3 7 6

2A + 5B represents the number of items produced by one unit in two days and another in 5 days together. 3. A man buys 3 kgs of dal, 2 kgs of rice and 5 kgs of oil. If the cost of each kg is Rs. 35, Rs. 20 and Rs. 75 respectively. Then find the total cost by matrix method. Solution: Matrix representing the commodities = [3 2 5]

LM MM N

OP PP Q

35 Matrix representing the costs = 20 75


Total cost = 3

 →

2

165

LM35OP 5 20 B MM75PP N Q

= 3 × 35 + 2 × 20 + 5 × 75 = 105 + 40 + 375 = 520

∴

Total cost = Rs. 520. 4. A company sold 30 metal chairs, 40 wooden chairs and 25 plastic chairs in February and 60, 50 and 75 respectively in March. The selling price of a metal chair is Rs. 150, that of wooden chair is Rs. 500 and plastic chair is Rs. 300. Find the total revenue in February and March using matrix method. Solution: Metal chair February: 30 March: 60 ∴

Wooden Chair 40 50

LM N

30 Corresponding matrix = 60

Plastic chair 25 75 40 50

OP Q

25 75

LM OP MM PP N Q OP L30 40 25OP LM150 Total revenue = M PB N60 50 75Q MMN500 300PQ L30 × 150 + 40 × 500 + 25 × 300OP =M N60 × 150 + 50 × 500 + 75 × 300Q L 4500 + 20000 + 7500 OP =M N9000 + 25000 + 22500Q L32,000OP =M N56,500Q 150 Matrix representing the price = 500 300 →

∴

∴ ∴

Total revenue in February = Rs. 32,000/and Total revenue in March = Rs. 56,500/Total revenue = 32,000 + 56,500 = Rs. 88,500/-

166

Basic Mathematics

5. Matrix A and matrix B give the daily sales and selling price of soft drinks for a shopkeeper. Pepsi Coke Thumbsup

Mon Tue A= Wed Thu

LM6 MM21 MN2

3 3 1 1

1 0 7 5

OP PP PQ

LM OP MM PP NQ

Pepsi 7 6 B = Coke Thumbsup 6

Find the total revenue from four days. Solution: Revenue matrix = AB.

LM6 2 =M MM 1 N2

 →

3 3 1 1

OP L7O PP MM6PP B PQ MN6PQ

1 0 7 5

LM 6 × 7 + 3 × 6 + 1 × 6 OP LM42 + 18 + 6OP LM66OP Mon 2×7+3×6+0×6 14 + 18 + 0 32 Tue P P =M =M =M P MM 1 × 7 + 1 × 6 + 7 × 6 PP MM 7 + 6 + 42 PP MM55PP Wed N 2 × 7 + 1 × 6 + 5 × 6 Q N14 + 6 + 30Q N50Q Thu − Total revenue for 4 days = 66 + 32 + 55 + 50 = Rs. 203. 5. In a certain town there are 4 colleges and 12 schools. Each school has 8 peons, 5 clerks and 2 cashiers. Each college has 10 peons, 7 clerks and 3 cashiers. In addition, each college has 1 section officer and one librarian. The monthly salary of each of them is as follows: Peon: Rs. 2000; Clerk Rs. 3000, Cashier Rs. 5000, Section Officer Rs. 6000 and Librarian Rs. 4500. Using matrix notation find (1) Total number of posts of each kind in schools and colleges taken together (2) Monthly salary bill of all the schools and colleges taken together. Solution: Let A = [5 12] represents the number of colleges and schools in that order. Peon Clerk Cashier S. off. Librarian

Let

B=

LM N

College 10 School 8

7 5

Peon Clerk C = Cashier S. off. Librarian

3 2

1 0

LM2000OP MM3000 P 5000 P MM6000PP N4500Q

1 0

OP Q


167

Number of employees = AB  →

5

i.e.,

= 5 × 10 + 12 × 8

12

LM10 N8

5 × 7 + 12 × 5

= 50 + 96

7 5

1 0

1 0

5 × 3 + 12 × 2

35 + 60 = 146

3 2

15 + 24

95

39

OP B Q

5 × 1 + 12 × 0 5+0

5

5×1+1× 0

5+0

5

1st element i.e. 146 represent number of peons, 2nd element, 95 represent number of clerks and so on. 6. Total monthly salary bill of each school and college is given by matrix BC.

LM10 N8

LM2000OP 3000 P 1O M P0Q MM5000PP B MM6000PP N4500Q

 →

7 5

3 2

1 0

LM10 × 2000 + 7 × 3000 + 3 × 5000 + 1 × 6000 + 1 × 4500OP 8 × 2000 + 5 × 3000 + 2 × 5000 + 0 + 0 Q N L20,000 + 21,000 + 15,000 + 6,000 + 4,500OP =M N 16,000 + 15,000 + 10,000 + 0 + 0 Q L66,500OP =M N41,000 Q

∴ Total monthly salary bill of all colleges and schools taken together

a f

= A BC = 5

12

LM66,500OP N41,000Q

= 5 × 66,500 + 12 × 41,000 = 332500 + 492000

= [8,24,500] ∴ Total salary = Rs. 8,24,500. Alieter: Total monthly salary of all colleges and schools

168

Basic Mathematics

= Total number of employees ⋅ salary

= 146

  →

95

39

5

LM2000OP 3000 M P 5 M5000 P B MM6000PP MN4500PQ

= 146 × 2000 + 95 × 3000 + 39 × 5000 + 5 × 6000 + 5 × 4500 = 292000 + 285000 + 195000 + 30000 + 22500 = 8,24,500

Hence total salary = Rs. 8,24,500. 6. A salesman has the following record of sales during 3 months for 3 items A, B, C which have different rates of commission. Month

Jan Feb Mar

Sales of units

Total Commission

A

B

C

100 300 100

100 200 200

200 100 300

900 1000 1400

Find out the rates of commissions on items A, B and C. Solution: Let x, y and z denote the rates of commission in Rupees per unit for A, B and C items respectively. Then the data given can be expressed as a system of linear equation.

100 x + 100 y + 200 z = 900 300 x + 200 y + 100 z = 1000

100 x + 200 y + 300z = 1400 or x + y + 2z = 9

3x + 2 y + z = 10 x + 2 y + 3z = 14

Solving these equations by matrix method:

Let

LM1 A= 3 MM1 N

1 2 2

OP PP Q

LM OP MM PP NQ

LM MM N

9 2 x 1 , X = y and D = 10 14 3 z

OP PP Q


AX = D ⇒

X = A −1 D A −1 =

Now

adj A A

1

1

2

A=3 1

2 2

1 = 1 2 × 3 − 2 −1 3 × 3 −1 + 2 6 − 2 3

a

f a

f a f

a f af

= 6 − 2 − 9−1 + 2 4

∴ A−1 exist.

4 − 8 + 8 = 4 ≠ 0.

LM 2 MM 23 Now adj A = − MM 1 MM 31 N

1 3 1 3 2 2

1 2 1 1 1 − 1

−

2 3 2 3 1 2

1 2 1 − 3 1 3

AX = D X = A−1D

⇒

LM 4 adj A = −8 MM 4 N

Now

A

−1

1 1 −1

LM MM N

4 adj A 1 = = −8 4 A 4

LM MM N

OP PP Q

−3 −15 −1 1 1 −1

OP PP Q

−3 +5 −1

OP LM 9 OP PP MM1014PP QN Q L 4 × 9 + 1 × 10 − 3 × 14 O 1 X = M−8 × 9 + 1 × 10 + 5 × 14P 4M MN 4 × 9 − 1 × 10 − 1 × 14 PPQ L 36 + 10 − 42 OP 1 LM 4 OP 1M −72 + 10 + 70 = 8 X= 4M MN 36 − 10 − 14 PPQ 4 MMN12PPQ

X = A −1 D =

4 1 −8 4 4

1 1 −1

−3 +5 −1

2 1 2 1 1 2

OP PP PP PP Q

169

170

Basic Mathematics

LM1OP X= 2 MM3PP NQ LM xOP LM1OP MM yzPP = MM23PP NQ NQ ⇒

x = 1, y = 2 and z = 3.

i.e., Rate of commission per unit for the 3 items is Re. 1, Rs. 2 and Rs. 3 respectively. 7. The prices of 3 commodities X, Y and Z are x, y and z respectively. A sells 1 unit of X, 1 unit of Y and 1 unit of Z. B sells 3 units of X, 1 unit of Y and purchases 1 unit of z. C sells 1 unit of X, 3 units of Y and purchases 1 unit of Z. In the process A, B and C earns Rs. 9000, Rs. 1000 and Rs. 5000 respectively. Using matrices find the prices per unit of the commodities. (Note that selling the units is positive earning and buying the units is negative). Solution. The above data can be written in the form of simultaneous equations: A: x + y + z = 9000 B: 3x + y − z = 1000 C: x + 3y − z = 5000. Solving these equations, by Cramer’s rule:

Let

1 ∆= 3 1

1 1 3

1 −1 = 1 −1 + 3 − 1 −3 + 1 + 1 9 − 1 −1

a

f a

f a f

a f af

2 − 1 −2 + 1 8

2 + 2 + 8 = 12.

9000 ∆ x = 1000 5000

a

1 1 3

1 −1 −1

f a f a 9000 a2f − 1 a 4000 f + 1 a −2000 f

9000 −1 + 3 − 1 −1000 + 5000 + 1 3000 − 5000

18000 − 4000 − 2000 = 12000 1 ∆y = 3 1

9000 1000 5000

1 −1 −1

f


a

f

a

f a 4000 − 9000 a −2 f + 1 a14000 f

= 1 −1000 + 5000 − 9000 −3 + 1 + 1 15000 − 1000

171

f

4000 + 18000 + 14000 = 36000.

1 ∆z = 3 1

a

1 1 3

9000 1000 5000

f a

f

a f

= 1 5000 − 3000 − 1 15000 − 1000 + 9000 9 − 1 2000 − 14000 + 72000 ∆ z = 60000.

x=

Hence

y= z=

∆ x 12000 = = 1000 12 ∆ ∆y ∆

=

36000 = 3000 12

∆ z 60000 = = 5000 ∆ 12

Hence price per unit of x = Rs. 1000, that for y = Rs. 3000 and for z = Rs. 5000. 8. A company is considering which of the 3 methods of production it should use in producing 3 products X, Y and Z. The amount of each product and produced by each method is as shown below. Method

Product X

Product Y

Product Z

I II III

4 5 3

8 7 6

2 2 5

Further information relating to profit per unit is as follows: Product

Profit/Unit

X Y Z

10 5 7

Using matrix multiplication find which method maximises the total profit:

172

Basic Mathematics

LM4 A= 5 MM3 N

Let

8 7 6

OP PP Q

LM MM N

2 10 2 B= 5 5 7

OP PP Q

Total profit = AB

LM4 8 2OP LM10OP = 5 7 2 MM3 6 5PP MM 57 PP N QN Q LM4 × 10 + 8 × 5 + 2 × 7OP = 5 × 10 + 7 × 5 + 2 × 7 MM3 × 10 + 6 × 5 + 5 × 7PP N Q L40 + 40 + 14O L94O A = M 50 + 35 + 14 P = M99P MM30 + 30 + 35PP MM95PP N Q N Q ∴

Total profit in method I = 94. That in method II = 99 and in method III = 95. Method II maximises the profit.

∴

REMEMBER: • Matrix is an arrangement of numbers in horizontal rows and vertical columns. • Two matrices of the same order are said to be equal if and only if the corresponding elements are equal. • Two matrices can be added subtracted if they have same order. It is obtained by adding/subtracting the corresponding elements. • Multiplication of a matrix by a scalar is obtained by multiplying each and every element by a scalar. • It A is of order m × n and B is of order n × p then AB is of order m × p, i.e., matrix multiplication is possible only when number of column in 1st matrix is equal to number of rows in the 2nd matrix. • Transpose of a matrix is obtained by interchanging rows and columns. • A unique value associated with every square matrix is called its determinant value. • If det A = 0, i.e., |A| = 0 for a matrix A, then A is called singular matrix. Otherwise it is called nonsingular matrix. • The value of the determinant is unaltered if its rows and columns are interchanged. • If 2 rows or columns are interchanged the value of the determinant changes its sign. • If in a determinant 2 rows or columns are identical then the value of the determinant is zero.


173

• If the elements of any row (or column) is multiplied by k, the value of the determinant is multiplied by k. • If to the elements of any row (or column) of a determinant the same multiples of the corresponding elements of other rows (or columns) of the determinant are added the value of the determinant remains the same. • The determinant obtained by deleting the row and column containing the element is called minor of that element. If the minors are multiplied with (−1)i + j [where i = row number and j = column number of the element] we get co-factors. The transpose of the co-factor matrix is called adjoint of the matrix. • For a non-singular matrix A.

A −1 =

adj A . A

• If A is a square matrix and I is the identity matrix of the same order. • The characteristic equation: |A − λI| = 0. The values of λ obtained is called eigen values or characteristic roots. • Every square matrix satisfies its characteristic equation, |A − λI| = 0. This is Cayley Hamilton theorem. • The solution of system of equations

a1 x + b1 y + c1 z = d1 a2 x + b2 y + c2 z = d 2

a3 x + b3 y + c3 z = d3 By Cramer’s rule:

Let

a1 ∆ = a2 a3

b1 b2 b3

c1 d1 c2 , ∆ x = d 2 c3 d3

b1 b2 b3

c1 c2 c3

a1 ∆ y = a2 a3

d1 d2 d3

c1 a1 c2 , ∆ z = a2 c3 a3

b1 b2 b3

d1 d2 d3

x=

Then

∆y ∆x ∆ , y= and z = z . ∆ ∆ ∆

By matrix method:

LM a A= a MMa N

1

Let

2 3

b1 b2 b3

OP PP Q

LM OP MM PP NQ

LM MM N

c1 x d1 c2 , X = y and D = d2 c3 z d3

OP PP Q

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Basic Mathematics

Then matrix equation

AX = D X = A−1D. and A −1 =

adj A or A−1 can be found by using Cayley Hamilton theorem. A

EXERCISE

LM 1 N2 L1 2. If A = M N1

OP Q

LM N

3 6 and B = −1 1

1. If A =

6 7

OP Q

OP Q

5 , find A + B ′. 0

LM N

0 0 and B = −1 6

1 5

OP Q

−7 , 2

find A + 3B.

OP LM 1OP , find matrix X such that A + X = B. Q N −1Q L4 7OP + LM y + 2 1OP = LM 1 8OP . 4. Find x and y if M N x 7 Q N 5 0 Q N6 7Q L 1 2OP and B = LM 1 3OP, verify that (A + B)′ = A′ + B′. 5. If A = M N−7 6 Q N −7 8Q L x 4OP + L2 x 5O = L3 9O 6. Solve for x: M N 1 7Q MN 8 3PQ MN9 10PQ 3. If A =

LM 1 N2

−1 0 and B = −1 1

2

7. If A is of order 4 × 5 and B is of order 5 × 3, does AB and BA exists? If so what are their order?

LM1 N5 L1 9. If A = M N2 L1 0 10. If M 0 1 MM0 0 N L1 11. If A = M N−0 8. If A =

OP LM 0OP, then find AB. Q N −5Q 2O , find AA′ and A′A. Is AA′ = A′A? 4 PQ 0 O L x O L1 O 0 P M y P = M2 P , find x, y and z. PM P M P 1 PQ MN z PQ MN 3PQ 0O , then prove that AA′ = A′A = I. 1PQ

7 −1 and B = 2 6


175

OP Q L 1 3OP LM 2OP = LM x OP , then find x and y. 13. If M N7 1Q N−3Q N y Q LM 1 2 3OP LM 1 7 2OP 14. If A = −1 5 3 , B = −1 MM 1 2 7PP MM 5 −20 61PP , then verify A (B + C) = AB + AC. N Q N Q L1 2 3OP , B = LM13OP and C = 2 1 , then verify A (BC) = (AB) C. 15. If A = M N1 5 2Q MMN2PPQ L 1 2 3OP 16. If A = M0 MM 1 11 −21PP , then find A . N Q L 1 −2OP , then prove that A − 4A + 17I = 0. Where I is identity matrix of order 2 × 2. 17. If A = M N7 3Q LM 1 2 0OP 18. If A = 2 MM 1 31 −−66PP , then prove that A − A + 13A − 9I = 0. N Q L1 2OP , B = LM2 −1OP , then prove that (AB)′ = B′A′. 19. If A = M N3 −1Q N3 2Q LM 1 2 2OP 20. Prove that A = 2 MM2 21 21PP satisfies A − 4A − 5I = 0. N Q 12. If A =

LM0 Ni

i , then find A2. 0

3

2

3

2

2

21. Evaluate: 1 (a) 7 5 (d) −6

2 5

(b) −3 −7

−1 6

−7 (e) −5

3 5 5 −2

(c)

−1 2

7 −6

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Basic Mathematics

22. Evaluate:

1 (a) 2 3

2 3 7

−1 (b) 3 2

4 5 8

2 −5 0

1 (c) 2 4

7 4 −8

23. Find x if the matrix A is singular. (a) A =

LM x N20

5 x

OP Q

(b)

LM 2 A = −4 MM 5 N

3 x 6

OP PP Q

4 −8 7

24. Solve for x: 5 (a) 2

3 (b) 5

x =3 7

x 2 (c) −1

x =8 x

25. Using the properties of determinants find 3485 (a) 3487 241 (c) 244 247

3486 3488 242 245 248

243 246 249

a−b 26. Prove that b − c c−a

b−c c−a a−b

6001 (b) 6004

6002 6005

2100 (d) 2110 2120

2101 2111 2121

c−a a − b = 0. b−a

1 27. Prove that 1 1

a b c

a3 b3 = a − b b − c c − a a + b + c c3

1 28. Prove that 1 1

a b c

a2 b2 = a − b b − c c − a c2

− a2 29. Prove that ab ac

a

a

ab −b2 bc

2102 2112 2122

fa fa fa fa fa f

ac bc = 4a 2 b 2 c 2 −c 2

f

2 5 2

−1 x =0 x

−1 1 −1

2 1 −2


x 30. Prove that p p

p x q

1+ a 31. Prove that a a

q q = x− p x−q x+ p+q x

a

b 1+ b b

fa

fa

f

c c = 1+ a + b + c 1+ c

32. Prove that x = −9 is a root of equation x +1 2 2

a + b + 2c c c

33. Prove that

34. Find the adjoint of: (a)

LM1 2OP N3 4 Q

(b )

LM 1 N −7

−2 6

3 x+3 3

a b + c + 2a a

OP Q

(c )

LM−1 N5

5 5 =0 x+4

b b =2 a+b+c c + a + 2b

a

7 −6

OP Q

(d )

LM−6 N0

5 4

OP Q

( e)

f

3

LM−5 N −2

OP Q

−3 −1

35. Find the inverse of:

LM6 5OP L−5 (c) M 6 0 7 N Q N 3O L−1 2OP , verify (AB) = B A . and B = M 7PQ N 2 7Q LM3 2 5OP 37. Find the adjoint of A = 4 5 6 and hence find A . MM1 3 5PP N Q L 0 1 1O 38. Find the inverse of M −4 − 2 0 P . MM 3 −1 4PP N Q

LM−4 N1 L1 36. If A = M N4 (a)

OP Q

3 1

(b)

−1

OP Q

1 −2

−1 −1

−1

39. Find the characteristic polynomial and characteristic roots of the matrix

LM1 N2

OP Q

4 . 3

177

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Basic Mathematics

LM1 −7OP . N5 2 Q LM 1 2 0OP 41. Verify Cayley Hamilton theorem for the matrix −1 3 MM 1 6 −25PP . N Q L1 0 42. Find A using Cayley Hamilton theorem for the matrix M2 MM3 21 N LM 1 2 3OP 43. Find A using Cayley Hamilton theorem if A = 0 MM 1 11 −21PP . N Q

40. Verify Cayley Hamilton theorem for the matrix

−1

OP PP Q

2 0 . 1

3

44. Solve the following system of equations (a) By Cramer’s rule (b) By matrix method: (i) 7 x − y = 16

2 x + 3y = −2 (iii) 2 x − 3y = 4

3x + 2 y = 5 (v) 3 x + 4 y = 7

7x − y = 6

(ii) 3x + 4 y = 10

4 x − 5y = 3 (iv) x + y = 7 2x + y = 8 (vi) x + y + 2 z = 9 3x + 2 y + z = 10

x + 2 y + 3z = 14 (vii) x − 4 y − 3z = 9 5 x + y = 19 2 x − 5z = 3

(viii) x − y − 2 z = 3 2x + y + z = 5 4 x − y − 2 z = 11

(ix) x + y + 2 z = 9 3x + 2 y + z = 10

(x) x + y = 1 y+z=7

x + 2 y + 3z = 14

z+x=2

45. A man buys 8 dozens of mangoes; 10 dozens of apples and 4 dozen of bananas. Mangoes cost Rs. 18 per dozen, apple Rs. 9 per dozen and bananas Rs. 6 per dozen. Represent the quantities bought by a row matrix and prices by column matrix and hence find the total cost. 46. A company is considering which of the 3 methods of production it should use in producing 3 goods X, Y and Z. The amount of each good produced by each method is shown in the matrix.


X

Y

Z

I 4 II 5 III 5

8 7 3

2 1 9

LM MM N

179

OP PP Q

The vector [10 4 6] represents the profit per unit for the goods X, Y and Z in that order. Find which method maximises profit. 47. Matrix A and B give the daily sales and sale price of chocolates for a shopkeeper. Bar one Dairy milk Five star

Mon Tue A= Wed Thu

LM6 MM21 MN1

3 3 2 4

1 0 4 3

OP PP PQ

LM OP MM PP NQ

Bar one 5 B = Dairy milk 6 Five star 7

Find the total revenue for four days. 48. A salesman has the following record of sales during 3 months for 3 items, A, B and C which have different rates of commission. Months

Sales of units

Total commission in Rs.

A

B

C

January

90

100

20

800

February

130

50

40

900

March

60

100

30

850

Find out the rates of commission on A, B and C. 49. The prices of the 3 commodities X, Y and Z are x, y and z per unit respectively. A purchases 4 units of z and 3 units of x and 5 units of y. B purchases 3 units of y and sells 2 units of x and 1 unit of z. C purchases 1 unit of x and sell 4 units of y and 6 units of z. In the process A, B and C earn Rs. 6000, 5000 and 13,000 respectively. Using matrices, find the prices per unit of the 3 commodities. (Note that selling the unit is positive earnings and buying the unit is negative earning). 50. Suppose the matrices X and Y represent the number of items of different kinds produced by 2 manufacturing units in one day.

LM OP MM PP NQ

LM OP MM PP NQ

4 3 5 and = X Y= 4 . 6 5

Compute 2X + 3Y, what does 2X + 3Y represent?

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Basic Mathematics

ANSWERS 1.

OP Q

LM7 N7

4 −1

LM 1 N19

2.

4. x = −1, y = −5

9 22

OP Q

−21 5

3.

LM−1 2OP N −1 0 Q

6. x = 1 or x = −3

7. AB exists and is of order 4 × 3. BA does not exist. 8.

12.

LM1 0OP N1 0Q LM−1 N0

9. AA ′ =

OP Q

0 −1

LM 5 N10

OP Q

10 = A′A 20

13. x = −7, y = 11

16.

21. ( a ) − 9 ( b ) − 23 (c ) − 8 ( d ) − 53 (e) 39. 23. ( a) x = 10, (b) x = −6

10. x = 1, y = 2 and z = 3.

LM 11 MM−49 N

24. ( a) 19 (b) − 4 (c ) 3 or − 1

LM 4 −2OP (b) LM6 2OP (c) LM−6 −7OP (d ) LM4 −5OP (e) LM−1 N−3 1Q N7 1Q N −5 −1Q N0 −6Q N 2 −1O 1 L 1 −3O 1 L 7 −5O 1 L −2 (a) ( b) ( c) M P M P M 6Q 42 N 0 −7 N −1 −4 Q 4 N −6 −5PQ L 7 5 −13OP L−8 5 2OP 1 M 1 M −14 10 2 3 −1 38. 11 M 4 P 28 M MN 7 −7 7PQ MN 7 −3 1PPQ

34. ( a )

37.

39. λ2 − 4λ − 5 = 0, roots : 5 and − 1.

LM MM N

1 1 42. 3 −2 1

4 −5 −2

OP PP Q

−2 4 1

43.

LM 11 MM−49 N

44. (i ) x = 2, y = −2 (ii ) x = 2, y = 1 (iii ) x =

19 −9 16

OP PP Q

22. ( a ) 7 ( b ) 94 (c) 3

25. ( a ) − 2 ( b ) − 3 (c ) 0 ( d ) 0

35.

22 −5 17

22 −5 +17

19 −9 16

23 −2 ,y= 13 13

(iv ) x = 1, y = 6 (v ) x = 1, y = 1 ( vi ) x = 1, y = 2, z = 3

( vii) x = 4, y = −1, z = 1 ( viii ) x = 8 3, y = − 1 3 , z = 0

OP PP Q

3 −5

OP Q


181

(ix ) x = 1, y = 2, z = 3 ( x ) x = −2, y = 3, z = 4.

45. 8

10

4

LM18OP MM 69 PP = 258 . N Q

46. 47. 48. 49.

Cost = Rs. 258. Method III. Total profit 116. Rs. 178. Rs. 2, 4 and 11 respectively. Price per unit of X = Rs. 3000 Y = Rs. 1000 and Z = Rs. 2000.

50.

LM17 OP MM2227PP It represent the number of items produced by one unit in 2 days and another in 3 days N Q together.

182

Basic Mathematics

7 Ratio and Proportions, Variations 7.1

INTRODUCTION:

Two quantities of the same kind can be compared either by subtraction method or by division method. In subtraction method we find how much more (or less) is one quantity than the other, and in division method, we find how many times (or what fractional part) is one quantity of the other. The quotient here is nothing but the ratio of the two quantities. For example, if I have Rs. 100 and you have Rs. 600 then we can compare the money by subtraction method and say ‘You have Rs. 500 more than what I have’. Or we can compare by division method and say you have 6 times the money what I have. If the ratio of 2 mutual quantities are equal then they are said to be proportional.

7.2

RATIO:

A ratio is a relation or comparison between two quantities of the same kind. The comparison is made by considering what multiple, part or parts the first quantity is of the second. The ratio of 2 quantities x and y is denoted by x : y or

N . The first term x is called antecedent and O

the second term y is called consequent. Note: 1. A ratio is a pure number. Hence it has no units. 2. When the terms of the ratio are multiplied or divided by the same quantity the ratio is not altered. For instance 2 : 3 = 4 : 6 = 40 : 60 = 80 : 120... 3. If a : b and c : d are two ratios, then the ratio ac : bd is called their compound ratio. Example : The compound ratio of 5 : 2 and 3 : 7 is 5 × 3 : 2 × 7 i.e., 15 : 14. 4. If a : b is the given ratio then the ratio

Ratio and Proportions, Variations

183

(i) a2 : b2 is called its duplicate ratio. = : > is called its subduplicate ratio.

(ii)

(iii) a3 : b3 is called its triplicate ratio. (iv)

3

= : 3 > is called its sub-triplicate ratio.

WORKED EXAMPLES: 1. Express the following ratios in their simplest form (a) 16 : 26 (b) 16 : 64 (c) 6 : 90 Solution: (a) 16 : 26 =

16 8 = = 8 : 13 26 13

(b) 16 : 64 =

16 1 = = 1: 4 64 4

(c) 6 : 90 =

(d) 90 : 10

6 2 1 = = = 1 : 15 90 30 15

(d) 90 : 10 =

90 9 = = 9 :1 10 1

2. Compare the following ratios : (a) 2 : 5 and 3 : 7 (b) 12 : 13 and 13 : 2 (c) 5 : 11 and 11 : 6 (d) 8 : 3 and 4 : 7 Solution: (a) To compare 2 : 5 and 3 : 7

2:5 =

2 and 5

3: 7 =

3 7

To compare 2 fractions first we make denominators equal by multiplying with suitable numbers.

2 7 14 × = 5 7 35 3 5 15 × = 7 5 35 Clearly i.e. i.e.,

14 15 is less than 35 35 14 : 35 < 15 : 35 2:7 21 21 8 : 3 > 4 : 7.

3. Find the ratio between (a) 1 hr 10 min and 140 min. (b) 3 kg 30 gm and 1 kg 260 gms. (c) 6 Rs. 50 ps. and 8 Rs. 75 ps. Solution: (a) 1 hr 10 min = 60 + 10 min. = 70 min. ∴

Ratio between 70 min. and 140 min. =

70 140

= 70 : 140 = 1 : 2 (b) 3 kg 30 gm and 1 kg 260 gms. 3 kg 30 gm = 3000 + 30 gm = 3030 gm


∴

185

1 kg 260 gms = 1000 + 260 = 1260 gm Ratio between 3030 and 1260

=

3030 101 = 1260 42

= 101 : 42. (c) 6 Rs. 50 ps. and 8 Rs. 75 ps. 6 Rs. 50 ps = 650 ps. 8 Rs. 75 ps = 875 ps. Ratio between 650ps and 875 ps

=

650 26 = = 26 : 35. 875 35

4. Write the duplicate and subduplicate of the ratio 1 : 9 Solution: Duplicate of a : b is a2 : b2 ∴ Duplicate of 1 : 9 is 12 : 92 = 1 : 81 Subduplicate of a : b =

a: b

∴ Subduplicate of 1 : 9 =

1 : 9 = 1: 3.

5. Write the triplicate and subtriplicate of the ratio 1 : 8. Solution: Triplicate of a : b is a3 : b3 ∴ Triplicate of 1 : 8 is 13 : 83 = 1 : 512. Subtriplicate of a : b is

3

a: 3 b

Subtriplicate of 1 : 8 is

3

1: 3 8

= 1 : 2. 6. Raju’s monthly salary is Rs. 3000 and Rama’s annual income is Rs. 60,000. What is the ratio of their incomes? Solution: Raju’s monthly salary = Rs. 3000 Raju’s annual income = 3,000 × 12 = 36,000 Given Rama’s annual income = Rs. 60,000 ∴ Ratio of their income = 36,000 : 60,000 = 6 : 10 =3:5

OR Given

Raju’s monthly salary

= Rs. 3000

186

Basic Mathematics

Rama’s annual income = Rs. 60,000 Rama’s monthly income = 60,000/12 = Rs. 5,000 Ratio of their income = 3000 : 5000 = 3 : 5. 7. A number is divided into 3 parts in the ratio 2 : 3 : 4. If the 3rd part is 32. Find the other 2 parts. Solution: Let 2x, 3x and 4x be the parts of the number. Given 3rd part = 32 4x = 32 ⇒ x=8 ∴ 1st part = 2x = 2 (8) = 16 2nd part = 3x = 3 (8) = 24. 8. A bag contains rupee, 50 paise and 25 paise coins in the ratio 5 : 6 : 8. If the total amount is Rs. 840, find the number of coins of each type. Solution: Ratio of 1 Re., 50 ps., and 25 ps. coins = 5 : 6 : 8 ∴ ∴

∴

Ratio of values =

5 6 8 : : 1 2 4

Ratio of values = 5 : 3 : 2 Now Divide Rs. 840 in the ratio 5 : 3 : 2 Sum of the terms of ratio = 5 + 3 + 2 = 10 ∴

1st part = Rs.

5 × 840 = Rs. 420. 10

2nd part = Rs.

3 × 840 = Rs. 252. 10

3rd part = Rs.

2 × 840 = Rs. 168. 10

∴

Number of 1 Re. coins = 420 Number of 50 ps. coins = 252 × 2 = 504 Number of 25 ps. coins = 168 × 4 = 672. 9. In a mixture of 35 litres, the ratio of milk and water is 4 : 1. If 7 litres of water is added to the mixture, then find the ratio of milk and water in the new mixture. Solution: Given milk : water = 4 : 1 Sum of terms = 4 + 1 = 5 ∴

Milk in 35 litres mixture =

4 × 35 5

= 28 litres.


187

Water in mixture = 35 − 28 = 7 litres. Now if 7 litres of water is added. Water in mixture = 7 + 7 = 14. Milk in the mixture = 28. ∴ Milk : Water = 28 : 14 = 2 : 1. 10. A mixture contains nuts and screws in the ratio 4 : 3. If 7 screws are added to the mixture, the ratio becomes 3 : 4. Find the number of nuts in the mixture. Solution: Let the number of nuts and screws be 4x and 3x. Given : If 7 screws are added the ratio becomes 3 : 4. i.e., 4x : 3x + 7 = 3 : 4 i.e.,

4x 3 = 3x + 7 4

Cross multiplying 16x = 3 (3x + 7) 16x = 9x + 21 16x − 9x = 21 ⇒ 7x = 21 ⇒ x = 3. ∴ Number of nuts = 4x = 4 (3) = 12. 11. 5 years ago, Arun’s father’s age was 5 times his son’s age. After 2 years he will be 3 times Arun’s age. Find the ratio of their present ages. Solution: Let Arun’s age 5 years ago be x yrs. Then his father’s age = 5x years. After 2 years, Arun’s age = x + 5 + 2 = x + 7 His father’s age = 5x + 5 + 2 = 5x + 7 Given :

5x + 7 = 3 (x + 7) 5x + 7 = 3x + 21 5x − 3x = 21 − 7 2x = 14 ⇒ x = 7.

∴

Father’s age now = 5x + 5 = 5 (7) + 5 = 40 yrs. Arun’s age now = x + 5 = 7 + 5 = 12 yrs.

∴

Ratio of their ages = 40 : 12 = 10 : 3.

12. If 4 kgs of tea is worth 3 kgs of sugar, 5 kgs of sugar is worth 16 kgs of flour and 7 kgs of flour is worth 2 kgs of coffee. How many kgs of tea is worth 24 kgs of coffee. Solution: Given Tea : Sugar = 4 : 3 Sugar : Flour = 5 : 16

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Basic Mathematics

Flour : Coffee = 7 : 2 Tea : Coffee = x : 24

Let

Tea Tea Sugar Flour = × × Coffee Sugar Flour Coffee

Now

x 4 5 7 = × × 24 3 16 2 x=

∴

24 × 4 × 5 × 7 = 35 3 × 16 × 2

∴ x = 35 ∴ Tea : Coffee = 35 : 24. ∴ 35 kgs of tea is worth 24 kgs of coffee. 13. A, B and C starts a business with a capital of Rs. 10,500 of this Rs. 4400 is contributed by A. Rs. 3700 is contributed by B and the rest by C. After 5 months C withdraws Rs. 800 capital, while A and B each adds Rs. 400. At the end of the year, profit of the original capital is shared in the ratio of capitals. Find to the nearest rupee the amount to be received by each Solution: Total capital

: Rs. 10,500

A’s contribution : Rs. 4400 B’s contribution : Rs. 3700 C’s contribution : Rs. 10,500 − (4400 + 3700) = Rs. 2400. A’s share in capital = Rs. 4400 used for 12 months and Rs. 400 used for (12 − 5) months. i.e.,

4400 × 12 + 400 × 7 = Rs. 55,600.

Similarly B’s share in capital Rs. 3700 used for 12 months and Rs. 400 used for (12 − 5) months Rs. 3700 × 12 + 400 × 7 = Rs. 47,200. C’s share in capital : Rs. 2400 used for 5 months and (Rs. 2400 − Rs. 800) used for (12 − 5) months i.e.,

2400 × 5 + 1600 × 7 = Rs. 23,200

Now Ratio of capital of A, B and C = 55600 : 47200 : 23200 = 139 : 118 : 58 Total : 139 + 118 + 58 = 315 Now profit = 13¾% of capital


=

189

55 × 10,500 = 1443.75 4 × 100

A’s share in profit =

139 × 1443.75 = Rs. 637 315

B’s share in profit =

118 × 1443.75 = Rs. 541 315

C’s share in profit =

58 × 1443.75 = Rs. 266. 315

14. Three utensils contains equal mixture of milk and water in the ratio 6 : 1, 5 : 2, and 3 : 1 respectively. If all the solutions are mixed together, find the ratio of milk and water in the final mixture. Solution: In 1st utensil milk : water = 6 : 1 ∴

Quantity of milk in 1st utensil =

6 7

and

Quantity of water in 1st utensil =

1 7

Similarly 2nd utensil contains

5 2 milk and water. 7 7

Similarly quantity of milk in 3rd utensil =

3 4

and water

1 . 4

=

∴ If all solutions are mixed, quantity of milk =

=

quantity of water =

=

6 5 3 + + 7 7 4 24 + 20 + 21 65 = 28 28 1 2 1 + + 7 7 4 4 + 8 + 7 19 = 28 28 65 19 : 28 28

∴

Milk : Water =

i.e.,

Milk : Water = 65 : 19.

190

7.3

Basic Mathematics

PROPORTION:

If 2 ratios are equal then the 4 quantities comprising them form a proportion i.e. if the ratio a : b is equal to c : d, then 4 quantities a, b, c, d are in proportion. Example: 1, 2, 4, 8 are in proportion since 1 : 2 = 4 : 8. Example Extremes Note : 1. a : b = c : d is also denoted by a : b : : c : d. 2. In a proportion a : b = c : d, the first and the last terms i.e., a and d are called exa : b = c : d tremes and the second and 3rd terms i.e. b and c are called means. means 3. In every proportion, the product of the means is equal to the product of the exFig. 7.1 tremes. i.e., ad = bc Conversely, if 4 quantities a, b, c, d are such that ad = bc, then they are said to be in proportion. a : b = c : d ⇔ ad = bc is called rule of 3. This rule is used to solve a proportion when one of the terms is unknown. For example: 4 : 5 = x : 15 ⇒

5 x = 4 × 15

x=

4 × 15 = 12. 5

ver tendo: If a : b = c : d, then prove that b : a = d : c 1. In Inv ertendo: Proof: Pr oof: Given a : b = c : d ⇔ bc = ad

⇔

b d = a c

⇔ b: a = d :c

Hence proved. nendo: If a : b = c : d, then prove that a : c = b : d 2. Alter Alternendo: Proof: Pr oof: Given a : b = c : d

⇔ bc = ad ⇔

b a = d c

⇔ a:c = b: d Hence proved. Componendo: 3. If a : b = c : d, then prove that a + b : b = c + d : d Proof: Pr oof: Given a : b = c : d

⇔ bc = ad


191

Adding bd to both sides bc + bd = ad + bd

a

f a

⇔ b c+d =d a+b

f

c+d a+b = d b i.e,

a+b c+d = b d

⇔ a + b:b = c + d :d Hence proved. videndo: If a : b = c : d, then prove that a − b : b = c − d : d. 4. Di Dividendo: Pr oof: Given a : b = c : d Proof: ⇔ ad = bc Subtracting bd on both sides. ad − bd = bc − bd

a

f a

d a−b =b c−d

⇔

f

a−b c−d = b d

⇔ a − b:b = c − d : d Hence proved. videndo: If a : b = c : d, then prove that a + b : a − b = c + d : c − d 5. Componendo and Di Dividendo: Proof: Pr oof: Given a : b = c : d ⇒a+b:b=c+d:d

a+b c+d = b d Also

a − b:b = c − d :d

i.e.,

a−b c−d = b d

Dividing (1) by (2) we get a+b c+d b = d a−b c−d b d

⇒

a+b c+d = a−b c−d

(Componendo) ...(1) (Dividendo) ...(2)

192

Basic Mathematics

a + b:a − b = c + d :c − d

⇒

Hence proved.

CONTINUED PROPORTION: Quantities a, b, c, d, ... are said to be in continued proportion if

a b c = = ... . b c d

i.e., a : b = b : c = c : d = ... Three quantities a, b, c are in continued proportion if a : b = b : c. Here ‘b’ is called mean proportional and c is called 3rd proportional. For instance 7, 14, 28, 56, ... are in continued proportion since 7 : 14 = 14 : 28 = 28 : 56 ...

7.4

DIRECT PROPORTION OR DIRECT VARIATION:

Quantities are said to be in direct proportion when an increase (or decrease) in one kind is accompanied by an increase (or decrease) in the other. 2 3 4 ... Example: Number of chocolates 1 Cost of chocolates 3 6 9 12 ... Here we notice as the number of chocolates increases cost of chocolates also increases. So number of chocolates is directly proportional to cost of chocolates.

INVERSE PROPORTION OR INVERSE VARIATION: Quantities are said to be in inverse proportion when an increase (or decrease) in one kind is accompanied by decrease (or increase) in the other. 1 2 3 4... Example: Number of workers Number of days 12 6 4 3... Here we notice as the number of workers increase, number of days required to finish the work decreases and vice versa. So it is inverse proportion. Note: If a : b = c : d represent a direct proportion, then a : b = d : c or (b : a = c : d) represent an inverse proportion.

WORKED EXAMPLES: 1. Find the missing term in the proportion: (a) x : 4 = 27 : 12 (b) 10 : 50 = ? : 250. Solution: (a) Given:

x : 4 = 27 : 12 12 x = 27 × 4


x=

27 × 4 12

x = 9.

(b) Given:

10 : 50 = x : 250 50 x = 10 × 250

x=

10 × 250 50

x = 50. 2. Find the 3rd proportional to (a) 5 : 10 (b) 3 : 27 (a) Let the 3rd proportional be x. 5 : 10 = 10 : x

Then

5 x = 10 × 10

10 × 10 = 20. 5

x= (b) Let the 3rd proportional be y. Then

3 : 27 = 27 : y 3 y = 27 × 27

y=

27 × 27 = 243. 3

3. Find the fourth proportional to (a) 1 : 2 : 3 (b) 9 : 8 : 18 Solution: (a) Let the 4th proportional be x. 1: 2 = 3: x

Then

x = 2 × 3 = 6.

(b) Let the 4th proportional be y. Then

9 : 8 = 18 : y 9 y = 8 × 18

y=

8 × 18 = 16. 8

4. Find the mean proportional to (a) 2 : 8 (b) 5 : 45.

193

194

Basic Mathematics

Solution: (a) Let the mean proportional be x. 2: x = x :8

Then

x2 = 8 × 2 x 2 = 16 x=4.

(b) Let the mean proportional be y. 5 : y = y : 45

Then

y 2 = 5 × 45 y 2 = 225 y = 15.

5. If 15x = 12y. Then what is x : y? Given: 15x = 12y

x 12 = y 15

⇒

x 4 = y 5 ⇒

x : y = 4 : 5.

6. If 4x − 7y : 3x + y = 2 : 3. Then find x : y. Given:

4 x − 7y : 3x + y = 2 : 3

a

f a

3 4 x − 7 y = 2 3x + y 12 x − 21y = 6 x + 2 y 12 x − 6 x = 2 y + 21y 6 x = 23 y

⇒

x 23 = y 6

⇒

x : y = 23 : 6.

7. If x : y = 7 : 5. Then find 4x − 2y : x + 3y Given : x : y = 7 : 5

f


195

x 7 = y 5

⇒

4x − 2y : x + 3y

Consider

=

4x − 2y = x + 3y

4 =

LM FG x IJ − 2OP N H yK Q Lx O y M + 3P Ny Q

y 4

F 7I − 2 H 5K 7 +3 5

28 28 − 10 −2 5 = 5 = 7 7 + 15 +3 5 5

=

18 9 = . 22 11

4 x − 2b : a + 3b = 9 : 11. 8. What number must be subtracted from each of 9, 11, 15 and 19, so that the difference will be proportional. Solution: Let the number subtracted be x, So that 9 − x, 11 − x, 15 − x and 19 − x will be in proportion. ∴

9 − x : 11 − x = 15 − x : 19 − x

a9 − x f a19 − x f = a11 − xf a15 − xf 171 − 19 x − 9 x + x 2 = 165 − 15 x − 11x + x 2 171 − 28 x + x 2 − 165 + 26 x − x 2 = 0

⇒

6 − 2x = 0

2 x = 6 ⇒ x = 3. 9. What must be added to each of 9, 42, 3 and 18 so that the sums will be in proportion. Solution: Let the number added be x. ∴

9 + x, 42 + x, 3 + x and 18 + x are in proportion.

196

Basic Mathematics

9 + x : 42 + x = 3 + x :18 + x

a9 + x f a18 + x f = a42 + x fa3 + x f 162 + 18 x + 9 x + x 2 = 126 + 3 x + 42 x + x 2

162 + 27 x + x 2 − 126 − 45x − x 2 = 0 ⇒

36 − 18 x = 0

⇒ x = 2. 10. If a : b = 2 : 1, b : c = 3 : 2, then find a : b : c. Given:

a:b b:c

= 2 : 1 = 3 : 2

Multiplying 1st ratio by 3 and 2nd ratio by 1 to make the value of b same. ∴

a:b= 6 :3 b:c= 3:2

⇒ a : b : c = 6 : 3 : 2. 11. If x : y = 1 : 2, y : z = 3 : 4 and z : w = 5 : 1, then find x : y : z : w.

x: y = 1:2 y : z = 3: 4 To make the value of y same, multiplying 1st ratio by 3 and 2nd ratio by 2.

x:y = 3 :6 y:z = 6:8 ∴

x : y: z=3:6:8

Given:

z:w=

5 : 1.

To make the value of z same, multiplying first ratio by 5 and 2nd ratio by 8.

x : y : z = 15 : 30 : 40

z:w=

40 : 8

∴ x : y : z : w = 15 : 30 : 40 : 8. 12. Divide Rs. 2360 among A, B, C so that A : B = 3 : 4, B : C = 5 : 6. Given:

A: B = 3 :4 B:C = 5:6

⇒

A : B : C = 15 : 20 : 24.


197

Sum = 15 + 20 + 24 =59. Out of 59. A’s share = 15 ∴

Out of 2360 A’s share =

2360 × 15 59

= Rs. 600. Out of 59, B’s share = 20 Out of 2360 B’s share =

20 × 2360 59

= Rs. 800. Similarly C’s share =

24 × 2360 = Rs. 960. 59

Verification: A’s share + B’s share + C’s share = Rs. 600 + Rs. 800 + Rs. 960 = Rs. 2360 = Total. 13. Divide 166 into 3 parts such that 4 times the first part, 5 times the 2nd part and 7 times the 3rd part are equal. Solution: Let 1st, 2nd and 3rd part be a, b and c. Given: 4a = 5b = 7c = x (say) Then 4a = x, 5b = x ⇒ ∴ i.e.,

∴

a=

x x x , b = and c = . 4 5 7

a:b:c=

x x x : : . 4 5 7

a : b : c = 35 : 28 : 20. Sum = 35 + 28 + 20 = 83. 1st part =

35 × 166 = 70 83

2nd part =

28 × 166 = 56 83

3rd part =

20 × 166 = 40. 83

Ver if ica tion: 70 + 56 + 40 = 166. erif ifica ication: 14. If the cost of 10 metres of cloth is Rs. 225, find the cost of 22 metres of cloth.

198

Basic Mathematics

Solution:

a f

Length mts.

a f

Cost Rs.

10

225

22

x

As the length of the cloth increases. Cost also increases. ∴ Length and Cost are directly proportional. To denote direct proportion, we use 2 arrows, with same direction.

10 : 22 = 225 : x

∴

10 x = 22 × 225

x=

22 × 225 10

x = 495. ∴ 22 mts of cloth costs Rs. 495. 15. If 60 men can complete a job in 12 days, how many days will 36 men take to complete the same job? Solution:

Men 60

days 12

36

x

As the number of men increases, the days required to complete the job decreases ∴ Men and days are inversely proportional. To denote this we use 2 arrows with opposite direction. 60 : 36 = x : 12 36 x = 60 × 12

x=

60 × 12 = 20. 36

∴ 36 men can complete a job in 20 days. 16. If 10 men can earn Rs. 105 in 7 days, in how many days will 15 men earn Rs. 225? Solution:

Men 10 15

Money 105 225

days 7 x

Here number of days is unknown. Leaving money, or keeping money constant, let us first consider men and days. As the number of men increases, days required to complete the job decreases. ∴ Men and days are inversely proportional. Now, leaving men, let us consider money and days. As the days increase, money earned also increases. ∴ Money and days are directly proportional.


199

∴ Corresponding compound proportion is

10 : 15 =x:7 225 : 105 15 × 105 × x = 7 × 10 × 225

x=

7 × 10 × 225 15 × 105

x = 10 ∴ 15 men earn Rs. 225 in 10 days. 17. 5 carpenters can earn Rs. 3600 in 6 days, working 9 hrs. a day. How much will 8 carpenters earn in 12 days working 6 hrs. a day. Solution: Carpenter 5

days 6

8

hrs. 9

12

6

Money 3600 x

Here money is unknown. Taking carpenter and money alone, as the carpenters increase, the money earned by them also increases. So it is direct proportion. Taking days and money, as the day increases money also increases. So it is direct proportion. Now taking hours and money, as the number of hours increases, money also increases. ∴ It is direct proportion. So corresponding compound proportion is

5:8 6 : 12 = 3600 : x 9:6 5 × 6 × 9 × x = 8 × 12 × 6 × 3600

x=

8 × 12 × 6 × 3600 5×6×9

x = 7680. ∴ 8 carpenters earn Rs. 7680 in 12 days working 6 hours a day. 18. A contractor undertook to make 15 kms of roadways in 40 weeks. In 10 weeks 3 kms were completed by 180 men working 8 hours a day. Then the men agreed to work 1 hour a day overtime and some boys were engaged to assist them. The work was finished in the stipulated time (that is 40 weeks). How many boys were employed if the work of 3 boys is equal to that of 2 men. Solution: Men

weeks

hrs.

kms.

180

a40 − 10f = 30

10

a8 + 1f = 9

8

a15 − 3f = 12

x

3

200

Basic Mathematics

men and Kms are directly proportional. men and hrs. are inversely proportional. men and week are inversely proportional. ∴ Required compound proportion is

180 : x = 30 : 10 9: 8 3 : 12 x × 30 × 9 × 3 = 180 × 10 × 8 × 12

x=

180 × 10 × 8 × 12 640 . = 30 × 9 × 2 3

Already 180 men are there, ∴

640 100 − 180 = 3 3

Given

3 boys ≡ 2 men.

⇒

1 man ≡

3 boys 2

100 100 3 men ≡ × boys 3 3 2

∴

= 50 boys. ∴

50 boys were engaged to assist them.

19. If 15 men build a wall 40 ft long, 2 and 1/2 ft. thick and 21 ft. height in 18 days working 10 and 1/2 hrs. each day. In how many days working 15 hrs. a day will 45 men build a wall 200 ft. long, 5 ft thick and 20 ft. height. Solution:

Men

Length

15

40

45

200

Breadth (thickness) 1 2 2 5

Height 21 20

hours 1 2 15

10

days 18 x

As the men increase, days decrease, so inverse proportion. As length increase, days also increase, as breadth increase, days also increase. As height increase, days also increase. So, It is direct proportion. As the number of hrs. increase, days required to construct decrease. So it is inverse proportion. ∴ Corresponding Compound proportion is:


U| 200 : 40 | 1| 5 : 2 |V 2 | 20 : 21 | 1 | 10 : 15 | 2 W

201

15 : 45

x × 45 × 40 × 2

1 1 × 21 × 15 = 18 × 15 × 200 × 5 × 20 × 10 . 2 2

x=

∴

= x : 18

18 × 15 × 200 × 5 × 20 × 10

1 2

1 45 × 40 × 2 × 21 × 15 2

x = 40 ∴ 40 days are required. 20. If 12 pumps working 6 hours a day can draw 2000 gallons of water in 20 days, find in how many days will 20 pumps working 9 hours a day draw 3000 gallons of water? Solution:

Pumps 12

Hours 6

20

Quantity 2000

9

3000

Days 20 x

As the number of pumps increase, the days required decreases. So it is inverse proportion. As the number of hours increases the days required to draw water decreases. So it is inverse proportion. As the number of gallons increase, number of days required also increases, so it is direct proportion. ∴ Corresponding compound proportion is

12 : 20 6: 9 3000 : 2000

U| V| W

= x : 20

x × 20 × 9 × 2000 = 20 × 12 × 6 × 3000

x=

20 × 12 × 6 × 3000 20 × 9 × 2000

x = 12 ∴

12 days are required.

202

7.5

Basic Mathematics

PROBLEMS ON TIME AND WORK:

Note: 1. If A can do a piece of work in n days. Then work done by A in 1 day = 2. If B’s 1 day’s work =

1 . n

1 . Then B can finish the work in x days. x

3. If A is twice as good a workman as B, then Ratio of work done by A and B = 2 : 1 Ratio of time taken by A and B = 1 : 2 4. If A can do a piece of work in x days, and B can do it in y days, then A and B working together

xy will do the same work in x + y days. 5. If A and B together can do a piece of work in z days and A alone can do it in x days then B alone can do it in

zx . x−z

SOLVED EXAMPLES: 1. Ram can reap a field in 6 days which Raju alone can reap in 8 days. In how many days both together can reap this field? Solution:

Ram’s 1 day’s work =

1 6

Raju’s 1 day’s work =

1 8

Ram and Raju’s 1 day’s work =

= ∴

1 1 + 6 8

4+3 7 = . 24 24

Both together can reap the field in

24 3 = 3 days. 7 7

OR We know, if A can do a piece of work in x days and B can do it in y days then A and B together can do it in

xy days. x+y


∴

Ram and Raju together can reap the field in

=

203

6×8 days. 6+8

48 24 3 = = 3 days. 14 7 7

2. X and Y together can dig a trench in 10 days which X alone can dig in 30 days. In how many days Y alone can dig it? Solution: X and Y’s 1 day’s work =

1 10

X’s 1 day’s work = ∴

Y’s 1 day’s work =

1 30

1 1 − 10 30

3 −1 2 = 30 30 ∴

Y alone can dig a trench in

30 = 15 days. 2

OR We know, if A and B together can do a piece of work in z days and A alone can do it in x days then B alone can do it in ∴

zx days. x−z Y alone can dig a trench in

=

30 × 10 30 − 10

300 = 15 days. 20

3. A and B can do a piece of work in 12 days; B and C in 15 days; C and A in 20 days. In how many days will they finish it together and separately? (A + B)’s 1 day’s work =

1 12

(B + C)’s 1 day’s work =

1 15

(C + A)’s 1 day’s work =

1 20

204

Basic Mathematics

Adding 2 [A + B + C]’s 1 day’s work =

1 1 1 + + 12 15 20

2 (A + B + C)’s 1 day’s work =

5+4+3 60

2 (A + B + C)’s 1 day’s work =

1 5

(A + B + C)’s 1 day’s work =

1 10

So A, B and C together finish the work in 10 days. Now, A’s 1 day’s work = (A + B + C)’s 1 day’s work − (B + C)’s 1 day’s work.

=

1 1 1 − = 10 15 30

∴A alone can finish the work in 30 days. Similarly, B’s 1 day’s work =

1 1 1 − = 10 20 20

∴ B alone can finish the work in 20 days. Similarly, C’s one day’s work = ∴

1 1 1 − = 10 12 60

C alone can finish the work in 60 days.

OR A, B and C can do together in

2xyz days xy + yz + zx

a

f

=

2 12 × 15 × 20 12 × 15 + 15 × 20 + 20 × 12

=

2 3600 2 3600 = = 10 days. 180 + 300 + 240 720

a

f

a f a f

A, B and C can together finish the work in 10 days. 4. A can do a piece of work in 25 days which B alone can finish in 20 days. Both work for 5 days and then A leaves off. How many days will B take to finish the remaining work.


Solution: A’s work in 1 day =

1 25

B’s work in 1 day =

∴

1 20

(A + B)’s 1 day’s work =

1 1 + 25 20

(A + B)’s 5 day' s work = 5

=

=

Now

LM 1 + 1 OP N 25 20 Q

9 20

a

9 Note this step 20

Remaining work = 1 −

∴

205

f

11 20

1 work is done by B in 1 day. 20

11 ×1 11 20 work will be done by B in = 11 days. 1 20 20

∴

B takes 11 days to finish the remaining work. 5. X is thrice as good a work man as Y and is therefore able to finish the piece of work in 60 days less than Y. Find the time in which they can do it, working together. Solution: Ratio of work done by x and y in same time = 3 : 1 ∴ Ratio of time taken = 1 : 3 If Y takes y days to finish a work. Then X takes y − 60 days to finish. Now

y − 60 : y = 1 : 3

a

f

3 y − 60 = y 3 y − y = 180

2 y = 180 ⇒ y = 90

206

Basic Mathematics

∴ Time taken by Y to finish the work = 90 days and time taken by X to finish the work = 90 − 60 = 30 days. ∴

X’s 1 day’s work =

1 30

Y’s 1 day’s work =

1 90

(X + Y)’s 1 day’s work = ∴ Both X and Y can finish the work in

1 1 2 + = 90 30 45

45 1 = 22 days. 2 2

6. A can build a wall in 30 days which B alone can build in 40 days. If they build it together and get a payment of Rs. 1400 what is A’s share and B’s share? Solution:

∴

A’s 1 day’s work =

1 30

B’s 1 day’s work =

1 40 1 1 : 30 40

Ratio of their work =

=4:3 Total = 4 + 3 =7 A’s share =

4 × 1400 = Rs. 800 7

B’s share =

3 × 1400 = Rs. 600 7

7. A can do a piece of work in 10 days, while B alone can do it in 15 days. They work together for 5 days and the rest of the work is done by C in 2 days. If they get Rs. 1200 for the whole work how should they divide the money? Solution: A’s 1 day’s work =

1 10

B’s 1 day’s work =

1 15

(A + B)’s 1 day’s work =

1 1 1 + = 10 15 6


(A + B)’s 5 day’s work = 5 Remaining work = 1 − Given: ∴

207

F 1I = 5 H 6K 6

5 1 = 6 6

(Note this step)

C completes rest of the work in 2 days. C’s 2 day’s work =

1 6

Now Consider A’s 5 day’s work : B’s 5 day’s work : C’s 2 day’s work.

5

F 1 I : 5F 1 I : 1 H 10 K H 15K 6 1 1 1 : : 2 3 6

=3:2:1 Total = 3 + 2 + 1 = 6. A’s share =

3 × 1200 = Rs. 600 6

B’s share =

2 × 1200 = Rs. 400 6

C’s share =

1 × 1200 = Rs. 200. 6

8. A certain number of men complete a piece of work in 60 days. If there were 8 men more the work could be finished in 10 days less. How many men were originally there? Solution: Let number of men = x 8 men more means x + 8 Given:

Men x x +8

days 60 60 − 10 = 50

As the number of men increases, the days required to complete the job decreases. ∴ It is inverse proportion. x : x + 8 = 50 : 60

a f a

f

x 60 = x + 8 50 60 x − 50 x = 400

208

Basic Mathematics

10 x = 400 ⇒ x = 40.

∴ 40 men were originally there. 9. 16 men or 28 boys can fence a farm in 40 days. In how many days will 24 men and 14 boys complete the same work? Solution: Given, 16 men’s work ≡ 28 boy’s work. i.e., 16 : 28 i.e., 8 : 14 ∴ 8 men ≡ 14 boys. Now 24 men and 14 boys ≡ 24 men + 8 men ≡ 32 men. Given:

Men 16

days 40

32

x

As the number of men increases, the days required to fence a farm decreases. ∴ It is inverse proportion. ∴

16 : 32 = x : 40 32 × x = 16 × 40

x=

16 × 40 = 20. 32

∴ 20 days are required to complete the work. 10. 2 men and 4 boys can do a work in 33 days. 3 men and 5 boys can do the same work in 24 days. How long shall 5 men and 2 boys take to finish it? Given: 2 men and 4 boys can do the work in 33 days. ⇒ 2 × 33 men and 4 × 33 boys can do it in 1 day. ∴ 66 men and 132 boys can do it in 1 day ...(1) Similarly 3 men and 5 boys can do the work in 24 days. ⇒

3 × 24 men and 5 × 24 boys can do it in 1 day

i.e.,

72 men and 120 boys can do it in 1 day

From (1) and (2) 66M + 132B ≡ 72M + 120B 132B − 120B ≡ 72M − 66M

...(2)


209

12B ≡ 6M M ≡ 2B.

∴ Now given: i.e.,

One man’s work is equivalent to 2 boy’s work. 2 men and 4 boys can finish the work in 33 days. 2M + 2M can finish the work in 33 days. 4 men can finish the work in 33 days.

Now, 5 men and 2 boys ≡ 5M + 1M = 6 Men.

Men 4

days 33

6

x

As men increases, days required to finish the work decreases so it is inverse proportion.

4 : 6 = x : 33

∴

6 x = 33 × 4 x=

33 × 4 == 22. 6

∴ 5 men and 2 boys can finish the work in 22 days.

7.6

PROBLEM ON TIME AND DISTANCE:

Note : (1) Speed =

Distance Time

F H

(2) x km hr = x ×

F H

I K

5 mts sec 18

(3) x mts sec = x ×

I K

5 km hr. 18

WORKED EXAMPLES: 1. The distance between 2 stations A and B is 450 kms. A train starts at 4 p.m. from A and moves towards B at an average speed of 60 km/hr. Another train starts from B at 3 : 20 p.m. and moves towards A at an average speed of 80 km/hr. How far from A will the two trains meet and at what time?

210

Basic Mathematics

Solution: Let the trains meet at a distance x kms. from A. Let trains from A to B and B to A be X and Y respectively. Given: Speed of X = 60 km/hr Speed of Y = 80 km/hr. Time required to cover x kms. by X =

x 60

Time required to cover (450 − x) kms. by Y =

450 − x 80

Difference between time =

450 − x x − = 3:20 p.m. to 4 p.m. 80 60 450 − x x − = 40 min. 80 60 450 − x x 40 hrs. − = 80 60 60 450 − x x 40 − = 80 60 60

a

f

3 450 − x − 4 x 2 = 240 3 cross multiplying,

a

f

9 450 − x − 12 x = 480 4050 − 9 x − 12 x − 480 = 0 3570 − 21x = 0

x=

3570 = 170. 21

∴ The trains meet at a distance of 170 kms from A. Time taken by X to cover 170 kms. =

170 hrs. 60

= 2hrs. 50 min. So the trains meet at 4 pm + 2 hrs. 50 min. = 6 : 50 pm.

3 of his usual speed, a student is 10 min. late to his class. Find his usual time to cover 4 the distance.

2. Cycling


211

Let the usual time taken be x min. Time taken at

F I H K

3 4 x min. of the usual speed = 4 3

LM3 Distance = Speed, Time = Distance OP Speed Q N Time

4 x − x = 10 3

Given: ⇒

4 x − 3 x = 30 x = 30.

∴

Usual time taken = 30 min. 3. A bullock cart has to cover a distance of 80 km in 10 hrs. If it covers half of the journey in

F 3I H 5K

th

time. What should be its speed to cover the remaining distance in the time left? Total distance = 80 kms. Solution: Total time = 10 hrs. Distance left =

Given: Time taken to cover

80 = 40 kms. 2

1 3 distance = × 10 hrs. = 6 hrs. 2 5 Remaining time = 10 − 6 = 4 hrs.

Speed =

Distance 40 = km hr 4 Time

Speed = 10 km/hr. 4. A man travels 360 km in 4 hrs. partly by air and partly by train. If he had travelled all the way by

4 of the time he was in train and would have arrived at his destination 5 2 hrs. early. Find the distance he travelled by air and train. Solution: Total time = 4 hrs. air, he would have saved

Given:

4 of total time in train = 2 hrs. 5

Total time in train =

2×5 5 = hrs. 4 2

Given: If 360 km is covered by air then time taken is 4 − 2 = 2 hrs. ∴

When

3 is spent in air, 2

212

Basic Mathematics

Distance covered =

360 3 × = 270 kms. 2 2

Distance covered in train = 360 − 270 = 90 km. 5. An aeroplane started 30 minutes later than the scheduled time from a place 1500 km away from its destination. To reach the destination at the scheduled time, the pilot had to increase the speed by 250 km/hr. What was the speed of the aeroplane per hour during the journey? Solution: Let the time taken by aeroplane in later case = x hrs.

Speed =

We know

Distance Time

Distance = 1500 kms.

1500 1500 = + 250 1 x x+ 2

Given:

a f

1500 2 1500 = + 250 x 2x + 1

a

f

1500 3000 + 250 2 x + 1 = x 2x + 1 Cross multiplying

a

f a

1500 2 x + 1 = x 3000 + 500 x + 250

f

3000 x + 1500 − 3000 x − 500 x 2 − 250 x = 0

500 x 2 + 250 x − 1500 = 0 ÷ by 250

2x2 + x − 6 = 0 2 x 2 + 4 x − 3x − 6 = 0

a f a f 2 x a x + 2 f − 3 a x + 2f = 0 a2 x − 3fa x + 2f = 0

2x x + 2 − 3 x + 2 = 0

2x = 3

x= 3 x cannot be negative, x =

3 2

3 or x = −2 2

−12 x 2

+4 x −3x +x


∴ The plane takes

213

3 1 hrs. = 1 hrs. in later case 2 2

1 1 So in Normal case it takes 1 + = 2 hrs. 2 2 ∴ Normal speed =

7.7

1500 = 750 km hr. 2

PROBLEMS ON MIXTURE:

Note: 1. The word Alligation literally means linking. The rule takes its name from the lines or links used in working out questions on mixture. 2. Alligation method is applied for percentage value, ratio, rate, prices, speed etc. and not for absolute values. 3. Alligation is the rule that enables us to find the proportion in which two or more ingredients at the given price must be mixed to produce a mixture at a given price. Cost price of unit quantity of the mixture is called the mean price. 4. Rule of alligation: If 2 quantities are mixed in a ratio, then

a

f a

f

C. P. of dearer − Mean Price Quantity of cheaper = Quantity of dearer Mean Price − C.P. of Cheaper

a

f a

i.e., cheaper quantity: dearer quantity = d − m: m − c. We represent it as Cost price of cheaper (c)

f C.P. of dearer (d)

Mean Price (m) (d − m)

m−c

WORKED EXAMPLES: 1. The price of first quality of rice is Rs. 16 per kg and that of second quantity rice is Rs. 10. In what ratio these two should be mixed so that the mixture can be sold for Rs. 12 per kg. Solution: Cost price of 1 kg cheaper Rice 10 (c)

C.P. of 1 kg dearer rice 16 (d) Mean Price 12 (m)

(d − m) = 16 − 12 =4

m−c = 12 − 10 =2

214

Basic Mathematics

Quantity of cheaper rice : Quantity of dearer rice. =4:2 =2:1

OR Let us mix these 2 types of rice in the ratio x : y. The price of x kgs. cheaper rice = 10 x Price of y kgs. dearer rice = 16 y The price of (x + y) kgs. of mixture = 10x + 16y Price of 1 kg of mixture =

10 x + 16 y x+y

Given: Price of 1 kg of mixture = 12

10 x + 16 y = 12 x+y

∴

10 x + 16 y = 12 x + 12 y

16 y − 12 y = 12 x − 10 x 4y = 2 x 4 x = 2 y ⇒ x: y=4:2 ∴ x : y = 2 :1. 2. Arjun travelled a distance of 80 km in 7 hrs. partly in bullock cart at the rate of 8 km/hr and partly in tonga at 16 km/hr. Find the distance travelled in bullock cart. Solution: Average distance travelled in

1 hr =

80 km 80 = km hr. 7 hrs. 7

Distance travelled in 1 hr in bullock cart 8 km (c)

Distance travelled in 1 hr in tonga 16 km (d) Average in 1 hr

a f

80 m 7 (d − m)

16 −

80 32 = 7 7

(m − c)

80 24 −8= 7 7


215

Time taken by bullock cart d − m = m−c Time taken by tonga 32 = 7 24 7 32 = 4 : 3. 24 ∴ Out of 7 hrs., he took 4 hrs. to travel in bullock cart and 3 hrs. in tonga. Distance covered by bullock cart = 4 × 8 km = 32 km. 3. In what ratio must a person mix three kinds of rice costing Rs. 12.00, Rs. 14.00 and Rs. 17.40 per kg. So that mixture may be worth Rs. 14.10 per kg. Solution: Step 1: Mix 1st and 3rd kind of rice to get a mixture worth Rs. 14.10. C.P. of 1 kg rice C.P. of 1 kg of rice of 1st kind of 3rd kind Rs. 12.00 (c) Rs. 17.40 (d) Mean Price Rs. 14.10 (d−m) = Rs. 3.30 By alligation rule

(m−c) Rs. 2.10

Quantity of 1st kind of rice d − m = Quantity of 3rd kind of rice m − c =

3.30 11 = . 2.10 7

∴ They must be mixed in the ratio 11 : 7. Step II: Mix rice of 1st and 2nd kind to obtain a mixture worth Rs. 14.10 per kg. C.P. of 1 kg rice C.P. of 1 kg of rice of 1st kind of 2nd kind Rs. 12.00 (c) Rs. 14.40 (d) Mean Price (m) Rs. 14.10

(d−m) = Rs. 0.30

(m−c) = Rs. 2.10

216

Basic Mathematics

By alligation rule,

Quantity of 1st kind of rice d − m 0.30 = = Quantity of 2nd kind of rice m − c 2.10 =

1 . 7

∴ They must be mixed in the ratio 1 : 7. Now 1st kind : 2nd kind = 1 : 7 1st kind : 3rd kind = 11 : 7 To make the value of 1st kind equal, Multiply 1st ratio by 11 and 2nd ratio by 1. ∴ 1st kind : 2nd kind = 11 : 77 1st kind : 3rd kind = 11 : 7 ∴ 1st kind : 2nd kind : 3rd kind = 11 : 77 : 7. 4. Gauri possessing Rs. 84,000 lent a part of it at 8% simple interest and the remaining at 6 and 2/3% simple interest. Her total income after 1 and 1/2 years was Rs. 8820. Find the sum lent at different rates. Solution: Given: P = 84000 I = 8820 T= 1

1 3 yrs.= yrs. 2 2

R=? We know,

I= R=

PRT 100 I ⇒R= 100 PT 100 × 8820 = 7% 3 84000 × 2

Rate of Interest cheaper (c) 6 and 2/3%

Rate of Interest dearer (d) 8% (m) Average rate 7%

(d − m) = 8 − 7 = 1

m − c 7−6

2 1 = 3 3


217

By alligation rule,

2 Money given at 6 % SI 1 3 3 = = . 1 1 Money given at 8% SI 3 =3:1 Sum = 3 + 1 = 4. ∴ Money lent at 6 and 2/3% SI =

3 × 84000 4

= Rs. 63000. Money lent at 8% SI =

1 × 84000 4

= Rs. 21000. 5. Adarsh buys 2 horses for Rs. 1350 and sells one at 6% loss and other at 7.5% gain and on the whole, he neither gains, no loses. What does each horse cost? Solution: Cheaper horse −6% (c)

Dearer horse 7.5% (d) Mean O

(d − m) 7.5

(m − c) 6

Cost of 1st horse 7.5 75 5 = = = Cost of 2nd horse 6 60 4 1st horse : 2nd horse = 5 : 4 Sum = 5 + 4 = 9 ∴

Cost of 1st horse =

5 × 1350 = Rs. 750 9

Cost of 2nd horse =

4 × 1350 = Rs. 600. 9

REMEMBER: • For the given ratio a : b, duplicate ratio is a2 : b2, subduplicate ratio is is a3 : b3 and subtriplicate ratio is

3

a :3 b .

a : b , Triplicate ratio

218

Basic Mathematics

a : b : : c : d iff ad = bc. If a : b = b : c. Then b is called mean proportional and c is called 3rd proportional. If a : b = c : d. Then d is called fourth proportional. If a : b = c : d. Then (i) b : a = d : c (Invertendo) (ii) a : c = b : d (Alternendo) (iii) a + b : b = c + d : d (Componendo) (iv) a − b : b = c − d : d (Dividendo) (v) a + b : a − b = c + d : c − d (Componendo and dividendo) • If a : b = c : d represent a direct proportion then a : b = d : c or b : a = c : d represent an inverse proportion and vice-versa.

• • • •

Quantity of cheaper C.P. of Dearer − Mean • Rule of alligation: Quantity of dearer = Mean − C. P. of cheaper

acf d−m

amf

adf i.e., cheaper : dearer = d − m : m − c.

m−c

EXERCISE 1. Express the following ratios in their simplest form : (a) 8 : 24 (b) 24 : 88 (c) 9 : 900 2. Compare the following ratios : (a) 6 : 5 and 3 : 2 (b) 1 : 5 and 6 : 7 (c) 3 : 2 and 2 : 7 3. Find the ratio between : (a) 1 hr 15 min and 105 min (b) 3 kg 50 gm and 4 kg 500 gm 4. A number is divided into 3 parts in the ratio 2 : 3 : 4. If the 3rd part is 32. Find the other parts. 5. Find the third proportional to 2 : 6. 6. Find the mean proportional between 49 and 64. 7. Find the fourth proportional to 4, 5 and 12. 8. What number must be subtracted from each of 9, 11, 15 and 19. So that the difference will be proportional. 9. (a) If x : y = 2 : 3, y : z = 4 : 5. Then find z : x. (b) If x : y = 2 : 3, y : z = 4 : 5 and z : w = 6 : 7. Then find x : w. (c) If a : b = 2 : 3, b : c = 4 : 5; c : d = 6 : 7. Then find a : b : c : d.


219

10. Two numbers are in the ratio 5 : 8. If 9 is added to each then they are in the ratio 8 : 11. Find the numbers. 11. The ratio between the ages of Khan and Ranjith is 6 : 5 and the sum of their ages is 44 years. Find the ratio of their ages after 8 years. 12. One year ago the ratio of between Sarala and Saraswathi’s salary was 3 : 4. The ratio of their individual salaries between last year’s and this year’s salaries are 4 : 5 and 2 : 3 respectively. At present the total of their salary is Rs. 4160. Find the salary of Sarala now. 13. The ratio between Sumit’s and Prakash’s age at present is 2 : 3. Sumit is 6 years younger than Prakash. Find the ratio of Sumit’s age to Prakash’s age after 6 years. 14. 8 labourers can build a wall in 6 days. In how many days, 12 labourers can do the same work? 15. 6 carpenters working 7 hrs. a day can complete 24 tables in 20 days. How many days will 12 carpenters working 6 hrs. a day take to complete 36 tables. 16. A, B, C start a business with investments of Rs. 25,000, Rs. 16,000 and Rs. 12,000 respectively. If the profit for the year amounts to Rs. 6850. Find the share of each partner. 17. Ram can reap a field in 9 days which Deepak alone can reap in 12 days. In how many days both together can reap this field. 1 18. A can do F I H 3K

19.

20. 21. 22. 23. 24. 25. 26. 27. 28.

rd

F 2I of the work in 5 days and B can do H K 5

th

of the work in 10 days. In how many

days both A and B together can do the work? Sunil completes a work in 4 days, whereas Dinesh completes the work in 6 days. Ramesh works 1 and 1/2 times as fast as Sunil. How many days it will take for the 3 together to complete the work. A can complete a job in 9 days, B in 10 days and C in 15 days. B and C start the work and are forced to leave after 2 days. Find the time taken to complete the remaining work. If 3 men or 4 women can construct a wall in 43 days. Then find the number of days that 7 men and 5 women take to construct it. 8 men can dig a pit in 20 days. If a man works half as much again as a boy then in how many days 4 men and 9 boys can dig a similar pit. A train leaves Meerut at 6 a.m. and reaches Delhi at 10 a.m. Another train leaves Delhi at 8 am. and reaches Meerut at 11 : 30 a.m. At what time do the 2 trains cross each another. A boy goes to school with a speed of 3 km/hr and returns to the village with a speed of 2 km/hr. If he takes 5 hrs. in all, find the distance between village and the school. In what proportion must Ragi at Rs. 3.10 per kg be mixed with Ragi at Rs. 3.60 per kg so that the mixture be worth Rs. 3.25 a kg. A man possessing Rs. 1000 lent a part of it at 6% S.I. and the other at 8% SI the yearly income is Rs. 75. Find the sum lent at 8% SI. Kantilal mixes 80 kgs of sugar worth Rs. 6.75 per kg with 120 kg worth Rs. 8 per kg. At what rate he sell the mixture to gain 20%. A jar contains a mixture of 2 liquids A and B in the ratio 7 : 5. When 9 litres of mixture is drawn off and the jar is filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid A was contained by the jar initially?

220

Basic Mathematics

29. A sum of Rs. 41 was divided among 50 boys and girls each boy gets 90 paise and a girl 65 paise. How many boys are there?

ANSWERS 1. 2. 3. 4. 9. 10. 14.

(a) 1 : 3 (b) 3 : 11 (c) 1 : 100 (a) 3 : 2 is greater than 6 : 5 (b) 6 : 7 is greater than 1 : 5 (c) 3 : 2 is greater than 2 : 7. (a) 5 : 7 (b) 61 : 90 16 and 24 5. 18 6. 56 7. 15 8. 3 (a) 15 : 8 (b) 16 : 35 (c) 16 : 24 : 30 : 35 15 and 24 11. 8 : 11 12. Rs. 1600 13. 3 : 4 4 15. 105 16. Rs. 2650, Rs. 2200, Rs. 2000.

17. 5

1 days 7

18. 9

3 days 8

22. 16 days 23. 8.56 a.m. 27. Rs. 9 per kg

19. 1

5 days 19

24. 6 km. 28. 21 litres

20. 6 days

21. 12 days

25. 7 : 3

26. Rs. 750 29. 34.

8 Averages 8.1 INTRODUCTION: Condensation of data is necessary in statistical analysis because a large number of big figures are not only confusing to mind but also difficult to analyse. In order to reduce the complexity of data and to make them comparable it is essential that the various phenomena which are being compared are reduced to one figure each. It is obvious that a figure which is used to represent a whole series should neither have the lowest value in the series nor the highest value but a value somewhere between these two limits, possibly in the centre where most of the items of the series cluster. Such figures are called measures of central tendencies or averages. Averages are usually of the following types: (a) Mathematical average (i) Arithmetic average or mean (ii) Geometric mean and (iii) Harmonic mean. (b) Average of position: (iv) Median and (v) Mode. Of the above mentioned five important averages, we are going to discuss in this book, about arithmetic average or Mean in detail.

8.2 ARITHMETIC AVERAGE OR MEAN: Mean of the certain number of quantities which are all of equal weightage or importance is the figure obtained by dividing the total values of various items by their number. If the marks obtained by a student in five subjects are 56, 67, 82, 43 and 52, then to find the mean of these marks we shall add these marks and divide the total so obtained by the number of items which is 5. ∴

Mean =

56 + 67 + 82 + 43 + 52 5

222

Basic Mathematics

Mean =

300 = 60. 5

In general, if x1, x2, x3, ... xn are the values in a data containing n items then their mean, denoted by

X is X =

x1 + x 2 + x 3 + ... + x n n n

i.e.,

X=

∑ xi i =1

n

If all the quantities are not of equal importance i.e. equal weight, we compute the weighted average as follows: If w1, w2, w3, ... wn are the weights associated to the values x1, x2 ... xn respectively the weighted average Xw is given by

Xw =

x1w1 + x 2 w2 + ... + x n wn w1 + w2 + ... + wn n

∑x w i

Xw =

i

i =1 n

∑w

i

i =1

For example if Rama scores 35 in 2 subjects, 42 in 4 subjects and 72 in remaining 4 subjects. Then average marks scored by Rama

=

35 × 2 + 42 × 4 + 72 × 4 2+4+4 =

70 + 168 + 288 10 =

526 = 52.6. 10

Note: To compute arithmetic mean in a continuous series, the midpoints of the various class intervals are written down to replace the class intervals. Once it is done, there is no difference between a continuous series and discrete series.

8.3 COMBINED AVERAGE: It is nothing but average of averages. Combined average is computed when the data set consists of different groups and average for each group is known.

Averages

223

If X1 , X2 ... Xn , are the average of groups 1, 2, ..., n and N1, N2, ... Nn are number of quantities in groups 1, 2, 3, ... n. Then combined average X1, 2, 3 ... n is given by

X1, 2, 3 ... n =

X1 N1 + X2 N2 + Xn N n N1 + N 2 + ... + N n

For example if the average marks of group of 5 students is 56 and average marks of another group of 6 students is 72 then combined average marks of 11 students.

=

56 × 5 + 72 × 6 5+6

=

280 + 432 712 = ≈ 64.72. 11 11

WORKED EXAMPLES: 1. Find the arithmetic mean of 56, 87, 36, 72 and 44: Solution: Arithmetic mean X = x1 + x 2 + x3 + x 4 + x 5 5

=

56 + 87 + 36 + 72 + 44 5 =

295 = 59. 5

2. A market survey on demands of chocolates at a local shop provided the following distribution of daily demand: Daily demand Frequency (Number of chocolates) (Number of days) 20 5 30 27 40 62 50 4 60 2 Total: 100 Find the average demand of chocolates in numbers/day. Solution:

Mean =

x1w1 + x 2 w2 + x 3 w3 + ... + x n wn w1 + w2 + w3 + ... + wn

224

Basic Mathematics

Mean =

20 × 5 + 30 × 27 + 40 × 62 + 50 × 4 + 60 × 2 5 + 27 + 62 + 4 + 2

Mean =

100 + 810 + 2480 + 200 + 120 100

X = 37.1. 3. Find the arithmetic mean of the following data: Class Interval: 0-20 20-50 50-90 Frequency: 10 20 40 C.I. Mid value Frequency (x) (w) 0-20 10 10 20-50 35 20 50-90 70 40 90-140 115 15 140-200 170 15 ΣW = 100

Xw =

90-140 15 X.W

140-200 15

100 700 2800 1725 2500 ΣXW = 7825

x1w1 + x 2 w2 + ... + x n wn w1 + w2 + ... + wn

Xw =

ΣXw 7825 = = 78.25. 100 Σw

4. If 10 books are purchased at the rate of Rs. 7 each, 15 books are purchased at the rate of Rs. 9.50 and 20 books are purchased at the rate of Rs. 13.50, find the average price of a book. Solution: Given: N1 = 10, N2 = 15 and N3 = 20 X1 = 7, X2 = 9.50 and X3 = 13.50. Combined average =

∴

X1 N1 + X2 N 2 + X3 N3 N1 + N 2 + N3

=

7 × 10 + 9.50 × 15 + 13.50 × 20 10 + 15 + 20

=

70 + 142.50 + 270 45

= 10.7222. Average price of a book = Rs. 10.72. 5. The average marks of 15 students of class is 45. A student who has secured 17 marks leaves the class. Find the average marks of the remaining 14 students.

Averages

225

Solution: Average marks of 15 students = 45 ∴ Total marks = 45 × 15 = 675 If the one student with 17 marks leaves the class then total marks of remaining 14 Students = 675 − 17 = 658 Average marks of 14 students =

658 = 47. 14

6. The average age of 10 students is 6 years. The sum of the ages of 9 of them is 52 years. Find the age of the 10th student. Solution: Given: Average age of 10 students = 6 years Total age of 10 students = 10 × 6 = 60 years Given, the sum of ages of 9 of them = 52 years. ∴ Age of the 10th student = 60 − 52 = 8 years. 7. The average age of 10 students is 14 years. Among them, the average age of 5 students is 12 years. Find the average age of remaining students: Solution: Given: Combined Average X = 14.

N1 + N2 = 10 N1 = 5, N2 = 5, X1 = 12, X2 = ? We have

X=

X1 N1 + X2 N 2 N1 + N 2

14 =

12 × 5 + X2 × 5 5+5

14 =

60 + 5 X2 10

140 = 60 + 5 X2 5 X2 = 140 − 60 = 80 X2 =

80 = 16. 5

∴ Average age of remaining 5 students = 16 years. 8. A shopkeeper purchased a certain number of dress materials at an average price of Rs. 190 each. The average price of 10 dress materials was Rs. 175 and that of remaining dress materials was Rs. 200. Find the total number of dress materials purchased. Solution: Combined mean X = 190

(Given)

226

Basic Mathematics

N1 = 10, X1 = 175 X2 = 200 Let total number of dress materials purchased = x N1 + N 2 = x

10 + N 2 = x N 2 = x − 10

We have

X=

190 =

N1 X1 + N2 X2 N1 + N 2

a f a

f

10 175 + x − 10 200 x

190 x = 1750 + 200 x − 2000

190 x = −250 + 200 x 250 = 200 x − 190 x 250 = 10 x

⇒

x=

250 = 25 10

∴ Total number of dress materials purchased = 25. 9. The average weight of a group containing 26 persons is 70 kg. 6 persons with average weight 67 kg leave the group and 5 persons with weights 68, 72, 82, 56 and 54 kgs. joins the group. Find the average weight of the group now. Solution: Average weight of 26 persons = 70 kgs. (Given) Total weight of 26 persons = 26 × 70 = 1820 6 persons with average weight 67 kgs leave the group. (Given) i.e., Total weight of 6 persons = 67 × 6 = 402 Total weight of the remaining 20 persons = 1820 − 402 = 1418 5 persons with weight 68, 72, 82, 56 and 54 kgs. join the group. ∴ Total weight of (20 + 5) persons = 1418 + 332 = 1750 Average weight of the group now =

1750 = 70 kgs. 25

Averages

227

10. A batsman realises that by scoring a century in the 11th innings of his test matches he has bettered his average of the previous 10 innings by 5 runs. What is his average after the 11th inning: Solution. Let the average runs in 10 innings be x Then total runs in 10 innings = 10x Average runs after 11th innings = x + 5 (Given) Total runs in 11th innings = (x + 5) 11 Also given a batsman scores a century in the 11th innings. ∴ Runs in 11 innings — Runs in 10 innings. = Runs in 11th innings = 100.

a x + 5f11 − 10 x = 100 11x + 55 − 10 x = 100

x = 100 − 55 x = 45. ∴ Average runs after 11th innings = x + 5 = 45 + 5 = 50. 11. Ms. Vani bought 17 books in a discount sale. The average price of books being Rs. 53. The average price of the eleven Kannada books is Rs. 71. If the prices of the remaining 6 English books form an increasing arithmetic progression with last term Rs. 25. Find the price of cheapest English book. Solution. X = 53

Given:

X1 = 71, N1 = 11 X2 = ?, N2 = 6 We have

X=

53 =

53 = ⇒

N1 X1 + N 2 X2 N1 + N2

a f d i

11 71 + 6 X2 11 + 6

781 + 6 X2 17

781 + 6 X2 = 53 × 17 6 X2 = 901 − 781 6 X2 = 120 X2 = 20.

228

Basic Mathematics

∴ Total cost of 6 English books = 6 × 20 = Rs. 120. Given the cost of 6 English books form an increasing A.P. ∴ Costs are a, a + d, a + 2d, a + 3d, a + 4d and a + 5d. Given last term = 25 a + 5d = 25 5d = 25 − a Also sum of first n terms in an A.P. is

∴

a f

Sn =

n 2a + n − 1 d 2

S6 =

6 2a + 6 − 1 d 2

a f

S6 = 3 2 a + 5d

Substituting we get 120 = 3 2 a + 25 − a

120 = a + 25 3 40 = a + 25 40 − 25 = a ⇒ a = 15. ∴ Cost of the cheapest English book = Rs. 15. 12. At a place, the average temperatures from Monday to Thursday was 35°C and from Tuesday to Friday was 38°C. Find the day temperatures on Monday and Friday if the ratio of temperatures on Monday and Friday is 5:7. Solution: Average temperatures from Monday to Thursday = 35° (Given)

i.e., ∴

Temp. on Mon + Tue + Wed + Thu = 35° 4

Temperature on Mon + Tue + Wed + Thu = 35° × 4 = 140° Average temperature from tuesday to friday = 38° (Given)

...(1)

Temp. on Tue + Wed + Thu + Fri = 38° 4 ∴

Temperature on Tue + Wed + Thu + Fri = 38° × 4 = 152° Equation (2) − Equation (1) = Temp on Fri − Temp. on Monday = 152° − 140°

...(2)

Averages

Temp. on Fri − Temp. on Monday = 12° ∴ Temp. on Fri = 12° + Temp. on Monday Also given ratio of temperature on Monday by Friday = 5:7

229

...(3)

Temp. on Mon. 5 = . Temp. of Fri. 7

∴

7 (Temperature on Monday) = 5 (Temperature on Friday) 7 (Temperature on Monday) = 5 (12 + Temperature on Monday) 7 x = 60 + 5 x

7 x − 5 x = 60 ⇒ 2 x = 60 ⇒ x = 30 ∴ Temperature on Monday = 30° From (3) Temperature on Friday = 12 + Temperature on Monday = 12° + 30° = 42°. 13. In a hostel of 80 students, the total monthly expenses were increased by Rs. 8000 when 20 more join the hostel. If the average monthly expenses thereby reduced by Rs. 40 per head, Find the average monthly expenses per student. Solution. Let the average monthly expenses per student be x. ∴ Total monthly expenses for 80 students = 80 × x = 80x When 20 more join the hostel, monthly expenses increases by Rs. 8000. ∴ Total monthly expenses of 100 students = 80x + 8000 Monthly average is reduced by 40 ⇒ x − 40. ∴ Monthly expenses of 100 students = 100 (x − 40)

a

∴

f

100 x − 40 = 80 x + 8000 100 x − 4000 = 80 x + 8000

100 x − 80 x = 8000 + 4000 20 x = 12000

⇒ x=

12000 = 600. 20

∴ Monthly expenses per student = Rs. 600. 14. Out of 3 numbers, the first is twice the second and is half of the third. If the average of the three numbers is 56, Find the 3 numbers. Solution. Let the 3 number be x, y and z. Given:

x = 2y =

z x = 2y i.e., z = 4y 2

Also given average = 56. ∴

x+ y+ z = 56 3

230

Basic Mathematics

x + y + z = 56 × 3 = 168 2 y + y + 4 y = 168

7 y = 168 ⇒

y=

168 = 24 7

∴

The 3 numbers are x = 2y = 2 (24) = 48 y = 24 and z = 4y = 4 (24) = 96. 15. The average expenditure of a man for the first five months is Rs. 1200 and for the next seven months it is Rs. 1300. If he saves Rs. 2900 in that year, then find his average monthly income. Solution. Given average expenditure for 5 months = Rs. 1200 ∴ Total expenditure for 5 months = 5 × 1200 = Rs. 6000. Also given average expenditure for 7 months = Rs. 1300 ∴ Total expenditure for 7 months = 7 × 1300 = Rs. 9100 ∴ Total expenditure for 12 months = Rs. 6000 + Rs. 9100 = Rs. 15,100 Savings = Rs. 2900 (Given) We know Saving = Income − Expenditure ⇒ Income = Expenditure + Saving ∴ Total income = Rs. 15100 + 2900 = Rs. 18,000 Average monthly income =

18000 = Rs. 1500. 12

16. Ten years ago, the average age of a family of 4 members was 24 years. Two children having been born, the average age of the family is same today. What is the present age of the youngest child if they differ in age by 2 years. Solution: Let the age of youngest child be x yrs. ∴ Age of next child = x + 2 years. 10 years ago Average age of 4 members = 24 Total age of 4 members = 24 × 4 = 96. After 10 years, Total age of 4 members = 96 + 4 × 10 [Each member’s age increases by 10 years] = 96 + 40 = 136.

Averages

231

Now average of 6 members

=

a

136 + x + x + 2 = 24 given 6

f

136 + 2 x + 2 = 24 × 6 138 + 2 x = 144

2 x = 144 − 138 2x = 6 x=3

∴ Present age of the youngest child = 3 yrs. 17. The average ages of A and B is 42 yrs., that of B and C is 28 yrs. and that of C and A is 40 yrs. Find the ages of A, B and C. Solution: Let the ages of A, B and C be a, b and c. Given average of A and B = 42

a+b = 42 2

⇒

a + b = 84 Given: Average age of B and C = 28 years.

...(1)

b+c = 28 2

b + c = 56

...(2)

Also given average age of C and A is i.e., ⇒ Solving (1) and (2) we get

c+a = 20 2

c + a = 40 a + b = 84 b + c = 56 − − − a − c = 28

a fa fa f Solving (3) and (4) we get

c + a = 40 a − c = 28 ⇒ 2a = 68 ⇒

a = 34 a + c = 40 ⇒ 34 + c = 40 ⇒ c = 6

...(3)

...(4)

232

Basic Mathematics

b + c = 46 ⇒ b + 6 = 46 ⇒ b = 40

∴

Age of A = 34 yrs. B = 40 yrs. and C = 6 yrs.

REMEMBER: • Mean X = • Xw =

x1 + x 2 + ... + x n n

w1 x1 + w2 x 2 + ... + wn x n w1 + w2 + ... + wn

• X123 =

X1 N1 + X2 N2 + ... + XnN n N1 + N 2 + ... + N n

EXERCISES 1. A cricketer makes 72, 59, 101, 18 and 10 runs respectively in 5 matches played by him. Find his average score. 2. The average weight of a class of 24 students is 35 kgs. If the weight of the teacher is included, the average rises by 400 gms. Find the weight of the teacher. 3. The average marks of 15 boys of a class is 65 and 11 girls of the same class is 78. Find the average marks of the students. 4. 3 tests in English, 2 in Hindi, 4 in Kannada and 5 in Sociology are conducted. The average marks scored by Raju in English is 60, that in Hindi is 56 and in Kannada is 45. If the average marks of all the subjects taken together is 48. Then find the average marks scored by him in Sociology. 5. 10 shirts and 5 pants were bought for Rs. 6000. If the average price of a shirt is Rs. 450, then find the average price of a pant. 6. The average of 25 results is 18, that of first 12 is 14 and of the last 12 is 17. Then find the 13th result. 7. The average age of A, B, C, D five years ago was 45 years. By including X, the present age of all the five is 49 years. Find the present age of X. 8. A batsman makes a score of 87 runs in the 17th innings and thus increased his average by 3. Find his average after 17th inning. 9. Miss Radha bought 51 dress materials in a discount sale. The average price of a dress material being Rs. 318. The average price of 33 polyster dress materials is Rs. 426. If the prices of the remaining cotton dress materials form an increasing arithmetic progression with last term 150. Find the price of the cheapest cotton dress material. 10. The average temperature for Monday, Tuesday and Wednesday was 40°C. The average temperature for Tuesday, Wednesday and Thursday was 41°C. If the temperature on thursday was 42°C. Then find the temperature on Monday.

Averages

233

11. The average age of a husband and a wife was 23 years when they were married 5 years ago. The average age of the husband, the wife and a child who was born during the interval is 20 years now. How old is the child now? 12. The average age of 5 members of a committee is the same as it as 3 years ago, because an old member has been replaced by a new member. Find the difference between the ages of old and new member. 13. There were 35 students in a hostel. If the number of students is increased by 7, the expenses of the mess were increased by Rs. 42 per day while the average expenditure per head diminished by Re 1. Find the original expenditure of the mess. 14. The average weight of A, B and C is 45 kg. If the average weight of A and B is 40 kgs and that of B and C is 43 kg. Then find the weight of B. 15. The average salary of 20 workers in an office is Rs. 1900 per month. If the manager’s salary is added, the average salary becomes Rs. 2000 per month. What is the manager’s annual salary?

ANSWERS 1. 52 runs 7. 45 years 13. Rs. 420

2. 45 kgs 8. 39 14. 31 kg

3. 70.5 4. 40 9. Rs. 39 10. 39°C 15. Rs. 48,000.

5. Rs. 300 11. 4 years

6. 78 12. 15 years

234

Basic Mathematics

9 Bill Discounting 9.1 INTRODUCTION: Suppose a merchant A purchases goods worth say Rs. 50,000 from another merchant B at a credit for certain period say 6 months. Then B draws up a draft i.e. prepares a special bill called ‘Hundi’ or bill of exchange. On the receipt of the goods, A gives an agreement dually signed on the bill stating that he has accepted the bill and the money can be withdrawn from his bank account after 6 months of the date of bill. On this bill, there is an order from A to his bank asking to pay Rs. 50,000 to B after 6 months. If B needs the money of this bill earlier than 6 months, then B can sell the bill to a banker or a broker who pays him the money against the bill but somewhat less than the face value. Bill discounting is essentially lending by the banker against the bills and the bankers charges a certain interest for doing this service.

9.2 TERMINOLOGY: Bill: The document which required an individual to pay a fixed amount after a fixed later date is called a bill. Discount: When a bill is cashed in advance of its date of maturity an amount is deducted by the money lender or bank from the amount of the bill due. This amount deducted is called Discount. True discount: The present value of a sum of money is that principal which, if placed on stipulated rate for a specified period will amount to that sum of money at the end of the specified period. The interest on the present value of the bill is called true discount. Bank er’ Banker’ er’ss discount: The interest on the face value or amount of the bill is called Banker’s discount. Bank er’ ain: The difference between banker’s discount and true discount is called Banker’s Banker’ er’ss ggain: gain.

Bill Discounting 235

Illustration: Bill of Exchange: Stamp

No. 1843, K.R. Road, Bangalore 10.2.2005

To,

Anil, BTM Layout, Bangalore Six months later pay me or my order the sum of Rs. 50,000 (Rupees fifty thousand) for value received. Anil Akram In the above bill of exchange. Drawer is Mr Akram so payee is Akram. Drawee or Acceptor is Mr. Anil. Drawing date: 10.2.2005 Bill amount or Face value of Bill = Rs. 50,000 = Nominal value of the bill. Legally due date: drawing date + Bill period + grace period of 3 days (It is customary to give 3 days grace period). So In the above bill, Legally due date is 13.8.2005

Date of drawing : 10 - 2 - 2005 Bill period : 0 - 6 - 0 Q

Grace period : 3 - 0 - 0 Legally due date : 13 - 8 - 2005

9.3 FORMULAE: If F is the face value of the bill, BD is the banker’s discount, t is the time in years, R = Rate of interest

r=

R ; TD = True discount; BG = Banker’s gain, P = Present value of the bill, then 100

1. 2. 3. 4.

BD = Ftr TD = Ptr BG = BD − TD Banker’s present worth or the discounted value of bill = F − BD F − Ftr

a

F 1 − tr

5. True present worth Pw =

F 1 + tr

f

236

Basic Mathematics

6. F =

BD × TD BG

7. BG = TD ⋅ t ⋅ r

aTDf where P bP g TD = a P. w.f × a B. G.f 2

8. BG =

w

9.

w

= True present worth.

WORKED EXAMPLES: 1. Find the present value, true discount, Banker’s discount and Banker’s gain on a bill of Rs. 10,450 due in 9 months at 6% per annum. Solution: Given: F = Rs. 10,450

9 3 yrs. = yrs. 12 4

t = 9 months = R = 6% ⇒ r =

6 = 0.06 100

Banker’s discount = BD = Ftr

BD = 10,450 ×

3 × 0.06 4

BD = 470.25

∴ True present worth,

Banker’s discount = Rs. 470.25.

P=

F , 1 + tr

P=

10,450 3 1 + × 0.06 4

P=

∴

10,450 × 4 4 + 0.18

P = 10,000 Present worth = Rs. 10,000. True discount = Ptr = 10,000 × = 450.

3 × 0.06 4


TD = Rs. 450. Banker’s gain = BD − TD = 470.25 − 450 = 20.25. ∴ Banker’s gain = Rs. 20.25. 2. A bill for Rs. 13,000 was drawn on 3rd Feb 2005 at 6 months date and discounted on 13th Mar 2005 at the rate of 8% p.a. For what sum was the bill discounted and how much did the banker gain on this? ∴

Date of Bill : 3 - 2 - 2005 Bill period : 0 - 6 - 0 Grace period : 3 - 0 - 0 Legally due date : 6 - 8 - 2005

Solution:

But bill was discounted on 13th Mar 2005. ∴

Period = 18 days + 30 days + 31 + 30 + 31 +

B

B

B

B

B

B 6

aMarchf aAprilf aMayf aJunef aJulyf aAugf t = 146 days = R = 8% ⇒ r =

146 yrs. 365

8 = 0.08 100

F = Rs. 13,000 BD = Ftr = 13000 ×

146 × 0.08 365

BD = 416.

∴ Present worth,

Banker’s discount = Rs. 416.

P=

F 1 + tr

P=

13000 146 1+ × 0.08 365

P=

13000 = 12596.899 1.032

P ≈ 12596.90

True discount = Ptr = 12596.90 ×

146 × 0.08 365

238

Basic Mathematics

TD = 403.10 True discount = Rs. 403.10 Banker’s gain = BD − TD = 416 − 403.10 = 12.90 Banker’s gain = Rs. 12.90. 3. A banker pays Rs. 2440 on a bill of Rs. 2500, 73 days before the legally due date. Find the rate of discount charged by the banker? Solution: Given: F = Rs. 2500 Banker’s discount = Rs. 2500 − 2440 BD = Rs. 60 t=

73 yrs. 365

R=? BD = Ftr

We have

60 = 2500 × r=

⇒

73 ×r 365

60 × 365 2500 × 73

r = 0.12 R = 100r = 0.12 × 100 R = 12%. Rate of discount charged by banker = 12%.

∴

1 of the banker’s discount and the rate of interest is 20% p.a. Find 5 the unexpired period of the bill.

4. The banker’s gain on a bill is

Solution: Given: B. G. =

1 × BD 5 R = 20% ⇒ r =

20 = 0.2 100

BG = BD − TD

1 BD = BD − TD 5 TD = BD −

1 BD 5


TD =

F 4 BD Q P = 1 + tr 5

F F I tr = 4 ⋅ Ftr H 1 + tr K 5 ⇒

1 4 = 1 + tr 5

⇒

1 4 = 1 + 0.2 t 5

a f

5 = 4 + 0.8t

Cross multiplying

5 − 4 = 0.8t

t=

⇒

1 = 1.25 yrs. 0.8

t = 12 + 3 = 15 months. 5. The difference between the banker’s discount and true discount on a certain sum of money due in 4 months is Rs. 10. Find the amount of the bill if the rate of interest is 3% p.a. Solution: Given BD − TD = Rs. 10 t = 4 months = R = 3% ⇒ r = Now

4 1 yrs. = yrs. 12 3

3 = 0.03 100

BD − TD = 10

F F I tr = 10 H 1 + tr K 1 O Ftr LM1 − N 1 + tr PQ = 10 L OP 1M 1 F ⋅ 0.03 × M1 − P = 10 3 M 1 + 0.03 × 1 P 3Q N Ftr −

a f LMN 1.101OPQ = 10 0.0001 O F LM N 1.01 PQ = 10

F 0.01 1 −

240

Basic Mathematics

F=

10 × 1.01 0.0001

F = 1,01,000 ∴

Amount of the bill = Rs. 1,01,000. 6. A bill for Rs. 2920 drawn at 6 months was discounted on 10.4.2000 for Rs. 2916. If the discount rate is 5% p.a. On what date was the bill drawn. F = Rs. 2920. Solution: Given: Discounted value Rs. 2916.

5 = 0.05 100 Discounted values = F (1 − tr)

Rate = 5% ⇒ r =

b a fg 2916 = 1 − t a0.05f 2920

2916 = 2920 1 − t 0.05

a f

t 0.05 = 1 −

2916 2920

⇒

t = 0.027398 yrs.

⇒

t = 0.027398 × 365 days

t = 10 days ∴ Number of days from legally due date to discounted date = 10 days. Now Bill was discounted on 10.4.2000 Legally due date = 10.4.2000 + 10 days = 20.4.2000

Bill period = 6 months Grace period = 3 days.

∴

Bill drawing date = 20.4.2000 3.6.0 − Bill drawing date = 17.10.1999

af

7. A bill was drawn on March 8th at 7 months period and was discounted on May 18th at 5%. If the banker’s gain is Rs. 3 find the true discount, the banker’s discount, and the sum of the bill. Solution: Date on which the bill was drawn = March 8th i.e., 8−3 Period (7 months): 7 Grace period: 3−0


Legally due date: 11 − 10 i.e., Oct. 11th. Date on which the bill was discounted: May 18th Time for which the bill has yet to run:

May + June + July + Aug + Sep + Oct 13 30 31 31 30 11 = 146 days. t=

∴

146 2 = yrs. 365 5

Now Banker’s gain = Simple interest on true discount i.e. Rs. 3 is the S.I. on T.D. for

Then

3=

⇒

P=

P×5×

2 at 5% 5

2 5

100 100 × 3 = Rs. 150 2

Banker’s discount = T.D. + S.I. on TD

= Rs. 150 + SI on 150 for

150 +

2 yrs. at 5% 5

2 ×5 5 100

150 ×

= 150 + 3 = Rs. 153. OR B. D.= TD + B. G. BD = 150 + 3 = Rs. 153.

Now

Sum =

BD × TD 153 × 150 = = Rs. 7650. BD − TD 153 − 150

8. A bill was drawn on the 10th July 1960 at 3 months after signed and was accepted on presentation on 1st Aug. 1960. It was discounted on 23rd Aug ’69 at 5% p.a. simple interest to realise Rs. 2475. Find the face values of thee bill and banker’s discount.

Drawing date : 10 - 7 -1960 Bill period : 0 - 3 - 0 Solution:

Grace period : 3 - 0 - 0 Legally due date : 13 -10 -1960

242

Basic Mathematics

Given: Bill was present on 1st Aug. 1969. Number of days for which banker’s discount is paid =

30 days + 30 + 13 = 73 days. Aug Sept Oct

a f a fa f t=

73 = 0.2 yrs. 365

R = 5% ⇒ r = 0.05

Let the face value of bill be Rs. 100. Then BD = Ftr BD = 100 × 0.2 × 0.05 BD = 1. For Rs. 100 face value bill, BD = Re 1 i.e., Rs. 99 is realised. i.e., Rs 99 is realised when face value is Rs. 100 Rs. 2475 is realised when face value is x. ∴

x=

2475 × 100 99

x = 2500.

∴ Now,

Face value of the bill = Rs. 2500. BD = Ftr = 2500 × 0.2 × 0.05 BD = 25.

OR For 100,

BD = 1

For 2500,

BD =

2500 × 10 = Rs. 25. 100

9. The present worth of a bill due sometime hence is Rs. 1100 and the true discount on the bill is Rs. 110. Find the banker’s discount and the extra gain the banker would make in the transaction. Solution: Here time and rate of interest are not given. Given: PW = Rs. 1100 TD = Rs. 110 We have Squaring

a PW f × a BGf

TD =

aTDf

2

= PW × BG


aTDf = a110f 2

⇒

BG =

Now

BG = BD − TD

⇒

BD = BG + TD

2

1100

Pw

= Rs. 11.

BD = Rs. 11 + 110

BD = Rs. 121. 10. The true discount on a bill of Rs. 1860 due after 8 months is Rs. 60. Find the rate, the banker’s discount and the banker’s gain. F = Rs. 1860 Solution: Given: TD = Rs. 60

t = 8 months =

8 2 = yrs. 12 3

r=? BD = ? and BG = ?

Present worth,

P = 1860 − 60 = Rs. 1800. TD = Ptr 60 = 1800 ×

2 ×r 3

60 × 3 =r 1800 × 2 ⇒ Hence

r = 0.05 R = 5%. BD = Ftr BD = 1860 ×

2 × 0.05 3

BD = Rs. 62. BG = BD − TD = 62 − 60 = Rs. 2. OR

BG =

aTDf = a60f 2

Pw

2

1800

= Rs. 2.

BD = TD + BG = 60 + 2 = Rs. 62.

244

Basic Mathematics

REMEMBER: • • • •

Legally due date = Bill Drawing date + Bil period + 3 days (Grace period). BD = Ftr [Simple interest on face value of the bill] TD = Ptr [Simple interest on present worth of the bill] BG = BD − TD

• Present worth =

F 1 + tr

• BG = TD × tr • BG =

aTDf Pw

2

If r and t are not given

• TD = B ⋅ G ⋅ Pw

EXERCISE 1. A bill for Rs. 3500 due for 3 months was drawn on 27th March 2000 and was discounted at the rate of 7% on 18th April 2000. Find the banker’s discount and discounted value of the bill. 2. The banker’s discount and true discount on a sum of money due four months are respectively Rs. 510 and Rs. 500. Find the seem and the rate of interest. 3. The difference between BD and TD on a bill due after 6 months at 4% interest per annum is Rs. 20. Find the true discount bill discount and face value of the bill. 4. The banker’s gain on a certain bill due 6 months hence is Rs. 10, the rate of interest being 10% p.a. Find the face value of the bill. 5. A banker pays Rs. 2340 on a bill of Rs. 2500, 146 days before the legally due date. What is the rate of discount charged by banker? 6. A bill for Rs. 1460 drawn at 3 months was discounted at 4% p.a. on 9th November for 1454.40. On what date the bill was drawn? 7. A bill was drawn on April 14th at 8 months after date and was discounted on July 24th at 5% p.a. If the banker’s gain is Rs. 2, what is the face value of the bill. 8. Find the banker’s discount and cash value of a bill for Rs. 3400/- drawn on April 25th 1996 at 7 months and discounted on September 16th, 1996 at 5%. 9. The banker’s gain of a certain sum due 2 years hence at 5% per annum is Rs. 8. Find the present worth. 10. The present worth of a sum due sometimes hence is Rs. 576 and banker’s gain is Re. 1. Find the true discount. 11. The banker’s gain on a sum due 3 years hence at 5% is Rs. 90. Find the banker’s discount. 12. The banker’s discount on a bill due 1 year 8 months hence is Rs. 50 and true discount on the same sum at the same rate percent is Rs. 45. Find the rate of interest.


13. A bill for Rs. 3500 due for 3 months was drawn on 27th March 2000 and was discounted on 18th April 2000 at 7% rate of interest. Find the banker’s discount and discounted value of the bill. 14. A bill for Rs. 2920 drawn at 6 months was discounted on 10.4.97 for Rs. 2916. If the discount rate is 5% per annum, on what date was the bill drawn? 15. If the difference of simple interest and true discount of a sum due one year 6 months, hence at 8% p.a. is Rs. 81.45. Find the sum.

ANSWERS 1. 3. 5. 7. 9. 11. 13.

Rs. 3451, Rs. 49 T.D. = Rs. 1000, B.D. Rs. 1,020 F = 51,000 16% Rs. 5,100/Rs. 800 Rs. 690 Rs. 49, Rs. 3451

2. 4. 6. 8. 10. 12. 14. 15.

6%, Rs. 255 Rs. 4200 September 11th Rs.34, Rs. 3,366 Rs, 24 6 and 2/3% 17.10.96 Rs. 6335.

246

Basic Mathematics

10 Stocks and Shares 10.1 STOCK: In order to meet the expenses of a certain plan or a big project. Loan is raised from the public at a certain fixed rate of interest. Bonds or promissory notes of a fixed value are issued for sale to the public. If a man purchases a bond of Rs. 1000 at which 5% interest has been fixed. Then the holder of such bond is said to have ‘a Rs. 1000 stock at 5%’. Here Rs. 1000 is called the face value of the stock. Usually a period is fixed for the repayment of the loan i.e., the stock matures at a fixed date only. Now if the person holding a stock is in need of the money before the date of maturity of stock, he can sell the bond to some other person, where by the claim of interest is transferred to that person.

10.2 SHARES: To start a big concern or a business a large amount of money is needed. This is usually beyond the capacity of one or two individuals. Therefore a group of individuals get together and form the company. The company issues a prospectus and invites the public to subscribe. The required capital is divided into equal small parts called shares, each of a particular fixed value. The person who possesses one or more share is called a share holder. Sometimes the company asks its share holders to pay some amount immediately and balance after some period. The total money raised immediately is called the paid up capital.

10.3 DISTINCTION BETWEEN STOCK AND SHARES: The main distinction between stocks and shares is (i) Shares can be issued directly, but stocks cannot be issued directly. Firstly amount is collected on shares from the public. Stock certificates are issued only after collecting full amount of shares. Stocks are never issued unless the shares are issued and subscribed in full by the public. (ii) Shares need not be fully paid, but stocks must be fully paid. For example the value of a share is Rs. 100. This is known as face value of the share. A share of Rs. 100 each may be called up at Rs. 50 each by the company and the balance of Rs. 50 may be paid at later stage by the public as and when demanded by the company. Stock must be fully paid means face value of Rs. 100 on each share must be paid in full by the public in order to get the stock certificate from the company.

Stocks and Shares

247

10.4 TERMINOLOGY: De bentur es: Debentures are long term loans taken by the company from the public. Every person who Debentur bentures: lends such an amount is given a certificate of loan called debentures. Face Value of Shar es: It is the price at which shares are first issued by a company. It is the price printed Shares: on the share certificate. Mar ket Price: It is the price at which the share can be brought or sold on the stock/share market. Mark Par Value of Shar es: When the shares are issued to the public at the face value, it is called par value Shares: of share. Example: When Rs. 10 share is issued at its face value, it is called par value. Abo ve Par: If the market value of the share is more than the face value, it is said to be above par or Abov at premium. Example: When Rs. 10 shares are issued at Rs. 12 the shares are said to be issued at Rs. 2 premium. Belo w Par: If the market value of the share is less than the face value, it is said to be below par or it Below is said to be at discount. Example: When Rs. 100 shares are issued at Rs. 90. Then the shares are said to be issued at 10% discount. Di vidend: It is the portion of the profit of the company which is distributed to the share holders. The Dividend: dividend is always calculated on the face value of the share. Dividends may be cash dividends or share dividends. Bonus shares are known as stock dividends. Ex-Di vidend and Cum-di vidend Prices: Interest on bond is payable on pre-determined dates. If the Ex-Dividend Cum-dividend bond is bought or sold on a date closer to the interest due date, the prices may be quoted Ex-interest or Cum-interest. If the price is ex-interest the selling price of it is not inclusive of interest. If it is quoted cum interest, the buyers will receive the interest amount. In the case of shares, share dividend are paid instead of interest. Cum dividend price quotations are usually higher than the Ex-dividend quotations. Yield: Actual dividend received by the actual amount invested in a stock or shares called yield. i.e.,

Yield =

Dividend Amount invested

i.e.,

=

Nominal interest Amount invested

Br ok er ok er age: Buying and selling of stocks or shares is done through the person called Brok oker erss and Br Brok oker era ‘brokers’ at stock exchange. They charge certain amount called brokerage. Note that when stock/share is purchased, brokerage is added to the cost price and when stock or share is sold brokerage is subtracted from the selling price. hares: Kinds of Shar es: A company may issue two kinds of shares. They are (i) Preference shares (ii) Equity shares (i) Pr Pref efer erence shares: ef er ence shar es: A preference share holder enjoys a preferential claim with regard to the payments of dividend and repayment of capital. The rate of dividend is fixed, but it is paid before profit is distributed to other members.

248

Basic Mathematics

(ii) Equity shar es: An equity share holder has no special rights. The rate of dividend is not fixed. It shares: varies from year to year. An equity share holder is paid dividend only after the claims of preference share holders are satisfied. Quotation: Consider the statement “Government paper mills 11% shares at 110”. This is a quotation. This means a share of the mill having face value Rs. 100 is available for sale at Rs. 110. This share fetches him a dividend of Rs. 11 every year.

WORKED EXAMPLES: 1. Find the cost of 80 shares at 5% if the market value of the share is 93 and its par-value is Rs. 100. If a person invests Rs. 37200 in such shares then find his annual income. Solution. Cost of 1 share = Rs. 93 Cost of 80 such shares = 80 × 93 = Rs. 7440. Now, the person has invested Rs. 37,200. By investing Rs. 93, the person gets 1 share. ∴ By investing Rs. 37200 the person gets

=

37200 × 1 = 400. 93

∴ The person possesses 400 shares. Face value of 1 share = Rs. 100 Face value of 400 shares = 400 × Rs. 100 = Rs. 40000 Annual income = 5% of face value of shares

=

5 × 40,000 100

= Rs. 2,000. 2. Find the yield by investing Rs. 1140 on 15% stock quoted at Rs. 95. Solution.

Yield =


For Rs. 100 stock, Rs. 95 is the amount invested and Rs. 15 is the nominal interest. ∴

Yield =

15 = 0.1578 ≈ 0.16. 95 OR

By investing Rs. 95, 1 stock is obtained By investing Rs. 1140,

1140 × 1 = 12 stocks are obtained. 95

Stocks and Shares

249

1 stock has face value 100/12 stock has face value 1200/Nominal interest = 15% of Face value of stock = 15% of 1200

= Yield =

15 × 1200 = 180. 100 Nominal interest 180 = Amount invested 1140

= 0.1578 ≈ 0.16.

3. Which of the following is better investment? 5 ½% stock at 102 or 4 and 3/4% stock at 106.

1 Solution. Yield from 5 and ½% stock at 102 = 2 102 5

=

11 11 = = 0.05392. 2 × 102 204

3 Yield from 4 and 3/4% stock at 106 = 4 106 4

=

19 19 = 4 × 106 424

= 0.044811. So yield from 5 and ½% stock at 102 is greater. Hence it is a better investment. 4. Find the cash required to purchase Rs. 20,000 stock at 105 (brokerage ½%). Also find the annual dividend received if the company declares dividend of 8 and ½%. Cash required to purchase Rs. 100 stock. Solution.

F H

= 105 +

I K

1 211 = 2 2

∴ Cash required to purchase Rs. 20,000 stock =

=

20,000 × 100

20,000 × 211 = Rs. 21,100. 2 × 100

Annual dividend for Rs. 100 stock = 8 and ½ Rs.

211 2

250

Basic Mathematics

=

Annual dividend for Rs. 20,000 stock

20,000 × 8

1 2

100

= 200 ×

17 = Rs. 1,700. 2

5. A person has invested a certain sum of money in 13% stock at 96. He sold the investment when the market value went up to 101.5. He gained Rs. 1470 in this process. If he has paid the brokerage at 2% for all the transaction, what was the amount of cash investment and what was the stock value of the investment in the first instance. Solution. Let the amount invested = Rs. x In the first instance, Cost of 1 share = 96 + 2 (Brokerage) i.e., Cost of 1 share = Rs. 98. Now For Rs. 100 share — Rs. 98 is the amount received. For Rs. x share —

98 x is the amount received. 100

When the market value went upto Rs. 101.5. Cost of 1 share = 101.5 − 2 (Brokerage) = 99.5. Amount received from Rs. 100 − 99.5 Amount received for Rs. x —

99.5 x 100

The person gained Rs. 1470 in this process. (Given) ∴

−

98 x 99.5 x + = 1470 100 100 1.5 x = 1470 100 x=

470 × 100 1.5

x = 98,000 ∴ Amount invested by a person = Rs. 98,000. In the first instance, For Rs. 98 he gets Rs. 100 worth stock

Stocks and Shares

251

98000 × 100 98 Rs. 1,00,000. ∴ Stock value of the investment in 1st instance = Rs. 1,00,000 and cash investment = Rs. 98,000. For Rs. 98000 he gets

6. Vivek has Rs. 16,500 stock in 3%. He sells it out at 101

1 and invests the proceeds in 4% railway 8

7 1 debentures at 131 . Find the change in his income a brokerage of % being charged on each 8 8 transaction. Vivek has Rs. 16,500 stock in 3%. Income from Rs. 100 stock = Rs. 3 ∴

Income from Rs. 16,500 stock =

16500 × 3 100

= Rs. 495. He sells it out at 101 and 1/8 (brokerage 1/8%)

1 1 S.P. of Rs. 100 stock = 101 − 8 8

∴

S.P. of Rs. 100 stock = Rs. 101 ∴

Selling price of Rs. 16500 stock =

16500 × 101 100

= 16,665.

7 1 Now he invests this Rs. 16665 in 4% railway debentures at 131 . (Brokerage = % ) 8 8 i.e,

F H

I K

7 1 By investing Rs. 131 + , income derived = Rs. 4 8 8 ∴ By investing Rs. 16665, income derived =

16665 × 4 132

= Rs 505. ∴ Change in income = Rs. 505 − Rs. 495 = Rs. 10. ∴ Income is increased by Rs. 10. 7. Tulasi has invested Rs. 1,00,000 partly in 12% stock at 120 and partly in 15% stock at 75. If the total income from both is Rs. 15,000. Find the amount invested in 2 types of stocks. Solution. Let the amount invested in 12% stock at 120 be x. Then Amount invested in 15% stock at 75 = 1,00,000 − x

252

Basic Mathematics

x 12 x = 120 10

Income from 12% stock at 120 = Income from 15% stock at 75 =

=

a

15 1,00,000 − x 75

f

1,00,000 − x 5

Given total income = 15,000 ∴

x 1,00,000 − x + = 15,000 10 5 1x − 2,00,000 − 2 x = 15,000 10

− x + 2,00,000 = 1,50,000 ⇒ x = 2,00,000 − 1,50,000 x = 50,000 ∴ Amount invested in 12% stock at 120 = Rs. 50,000 and that invested in 15% stock at 75 = Rs. 1,00,000 − 50,000 = Rs. 50,000. 8. Mr. Gauriprasad sold Rs. 2250 stock at 75 and bought stock at 88.50 with proceeds. How much stock worth does he buy if the brokerage is 2% for selling and 1.5% for buying. Solution: Cost of 1 stock = Rs. 75 − 2 (Brokerage) Cost of 1 stock = Rs. 73 ∴ Amount received by selling stock worth Rs. 100 is Rs. 73. ∴ Amount received by selling stock worth Rs. 2250 =

∴

Rs. 2250 × 73 100

= Rs. 1642.50. Now Brokerage for buying = 1.5% Cost of 1 stock = 88.50 + 1.50 = 90 By investing Rs. 90, stock worth Rs. 100 can be purchased.

∴ By investing Rs. 1642.50, Stock worth −

1642.50 × 100 can be purchased. 90

= Rs. 1825. ∴ Mr. Gauriprasad bought stock worth Rs. 1825. 9. Pusphak buys Rs. 2,000 shares paying 9% dividend. If he wants to have an interest of 12% on his money, then find the market value of each share.

Stocks and Shares

253

Solution. Dividend on Rs. 2000 share in 9% of 2000.

=

9 × 2000 = 180. 100

He wants to have 12% on his money. Rs. 12 is an income on Rs. 100 share. Rs. 180 is income on

180 × 100 = 1500. 12

∴ Market value of each share must be Rs. 1500 in order to have interest of 12% on his money. 10. A man invests some money partly in 3% stock at 96 and partly in 4% stock at 120. What is the ratio of money he must invest in order to get equal dividends from both. Solution. For an income of Re 1 in 3% stock at 96, investment = Rs. For an income of Re 1 in 4% stock at 120 investment = Rs.

96 = Rs.32. 3

120 4

= Rs. 30. Ratio of investments = 32:30 = 16:15.

∴

REMEMBER: • Yield =


• • • •

When stock is purchased, brokerage is added to cost price. When stock is sold, brokerage is subtracted from selling price. Interest or dividend is paid on the face value of the stock or share not the market value. 4 and ½% stock at 96 means a stock whose face value is Rs. 100 is available at Rs. 96. Interest earned in 4 and ½. • Shares need not be fully paid but stock must be fully paid.

EXERCISE 1. Find the cost of (a) Rs. 8750 8 and 3/4% stock at 92. (b) Rs. 8500, 9 and ½% stock at 6 premium. (c) Rs. 7200, 10% stock at 7 discount. (d) Rs. 6400, 8% stock at par (brokerage 1/8%). 2. Find the cash required to purchase Rs. 1600, 8 and ½% stock at 105 (brokerage ½%).

254

Basic Mathematics

3. Find the cash realised by selling Rs. 2400, 5 and ½% stock at 5 premium, brokerage being 1/4%. 4. Which of the following is better investment? 6% at 94 or 8% at 110. 5. Ramu possesses 150 shares of Rs. 25 each, the dividend declared by the company is 12%. What is the dividend earned by him. If he sells the shares at Rs. 40 and reinvest the proceeds in 7% shares of par value Rs. 100 at Rs. 80, Find the change in his dividend income. 6. A man invested Rs. 6750 partly in 6% stock at 140 and partly in 5% stock at 125. Find his investment in cash if the income derived from both the investments is Rs. 280. 7. Mr. Vivek invested Rs. 2200 partly in 10% stock at 120 and partly in 12% stock at 96. Find his investment in each if the income derived from both the investments is Rs. 200. 8. A man invests some money partly in 6% stock at 96 and partly in 5% stock at 120. In what ratio, he must invest the money so as to get equal dividends from both. 9. Rs. 2780 is invested partly in 4% stock at 75 and 5% stock at 80. If the income from both investments are equal, find the investment in 5% stock. 10. Mr. Harish has invested a certain amount of money in 13% stock at 101. He sold it when market value went down to 96.5. He lost Rs. 3564 on this process. If he has paid the brokerage at 1 and ½% for all transactions, what was the amount of cash investment? What was the stock value of the investment in first instance.

ANSWERS 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

(a) Rs. 8050 (b) Rs. 9010 Rs. 1688 Rs. 2514 8% at 110 Rs. 75 Rs. 3500 and Rs. 3250 Rs. 1800 and Rs. 400 2:3 Rs. 1280 Rs. 73,062 and Rs. 72338.60.

(c) Rs. 6696

(d) Rs. 6408

11 Learning Curve 11.1 INTRODUCTION: When an individual performs the same task repeatedly the second and subsequent time will provide experience to the individual and there by there will be an increase in the degree of efficiency in performance. This is due to the learning process. The learning effect improves productivity of individuals especially that of workers engaged in factories and industries.

11.2 LEARNING CURVE: It is the curvilinear relationship between the decrease in average labour hours per unit with increase in the cumulative output. The assumption of learning curve theory is that every time total output of a product doubles, the cumulative average time taken to produce one unit decreases by a constant percentage. For example: 80% learning effect means that when the cumulative output is doubled, the cumulative average hours per unit will be 80% of the previous level.

11.3 THE LEARNING CURVE RATIO: It is the ratio of average labour cost of first 2N units and average labour cost of first N units. It is also known as experience ratio or improvement ratio or efficiency ratio. ∴

Experience ratio =

Average labour cost of first 2 N units Average labour cost of N units

11.4 GRAPHICAL REPRESENTATION OF LEARNING CURVE: We know 80% learning effect means that when the cumulative output is doubled the cumulative average hours per unit will be 80% of the previous level. Let us consider the effect of 80% learning in the form of a table taking cumulative time per unit as 100.

256

Basic Mathematics

Units Produced

Total Output

Cumulative time percent

Total number of hours

Average hour per additional unit

1

1 unit

100

100

100

1

2 units

80% of 100 = 80

80 × 2 = 160

160 − 100 = 60 1

2

4 units

80% of 80 = 64

64 × 4 =256

256 − 160 = 48 2

4

8 units

80% of 64 = 51.2

51.2 × 8 = 409.6

409.6 − 256 = 38.4 4

8

16 units

80% of 51.2 = 40.96

40.96 × 16 = 655.36

655.36 − 409.6 = 30.72 8

16

32 units

80% of 40.96 = 32.76

32.76 × 32 = 1048.57

32

64

80% of ...

1048.57 − 655.36 = 24.57 16 ...

M

M

...

M

M

M

Taking the total output on x-axis and cumulative average time per unit on y-axis, we get the learning curve as shown in the figure.

x

1

2

4

8

16

32

y

100

60

48

38.4

30.72

24.57

y Scale: x - axis : 1 Unit = 1 cm y - axis : 20 units = 1 cm

100 80 60 40 20 0

x 1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

Fig. 11.1

The curve clearly indicates that there will be fast learning effect in the initial stages and after sometime there will be a steady state phase in which there is not any significant learning effect.

Learning Curve

257

11.5 LEARNING CURVE EQUATION: If b = logarithmic of learning ratio to base 2, y = cumulative average time per unit, x = cumulative total number of units produced a = time for first unit, then the learning curve equation is given by y = axb. By considering log on both sides we get

@ E

log y = log ax b

log y = log a + log x b log y = log a + b log x.

WORKED EXAMPLES: 1. Find the index of learning for 60% learning effect. Solution: Index of learning b = log260%

b=

log10 60% log10 2

b=

log10 0.6 1.7782 = 0.3010 log10 2

=

−1 + 0.7782 −0.2218 = 0.3010 0.3010

b = −0.7368. 2. A worker takes 10 hrs. to produce the first unit of product. What is the cumulative average time per unit taken by him for the production of first two units? (Assuming the learning effect to be 80%). Also find the total time for producing the first two units. Solution: Given: a = 10 hrs. x=2 80% learning effect. We have learning curve equation y = axb where

b = log 2 80% =

log 0.8 log 2

b=

1.9031 −1 + 0.9031 = 0.3010 0.3010

b=

−0.0969 = −0.3219 0.3010

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Basic Mathematics

Now

y = 10 × 2 −0.3219 y= y=

10 2

0.3219

10 1.249

y = 8.0064 ≈ 8 ∴

Cumulative average time = 8 hrs. Let x = 20.3219 Consider log on b.s.

log x = log 2 0.3219 log x = 0.3219 log 2 log x = 0.3219 × 0.3010

log x = 0.09689 x = Antilog 0.09689

= 1.249. ∴

Time taken for producing the first 2 units = 8 × 2 = 16 hrs. 3. The time required to produce the first unit of a product is 1000 hrs. If the manufacturers experiences 80% learning effect, calculate the average time per unit and the time taken to produce altogether 8 units. Also find the total labour charges for the production of 8 units at the rate of Rs. 12.50 per hour. a = 1000 Solution: Given: x=8

b = log280% = Formula:

log 0.8 = −0.3219 log 2

y = axb = 1000 × 8 −0.3219

= y= ∴

1000 8 0.3219 1000 = 512. 1.953

Average time/unit = 512 hrs. To find x = 80.3219 log x = log80.3219

Learning Curve

259

0.3219 × log 8 0.3219 × 0.9031 log x = 0.2907

x = antilog 0.2907 = 1.953.

∴

Average time/unit = 512 hrs. Total time to produce 8 units = 512 × 8 = 4096 hrs.

∴

Labour charges for 4096 hrs. = 4096 × 12.50 [Q Labour charges per hour = Rs. 12.50 (given)] = Rs. 51,200. OR

Units Produced

Total Output

Cumulative time percent

Total number of hours


1

1

1000

1000

1000

1

2

80% of 1000 = 800

800 × 2 = 1600

1600 − 1000 = 300 2

2

4

80% of 800 = 640

640 × 4 = 2560

2560 − 1600 = 240 4

4

8

80% of 640 = 512

512 × 8 = 4096

4096 − 2560 = 192 8

Total time to produce 8 units = 4096 hrs. Cumulative average time per unit = 512 Total time to produce 8 units = 512 × 8 = 4096. Given Labour charges for 1 hr. = Rs. 12.50 ∴ Labour charges for 4096 hrs. = 4096 × 12.50 = Rs. 51,200. 4. An engineering company has won the contract for supplying aircraft engines of a new type. The prototype constructed to win the contract took 500 hrs. It is expected that there will be 90% learning effect. Estimate the labour cost of producing 8 engines of new order, if the labour cost is Rs. 40 per hour.

260

Basic Mathematics

Solution: Units

Total Output

Cumulative average time

Total hours


1

1

500

500

500

1

2

90% of 500 = 450

450 × 2 = 900

900 − 500 = 200 2

2

4

90% of 450 = 405

405 × 4 = 1620

4

8

90% of 405 = 364.5

364.5 × 8 = 2916

1620 − 900 = 180 4 2916 − 1620/8 = 162

Total time to produce 8 units = 2916 Given Labour charges per hour = Rs. 40 Labour charge for 2916 hrs. = 2916 × 40 = Rs. 1,16,640.

OR Given: a = 500 x=8

b = log290% = b=

log 0.9 log 2

1.9542 −1 + 0.9542 = 0.3010 0.3010

b=

−0.0458 = −0.1521. 0.3010

y = axb

Formula:

= 500 × 8 −0.1521

=

500 8

0 .1521

=

500 1.372

y = 364.43 To find 80.1521 x = 80.1521 logx = 0.1521 × log8 = 0.152 × 0.9031. Average time/unit = 364.43. Total time to produce 8 units.

Learning Curve

261

= 364.43 × 8 = 2915.4518 ≈ 2916.

Given labour charges/hr = Rs. 40 ∴ Labour charges/2916 hrs. = 2916 × 40 = 1,16,640.

REMEMBER: • The learning curve ratio =

=Average labour cost of first 2N unitsB =Average labour cost of first N unitsB

• The learning curve equation is

y = axb where y = cumulative average time/unit. a = Time for first unit. b = log of learning ratio to base 2. x = Cumulative total number of units produced. • Index of learning, b = log of learning ratio to base 2.

EXERCISE 1. 2. 3. 4. 5.

Define the term learning curve ratio. What is learning curve? What do you mean by 80% learning effect. Find the index of learning for 80% learning effect. A worker requires 40 hrs. to produce first unit of a product. How much time is required for him to produce a total of 4 units. 6. A worker requires 20 hrs. to produce the first unit of a product. If the learning effect is 80% calculate how much time is required to produce 4 units. Also find the labour charges for the production of 4 units at the rate of Rs. 20 per hr. 7. A company requires 2134.2 hrs. to produce the first 40 machines. If the learning effect is 80%, find the number of hrs. required to produce next 120 machines. 8. The first unit of a product took 80 hrs. to manufacture. If the workers show 80% learning effect, find the total time taken to produce 8 units.

ANSWERS 1.

Average labour cost of first 2N units . Average labour cost of first N units

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Basic Mathematics

2. Curvilinear relationship between the decrease in average labour hrs. per unit with increase in curve ratio. 3. When the cumulative output is doubled, cumulative average labour hrs./unit will be 80% of the previous level. 4. −0.3219 5. 108.4 hrs. 6. 48.8 hrs. 7. 3329.35 hrs. 8. 32.768 hrs.

12 Linear Programming 12.1 INTRODUCTION: Linear programming is a powerful technique which can indicate a definite conclusion as to the best utilisation of available resources under given circumstances. Any industrial process may consists of a number of activities relating to capital to be employed, products to be made and sold, materials to be used: machines to be run; inventories to the stored and consumed or a combination of the above. Since utilisation of one affects the utilisation of another and due to the limitation of the total available resources, these activities are interdependent or interlocking. In such a situation, a large number of ways exists in which the available resources can be allocated to the competing demands. Linear programming enables us to arrange for that combination of resources which optimise the cost, production, profit etc. This technique was evolved by George B. Dantzig as a tool for planning the diversification activities of U.S. Airforce in 1947.

12.2 LINEAR PROGRAMMING: Def inition: Linear programming is a mathematical technique for determining the optimal allocation of Definition: resources and obtaining a particular objective when there are alternative uses of the resources: money, manpower, material, machine and other facilities. The objective in resource allocation may be cost minimisation or profit maximisation. The technique of linear programming is applicable to problems in which the total effectiveness can be expressed as a linear function of individual allocations and the limitations on resources give rise to linear equalities or inequalities of the individual allocations. Objecti ve function: The objective function is a quantified statement in linear programming model of Objectiv what the best results or the best advantage which is aimed for as the objective of the resource allocation decision. Objective function will either be to maximise a value or to minimise a value. The objective function is expressed as a linear function of decision variables. It is usually stated as maximise or minimise a1x1 + a2x2 + ... +anxn where x1 x2 ... xn are the decision variables in the problem and a1, a2, ..., an are constant values for each variable. aria iab Decision vvar ar ia bles: The decision variables in any activity is a variable which is competing with other activities for limited resources. These variables are inter-related linearly in terms of utilisation of resources.

264

Basic Mathematics

Constr aints: The resources like production capacity, manpower, time, space, technology, etc. are Constraints: scarce and there are limitations on what can be achieved. These restrictions are a set of conditions which an optimal solution must satisfy. They are known as constraints. These are expressed as linear inequalities or equalities in terms of decision variables. Non-neg tivity Non-ne gati vity conditions: All decision variables must assume non-negative values. If any of the variable is unrestricted in sign, a trick can be employed which will enforce the non-negativity without changing the original information of the problem.

12.3 SOLUTION TO LINEAR PROGRAMMING PROBLEM: A set of values of decision variables x1 x2 ... xn which satisfies the constraints of linear programming problem is called solution. easible Feasib le solution: Any solution to a linear programming problem which satisfies the non-negativity restriction of the problem is called a feasible solution to the Linear programming problem. Optimal solution: Any feasible solution which optimises (maximises or minimises) the objective function of a linear programming problem is called an optimal solution. Solution of linear programming by graphical method: If the objective function is a function of 2 variables only then the LPP can be solved by graphical method. One variable is taken along x-axis and another along y-axis. Since negative values are not allowed, the graph contains only first quadrant. That is, 0 and positive values of x and y are considered.

3 I quadrant

2 1

1

2

3

Fig. 12.1

Illustration to find feasible region: Consider the graph of y = 2. 3 2 1 0

1

2

Fig. 12.2

3

Linear Programming

265

This is a straight line parallel to x-axis. All points below this line are represented by the inequality y < 2 and all points above this line are represented by y > 2 and the corresponding graphs are y>2 y=2

2

2 1

y 0 or

x12 + y12 + 2gx1 + 2 fy1 + c > 0.

Fig. 13.20

If P coincide with A, then AP = 0 and the point P lie on the circle. ∴

x12 + y12 + 2gx1 + 2 fy1 + c = 0. If P lie inside the circle then

x12 + y12 + 2gx1 + 2 fy1 + c < 0.

WORKED EXAMPLES: 1. Find the equation of tangent to the circle 2 x 2 + 2 y 2 + x − 3y − 12 = 0 at (1, 3). Solution: Equation of circle:

2 x 2 + 2 y 2 + x − 3y − 12 = 0 ÷ by 2. x 2 + y2 + 2g =

1 2

x 3 − y−6 = 0 2 2

2f = −

3 2

c = −6

Circles 303

g=

⇒

1 4

3 f =− . 4

b x , y g = a1, 3f. 1

Equation of tangent

1

b g b g 1 −3 x a1f + y a3f + a x + 1f + F I a y + 3f − 6 = 0 H 4 4K xx1 + yy1 + g x + x1 + f y + y1 + c = 0

x + 3y +

1 1 3y 9 x+ − − −6=0 4 4 4 4

4 x + 12 y + x + 1 − 3 y − 9 − 24 =0 4 Crossmultiplying 5 x + 9 y − 32 = 0.

2. Find the equation of the tangent to the circle x 2 + y 2 + 8 x − 6 y − 11 = 0 parallel to 4x − 3y + 1 = 0. Solution: Any line parallel to 4x − 3y + 1 will have the equation 4x − 3y + k = 0. Now 4x − 3y + k = 0 will be a tangent to the circle if the length of the perpendicular from the centre to the line = Radius. Now Equation of circle = x 2 + y 2 + 8 x − 6 y − 11 = 0 Centre = (−g, −f) = (−4, 3) 2 2 Radius of x + y + 8 x − 6 y − 11 = 0 is

4 2 + 3 2 + 11

= 25 + 11 = 6 units.

4x − 3y + 1

Length of perpendicular from (−4, 3) to 4x − 3y + k = 0 = Radius

a f af

4 −4 − 3 3 + k 4 2 + 32

=6

−16 − 9 + k =6 5 Fig. 13.21

−25 + k = 30 −25 + k = ±30

[Note: We get 2 tangents parallel to given line]

304

Basic Mathematics

k = ±30 + 25

∴

k = 55 or k = −5.

∴ Equations of tangents = 4 x − 3 y + 55 = 0 and 4 x − 3 y − 5 = 0

3. Find the equation of tangent to the circle x 2 + y 2 − 2 x − 4 y − 4 = 0 which are perpendicular to the line 4x + 3y − 7 = 0. Any line perpendicular to 4x + 3y − 7 will have the equation 3x − 4y + k = 0. 3x − 4y + k = 0 will be a tangent to the circle if the length of the perpendicular from the centre = Radius. Now equation of circle = x 2 + y 2 − 2 x − 4 y − 4 = 0 Centre = (−g, − f) = (+1, 2) Radius =

a f

4x + 3y − 7 = 0

g 2 + f 2 − c = 12 + 2 2 − −4 = 1 + 4 + 4 = 3 units.

Length of the perpendicular from (1, 2) to 3x − 4y + k = 0 (Radius)

af a f

3 1 −4 2 +k

i.e.,

32 + 4 2

=

=3

3−8+ k =3 5

= −5 + k = 15 −5 + k = ± 15

Fig. 13.22

[Note: We get 2 tangents perpendicular to given line]

k = ±15 + 5

k = 20 or k = −10. ∴ Equation of tangents are 3 x − 4 y + 20 = 0 and 3 x − 4 y − 10 = 0.

4. Find the value of k if x + y + k = 0 touches the circle x 2 + y 2 − 7 x − 5 y + 18 = 0. Solution: x + y + k touches the circle if the length of the perpendicular from the centre to the line = Radius. Now Equation of circle = x 2 + y 2 − 7 x − 5 y + 18 = 0

a

f FH 72 , 25 IK

Centre = − g, − f =

Circles 305

Radius = g 2 + f 2 − c =

=

Radius =

F 7I + F 5I H 2K H 2K 2

2

− 18

49 25 + − 18 4 4 74 − 72 = 4

Length of the perpendicular from

F 7 , 5I H 2 2K

2 1 = units. 4 2

to x + y + k = 0 = Radius

7 5 + +k 1 2 2 = 2 12 + 12

12 +k 1 2 = 2 2 2 1 +1

12 +k 1 2 = 2 2 6+ k =1

6 + k = ±1 k = ±1 − 6 k = −7 or k = −5

Hence

k = −7 or −5.

5. Find the length of the tangent to the circle 3x 2 + 3y 2 − 7 x − 6 y − 12 = 0 from (6, −7) Solution: Equation of the circle

3x 2 + 3y 2 − 7 x − 6 y − 12 = 0 ÷ by 3. x 2 + y2 −

7 x − 2 y − 4 = 0. 3

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Basic Mathematics

Length of the tangent from (x1, y1)

= x12 + y12 + 2gx1 + 2 fy1 + c Length of tangent from (6, −7)

a f

= 6 2 + −7

2

−

af a f

7 6 − 2 −7 − 4 3

= 36 + 49 − 14 + 14 − 4 = 81 = 9 units.

6. Find whether the origin is inside or on or outside the circle (i) x 2 + y 2 + 4 x − 7 y + 8 = 0 (ii) x 2 + y 2 = 7 (iii) x 2 + y 2 + 2 y + 6 x = 0 Solution: (x1, y1) lie inside the circle

x 2 + y 2 + 2gx + 2 fy + c = 0 if x12 + y12 + 2 gx1 + 2 fy1 + c < 0 Outside if x12 + y12 + 2 gx1 + 2 fy1 + c > 0 and on the circle if

x12 + y12 + 2 gx1 + 2 fy1 + c = 0 (i) Clearly (0, 0) lie outside the circle x 2 + y 2 + 4 x − 7 y + 8 = 0

af af

Q 0+0+4 0 −7 0 +8=8> 0

(ii) Clearly (0, 0) lie inside the circle x 2 + y 2 − 7 = 0 . Since 0 + 0 − 7 = −7 < 0 (iii) (0, 0) lie on the circle x 2 + y 2 + 2 y + 6 x = 0

af af

Q 0+0+2 0 +6 0 = 0

7. Find the equation of the circle so that the lengths of the tangents from (−1, 0), (0, 2) and (−2, 1) are respectively 3,

10 and 3 3.

Solution: Let the equation of circle be

x 2 + y 2 + 2gx + 2 fy + c = 0 Given: length of tangent from (−1, 0) to circle is 3 units. ∴

a−1f

2

a f

af

+ 0 2 + 2 g −1 + 2 f 0 + c = 3

...(1)

Circles 307

1 − 2g + c = 3 Squaring 1 − 2g + c = 9

−2g + c = 8

...(2)

Given: Length of tangent from (0, 2) to the circle =

af

10 units.

af

0 + 2 2 + 2 g 0 + 2 f 2 + c = 10 Squaring 4 + 4 f + c = 10

4f +c = 6

...(3)

Given: Length of the tangent from (−2, 1) is 3 3

a−2f

∴

2

a f

af

+ 12 + 2g −2 + 2 f 1 + c = 3 3

Squaring,

d i

4 + 1 − 4g + 2 f + c = 3 3

2

5 − 4 g + 2 f + c = 27 −4 g + 2 f + c = 22

...(4)

Solving (3) and (4)

4f +c = 6 − 4 g + 2 f + c = 22

a + f a −f a −f a −f 4g + 2 f

= −16 ÷ by 2

2 g + f = −8

...(5)

Solving (2) and (3)

− 2g + c = 8 4f +c= 6 ( − ) ( −) ( − ) − 2g − 4 f = 2

...(6)

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Basic Mathematics

Solving (5) and (6)

(5): ( 6):

2g + f = −8 − 2g − 4 f = 2 − 3 f = −6 f = 2.

⇒ Substituting f = 2 in (5)

2 g + 2 = −8 2 g = −8 − 2

2 g = −10 g = −5

⇒ Substituting g = −5, f = 2 in (2)

−2g + c = 8

a f

−2 −5 + c = 8 c = 8 − 10

c = −2 Substituting g = −5, f = 2 and c = −2 in (1). Equation of circle

a f

af a f

= x 2 + y 2 + 2 −5 x + 2 2 y + − 2 = 0 x 2 + y 2 − 10 x + 4 y − 2 = 0. 8. Find the equation of the circle passing through the points (1, 2) and (3, 4) and touching the line 3x + y − 3 = 0. ...(1) Solution: Let the equation of the circle be x2 + y2 + 2gx + 2fy + c = 0 Given: (1) passes through (1, 2) ∴

af

af

12 + 2 2 + 2 g 1 + 2 f 2 + c = 0 5 + 2g + 4 f + c = 0 2 g + 4 f + c = −5 Given (1) passes through (3, 4)

af

...(2)

af

32 + 4 2 + 2g 3 + 2 f 4 + c = 0 25 + 6 g + 8 f + c = 0 6 g + 8 f + c = −25

...(3)

Circles 309

Solving (2) and (3) we get

2g + 4 f + c = −5 6g + 8 f + c = −25

a − f a − f a −f a + f − 4g − 4 f

= 20 ÷ by − 4

g + f = −5

f = −5 − g

...(4)

Given: 3x + y − 3 = 0 touches the circle. ∴ Length of the perpendicular from centre (−g, −f) to 3x + y − 3 = 0 (Radius)

a f a f

3 −g + − f − 3

∴

3 2 + 12

a1 + gf + a2 + f f

−3g − f − 3 = 10

2

= Radius

LMQ Radius = Distance between centreOP N and any point on circumference Q

2

Substituting f = −5 −g we get

a

(1, 2)

f

−3g − −5 − g − 3

=

10

a

f

−3g − −5 − g − 3 10

a1 + gf + a2 − 5 − gf 2

2

= 1 + g2 + 2g + 9 + 6g + g2

C (−g, −f)

Squaring,

FG −2g + 2 IJ H 10 K 4 ag − 1f 10

2

2

ag − 1f 5

= 10 + 2 g 2 + 8g = 2 g 2 + 8g + 10

2

= g2 + 4g + 5

g 2 − 2 g + 1 = 5g 2 + 20 g + 25 ⇒

4 g 2 + 22 g + 24 = 0 ÷ by 2

Fig. 13.23

310

Basic Mathematics

2g 2 + 11g + 12 = 0 2 g 2 + 8g + 3g + 12 = 0

a

f a

2 g = −3 3 g=− 2

or

24g 2

f

2g g + 4 + 3 g + 4 = 0

+8g +3g +11g

g = −4

g = −4,

when Substituting g = −4 in (4)

f = −5 − g

a f

f = −5 − −4

f = −1. Substituting g = −4, f = −1 in (2) 2 g + 4 f + c = −5

a f a f

2 −4 + 4 −1 + c = − 5 c = −5 + 12

c=7 Similarly we can get f = −7 2 and c = 12 when g = −3/2. ∴ Equation of required circle

a f

a f

x 2 + y 2 + 2 −4 x + 2 −1 y + 7 = 0 x 2 + y 2 − 8 x − 2 y + 7 = 0 or

LM N

3 7 x 2 + y 2 − 3x − 7 y + 12 = 0. By putting g = − , f = − and c = 12 2 2

OP Q

9. Find the equation of the circle which is concentric with the circle x 2 + y 2 − 8 x + 12 y − 20 = 0 and which touches the line 4x − 3y − 14 = 0. ...(1) Solution: Let the equation of the circle be x2 + y2 + 2gx + 2fy + c = 0 Given (1) is concentric with

x 2 + y 2 − 8 x + 12 y − 20 = 0 ∴

Centre of (1) = Centre of x 2 + y 2 − 8 x + 12 y − 20 = 0

Circles 311

a−g, − f f = a4, − 6f g = −4 and f = 6.

∴

Also given (1) touches 4x − 3y − 14 = 0 ∴ Length of the perpendicular from centre = Radius

af a f

4 4 − 3 −6 − 14

∴

4 +3 2

2

= g2 + f 2 − c

16 + 18 − 14 = 42 + 62 − c 5

=

20 = 16 + 36 − c 5 Squaring,

4 2 = 52 − c 16 = 52 − c c = 52 − 16 = 36

∴ Equation of the required circle

a f

af

x 2 + y 2 + 2 −4 x + 2 6 y + 36 = 0 x 2 + y 2 − 8 x + 12 y + 36 = 0. 10. Find the equation of the circle passing the points (4, 1) and (6, 5) and having its centre on the line 4x + y = 16. Solution: Let the equation of the circle be x2 + y2 + 2gx + 2fy + c = 0 ...(1) Given: (1) passes through (4, 1)

af

af

4 2 + 12 + 2 g 4 + 2 f 1 + c = 0 16 + 1 + 8g + 2 f + c = 0 8g + 2 f + c = −17

Given: (1) passes through (6, 5)

af

...(2)

af

6 2 + 52 + 2g 6 + 2 f 5 + c = 0 12g + 10 f + c = −61

...(3)

Also given (1) has centre on 4x + y = 16 i.e., (−g, −f) lie on 4x + y = 16. ∴

a f a f

4 − g + − f = 16 −4 g − f = 16

...(4)

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Basic Mathematics

Solving (2) and (3)

8g + 2 f + c = − 17 12 g + 10 f + c = − 61 ( −) ( − ) −4 g − 8 f

( −)

(+ )

...(5)

= 44

Solving (4) and (5)

−4 g − f = 16 −4g − 8 f = 44 ( +) ( +) ( − ) 7 f = −28 f =−4 Substituting f = −4 in (4)

a f

−4 g − −4 = 16 −4 g + 4 = 16

−4 g = 16 − 4 −4 g = 12

g = 12 −4 ⇒ g = −3 Substituting g = −3 and f = −4 in (2)

a f a f

8 −3 + 2 −4 + c = −17. −24 − 8 + c = −17

c = −17 + 32 c = 15 ∴ Equation of the required circle

a f

a f

= x 2 + y 2 + 2 −3 x + 2 −4 y + 15 = 0 x 2 + y 2 − 6 x − 8 y + 15 = 0.

REMEMBER: • Distance between 2 points (x1, y1) and (x2, y2)

bx

2

− x1

g + by 2

2

g

2

− y1 .

Circles 313

• Slope of the line joining (x1, y1) and (x2, y2)

=

y2 − y1 . x 2 − x1

• 2 lines are perpendicular iff product of their slopes = −1. • Equation of circle with centre (0, 0) and radius r units: x2 + y2 = r2.

a

• Equation of circle with centre (h, k) and radius r units: x − h

f + a y − kf 2

2

= r2.

• Equation of circle which is described on line joining (x1, y1) and (x2, y2) as diameter =

b x − x gb x − x g + b y − y gb y − y g = 0. 1

2

1

2

• General equation of the circle x2 + y2 + 2gx + 2fy + c = 0 Centre = (−g, −f), Radius =

g2 + f 2 − c .

• Centre is the mid point of diameter. • Co-ordinates of mid point of line joining (x1, y1) and (x2, y2) =

Fx +x H 2 1

2

,

y1 + y2 2

I K

• The point of intersection of 2 diameters of a circle = centre. • If a circle x2 + y2 + 2gx + 2fy + c = 0 touches x-axis, then, radius = − f and g2 = c. • If a circle x2 + y2 + 2gx + 2fy + c = 0 touches y-axis, then Radius = − g and f 2 = c. • Length of the perpendicular from (x1, y1) to the line ax + by + c = 0 is ax1 + by1 + c a 2 + b2

• Equation of tangent at (x1, y1) on the circle x2 + y2 + 2gx + 2fy + c = 0 is

b

g b

g

xx1 + yy1 + g x + x1 + f y + y1 + c = 0. • Length of the tangent from (x1, y1) to the circle x2 + y2 + 2gx + 2fy + c = 0 is

x12 + y12 + 2 gx1 + 2 fy1 + c. • Any line is a tangent to the given circle if the length of the perpendicular from centre to the line = Radius of the circle.

d

i

2 2 2 • Condition for the line y = mx + c to be a tangent to the circle x2 + y2 = a2 is c = ± a m + 1 and

point of contact is

F GH

am m2 + 1

,

I or F J GH + 1K

−a m2

− am m2 + 1

,

a m2

I.. J +1K

• Condition for the line lx + my + n = 0 to be a tangent to the circle x2 + y2 + 2gx + 2fy + c = 0 is

an − lg− mf f = dl 2

2

id

i

+ m2 g 2 + f 2 − c .

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• The point (x1, y1) lie inside the circle x2 + y2 + 2gx + 2fy + c = 0 if x12 + y12 + 2gx1 + 2 fy1 + c < 0. Outside the circle if x12 + y12 + 2 gx1 + 2 fy1 + c > 0. On the circle if x12 + y12 + 2 gx1 + 2 fy1 + c = 0. • Any line parallel to ax + by + c = 0 will have the equation ax + by + k = 0. [Take x and y co-efficients same]. • Any line perpendicular to ax + by + c = 0 will have the equation bx − ay + k = 0 [interchange x and y co-efficients and change the sign of anyone].

EXERCISE I. 1. 2. 3.

Find the equa tion of cir h of the ffollo ollo wing: equation circcle in eac each ollowing: Centre: Origin, Radius: 3 Units. Centre: (0, 2), Radius: 2 Units. Centre: (3, −4), Radius: 5 Units.

4. Centre: 5. 6. 7. 8. 9. 10. II.

F 4 , 7 I , Radius: H 3 3K

15 units.

Two diameters are x + y = 8 and 2x − y = 4 and radius 4 units. Diameter is the line joining (5, 0) and (0, 5). Centre: (3, 4), touching the x-axis. Centre: (−5, 1), touching the y-axis. Centre: (4, 4), touching the axes. Centre: (−3, 2), passes through the origin. centree and rradius ollowing circcles: wing cir Find the centr adius of the ffollo ollo

1. x 2 + y 2 − 8 x + 4 y + 2 = 0 2. 2 x 2 + 2 y 2 − 6 x − 10 y − 15 = 0 3. 3x 2 + 3y 2 − 6 x − 12 y − 2 = 0. III. 1. If one end of a diameter of a circle x 2 + y 2 + 4 x − 6 y − 14 = 0 is (−5, −1), find the other end. 2. Find the equation of the tangent to the circle 2 x 2 + 2 y 2 + x − 3y − 12 = 0 at (1, 3). 3. Find the equation of tangent at the point (−3, −5) on the circle x 2 + y 2 + 4 x + 6 y + 8 = 0. 4. Find the value of k if 3x + 4y = k touches the circle x2 + y2 = 16. Also find the point of contact. 5. Find the equation of tangent to the circle x 2 + y 2 − 2 x + 6 y − 15 = 0 which are parallel to the line 3x + 4y − 8 = 0.

Circles 315

6. Find the equation of tangent to the circle x 2 + y 2 − 2 x − 4 y + 1 = 0 which are perpendicular to the line 3x − 4y + 8 = 0. 7. Find the length of the chord 4x − 3y = 5 of the circle x 2 + y 2 + 3x − y − 10 = 0. 8. Find the length of the tangent to the circle 3x 2 + 3y 2 − 7 x − 6 y − 12 = 0 from (6, −7). 9. Find whether (1, −2) lie inside or on or outside the circle x 2 + y 2 − 4 x + 8 y − 8 = 0 . 10. Find the equation of the circle so that the lengths of the tangents from (−1, 2), (−2, 3) and (2, 4) are respectively 2 3, 30 and 17 units. 11. Find the equation of the circle which passes through (1, 1), (2, 2) and whose radius is unity. 12. Find the equation of circle which has centre on x-axis, passes through (1, 3) and whose radius is 18 units.

13. Find the equation of circle which makes an intercepts −8 and 6 on X and Y-axes and which passes through the origin. 14. Find the equation of circle passing through the point (1, −2), (4, −3) and whose centre lies on 3x + 4y − 7 = 0. 2 2 15. Find the equation of circle concentric with x + y − 6 x + 8 y + 16 = 0 and touching the line

5 x − 12 y + 15 = 0.

16. Find the equation of circle which touches the y-axis at (0, 9) and cuts x-axis at (−3, 0) and (−9, 0).

ANSWERS I.

1. x 2 + y 2 = 9

2. x 2 + y 2 − 4 y = 0

3. x 2 + y 2 − 6 x + 8 y = 0

4. 9 x 2 + 9 y 2 − 24 x − 42 y − 70 = 0

5. x 2 + y 2 − 8 x − 8 y + 16 = 0

6. x 2 + y 2 − 5 x − 5 y = 0.

7. x 2 + y 2 − 6 x − 8 y + 9 = 0

8. x 2 + y 2 − 10 x + 2 y + 1 = 0

9. x 2 + y 2 − 8 x − 8 y + 16 = 0 II.

a f C ≡ a1, 2f, r = a1, 7f

1. C ≡ 4, − 2 , r = 3 2 units. 3.

III. 1.

10. x 2 + y 2 + 6 x − 4 y = 0 2. C ≡

F 3 , 5 I , r = 4 units. H 2 2K

17 3 units. 2. 5 x + 9 y − 32 = 0

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Basic Mathematics

LM N

12 16 ,− 5 5

OP Q

3. x + 2 y + 13 = 0

4. k = 20, −

5. 3x + 4 y + 34 = 0 and 2 x + 4 y − 16 = 0.

6. 4 x + 3y − 20 = 0; 4 x + 3y = 0

7. 5 units.

8. 9 units.

9. Inside

10. x 2 + y 2 − 6 x + 4 y − 7 = 0

11. x 2 + y 2 − 4 x − 2 y + 4 = 0

12. x 2 + y 2 + 4 x − 14 = 0

13. x 2 + y 2 + 8 x − 6 y = 0

14. 15x 2 + 15 y 2 − 94 x + 18 y + 55 = 0

15. x 2 + y 2 − 6 x + 8 y − 11 = 0

16. x 2 + y 2 + 12 x − 18 y + 27 = 0.

14 Parabola 14.1 INTRODUCTION: When a solid cone is cut by a plane, the curves which lies on the surface of the cone and the plane, are the curves – circle, parabola, ellipse and hyperbola. These curves are called as conic sections. Definitions: Let l be a fixed line and S be fixed point. A point P moves in a plane containing S and l such that its distance from S bears a constant ratio to its distance from the line l, i.e.,

P

M

SP = a constant. PM The locus of the point P is called a conic. A conic is the locus of the point which moves such that the ratio of its distance from a fixed point in the plane to its distance from a fixed line in a plane is constant.

S

l

Fig. 14.1

The fixed point is called focus and the fixed line is called directrix and the constant ratio

SP is PM

called eccentricity of the conic. If the eccentricity of the conic is less than 1 then the conic is called ellipse. If the eccentricity is greater than 1 then it is called hyperbola. If the eccentricity is equal to one then the conic is called parabola. Circle is regarded as the conic of eccentricity zero. It is a particular case of an ellipse.

14.2 PARABOLA: Parabola is the locus of point which moves in a plane such that its distance from a fixed point is equal to its distance from a fixed line. The fixed point is called focus. The fixed line is called directrix.

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Basic Mathematics

14.3 EQUATION OF THE PARABOLA IN THE STANDARD FORM: Let S be the focus, l be the directrix. Draw SZ ⊥ l. Let ‘O’ be the midpoint of SZ. Let SZ = 2a. So that OS = OZ = a. Choose O as origin the 2 mutually perpendicular lines OX and OY as co-ordinate axes. Let P (x, y) be any point on the parabola. Draw PM ⊥r to l and PN ⊥r X-axis. Join PS. y

l

P M

x′

z

O

x

S

N

y′

Fig. 14.1

SP =1 PM

From definition

SP = PM

...(1)

Since S is on X-axis and OS = a, Co-ordinates of S = (a, 0) SP = Distance between S (a, 0) and P (x, y).

F Distance formula: b x − x g + b y − y g I H K SP = a x − af + a y − 0f = a x − af + y 2

2

1

2

2

2

2

1

2

2

...(2)

Now

PM = NZ = ON + OZ PM = x + a Substituting (2) and (3) in (1) we get

a x − af

2

+ y2 = x + a

...(3)

Parabola

319

Squaring,

x 2 + a 2 − 2ax + y 2 = x 2 + a 2 + 2 ax ⇒

y 2 = 2ax + 2ax

⇒

y 2 = 4 ax.

This is the standard equation of parabola.

THE PARABOLA y2 = 4ax. For the parabola y2 = 4ax we have the following: 1. Shape: • Since (0, 0) satisfy the equation y2 = 4ax, the curve passes through the origin. • The equation y2 = 4ax remains unchanged if we replace y by −y. So the curve is symmetric about x-axis. • For negative values of x, we get imaginary values for y. So the curve entirely lies on the right side of y-axis if a is +ve or a > 0. (and if ‘a’ is −ve, the curve entirely lies on the left side of y-axis). Hence the shape of the parabola is

L l

S

O

x

L′

Fig. 14.2

2. 3. 4. 5. 6.

The point O(0, 0) is called vertex of the parabola. The line OX, the +ve x-axis is called the axis of the parabola. S is the focus and its co-ordinate = (a, 0). The line l is the directrix and its equation is x = −a. The line LSL′ which is perpendicular to axis and passing through the focus is called Latus rectum. L and L′ are ends of latus rectum. L and L′ has x-co-ordinate ‘a’. To get y-co-ordinate, since L and L′ lie on the parabola y2 = 4ax it satisfies the equation. ∴

y 2 = 4 ax

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Basic Mathematics

af

∴

y 2 = 4a a

3x = a

y 2 = 4 a 2 ⇒ y = ± 4a 2 y = ±2 a.

∴

Ends of latus rectum = (a, ±2a). Equation of latus rectum: x = a Length of latus rectum = 2a + 2a = 4a.

14.4 DIFFERENT FORMS OF PARABOLA WITH VERTEX (0, 0): 1. The parabola y2 = −4ax. Shape: y

L

x′

x

S

L′ l

y′

Fig. 14.3

Vertex: (0, 0) Axis: Negative x-axis. Co-ordinates of focus = (−a, 0) Equation of directrix = x = +a Ends of latus rectum = (−a, ±2a) Equation of latus rectum: x = −a Length of latus rectum: 4a. 2. The parabola x2 = 4ay Shape: Vertex: (0, 0) Axis: Positive y-axis. Co-ordinates of focus = (0, a) Equation of directrix = y = −a

y

L

L′

S

x

x′

l

y′

Fig. 14.4

Parabola

321

y

Ends of latus rectum = (±2a, a) Equation of latus rectum = y = a Length of latus rectum = 4a. 3. The parabola x2 = −4ay Shape: Vertex: (0, 0) Axis: Negative y-axis Co-ordinate of focus = (0, −a) Equation of directrix = y = a Ends of latus rectum = (±2a, −a) Equation of latus rectum = y =−a Length of latus rectum = 4a

l S

x′

L

x

L′

y′

Fig. 14.5

14.5 DIFFERENT FORMS OF PARABOLA WITH VERTEX (h, k): 1.

a y − kf

2

a

= 4a x − h

f y (h, k)

s

O

x

Fig. 14.6

Let OX and OY be x and y-axis respectively change the origin to the point (h, k) without changing the direction. Then the new co-ordinates (X, Y) are given by

x = X + h and y = Y + k X = x − h and Y = y − k

⇒

a y − kf

∴

2

a

= 4a x − h

f

⇒ = 4aX which is the equation of parabola, which opens to the right.

Y2

a

For the parabola y − k Vertex = (h, k) Axis: y = k

f

2

a

f

= 4 a x − h we have the following

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Basic Mathematics

Co-ordinates of focus = (a + h, k) Equation of directrix = x = −a + h. Ends of Latus rectum = (a + h, ±2a + k) Equation of Latus rectum = x = a + h Length of Latus rectum = 4a. 2.

a y − kf

2

a

= −4 a x − h

f

Shape:

h, k

Fig. 14.7

Parabola which opens to the left in any quadrant depending on value of h and k. Vertex: (h, k) Axis: y = k Co-ordinates of focus = (−a + h, k) Equation of directrix: x = a + h Ends of latus rectum = (− a + h, ±2a +k) Equation of Latus rectum: x = −a + h Length of Latus rectum = 4a 3.

a x − hf

2

a

= 4a y − k

f

Shape:

(h, k)

Fig. 14.8

Parabola which opens upward in any quadrant depending on value of h and k.

Parabola

323

Vertex : (h, k) Axis: x = h Focus: (h, a + k) Ends of latus rectum = (±2a + h, a + k) Equation of latus rectum = y = a + k Length of latus rectum = 4a 4.

a x − hf

2

a

= −4 a y − k

f

Shape:

(h, k)

Fig. 14.8

Parabola which open downward is any quadrant depending on value of h and k. Vertex : (h, k) Axis : x = h Focus = (h, −a + k) Equation of directrix: y = a + k Ends of Latus rectum: (±2a + h, −a + k) Equation of latus rectum: y = −a + k Length of latus rectum: 4a. Note: In any parabola, the focus is inside the curve (i.e. on the axis of the parabola) and directrix is away from the curve (i.e. directrix never meet the parabola). 2. The distance between the vertex and the focus is equal to the distance between the vertex and the directrix is equal to a. 3. The distance of a point on the parabola from the focus is called the focal distance of the point. 4. A chord drawn through the focus is called focal chord and the focal chord perpendicular to the axis is Latus rectum.

WORKED EXAMPLES: I. Find the vertex, axis, focus, equation of directrix, Ends of latus rectum, Length of latus rectum, equation of latus rectum of the following parabolas. 1. x2 = 8y

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Basic Mathematics

Comparing with x2 = 4ay we get 4a = 8 ⇒ a = 2. Vertex = (0, 0) Axis: y-axis or x = 0. Focus: (0, a) = (0, 2) Equation of directrix: y = −a y = −2 Ends of latus rectum = (±2a, a) = (±2(2), 2) = (±4, 2). Length of latus rectum = 4a = 4 (2) = 8. Equation of latus rectum = y = a y = 2. 2 2. y = −6x. Comparing with y2 = −4ax we get 4a = 6

a=

6 3 = . 4 2

Vertex: (0, 0) Axis: x-axis or y = 0

F H

I K

3 Focus = (−a, 0) = − , 0 . 2 Equation of directrix = x = a

3 x= . 2 Ends of Latus rectum: (−a, ±2a)

FG F I IJ H KK H 3 = F − , ± 3I H 2 K 3 3 = − , ±2 2 2

Equation of latus rectum x = −a

3 x=− . 2

Parabola

Length of latus rectum = 4a

=4

F 3 I = 6. H 2K

3. x2 = −y Comparing with x2 = −4ay we get

4a = 1 ⇒ a = Vertex = (0, 0) Axis: x-axis or x = 0.

F H

Focus = (0, −a) = 0, −

1 4

1 4

I K

Equation of directrix : y = a

1 y= . 4

a f F F 1 I 1 I F 1 −1 I = GH ±2 H K , − JK = H ± , K 4 4 2 4

Ends of latus rectum = ±2a, − a

Equation of latus rectum: y = −a

y=−

1 4

Length of latus rectum = 4a =4 4.

F 1 I = 1. H 4K

y2 = 4 (x + 1)

a

⇒ y−0

f

2

b a fg

= 4 x − −1

Comparing with (y − k)2 = 4a (x − h) We get (h, k) = (−1, 0) 4a = 4 ⇒ a = 1 Vertex = (h, k) = (−1, 0) Axis: y = k y = 0. Focus: (a + h, k) = (1 + (−1), 0) = (0, 0).

325

326

Basic Mathematics

Equation of directrix x = −a + h x = −1 −1 x = −2 Ends of latus rectum = (a + h, ±2a + k) = (1 + (−1), ±2 (1) + 0) = (0, ±2). Equation of latus rectum = x = a + h x = 1 + (−1) x = 0. Length of latus rectum = 4a = 4 (1) = 4. 5.

a y + 2f

2

= 3x + 1

a y + 2f

2

=3 x+

a f

y − −2

F H

2

1 3

I K

LM F 1I OP N H 3K Q

=3 x− −

Comparing with (y − k)2 = 4a (x − h) We get

ah, kf = FH − 13 , − 2IK 4a = 3 ⇒ a =

3 4

a f FH

1 Vertex = h, k = − , − 2 3

I K

Axis: y = k, y = −2 Focus: (a + h, k)

F 3 − 1 , − 2I H4 3 K 9−4 I F5 I =F H 12 , − 2K = H 12 , − 2K =

Equation of directrix: x = −a + h

x=−

3 1 − . 4 3

Parabola

x=−

13 . 12

Ends of latus rectum = (a + h, ±2a + k)

FG 3 − 1 , ± 2 F 3 I − 2IJ H 4 3 H 4K K F 5 , ± 3 − 2I = F 5 , − 7 I and F 5 , − 1 I . H 12 2 K H 12 2 K H 12 2 K =

Equation of Latus rectum x = a + h

x=

3 1 5 − = . 4 3 12

Length of latus rectum = 4a = 4 6.

a x + 3f = −24 a y − 1f ⇒ b x − a −3fg = −24 a y − 1f

F 3 I = 3. H 4K

2

2

Comparing with (x − h)2 = −4a (y − k) We get (h, k) = (−3, 1)

4a = 24 ⇒ a = 6. Vertex = (h, k) = (−3, 1) Axis: x = h, x = −3. Focus = (h, −a + k) = (−3, −6 + 1) = (−3, −5) Equation of directrix: y = a + k y=6+1=7 Ends of latus rectum = (±2a + h, −a + k)

b af a f g = a ±12 − 3, − 5f = a −15, − 5f and a9, − 5f. = ±2 6 + −3 , − 6 + 1

Equation of latus rectum: y = −a + k y = −6 + 1 = −5 Length of latus rectum = 4a = 4 (6) = 24.

327

328

Basic Mathematics

7. x 2 + 8 x + 12 y + 4 = 0 Consider

LMCo - efficient of x → 8 OP 8 + 12 y + 4 = 0 M = 4 → Square it, Add & Subtract P N2 Q

x 2 + 8 x + 12 y + 4 = 0 x 2 + 8x + 4 2 − 4 2

a x + 4f

− 16 + 12 y + 4 = 0

2

Comparing with (x − h)2 = −4a (y − k) We get

ah, kf = a−4, 1f

4 a = 12 ⇒ a = 3.

a f a f

Vertex: h, k = −4, 1 Axis: x = h, x = −4.

a

f a

f a

Focus: h, − a + k = −4, − 3 + 1 = −4, − 2 Equation of directrix: y = a + k y=3+1=4 Ends of latus rectum: (±2a + h, −a + k)

b af a f g = a ±6 − 4, − 2f = a −10, − 2f & a2, − 2f. = ±2 3 + −4 , − 3 + 1

Equation of latus rectum, y = −a + k y = −3 + 1 y = −2 Length of latus rectum = 4a = 4 (3) = 12. 8. y 2 − 4 y − 6 x + 13 = 0 Consider

y 2 − 4 y − 6 x + 13 = 0 y 2 − 4 y + 2 2 − 2 2 − 6 x + 13 = 0

a y − 2f

2

− 4 − 6 x + 13 = 0

f

Parabola

a y − 2f − 6 x + 9 = 0 a y − 2f = 6 x − 9. a y − 2f = 6 FH x − 96 IK a y − 2f = 6 FH x − 23 IK 2

2

2

2

Comparing this with (y − k)2 = 4a (x − h)

ah, kf = FH 32 , 2IK

We get

4a = 6 ⇒ a =

6 3 = 4 2

a f FH 32 , 2IK

Vertex: h, k =

Axis: y = k , y = 2.

a

f FH 23 + 32 , 2IK 6 = F , 2I = a3, 2f. H2 K

Focus: a + h, k =

Equation of directrix: x = −a + h

x=−

3 3 + 2 2

x = 0. Ends of Latus rectum = (a + h, ±2a + k)

=

F 3 + 3 , ± 2 ⋅ 3 + 2I H2 2 2 K

a

f a f a

f

= 3, ± 3 + 2 = 3, 5 & 3, − 1 . Equation of latus rectum: x = a + h

x=

3 3 6 + = = 3. 2 2 2

329

330

Basic Mathematics

Length of latus rectum: 4a

=4

F 3 I = 6. H 2K

9. 3 y 2 + 6 y + 8x − 5 = 0. Consider

3 y 2 + 6 y + 8 x − 5 = 0.

d

i

3 y2 + 2 y + 8x − 5 = 0 3 y 2 + 2 y + 12 − 12 + 8x − 5 = 0

a f 3 a y + 1f − 3 + 8 x − 5 = 0 3 a y + 1f + 8 x − 8 = 0 3 a y + 1f = −8 x + 8 3 a y + 1f = −8 a x − 1f a y + 1f = − 83 a x − 1f 2

3 y + 1 − 1 + 8x − 5 = 0 2

2

2

2

2

b y − a−1fg

2

=−

8 x −1 3

Comparing with (y − k)2 = −4a (x − h) we get (h, k) = (1, −1), 4a = Vertex = (h, k) = (1, −1) Axis: y = k, y = −1

F 2 + 1, − 1I H 3 K 1 = F , − 1I H3 K

Focus: (−a + h, k) = −

Equation of directrix : x = a + h

x=

2 +1 3

5 x= . 3

8 2 ⇒a= 3 3

Parabola

Ends of latus rectum:

a−a + h, ± 2a + kf 2 F 2 I = G − + 1, ± 2 F I − 1J H 3K K H 3 1 4 1 7 1 1 = F , ± − 1I = F , − I and F , I . H 3 3 K H 3 3 K H 3 3K

Equation of latus rectum

x = −a + h , 2 1 +1= . 3 3

x=−

Length of latus rectum: 4

F 2I = 8 . H 3K 3

10. 2 x 2 − 5 x + 3y + 4 = 0 Consider 2 x 2 − 5 x + 3y + 4 = 0

F H

2 x2 −

LM MN

I K

5 x + 3y + 4 = 0 2

F I − F 5 I OP + 3y + 4 = 0 H K H 4 K PQ L 5 25 O 2 MF x − I − P + 3y + 4 = 0 MNH 4 K 16 PQ 5 25 2F x − I − H 4 K 8 + 3y + 4 = 0 5 7 2 F x − I + 3y + = 0 H 4K 8 7 5 2 LM x − OP = −3y − 8 4 N Q 7 5 2 LM x − OP = −3 LM y + OP N 4 Q N 24 Q

2 x2 −

5 5 x+ 2 4

2

2

2

2

2

2

2

331

332

Basic Mathematics

F x − 5I H 4K

2

=−

FG F H H

3 7 y− − 2 24

I IJ KK

Comparing with (x − h)2 = −4a (y − k) we get

ah, kf = FH 45 , − 247 IK and 4a = 23 , a = 83 .

Vertex = (h, k) =

F5, − 7 I H 4 24 K

Axis: x = h

x=

5 4

Focus: (h, −a + k)

F 5, − 3 − 7 I H 4 8 24 K 5 −9 − 7 I =F , H 4 24 K 5 16 5 2 = F , − I = F , − I. H 4 24 K H 4 3 K =

Equation of directrix: y=a+k

y=

3 7 − 8 24

y=

9−7 2 1 = = . 24 24 12

Ends of latus rectum

a±2a + h, − a + k f = FGH ±2 FH 83IK + 45 , − 83 − 247 IJK

F H

= ±

I K

3 5 16 + ,− . 4 4 24

F 2, − 2 I and F 1 , − 2 I . H 3K H 2 3K

Parabola

Equation of latus rectum,

3 7 y = −a + h = − − 8 24 2 y=− . 3 Length of latus rectum = 4a

=4

F 3I = 3 . H 8K 2

II. Find the Equation of Parabola given that: 1. Vertex: (0, 0), Focus: (5, 0) Solution: Since the focus lies inside the parabola and vertex is given as origin. The parabola opens to the right. ∴ Equation of the parabola with vertex (0, 0) and which opens to right = y2 = 4ax Co-ordinates of focus = (a, 0) = (5, 0) [given] ⇒ a = 5.

(5, 0)

Fig. 14.9

Equation of the parabola = y2 = 4 (5) x

y2 = 20x 2. Vertex (0, 0), Focus (0, 4) Solution: Since focus = (0, 4) and vertex: (0, 0) and focus lies inside the parabola, the parabola opens upward. Equation of parabola: x2 = 4ay Given focus = (0, 4) = (0, a) ⇒ a=4 ∴ Equation of parabola: x2 = 4 (4) y x2 = 16y 3. Vertex: (0, 0), Axis = Negative Y-axis and length of latus rectum = 5.

(0, 4)

Fig. 14.10

Since axis = Negative Y-axis, and vertex = (0, 0) parabola opens downward as shown in fig. Equation of parabola: x2 = −4ay Given length of latus rectum = 5 4a = 5 Fig. 14.11

333

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Basic Mathematics

∴ Equation of required parabola.

Y

l

x2 = −5y. 4. Vertex (0, 0), Directrix x = 4. Since directrix x = 4, and directrix is away from the parabola and vertex = (0, 0), Parabola opens to the left. Equation of parabola: y2 = −4ax. For y2 = −4ax parabola, Equation of directrix is x = a. a = 4 (3 x = 4 is given)

x x=4

Fig. 14.12

af

∴ Equation of the parabola: y 2 = −4 4 x ⇒ y 2 = −16 x 5. Vertex (1, −2), Focus (−1, −1) Since vertex = (1, −2) and focus = (1, −1) and focus lie inside the parabola, the parabola open upwards. Equation of

a

parabola = x − h

f

2

a

= 4a y − k

Y

f

Given vertex (1, −2) ∴ (h, k) = (1, −2)

x′

x

0 S

⇒

h = 1, k = −2. V

Focus = (h, a + k) = (1, −1) ∴

a + k = −1 a + (−2) = −1 a = −1 + 2 a=1 a = +1.

⇒ ∴ Equation of parabola

a x − 1f = 4 a1f b y − a−2fg a x − 1f = 4 a y + 2f 2

2

x2 + 1 − 2x = 4y + 8 x 2 − 2 x − 4 y − 7 = 0. Alieter: Given: Vertex = (1, −2) = (h, k) ⇒ h = 1, k = −2

Y′

Fig. 14.13

Parabola

We know, Distance between vertex and focus = a i.e. distance between (1, −2) and (1, −1) = a

a1 − 1f + a−1 + 2f 2

2

=a

12 = a ⇒ a = 1. ∴ Equation of parabola

a x − hf = 4a a y − k f a x − 1f = 4 a1fa y + 2f 2

2

⇒

x2 − 2x − 4y − 7 = 0 .

6. Focus (1, 1), directrix x + 8 = 0. Let S (1, 1) be the focus and l: x + 8 = 0 be the directrix. Let P (x, y) be any point on the parabola. Join SP, draw PM ⊥ l. We know from definition,

M

P

l

SP =1 PM

S

SP = PM

SP = Distance between S (1, 1) and P (x, y) =

a x − 1f + a y − 1f 2

2

Fig. 14.14

[Using distance formula] PM = perpendicular distance of P (x, y) from x + 8 = 0 =

∴

x+8 12 + 0

a x − 1f + a y − 1f 2

2

=

x+8 1

Squaring,

a x − 1f + a y − 1f = a x +18f 2

2

2

x 2 + 1 − 2 x + y 2 + 1 − 2 y = x 2 + 64 + 16 x y 2 − 2 y − 18x − 62 = 0.

335

336

Basic Mathematics

7. Focus (4, 0) and directrix x = 6. Let P (x, y) be any point on the parabola whose focus is S (4, 0) and directrix is x − 6 = 0. Join, SP and draw PM ⊥ l. From definition

SP =1 PM SP = PM

P (x, y)

M

a f a f a x − 4f + a y − 0f = a x − 4f + y

SP = Distance between S 4, 0 and P x, y =

2

2

2

x−6=0

2

S (4, 0)

PM = perpendicular distance of P (x, y) from x − 6 = 0. = ∴

x−6

l

x−6 = = x − 6. 1

12

a x − 4f

2

Fig. 14.15

+ y2 = x − 6

Squaring,

a x − 4f

2

a

+ y2 = x − 6

f

2

x 2 + 16 − 8 x + y 2 = x 2 + 36 − 12 x y 2 + 4 x − 20 = 0. 8. Ends of latus rectum (−5, 2) and (−3, 2) Y

(−5, 2)

S

(−3, 2) L

L′

x

x′

Y′

Fig. 14.16

Since the ends of latus rectum are (−5, 2) and (−3, 2). There are 2 parabolas, open upwards and open downwards. So equations are

a x − hf

2

a

f

= ±4 a y − k .

Parabola

To find h, k and a: Length of latus rectum = 4a i.e., Distance between (−3, 2) and (−5, 2) = 4a.

b−5 − a−3fg + a2 − 2f = 4a a −5 + 3f + 0 = 4a a−2f = 4a 2

2

2

2

4 = 4a 1 a= . 2 Focus = Mid point of latus rectum Focus = mid point of (−3, 2) and (−5, 2)

FG −3 + a−5f , 2 + 2 IJ H 2 2 K 8 4 = F − , I = a −4, 2f. H 2 2K

Focus =

But Focus = (h, ± a + k) ∴ (h, ± a + h) = (−4, 2)

h = −4, ± a + k = 2

⇒

±

1 +k=2 2

k =2±

1 . 2

5 3 or k = 2 2 (for −ve a) (for +ve a) ∴ Equations of parabolas: k=

(i)

F 1 I F y − 3I H 2K H 2K a x + 4f = 2 FH y − 23 IK a f

x − −4

2

=4

2

x 2 + 16 + 8 x = 2 y − 3 x 2 + 8 x − 2 y + 19 = 0 .

337

338

(ii)

Basic Mathematics

F 1 I F y − 5I H 2 K H 2K a x + 4f = −2 FH y − 25 IK a x + 4f = −2 y + 5 a f

x − −4

2

= −4

2

2

x 2 + 8 x + 16 + 2 y − 5 = 0 x 2 + 8 x + 2 y + 11 = 0 9. Vertex: (3, 4) Directrix y = 1 Given: Vertex = (3, 4) ∴ (h, k) = (3, 4) ⇒ h = 3 and k = 4. Since directrix is away from the parabola and perpendicular to axis, the parabola opens upwards. Equation of parabola is (x − h)2 = 4a (y − k) Equation of directrix: y = −a + k 3 y = 1 is directrix.

(3, 4)

y=1

1 = −a + k 1 = −a + 4

Fig. 14.17

a = 4 −1

a = 3.

a

f

∴ Equation of parabola: x − 3

2

a

= 4.3 y − 4

f

x 2 + 9 − 6 x = 12 y − 48 x 2 − 6 x − 12 y + 57 = 0. 10. Directrix: x + 3 = 0, axis y = 2 and length of latus rectum: 6. Given: Axis y = 2, and directrix x + 3 = 0, x = −3. From figure, l Since directrix is away from parabola, the parabola opens to the right. ∴ Equation of parabola.

a y − kf

Axis: ∴

2

a

= 4a x − h

x=3

axis y=2

f

y = k , 3 y = 2.

k = 2.

Fig. 14.18

Parabola

339

Length of latus rectum = 4a 4a = 6

a=

6 4

3 a= . 2

x = −a + h

Directrix:

x=−

3 +h 2

−3 = − −3 +

3 +h 2

3 =h 2

−6 + 3 =h 2 3 h=− . 2 ∴ Equation of parabola:

F 3 I FG x − F − 3 I IJ H 2K H H 2KK a y − 2f = 6 FH x + 23 IK

a y − 2f

2

=4

2

y2 + 4 − 4y = 6x + 9 y2 − 4y − 6x − 5 = 0

(Given)

Figure

Vertex

Focus

Equation of directrix

Axis

Ends of Latus rectum

Equation Length of of Latus rectum Latus rectum

y2 = 4ax

(0, 0)

(a, 0)

x = −a

x-axis y=0

(a, ±2a)

x=a

4a

y2 = −4ax

(0, 0)

(−a, 0)

x=a

x-axis y=0

(−a, ±2a)

x = −a

4a

x2 = 4ay

(0, 0)

(0, a)

y = −a

y-axis x=0

(± 2a, a)

y=a

4a

x2 = −4ay

(0, 0)

(0, −a)

y=a

y-axis x=0

(± 2a, −a)

y = −a

4a

(h, k)

(a + h, k)

x = −a + h

y=k

(a + h, ±2a + k)

x=a+h

4a

(h, k)

(−a + h, k)

x=a+h

y=k

(−a + h, ±2a + k)

x = −a + h

4a

(h, k)

(h, a + k)

y = −a + k

x=h

(± 2a + h, a + k)

y=a+k

4a

(h, k)

(h, −a + k)

y=a+k

x=h

(± 2a + h, −a + k)

y = −a + k

4a

(h, k)

(y − k)2 = 4a (x − h) (h, k)

(y − k)2 = −4a (x − h)

(x − h)2 = 4a (y − k)

(x − h)2 = −4a (y − k)

(h, k)

(h, k)

Basic Mathematics

Equation

340

REMEMBER:

Parabola

• • • • •

341

In any parabola, focus is inside the curve and directrix is away from the parabola. Distance between vertex and focus = a. For the given ends of latus rectum, there are 2 possible parabolas. Focus is the mid point of latus rectum. Axis is perpendicular to the directrix. Distance between directrix and vertex = a.

EXERCISE er te x, F ocus, equa tion of dir ectr ix, axis, length of la tus rrectum, ectum, ends of la tus I. Find the vver erte tex, Focus, equation directr ectrix, latus latus rectum and equa tion of la tus rrectum ectum of the ffollo ollo wing par abolas: equation latus ollowing para 1. y 2 = 4 x

2. y 2 = −8 x

3. x 2 = 3 y

4. x 2 = −7 y

5.

a y − 1f

2

= 4x

a f

7. x 2 = −12 y − 7

9. y 2 − 20 x − 10 y + 15 = 0

6. 8.

a x + 1f a y + 3f

2 2

a f = −8 a x − 7f

= 6 y +1

10. y 2 − 4 y + 6 x − 8 = 0

11. 2 y 2 − 5 y + 3x + 4 = 0

12. x 2 + 4 x + 10 y + 8 = 0

13. x 2 + 4 y − 2 x + 3 = 0

14. x 2 − 8 x + 16 y = 0

15. x 2 − 4 x − 5y − 1 = 0 tion of par abola ggiiven tha t: II. Find the equa equation para that: 1. Vertex: (0, 0), Focus (−3, 0) 2. Vertex: (0, 0), Directrix : y = 8 3. Vertex : (0, 0), Length of Latus rectum = 3, axis positive x-axis. 4. Vertex: (0, 0), Equation of latus rectum x = 4. 5. Vertex : (0, 0), Axis x-axis and passing through (3, −1) 6. Focus (0, 5) and directrix y = −4 7. Focus (4, 0), Directrix x = 6 8. Vertex: (1, 1), Focus (4, 1) 9. Ends of latus rectum (−3, 1) and (1, 1) 10. Directrix x + 2 = 0, axis y = 3 and length of latus rectum = 6. 11. Vertex: (3, 4), Directrix y = 1 12. Vertex: (3, −3), Directrix y = −7 and axis parallel to y-axis. 13. Directrix 4x − 1 = 0, Axis y = 2 and length of latus rectum 8. 14. Ends of latus rectum (8, 4) and (8, −2) 15. Directrix: 4x + 1 = 0, axis 2y + 1 = 0 and length of latus rectum 24.

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Basic Mathematics

ANSWERS I. Question No.

Vertex

Focus

Directrix

Axis

Length of

Ends of

Equation of

1

(0, 0)

(1, 0)

x = −1

x-axis y = 0

4

(1, ± 2)

x=1

2

(0, 0)

(−2, 0)

x=2

x-axis y = 0

8

(−2, ± 4)

x = −2

3

(0, 0)

(0, 3/4)

y = −3/4

y-axis x = 0

3

(±3/2, 3/4)

y = 3/4

4

(0, 0)

(0, −7/4)

y = 7/4

y-axis x = 0

7

(±7/2, −7/4)

y = −7/4

5

(0, 1)

(1, 1)

x = −1

y=1

4

(1, ±2 + 1)

x=1 y = 1/2

Latus rectum Latus rectum Latus rectum

6

(−1, −1)

(−1, 1/2)

y = −5/2

x = −1

6

(−1±3, 1/2)

7

(0, 7)

(0, 10)

y=4

x=0

12

(±8, 10)

8

(7, −3)

(5, −3)

x = −5

y = −3

8

(5, ±4 −3)

x=5

9

(1/2, 5)

(9/2, 5)

x = −11/2

y=5

20

(9/2, 25) (9/2, −15)

x = 9/2

10

(2, 2)

(1/2, 2)

x = 7/2

y =2

6

(1/2, 5) (1/2, −1)

x = 1/2

11

(−7/24, 5/4)

(−2/3, 5/4)

x = 1/12

y = 5/4

3/2

(−2/3, 2) (−2/3, 1/2)

x = −2/3

12

(2, −2/5)

x=2

10

(3, −29/10) (−7, 29/10)

y = −29/10

13

(1, −1/2)

(1, −3/2)

y = −1/2

x=1

4

(3, −3/2) (−1, −3/2)

y = −3/2

14

(4, 1)

(4, −3)

y=5

x=4

16

(12, 1) (−4, 1)

y=1

15

(2, −1)

(2, 1/4)

y = −9/4

x=2

5

(9/2, 1/4) (−1/2, 1/4)

y = 1/4

(−2, −29/10) y = −21/10

II. 1. y 2 = −12 x

2. x 2 = −24 y

4. y 2 = 16 x

2 5. y =

a f a y − 3f = 6 FH x + 12 IK a y − 2f = 8 FH x + 74 IK

7. y 2 = −4 x − 5 10.

13.

2

2

1 x 3

8.

a y − 1f

11.

a x − 3f

14.

3. y 2 = 3x

2

= 12 x − 1

a f

9.

a x + 1f

a

12.

a x − 3f

f a y − 1f = 6 FH x − 132IK 2

2

F H

2 6. x = 18 y −

= 12 y − 4

15.

2

1 2

a

I K

= −4 y − 2

f

a f a y − 2f = 8 FH x + 74 IK 2

2

= 16 y + 3

15 Limits and Continuity 15.1 INTRODUCTION: The discovery of calculus was done independently and almost during the same time by Sir Isaac Newton and Gottfried Wilhelm Leibnitz. Calculus is developed on the basis of a more fundamental concept called the Limit.

15.2 CONSTANTS AND VARIABLES: A quantity which remains the same throughout any mathematical discussion is called a constant. A quantity which takes different values in any mathematical discussion is called a variable. For instance, if we increase the production by using more raw materials, the cost of the machine doesn’t change, but the costs of raw materials, Labour sales change. So cost of the machine which remains the same value throughout is constant and cost of raw materials and labour which assumes any numerical value out of given set of values are variables. Constants are represented by a, b, c, d, e and variables are represented by u, v, w, x, y, z.

15.3 FUNCTION: Let A and B be 2 non empty sets. A rule f which associates each element x of A to an unique element y of B is called a function. It is denoted by f : A → B. If x ∈ A is related to y ∈ B, then we write y = f (x). Algebraic function: A function y = f(x) is said to be an algebraic function if f(x) is a polynomial function (eg: x2 + 3x − 1)

F 3x + 8 I or rational function G H 4 x − 1 JK 2

3

or irrational function (e.g.: (1 − x)2/3).

Note: The functions other than algebraic functions are called transcendental functions. They include, trigonometric functions, logarithmic function, inverse trigonometric functions, exponential functions, hyperbolic functions etc.

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Basic Mathematics

Exponential function: The function which associates every real number x to the real number ex is called exponential function. Where e is the sum of the infinite series

1+

1 1 1 1 + + + + ... 1! 2 ! 3! 4 !

Its value is 2.7182818284.... e is an irrational number. Logarithmic function: The function which associates every positive real number x to the real number loge x is called the logarithmic function.

15.4 LIMITS: Consider the function

y= when x = 1, y =

x2 − 1 x −1

12 − 1 0 = which is not defined. 1−1 0

Instead of giving x = 1, Let us give, x a value which is slightly less than 1 or slightly greater than 1. Then when x = 0.9, y = 1.9

when

x = 0.99,

y = 1.99

x = 0.999,

y = 1.999

x = 1.0001,

y = 2.0001

x = 1.001,

y = 2.001

x = 1.01,

y = 2.01

x = 1.1,

y = 2.1

From the above set of values, we can observe that the value of y is slightly less than 2 or slightly more than 2 when the value of x is slightly less than 1 or slightly more than 1. But when x = 1, y has no value. In otherwords, when x is very nearly equal to 1, y is very nearly equal to 2. In symbols this is expressed as lim

x →1

i.e.

x2 − 1 =2 x −1

lim y = 2 x →1

Read it as the limit of y as x tends to 1 is 2. The statement simply imply when x ≈ 1 y ≈ 2 even though it doesn’t exist when x = 1.

Limits and Continuity 345

Definition of Limit: A function f(x) is said to tend to a limit l as x tends to a if the numerical difference between f(x) and l can be made as small as we please by taking the numerical difference between x and a as very small. In other words f(x) is very nearly equal to l when x is very nearly equal to a.

af

It is denoted by lim f x = l or x→a

af

lt f x = l.

x →a

Left Hand and Right Hand Limits: When x → a, through the values which are smaller than ‘a’ then we write x → a − 0 this

af

a

f

lim f x or lim f a − h is called Left Hand Limit [LHL] of a function f (x).

x →a− 0

h→ 0

For example: if x → 3 by taking the values 2.9, 2.99, 2.999, 2.9999... then x → 3 − 0 or Simply x → 3−

af

When x → a, through the values which are greater than ‘a’ then we write x → a + 0 this lim f x

a

x → a +0

f

or lim f a + h is called Right Hand Limit [RHL] of a function f(x). h→0

For example: if x → 3 by taking the values 3.1, 3.01, 3.001... then x → 3 + 0 or Simply x → 3+

15.5 STANDARD LIMITS: 1. Prove that lim

x →a

x n − an = na n −1 , where n is any rational number. x−a

Proof. Case (i): Let n be a positive integer. We know

a

fd

x n − a n = x − a x n −1 + x n −2 ⋅ a + x n −3 ⋅ a 2 + ... + a n −1

a

i

fd

x − a x n −1 + x n − 2 ⋅ a + x n − 3 ⋅ a 2 + ... + a n −1 x n − an lim = lim x→a x − a x→ a x−a = lim x n −1 + x n − 2 ⋅ a + x n − 3 ⋅ a 2 + ... + a n −1 x→ a

i (n terms)

= a n −1 + a n − 2 ⋅ a + a n −3 ⋅ a 2 + ... + a n −1 = a n −1 + a n −1 + a n −1 + ...

x n − an = na n −1 x →a x − a lim

Case (ii): Let n be a negative integer, say n = −m, where m is a +ve integer.

(n terms)

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Basic Mathematics

Consider

x n − an x − m − a −m = lim x →a x − a x →a x−a lim

1 1 m − m x a = lim x→ a x−a

As a − m =

1 am

am − xm m m = lim x a x→ a x−a = lim

x→a

am − xm x m ⋅ am x − a

a

= lim − x→a

f

LM x − a OP ⋅ N x−a Q x m

m

1 x n − an = − ma m −1 m m x →a x − a a ⋅a lim

=−

ma m −1 a2m

= − ma m −1− 2 m = − ma − m −1

∵ n = −m x n − an = na n −1 x →a x − a lim

Case (iii) Let be a fraction, positive or negative. Take n =

p where p is an integer and q ≠ 0. q

Now

x n − an xp q − ap q = lim x →a x − a x→a x−a lim

Taking ⇒

x = yq and a = bq x1/q = y and a1/q = b

m

1 ⋅ am

case (i) as O LMmUsing N is + ve integerQP


So that

x → a ⇒ yq → bq y → b. xp q − ap q yp − bp = lim q x →a y→b y − b q x−a lim

Dividing both numerator and denominator by y − b

yp − bp pb p −1 y−b lim q = x →a y − b q qb q −1 y−b

LM∵ N

lim x =a

x n − an = na n −1 x→a

OP Q

p P−1− q +1 ⋅b q p p− q ⋅b q But

b=a

1 q

x p q − ap q p 1 q = ⋅ a x →a x−a q

d i

lim

p− q

p

x p q − a p q p q −1 = a lim x →a x−a q x n − an = na n −1 x →a x − a

∴

lim

for all rational values of n.

Some Standard Limits: Here we state some standard limits (without proof) and use them while solving problems.

F H

1. (a) lim 1 + n→∞

2.

3.

F H

lim 1 +

n→∞

k n

I K

1 n

n

ex − 1 =1 x→0 x lim

I K

n

=e

= ek

a f

(b) lim 1 + n n→ 0

1 n

=e

348

4.

Basic Mathematics

lim

x →0

ax −1 = log e a where a > 0. x

Algebra of Limits: If f(x) and g(x) are 2 functions of x and k is any scalar Then

af af af af lim k ⋅ f a x f = k ⋅ lim f a x f lim f a x f ⋅ g a x f = lim f a x f ⋅ lim g a x f f a gf lim f a x f = lim provided lim g a x f ≠ 0 . g a x f lim g a x f

1. lim f x ± g x = lim f x ± lim g x x→a

2. 3.

4.

x→ a

x→a

x→ a

x→a

x →a

x→a

x →a

x→ a

x→a

x→a

x→a

Indeterminate Forms: In mathematics the forms like

Ex:

0 ∞ 0 ∞ , , 0 , 1 ... are called indeterminate forms. 0 ∞

0 12 − 1 0 x2 − 1 then the value of y = when x = 1 ∵ y = = 0 1−1 0 x −1

1. If y =

a f

2. If y = 1 + n

1 n

a f

then value of y = 1 + 0

1 0

y = 1∞ when n = 0.

Evaluation of Limits:

af

To evaluate lim f x , first find f(a). x →a

(i) If f (a) ≠ an indeterminate form then f (a) itself is the limit. (ii) If f (a) = an indeterminate form then suitable method / formula is applied to reduce f (x) so that it will not take an indeterminate form when x is replaced by a.

WORKED EXAMPLES: 1. Evaluate: lim

x→0

4 x 2 + 3x + 1 3x 2 − 4 x − 1

Solution: By Putting x = 0. We get

lim

x →0

af af af af

4 x 2 + 3x + 1 4 0 + 3 0 + 1 1 = = = −1 ≠ an intermediate. 3 x 2 − 4 x − 1 3 0 − 4 0 − 1 −1


Hence lim

x →0

4 x 2 + 3x + 1 = −1 3x 2 − 4 x − 1

2. Evaluate: lim

x →1

x 2 − 5x + 4 x 2 − 4x + 3

Solution: By putting x = 1 we get

af af

12 − 5 1 + 4 0 = = an indeterminate 12 − 4 1 + 3 0

Hence consider

lim

x →1

x 2 − 5x + 4 by factorising both numerator and denominator we get x 2 − 4x + 3 lim

x →1

a f a f a f a f

x 2 − 4 x − 1x + 4 x x − 4 −1 x − 4 = lim 2 x − 3 x − 1x + 3 x→1 x x − 3 − 1 x − 3

= lim x →1

a x − 1fa x − 4f = lim x − 4 x−3 a x − 1fa x − 3f x →1

Now substituting x = 1 or applying lim we get x→1

1 − 4 −3 3 = = . 1 − 3 −2 2 ∴ 3. Evaluate: lim

x →0

x 2 − 5x + 4 3 = . x2 − 4x + 3 2

lim

x →1

3+ x − 3− x x

Solution: By putting x = 0 we get

3+0 − 3−0 3− 3 0 = = = an indeterminate 0 x 0 Consider

lim

x →0

3+ x − 3− x By rationalising the numerator, x lim

x →0

3+ x − 3− x 3+ x + 3− x × x 3+ x + 3− x

d lim

x →0

3+ x x

d

i −d 2

3+ x +

i 3− xi 3− x

2

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= lim

x→ 0

= lim

x→0

= lim

x→ 0

x

x

3+ x −3+ x

d

3+ x + 3− x 2x

d

3+ x + 3− x

i

i

2 3+ x + 3− x

Now applying lim , x→0

2 = 3+ 0 + 3− 0

=

∴

lim

x→0

2 2 3

=

2 3+ 3

1 3

3+ x − 3− x 1 = , x 3

x 5 − 243 x →3 x − 3

4. Evaluate: lim

LMFormula: lim x − a x−a N n

x 5 − 243 x 5 − 35 = lim Solution: lim x →3 x − 3 x →3 x − 3

x→a

n

= na n −1

OP Q

a f

= 5 ⋅ 35−1 = 5 ⋅ 34 = 5 81 = 405 ∴ 5. Evaluate: lim

x →5

3

x 5 − 243 = 405 x →3 x − 3 lim

x −3 5 x− 5

Consider

x 1 3 − 51 3 x →5 x 1 2 − 51 2 lim

Dividing both numerator and denominator by x − 5.

x 1 3 − 51 3 x 1 3 − 51 3 lim x−5 lim = x→ 5 1 x2 − 51 2 x →5 x 1 2 − 51 2 x −5 lim x→5 x−5 x−5

[using algebra of limits]


Now applying the formula,

lim

x →a

x n − an = na n −1 for numerator and denominator separately we get x−a −2

1

1 3 1 3 −1 5 5 3 3 = 1 1 1 2 −1 1 − 2 5 5 2 2 2 1

= 2 5 3

lim

∴ 6. Evaluate: lim

x →−2

−4 + 3 6

x →5

3

2 −3+2 5 3 1

2 − 2 = 5 6 = 6 3 3⋅ 5 2 x −3 5 = 6 . x − 5 3⋅ 5

x 5 + 32 x8 − 28 lim

x →−2

a f a f

x 5 − −2 x 5 + 32 = lim x 8 − 2 8 x →−2 x 8 − −2

5 8

a f = −32 a−2f = +2

∵ −2

Dividing both Numerator and Denominator by

5

8

x − (−2) we get

lim

x →−2

a f a f x − a−2f x − a−2f 5 a −2f = 8 a −2f 5 = ⋅ a −2f 8

x 5 − −2 x − −2 8

5

a f a f = x − a−2f lim x − a−2f 5 a −2f = 8 a −2f 5 = ⋅ a −2f 8 lim

x →−2

8

x 5 − −2 x − −2 8

x →−2

5 −1

4

8 −1

7

4−7

5 1 5 ⋅ = 8 −8 −64

a f

−3

5

8

8

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Basic Mathematics

7. Evaluate: lim

n→∞

Solution: lim

n→∞

n2 + n + 1 . 3n 2 + 2n − 1

∞ n2 + n + 1 = = an indeterminate . 2 3n + 2 n − 1 ∞

Consider

n2 + n + 1 = lim lim n→∞ 3n 2 + 2 n − 1 n→∞

F 1+ 1I H n nK F3 + 2 − 1 I H n nK

n2 1 +

2

n2

2

1 1 + 2 n n = lim 2 1 n→∞ 3+ − 2 n n 1+

Now applying the lim we get n→∞

1+ 0 + 0 1 = 3+ 0+ 0 3 lim

∴

n→∞

8. Evaluate: lim

n→∞

n 12 + 2 2 + 32 + ... n 2 13 + 2 3 + 33 + ... n 3 n

lim

n→∞

1 n2 + n + 1 = . 3n 2 + 2n − 1 3 .

LM n an + 1fa2n + 1f OP N 6 Q n an + 1f

∵12 + 2 2 + 32 + ...n 2 =

2

2

4 lim

n →∞

a fa F H

n2 ⋅ n 1 + lim

n →∞

f

n 2 n + 1 2n + 1 4 × 2 6 n n +1 2

I F K H

3 3 3 and 1 + 2 + ...n =

a f

I K

1 1 n 2+ n n ⋅ 6

a fa

4

F H

n2 ⋅ n2 1 +

1 n

I K

2

f

n n + 1 2n + 1 6

a f

n2 n + 1 4

2


F1 + 1 I F 2 + 1 I H nK H nK ⋅ lim 6

n→∞

F H

4 1 1+ n

I K

2

Applying the lim

n→∞

a1 + 0fa2 + 0f × 6

∴

lim

n→∞

9. Evaluate: nlim →∞

4 1+ 0

a f

=

2

2 4 4 × = 6 1 3

n 12 + 2 2 + ... n 2 3

3

1 + 2 + ... n

=

3

4 . 3

n +1 4n 2 − 3

n +1

lim

4n2 − 3

n→∞

F 1I H nK 3 n F4 − I H nK n 1+

= lim

n→∞

2

F H

2

I K

1 n = lim n→∞ 3 n 4− 2 n n 1+

Applying the limit, we get 1+ 0 1 1 = = . 4−0 4 2 lim

∴

10. Evaluate: lim

n →∞

F n + 1I H nK

n →∞

n +1 2

4n − 3

=

1 2

4n

F n + 1 I = lim F1 + 1 I H n nK H nK LF 1 I OP = e . = lim M 1 + MNH n K PQ 4n

lim

n →∞

4n

n →∞

n 4

n→∞

4

F H

∵ lim 1 + n →∞

1 n

I K

n

=e

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Basic Mathematics

11. Evaluate: lim

n →∞

F nI H n + 1K

6n

Dividing both numerator and denominator by n

F n I = lim GG n JJ GH n n+ 1 JK

6n

n→∞

F I 1 J lim GG GH 1 + 1n JJK

F I 1 J = lim GG n 1J GH n + n JK n→∞

6n

=

n→∞

= lim = n→∞

12. Evaluate: lim

n→∞

6n

16 n

F1 + 1 I H nK

16 n

LMF1 + 1 I OP NH n K Q

n 6

=

6n

1 = e −6 . e6

1 1 1 1 + 2 + 3 + ... + n . 6 6 6 6

1 2 1 1 1 1 1 1 Solution: + 2 + 3 + ... + n is a G.P. with a = , r = 6 = 1 6 6 6 6 6 6 6 Sn =

lim

n→∞

d

a 1− rn

i

1− r

[formula]

1 1 1 1 + 2 + 3 + ... + n . 6 6 6 6

LM F I OP 1 L F 1I O 1 1− M P 1− F I H K H K 6 6 H N Q = lim N Q = 6K

1 1 1− 6 6 = lim 1 n→∞ 1− 6

n

n

n→∞

Applying the lim, we get

1− 0 1 1 = = . 5 5 5

5 6

5

n


13. Evaluate: lim 1 − n→∞

Solution: 1 −

1 1 1 + − + ... n terms. 2 4 8

1 1 1 + − ...n terms is a G.P. 2 4 8 −1 1 with a = 1, r = 2 = − 1 2 Sn = ∴

lim 1 −

n→∞

d

a 1 − rn 1− r

lim

Formula

1 1 1 + − + ... n terms. 2 4 8

F F 1I I 1 1− F− I GH H 2 K JK H 2K = lim 1 1 1+ 1− F− I H 2K 2

1 1− − n→∞

i

n

n→∞

Applying the lim

n→∞

14. Evaluate: lim

x →0

n

1− 0 2 = . 3 3 2

ax − bx x lim

x →0

= lim

ax − bx +1 −1 x

d

Adding and Subtracting 1

i

ax −1 − 1 bx − 1 x

x→ 0

ax −1 bx − 1 − x →0 x x lim

ax −1 bx −1 − lim x →0 x→0 x x lim

= log e a − log e b = log e

F aI H bK

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Basic Mathematics

3x − 1 x +1 −1

15. Evaluate lim

x →0

Consider 3x − 1 Rationalising the denominator, x +1 −1 3x − 1 × x +1 −1

x +1 +1 x +1 +1

d3 − 1id =

i

x

x +1 +1

x +1−1

d3 − 1id

i

x

x +1 +1 x

∴

lim

x →0

d

id

i

3x − 1 x + 1 + 1 3x − 1 = lim x x + 1 − 1 x →0

Applying the lim

x→0

log e 3

d

i

0 +1 +1

∵ lim

x →0

3x − 1 = log e 3 x

af

log e 3 2

= 2 log e 3

15.6 CONTINUOUS FUNCTIONS:

af af

af

A function y = f (x) is said to be continuous at x = a if lim− f x = f a = lim+ f x x→a

x→ a

i.e., Left Hand Limit = f (a) = Right Hand Limit or

a

f af

a

f

lim f a − h = f a = lim f a + h

h→ 0

h→ 0

af af

In other words, y = f(x) is continuous at x = a if lim f x = f a x→a

Geometrically y = f (x) is continuous at x = a means that there is no break in the graph of y = f (x) at x = a. Note: A function which is not continuous at x = a is said to be discontinuous at x = a.


WORKED EXAMPLES:

Rx + 3 when x ≤ 2 a f |S 5 when x = 2 is continuous at x = 2. T|3x − 1 when x ≥ 2

1. Prove that the function f x =

Proof. For a function y = f (x) to be continuous at x = a we have LHL = f (a) = RHL. Here a = 2.

af

LHL = lim− f x = lim x + 3 = 2 + 3 = 5 x →2

x→2

f(a) = f(2) = 5 (Given)

af

af

RHL = lim+ f x = lim 3 x − 1 = 3 2 − 1 = 5 x →2

x→ 2

af

∴

LHL = f a = RHL

Hence f(x) is continuous at x = 2.

af

2. Verify whether the function f x =

when x > 1 R|x +2 3 when S|2 x − 1 when xx 4

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Basic Mathematics

x 3 − 43 3.4 3−1 3 × 4 2 3 × 4 = lim x2 − 4 2 = = = =6 x→4 x − 4 2×4 2 2.4 2 −1 x−4 Given

f(a) = f(4) = 26 ∴ LHL = RHL ≠ f(a). ∴ The function f(x) is not continuous at x = 4.

4. Prove that

R| f a x f = Sa1 + 2 x f |Te 2

1 x

for x ≠ 0 is continuous at x = 0. for x = 0

Proof. For a function to be continuous at x = a we have LHL = f(a) = RHL. Here a = 0

a

LHL = RHL = lim 1 + 2 x x →0

a

lim 1 + 2 x

x →0

f

f

1 x

2 2x

L O = lim Ma1 + 2 x f P N Q

1 2 2x

x→0

=e

LM∵ lim a1 + nf N

2

af af

x→ 0

a

f a = f 0 = e 2 given

1 n

=e

OP Q

f

af

∴ LHL = RHL = f 0

Hence f (x) is continuous at x = 0. 5. If f (x) = + 3x − 1, then prove that f (x) is continuous at x = 1. Proof. For a function f (x) to be continuous at x = a we have LHL = f (a) = RHL. Here a = 1. x2

af = lim a1 − hf

a f + 3 a1 − hf − 1

LHL = lim− f x = lim f 1 − h h →0

x →1

h →0

2

= lim 1 + h 2 − 2 h + 3 − 3h − 1 h →0

= lim h 2 − 5h + 3 h →0

af

∵ f x = x 2 + 3x − 1


Applying the lim we get h→0

LHL = 0 − 5(0) + 3 = 3.

af

a f

RHL: lim+ f x = lim f 1 + h x →1

h→ 0

i a f

d

= lim 1 + h 2 + 3 1 + h − 1 h→ 0

= lim 1 + h 2 + 2 h + 3 + 3h − 1 h→ 0

= h 2 + 5h + 3

Applying the lim we get h→0

RHL = 3.

a f af

af LHL = f a af = RHL.

f a = f 1 = 13 + 3 1 − 1 = 1 + 3 − 1 = 3 . ∴

Hence the function y = f(x) is continuous at x = 1. 6. Find K if the function

R| x f axf = S |T

2

− 5x + 6 if x ≠ 2 is continuous at x = 2. x−2 K if x = 2

Given f (x) is continuous at x = 2. LHL = f (a) = RHL.

x 2 − 5x + 6 =K x →2 x−2 lim

i.e.

lim

x →2

ax − 3fa x − 2f = K a x − 2f lim x − 3 = K

x →2

2−3= K

⇒

af a

7. Define f (0) so that f x = 1 + 3 x

K = −1

f

1 x

x ≠ 0 is continuous at x = 0.

Given: f (x) is continuous at x = 0.

af af

lim f x = f 0

x →0

a

lim 1 + 3 x

x →0

f

1 x

af

= f 0

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Basic Mathematics

a

lim 1 + 3x

x →0

LMa N

f

e3

⇒

8. Find K if

R| f a xf = S |T

af

= f 0

f OPQ = f a0f = f a0 f 1 3 3x

lim 1 + 3 x

x→0

3 3x

1+ x − x if x ≠ 0 x is continuous at x = 0. K +3 if x = 0

Given f (x) is continuous at x = 0 x →0

lim

x →0

lim

x →0

af af

lim f x = f 0

∴

1+ x −1 = K+3 x

1+ x −1 1+ x +1 = K +3 × x 1+ x +1

d 1+ x i −1 = K + 3 lim x d 1 + x + 1i 2

2

x →0

lim

x →0

lim

x →0

1+ x −1

x

i

= K +3

d 1 + x + 1i

= K +3

d

1+ x +1 x

x

Applying the lim we get x→0

1 = K+3 1+ 0 +1

1 = K +3 1+1 1 −3= K 2

a f

∵ lim 1 + n n→ 0

1 n

=e


1− 6 =K 2

⇒

K=

⇒

−5 2

af {

2 x + a if x ≤ 2 9. Find a if f a = x − 1 if x > 2 is continuous at x = 2. Given: f(x) is continuous at x = 2. LHL = RHL = f(a)

af af

af

lim f x = f 2 = lim+ f x

x →2 −

x →2

lim 2 x + a = lim x − 1

x →2

x→2

af

2 2 + a = 2 −1

4+ a =1 a =1− 4 a = −3

REMEMBER: • lim

x →a

x n − an = na n −1 x−a

ex − 1 =1 x→0 x

• lim

• lim

x →0

ax − 1 = log e a x

F H

1 n

I K

a f

1 n

• lim 1 + n→∞

• lim 1 + n n→ 0

n

=e

=e

af af

af

• A function y = f (x) is said to be continuous at x = a if lim− f x = f a = lim+ f x x →a

i.e., LHL = f (a) = RHL. • Limit of a function exists at x = a if

x→a

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Basic Mathematics

af

af

lim f x = lim+ f x

x →a−

∴

x→a

f(x) is continuous at x = a iff

af af

lim f x = f a

x →a

EXERCISE Evaluate:

a

fa

2 x + 1 3x + 2 1. xlim →2

f

x 3 + 4 x 2 − 3x + 3 lim 2. x →1 7x 2 − x − 1

13.

lim

x→ −2

x 5 + 32 x+2

1 1 3 − 64 x 14. lim x→4 x−4

3. lim

3x 2 + 8 x − 9 4 x2 − 9x − 3

15. lim

4. lim

x 3 + 3x 2 + 2 x − 1 x 3 + 3x 2 + 6

16. lim

x 4 − 81 x −3

5. lim

x 2 − 5x + 6 3x 2 + 3x − 7

17. lim

x3 2 − 1 x −1

6. lim

3+ x − 3 x

18. lim

x 3 − 125 x2 − 6x + 5

7. lim

x →0

a− x − a+x x

19.

8. lim x →3

x −3 x−2 − 4−x

20. lim

x →0

x →1

x →3

x →0

x − 16 x − 17

x → 25

x →3

x →1

x →5

lim z→

3 2

n →−1

3− 5+ x 9. lim x →4 x−4 10. lim

x →3

x 3 − 27 3x − 4 − 2 x − 1

6 z 2 − 5z − 6 8 z 2 − 14 z + 3

a n + 2f d n

x −4 3x − 2 − x + 2

12. lim

3x − 4 − 4 − x x−2

x →2

x →2

i

+ 4n + 3

2

n −1

21. ylim →−1

2 y 3 − 3y 2 − 3 y + 2 3 y 3 + 2 y 2 − 11y − 10

22. lim

a 1 − 2 x − a x − ax

x →a

2

11. lim

2

1 1 3 − 27 x lim 23. x →3 1 1 4 − 81 x


24. lim

x →3

25. lim

x →5

x 3 − 27 x x −3 3

a x − 3f

5

− 32 x−5

26. lim

ax − bx x

27. lim

e x − e−x x

28. lim

2x −1 x

x →0

x →0

x→0

29. lim

2x −1 x +1 −1

30. lim

3n 2 − 6n + 8 4n 2 − 8

x →0

n→∞

31. lim

a1 + 2 + 3 + ...+nf 3n 2

n→∞

32. lim

n→∞

12 + 2 2 + 32 + ...+ n 2 3n 3 − 8n + 1

2 33. lim n n − 4 − n n→∞

34. lim

n→∞

1 1 1 + + ... + n 3 32 3

35. nlim →∞

13 + 2 3 + ... + n 3 4 n+4

36. lim

3 n +1 + 1 3n + 2 + 2

n→∞

a

f

af {

af

3x + 2 if x < 2 37. If f x = 7 x − 6 if x > 2 , then find lim f x . x →2 38. Prove that lim

x →0

x does not exist x

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Basic Mathematics

a

f

1+ 2 n 39. Evaluate (a) nlim →∞

3n

a

1 + 3x (b) nlim →0

f

4x

40. Examine the Continuity of the following functions: (a) f(x) = x2 + x + 1 at x = 1

R|4 x1+ 1 ifif xx =< 00 at x = 0 S|2 + 3x if x > 0 T R| x − 9 f a x f = S x − 4 x + 3 for x ≠ 3 at x = 3. |T 3 for x = 3 R| x for x ≠ 0 f a xf = S 1 + x − 1 at x = 0 |T 3 for x = 0 R| log a1 + bxf when x ≠ 0 f a xf = S |T xb when x = 0 at x = 0 af

(b) f x =

2

(c)

(d)

(e)

2

41. Prove that the function f(x) = |x| is continuous at x = 0.

Rx a f |S x T|0

42. Prove that f x =

for x ≠ 0 for x = 0

is discontinuous at x = 0.

R|a f for x ≠ 0 S| is continuous at x = 0. for x = 0 T R| a + x − a − x for x ≠ 0 x | f axf = S is continuous at x = 0. 1 || for x = 0 K T af

1 + 4x 43. Find K if the function f x = e2 K

44. Find K if the function

R| 1 − 1 81 x | 1 1 f a xf = S − 27 || x T K 4

45. Find a if

for x ≠ 3

3

for x = 3

1 x

is continuous at x = 3.


ANSWERS 1. 40 7. − 13. 80 19.

5 2

25. 80 31.

1 6

a 2

2. 1

3. 3

8. 1

9. −

14.

−3 256

20. −1 26. log 32.

1 9

a b

1 6

1 6. 2 3

1 2

5. 0

10. 54 5

11. 8

12.

2

17.

3 2

18.

75 4

23.

9 4

24. 6 3

4.

15.

1 2

16. 108

21.

−3 2

22.

1 a

27. 2

28. log 2

33. −2

34.

1 2

29. log 4 35.

1 4

30.

3 4

36.

4 11

37. 8 39. (a) e6 (b) e12 40. (a) Continuous (b) discontinuous (c) Continuous (d) discontinuous (e) Continuous. 43. e2

44. a

45.

4 9

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16 Differential Calculus 16.1 INTRODUCTION: Differential calculus was discovered by Sir Isaac Newton of England and Wilhelm Leibnitz of Germany. It deals with the study of rate of change of one quantity with another.

16.2 DERIVATIVE OF A FUNCTION: Let y = f(x). Let δx be an increment given to x. δy be the corresponding change in y. y + δy = f(x + δx) δy = f(x + δx) − y δy = f(x + δx) − f(x) Dividing by δx and taking lim . δx→0

lim

δx →0

If lim

δx →0

and lim

δx →0

a

f af

a

f af

a

f af

f x + δx − f x δy = lim δx δx→ 0 δx

f x + δx − f x exists and finite then the function y = f (x) is said to be differentiable at x δx f x + δx − f x is called derivative or differential co-efficient of y with respect to x. It is δx

denoted by

dy or y′ or y1 or f′(x). dx ∴

lim

δx →0

a

f af

f x + δx − f x δy dy = = lim δx dx δx→ 0 δx

Differential Calculus

367

16.3 DERIVATIVE OF SOME STANDARD FUNCTIONS FROM FIRST PRINCIPLES: 1. xn. Let y = xn. Let δx be an increment given to x. δy be the corresponding increment in y y + δy = (x + δx)n δy = (x + δx)n − y δy = (x + δx)n − xn. Divide by δx and take lim

δx→0

a

f

n

f

n

x + δx − x n δy = lim lim δx →0 δx δx → 0 δx Add and subtract x in the denominator of RHS.

lim

δx →0

a

x + δx − x n δy = lim δx δx→ 0 x + δx − x

As lim δx → 0, lim x + δx → x.

lim

Also

x →a

∴ lim

δx →0


a

f

n

x + δx − x n δy = lim = nx n −1 δx δx→ 0 x + δx − x

⇒

dy = nx n −1 dx

i.e.,

d n x = nx n −1 dx

d i

Hence derivative or differential co-efficient of xn is nxn−1. 2. ex. Let y = ex. Let δx be an increment given to x. δy be the corresponding increment in y.

y + δy = e x +δx δy = e x + δx − y

[formula].

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Basic Mathematics

δy = e x + δx − e x ∵ a m+n = a m ⋅ a n

δy = e x ⋅ e δx − e x

e

j

δy = e x e δx − 1

Dividing by δx and taking lim . δx→0

e

j

e

j

e x e δx − 1 δy = lim lim δx →0 δx δx → 0 δx But we have lim

x →0

ex − 1 = 1 (formula) x ∴

e x e δx − 1 δy = lim lim δx →0 δx δx → 0 δx dy = e x ⋅1 dx

⇒

dy = ex dx d x e = ex. dx

d i

i.e.,

Hence derivative or differential co-efficient of ex is ex. 3. ax. Let y = ax. Let δx be an increment given to x. δy be the corresponding increment in y.

y + δy = a x +δx δy = a x +δx − y δy = a x + δx − a x δy = a x ⋅ a δx − a x

e

j

δy = a x a δx − 1

Dividing by δx and taking lim

δx→0

∵ a m+n = a m ⋅ a n


e

369

j

a x a δx − 1 δy = lim δx →0 δx δx → 0 δx lim

∵

∴

ax − 1 = log e a x →0 x lim

(formula)

e

j

a x ⋅ a δx − 1 δy = lim lim δx →0 δx δx → 0 δx

⇒

dy = a x ⋅ log e a. dx

∴

d x a = a x ⋅ log e a. dx

d i

Hence derivative or differential co-efficient of ax is ax.logea. 4. logex. Let y = logex Let δx be an increment given to x. δy be the corresponding increment in y.

a f δy = log a x + δx f − y δy = log a x + δx f − log

y + δy = log e x + δx e

e

FG x + δx IJ H x K F x δx I δy = log G + J Hx x K F δx I δy = log G1 + J H xK δy = log e

∴

e

e

Divide by δx and take lim

δx→0

δy = lim lim δx → 0 δ x δx → 0

FG H

log e 1 + x

δx x

IJ K

e

x

but log m − log n = log

F mI H nK

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Basic Mathematics

FG H

IJ K

δy δx 1 = lim log e 1 + δx → 0 δ x δx → 0 δ x x lim

Multiplying and dividing by x in RHS.

FG H

δy δx 1 x = lim ⋅ ⋅ log 1 + δx → 0 δ x δx → 0 x δx x lim

IJ K

But n log m = log m n

FG H

δy δx 1 = lim ⋅ log e 1 + lim δx →0 δx δx → 0 x x

F δx I lim G1 + J H xK

As

x δx

δx →0

IJ K

x δx

=e

FG H

δy δx 1 = lim log e 1 + lim δx →0 δx δx → 0 x x ⇒

dy 1 = ⋅ log e e dx x

∴

dy 1 = ⋅1 dx x

IJ K

x δx

But logee = 1

dy 1 = . dx x i.e.,

a f

1 d log x = dx x

∴ Derivative or differential Co-efficient of logx is

1 . x

5. Constant function: Let y = c where c is a constant. Let δx be an increment given to x. δy be the corresponding increment in y. But Since c is a constant function any change in x will not cause change in y. In other words δy = 0. ∴ ∴

lim

δx → 0

δy 0 = =0 δx δx dy =0 dx


Hence

a

371

f

d constant = 0. dx

∴ derivative of a constant is zero.

F I H K

af

af

5 d d d − a = 0 if a is constant. =0; 5 =0 2 dx dx dx

Ex.:

16.4 RULES OF DIFFERENTIATION: 1. Derivative of Product of Constant and a Function: Let y = Ku where K is a constant and u is a function of x. Let δx be an increment given to x. δu be the increment in u and δy be the corresponding increment in y.

a

f

y + δy = K u + δu

∴

y + δy = Ku + Kδu δy = Ku + Kδu − y

δy = Ku + Kδu − Ku δy = Kδu


δx→0

lim

δx → 0

δy δu = lim K ⋅ δx δx → 0 δx

dy du =K⋅ dx dx

⇒

a f

d du Ku = K ⋅ dx dx

i.e.,

Hence derivative or differential co-efficient of constant multiple of a function is constant into derivative of the function. Examples: (a)

a

f

a f

d d 4 log x = 4 ⋅ log x dx dx = 4⋅

1 4 = . x x

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Basic Mathematics

(b)

d d x 8e x = 8 ⋅ e dx dx

d i

d i = 8⋅ex

2. Derivative of Sum of 2 Functions: Let y = u + v where u and v are functions of x. Let δx be an increment given to x. δu, δv be the increments in u and v. δy be the corresponding increment in y.

y + δy = u + δu + v + δv

∴

δy = u + δu + v + δv − y

a f

δy = u + δu + v + δu − u + v δy = u/ + δu + v/ + δv − u/ − v/

δy = δu + δv Divide by δx and take lim

δx→0

lim

δy δu + δv = lim δx δx → 0 δx

lim

δy δu δv = lim + δx δx → 0 δx δx

δx → 0

δx → 0

⇒

dy du dv = + dx dx dx

i.e.,

d du dv + u+v = dx dx dx

a f

∴ Derivative or differential co-efficient of sum of 2 functions is derivative of first function plus the derivative of the second function. Examples: (a) If y = x4 + ex, then

dy d 4 = x + ex dx dx

d

=

d i

i

d i

d 4 d x x + e dx dx


dy = 4 x 4 −1 + e x dx dy = 4x3 + e x dx (b)

d

d log x + x 2 dx

a f

i

=

d d 2 log x + x dx dx

=

1 + 2 x 2 −1 x

=

1 + 2x x

d i

3. Derivative of Difference of 2 Functions: Let y = u − v where u and v are functions of x. Let δx be an increment given to x. δu, δv be the increments in u and v. δy be the corresponding increment in y.

a f δy = u + δu − av + δv f − y δy = u + δu − av + δvf − au − v f

y + δy = u + δu − v + δv

δy = u + δu − v − δv − u + v

δy = δu − δv Divide by δx and take lim

δx→0

lim

δy δu − δv = lim δx δx → 0 δx

lim

δy δu δv = lim − δx δx → 0 δx δx

lim

δy δu δv = lim − lim δx δx → 0 δx δx → 0 δx

δx → 0

δx → 0

δx → 0

⇒

dy du dv = − dx dx dx

373

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Basic Mathematics

a f

d du dv − u−v = dx dx dx

i.e.,

∴ Derivative of difference of 2 functions is the derivative of first function minus the derivative of the second function. Examples: (a)

d

d 3 x − x2 dx

i =

d 3 d 2 x − x dx dx

d i

d i

= 3 x 3−1 − 2 x 2 −1 3x 2 − 2 x

(b) If y = logx − ex, then

d

dy d = log x − e x dx dx

a f

i

=

d d x log x − e dx dx

=

1 − ex . x

d i

4. Derivative of Product of 2 Functions: Let y = uv where u and v are functions of x. Let δx be an increment given to x, δu, δv be the increments in u and v. Let δy be the corresponding increment in y.

a fa f δy = au + δufav + δv f − y

y + δy = u + δu v + δv

δy = uv + uδv + vδu + δuδv − uv

δy = uδv + vδu + δuδv Divide by δx and take lim

δx→0

lim

δx → 0

δy uδv + vδu + δuδv = lim δ x δx → 0 δx = lim u δx → 0

δv δu δuδv +v + δx δx δx


As lim , δu and δv are small. Hence the product δu × δv is very very small. So the term δx→0

375

δuδv can δx

be neglected.

δy δv δu = lim u + v δx δx → 0 δx δx

lim

∴

δx → 0

⇒

dy dv du =u +v dx dx dx

i.e.,

d dv du uv = u + v dx dx dx

a f

So Differential Co-efficient or derivative of product of 2 function is first function into derivative of 2nd function plus second function into derivation of the first function.

d [I function ⋅ II function] dx

i.e., = (I function) ⋅

d d (II function) + (II function) ⋅ (I function). dx dx

This rule is known as product rule. Examples: (a)

d 3 x d d 3 x x e = x3 ⋅ ex + ex ⋅ dx dx dx

d

i

d i = x 3 ⋅ e x + e x ⋅ 3 x 3−1 = x 3e x + e x ⋅ 3 x 2

(b) If y = x4 log x, then

d

dy d 4 = x log x dx dx = x4 ⋅

i

a f a f d i

d d 4 x log x + log x ⋅ dx dx = x4 ⋅

1 + log x ⋅ 4 x 4 −1 x

x 3 + log x 4 x 3 x 3 + 4 x 3 log x

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Basic Mathematics

5. Derivative of Quotient of 2 Functions: Let y =

u where u and v are functions of x. Let δx be an increment given to x. v

δu, δv be the increments in u and v. δy be the corresponding increment in y.

y + δy =

u + δu v + δv

δy =

u + δu −y v + δv

δy =

u + δu u − v + δv v

δy =

v u + δu − u v + δv v + δv v

δy =

uv + vδu − uv − uδv v v + δv

δy =

vδu − uδv v v + δv

a

f a f a fa f a

f

a

f


δx→0

lim

δy

δx → 0 δ x

= lim

δx → 0

vδu − uδv v v + δv ⋅ δx

a

f

δu δv −u v δy δx = lim δx lim δx → 0 δ x δx → 0 v v + δ v

a

f

As lim , δv → 0 δx→0

∴

⇒

δu δv −u v δy δ δx x = lim lim δx → 0 δ x δx → 0 v v + δ v

a

dy = dx

v

du dv −u dx dx v v+0

a f

f

(note this step)


dy = dx

v

FI HK

d u = dx v

i.e.,

377

du dv −u dx dx v2 v

du dv −u dx dx v2

Hence Derivative or differential co-efficient of quotient of 2 functions is Denominator into differential co-efficient of Numerator, minus Numerator into differential co-efficient of Denominator, whole divided by square of the denominator.

d d a Dr.f aNr.f − aNr.f aDr.f F I dx dx H K aDr.f

d Nr. = dx Dr.

i.e.,

2

Nr.: Numerator Dr.: Denominator This rule is known as quotient rule. Examples:

F H

I K

d log x = dx e x

(a)

=

(b) If y =

ex ⋅

a f a f d i de i

d d x e log x − log x ⋅ dx dx x 2

ex ⋅

1 − log x ⋅ e x x

de i

x 2

x3 , then log x

F GH

I JK

dy d x3 = = dx dx log x

=

=

log x ⋅

d i a f

a f

d 3 d log x x − x3 ⋅ dx dx 2 log x

log x ⋅ 3 x 2 − x 3 ⋅

alog x f

2

log x ⋅ 3 x 2 − x 2

alog x f

2

1 x

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Basic Mathematics

Some particular cases of

d i

d n x = nx n −1 dx

1. When n = 0

d i

d 0 x = 0 x 0 −1 dx

af

d 1 =0 dx

∴ Also we have

a f

d du K ⋅u = K ⋅ dx dx

a f

af

∴

d d K ⋅1 = K ⋅ 1 = K ⋅0 = 0 dx dx

∴

d K =0 dx

af

2. When n = 1,

d 1 x = 1 x 1−1 = 1x 0 = 1 dx

d i

af

d x =1 dx

∴ 3. When n = 2,

d 2 x = 2 x 2 −1 = 2 x dx

d i

d i

d 2 x = 2x dx

∴ 4. When n = 3,

d 3 x = 3x 3−1 dx

d i

d i

d 3 x = 3x 2 dx Similarly, 5. When n =

d i

d i

d i

d 4 d 5 d 100 = 100 x 99 and so on. x = 4x 3, x = 5 x 4 ... x dx dx dx 1 , 2

But x0 = 1.


F I GH JK 1

379

1

1 −1 d x2 = x2 2 dx 1

1 − x = x 2 2

d dx

d i

d dx

d xi = 2 1x

1 d x= dx 2 x 6. When n = −1

d −1 x = −1 x −1−1 dx

d i d F 1I = −1 x dx H x K d F 1 I −1 = dx H x K x

−2

2

7. When n = −2

d −2 = −2 x −2 −1 = −2 x −3 x dx

d i d F 1I 2 =− dx H x K x 2

Similarly

F I H K

3

F I H K

d 1 3 d 1 4 =− 4, =− 5 3 4 dx x x dx x x

F I H K

d 1 5 = − 6 and so on. 5 dx x x

1

∵ x2 = x a−m =

1 am

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Basic Mathematics

List of Formulae:

dy dx

y (1) x n

nx n−1

(a) x

1

(b) x 2

2x

(c) x 3

3x 2

(d) x 4

4x3

(e) x 5

5x 4

(f)

1 2 x

x

1 x2

(g)

1 x

−

(h)

1 x2

−2 x3

(i)

1 x3

−3 x4

(j)

1 x4

−4 x5

(k) K (Constant)

0

x 2. e

ex

3. a x

a x log e a

4. log x

1 x

5. u ± v

du dv ± dx dx

6. (I function) (II function)

I function ⋅

Nr. 7. Dr.

Dr.

af

af

d d II + II ⋅ I dx dx

a f a f a f a f

d d Nr. − Nr. ⋅ Dr. dx dx 2 Dr


Note: Sum, Difference and product rule can be extended i.e.,

a

f

d du dv dw ± ± ± ... u ± v ± w ± ... = dx dx du du

a

f

d du dv dw ⋅ vw ...+ ⋅ uw ...+ uvw ... = uv ... + ... dx dx dx dx

WORKED EXAMPLES: 1. Find

dy if y = x3 − 3x + 7 dx

Consider y = x 3 − 3x + 7, differentiate with respect to (w.r.t) x.

dy d 3 = x − 3x + 7 dx dx =

a f

af

d 3 d d 3x + 7 x − dx dx dx

d i

af − 3 ⋅ a1f + 0

3x 2 − 3 3x 2

d x +0 dx

dy = 3x 2 − 3. dx 2. Find y′ if y = 7e x − 4 log e x. x x Given: y = 7e − 4 log e

diff. w.r.t.x. (differentiate with respect to x)

y′ =

d i

y′ = 7 ⋅ e x − 4 ⋅ y ′ = 7e x − 3. Find f ′(x) if f(x) = 7x + 8 ex − 9 Consider f(x) = 7x + 8 ex − 9 diff. w.r.t. x.

b

d d 7e x − 4 log e x dx dx

4 x

1 x

g

381

382

Basic Mathematics

a f dxd d7 i + 8 ⋅ dxd e − dxd a9f f ′a x f = 7 log 7 + 8 ⋅ e − 0 f ′a x f = 7 log 7 + 8e x

f′ x =

x

x

x

x

4. If y = x +

x

1 , then find y1 x

y= x +

1 x

diff. w.r.t. x. y1 =

d dx

d x i + dxd FGH 1x IJK

d i d i F 1 I 1 = x + G− x 2 H 2 JK =

d 12 d −1 2 + x x dx dx 1 −1 2

y1 = y1 =

1 − −1 2

1 −1 2 1 −3 2 x − x 2 2 1 2 x

−

1 2x3 2

Alieter:

y= x + y=

1 x

x +1 x

diff. w.r.t. x. using quotient rule.

y1 =

a f a f d xi d xi 1 x 1 + 0 − a x + 1f ⋅ 2 x

x⋅

d d x +1 − x +1 ⋅ dx dx 2

x


x−

−

2 x x

1 2 x

=

x x 1 − − x 2 x ⋅x 2 x ⋅x

=

1 1 1 − − x 2 x 2x x

=

2 −1 1 − 1+1 2 2 x 2x

y1 =

5. Find

x

1 2 x

−

1 . 2x3 2

dy if y = x38x. dx y = x 3 8x .

diff. w.r.t. x using product rule.

d i

d i

dy d x d 3 = x3 ⋅ 8 + 8x ⋅ x dx dx dx dy = x 3 ⋅ 8 x log 8 + 8 x ⋅ 3 x 2 dx

af

6. If f x =

x2 + 1 , then find f ′(x) x3 − 7

af

f x =

x2 + 1 x3 − 7

diff. w.r.t. x using quotient rule.

dx − 7i ⋅ dxd dx + 1i − d x + 1i ⋅ dxd dx − 7i f ′a x f = d x − 7i d x − 7i LMN dxd d x i + dxd a1fOPQ − d x + 1i LMN dxd d x i − dxd a7fOPQ = d x − 7i 3

2

2

3

3

3

2

2

2

3

2

3

383

384

Basic Mathematics

d x − 7i a2 xf − dx + 1id3x i d x − 7i 3

2

2

3

af

2 x 4 − 14 x − 3 x 4 − 3 x 2

af

− x 4 − 3 x 2 − 14 x

f′ x =

f′ x =

7. If y =

2

d x − 7i 3

d x − 7i 3

2

2

x 2 dy . + − 2 x 2 , then find 2 x dx y=

x 2 + − 2x 2 2 x

diff. w.r.t.x.

F I H K

F I d i H K 1 d = a xf + 2 dxd FH 1x IK − 2 dxd dx i 2 dx 1 F 1 I − 2 ⋅ a2 x f = ⋅1 + 2 − H xK 2

dy d x d 2 d = + − 2x2 dx dx 2 dx x dx

2

2

dy 1 2 = − − 4 x. dx 2 x 2 8. If y =

a

dy xe x , then find . dx log x + 7

f

y=

xe x . log x + 7


alog x + 7f ⋅ dxd dxe i − d xe i ⋅ dxd alog x + 7f alog x + 7f alog x + 7f LMNx ⋅ dxd e + e ⋅ dxd a xfOPQ − xe LMN dxd alog xf + dxd a7fOPQ = alog x + 7f

dy = dx

x

x

2

x

x

x

2


385

alog x + 7f x ⋅ e + e ⋅1 − xe LMN 1x + 0OPQ = alog x + 7f dy alog x + 7f d xe + e i − e = dx alog x + 7f x

x

x

2

x

x

x

2

9. If y =

dy xn − nx , then find . x dx e y=

xn − nx ex


dy = dx

ex ⋅

=

=

i d de i

i d i

x 2

ex =

d

d n d x x − nx − xn − nx e dx dx

LM d dx i − d dn iOP − d x dx Q N dx de i n

x

n

i

− nx ⋅ex

x 2

d

i

e x nx n −1 − n x log n − x n − n x e x

de i

x 2

e x nx n −1 − n x log n − x n + n x

de i

x 2

dy nx n −1 − n x log n − x n + n x = . dx ex 10. If f(x) = (x2 + 1) (x3 + 7x + 8) ex, then find f ′(0).

af d

id

i

f x = x 2 + 1 x 3 + 7x + 8 ⋅ e x diff. w.r.t. x using extended product rule.

af d

id

i dxd de i + dx + 1i e

f ′ x = x2 + 1 x3 + 7x + 8 ⋅

x

2

x

⋅

d 3 d 2 x + 7x + 8 + x 3 + 7x + 8 e x ⋅ x +1 dx dx

d

i d

i

d

i

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Basic Mathematics

a f d id i d i d i d i f ′a0f = a0 + 1fa0 + 8f e + a0 + 1f e a0 + 7f + a0 + 0 + 8f e a2 ⋅ 0f

a f

f ′ x = x 2 + 1 x 3 + 7 x + 8 ⋅ e x + x 2 + 1 e x ⋅ 3x 2 + 7 + x 3 + 7x + 8 ⋅ e x ⋅ 2 x 0

∴

0

0

=8+7+0

af

f ′ 0 = 15

16.5 DIFFERENTIATION OF COMPOSITE FUNCTIONS: Chain rule: If y = g(u) and u = f (x) are 2 differentiable functions, then the composite function y = g[f (x)] can be differentiated by using chain rule, which can be stated as

dy dy du = ⋅ dx du dx

af af

dy = g′ u ⋅ f ′ x dx

i.e. i.e.

b a f g = g ′ f a x f ⋅ f ′a x f

d g f x dx

Examples: 1. If y = log (x2 − 4x + 8), then

dy 1 d 2 = 2 ⋅ x − 4x + 8 dx x − 4 x + 8 dx

d

a

=

1 ⋅ 2x − 4 x − 4x + 8

=

2x − 4 x2 − 4x + 8

2

i

f

Alieter: Consider y = log (x2 − 4x + 8) Let y = logu where u = x2 − 4x + 8 diff. w.r.t. x.

dy 1 du = ⋅ dx u dx

u = x 2 − 4 x + 8 diff. w.r.t. x

a

f

a

f

dy 1 = ⋅ 2x − 4 dx u dy 1 = ⋅ 2x − 4 dx u

∴

du = 2x − 4 dx


a

dy 1 = ⋅ 2x − 4 dx x 2 − 4 x + 8

f

dy 2x − 4 = 2 dx x − 4 x + 8

af

2

2. If f x = e x , then

af

d i

d 2 x dx

2

f ′ x = ex ⋅

af

2

f ′ x = e x ⋅ 2 x.

Note: Chain rule can be extended. i.e., If y = f[g(h(x))], then

b a fg ⋅ g′ bha xfg ⋅ h′axf

dy = f′ g h x dx

Example. If y = e d

x2 +4

i

8

, then

dy d x 2 + 4i ⋅ d x 2 + 4 =e dx dx

d

8

i

8

dy d x 2 + 4i ⋅ 8 x 2 + 4 7 ⋅ d x 2 + 4 =e dx dx

d

8

=e

d x + 4i 2

i

d

8

i

d

7

⋅ 8 x 2 + 4 ⋅ 2 x.


dy if y = log (xn − ex) dx

d

y = log x n − e x

i

diff. w.r.t. x.

d

dy d n 1 x − ex = n ⋅ x dx x − e dx =

1 ⋅ nx n −1 − e x x x −e

=

nx n −1 − e x . xn − ex

n

i

i

387

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Basic Mathematics

2. Find

dy if y = x 2 + 4 x − 9 . dx

2 Consider y = x + 4 x − 9

diff. w.r.t x.

d

1 dy d 2 = ⋅ x + 4x − 9 2 dx 2 x + 4 x − 9 dx

i

dy 1 2x + 4 − 0 = 2 dx 2 x + 4 x − 9

a

1 dy = ⋅2 x + 2 2 dx 2 x + 4 x − 9 dy = dx

3. If y = 7

d x + 9 x −6 i , then find 3

x+2 x + 4x − 9 2

f

.

dy . dx

d x + 9 x −6i 3

y=7

diff. w.r.t. x.

dy d x 3 + 9 x −6 i ⋅ log 7 ⋅ d x 3 + 9 x − 6 =7 dx dx

d

=7

i

d x + 9 x −6 i ⋅ log 7 3x 2 + 9 a1f − 0 3

d

i

3 dy = 7 x +9 x − 6 ⋅ log 7 ⋅ 3 ⋅ x 2 + 3 . dx

af

4. If f x = x 3 e 5 x

af

2

+8

, then find f ′(x)

Consider f x = x 3 e 5 x

2

+8


af

f ′ x = x3 ⋅

e 2

+8

2

+8

= x 3 ⋅ e5 x

af

d i

j

2 d 5 x 2 +8 d 3 e x + e5 x +8 ⋅ dx dx

f ′ x = x 3 ⋅ e5 x

⋅

d

i

d i

2 d 5x 2 + 8 + e 5 x +8 ⋅ 3x 2 dx

a10 x + 0f + e

5x 2 +8

⋅ 3x 2


5. If y =

dx

x +1 2

+ 8x − 9

i

dy . dx

, then find

3

Consider

y=

dx

x +1 + 8x − 9

2

i

3


dy = dx

=

dx

2

dx

2

i

3

+ 8x − 9 ⋅

a f a f d LMd x + 8x − 9i OP N Q

d d 2 x +1 − x +1 ⋅ x + 8x − 9 dx dx

3

3 2

2

i a f a f LMN d i dxd d x + 8x − 9iOPQ d x + 8x − 9i x + 8 x − 9i − a x + 1f L3 d x + 8x − 9i a2 x + 8fO MN dy d QP . = dx d x + 8x − 9i 3

2

+ 8x − 9 ⋅ 1 + 0 − x + 1 ⋅ 3 x 2 + 8x − 9 ⋅

2

6

2

3

2

2

2

6

2

6. If y =

i

xe 3 x + 7 dy , then find . log 7 x − 6 dx

a

f

y=

xe 3 x + 7 log 7 x − 6

a

f


dy = dx

=

a

3x+7

a

f LNM

log 7 x − 6 x ⋅ e 3 x + 7 ⋅

a

f

d d xe 3 x + 7 − xe 3 x + 7 ⋅ log 7 x − 6 dx dx 2 log 7 x − 6

a

f LMN dxd de i + e

a

log 7 x − 6 x

=

f

log 7 x − 6 ⋅

f

a fOPQ a f

LM N

a

d d 1 x − xe 3 x + 7 ⋅ ⋅ 7x − 6 dx 7 x − 6 dx 2 log 7 x − 6

3x + 7

a

⋅

f

OP Q

LM N

a fOQP

d 1 3x + 7 + e 3 x + 7 ⋅ 1 − xe 3 x + 7 ⋅7 dx 7x − 6 2 log 7 x − 6

a

f

fOPQ

389

390

Basic Mathematics

dy = dx

a

f

log 7 x − 6 xe 3 x + 7 ⋅ 3 + e 3 x + 7 − xe 3 x + 7 ⋅

a

log 7 x − 6

f

7 7x − 6

2

16.6 DIFFERENTIATION OF IMPLICIT FUNCTIONS: Consider the function, y = x2 +ex − logx. Here y is expressed as a function of x. i.e., y = f (x). Such functions are called explicit functions. Now consider the function xy + log y + ex = 0. Here y is not expressed explicitly as a function of x. Such functions are called implicit functions. They are of the form f (x, y) = 0.

dy dy , differentiate the given function shift all the terms containing to Left Hand Side and dx dx dy dy common and shift the co-efficient of to Right the remaining terms to Right Hand Side. Take dx dx Hand Side. To find


dy if x3 + y3 = a3. dx

Consider x3 + y3 = a3 diff. w.r.t. x.

LMQ a is constant d da i = 0OP dx MM d d a yf P y i = 3y ⋅ d MM dxausing chain ruledxf PPP MN PQ 3

3x 2 + 3y 2 ⋅

dy =0 dx

3 x 2 + 3y 2

dy =0 dx

3y 2

dy = −3x 2 dx

dy −3x 2 = dx 3y 2 dy − x 2 = 2 . dx y

3

2


dy . dx

2. If y + x2 + ey = 0, then find Solution: Consider y + x2 + ey = 0

diff. w.r.t. x.

dy dy + 2x + ey ⋅ =0 dx dx dy dy + ey ⋅ = −2 x . dx dx dy 1 + e y = −2 x . dx dy −2 x . = dx 1 + e y 3. If ex + ey = logx, then find

dy when x = 1 and y = 0 dx

Solution: Consider ex + ey = logx. diff. w.r.t. x.

ex + ey ey ⋅

dy 1 = dx x

dy 1 = − ex dx x

1 − ex dy x = dx ey 1 1 −e dy when x = 1 and y = 0 = 1 0 dx e dy 1− e = =1− e . 1 dx a1, 0 f 4. If y = x + x + x + x + ... ∞ Then find

dy . dx

391

392

Basic Mathematics

Solution: y = x + x + x + x + ... ∞

y= x+y Squaring,

y2 = x + y diff. w.r.t. x.

2y ⋅

dy dy =1+ dx dx

2y ⋅

dy dy − =1 dx dx

a

f

dy 1 = dx 2 y − 1

dy 2y −1 = 1 ⇒ dx dy . dx

5. If y = ax ax ax ... ∞ , then find

y = ax ax ax ... ∞

y = axy Squaring

y 2 = axy. diff. w.r.t. x (using product rule in RHS, Chain rule in LHS)

2y

LM N

a fOPQ

d dy dy x =a x⋅ + y⋅ dx dx dx 2y

LM N

dy dy =a x +y dx dx

OP Q

2y

dy dy = ax + ay dx dx

2y

dy dy − ax = ay dx dx

a

f

dy 2 y − ax = ay dx dy ay = . dx 2 y − ax


393

16.7 DIFFERENTIATION OF PARAMETRIC FUNCTIONS: If both x and y are expressed as a function of another variable say t then the function y = f (x) is said to be in parametric form. The variable t is called a parameter. To find

dy we use. dx dy dy dt = dx dx dt


dy if x = at2 and y = 2 at. dx

Consider y = 2 at diff. w.r.t. t.

af

dy = 2a 1 dt dy = 2a dt Now Consider x = at2

diff. w.r.t. t.

a f

dx = a 2t dt dx = 2 at dt Now

dy dy dt 2a = = dx dx 2 at dt dy 1 = . dx t 2. If x = 7t + et and y = et − 7t, then find Consider y = et − 7t.

dy when t = 0. dx

394

Basic Mathematics

diff. w.r.t. t.

dy = e t − 7t log 7. dt Now consider

x = 7t + e t diff. w.r.t. t.

dx = 7t log 7 + e t dt dy dy dt = dx dx dt =

e t − 7 t log 7 7t log 7 + e t

dy e 0 − 7 0 log 7 = 0 dx when t =0 7 log 7 + e 0 1 − log 7 dy = dx when t = 0 log 7 + 1 3

3. Differentiate e x with respect to log x. 3

Solution. Let u = e x and v = log x.

du dv

To find:

du du dx = dv dv dx x Consider u = e

3

diff. w.r.t. x.

d i

3 du d 3 x = ex ⋅ dx dx 3 du = e x ⋅ 3x 2 dx


v = log x diff. w.r.t. x

dv 1 = dx x 3

du e x ⋅ 3 x 2 = 1 dv x

∴

3 du = e x ⋅ 3x 3 . dv

4. Differentiate (x2 + 8x − 1)4 with respect to e x Solution. Let u = (x2 + 8x − 1)4 and v = e x To find:

2

2

−9

−9

du dv du du dx = dv dv dx

Consider u = (x2 + 8x − 1)4 diff. w.r.t. x.

d

i

d

d

i a

i

3 d du x 2 + 8x − 1 = 4 x 2 + 8x − 1 ⋅ dx dx

f

3 du = 4 x 2 + 8x − 1 ⋅ 2 x + 8 dx

Now,

v = ex

2

−9

diff. w.r.t. x.

d

i

2 dv d 2 x −9 = e x −9 ⋅ dx dx 2 dv = e x −9 ⋅ 2 x dx

Hence

d

ia

f

du 3 2 du dx 4 x + 8 x − 1 2 x + 8 = = . 2 dv dv e x −9 ⋅ 2 x dx

395

396

Basic Mathematics

5. Differentiate

4 + log x with respect to e

Let u = 4 + log x and v = e

d

log x 2 + 9

i

i

⇒ v = x2 + 9 e To find:

d

log x 2 + 9

b a fg = f a x f aformula f

log f h

du dv du du dx = dv dv dx

Consider u = 4 + log x diff. w.r.t. x.

a

du d 1 4 + log x = ⋅ dx 2 4 + log x dx du 1 1 = ⋅ dx 2 4 + log x x du 1 = dx 2 x 4 + log x Now

v = x2 + 9. diff. w.r.t. x. dv = 2x + 0 dx dv = 2x dx

∴

1 du du dx 2 x 4 + log x = = dv dv 2x dx

∴

du 1 . = 2 dv 4 x 4 + log x

f


397

16.8 LOGARITHMIC DIFFERENTIATION: If it is required to differentiate (function)function or (constant)function then we consider log on both sides, apply logmn = nlogm and then differentiate.


dy if y = xx dx y = xx

Consider log on both sides

log y = log x x log y = x log x

diff. w.r.t. x.

a f

af

d d 1 dy x ⋅ = x⋅ log x + log x ⋅ dx dx y dx 1 dy 1 ⋅ = x ⋅ + log x ⋅ 1 y dx x 1 dy ⋅ = 1 + log x y dx

a

dy = y 1 + log x dx

d i

f

a

d x x = x x 1 + log x dx

∴ 2. Find

f

dy if x y = y x . dx xy = yx


log x y = log y x y log x = x log y diff. w.r.t. x.

y⋅

a f

a f

af

d dy d d x log x + log x ⋅ = x⋅ log y + log y ⋅ dx dx dx dx

398

Basic Mathematics

y⋅

dy 1 1 dy + log x ⋅ = x⋅ ⋅ + log y ⋅ 1 x dx y dx dy x dy y + log x ⋅ = ⋅ + log y dx y dx x log x ⋅

FG H

y dy x dy − = log y − x dx y dx

IJ K

dy x y log x − = log y − dx y x y dy x. = dx log x − x y log y −

3. If y = y=

1+ x dy , then find . 1− x dx 1+ x 1− x

Consider log on both sides. log y = log

1+ x 1− x

1+ xI log y = logF H1− xK 1+ xI 1 log y = logF 2 H1− xK 1 log y = loga1 + x f − loga1 − x f 2 1 2

diff. w.r.t. x.

LM a fOPQ N 1 dy 1 L 1 1 ⋅ = M − a−1fOPQ y dx 2 N 1 + x 1 − x

d 1 dy 1 1 1 1− x ⋅ = − ⋅ y dx 2 1 + x 1 − x dx


LM OP N Q dy y L 1 − x + 1 + x O = M P dx 2 N a1 + x fa1 − x f Q dy y 1 1 = + dx 2 1 + x 1 − x

1+ x 2 1− x ⋅ 2 1+ x 1− x

OP LM N a fa f Q

dy = dx

dy 1+ x 1 1 ⋅ = = dx 1− x 1+ x 1− x 1+ x ⋅ 1− x

a fa f

a f

32

Alieter: y=

1+ x 1− x

diff. w.r.t. x.

dy = dx

dy = dx

1 ⋅ 1+ x 2 1− x

F H

d 1+ x 1 ⋅ 1 + x dx 1 − x 2 1− x

I K

a1 − xf ⋅ dxd a1 + xf − a1 + xf ⋅ dxd a1 − xf a1 − xf 2

a fa f a fa f a f

1 − x 1 − x 1 − 1 + x −1 ⋅ 2 1+ x 1− x 2 1− x 1− x +1+ x ⋅ 2 1+ x 1− x 2

a f

1− x 2 ⋅ 2 1+ x 1− x

a f

dy = dx

1

a f

1+ x ⋅ 1− x

1 2− 2

=

2

1

a1 + xf ⋅ a1 − xf

3 2

399

400

Basic Mathematics

4. If yey = xx, then find

dy . dx

Consider yey = xx Consider log on both sides

d i

log ye y = log x x

Q log mn = log m + log n

log y + log e y = log x x log y + y log e = x log x log y + y = x log x

Q log e = 1

diff. w.r.t. x.

a f

d d 1 dy dy ⋅ + = x⋅ log x + log x x dx dx y dx dx 1 dy dy 1 ⋅ + = x ⋅ + log x ⋅ 1 y dx dx x

LM N

OP Q

dy 1 + 1 = 1 + log x dx y dy 1 + log x = 1 dx +1 y

a

f

dy y 1 + log x . = dx 1+ y 5. If ey = ax+y, then find

dy . dx ey = a x+y


log e y = log a x + y

a f y ⋅ 1 = a x + y f log a

y log e = x + y log a

diff. w.r.t.x.

LM N

dy dy = log a 1 + dx dx

OP Q

Q log a is a constant.


401

dy dy = log a + log a ⋅ dx dx dy dy − log a ⋅ = log a. dx dx

a

f

dy 1 − log a = log a. dx dy log a . = dx 1 − log a

16.9 SUCCESSIVE DIFFERENTIATION: If y = f (x), then

af

dy = y1 = y ′ = f ′ x is first order derivative of y with respect to x. It is a function of x. The derivadx tive of this function with respect to x is

i.e.,

F I H K

af

d2 y = y2 = y ′′ = f ′′ x and is called second order derivative. dx 2

af

d dy d2y = 2 = f ′′ x = y ′′ = y2 . dx dx dx


d2y if y = x2 + 3x + 8. dx 2

Solution: y = x 2 + 3 x + 8 diff. w.r.t. x.

af

dy = 2x + 3 1 + 0 dx dy = 2x + 3 dx diff. again w.r.t. x.

af

d2y =2 1 +0 dx 2 d2y = 2. dx 2

402

Basic Mathematics

2 x 2. If y = x e , then find

d2y . dx 2 y = x 2e x .


d i

d i

dy d x d 2 e + ex ⋅ x = x2 ⋅ dx dx dx dy = x 2 ⋅ e x + e x ⋅ 2x dx

d

dy = e x x 2 + 2x dx

i

diff. again w.r.t. x using product rule,

d

i d i

d 2 d x d2 y x + 2x + x2 + 2 x e = ex ⋅ 2 dx dx dx

d

i

= ex 2x + 2 + x2 + 2x ⋅ ex d2 y = ex 2x + 2 + x2 + 2 x 2 dx d2 y = e x x2 + 4x + 2 . 2 dx 3. If x = t2 and y = 4t, then find

d2y at t = 1. dx 2

Consider x = t2 diff. w.r.t. t.

dx = 2t dt Consider y = 4t. diff. w.r.t. t.

af

dy = 4⋅ 1 dt dy = 4. dt

dy dy dt 4 = = dx dx 2t dt


403

dy 2 = dx t diff. again w.r.t. x.

F I af H K

1 d2 y d =2 − 2 ⋅ t 2 dx dx t

2 1 d2y =− 2 ⋅ 2 dx t 2t 1 d2 y =− 3 2 dx t 1 d2y =− 3. dx 2 when t =1 1 d2 y = −1. dx 2 when t =1

4. If y = e2t and x = log 3t, then Find

d2y . dx 2 y = e 2t diff. w.r.t. t

a f

dy d 2t = e2t ⋅ dt dt

dy = e 2 t ⋅ 2 = 2e 2t dt x = log 3t diff. w.r.t. t.

a f

dx 1 d 3t = ⋅ dt 3t dt dx 1 = ⋅3 dt 3t dx 1 = dt t

∴

dy dy dt 2e 2 t = = 1 dx dx dt t

[Note this step]

Q

dx = 2t dt

dt 1 = . dx 2t

404

Basic Mathematics

dy = 2te 2 t dx diff. w.r.t. t.

LM d i OP a f N Q d y d dt = 2 LMt ⋅ e ⋅ a2t f + e ⋅ OP dx Q dx N dx d y dt dt = 2 LMte ⋅ 2 + e ⋅ OP dx N dx dx Q d2 y d 2t d = 2 t⋅ e + e 2t ⋅ t 2 dx dx dx 2

2t

2t

2

2

2t

2t

2

dt d2 y 2t + 1 = 2 e 2t ⋅ 2 dx dx

Q

d2 y = 2e 2 t ⋅ t 2t + 1 2 dx 5. If y = ae mx + be − mx , then prove that y2 − m 2 y = 0

y = ae mx + be − mx diff. w.r.t. x.

y1 = ae mx ⋅

a f

a f

d d mx + be − mx ⋅ mx dx dx

af

a f

y1 = ae mx m + be − mx − m y1 = m ae mx − be − mx

diff. again w.r.t. x.

LM N

y2 = m ae mx ⋅

a f

af a f amf + be m

y2 = m ae mx ⋅ m − be − mx − m y2 = m ae mx

− mx

y2 = m 2 ae mx + be − mx y2 = m 2 y

a fOPQ

d d mx − be − mx ⋅ − mx dx dx

dx 1 = dt t dt =t dx


y2 − m 2 y = 0 Hence proved.

MISCELLANEOUS PROBLEMS: I. One Mark Problems: 1. Differentiate xe + ex − ee with respect to x. Solution: Let y = xe + ex − ee diff. w.r.t. x.

dy = ex e−1 + e x − 0 dx dy = ex e −1 + e x dx 2. Differentiate 2x + x2 − logx with respect to x. Let y = 2x + x2 − logx diff. w.r.t. x.

dy 1 = 2 x log 2 + 2 x − dx x 3. Differentiate e Let y = e

d

d

2 log x + x 2

2 log x + x 2

i

w.r.t. x.

i diff. w.r.t. x.

dy 2 d log x + x 2 i d 2 log x + x 2 =e ⋅ dx dx

ed

=e =e 4. If y = 5 x

2

+4 x−7

d

i ⋅ 2 F 1 + 2 xI

d

i F 2 + 4 xI .

2 log x + x 2

Hx

2 log x + x 2

Hx

, then find y1 y = 5x

2

+4 x−7

diff. w.r.t. x.

K

K

ij

405

406

Basic Mathematics

5. Differentiate e Let y = e

d

d

log x 2 + 4 x

log x 2 + 4 x

i

y1 = 5 x

2

+ 4x −7

y1 = 5 x

2

+4 x−7

log 5

d

i

d 2 x + 4x − 7 dx

a

⋅ log 5 ⋅ 2 x + 4

f

w.r.t. x.

i

af

Q e log f a x f = f x

y = x2 + 4x diff. w.r.t. x.

dy = 2 x + 4. dx 6. If y =

1 3

x

2

, then find

dy . dx y=

1 3

x2

=

1

d i x2

y=x

−

1 3

=x

−

2 3.

2 3.

diff. w.r.t. x. 2

dy 2 − −1 −2 x =− ⋅x 3 = dx 3 3 =− 7. If y = 3−8x, then find

−5 3

2 . 3x 5 3

dy dx y = 3−8 x diff. w.r.t. x.

a f

dy d = 3−8 x ⋅ log 3 ⋅ −8 x dx dx

a f

dy = 3−8 x ⋅ log 3 ⋅ −8 dx


8. If y = x 2 − 4e x , then find y′

y = x 2 − 4e x diff. w.r.t. x.

y′ =

y′ =

1 2 x 2 − 4e x 1 2 x − 4e

y′ =

af

9. If f x =

2

x

⋅

d

d 2 x − 4e x dx

i

⋅ 2 x − 4e x

x − 2e x x 2 − 4e x

4 , then find f ′(x) x −9 2

af d

f x = 4 x2 − 9

i

−1

diff. w.r.t. x.

a f LNM d i OQP ⋅ dxd d x f ′a x f = 4 L −d x − 9i O 2 x PQ MN −8 x f ′a x f = d x − 9i f ′ x = 4 −1 x 2 − 9

−2

2

2

2

10. If y =

1 3

3x

−1−1

, then find y1

y=

1

d3 i

x 13

=

1 3

x3

= 3− x 3

y = 3− x 3 diff. w.r.t. x.

F I H K 1 ⋅ log 3 ⋅ F − I H 3K

y1 = 3− x 3 ⋅ log 3 ⋅ y1 = 3 − x 3

d x − 3 dx

2

−9

i

407

408

Basic Mathematics

y1 = −

log 3 log 3 = − 1− x 3 . −x 3 3⋅3 3

II. Two Marks Problems: 1. If x2 + y2 = 10, then find

dy at (1, − 1) dx

Consider x2 + y2 = 10 diff. w.r.t. x.

2x + 2y ⋅ 2y

dy =0 dx dy = −2 x. dx dy −2 x = dx 2y dy x =− dx y

dy −1 = dx a1, −1f −1

a f

dy = 1. dx a1, −1f 2. If x = 4t and y = 5t2, then find

dy . dx

dy dy dt = Solution: dx dx dt Now

y = 5t2 diff. w.r.t. t.

a f

dy = 5 2t dt dy = 10t dt


Next, x = 4 t diff. w.r.t. t.

dx = 4 ⋅ 1 = 4. dt dy dy dt 10t 5t = = = . dx dx 4 2 dt

af

3. If f x = log x 2 a, then find f ′(a)

af

log e a log e x 2

f x = log x 2 a =

=

af

f x =

log e a 2 log e x log e a 1 ⋅ 2 log e x

diff. w.r.t. x.

af

f′ x =

af

f′ x = −

log e a 1 ⋅ 2 log e x

b

af

log e a 4 log e a ⋅ a

af

1 4a

f′ a =− f′ a =− 4. If y = log

F GH b

log e a 1 ⋅ − 2 log e x

dy 1− x , then find dx 1+ x 1− xI y = log F H1+ xK 1 1− xI y = log F H1+ xK 2 1 2

g

I d log x g JK dx a f

2

2

⋅

1 x

409

410

Basic Mathematics

y=

a f

a f

1 log 1 − x − log 1 + x 2 diff. w.r.t. x.

LM a f N dy 1 L 1 1 = M −1f − ⋅ 1O a dx 2 N 1 − x 1 + x PQ dy 1 L −1 1 O − = dx 2 MN1 − x 1 + x PQ 1 L −1 − x − 1 + x O = M P 2 N a1 − x fa1 + x f Q dy 1 L −2 O = dx 2 MN1 − x PQ

a fOPQ

dy 1 1 d d 1 = ⋅ ⋅ 1− x − 1+ x dx 2 1 − x dx 1 + x dx

2

dy −1 . = dx 1 − x 2

e

j

2 5. If y = log x + 1 + x , then Prove that

e

2 Pr oof: Consider y = log x + 1 + x Proof:

1 dy = . dx 1 + x2

j diff. w.r.t. x.

e

1 dy d = ⋅ x + 1+ x2 2 dx x + 1 + x dx

j

LM OP d i MN QP LM1 + 1 ⋅ a2 xfOP MN 2 1 + x PQ LM 1 + x + x OP MN 1 + x PQ

1 1 dy d = ⋅ 1+ 1 + x2 2 dx dx x + 1 + x 2 2 1+ x

dy 1 = dx x + 1 + x 2 dy 1 = dx x + 1 + x 2

2

2

2


411

1 dy = . dx 1 + x2

Hence proved. 6. If xmyn = am+n, then prove that

dy my =− . dx nx Consider xmyn = am+n. Taking log on both sides log (xmyn) = log am+n

log x m + log y n = log a m + n

a

f

m log x + n log y = m + n log a.

diff. w.r.t. x.

m⋅

1 1 dy + n⋅ ⋅ =0 x y dx

m n dy + ⋅ =0 x y dx n dy m =− y dx x dy my =− . dx nx Hence proved.

dy y log a 7. If y = ax+y, then prove that dx = 1 − y log a Consider y = ax+y Consider log on both sides

log y = log a x + y

a

f

log y = x + y log a diff. w.r.t. x.

LM N

dy 1 dy = log a 1 + y dx dx

OP Q

dy 1 dy = log a + log a ⋅ dx y dx

a

f

Q m + n log a is constant

412

Basic Mathematics

dy 1 dy − log a ⋅ = log a y dx dx

LM N

OP Q

dy 1 − log a = log a dx y dy log a = 1 dx − log a y dy log a = y log a 1 − dx y dy y log a = dx 1 − y log a Hence proved. 8. If yey = x, then prove that

dy y = dx x 1 + y

a f

Consider

yey = x. taking log on both sides

d

i

log y ⋅ e y = log x log y + log e y = log x log y + y ⋅ log e = log x

log y + y = log x diff. w.r.t. x.

1 dy dy 1 + = y dx dx x

FG IJ H K dy F 1 + y I 1 G J= dx H y K x

dy 1 1 +1 = dx y x

Q log e = 1


413

dy y = dx x 1 + y

a f

⇒ Hence proved. Alieter: Consider yey = x. diff. w.r.t. x. using product rule.

y⋅

d i

d y dy e + ey ⋅ =1 dx dx

y ⋅ ey ⋅

dy dy + ey ⋅ =1 dx dx

d

i

dy y e ⋅ y + ey = 1 dx dy 1 = dx e y y + 1

a f

But ye y = x , e y =

dy 1 = dx x y + 1 y

a f

dy y = dx x y + 1

a f

Hence proved. III. 4 Marks Problems: 1. If x 1 + y + y 1 + x = 0 and x ≠ y , prove that

dy −1 = dx 1 + x

a f

2

Consider

x 1+ y + y 1+ x = 0 x 1 + y = −y 1 + x Squaring

dx

1+ y

i = d− y 2

1+ x

i

2

x y

414

Basic Mathematics

a f

a f

x 2 1 + y = y2 1 + x

x 2 + x 2 y = y2 + y 2 x x 2 + x 2 y − y2 − y 2 x = 0

a

f

x 2 − y 2 + xy x − y = 0

a x − yfa x + yf + xy ax − yf = 0 a x − yf x + y + xy = 0 x − y = 0 or x + y + xy = 0

⇒

Q x ≠ y, x + y + xy = 0 y + xy = − x

a f

y 1 + x = −x

y=

−x 1+ x

diff. w.r.t. x.

a1 + xf ⋅ FH dxd a− xf − a− xf ⋅ dxd a1 + xfIK a1 + xf dy a1 + x fa −1f + x a1f −1 − x + x −1 = = = dx a1 + xf a1 + xf a1 + xf dy = dx

2

2

2

2

Hence proved. 2. If x = at2, y = 2at, then prove that

1 d2y =− 2 2 at 3 dx Consider x = at2 diff. w.r.t. t.

a f

dx = a ⋅ 2t dt y = 2at

diff. w.r.t. t.

af

dy = 2a 1 dx

...(1)


dy = 2a dt dy dy dt 2a 1 = = = dx dx 2 at t dt

∴

dy 1 = dx t diff. w.r.t. x.

1 dt d2 y =− 2 ⋅ 2 dx t dx

F I H K

1 1 d2y =− 2 ⋅ 2 at dx 2 t

1 d2y =− 2 2 at 3 dx Hence proved. 3. Differentiate

x from 1st principles:

Let y = x Let δx be an increment given to x. δy be the corresponding increment in y

y + δy = x + δx δy = x + δx − y δy = x + δx − x Divide by δx and take lim

δx→0

lim

δx →0

x + δx − x δy = lim δ → x 0 δx δx

Add and subtract x in the Denominator of RHS.

x + δx − x δy = lim δx →0 δx δx → 0 x + δx − x lim

415

...(2)

416

Basic Mathematics

δy

lim

δx → 0 δx

= lim

δx → 0

a x + δx f

1 2

1 x2

− x + δx − x

x n − an x→a x − a

RHS is of the form lim

with x + δx in place of x x in place of a. and 1/2 in place of n. As δx → 0 x + δx → x. ∴ By applying the formula.

x n − an = na n −1 x →a x − a lim

we get

lim

δx →0

a

f

1 2

1 x2

x + δx − δy = lim δx δx →0 x + δx − x 1

dy 1 2 −1 = x dx 2

⇒

1

dy 1 − 2 1 = x = dx 2 2 x d dx

∴

d x i = 2 1x .

4. If xmyn = (x + y)m+n, then prove that

dy y = . dx x Pr oof: xmyn = (x + y)m+n Proof: Consider log on both sides

i a f a f log x + log y = a m + nf log a x + y f m log x + n log y = a m + nf log a x + yf d

log x m y n = m + n log x + y m

n

diff. w.r.t. x.


m⋅

a

f

a

d 1 1 dy 1 x+y + n⋅ = m+n ⋅ ⋅ x y dx x + y dx

LM N

m n dy m + n dy 1+ + ⋅ = x y dx x + y dx

f

OP Q

m n dy m + n m + n dy ⋅ + + ⋅ = x y dx x + y x + y dx n dy m + n dy m + n m − ⋅ = − y dx x + y dx x + y x

⇒

OP a f a f a f Q dy L n a x + yf − a m + nf y O mx + nx − mx − my M y a x + yf PQ = x a x + yf dx N dy L nx + ny − my − ny O nx − my PQ = x dx MN y dy L nx − my O nx − my = dx MN y PQ x LM N

m+n x−m x+y dy n m + n − = dx y x + y x x+y

⇒

dy y = dx x Hence proved.

e

5. If y = x + x 2 + 1

j

m

d

i

, then prove that 1 + x 2 y2 + xy1 − m 2 y = 0 .

Consider

e

y = x + x2 + 1

j

m

diff. w.r.t. x

e

j

e

L j ⋅ MM1 + 2 N

y1 = m x + x 2 + 1

y1 = m x + x 2 + 1

m −1

m −1

⋅

e

d x + x2 + 1 dx 1

j d

O iPP Q

d 2 x +1 2 x + 1 dx ⋅

417

418

Basic Mathematics

L O j MMN1 + 2 x1 + 1 ⋅ 2 x PPQ F x +1 + xI + 1j ⋅G H x + 1 JK

e

m −1

y1 = m x + x 2 + 1

e

m −1

y1 = m x + x 2

y1

ex + =m

y1 =

2

2

2

x2 + 1

j

m −1+1

x2 +1

e

m x + x2 + 1

j

m

e

But x + x 2 + 1

x2 + 1 my

y1 =

x2 + 1

Cross multiplying

y1

e

j

x 2 + 1 = my

Squarring

d

i

y12 x 2 + 1 = m 2 y 2 Diff. again w.r.t. x.

d

i

y12 ⋅ 2 x + x 2 + 1 ⋅ 2 y1 y2 = m 2 ⋅ 2 yy1

d

i

d i

2 y1 xy1 + x 2 + 1 y2 = 2 y1 ym 2

d

i

xy1 + x 2 + 1 y2 = m 2 y Rearranging

d x + 1i y 2

2

Hence proved. 6. If y = ex log x, then prove that

a

+ xy1 − m 2 y = 0

f a f

xy2 − 2 x − 1 y1 + x − 1 y = 0 x Given: y = e log x

diff. w.r.t. x.

j

m

=y


y1 = e x ⋅

419

1 + log x ⋅ e x x

y1 =

ex +y x

y1 =

e x + xy x

Q y = e x log x

Cross multiplying

xy1 = e x + xy

...(1)


af

xy2 + y1 ⋅ 1 = e x + x ⋅ y1 + y ⋅ 1 xy2 + y1 − e x − xy1 − y = 0 But from (1) xy1 = ex + xy

e x = xy1 − xy

b

g

xy2 + y1 − xy1 − xy − xy1 − y = 0 xy2 + y1 − xy1 + xy − xy1 − y = 0

xy2 + y1 − 2 xy1 + xy − y = 0

a

f a f − y a2 x − 1f + y a x − 1f = 0

xy2 − y1 −1 + 2 x + y x − 1 = 0 xy2

1

Hence proved. 7. Differentiate eax from first principles: Let y = eax Let δx be an increment give to x. δy be the corresponding increment in y y + δy = e

b

a x + δx

g

δy = e ax + aδx − y δy = e ax + aδx − e ax δy = e ax ⋅ e aδx − e ax δy = e ax e aδx − 1

420

Basic Mathematics


δx→0

e

j

ax aδx δy e e − 1 = δx →0 δx δx

lim

Multiply and divide by a in RHS.

ex − 1 =1 x→0 x

Q

lim

e

j

ax aδx δy ae ⋅ e − 1 = δx →0 δx aδx

lim

⇒

dy = a ⋅ e ax ⋅1 dx

∴

d ax e = ae ax dx

d i

e

j

d

i

2 2 2 2 8. If y = log x + a + x , then prove that a + x y2 + xy1 = 0.

e

2 2 Consider y = log x + a + x

j diff. w.r.t. x.

y1 =

y1 =

y1 =

y1 =

y1 =

1 x+ a +x 2

2

1

e

d x + a2 + x 2 dx

j

LM1 + 1 ⋅ d a + x OP iP MN 2 a + x dx d Q LM1 + 1 ⋅ a0 + 2 xfOP MN 2 a + x PQ LM a + x + x OP MN a + x PQ 2

x+ a +x 2

2

1 x + a2 + x 2

2

1 x + a2 + x 2 1 a + x2 2

⋅

2

2

2

2

2

2

.

a 2 + x 2 y1 = 1

2

2


Squaring

da

diff. again with respect to x.

da

2

2

421

i

+ x 2 y12 = 1

a f

i

+ x 2 ⋅ 2 y1 y2 + y12 ⋅ 2 x = 0

d

i

2 y1 a 2 + x 2 y2 + xy1 = 0

da

2

i

+ x 2 y1 + xy1 = 0

Hence proved. 9. If xy = ex−y, then prove that log x dy = dx 1 + log x

a

f

2

Pr oof: x y = e x − y Proof: Consider log on both sides

log x y = log e x − y

a

f

y log x = x − y log e y log x = x − y ...(1)

Q log e = 1

diff. w.r.t. x.

y⋅

dy dy 1 + log x ⋅ = 1− x dx dx

y dy dy + log x + =1 x dx dx

a

f

a

f

dy y log x + 1 = 1 − dx x dy x−y log x + 1 = dx x dy x−y = dx x 1 + log x

a

f

dy y log x = dx x 1 + log x

f

a

But x − y = y logx from (1) ∴

log x dy = dx 1 + log x

a

f

2

Hence proved.

LM Q MM MM MM MM MM N

OP P x = y log x + y P PP x = y a1 + log x fP P x = a1 + log x f P PP y y 1 = P x 1 + log x Q x − y = y log x

422

Basic Mathematics

x 10. Differentiate x

Let u = x x

x ... ∞

x...∞

with respect to e8x.

and v = e 8 x .

du du dx = To find dv dv dx Now u = x x

x ... ∞

u = xu Consider log on both sides

log u = log x u log u = u log x

diff. w.r.t. x.

du 1 du 1 ⋅ = u ⋅ + log x ⋅ u dx x dx du u 1 du ⋅ − log x ⋅ = u dx dx x

LM N

OP Q

du 1 u − log x = . dx u x u du u2 x = = dx 1 − log x x 1 − u log x u

a

Now v = e8x diff. w.r.t. x.

a f

dv d 8x = e8 x ⋅ dx dx dv = e 8 x ⋅ 8. dx Hence

u2 du x 1 − u log x du dx = = dv dv 8e 8 x dx

a

f

f


du u2 N∞ xx = 8 x where u = x dv 8 x 1 − u log x e

a

f

REMEMBER: y

dy dx

xn x x2 x3

nxn−1 1 2x 3x 2 1

x

2 x

1 x 1 x2

−2 x3

1 x3 ex ax

−3 x4 ex x a loga

logx

1 x C⋅

Cu

I⋅

I ⋅ II Nr.* Dr.**

**: Denominator

du dx

du dv ± dx dx

u±v

*: Numerator

1 x2

−

Dr.

af

af


a f a f a f a f

d d Nr − Nr ⋅ Dr. dx dx Dr. 2

423

424

Basic Mathematics

• Chain rule: If y = g(u) & u = f(x) Then y = g[f(x)] is differentiated by chain rule

dy dy du = ⋅ dx du dx entia tion: Given function f(x, y) = 0. • Implicit dif difffer erentia entiation: diff. w.r.t. x., take RHS. Then find

dy dy common among the terms containing and shift the remaining terms to dx dx

dy . dx

ametr ic dif entia tion: • Par arametr ametric difffer erentia entiation: Given x = f (t) and y = g(t).

dy dy dy dt Then to find , use dx = dx dx dt der der ve: • Second or order deriivati tiv If y = f (x), then by differentiating we get

dy or y′ or y1 or f ′(x). This is a function of x. By dx

d2y differentiating this again with respect to x we get or y″ or f ″(x) or y2. dx 2

EXERCISE (1) Dif entia te the ffollo ollo wing functions fr om ffir ir st pr inciples: Difffer erentia entiate ollowing from irst principles: n ax (a) x (b) e (c) logax

(d)

x

(e) 7x (2) Dif entia te the ffollo ollo wing with rrespect espect to x: Difffer erentia entiate ollowing (a) x 4 + 3e x − 7 (c)

x+

1 1 −4 3x x

(e) a x + x a − e a

(b) 3x 3 + 8 x + 9 (d) e x + e e − e π + π e (f) 7 x + x 7 + e 7 − 7 e


(3) Find

dy if dx

d

id

2 3 (a) y = 3x + 8 5x + 7

F H

(c) y = x 5 x + (e) y =

(4) Find

1 12

I K

e x − e−x e x + e− x

(b) y =

d

id

(d) y =

x2 + x + 2 x2 − 2x + 3

(f) y =

1 + x2 1 − x2

x +7

3

x −6

i

dy if dx

(a) y = e

5x

(c) y = e

1 + ex

6x

+ e πx

d

d

x (g) y = e log

(i) y = log

(b) y = e −7 x + 86 x + 7 π

i

− log x 3 + 8

−x x (e) y = log e + e

5. Find

i

i

e

2 2 (d) y = log x + a + x

(f) y =

F e + 1I GH e − 1JK

F H

x−a 1 log x+a 2a

j

I K

x

e

2 2 2 2 2 (h) y = x x + a + a log x + x + a

x

Fx− GH x +

1 − x2 1 − x2

I JK

LM F x − 2 I OP MN H x + 2 K PQ

x (j) y = log e

34

dy if dx

(a) x 2 + y 2 = 10 (c) y = x x

a

x ...

(b) y = log x + log + log x + ... ∞

∞

f

6. If (a) y 2 2 a − x = x 3 . Then pr ove tha pro thatt

a f

dy at a, a = 2. dx

dy 2 x − 7 y (b) If x2 + 3y2 − 7xy = 5 Then prove that dx = 7 x − 6 y dy 1 (c) If x ≠ y, x 1 + y + y 1 + x = 0 . Then prove that dx = − 1 + x

a f

2

j

425

426

Basic Mathematics

(d) If ax2/3 + by2/3 = (a2 − b2)2/3

F I H K

dy y =− Then prove that dx x 7. Find

1 3

.

dy if dx (b) x =

(a) x = t 3 , y = t 2 + 1 (c) x =

1− t2 2t , y= 2 1+ t 1 + t2

3at 3at 2 y = , 1 + t3 1 + t3

8. Dif entia te Difffer erentia entiate (a) log e x with respect to ex (b) log10 x with respect to x2 (c) 10x with respect to 5x (d) If xm⋅yn = (x + y)m+n, then prove that (e) If ex + ey = ex+y, then prove that (f) If xy = ex−y, then prove that

9. Find

dy = −e y − x dx

log x dy = dx 1 + log x

a

f

2

.

d2y if dx 2

(a) x 2 + y 2 = a 2 (c)

dy y = . dx x

(b) y = x 2 e x

x 2 y2 + =1 a2 b2

(d) x 3 y 3 = a 5

e

10. (a) If y = x + 1 + x 2

e

j

m

d

i

2 2 , then prove that 1 + x y2 + xy1 − m y = 0 .

j

d

i

2 2 2 2 (b) If y = log x + a + x , prove that a + x y2 + xy1 = 0 .

(c) If y = aemx + be−mx, then prove that y2 − m2y = 0.

a

f a f − 1) , prove that d x − 1i y + 2 x a1 − nf y − 2ny = 0.

(d) If y = exlogx, then prove that xy2 − 2 x − 1 y1 + x − 1 y = 0. (e) If y = (x2

n

2

2

1


d

2 2 (f) If y = a + x n +1 (g) If y = ax +

i , prove that d x 6

2

427

i

+ a 2 y2 − 10 xy1 − 12 y = 0.

a f

b , then prove that x 2 y2 + n n + 1 y = 0 . xn

(h) If y = (a + bt)e−nt, then prove that

e

2 (i) If y = a x + x − 1

j

n

e

dy d2 y + 2n ⋅ + n 2 y = 0. 2 dt dt

+ b x − x2 − 1

j

n

d

i

2 2 , then prove that x − 1 y2 + xy1 − n y = 0.

ANSWERS 1. (a) nxn−1

2. (a)

4x3

+

(b) aeax

3ex

(b)

9x2

+8

(e) ax loga + axa−1 (3) (a)

(c)

(e)

2

3

1 1 x a 5f + F 5 x + I H 12 K 2 x

de

4 x

−e

4. (a) 5e 5 x −

i

x

2

1 2 x

(d) −

1

(e) 7x log 7.

5

1 3 4 x4 (c) − + 4 2 x 2x x 1

(d) ex

F I d x + 7i GH 13 x JK + d x − 6i FGH 2 1 x IJK d x − 2 x + 3ia2 x + 1f − d x + x + 2i a2 x − 2f d x − 2 x + 3i −

(b)

2 3

3

2

(d)

2

2

4x

−x 2

1 ex

1 x

(f) 7x log7 + 7x6.

d3x + 8i15x + d5x + 7i 6 x 2

(c)

+ πe πx

(f)

d1 − x i

2 2

(b) −7e −7 x + 6 ⋅ 86 x ⋅ log 8 1

(c)

3x 2 e 6x ⋅6 − 3 2 6x x +8

(d)

(e)

ex + e−x e x − e−x

(f)

x 2 + a2

1 x − a2 2

2

428

Basic Mathematics

LM N

x (g) e log

OP Q

ex + 1 2e 2 x − 2x x e −1 e −1

(h) 2 x 2 + a 2

2

(i)

d2 x − 1i 2

x 5. (a) − y

1− x

(j)

2

f

2 3t

(b)

8. (a)

e− x x

1 (b) 2 x 2 log 10 e

a2 y3

a

t2 − 1 x or − 2t y

7. (a)

9. (a) −

y2 (c) 1 x − y log x

1 (b) x 2 y − 1

a

(b) e

x

dx

2

+ 4x + 2

x2 − 1 x2 − 4

i

(c)

2t − t 4 1 − 2t 3

(c)

10 x log 10 5 x log 5

−b 4 (c) 2 3 a y

f

( d)

15 y . 4x2

17 Application of Derivatives Derivatives have many applications. To mention a few, we use differentiation in finding 1. Equation of tangent and normal. 2. Length of subtangent and subnormal 3. Angle between the curves 4. Rate measure i.e., variation with respect to time 5. Maximum and minimum values of the function. Derivatives are also used in finding marginal cost and marginal revenue.

17.1 DERIVATIVE AS A RATE MEASURE: Velocity: Rate of change of displacement is called velocity. If v is the velocity, ‘s’ is the displacement and s = f (t) Then

Velocity =

af

ds = f′ t dt

Acceler ation: The rate of change of velocity is called acceleration. If ‘a’ is the acceleration v is the Accelera velocity. Then a=

F I H K

dv d ds d 2s = 2 = dt dt dt dt

Note: 1. Rate means differentiation with respect to time, t. Rate of increase of area =

dA . dt

Rate of decrease of volume =

dV and so on. dt

2. Formulae:

(b) Area of circle = π(radius)2 = πr2.

4 3 πr . 3 (e) Volume of a cylinder = πr2h.

(c) Surface area of Sphere = 4πr2

(f) Volume of a cone =

(a) Area of square = (side)2 = S2.

(d) Volume of sphere =

1 2 πr h. 3

430

Basic Mathematics

WORKED EXAMPLES: 1. If the displacement of a particle at time t seconds is s = 3t2 − 7t + 6, then find the velocity and acceleration at t = 1 second. Given: s = 3t2 − 7t + 6 diff. w.r.t. t.

a f af

ds = 3 2t − 7 1 dt Velocity =

ds = 6t − 7 dt

Velocity when t = 1 sec. = 6(1) − 7. = 6 − 7 = −1 unit/sec. Now Velocity = 6t − 7 v = 6t − 7 diff. w.r.t. t.

af

dv =6 1 −0 dt dv =6 dt acceleration = 6 units/sec2 acceleration when t = 1 is 6 units/sec2. 2. When the brakes are applied to the moving car, the car travels a distance s mts in t seconds given by s = 6t − 3t2, when and where does the car stops? Solution: Car stops when velocity = 0. Now s = 6t − 3t2. diff. w.r.t. t.

af a f

ds = 6 1 − 3 2t dt ds = 6 − 6t dt Velocity =

ds =0 dt

∴

0 = 6 − 6t

⇒

6 = 6t ⇒ t = 1 .

Application of Derivatives

431

∴ Car stops when t = 1 sec. Distance travelled in 1 sec.

s = 6t − 3t 2 when t = 1 sec.

af af

s=6 1 −3 1 2 s =6−3 s = 3 mts.

So car stops after travelling a distance of 3 mts. 3. With usual notation, if s = at + b. Where a and b are constants then prove that velocity is constant and acceleration is zero. Solution: Given s = at + b diff. w.r.t. t.

af

ds = a 1 + 0 3 a and b are constants dt ds = velocity = a = a constant. dt diff. w.r.t. t.

d 2s = 0 3 a is constant. dt 2 ∴ acceleration = 0 4. A square plate expands uniformly, the side is increasing at the rate of 6 mm/sec. What is the rate of increase in area when side is 13 mm. Solution: Given Rate of increase of side = 6 mm/sec. i.e.,

dS = 6 mm sec. dt

To find:

dA when S = 13 mm. dt

We have Area of square = side × side A = S2

diff. w.r.t. t.

dA dS = 2S ⋅ dt dt

a fa f

dA = 2 13 6 dt

432

Basic Mathematics

dA = 156 mm 2 sec. dt ∴ Rate of increase of area = 156 mm2/sec. 5. A stone is dropped into a pond, waves in the form of circles are generated and the radius of the outermost wave increase at the rate of 2 mm/sec. How fast is the area increasing (a) when the radius is 5 mm (b) after 3 sec. Solution: If r is the radius at time t; Then given: To find (a) ( b)

dr = 2 mm sec dt dA when r = 5 mm dt

dA when t = 3 sec dt

We have Area of circle = πr2

A = πr2 diff. w.r.t. t.

a f

dA dr = π 2r dt dt (a)

b a fg

dA = π 2 5 ⋅2 dt dA = 20π dt dA = 20π mm 2 sec dt

(b)

a f

dA dr = π 2r dt dt Now we have radius is increasing at the rate of 2 mm/sec. After 3 seconds, radius = 3 × 2 = 6 mm.

(3

dr = 2 mm sec dt

For 1 sec radius is 2 mm. For 3 sec. radius is 3 × 2 = 6 mm) From (1)

...(1)


433

a f

dA dr = π 2r ⋅ dt dt

a f

dA = π 2 × 6 ⋅2 dt after 3 seconds dA = 24π dt after 3 sec ∴ After 3 seconds, Area is increasing at the rate of 36 π mm2/sec. 6. Water flows into a cylindrical tank of radius 4 mts at 80,000 cc/hr. How fast water level is raising? Solution: If r is the radius, h is the height. The volume V is given by

V = πr 2 h dV = 80,000 cc hr. dt

Given If h is the height of the water level,

dh represent rate of increase in water level. dt

r = radius = 4 mts = 400 cm = a constant while filling the water in the cylindrical tank. Consider V = πr 2 h

diff. w.r.t. t.

d i

dV dh = π r2 ⋅ dt dt

a f

80,000 = π 400 2 ⋅

dh dt

dh 80,000 = 160000 π dt = ∴ Water level is rising at the rate of

dh 1 = 2π dt

1 cms/hr. 2π

7. Sand is poured at the rate of 30 cc/sec and it forms a conical pile in which the diameter of the circular base is always equal to one third the height. At what rate height of the pile is increasing when the height is 30 cm. Solution: 2 × Radius = Diameter

434

Basic Mathematics

Given Diameter = 2 × radius =

r=

1 ⋅h 3

1 h 3

1 h 6

...(1)

Fig. 17.1

dV = 30 cc sec dt To find

3h

dh when h = 30 cm. dt

We have volume of the cone =

1 2 πr h. 3 1 V = πr 2 h 3 V=

F I H K

1 1 π⋅ h 3 6

2

⋅h

1 h 2 ⋅ h πh 3 V = π⋅ = 3 36 108 V= diff. w.r.t. t.

πh3 108

LM OP N Q dh π L 30 = 3 × a30f ⋅ OP M dt Q 108 N dh dv π = 3h 2 ⋅ dt dt 108

2

dh 30 × 108 = 3 × 900 dt π dh 30 × 108 = 3 × 900 × π dt ⇒

dh 6 = dt 5π

[from (1)]


Height of the pile is increasing at the rate of

435

6 cm/sec. 5π

8. A spherical snow ball is forming so that its volume is increasing at the rate of 8 cm/sec. Find the rate at which the radius is increasing when the snow ball is 2 cm in diameter. Also find the rate of increase in surface area. Solution. Given: To find:

dV = 8 cm sec. dt dr when diameter = 2 cm. dt

ds 2 i.e. radius = 1 cm . dt 2 We have for a sphere,

and

V=

4 3 πr 3

diff. w.r.t. t.

dV 4 dr = π ⋅ 3/ r 2 ⋅ dt 3/ dt dV dr = 4 πr 2 ⋅ dt dt

af

8 = 4π ⋅ 1 2 ⋅

dr dt

⇒

dr 8 2 = = dt 4 π π

∴

dr 2 = cm sec. dt π Now surface area = 4 πr2. diff. w.r.t. t.

a f

ds dr = 4π 2r dt dt

af

ds 2 = 4π ⋅ 2 1 dt π ds = 16 sq cm sec. dt

436

Basic Mathematics

9. A ladder 13 ft long rests with its ends on a horizontal floor and against a smooth vertical wall. If the upper end is coming downwards at the rate of 1 ft/min. Find the rate at which the lower end moves, when the upper end is 5 ft from the ground. Solution: Let PQ be the ladder. At time t, Let OP = y, OQ = x From figure OP2 + OQ2 = PQ2 P y2 + x2 = 132 Given To find

dy = 1ft min. dt

y

dx when y = 5ft. dt

i.e., To find

0

dx when y = 5 ft. and x = 12 ft. dt

Q

x

Fig. 17.2

y 2 + x 2 = 132 5 2 + x 2 = 169

x 2 = 169 − 25 x 2 = 144

x = 144 x = 12

From fig.,

y 2 + x 2 = 132 diff. w.r.t. t.

2y

dy dx + 2x =0 dt dt

2x

dx dy = −2 y dt dt

dx 2 y dy =− ⋅ dt 2 x dt dx y dy =− dt x dt dx 5 = − ⋅1 dt 12


437

dx 5 =− dt 12 ∴ Lower end moves at the rate of

5 ft/min. Negative sign shows as x increases, y decreases. 12

10. An aeroplane at an altitude of 400 kms flying horizontally at 500 km/hr passes directly over an observer. Find the rate at which it is approaching the observer when it is 500 kms away from him. Solution. Let at time ‘t’ ‘A’ be the position of the A x B aeroplane, B be the point on the path directly above the observer O. Let AB = x. OA = y. From fig. AB 2 + OB 2 = OA 2

400 kms

y

x 2 + 400 2 = y 2 diff. w.r.t. t.

O

2x

dx dy + 0 = 2y dt dt

Fig. 17.3

...(1)

dx = −500 km hr dt

Given

[−ve Sign is taken since x is decreasing] To find

dy when y = 500 kms. dt

From fig.

x 2 + 400 2 = y 2

a3 y = 500f

x 2 + 400 2 = 500 2

x 2 = 500 2 − 400 2 x 2 = 250000 − 160000

x 2 = 90000 ⇒

x = 300

Substituting x = 300, y = 500 and

dx = −500 in (1) we get dt

a fa

f a f

2 300 −500 = 2 500 ⋅ −300 =

dy dt

dy dt

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Basic Mathematics

dy = −300 kms hr. dt

∴

∴ The aeroplane is approaching at the rate of 300 km/hr.

17.2 MAXIMA AND MINIMA: eases, if y also incr eases then y = f (x) is said to be an Let y = f (x) be a continuous function. As x incr increases, increases incr easing function. increasing Ex.: y = x + 3 is an increasing function   → increase

3

x y

0 3

1 4

2 5

3 6

....... .......

increase  →

Fig. 17.4

eases if y decr eases then the function y = f(x) is said to be a decr easing function As x incr increases decreases decreasing function. Ex.: y = 7 − x is a decreasing function.   → increase

3

x

0

1

y

7

6

2

3

.......

5

4

.......

decrease →

Note that for an increasing function, For a decreasing function,

dy > 0 i.e. positive. dx

Fig. 17.5

dy < 0 i.e., negative. dx

y is said to be stationary at a point if it neither increases nor decreases. At such point

dy =0. dx

If a continuous function increases upto a certain value and then decreases from that value, then that value is called a maximum value of a function. Similarly if continuous function decreases to a value and then increases, then that value is called minimum value of the function. Note that a continuous function may attain maxima/minima at several points or it may neither have maxima nor minima. In the figure, the points P, Q, R, S, are points of maxima. The points A, B, C, are points of minima. Between any 2 maxima there must be a minima and vice versa.


439

Y S

X

Fig. 17.6

DERIVATIVE TEST FOR MAXIMA AND MINIMA: Let y = f (x) be the given function. To find the maximum and / or minimum value of the function. We find

dy and equate it to zero. Let x = a, x = b and x = c be the points. dx

d2 y d2y , if is greater than zero, then x = a is a point of minima. Minimum value dx 2 at x = a dx 2 at x = a of the function is y = f (a).

We find

d2 y d2 y if is less than zero, then x = b is a point of maxima. Maximum Next we find dx 2 at x = b dx 2 at x = b value of the function is y = f (b).

d2y is equal to zero, then at x = c, the function neither attains maxima, nor minima. The If dx 2 at x = c point x = c is called a point of inflexion.

WORKED EXAMPLES: 1. Find the maximum and minimum value of the function 2x3 − 15x2 + 36x + 10. Solution: Let y = 2x3 − 15x2 + 36x + 10. diff. w.r.t. x.

d i a f

af

dy = 2 3 x 2 − 15 2 x + 36 1 + 0 dx dy = 6 x 2 − 30 x + 36. dx

...(1)

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Basic Mathematics

At extremum,

dy =0. dx 0 = 6 x 2 − 30 x + 36 ÷ by 6. x 2 − 5x + 6 = 0 x 2 − 3x − 2 x + 6 = 0

a f a f a x − 2fa x − 3f = 0

x x−3 −2 x−3 =0

∴

x = 2 or x = 3 .

Consider equation (1)

dy = 6 x 2 − 30 x + 36 dx diff. w.r.t. x.

a f

af

d2 y = 6 2 x − 30 1 + 0 dx 2 d2 y = 12 x − 30 dx 2

af

d2 y = 12 2 − 30 dx 2 at x = 2

= −6 < 0 → negative. ∴

x = 2 is a point of maxima. Maximum value of the function 2 x 3 − 15 x 2 + 36 x + 10

af

is

2 2

3

af

− 15 2

af

2

af

+ 36 2 + 10

2 8 − 60 + 72 + 10 16 − 60 + 72 + 10 = 98 − 60 = 38.

Now

d2 y = 12 x − 30 dx 2

af

d2 y = 12 3 − 30 = 6 > 0 → positive dx 2 at x = 3

∴

x =3 is a point of minima.

6x 2

−3 x −2 x −5 x


441

Minimum value of the function

2x 3 − 15 x 2 + 36 x + 10 is

af af af 2a27f − 15a9f + 108 + 10

2 3 3 − 15 3 2 + 36 3 + 10

54 − 135 + 118 172 − 135 = 37.

2. Find two numbers such that their sum is 20 and their product is maximum. Solution. Let the 2 numbers be x and y. Given: Their sum = 20.

x + y = 20 ⇒ y = 20 − x To find: x and y such that their product is maximum. Their product = xy

P = xy

a

P = x 20 − x

f

P = 20 x − x 2

For P to be extremum,

dP =0 dx P = 20 x − x 2

diff. w.r.t. x.

af

0 = 20 1 − 2 x 2 x = 20 x = 10.

So the numbers are x = 10 and y = 20 − 10 = 10. Reqd. Numbers are: 10 and 10. 3. Find two numbers whose sum is 16 and sum of their cubes is minimum. Solution: Let the 2 numbers be x and y. Given: Their Sum = 16.

x + y = 16 ⇒

y = 16 − x

To find: x and y such that sum of their cubes is minimum. i.e.,

x 3 + y 3 Should be minimum.

...(1)

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Basic Mathematics

P = x 3 + y3

Let

a

P = x 3 + 16 − x

f

3

[from (1)]

For P to be extremum,

dP =0 dx

a

P = x 3 + 16 − x

Now

f

3

diff. w.r.t. x.

a

dP = 3x 2 + 3 16 − x dx

f

2

⋅

a

d 16 − x dx

a f a−1f 0 = 3 x − 3 a16 − x f 0 = 3 x − d16 + x − 32 x i

dP = 3x 2 + 3 16 − x dx

2

2

2

2

2

2

0 = x 2 − 256 − x 2 + 32 x

0 = −256 + 32 x 32 x = 256

x=

256 16 = = 8. 32 2

So the numbers are x = 8 and y = 16 − 8 = 8 i.e. 8 and 8. 4. Prove that logx do not have maxima or minima. Proof: Pr oof: Let y = log x. For maxima or minima,

dy =0 dx y = log x dy 1 = dx x 0=

1 x

1 =0 x

f


⇒ x has no finite value. ∴ logx do not have maxima or minima. 5. The product of 2 positive numbers is 36. Find the minimum value of their sum. Solution: Let the 2 number be x and y. Given: Their product = 36.

xy = 36 y=

⇒

36 x

Their sum = x + y.

For S to be extremum

S=x+

36 x

S=x+

36 x

dS =0 dx

diff. w.r.t. x.

F I H K

dS −1 = 1 + 36 2 dx x 0 = 1− 0= ⇒

36 x2

x 2 − 36 x2

x 2 − 36 = 0 x 2 − 36 ⇒ x = ± 6 dS 36 =1− 2 dx x diff. w.r.t. x.

Now

d

d2S = −36 −2 x −3 dx 2

i

d 2 S 72 = dx 2 x 3 72 d2S = 3 > 0 → positive. 2 dx at x = 6 6

443

444

Basic Mathematics

∴ x = 6 is a point of minima.

y= ∴ Minimum value of their sum = 6 +

36 36 = = 6. x 6

36 6

= 6 + 6 = 12.

6. Prove that a maximum rectangle that can be drawn with a constant perimeter is a square. oof: Let x be the length and y be the breadth of a rectangle. Proof: Pr Given: Perimeter = constant. Sum of all sides = constant.

a f

x + y + x + y = 2k say

D

x

C

2 x + 2 y = 2k

a

f

2 x + y = 2k x+y=k

A

y=k−x

a

A=x k−x

f

A = kx − x 2

For area to be extremum,

dA =0 dx A = kx − x 2

diff. w.r.t. x.

af

dA = k 1 − 2x dx 0 = k − 2x

⇒

2x = k

x=

Consider

k 2

dA d2A = k − 2 x diff. again w.r.t. x, = −2 < 0 for all dx dx 2

values of x.

x

Fig. 17.7

Area of rectangle = Length × Breadth A = xy

y

y

B


∴ A attains maxima for all values of x. Now,

y=k−x=k−

k 2k − k k = = 2 2 2 k 2

x=y=

Hence Length

2x = 2 ⋅

k =k 2

Breadth

2y = 2 ⋅

k =k 2

∴ Rectangle has Length = Breadth. Hence it is a square. 7. Prove that the maximum rectangle that can be inscribed in a circle is a square. D

x

C

y

y

A

x

B

Fig. 17.8

Let ABCD be the rectangle with length x and breadth y inscribed in a circle of radius say a. ∴

OB = OD = a

∴

BD = 2 a

From right angled triangle, ABD,

AB 2 + AD2 = BD2

a f

x 2 + y 2 = 2a

2

x 2 + y2 = 4a 2 y 2 = 4a 2 − x 2

y = 4a 2 − x 2

445

446

Basic Mathematics

Now, Area of the rectangle = Length × Breadth

A= x×y

A = x 4a 2 − x 2 A = x 4a 2 − x 2 dA =0 dx

For A to be maximum/minimum

A = x 4a 2 − x 2 diff. w.r.t. x.

dA d = x⋅ dx dx

e

j

4 a 2 − x 2 + 4a 2 − x 2 ⋅

d

af

d x dx

af

i

dA d 1 4a 2 − x 2 + 4a 2 − x 2 ⋅ 1 = x⋅ ⋅ 2 2 dx dx 2 4a − x

a f

dA x = ⋅ −2 x + 4 a 2 − x 2 2 2 dx 2 4a − x 0=

−2 x 2 2 4a − x 2

0=

2

+ 4a2 − x 2

− x 2 + 4a 2 − x 2 4a 2 − x 2

0 = −2 x 2 + 4 a 2

⇒

2 x 2 = 4a2 x 2 = 2a 2

x= 2a d2 A dx 2 at x =

Also

2a

is < 0.

∴ A attains maxima at x = 2 a . Now

y = 4a 2 − x 2 = 4a 2 −

d 2 ai

y = 4a 2 − 2 a 2 = 2 a

2


447

∴ Rectangle has maximum area when its length = 2 a and Breadth = 2 a , i.e., when its length = Breadth = 2 a , i.e., when it is a square. 8. What is the largest size rectangle that can be inscribed in a semicircle of radius r unit so that 2 vertices lie on the diameter. Solution: Let AB be the diameter of a semicircle. S R 2x Let 2 vertices P and Q of the rectangle PQRS lie on the diameter. Let

PQ = RS = 2 x

2y

2y

PS = QR = 2 y

O

Join OR. Given: r is the radius of the semicircle. From fig.

A

P

OQ 2 + QR 2 = OR 2

a f

x2 + 2y

2

2x

Q

B

Fig. 17.9

= r2

x 2 + 4y2 = r 2 4y2 = r2 − x 2 y2 =

1 2 r − x2 4

...(1)

Now, Area of the rectangle = Length × Breadth

A = 2x × 2y A = 4 xy

A = 4x ⋅

1 2 r − x2 2

from (1)

A = 2x r2 − x2 Area is extremum when

dA =0 dx A = 2x r2 − x2

Now

LM e N

d dA = 2x dx dx

diff. w.r.t. x.

r2 − x2

jOQP +

r2 − x2 ⋅

a f

d 2x dx

448

Basic Mathematics

d

i

dA d 2 1 r − x2 + r2 − x2 ⋅ 2 = 2x ⋅ ⋅ 2 2 dx dx 2 r −x dA = dx

0=

a f

x

⋅ −2 x + 2 r 2 − x 2

r − x2 2

d

−2 x 2 + 2 r 2 − x 2 r −x 2

i

2

0 = −2 x 2 + 2r 2 − 2 x 2 4 x 2 = 2r 2

⇒

Also

d2A dx 2 at x =

r 2

x2 =

r2 2

x=

r . 2

is negative.

Hence A attains maxima when x = 2 r. ∴ Area of rectangle is maximum when its length = 2 x = 2 ⋅

Breadth = 2 y = 2 ⋅

Breadth = r 2 −

1 2 r − 2

FG r IJ H 2K

r = 2r 2

2

r2 r . = 2 2

Maximum area = Length × Breadth =r 2×

r 2

= r2. 9. A box is constructed from a square metal sheet of side 60 cm by cutting out identical squares from the four corners and turning up the sides. Find the length of the side of the square to be cut out so that the box is of maximum volume.


Solution: Let the side of the cut out square be x. Now length of the box = 60 − 2x Breadth of the box = 60 − 2x Height of the box = x. Volume = Length × Breadth × height.

a fa V = a60 − 2 xf ⋅ x

x

449

x

x

x 60 − 2x

f

V = 60 − 2 x 60 − 2 x x

x

x

2

V attains extrema when Consider

x

x

dV =0 dx 60 − 2x

a

f

Fig. 17.10

V = 60 − 2 x 2 x diff. w.r.t. x.

a

f

a

f

2

⋅ 1 + x ⋅ 2 60 − 2 x ⋅

a

f

2

+ 2 x 60 − 2 x −2

a

f

2

− 240 x + 8 x 2

dV = 60 − 2 x dx

dV = 60 − 2 x dx dV = 60 − 2 x dx dV = 60 − 2 x dx

2

⋅

af

a

d d x + x⋅ 60 − 2 x dx dx

a

f

2

f dxd a60 − 2 xf

a

fa f

dV = 60 2 + 4 x 2 − 240 x − 240 x + 8 x 2 dx 0 = 3600 + 12 x 2 − 480 x ÷ by 12. 300 + x 2 − 40 x = 0 x 2 − 40 x + 300 = 0

x − 30 x − 10 x + 300 = 0 2

a

f

a

f

x x − 30 − 10 x − 30 = 0

a x − 30fa x − 10f = 0 x = 30 or x = 10

Now

dV = 3600 + 12 x 2 − 480 x dx

300 x

2

−30 x −10 x −40 x

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Basic Mathematics

d 2V = 24 x − 480. dx 2 d 2V is negative. dx 2 at x =10

∴ V attains maxima when x = 10. ∴ Length of the square to be cut = 10 cm.

F 10. Prove that G ± H

1

− 1 ,e 2 2

I JK

are the points of inflection for the curve y = e − x . 2

Pr oof: Consider Proof:

y = e−x

2

diff. w.r.t. x.

d i

2 dy d = e−x ⋅ −x2 dx dx

a f

2 dy = e − x ⋅ −2 x dx


a f a f e j

2 d d − x2 d2 y e = e− x ⋅ −2 x + −2 x 2 dx dx dx

a f

e ja f

2 2 d2 y = e − x ⋅ −2 − 2 x ⋅ e − x ⋅ −2 x 2 dx 2 2 d2 y = −2e − x + 4 x 2 e − x 2 dx

y attains neither maxima nor minima when d2y = 0 i.e. dx 2 0 = −2e − x + 4 x 2 e − x 2

2

0 = 2 e − x −1 + 2 x 2 2

e − x = 0 or − 1 + 2 x 2 = 0 2

− x 2 = 0 or x=0

2x2 = 1 ±1 x= 2


451

Now

F 1 IJ 2K

−G 2 1 H , y = −e − x = e x= 2

when

2

=e

−

1 2

1

x=−

when

− 2 1 , y = e−x = e 2 . 2

∴ The points of inflection for the curve y = e − x are 2

F 1 , e I and F − GH 2 JK GH −

1 2

1

I JK

− 1 ,e 2 . 2

Hence proved.

REMEMBER: • Velocity =

ds dt

• Acceleration =

dV d 2 s = dt dt 2

• Rate means differentiation w.r.t. t. ∴ rate of change of area =

dA dt

• Area of Square = S2 • Area of circle = πr2 • Surface area of sphere = 4πr2 • Volume of Sphere =

4 3 πr 3

• Volume of a cylinder = πr2h • Volume of a cone =

1 2 πr h 3

• For an increasing function

dy dy < 0. > 0 and for a decreasing function dx dx

• To find maximum and/or minimum value of the function y = f (x), find

dy , equate it to zero. Let dx

d2 y d2y d2y , and x = a, x = b, x = c be the points. Find . dx 2 at x = a dx 2 at x = b dx 2 at x = c.

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Basic Mathematics

if

d2y > 0, then x = a is a point of minima. Minimum value of the function is y = f (a). dx 2 at x = a

d2 y is less than zero, x = b is a point of maxima. Maximum value of the function is dx 2 at x = b y = f (b).

if

d2y if is equal to zero, then x = c is called point of inflection. At x = c the function neither dx 2 at x = c attains maxima nor minima.

EXERCISE 1. The distance s is metres moved by a particle in ‘t’ seconds is given by s = 45t + 11 t2 − t3. Find the time when the particle comes to rest? 2. The displacement s of a particle at time ‘t’ seconds is given by 2t3 − 3t2 − 36 t + 90. Find the (a) velocity after 4 seconds (b) displacement and acceleration when the velocity vanishes (c) acceleration after 4 seconds. 3. With usual notation if. If s2 = at2 + 2bt + c, then prove that (a) the acceleration is inversely proportional to s3. (b) the acceleration is

a − v2 where v is the velocity. s

4. The equation of motion of a particle is given by s = 9t2 − t3. Find the displacement when velocity is zero and velocity when the acceleration is zero. 5. If s = at3 + bt, find a and b given that when t = 3 velocity is zero and acceleration is 14 units. 6. When breaks are applied to the moving car, the car travels a distance S feet in t seconds given by s = 20t − 40 t2. When does the car stop? 7. The side of a square sheet metal is increasing at 3 mm/min. At what rate is the area increasing when the side is 10 mm long. 8. A circular patch of oil spreads on water, the area is growing at the rate of 2 sq cm/hr. How fast are the radius and the circumference increasing when the diameter is 24 cm. 9. A drop of ink spreads over a blotting paper so that the circumference of the blot which is circular increases at the rate of 3 cm/min. Find the rate of increase of the radius and area when its circumference is 4π cm. 10. A stone is dropped into a pond, waves in the form of circles are generated and the radius of the outermost ripple increases at the rate of 2 mm/min. How fast is the area increasing when the radius is 5 mm, after 5 min? 11. A cylindrical tank is 10 mts in diameter, water is flowing in it at the rate of 24 m3/min. at what rate height of the water is rising?


453

12. Water is flowing into a right circular cylindrical tank of radius 50 cm at the rate 500 π cc/min. Find how fast is the level of water rising? 13. A ladder 20 ft long rests with its ends on a smooth horizontal floor and against a smooth vertical wall. If the lower end is moved at the rate of 5 ft/min. Find the rate at which the upper end moves when the lower end is 12 ft. from the wall. 14. The radius of the sphere is decreasing at the rate of 3 cm/sec. Find the rate at which surface area is decreasing when radius is 12 cm. 15. The height of circular cone is 30 cm and it is constant. The radius of the base is increasing at the rate of 0.25 cm/sec. Find the rate of increase of volume of the cone when the radius is 10 cm. 16. A man 6 ft tall is moving directly away from a lamp at a height of 10 ft above the floor. If he is moving at the rate of 6 ft/sec find the rate at which the length of his shadow is increasing? 17. Find the maximum and minimum value of the function 4x3 − 15x2 + 12 x + 7. 18. Find the maximum and minimum value of the function x3 − 3x2 − 9x + 17. 19. Prove that the function xe−x attains maxima at x = 1 and its maximum value is 20. Prove that xx is minimum when x =

21. Prove that the maximum value of 22. 23. 24. 25. 26. 27. 28. 29. 30.

1 . e

1 . e

F 1I H xK

x

is e1/e.

Prove that x3 − 3x2 − 9x + 9 has a point of inflection at x = 1 Prove that y = a−(x − b)2/5 has no point of inflection. Find two number whose sum is 24 and their product is maximum. The sum of two numbers is 20. If the product of square of the first and cube of the 2nd is maximum. Find the numbers. Prove that the area of the right angled triangle of given hypotenuse is maximum when the triangle is isoceles. Prove that among all rectangles of a given area the square has minimum perimeter. Prove that among all rectangles of given perimeter, the square is the one with shortest diagonal. What is the largest rectangle that can be inscribed in a semicircle of radius 1 so that two vertices lie on the diameter. A rectangular field is to be fenced off along the bank of a river, no fencing is required along the river. If the material for fence cost Rs. 8 per running foot for 2 ends and Rs. 12 per running foot for the side parallel to the river, find the dimension of the field of largest possible area that can be enclosed with Rs. 3600 worth of fences.

ANSWERS 1. 9 Seconds 2. (a) 0 (b) 9 cm; 30 cm/sec2

(c) 42 cm/sec2,

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Basic Mathematics

4. 108 units; 27 units/sec.

5. a =

7 , b = −21 9

6.

1 sec. 4

7. 60 mm2/min.

8.

1 1 cm/hr; cm/hr. 12π 6

9.

10. 20 π mm2/m’ 40 πmm2/min

11.

12.

1 cm/min 5

3 cm/min; 6 sq cm/min 2π 24 m/min 25π

13. −

15 ft/sec. 4

14. 288 π sq cm/sec.

15. 50 π cm3/sec.

16. 9 ft/sec.

17.

18. 22; −10

24. 12; 12

25. 8 and 12

1 30. 112 ft and 150 ft. 2

39 ;3 4

18 Integration If

af af

af

af

d d f x =g x , f x + c = g x . Then g(x) is called integrand and f(x) or f(x) + c is called dx dx

integral or primitive or antiderivative of g(x) with respect to x. It is denoted by

z af

af

g x dx = f x + c

∴

zaf

g x dx.

18.1 STANDARD INTEGRALS: By using the definition of integration as the reverse operation of differentiation, the following formulae may be obtained. 1.

z

x n dx =

x n +1 +c n +1

Provided n ≠ −1

3 if n = −1,

Since 2.

z

LM N

OP a f Q

n + 1 x n +1−1 d x n +1 + 0 = xn +c = dx n + 1 n +1

z

x −1 dx =

z

1 dx = log x + c x

d 1 1 log x + c = + 0 = dx x x

e x dx = e x + c 3

d

i

d x e + c = ex dx

456

3.

Basic Mathematics

z

a x dx =

ax +c log a 3

4.

z

LM N

OP Q

d ax a x log a +0 +c = dx log a log a = ax

k dx = kx + c 3

a

f af

d kx + c = k 1 + 0 = k dx

18.2 ALGEBRA OF INTEGRALS: 1. If K is constant, f(x) and g(x) are 2 integrable functions, then (i)

z af z af z b a f a fg z a f z a f k ⋅ f x dx = k ⋅ f x dx

i.e., Integral of constant multiplied by function is constant multiplied by integral of the function. (ii)

f x ± g x dx =

f x dx ± g x dx

i.e., Integral of sum or difference of 2 or more functions = Integral of the first function ± integral of the second function.

WORKED EXAMPLES: 1. Evaluate: Solution:

zd

zd

i

4 x 3 − 1 dx

i

z

z

4 x 3 − 1 dx = 4 x 3 dx − 1 dx

z z

= 4 x 3 dx − 1 dx 4⋅

x4 − x + c. 4

where c is the constant of integration

zd

2. Integrate

i

4 x 3 − 1 dx = x 4 − x + c.

1 + e x with respect to x. x

Integration

z FH 3. Evaluate: Solution:

z FH

z FH

I K

1 + e x dx = x

I K

1 1 4 dx − + x x2 x3

I K

z z 1 dx − x

log x −

1 dx + x2

1 2 − + c. x x2

2

dx

z

z FH z z

F x + 1I H xK

2

dx =

= x 2 dx +

x2 +

5. Evaluate:

Solution:

z

z

2

F x + 3x − GH x x 2

I JK 1 I J dx xK

z

1 dx + 2 dx x2

=

x 2 +1 x −2 +1 + + 2x + c 2 + 1 −2 + 1

=

x 3 x −1 + + 2x + c 3 −1

+ 3x − 1 dx x

I K

1 1 + 2 ⋅ x ⋅ dx x x2

x3 1 − + 2x + c . 3 x

Fx GH

4 dx x3

x −1 4 x −2 + +c −1 −2

log x +

F x + 1I H xK

z

x −2 +1 4 x −3+1 + +c −2 + 1 −3 + 1

log x −

z

1 dx + e x dx x

= log x + e x + c.

1 1 4 − 2 + 3 dx = x x x

4. Evaluate:

z z

457

458

Basic Mathematics

z z z z z z

= x

2−

1 2 dx

+ 3x

3

1−

1 2

dx − x

1

x 2 dx + 3 x 2 dx − x

3 +1 x2

3 +1 2 5 x2

=

5 2

+

+

1 +1 3x 2

−

1 +1 2

3 3x 2

3 2

−

1 x2

1 2

−

1 − +1 x 2

−

1 +1 2

1 2

−

1 2

dx

dx

+c

+c

5 3

1

2x 2 + 2 x 2 − 2 x 2 + c. 5 6. Integrate (x2 + 1) (2x3 − 6x) with respect to x.

zd

id

i

x 2 + 1 2 x 3 − 6 x dx = =

zd zd z z

i

2 x 5 − 6 x 3 + 2 x 3 − 6 x dx

i

2 x 5 − 4 x 3 − 6 x dx

z

= 2 x 5 dx − 4 x 3 dx − 6 x dx

7. Evaluate:

zd

i

=

2x 6 4x 4 6 x2 − − +c 6 4 2

=

x6 − x 4 − 3x 2 + c. 3

2 x − e x + 3x 2 dx

z z z

2 x dx − e x dx + 3x 2 dx 2x 3x 3 − ex + +c 3 log 2 2x − e x + x 3 + c. log 2

Integration

8. Evaluate:

z

F x − 1I dx Hx K 2

2

z

F x − 1I Hx K

2

z z z z z

dx =

2

= = =

F x − 1 I dx Hx x K F 1 − 1 I dx Hx x K F 1 + 1 − 2 ⋅ 1 ⋅ 1 I dx Hx x x x K 2

2

2

2

2

2

4

2

z

1 1 1 dx + 4 dx − 2 3 dx x2 x x

x −2 +1 x −4 +1 x −3+1 + −2⋅ +c −2 + 1 −4 + 1 −3 + 1 x −1 x − 3 2 x − 2 + − +c −1 −3 −2 − Note: If

z af

Examples

af

f x dx = g x + c , then

z z z za

f

∴

1 e 3 x + 7 dx = e 3 x + 7 ⋅ + c 3 x 3 dx =

2. ∴

x 3+1 x4 +c= +c 3+1 4

6 − 4xf

3

a6 − 4 x f dx =

WORKED EXAMPLES:

z

1 dx. 8x − 6

4

⋅

4

=

a

f

f ax + b dx = g ax + b ⋅

e x dx = e x + c

1.

1. Evaluate:

za

1 1 1 − 3 + 2 + c. x 3x x

a6 − 4 x f −16

4

1 +c −4

+ c.

1 + c where a and b are constants. a

459

460

Basic Mathematics

z 2. Evaluate:

zd

a f log a8 x − 6f = + c.

1 1 dx = log 8 x − 6 ⋅ + c 8x − 6 8

8

i

e 3 x − 76 x dx.

z z

= e 3 x dx − 76 x dx =

3. Evaluate:

4. Integrate Solution:

z

za

76 x 1 e 3x − ⋅ +c. 3 log 7 6

f

4 x − 6 7 dx

z

a 4 x − 6f

7

dx =

a 4 x − 6f

7+1

a4 x − 6f

8

=

7 +1

32

1 +c 4

+ c.

6 − 4 x with respect to x.

6 − 4 x dx =

za

6 − 4x

f

1 2

dx.

a6 − 4 x f =

1 +1 2

1 +1 2

a6 − 4 xf =−

3 2

3 2

a6 − 4xf =−

3 2

6

5. Integrate

⋅

1 e

8 x− 9

⋅

1 +c −4

⋅

1 +c 4

+c

with respect to x.

z

z

1 dx = e −a8 x −9 f dx e 8 x −9

Integration

z

= e −8 x + 9 dx =

e −8 x + 9 +c. −8

18.3 SUBSTITUTION METHOD:

z

To evaluate integrals of the type

af

f x

n

af

⋅ f ′ x dx, we put f (x) = t

Diff. w.r.t. x.

a f dxdt f ′a x f dx = dt. f′ x =

Substituting we get

z

af

f x

n

z

af

⋅ f ′ x dx = t n ⋅ dt

t n+1 + c provided n ≠ −1 n +1 if n = −1,

z z

t n dt = t −1dt = log t + c

z

∴

af

f x

n

af

⋅ f ′ x dx =

af

f x

n +1

n +1

+ c if n ≠ −1

af

log f x + c if n = −1 Note: To evaluate ∫ef(x)⋅f ′(x)dx, put f(x) = t and proceed the same way.

WORKED EXAMPLES: 1. Evaluate: Solution:

zd

z

dx

2

i a

f

4

+ x − 1 ⋅ 2 x + 1 dx

i a 4

f

x 2 + x − 1 ⋅ 2 x + 1 dx Put x2 + x + 1 = t diff. w.r.t. x.

2x + 1 =

dt dx

3

461

1 = a −m am

462

Basic Mathematics

zd

Substituting we get

a2 x + 1f dx = dt

i a2 x + 1f dx =

x2 + x + 1

4

=

z

t 4 ⋅ dt

t5 +c 5

dx = 2. Evaluate:

z

z

∴

dt dx

4 x 3 dx = dt

z

dt 4x3 dx = = log t + c 4 t x −9

d

i

= log x 4 − 9 + c. x 4x + 7 2

dx. Put 4x2 + 7 = t diff. w.r.t. x.

Solution:

8x =

dt dx

8x dx = dt 1 x dx = dt. 8 Substituting,

5

Put x4 − 9 = t diff. w.r.t. x.

4x3 =

z

i

+ x +1

4x3 dx. x4 − 9

Solution:

3. Evaluate:

2

z

x 4x + 7 2

dx =

z

1 dt 8t

1 = log t + c 8

5

+ c.

Integration

4. Evaluate:

z

d

9 dx x log x

Solution:

Put log x = t. diff. w.r.t. x.

1 dt = x dx dx = dt. x Substituting,

z

z

dt 9 dx = 9 ⋅ = 9 log t + c t x log x

a f

= 9 log log x + c. ex with respect to x. 5. Integrate 6 − 4e x

z

Solution:

ex dx 6 − 4e x

Put 6 − 4ex = t diff. w.r.t. x.

0 − 4e x =

dt dx

− 4e x dx = dt e x dx = Substituting,

6. Evaluate: Solution:

z

z

i

1 = log 4 x 2 + 7 + c. 8

ex dx = 6 − 4e x

z

dt −4

dt 1 = − log t + c −4t 4

d

i

1 − log 6 − 4e x + c. 4 3

e x ⋅ x 2 dx Put x3 = t

463

464

Basic Mathematics

diff. w.r.t. x.

3x 2 =

dt dx

3 x 2 dx = dt

x 2 dx =

z

∴

dt 3

z

3

e x x 2 dx = e t ⋅ dt = e t + c 3


z

= e x + c. e 4x

2

+8 x − 7

a x + 1f dx Put 4x2 + 8x − 7 = t diff. w.r.t. x.

8x + 8 =

dt dx

a8x + 8f dx = dt 8 a x + 1f dx = dt a x + 1f dx = dt8 Substituting,


z

z

e4x

2

+8x − 7

z

a f

⋅ x + 1 dx = e t ⋅

z

=

1 1 ⋅ e t dt = e t + c 8 8

=

1 4 x 2 +8x − 7 ⋅e +c. 8

2

xe 2 x dx Put 2x2 = t diff. w.r.t. x.

4x =

dt dx

4x dx = dt

dt 8

Integration

x dx = Substituting,

z

z

2

xe 2 x dx = e t ⋅

z

z

dt 1 t e dt = 4 4

1 t 1 2 e + c = e 2 x + c. 4 4

=

9. Evaluate:

dt 4

x e−1 + e x −1 dx xe + ex Put xe + ex = t diff. w.r.t. x.

Solution:

ex e −1 + e x =

LM N

e x e−1 +

LM N

e x e−1 +

dx Substituting

z

∴

e −1

dt dx

OP Q

ex dt = e dx

OP Q i dx = dte .

ex dx = dt e

+ e x −1

z af

dt x e−1 + e x −1 dx = e x e t x +e 1 = log t + c e

10. Evaluate:

Solution:

z

d

i

1 log x e + e x + c . e dx x − x log x

z

dx = x − x log x

za

dx x 1 − log x

Put 1 − logx = t

f

465

466

Basic Mathematics

diff. w.r.t. x.

−

1 dt = x dx

dx = − dt x Substituting

za

z

dx dt = − = − log t + c x 1 − log x t

f

a

f

= − log 1 − log x + c.

18.4 INTEGRATION BY PARTIAL FRACTION METHOD:

za

To evaluate integrals of the type

px + q or ax + b cx + d

fa

f

za

px + q , first resolve into partial ax + b 2 cx + d

fa

f

fractions. i.e., Express

or

a

px + q A B = + ax + b cx + d ax + b cx + d

a

px + q A B = + 2 ax + b ax + b ax + b x + d

fa

fa

f a f

f

a

f a 2

+

C cx + d

f

(where A, B and C are constants to be determined) and then integrate.

WORKED EXAMPLES: 1. Evaluate:

za

3x − 1 dx. x − 3 x +1

fa f

3x − 1 Consider x − 3 x + 1 ,

a fa f

Resolve into partial fractions

A B 3x − 1 = + x − 3 x +1 x − 3 x +1

a fa f a f 3x − 1 A a x + 1f + B a x − 3f = a x − 3fa x + 1f a x − 3fa x + 1f 3 x − 1 = A a x + 1f + B a x − 3f Put

x+1=0

...(1)

Integration

⇒

a f

x = −1

af a −4 = B a −4 f

467

f

3 −1 − 1 = A 0 + B − 1 − 3

⇒

B =1

a f a

f

3x − 1 = A x + 1 + B x − 3

x−3=0 x=3

Put

af

a f af 8 = A a4f

3 3 −1= A 3+1 + B 0

⇒

A = 2.

Substituting A = 2 and B = 1 in (1) we get

3x − 1

2

1

a x − 3fa x + 1f = x − 3 + x + 1 integrating with respect to x.

za

3x − 1 dx = x − 3 x +1

fa f

z

2 dx + x−3

z

1 dx. x +1

a f a f = log ea x − 3f a x + 1fj + c.

= 2 log x − 3 + log x + 1 + c. 2

2. Evaluate:

z

4x + 6 dx. x2 − 1

Consider

4x + 6 4x + 6 = 2 x −1 x +1 x −1

a fa f

Resolve into partial fractions.

4x + 6

⇒ Put

A

B

a x − 1fax + 1f = a x − 1f + a x + 1f . 4 x + 6 = A a x + 1f + B a x − 1f x+1=0 x = −1

...(1)

468

Basic Mathematics

a f

af a f −4 + 6 = B a −2 f 2 = B a −2 f ⇒ B = −1

4 −1 + 6 = A 0 + B − 1 − 1

Now

a f a f

4x + 6 = A x + 1 + B x − 1

x−1=0

Put

af

x =1

a f af 10 = A a2 f

4 1 + 6 = A 1+1 + B 0

⇒

A=5

Substituting A = 5 and B = −1 in (1)

−1 4x + 6 5 = + x −1 x +1 x −1 x +1

a fa f integrating,

za

4x + 6 dx = x −1 x +1

fa f

z

=5

3. Evaluate:

za

za

5 dx + x −1

z

z

1 dx − x −1

a f

−1 dx x +1

z

−1 dx x +1

a f

4x + 6 dx = 5 log x − 1 − log x + 1 + c. x −1 x +1

fa f

x2 + 2 dx . x + 1 2x + 3 4x − 1

fa

fa

f

Solution: Consider

x2 + 1 x + 1 2x + 3 4 x − 1

a fa

fa

f

Resolve into partial fractions,

x2 + 1 A B C = + + x + 1 2x + 3 4x − 1 x +1 2x + 3 4x −1

a fa

fa

f

...(1)

Integration

a

⇒

fa

f a fa

f a fa

Put x + 1 = 1 x = −1

a−1f

b a f gb a f g a f a f 2 = A a1fa −5f

+ 1 = A 2 −1 + 3 4 −1 − 1 + B 0 + C 0

2

A=

⇒

−2 5

Put 2x + 3 = 0 2 x = −3

x=

−3 2

a f FH −23 + 1IK FGH 4 FH −23IK − 1IJK + C a0f 9 −3 + 2 I F −12 − 2 I + 1 = BF H 4 2 KH 2 K 13 14 = BF I H 4K 4

F −3I H2K

2

+1 = A 0 + B

B=

⇒

13 14

Put 4x − 1 = 0 4x = 1 x = 1/4

F 1I H 4K

a f a f FH 14 + 1IK FH 2 ⋅ 14 + 3IK 1 5 14 +1 = CF I F I H 4K H 4 K 16 17 70 = CF I H 16 16 K

⇒

f

x2 + 1 = A 2x + 3 4x − 1 + B x + 1 4x − 1 + C x + 1 2 x + 3

2

+1= A 0 + B 0 + C

C=

17 70

469

470

Basic Mathematics

Substituting the values of A, B and C in (1)

2 − 13 14 17 70 x2 + 1 = 5 + + x + 1 2x + 3 4x − 1 x +1 2x + 3 4x −1

a fa

fa

f

integrating w.r.t. x.

za

x2 + 1 dx = x + 1 2 x + 3 4x − 1

fa

fa

f

=

4. Evaluate:

za

z

z

z

13 14 17 70 −2 5 dx + dx + dx 2x + 3 4x − 1 x +1

a

a f

f

2x − 3 dx x x +3 x −2

fa

f

Consider

2x − 3 , Resolve into partial fractions. x x +3 x −2

a fa

f

A B C 2x − 3 = + + x x+3 x−2 x x +3 x −2

a fa

f

a fa

f a fa f a fa f fa f a fa f 2 x − 3 = A a x + 3fa x − 2 f + Bx a x − 2 f + Cx a x + 3f

2x − 3 A x +3 x−2 + B x x−2 +C x x+3 = x x+3 x−2 x x+3 x−2

a

⇒ Put x + 3 = 0 x = −3

a f

a f a fa −6 − 3 = B a −3fa −5f −9 = B a15f

f af

2 −3 − 3 = A 0 + B −3 −3 − 2 + C 0

B=

⇒ Put x − 2 = 0 x=2

af

−9 15

a

f

13 log 2 x + 3 17 70 log 4 x − 1 −2 log x + 1 + + +c 5 28 4

⇒ B=

−3 5

a f a f a fa 4 − 3 = C a2 fa5f

f

2 2 −3= A 0 + B 0 +C 2 2+3

Integration

471

a f

1 = C 10

C=

⇒ Put x = 0,

af

1 10

a fa −3 = A a −6f

f af af

2 0 −3= A 0+3 0−2 + B 0 +C 0

A=

⇒ Substituting A =

−3 ⇒ −6

A=

1 2

af

1 1 −3 , B= and C = in 1 2 5 10

2x − 3 1 2 −3 5 1 10 + = + x x+3 x−2 x x+3 x−2

a fa

f

integrating with respect to x,

za

2x − 3 dx = x x+3 x−2

fa

f

= 5. Evaluate:

za

z z 1 2 dx − x

3 5 dx + x+3

z

1 10 dx x−2

a f

a

f

1 1 3 log x − log x + 3 + log x − 2 + c 2 5 10

3x + 4 x − 2 2 x +1

fa f

3x + 4 Consider x − 2 2 x + 1

a

fa f

Resolve into partial fractions:

a

A B 3x + 4 = + 2 x−2 x − 2 x +1 x−2

+

C x +1

fa f a f a

f

fa f a

Multiplying by (x − 2)2 (x + 1)

a

f a

f

2

3x + 4 = A x − 2 x + 1 + B x + 1 + C x − 2 Put x − 2 = 0 x=2

af

af a f af 10 6 + 4 = B a3f ⇒ B = 3

3 2 + 4 = A 0 + B 2 +1 + C 0

...(1)

2

472

Basic Mathematics

Put x + 1 = 0 x = −1

a f

af af a 1 1 = C a 9f ⇒ C = 9

3 −1 + 4 = A 0 + B 0 + C −1 − 2

Put x = 0, B =

f

2

10 1 and C = 3 9

a fa f

4 = A −2 1 + 4 = −2 A + 4−

af a f

10 1 1 + −2 3 9

2

10 4 + 3 9

10 4 − = 2A 3 9

36 − 30 − 4 = 2A 9 +

⇒ Substituting A =

2 1 = 2A ⇒ A = 9 9

af

1 10 1 , B= and C = in 1 9 3 9

1 1 10 3 9 = + + 9 2 − 2 x x +1 x +1 x−2

3x + 4

a x − 2f a f 2

integrating we get,

za

3x + 4 dx = x − 2 2 x +1

fa f

=

a

za

19 dx + x−2

a

f

a

f a

f

za

−2 +1

1 10 x − 2 log x − 2 + ⋅ 9 3 −2 + 1

=

6. Evaluate:

f

z

a

f

10 3 dx + x−2 2

f

+

z

a f

1 log x + 1 + c 9

a f

10 1 1 log x − 2 − + log x + 1 + c. 3 x−2 9 9

4x + 8 2x − 1 2 x + 3

fa f

19 dx x +1

Integration

473

Consider

a

4x + 8 2 x − 1 2 x + 3 Resolve into partial fractions

fa f

a

A B C 4x + 8 = + 2 2 + x +3 2x − 1 2x − 1 2x −1 x + 3

fa f

Multiplying by (2x − 1)2 (x + 3)

a

a

f

fa f a f a

...(1)

f

4x + 8 = A 2 x − 1 x + 3 + B x + 3 + C 2 x − 1 Put 2x − 1 = 0 2x = 1 x = 1/2.

4

F 1 I + 8 = A a0f + B F 1 + 3I + C a0f H 2K H2 K 7 20 10 = B F I ⇒ B = H 2K 7

Put x + 3 = 0 x = −3

a f

af af ba f g −12 + 8 = C a−7f −4 −4 = C a 49f ⇒ C = 49

4 −3 + 8 = A 0 + B 0 + C 2 −3 − 1 2

Substituting B =

20 −4 ,C= and x = 0 7 49

we get

a f a fa f

4 0 + 8 = A −1 3 +

af

32 = −3 A +

60 4 − 7 49

3 A = −32 +

60 4 − 7 49

3A =

a f

20 4 3 − −1 2 7 49

−1568 + 420 − 4 49

2

2

474

Basic Mathematics

3A = A=

−1152 49 384 −1152 =− 49 × 3 49

Substituting in (1) −384 20 49 7 = + − 2 1 x 2x − 1 x+3

4x + 8

a2 x − 1f a 2

f

a

f

za

20 7 2x −1

2

−4 + 49 x+3

integrating with respect to x.

za

4x + 8

f a x + 3f =

2x − 1

2

z

−384 49 dx + 2x − 1

f

a

f

a

f

2

dx +

a

=

−384 log 2 x − 1 20 4 ⋅ − − log x + 3 + c . 49 2 14 2 x − 1 49

a

a f

d dv du uv = u ⋅ +v⋅ dx dx dx

z z z z z z FH IK a f

d dv du uv = u dx + v dx dx dx dx uv = u dv + v du.

⇒

udv = uv −

⇒

a f

−384 log 2 x − 1 20 2 x − 1 −2 +1 1 −4 ⋅ + ⋅ ⋅ + ⋅ log x + 3 + c 49 2 7 2 49 −2 + 1

From product rule in differentiation we have

z

f

=

18.5 INTEGRATION BY PARTS:

Integrating w.r.t. x. we get

z

−4 49 dx x+3

By taking u as first function,

dv as second function du as differential of 1st function v as the integral of the 2nd function we get

v

du dx dx

f

a

f

Integration

za

fa

f

a

I function II function dx = I function

z

zz

f aII functionf dx − LMN

II function ⋅

a

475

fOPQ

d I function dx dx

This is known as integration by parts. The success of the method of integration by parts depends on choice of 1st function and 2nd function. Logarithmic, algebraic and exponential functions should be taken in the same order of priority. [LIATE: log, Inverse trig., Algebraic, Trigonometric and Exponential] For example, to integrate xex, x is first function and ex is 2nd function where as to integrate xlogx, logx is first function and x is 2nd function.

WORKED EXAMPLES: 1. Evaluate:

z

x e x dx.

z

z z FH z

x e x dx = x ⋅ e x − I II

ex ⋅

I K

d x dx dx

= xe x − e x ⋅ 1 dx

2. Evaluate:

z

xe x − e x + c. x log x dx .

z z z z LMN z FGH z

x log x dx = log x⋅ x dx

log x x dx −

= log x ⋅

I

II

a fOP Q x 1I ⋅ J dx 2 xK

x2 d log x dx. ⋅ 2 dx

x2 − 2

2

log x ⋅

x2 x dx − 2 2

log x ⋅

x2 1 x2 − ⋅ +c 2 2 2

log x ⋅

x2 x2 − + c. 2 4

476

Basic Mathematics

3. Evaluate:

z

log x dx.

z

z

log x dx = log x⋅ 1 dx II

I

z z LMN z z

log x ⋅ 1 dx −

x⋅

log x ⋅ x − x ⋅

1 dx x

a fOPQ

d log x dx dx

log x ⋅ x − 1 dx

4. Evaluate: Put

z

log x ⋅ x − x + c. 2

x 3 e x dx.

x2 = t. diff. w.r.t. x. 2x =

dt dx

2x dx = dt

x dx = Substituting we get

z

z

2

dt 2

2

x 3 e x dx = x 2 e x ⋅ x dx =

LM N

z z FH z

1 t et − 2

z

1 t e t ⋅ dt 2 I II et ⋅

1 t te − e t ⋅ dt 2 1 t te − e t + c 2 2

2

I OP K Q

d t dt dt

x 2e x − e x = + c. 2

Integration

5. Evaluate:

z

x n log x dx where n ≠ −1

z

log x⋅ x n dx II

I

z z

LM x d alog xfOP dx N n + 1 dx Q L x ⋅ 1 OP dx x = log x ⋅ − M n +1 Nn +1 xQ n +1

log x ⋅ x n dx −

z

n +1

log x ⋅ log x ⋅

LM N

OP Q

x n +1 x n +1 − + c. n +1 n +1 2

18.6 INTEGRALS OF THE TYPE:

z

z

x n +1 xn dx − n +1 n +1

x n +1 1 x n +1 − + c if n ≠ −1 n +1 n +1 n +1

= log x ⋅

Consider

n +1

z

a f

af af

e x f x + f ′ x dx

af af e ⋅ f a x f dx + e ⋅ f ′a x f dx

e x f x + f ′ x dx

z

=

x

I

II

z

x

z z RST LMN a fOPQUVW z af z af z af af

f x ⋅ e x dx −

ex

d f x dx

f x ⋅ e x − e x ⋅ f ′ x dx + e x ⋅ f ′ x dx ∴

af

= f x ⋅ ex

z

∴

z

ex

LM 1 − 1 OP dx MN x + 1 a1 + xf PQ 2

af

e x f x + f ′ x dx = e x ⋅ f x + c.

Examples:

1.

af af

af

dx + e x f ′ x dx

477

478

Basic Mathematics

ex ⋅

F H

−1 I K a f = a1 + x f LM 1 + log x OP dx Nx Q

1 1 d =− dx x + 1 x +1

Since

2.

1 +c x +1

z

ex

a1 + x log x f dx = x

z

ex

2

= e x ⋅ log x + c Since

3.

z

a f

d 1 log x = . dx x

LM a f 1 OP dx = e ⋅ log alog xf + c x log x Q N d d 1 log alog x f = ⋅ alog x f Since dx log x dx e x log log x +

x

1 1 1 . ⋅ = log x x x log x

=

4.

z

a f

e x x + 1 dx = e x ⋅ x + c

af

d x = 1. dx

3 5.

z

ex

z LMN z LMN

FG 2 x + 1IJ dx = H2 xK

ex

OP Q

2x 1 + dx 2 x 2 x

= ex

x+

1 2 x

OP dx Q

= ex ⋅ x + c Since

REMEMBER: •

z

x n dx =

x n +1 +c n +1

Provided n ≠ −1

d dx

d x i = 2 1x .

2

.

Integration

• if n = −1

z z z z z z af af z af z af z af af za f a f z af af af x −1dx =

•

• • •

1 dx = log x + c. x

e x dx = e x + c. a x dx =

ax +c log a

k dx = kx + c

k f x ± g x dx = k

f x dx ± k g x dx

f x dx = g x + c, then

• If

f ax + b dx =

f x

• To evaluate

af

f x

g ax + b +c a n

⋅ f ′ x dx , put f x = t and proceed to get the answer.

n +1

n +1

+ c for n ≠ −1

and log [f (x)] + c for n = −1 • To evaluate ∫ef(x) ⋅ f ′(x) dx, put f (x) = t and proceed to get e f(x) + c as answer. • To evaluate integrals of the type

za

px + q dx or ax + b cx + d

fa

f

za

px + q ax + b 2 cx + d

fa

f

First resolve into partial fractions

px + q

A

B

aax + bfacx + d f = ax + b + cx + d or

a

px + q A B = + 2 ax + b ax + b cx + d ax + b

fa

f a

Find A, B, C, then finally integrate. •

za

fa

f

I function II function dx

f a

f a 2

+

C cx + d

f

479

480

Basic Mathematics

z

= I function II function −

z LMNz

II function

fOPQ

a

d I function dx dx

I function and II function are kept by making use of LIATE rule. •

z

af af

af

e x f x + f ′ x dx = e x ⋅ f x + c.

EXERCISE I.

Ev alua te: Evalua aluate: 1.

2.

3. 4. 5. 6.

7.

8. 9.

10.

11.

12.

z FH z FH z FH zd zd ze z FGH z FH za F z GH z LMMNa z

x3 +

I K

4 dx x

13.

I K

4x 2 −

5 8 dx + x2 x

14.

ax 2 +

b c + + d dx x2 x

I K

15. 16.

i

x e + e x − e x dx π

x + x −π +e e

e

π

i dx

j

18.

I JK

6 x 4 + 8x − 9 dx x

I d x − 8i dx K 6 x + 4f dx 4 x − 9 x + 1I JK dx x

19.

20.

2

3

21.

2

1 4x + 7

f

2

a

22.

OP PQ

23.

dx

24.

1 dx − 8x + 7

1 1 + 3 − 4 x 8 − 5x

2x + 7 dx x + 7x − 9 2

fd

f

3

i

2

2 x − 9 x 2 − 9 x + 6 dx

d i + bx + ci a2 ax + bf dx

x 3 + 6 x − 1 ⋅ x 2 + 2 dx

ax 2

1 dx x log x x2

6 x − 6 x 2 dx

1 x+ x

17.

z za z zd z zd z z z aa z z z

2 − 7x3

i

4

dx

2

2 x e x dx e x + e− x dx e x − e− x

f

log 7 x − 6 dx 7x − 6

f

1 6 − log x dx x

6 log x dx x x +6 dx x

Integration

25.

26. 27. 28.

29.

30.

z z z z z a fa f z z z za f LM OP z MN a f PQ z z z z LMN a f z LMN a f

31.

1 dx x − 2x − 8

32.

2x + 1 dx x + 6x + 8

33.

3x + 7 dx 3x − 7x − 6

34.

2

2

2

2x + 1 x +1 x + 2

e4x

35.

2

e2x dx + 4e 2 x + 3

36.

2

37. 38.

za f a z z za f z z e x

1 dx x + 7 x + 12 2

x −1

f dx

x 4 e −2 x dx x x −1

20

dx

x 6 log x dx

j

log x + x 2 + a 2 dx

Hint: x 5 = x 3 ⋅ x 2 Put x 3 = t

x e−

Hint: Put x = t 2

dx

x+2

x 2 e x dx

x 5 e x dx x

2

xe x

39.

40.

41. 42. 43.

44.

45.

x +1 2

ex

x2 + 1 x +1 2

[Hint: Hint: Add and Subtract 1 in the Numerator and simplify]

log x − 1 dx 1− x dx 1+ x

2x + 3 dx 4x − 7

e x log log x +

OP Q

1 dx x log x

OP Q

log log x + e log x dx x

481

482

Basic Mathematics

ANSWERS 1.

x4 + 4 log x + c 4

15.

d

x π +1 x e+1 + − πe x + eπ x + c π +1 e +1 4

3x + 8 x − 9 log x + c 2

10.

a6 x + 4f 5

1 + 2x 2

+c

a

f

−1 log 8 x + 7 − +c 11. 4 4x + 7 8

a

12.

f

a

f

log 3 − 4 x 1 + −4 10 8 − 5x

d

i

a

2 13. log x + 7 x − 9 + c

14.

dx

2

− 9x + 6 2

2

i

2

+c

a

f

log 7 x − 6

2

f

2

+c

+c

14

−2 6 − log x 22. 3

+c

3 2 − 6x

i

+c

3

19. e x + c

4

24 5 2 8x

d

21.

x 4 7x 2 − − 8 log x + c 8. 4 2 9.

1 18. 63 2 − 7 x 3

−x x 20. log e − e + c

3x 3 − +c 4

4

7.

+c

2

17. log log x + c

x e+1 + ex − ex + c 4. e +1

6.

+c

2

ax 2 + bx + c

16.

ax 3 b − + c log x + dx + k 3. 3 x

3 2 4x

3 2

9

4x3 5 + + 8 log x + c 2. x 3

5.

i

2 x3 + 6x − 1

23.

6 log x +c log 6

24.

d

i

x +6

2

3 2

+c

+c

F x + 3I + c H x + 4K x − 2I 1 +c log F H x + 4K 6 7 3 log a x + 4f − log a x + 2f + c 2 2

25. log

26.

27.

a

f

a

f

5 16 28. − log 3x + 2 + log x − 3 + c 3 11

Integration

a f

a

f

1 1 3 log x + 1 − log x − 2 − +c 29. x−2 3 3

− 38. e

I JK

F GH

e2 x + 1 1 log +c 30. 4 e2 x + 3 31. −

a f

a f

a

32. x 2 e x − 2 xe x + e x + c

LM − x − x + 3x 2 N2 a x − 1f − ax − 1f x 4

2

3

21

34.

21

f

−

OP Q

3x 3 +c − 2 4

3

OP PQ

+ 3x + 6 x + 6 + c

462

+c

LM x − 1OP + c N x + 1Q 1 1 1 x log a x − 1f − x − log a x − 1f + c 2 2 2 a f

42. 2 log 1 + x − x + c

22

43.

LM N

F H

1 13 7 x + log x − 2 4 4

I OP + c KQ

a f

44. e x log log x + c

j

2 2 2 2 36. x log x + x + a − x + a + c

ex x3 − 1 + c 37. 3

3 2

x 40. e

41.

x7 x6 log x − +c 35. 7 42

e

LMx MN

ex +c 39. 1+ x

1 2 2 x − 1 + log x − 1 − log x + 2 + c 3 9 9

−2 x 33. e

x

a f

45. log x ⋅ log log x − log x +

x2 + c. 2

483

484

Basic Mathematics

19 Definite Integrals 19.1 INTRODUCTION: If

z z af af

af

f x dx = g x + c. Then

z

af

b

af

f x dx = g x

a

b

af af

=g b −g a.

a

z af b

f x dx is called indefinite integral and

f x dx [Read as integral from a to b f (x)dx] is called

a

definite integral of f (x) from a to b. Here a is called lo low limit. wer limit and b is called upper limit To evaluate definite integral, integrate the given function as usual. In the final answer substitute the upper limit value for x − lower limit value for x.

WORKED EXAMPLES: I. Ev alua te: Evalua aluate:

z z z FH 2

1.

x 2 dx =

1

x3 3

1

2.

e x dx = e x

1

0

=

1

1

2 3 13 8 1 7 − = − = . 3 3 3 3 3

= e1 − e 0 = e − 1.

0

0

3.

2

x+

1 x

z

I dx = F x K H 1

2

0

=

2

+

I K

1 + 2 dx x2

LM x + x + 2 xOP N 3 −2 + 1 Q 3

−2 +1

1 0

Definite Integrals

Lx 1 O = M − + 2 xP N3 x Q LM1 − 1 + 2 a1fOP − LM 0 − 1 + 2 a0fOP N3 1 Q N 3 0 Q 1

3

0

3

3

1 −1+ 2 − 0 3

zd

1 4 +1= . 3 3

2

4.

i

1 + 2 x dx

1

za

f

a1 + 2 xf

3 2

2

1 + 2x

1 2

dx

a1 + 2 xf =

1 +1 2

1

3 2

a1 + 2 x f

1

b1 + 2 a2fg

3 2

−

3 3 2

3 =

z

3 52

af

1+ 2 1

−

3 2

3

1+ 2 − 3

3 2 −3

3

1

3 2 2

3

a1 + 4f

2

1 2

⋅

3 2

5 5 −3 3 . 3

1

5.

x e x dx

−1

LM x ⋅ N

1 +1 2

z z

ex − ex ⋅

d x dx

OP Q

1 −1

⋅

1 2

2 1

485

486

Basic Mathematics 1

xe x − e x

−1

1e1 − e1 − −1e −1 − e −1 e − e − − e −1 − e −1 0 − −2e −1 2 . e

= 2 e −1 =

z e

6.

1

x log x dx II

I

z z z z

LMlog x xdx − L x ⋅ d alog xfO dx OP MN 2 dx PQ PQ MN LMlog x ⋅ x − x ⋅ 1 dxOP N 2 2 x Q LMlog x ⋅ x − x dx OP N 2 2 Q LMlog x ⋅ x − 1 ⋅ x OP N 2 2 2Q 2

e

1

2

e

2

1

e

2

1

2 e

2

1

x2 x2 log x ⋅ − 2 4

e

1

LMlog e ⋅ e − e OP − LMlog1 ⋅ 1 − 1 OP N 2 4Q N 2 4Q LM1 e − e OP − L0 − 1 O N 2 4 Q MN 4 PQ 2

2

2

2

2

2

e 2 e 2 1 2e 2 − e 2 + 1 e 2 + 1 . − + = = 2 4 4 4 4

Definite Integrals

z 1

7.

0

x+2 dx. x + 4x + 8 2

Put x2 + 4x + 8 = t diff. w.r.t. x.

2x + 4 =

dt dx

a2 x + 4f dx = dt 2 a x + 2 f dx = dt a x + 2f dx = dt2 . when x = 0,

x2 + 4x + 8 = t

af

0+4 0 +8=t

⇒ when x = 1,

t=8

x2 + 4x + 8 = t

af

1+ 4 1 + 8 = t

⇒

z 1

∴

0

t = 13

x+2 = 2 x + 4x + 8

1 log13 − log 8 2

F I H K

1 13 log . 2 8

z 2

8.

x 1 + x 2 dx

1

Put 1 + x2 = t diff. w.r.t. x.

2x =

dt dx

2x dx = dt

z

13

8

13

dt 1 = log t 2t 2 8

487

488

Basic Mathematics

x dx = when x = 1,

dt 2

1 + x2 = t 1+1 = t

⇒ t=2 when x = 2,

1 + 22 = t

⇒ t=5 Substituting,

z 2

z 5

x 1 + x dx = 2

1

t dt 2

2

5

1 t 1 2 +1 = 2 1 2 +1 2 5

t3 2 2t 3 2 = 3×2 2 3

5

2

1 32 1 5 − 23 2 = 5 ⋅ 5 − 2 2 3 3 =

z 1

9.

0

5 5−2 2 . 3

1− x dx 1+ x

za f z LM z MNz 1

−

0

x −1 dx x +1

1

=−

0

x +1−1−1 dx = − x +1

1

−

1

1 dx − 2

0

0

1 dx x +1

a f

− x − 2 log x + 1

1 0

OP PQ

LM MN

z 1

0

x +1 dx − x +1

z 1

0

2 dx x +1

OP PQ

Definite Integrals

a f a fr m − m 1 − 2 log 2 − 0 − 2 log 1 r − 1 − 2 log 1 + 1 − 0 − 2 log 0 + 1

−1 + 2 log 2 − 0 = 2 log 2 − 1.

z 3

10.

2

2x + 3 dx 5x − 7

z 3

2

2x + 3 dx = 5x − 7

2 = 5

2 5

LM MM N

z 3

2

2 5

z 3

2

x−

z 3

2

F 3I H 2 K dx 7 5F x − I H 5K

2 x+

7 7 3 + + 5 5 2 dx 7 x− 5

7 5 dx + 7 x− 5

z

LM MM 1 dx + N

29 10 dx 7 x− 5

x−

z z 3

3

2

2

3

2

7 3 + 5 2 dx 7 x− 5

OP PP Q

OP PP Q

LM F I OP H KQ N 7 OU 29 7 O L 29 2 RL 3+ log F 3 − I P − M2 + log F 2 − I P V S M H K H 5K QW 10 5 Q N 5 T N 10 2R S3 + 29 log FH 85IK − 2 − 1029 log FH 35 IK UVW 5 T 10 2 29 7 x+ log x − 5 10 5

3

2

489

490

Basic Mathematics

LM MM MN

F GG GH

8 2 29 1 + log 5 3 5 10 5

LM N

F H

2 29 8 log 1+ 5 10 3

I OP JJ P JK PPQ

I OP KQ

19.2 PROPERTIES OF DEFINITE INTEGRALS:

z z af z af z af z af b

1.

af

z

a

z af b

z z af za f za f c

f x dx =

a

af

b

f x dx +

a

a

f x dx where a < c < b.

c

a

f x dx =

0

f a − x dx

0

b

5.

a

a

b

4.

f z dz

f x dx = − f x dx

a

3.

z af b

f t dt =

a

b

2.

af

b

f x dx =

b

f x dx =

a

f a + b − x dx

a

WORKED EXAMPLES:

za za 1

1. Evaluate:

f

x x − 1 dx 5

0

1

Let

f

I = x x − 1 5 dx 0

using property (4),

za 1

I=

fa

f

1 − x 1 − x − 1 dx

0

5

Definite Integrals

za zd

fa f

1

I=

1− x −x

5

dx

0

1

I=

i

− x 5 + x 6 dx

0

1

− x 5+1 x 6 +1 I= + 5 +1 6 +1 0 −x 6 x 7 + I= 6 7

I=− I=

za 1

f

x x −1

∴

za a

2. Evaluate:

x−a

f

4

−7 + 6 42 1 42

=−

1 . 42

0

0

1 1 + −0 6 7

I=−

5

1

⋅ x dx

0

za a

I=

Let

x−a

f

4

⋅ x dx

0

using property (4),

za f a za f a f za f a

I=

a−x−a

4

a

I=

−x

4

a − x dx

0

a

I=

x 4 a − x dx

0

f

⋅ a − x dx

0

491

492

Basic Mathematics

z a

I=

x 4 ⋅ a − x 5 dx

0

x5 x6 ⋅a− I= 5 6

a

0

I=

a 5 ⋅ a a6 − −0 5 6

I=

a 6 a 6 6 a 6 − 5a 6 a 6 − = = 5 6 30 30

z a

∴

a x − af

4

x dx =

0

z z a

3. Evaluate:

0

a

Let

I=

0

a6 . 30

x dx. a−x x dx a− x

using property (4)

z a f z z z FH IK z z a

I=

0

a

=

0

a

I=

0

a

=

0

a−x dx a− a− x

a− x dx = a−a+x a x − dx x x a

a dx − 1 dx x 0

= a log x − x

a 0

a log a − a − 0 = a log a − a.

a

0

a− x x

Definite Integrals

Alieter:

z a

0

z a

−x x dx = − a− x a−x 0

z LMN OPQ LM z MNz LM z MNz a

−x − a + a dx a−x

=−

0

a

=−

0

a−x + a−x

a

−

a

1 ⋅ dx −

0

0

a

0

−a dx a−x

a dx a− x

OP PQ

OP PQ

aa − xf OP −1 Q − x + a log a a − x f − a + a log a a − af − l0 + a log a a − 0fq LM N

a

− x − a log

0

a 0

− a + a log 0 − a log a − a + a log a = a log a − a.

z 4

4. Evaluate:

0

z 4

Let I

0

x+4 dx x−4

x+4 dx using property (4) x−4

z z 4

I=

0

4

I=

0

4−x+4 4− x−4

8− x dx = − −x

LM MN

z z 4

0

8 dx − x

4

0

x dx x

OP PQ

493

494

Basic Mathematics 4

− 8 log x − x

a

f

0

− 8 log 4 − 4 = 4 − 8 log 4. Alieter:

z 4

0

x+4 dx x−4

z z 4

=

0

4

=

0

x+4−4+4 dx x−4 x−4 dx + x−4

z z 4

4

x −8

0

8 dx x−4

8 dx 4−x

1 dx −

0

z 4

0

a

log 4 − x −1

a

f

4 0

f

a

f

4 + 8 log 4 − 4 − 0 + 8 log 4 − 0 4 − 0 − 0 − 8 log 4 4 − 8 log 4.

19.3 APPLICATION OF DEFINITE INTEGRALS TO FIND AREA: y

One of the application of definite integrals is to find area. (i) Area enclosed by the curve y = f (x), the xaxis and the lines x = a and x = b is given by

y = f (x)

z z af b

b

A = y dx = a

f x dx.

a

0

WORKED EXAMPLES:

x=a

x=b

Fig. 19.1

1. Find the area bounded by the curve y = x2 − x, x-axis and the ordinates x = 1 and x = 2.

x

Definite Integrals

z

495

b

Solution: Area = y dx a

zd 2

A=

i

x 2 − x dx

1

x3 x2 − A= 3 2

A= = Area =

2

1

F 2 − 2 I − F1 − 1 I GH 3 2 JK GH 3 2 JK 3

2

3

2

8 1 1 16 − 12 − 2 + 3 −2− + = 3 3 2 6 5 square units. 6

2. Find the area bounded by the curve y = 3x2 − 8 with x-axis and ordinates x = 0 and x = 3.

z zd b

Area =

Solution:

y dx

a

3

=

i

3 x 2 − 8 dx

0

3

=3

x3 − 8x 3 0

= x 3 − 8x

3 0

af

33 − 8 3 − 0 27 − 24 = 3 Sq. units.

3. Find the area bounded by the parabola x2 = 4y, x-axis and x = 1 and x = 2.

z b

Solution: A = y dx a

z 2

A=

1

x2 dx 4

Q x 2 = 4 y, 4 y = x 2 , y =

x2 4

496

Basic Mathematics

1 x3 A= ⋅ 4 3

A=

2

1

LM N

OP L O Q MN PQ

1 2 3 − 13 1 8 −1 7 sq. units. = = 4 3 4 3 12

4. Find the area bounded by the x-axis and the curve y = x2 − 5x + 6 Solution. On X-axis, y = 0.

∴ x 2 − 5x + 6 = 0 x 2 − 3x − 2 x + 6 = 0

a f a f a x − 2fa x − 3f = 0

x x−3 −2 x−3 = 0

x = 2 or x = 3.

z 3

2 So Area bounded by the x-axis and the curve y = x − 5 x + 6 = y dx 2

zd 3

=

i

x 2 − 5 x + 6 dx

2

3

=

x3 x2 − 5⋅ + 6x 3 2 2

a f

33 − 2 3 32 − 2 2 − 5⋅ + 6⋅ 3− 2 3 2

F H

I af K

27 − 8 9−4 +6 1 −5 3 2

19 25 38 − 75 + 36 1 − +6= =− 3 2 6 6 Q

Area is +ve, A =

1 sq. units. 6

5. Find the area bounded by the curve y = x2 − x with x-axis. Solution: On X-axis, y = 0.

x2 − x = 0

Definite Integrals

497

a f

x x −1 = 0 x = 0 or x − 1 = 0 x = 0 or x = 1

So Area bounded by the curve y =

x2

− x with x-axis

z zd 1

b

= y dx =

i

x 2 − x dx

0

a

x3 x2 = − 3 2

1

= 0

1 1 − −0 3 2

2 − 3 −1 = 6 6 Area is non-negative,

Q Required Area =

1 sq. units. 6

6. Find the area between the parabolas y2 = x and x2 = y. Solution:

y

2

x =y

Point of intersection: y 2 = x (1, 1)

x2 = y

2

y =x

Squaring,

x 4 = y2 x4 = x

dQ

y2 = x

x

0 (0, 0)

i

x4 − x = 0

d

i

x x3 − 1 = 0

Fig. 19.2

x = 0 or x = 1

z b

Area bounded by the curve y2 = x, x-axis, x = 0 and x = 1 is

y dx .

a

z 1

=

0

x dx

Q y2 = x ∴ y= x

498

Basic Mathematics

=

1 +1 x2

1

1 +1 2 0 x3 2 32

1

0

13 2 2 −0= 32 3

z b

Area bounded by the curve

x2

= y, x-axis x = 0 and x = 1 is

y = x2

y dx .

a

z 1

= x 2 dx 0

=

x2

x3 3

1

0

1 = . 3

Area between the parabolas = Area bounded by the parabola y2 = x − Area bounded by the parabola = y with X-axis, x = 0 and x = 1.

2 1 1 − = sq units. 3 3 3 ∴

Reqd. area =

1 sq. units. 3

7. Find the area between the parabola y2 = 4x and the line y = x. Point of intersection:

y

y2 = 4x and y = x

y=

y2 = x 2 4x = x

2

y = 4x (4, 4)

2

x2 − 4x = 0

a

x

x

(0, 0)

f

x x−4 =0 x = 0 or x = 4. Fig. 19.3

Definite Integrals

499

z b

Area bounded by the parabola

y2

= 4x, the lines x = 0, x = 4 and X-axis = y dx . a

z

y2 = 4x y = 4x = 2 x

4

= 2 x dx 0

2⋅

2⋅

1 +1 x2

4

1 +1 2 0 3 x2

3 2

4 3 4

2×2 2 = ⋅x 3

0

0

d i

4 ⋅ 43 2 − 0 3 4 32 ⋅8 = . 3 5

z b

Area bounded by the line y = x, x = 0, x = 4 and x-axis = y dx a

z 4

= x dx = 0

x2 2

4

= 0

4 2 16 = = 8. 2 2

Required area = Area bounded by the parabola y2 = x − Area bounded by the line x =y with x-axis and lines x = 0 and x = 4

=

32 −8 3

=

32 − 24 8 = sq. units. 3 3

8. Find the area between the curves x2 = 5y and y = 2x. Solution: Required area = A1 ~ A2. Now: x2 = 5y and y = 2x.

a f

x2 = 5 2x

500

Basic Mathematics

x2 = 5y

y=

2x

A

A2

A1

0

B

Fig. 19.4

x 2 = 10 x x 2 − 10 x = 0

a

f

x x − 10 = 0 x = 0 or x = 10.

af

a f

y = 2 x ⇒ y = 2 0 or y = 2 10

y = 0 or y = 20. ∴ Point of intersection: (0, 0) and (10, 20). A1 = Area bounded by x2 = 5y, x-axis and lines x = 0 and x = 10

z z

10

b

y dx =

0

a

Q 5y = x 2 x2 y= 5

x2 dx 5

1 x3 = 5 3

10

0

1 1000 200 10 3 − 0 3 = = 15 15 3

A2 = Area bounded by y = 2x, x-axis and lines x = 0 and x = 10.

z z b

10

∴ = y dx = 2 x dx = a

0

2x 2 2

10

= 10 2 = 100. 0

Required area = A1 ∼ A2.

=

200 200 ~ 300 100 ~ 100 = = 3 3 3

Definite Integrals

∴ Required area =

100 sq. units. 3

REMEMBER: • If

z

af

z z af z af z af

f x dx =

a

z

af

f x dx = g x

b a

af af

=g b −g a

z af b

f z dz and so on.

a

z af a

f x dx = − f x dx

b

f x dx =

a a

•

f t dt =

a

a

•

af

b

b

•

z af b

a

af

b

•

af


f x dx =

0

b

z z af za f c

af

a

f x dx +

a

f x dx

c

a

f a − x dx

0

• Area enclosed by the curve y = f (x), X-axis and the lines x = a and x = b is given by

z z af b

b

A = y dx = a

f x dx .

a

EXERCISE I.

Evaluate:

z zd z FGH b

1.

a x + 1f dx

2.

3

i

3 x − 2 x + 1 dx 2

1

4.

5.

1

f

2 x − 1 dx 8

0

1

4

i

x 2 − 1 dx

1

a

3.

zd za zd 2

IJ K

1 x+ dx x

2

6.

0

i

x 3 + 2 x 2 − x + 1 dx

501

502

Basic Mathematics

zd zd za z za zd z 1

7.

i

−1

i

x 3 + 3 x 2 + x dx

−1

fa f

x x + 1 x − 1 dx

13.

10.

2

15.

f

12.

14.

0

16.

−x

i dx

18.

1

x 2 e − x dx

za z za −1 5

23.

2

e2

25.

x2 1 − x

5 2

0

20.

−1

21.

log x dx

1

1

0

1 dx x 1 + log x

1

0

19.

x + 1 x − 2 dx

e

dx

x

1+ e

dx

2

1

17.

f

1

e

x+2

log x

1

i

6 x −5 + 2 x 4 + 4 dx

0

x2 + 4x + 3

0

I K

1 − x dx x2

2

0

1

x2 +

1

1

11.

8.

1

2

9.

z FH zd z a fa z z za f z 2

x + 4 x − 2 dx 3

0

dx 1− x x + 2

fa

f

dx 2 x + 4x + 3

f

z z 1

22.

0

1

24.

0

x dx x +1

x dx x +1 2

x3 dx e2 x

2

log x dx

e

II. 1. Find the area bounded by the curve y = x3, the x-axis and the lines x = 1 and x = 2. 2. Find the area bounded by the curve y = x2 − 4x, x-axis and the lines x = 1 and x = 3. 3. Find the area enclosed by the curve y =

x x-axis and ordinates x = 0 and x = 1. x +1

Definite Integrals

4. 5. 6. 7. 8. 9. 10. 11. 12.

Find the area bounded by the x-axis and the curve y = x2 − 7x + 10. Find the area bounded by the curve y = 4x − x2 − 3 with x-axis. Find the area bounded by the curve y = 4x − x2 and x-axis. Find the area between the curves y2 = 4x and y = 2x. Find the area between the curves y2 = 2x − 2 and the line y = x − 5. Find the area between the parabolas y2 = 4x and x2 = 4y. Find the area between the parabola y2 = 4ax and the line y = 2x. find the area between x2 = y and y2 = 8x. Find the area between the curves y = 11x − 2y − x2 and y = x.

ANSWERS 1.

ab − af LMN b + 2a + 2 OPQ

2.

4 3

3. 20

4.

6.

28 3

8.

4 3

9.

57 4

10.

7. −4

4 3

13.

18.

16 693

19.

24.

−19 3 + 8e 2 8

2 25. 2e − e

12. −

8− 3

e2 − 5 e

14. 2

d

i

2 −1

20. 1 − log 2

15.

21.

alog 2f

1 9

5.

365 32

11. −

3

3 2 log 2 3

20 3 1 4

16. −1

17. 2 −

22. log 2

23.

b

g

1 5 log 2 4

15 sq units 4

2.

22 sq units 3

3. 1 − log e 2 sq units

4.

9 sq units 2

5.

4 sq units 3

6.

64 sq units 3

7.

1 2 a sq units 3

8. 18 sq units

9.

16 sq units 3

10.

1 2 a sq units 3

8 sq units 3

12.

4 sq units 3

II. 1.

11.

1 e

503

504

Basic Mathematics

20 Application of Calculus in Business 20.1 TERMINOLOGY: Cost Function: The outflows usually raw materials, rent, utilities, pay of salaries and so forth form the total cost. It is sum total of all costs. Economists and accountants often define total cost as sum of 2 components. Total variable cost and total fixed cost. Total variable cost varies with the level of output. (Eg.: raw materials) whereas total fixed cost remains the same (for example rent). Hence Total cost = Variable cost + Fixed cost era Aver age Cost: Average Cost is the Cost per unit of the output. It is obtained by dividing total cost by the total quantity produced. If TC is the total cost of producing x units or q units, then average cost AC is given by, AC =

TC TC or . x q

Mar ginal Cost: Marginal cost is the additional cost incurred as a result of producing and selling one Marg more unit of the product. If TC is the total cost of producing x units or q units then the derivative

d d TC or TC repredx dq

sents the instantaneous rate of change of total cost, given a change in number of units (x or q) produced. ∴

Marginal Cost, MC =

d d TC or TC . dx dq

Re ven ue function: The money which flows into an organisation from either giving service or selling Rev enue products is called as revenue. The most fundamental way of computing total revenue from selling a product is Total revenue = Price per unit × Quantity sold. If TR is the total revenue p is the price per unit and q or x is the quantity sold then TR = pq or TR = px.

Application of Calculus in Business

505

Mar ginal rreeven ue: It is the additional revenue derived from selling one more unit of a product. It is Marg enue: obtained by differentiating total revenue with respect to quantity demanded. If MR is the marginal revenue. TR is the total revenue and x or q is the No. of units produced. Then

MR =

a f

d TR dx

or

MR =

d TR . dq

Pr of it function: Profit for an organisation is the difference between total revenue and total cost. If TR Prof ofit is the total revenue and TC is the total cost then Profit = TR − TC. For many production situations, the marginal revenue exceeds the marginal cost, at lower level of output. As the level of output increases, the amount by which marginal revenue exceeds marginal cost becomes smaller. Eventually a level of output is reached at which marginal revenue = Marginal cost. Beyond this point marginal revenue is less than marginal cost and the total profit begins to decreases with added output. So from theoretical stand point, the profit maximization level of output can be identified by the criterion. Marginal revenue = Marginal cost. i.e., MR = MC.

a f

d d TR = TC dx dx

i.e.,

or

a f

d d TR = TC dq dq Note: We know if

af af

d f x =g x dx

Then

z af

af

g x dx = f x + c where c is the constant of integration.

By applying this to,

d TR = MR and dx d TC = MC dx We get

za

f

MR dx = TR and

za

f

MC dx = TC.

So if we know marginal cost/marginal revenue (MC/MR) we can calculate, total cost or Total revenue (TC/TR) by using

506

Basic Mathematics

z z

z z

TC = MC dx or MR dq TR = MR dx or MR dq

WORKED EXAMPLES: 1. If the total cost function C(x) of a firm is given by C(x) = x3 − 3x + 7. Then find the average cost and marginal cost when x = 6 units. Solution: Given c(x) = x3 − 3x + 7. We have Average Cost = Average Cost =

Total Cost x

=

x 3 − 3x + 7 x

=

7 x 3 3x 7 − + = x2 − 3 + . x x x x

Marginal Cost = =

af

d C x dx

d 3 x − 3x + 7 dx

= 3x 2 − 3 + 0 MC = 3 x 2 − 3

MC when x = 6 is 3(6)2 − 3 3(36) − 3 108 − 3 = 105. 2. If the marginal cost function is given by x3 + 3x − 7 and fixed cost is Rs 750 then find the total cost function. Solution: Given: Marginal cost function = x3 + 3x − 7. MC = x 3 + 3 x − 7

We know Total cost =

=

z zd

MC dx .

i

x 3 + 3 x − 7 dx


507

z z z

= x 3 dx + 3x dx − 7 dx . TC =

x 4 3x 2 + − 7x + c 4 2

where c is the constant of integration. Here it is fixed cost. Given fixed cost = Rs. 750 Total function cost =

∴

x 4 3x 2 + − 7 x + 750 4 2

3. The total revenue function is given by R = x + 3x2. Find the marginal revenue and demand function. Solution: Given TR = x + 3x2. diff. w.r.t. x.

d d x + 3x 2 TR = dx dx

af

d i

d d d 2 x x + 3⋅ TR = dx dx dx ⇒

Marginal revenue = 1 + 3.(2x) MR = 1 + 6x Demand function = Average revenue

=

Total revenue x

=

x + 3x 2 x

=

x 3x 2 + = 1 + 3x . x x

af

4. If the total revenue function is given by R q = 5 +

96 + 6q 2 where q is the number of units q

manufactured. Find the maximum value of total revenue.

af

Solution: Given R q = 5 +

96 + 6q 2 q diff. w.r.t. x.

af

R′ q =

LM N

d 96 5+ + 6q 2 dq q

OP Q

508

Basic Mathematics

FG H

af

R ′ q = 0 + 96 −

IJ a f K

1 + 6 2q q2

R(q) attains maxima when R′(q) = 0.

0=−

96 + 12q q2

96 = 12q q2 96 = 12q 3 q3 =

⇒

96 12

q3 = 8 q=2 ∴ Maximum value of the revenue = R(2)

af

R 2 =5+

af

96 +6 2 2

2

= 5 + 48 + 24 = 77.

5. The marginal revenue (in thousands of rupees) function for a particular, commodity is 4 + e−0.03x, where x is denotes the number of units sold. Determine the total revenue from the sale of 100 units given that e−3 = 0.05 (Approx). Solution: We know Total revenue =

za

f

Marginal revenue dx

Total revenue from the sale of

za zd

100

100 units =

f

MR dx

0

100

=

i

4 + e −0.03 x dx

0

e −0.03 x = 4x + −0.03

LM a f N

= 4 100 +

100

0

OP Q

LM a f N

e −0.03×100 e0 − 4 0 + − 0.03 − 0.03

OP Q


LM N

= 400 −

400 − ∴

OP L Q MN

e −3 1 − − 0.03 0.03

509

OP Q

0.05 1 + = 400 − 1.66 + 33.33 = 431.667 0.03 0.03

Total revenue = 431.667 thousands of rupees. = Rs. 4,31,667 6. Find the total cost of producing 1000 electric bulbs if the marginal cost (in Rs. per unit) is given

af

by C ′ x =

x + 2.5 where x is the output. 1000

Solution: We know, Total cost = ∫(Marginal cost)dx.

za

f

1000

∴ Total cost of producing 1000 electric bulbs =

MC dx

0

z FH

1000

=

0

I K

x + 2.5 dx 1000

LM N 1 L1000 = M 1000 N 2 =

1 x2 + 2.5 x 1000 2 2

OP Q

1000

0

a fOP Q

+ 2.5 1000 − 0

1000 + 2,500 = 500 + 2500 2 = 3000. ∴ Total cost of producing 1000 bulbs = Rs 3000. 7. If the total revenue function and total cost function are given by TR = 40x − x2 and TC = 2 + 4x respectively. Then find at what level of output profit is maximised. Solution: We know, profit is maximised when marginal cost = marginal revenue. Now Marginal cost =

MC =

d [Total cost] dx d 2 + 4x = 0 + 4 dx

MC = 4

510

Basic Mathematics

Marginal revenue =

=

d [Total revenue] dx d 40 x − x 2 dx

MR = 40 − 2x Profit is maximised when, MR = MC. ∴ 40 − 2x = 4 ⇒ 40 − 4 = 2x 36 = 2x ⇒ x = 18 ∴ Profit is maximised when 18 units are produced. 8. Find the maximum profit that a company can make, if the profit function P(x) = 41 + 24x − 18x2. Solution: Consider

af

P x = 41 + 24 x − 18 x 2 diff. w.r.t. x.

af a f

dP = 0 + 24 1 − 18 2 x dx dP = +24 − 36 x dx

P attains extrema when

dP =0 dx 0 = +24 − 36 x

x=

⇒ Again consider

2 +24 =+ . 36 3

dP = +24 − 36 x dx diff. w.r.t. x.

d2P = −36 < 0 dx 2 ∴ P attains maxima at x = 2/3. Maximum value of the profit = 41 + 24

F 2 I − 18 F 2 I H 3K H 3K

2


= 41 + 16 − 18 ⋅

511

4 9

= 57 − 8 = 49.

9. A T.V. manufacturer produces x sets per week at a total cost of Rs x2 + 1560x + 50,000. He is a monopolist and the demand function for this product is x =

12000 − P where P is the price 179

per set. What is the monopoly price in order to maximise the profit? Given: Total cost = x2 + 1560 x + 50,000 diff. w.r.t. x.

a f

d TC = 2 x + 1560 dx i.e., Now, also given

Marginal Cost = 2x + 1560

Demand function = x = ⇒

...(1)

12000 − P 179

P = 12000 − 179 x

Total revenue = Px.

TR = 12000 x − 179 x 2 diff. w.r.t. x.

a f

a f

d TR = 12000 − 179 2 x dx i.e.,

Marginal revenue = 12000 − 358x MR = 12000 − 358x

Profit is maximixed when Marginal cost = Marginal revenue. i.e., Equation (1) = (2) 2 x + 1560 = 12000 − 358 x 2 x + 358 x = 12000 − 1560 360 x = 10440

x=

10440 360

x = 29.

∴

Monopoly price P = 12000 − 179 x

...(2)

512

Basic Mathematics

a f

= 12000 − 179 29 = Rs. 6809.

10. The cost function of a company is given by C = 500 x − 20 x 2 +

x3 , where x stand for output. 3

Calculate the output when marginal cost is equal to average cost. Given: C = 500 x − 20 x 2 +

x3 3 diff. w.r.t. x.

a f

dC 3x 2 = 500 − 20 2 x + dx 3 i.e., Now,

Marginal Cost = 500 − 40x + x2

average cost =

=

Total cost x 500 x − 20 x 2 +

x3 3

x

= 500 − 20 x +

x2 3

Equating (1) & (2)

500 − 40 x + x 2 = 500 − 20 x + −40 x + x 2 + 20 x −

x2 =0 3

−20 x + x 2 −

x2 =0 3

−60 x + 3x 2 − x 2 =0 3 −60 x + 2 x 2 = 0

a

...(1)

f

2 x −30 + x = 0 2 x = 0 or − 30 + x = 0 x = 0 or x = 30.

x2 3

...(2)


513

∴ The output when the marginal cost = average cost is 30.

REMEMBER: • Total cost = Fixed cost + Variable cost. • Marginal cost =

a

f

a

d d Total cost or total cost dx dq

f

• Total cost = ∫(Marginal cost) dx or ∫(Marginal cost) dq. • Marginal revenue =

d d (Total revenue) or (Total revenue) dx dq

• Total revenue = ∫(Marginal revenue) dx or ∫(Marginal revenue) dq. • Average cost =

Total cost Total cost or x q

• Profit is maximised when marginal cost = Marginal revenue. • Average revenue is nothing but demand function =

Total revenue Total revenue or x q

EXERCISE 1. If the marginal cost of a product is 3x2 − 4x where x is the output, find the total cost of producing 10 units. 2. If the marginal revenue function is given by

2 , find the total revenue function. 2x + 1

3. A company has revenue function given by R = 100q − q2. Find the marginal revenue function (where q is the output). 4. Find the marginal revenue for the demand function 3x − x2. 5. Find the average cost function for the marginal cost function x2 + 2x. 6. The demand function of a firm is given by P = 50 − 2x. Where P is the price per unit for x units. Determine the marginal revenue. 7. The cost function C = 5 +

48 + 3 x 2 where x is the number of articles produced. Find the minix

mum value of C. 8. If the marginal cost is given by x2 + 7x + 6 and fixed cost is Rs 2500, determine the total cost of producing 6 units. 9. If C is the total cost for producing x units of a product and the average cost function is given by 0.003 x2 − 0.04 x + 6 +

3000 , find the marginal cost function. x

514

Basic Mathematics

10. The total cost function of a firm is given by C(x) = 2x3 − x2 + 5x. Find the average and marginal cost function. 11. If the marginal revenue is given by 30 −

x3 , then find the total revenue function. 30

af

1 3 x . Where C is the total cost for 3 x units. Calculate the output at which the average cost is minimum. 13. The unit demand function is x = 25 − 2p where x is the number of units and p is the price. Let the average cost per unit be Rs 20. Find the price per unit that maximises the profit function. 14. The demand function of a company is given by P(q) = 800 − 0.4q and the total cost function is given by C(q) = 80 q + 8000 where q is the level of output,P is the price per unit. Find the level of output and the price charged which maximises the profit. 15. Find the total revenue, marginal revenue, marginal revenue and average revenue when the demand function is given by Q = 30 − 4P + P2 where P is price and Q is the quantity demanded. Also calculate the marginal revenue when P = 3. 12. The cost function of a firm is given by C x = 300 x − 10 x 2 +

16. A manufacturer has a total cost function C = 100 + 32 x . He would like to know at what level of output his marginal cost will be Rs. 2.00. 17. If the total cost function is C = 8x3 − 12x2 + 20x where x is the level of output, find x at which the average cost is minimised find the total cost for this output. 18. The marginal cost function of a firm is 150 − 10x + 0.2x2, where x is the output, find the total cost function, if the fixed cost is Rs 750, what is the average cost? 19. If the total revenue (R) and total cost (C) functions are given by R = 30x − x2 and C = 20 + 6x, find x if marginal revenue = marginal cost. Also find average cost and average revenue. 20. If the marginal revenue is given by 15 +

q , find the total revenue obtained from an output of 15

30 units.

ANSWERS 1. 1200

2. log(2x + 1) + c 3. 100 − 2q

4. 6x − 3x2

6. 50 − 4x

7. 41

9. 0.009x2 − 0.08x + 6

10. 0.006 x − 0.04 −

14. Rs 440

3000 x2

8. 2734 11. 6x2 − 2x + 5

15. 30 − 4P + P2 16. 64

18. AC = 150 − 5 x +

0.2 x 2 750 + x 3

19. x = 12

12. 15

17.

3 4

20. 480.

5.

x2 c +x+ 3 x

13. Rs 12.34

Examination Corner

516

Pa Print Pre-Uni e-Univ ersity Hours] Question P aper Blue Pr int ffor or Second Year Pr e-Uni ver sity [90 Hour s]

No.

No. of Hours

Content ↓

Knowledge VSA

1. Mathematical Logic

8

1

2. Permutation Combination and Probability

15

1

3. Binomial Theorem

4

4. Partial Fractions

4

5. Matrices and Determinants

15

1

6. Ratio, Proportions and Variations

10

1

7. Averages

4

8. Bill Discounting

4

1

9. Stocks and Shares

4

1

10. Learning Curve

4

11. L.P.P.

6

12. Circles, Parabola

10

13. Limits and Continuity

8

14. Differentiation

10

15. Applications of Derivatives

8

16. Integration

10

17. Applications of Integration

6

Total

130

SA

Understanding

ET

VSA

1

SA

ET

Application VSA

SA

Skill

ET

1

1

1

1

1

VSA

SA

ET TOTAL 7

1

1

1

4

1 1

4

1

1

1

15

1

1

4

1

5 3

1

4

1 1 1 1

1

1

5

1

9

1

1

1

7

1

1

7

6

8

2

16

7

1

1

1

15

20

1

6

24

2

1

13

1

4

2

10

1

4

28

120

Basic Mathematics

Sl. Objectives →

Examination Corner

517

MODEL QUESTION PAPERS MODEL QUESTION PAPER 1 (Given by P.U. Board) I. Answer all questions 1. Negate “if two triangles are similar, then their areas are equal”. 2. If nc7 = nc3, find nc10. 3. Find the value of 4. 5. 6. 7. 8.

2400 2402

1 × 10 = 10

2401 . 2403

Find the fourth proportional of 2, 6, 5. A bill drawn for 4 months was legally due on 10.10.2003. Find the date of drawing the bill. Define yield. What is feasible region in L.P.P. Find the centre of the circle 4x2 + 4y2 − 15x − 8y + 11 = 0

e5 x − 1 . x →0 x

9. Evaluate Lt

10. Total cost of a commodity is given by C = 1/3 x3 − x2 + 3x + 10. Find the marginal cost. II. Answer any ten of the following: 11. If p is 2 is prime, q is 3 is even, r is

2 × 10 = 20. 2 + 3 = 5 , , find the truth value of (p → q) → r.

12. Find the number of words that can be formed using all the letters of the word “BOOKS”. How many of them begin with ‘B’. 13. There are 10 points out of which 3 are collinear. Find the number of straight lines that can be formed. 14. Two dice are thrown simultaneously. What is the probability of getting the sum 7?

F H

15. Expand 2 x −

1 x2

I K

5

using Binomial theorem.

F 16. Find the middle term or terms in the expansion of H x LM 2 −1OP 2 find AA . 17. If A = M −1 1 N 3PQ 1

3

+

3 x

I K

13

.

518

Basic Mathematics

18. Find the interest earned on Rs. 2448.35 cash invested in 15% Stock at 81.5, given that brokerage is 0.125%. 19. Find the equation of the parabola whose focus is (4, 0) and directrix is x + 4 = 0. Find the length of latus rectum.

af

20. If f x =

2x −1 , is continuous at x = 0, find f (0). x

e

j

21. If y = log e x + 1 + x 2 , prove that

22. Integrate w.r.t. x,

1 dy . = dx 1 + x2

x + 3x 2 + 4 . x

III. Answer any three of the following:

a

f a

4 × 3 = 12

f

23. Prove that p → q ↔ ~ p ∨ q is a tautology. 24. Prove that ncr + ncr − 1 =

n + 1c , r

25. Resolve into partial fractions

and verify this for n = 5 and n = 2.

1 . x x +1 2

a f

26. Prove that if each element of a row (or column) constant multiples of corresponding elements of other rows (or columns) are added, then the value of the determinant is unaltered. IV. Answer any three of the following: 4 × 3 = 12 27. If two men and four women can do a work in 33 days and 3 men and 5 women can do the same work in 24 days, how long shall 5 men and 2 women take to do the same work? 28. Calculate the total wages earned per week by 400 workers in a factory from the following data Daily wages

Number of Workers

12.50 – 17.50

51

17.50 – 22.50

38

22.50 – 27.50

42

27.50 – 32.50

59

32.50 – 37.50

80

37.50 – 42.50

60

42.50 – 47.50

50

47.50 – 52.50

20

29. A bill for Rs. 2725.75 was drawn on 03.06.1997 and made payable 3 months after due date. It was discounted on 15.06.1997 at 16% p.a. What is the discounted value of the bill and how much has the banker gained in this transaction. 30. A business men buys and sells chairs and tables. He has Rs. 3000/- to invest. A chair costs him

Examination Corner

519

Rs. 50/- and a table costs him Rs. 90/-. He has space which can accommodate at most 48 pieces. If he sells each chair for Rs. 200 and each table for Rs. 400/-. Find the number of chairs and tables he has to buy to obtain maximum profit. V. 31. Find the equation of the circle passing through the origin and having centre at (−3, 4). OR Derive the equation of a parabola in the form y2 = 4ax. VI. Answer any three of the following:

4

4 × 3 = 12

x n − an = na n −1 . for all rational values of n. x →a x − a

32. Prove that Lt

33. If x 1 + y + y 1 + x = 0 and x ≠ y, prove that

34. If x = at2, y = 2 at prove 35. Evaluate

z

dy −1 = dx 1+ x

a f

2

.

1 d2y =− . 2 at 3 dx 2

2

e x x 3 dx.

VII. Answer any two of the following: 10 × 2 = 20 36. (a) Two number are in the ratio 4:7. If 12 is added to each of them then the new ratio is 8:11. Find the numbers. (b) In how many ways 3 boys and 5 girls be arranged in a row so that (i) no two boys are together. (ii) all the girls are together. (c) Solve by matrix method. 2x + y + z = 7. x − y + 2z = 5 3x − 2y + 2z = 5. 37. (a) Find the equation of the circle with (−3, 5) and (6, 1) as the extremities of a diameter. (b) An engineering company has succeeded in winning a contract for supplying aircraft engines. The prototype constructed to win the contract took 4000 labour hours and the company experiences 80% learning effect. Find the total cost of 7 engines of new order if labour cost is Rs. 30/- per hour. (c) The radius of a circular blot of ink is increasing at the rate of 3 cm per minute. Find the rate of increases of its area when its radius is 2 cms. What is the rate of increases of its circumference? 38. (a) Find the range in which the function f (x) = x2 − 6x + 5 is (1) increasing, (2) decreasing. (b) Find the area bounded by the curve y = x2 − 7x + 10 with x-axis.

520

Basic Mathematics

1 (c) Prove without expanding 1 1

a2 b2 = a − b b − c c − a . c2

a b c

a

fa fa f

***

MODEL QUESTION PAPER 2: I. Answer all questions: 1 × 10 = 10 1. If the truth value of p and q are T and F respectively. Then what will be the truth value of

a p ∧ qf → a ~ p∨ ~ qf .

2. Find the number of permutation of the letters of the word BALLOON taken all together. 3. Find the mean proportional to 2:8. 4. Find x if 5. 6. 7. 8.

LM2 Nx

OP Q

7 is singular. 14

A bill was drawn on 14.3.1997 for 3 months. When does the bill fall legally due? Find the yield by investing Rs. 1140 on 15% stock quoted at Rs. 95. Write any one step to formulate an LP problem. Find the radius of the circle 4x2 + 4y2 + 8x + 9y − 7 = 0.

9. Evaluate: lim

x →1

d

3

id

x −1

4

i

x −1

10. If the marginal revenue function is given by 3x2 − 8, then find the total revenue function. II. Answer any ten of the following:

a

2 × 10 = 20

f

11. Construct the truth table for p∧ ~ q → ~ p . 12. How many four digits numbers greater than 2000 can be formed with the digits 0, 1, 2, 3, 4, 5. 13. Find the number of permutations of the letters of the word ‘BANANA’ taking all letters. How many of them begin with N. 14. Two cards are drawn at random from a pack of well shuffled 52 cards. What is the probability of getting king and queen cards.

F H

15. Expand using binomial theorem: 3 +

2 x

I K

4

.

16. Find the middle term or terms in the expansion of

LM MN

2 A 1 = − 17. If 1

0 6 3

OP PQ

1 2 , then find AA′. 2

F 1 + xI H x 2K 2

10

.

Examination Corner

521

18. Find the cash required to purchase Rs. 1600, 8 1/2 stock at 105, brokerage is 1/2%. 19. Find the equation of parabola whose vertex is (1, 1) and focus is (3, 1).

af

20. If f x =

e 3x − 1 is continuous at x = 0. Then find f (0). x

21. If x2 + y2 = 16. Then find 22. Integrate w.r.t. x: x 3 x +

dy when x = 3 and y = 4. dx

1 3x + 2

III. Answer any three of the following:

a

f a

4 × 3 = 12

f

23. Prove that p∧ ~ q ∧ ~ p ∨ q is a contradiction. 24. Prove that ncr = ncn − r and hence find n if nc18 = nc10. 25. Resolve into partial fractions:

2 x 2 + 3x + 2 . x2 − x − 2 26. If the elements of any row (or column) is multiplied by non zero constant K. Then prove that the value of the determinant is multiplied by K. IV. Answer any three of the following: 27. 16 men or 28 boys can fence a farm in 40 days. In how many days will 24 men and 14 boys complete the same work. 28. 3 tests in English, 2 in Hindi, 4 in Kannada and 5 in Sociology are conducted. The average marks scored by Rashmi in English is 60, that in Hindi is 56 and that in Kannada is 45. If the average marks of all the subjects and all the tests taken together is 48. Then find the averages marks scored by her in Sociology. 29. The difference between the bill discount and true discount on a certain sum of money due in 4 months in Rs. 10. Find the amount of the bill (or face value) if the rate of interest is 3% p.a. 30. Maximise Z = 7x + 4y subject to x + 3y ≤ 3, 6 x + 3 y ≤ 8 ; x ≥ 0, y ≥ 0. . V. 31. Derive the equation of circle which is described on the line joining (x1 y1) and (x2 y2) as the ends of diameter. OR Find the equation of parabola whose ends of latus rectum are (3, 2) and (3, −4). VI. Answer any three of the following:

12 + 2 2 + 32 + ... n 2 n→∞ n3 − 6

32. Evaluate: (i) lim

4 ×3 = 12

a

(ii) lim 1 + 3 x n →∞

f

1 x

522

Basic Mathematics

e

33. If y = x + 1 + x 2

j

m

d

i

, then prove that 1 + x 2 y2 + xy1 − m 2 y = 0.

d2y −24 = 34. If xy + 4y = 3x, then prove that 2 dx x+4 35. Evaluate: (a)

( b)

z

Fx GH x

e −1 e

z

a

f

3

.

x log x dx

I JK

+ e x −1 dx + ex

VII. Answer any two of the following: 10 × 2 = 20 36. (a) Divide Rs. 118 among A, B and C so that A:B = 3:4, B:C = 5:6. (b) From 8 lecturers and 4 students a committee of 6 is to be formed. In how many ways can this be done so that the committee contains (i) Exactly 2 students (ii) atleast 2 students. (c) Solve by matrix method: 5 x − y − 4 z = 4, x + 4 y + 2 z = 12, 3 x − y − z = 3.

37. (a) A circle with centre (2, −1) passes through (−1, 3). Find the equation of the circle. (b) The first sample batch is of 50 units of product. A took 80 hours to make. The company now wishes to estimate the total time taken to make 100 additional units. Solve the problem using 80% learning effect. (c) If the base of the triangle is 3 times the height and the height is decreasing at the rate of 4 cm/ sec. then find the rate of decrease of the area when height is 1 cm. 38. (a) Find 2 numbers whose sum is 20 and whose product is maximum. (b) Find the area bounded by the curve y2 = 4x and the line y = x. a2 (c) Prove without expanding − ab ac

ab −b 2 bc

ac bc = 4 a 2 b 2 c 2 . −c2

***

MODEL QUESTION PAPER 3: I. Answer all questions: 1 × 10 = 10 1. Write the contrapositive of ‘If the triangle is not right angled then its two sides are not perpendicular’. 2. If nc2 = 105, then find n. 3. Find x and y if 2

LM3 xOP + 3 LM y xOP = LM 0 1OP . N4 0Q N8 0Q N32 3Q

Examination Corner

523

4. If a : b = 2 : 3 and b : c = 5:7, then find a : c. 5. The present worth of a bill due sometime hence is Rs. 1100 and the true discount is Rs. 110. Find the banker’s gain. 6. Find the proceeds got by selling shares worth Rs. 3000 at 96 1/8 brokerage 1/8%. 7. What is an objective function in LPP. 8. Find the focus of the parabola x2 + 8y = 0. 9. Evaluate: lim

n →∞

d

n n2 + 2 6n − 8 3

i

10. If the marginal cost function is x3 + 2x − 1 and fixed cost is Rs. 2500, then find total cost function. II. Answer any ten questions:

a

2 × 10 = 20

f a f

11. The compound proposition p ∧ q → r ∨ s has truth value false. Find the truth values of p, q,

r and s. 12. In how many ways can 3 girls and 4 boys sit round the table so that no two girls sit next to each other. 13. Find the number of triangles that can be formed from 12 points of which 4 are given to be collinear. 14. 2 dice are thrown simultaneously what is the probability of getting the sum 8.

F a xI 15. Expand H + K x b

5

using binomial theorem.

16. Find the term independent of x in the expansion of

Fx − 1 I H xK 2

21

.

17. Solve by Cramer’s rule: 3x + y = 8 and 4x − 9y = 6. 18. Find the cash you would require to purchase a share worth Rs. 3000 of 4% at 102 8/9 Brokerage 1/9%. 19. Find the vertex focus, directrix and length of latus rectum of the parabola x2 − 8y + 24 = 0.

a f RST4kx +− 23 ifif xx 0. On the circle if x12 + y12 + 2 gx1 + 2 fy1 + c = 0. • Any line parallel to ax + by + c = 0 will have the equation ax + by + k = 0. [Take x and y co-efficients same]. • Any line perpendicular to ax + by + c = 0 will have the equation bx − ay + k = 0 [interchange x and y co-efficients and change the sign of anyone].

Figure

Vertex

Focus

Equation of directrix

Axis

Ends of of Latus rectum

Equation of LR

Length of LR

y2 = 4ax

(0, 0)

(a, 0)

x = −a

x-axis y=0

(a, ±2a)

x=a

4a

y2 = −4ax

(0, 0)

(−a, 0)

x=a

x-axis y=0

(−a, ±2a)

x = −a

4a

x2 = 4ay

(0, 0)

(0, a)

y = −a

y-axis x=0

(±2a, a)

y=a

4a

x2 = −4ay

(0, 0)

(0, −a)

y=a

y-axis x=0

(±2a, −a)

y = −a

4a

(h, k)

(a + h, k)

x = −a + h

y=k

(a + h, ±2a + k)

x=a+h

4a

(h, k)

(−a + h, k)

x=a+h

y=k

(−a + h, ±2a + k)

x = −a + h

4a

(h, k)

(h, a + k)

y = −a + k

x=h

(±2a + h, a + k)

y=a+k

4a

(h, k)

(h, −a + k)

y=a+k

x=h

(±2a + h, −a + k)

y = −a + k

4a

(h, k)

(y − k)2 = 4a (x − h) (h, k)

(y − k)2 = −4a (x − h)

(x − h)2 = 4a (y − h)

(x − k)2 = −4a (y − h)

(h, k)

(h, k)

Basic Mathematics

Equation

554

14. PARABOLA:

Examination Corner

• • • • •

In any parabola, focus is inside the curve and directrix is away from the parabola. Distance between vertex and focus = a. For the given ends of latus rectum, there are 2 possible parabolas. Focus is the mid point of latus rectum. Axis is ⊥r to the directrix. Distance between directrix and vertex = a.

15. LIMITS AND CONTINUITY: • lim

x →a


ex − 1 =1 x→0 x

• lim

• lim

x →0

ax − 1 = log e a x

F H

• lim 1 + n →∞

1 n

a f

• lim 1 + n n→ 0

I K 1 n

n

=e

=e

af af

af

• A function y = f(x) is said to be continuous at x = a if lim− f x = f a = lim+ f x x →a

i.e., LHL = f(a) = RHL. • Limit of a function exists at x = a if

af

x→a

af

lim f x = lim+ f x

x →a−

∴

x→a

f(x) is continuous at x = a iff

af af

lim f x = f a

x →a

16. DIFFERENTIAL CALCULUS: y

dy dx

xn x x2

nxn−1 1 2x

555

556

Basic Mathematics

x3

3x2

1 x

2 x

1 x

1 x2

−

1 x2

−2 x3

1 x3

−3 x4

ex ax

ex ax loga

logx

1 x C⋅

Cu

du dx

du dv ± dx dx

u±v I⋅

I ⋅ II Dr.

Nr.* Dr.**

af

af


a f a f a f a f

d d Nr − Nr ⋅ Dr. dx dx Dr. 2

• Chain rule: If y = g(u) & u = f (x), then y = g[f(x)] is differentiated by chain rule

dy dy du = ⋅ dx du dx • Implicit dif entia tion: Given function f (x, y) = 0. difffer erentia entiation: diff. w.r.t. x., take RHS. Then find

*: Numerator

dy dy common among the terms containing and shift the remaining terms to dx dx

dy . dx **: Denominator

Examination Corner

557

Parametric differentiation: Given x = f(t) and y = g(t).

dy dy dt dy Then to find , use dx = dx dx dt • Second or der der ve: order deriivati tiv If y = f(x), then by differentiating we get

dy or y1 or y1 or f1(x). This is a function of x. By dx

differentiating this again with respect to x we get

d2y or y″ or f ″(x) or y2. dx 2

17. APPLICATION OF DERIVATIVES: • Velocity =

dS dt

• Acceleration =

dv d 2 S = dt dt 2

• Rate means differentiation w.r.t. t. ∴ rate of change of area =

dA dt

• Area of Square = S2 • Area of circle = πr2 • Surface area of sphere = 4πr2 • Volume of Sphere =

4 3 πr 3

• Volume of a cylinder = πr2h • Volume of a cone =

1 2 πr h 3

• For an increasing function

dy dy < 0. > 0 and for a decreasing function dx dx

• To find maximum and/or minimum value of the function y = f(x), find

dy , equate it to zero. Let dx

d2 y d2y d2y , and x = a, x = b, x = c be the points. Find . dx 2 at x = a dx 2 at x = b dx 2 at x = c.

558

Basic Mathematics

if

d2y > 0 , then x = a is a point of minima. Minimum value of the function is y = f(a). dx 2 at x = a

d2 y is less than zero, x = b is a point of maxima. Maximum value of the function is dx 2 at x = b y = f(b).

if

d2y if is equal to zero, then x = c is called point of inflection. At x = c the function neither dx 2 at x = c attains maxima nor minima.

18. INTEGRATION: •

z

x n dx =

x n +1 +c n +1

Provided n ≠ −1 • if n = −1

z z z z z z af af z af z af z af af za f a f z af af af x −1dx =

•

• • •

1 dx = log x + c. x

e x dx = e x + c. a x dx =

ax +c log a

k dx = kx + c

k f x ± g x dx = k

• If

f x dx ± k g x dx


f ax + b dx =

• To evaluate

af

f x

f x

g ax + b +c a n

⋅ f ′ x dx, put f x = t and proceed to get the answer.

n +1

n +1

+ c for n ≠ −1

and log [f (x)] + c for n = −1

Examination Corner

• To evaluate ∫ef(x) ⋅ f′(x) dx, put f(x) = t and proceed to get ef(x) + c as answer. • To evaluate integrals of the type

za

px + q dx or ax + b cx + d

fa

f

za

px + q ax + b

f acx + d f 2

First resolve into partial fractions

a

px + q A B = + ax + b cx + d ax + b cx + d

fa

f

or

a

px + q A B = + 2 ax + b ax + b cx + d ax + b

f a

fa

f a

f a 2

+

C cx + d

f

Find A, B, C, then finally integrate. •

za

fa

f

I function II function dx

z

= I function II function −

z LMNz

II function

a

I function and II function are kept by making use of LIATE rule. •

z

af af

af

e x f x + f ′ x dx = e x ⋅ f x + c.

19. DEFINITE INTEGRALS: • If

z z af z af z af b

•

z

af

af

af

f x dx =

a

z

z af b

f z dz and so on.

a

z af a

f x dx = − f x dx

b

f x dx =

a a

•

f t dt =

a

a

•

af

b

f x dx =

0

b

z af z af za f c

a

f x dx +

a

f x dx

c

a

f a − x dx

0

af

f x dx = g x

a

b

•

z af b


b a

fOPQ

d I function dx dx

af af

=g b −g a

559

560

Basic Mathematics

• Area enclosed by the curve y = f(x), X-axis and the lines x = a and x = b is given by

z z af b

A=

b

y dx =

a

f x dx .

a

20. APPLICATION OF CALCULUS IN BUSINESS: • Total cost = Fixed cost + Variable cost. • Marginal cost =

a

f

a

d d Total cost or total cost dx dq

f

• Total cost = ∫(Marginal cost) dx or ∫(Marginal cost) dq. • Marginal revenue

d d (Total revenue) or (Total revenue) dx dq

• Total revenue = ∫(Marginal revenue) dx or ∫(Marginal revenue) dq. • Average cost =

Total cost Total cost or x q

• Profit is maximised when marginal cost = Marginal revenue. • Average revenue is nothing but demand function =

Total revenue Total revenue or x q

Examination Corner

Chapter

One mark questions (VSA)

Two mark questions (SA)

Mathematical Logic

1

1

1

7

Permutation, Combination and Probability

1

3

2

15

Binomial Theorem and Partial Fractions

–

2

1

8

Matrices and Determinants

1

1

3

15

Ratio, Proportion & Variation

1

1

1

7

Mathematics of Finance (Averages, B.D., Stock and Shares, L.C., LPP)

3

1

4

21

Circles, Parabola

1

2

1

9

Limits and Continuity

1

1

1

7

Differentiation and Its Application

1

2

3

17

Integration and Its Application

–

1

3

14

10

15

20

120

Total:

Four mark questions (ET)

561

Total marks

Comprehensive Basic Mathematics Vol 2

Basic Audio - VOL 2

Comprehensive Mathematics Practice: Book 2

Comprehensive mathematics for computer scientists 2

Dictionary of Basic Fuseki, Vol. 2

Dictionary of Basic Joseki, Vol. 2

Let's Learn Japanese Basic I Vol. 2

Basic Jazz Conception for Saxophone Vol 2 (Volume 2)

Wolf Prize in Mathematics, vol.2

SIAM Journal on Financial Mathematics Vol 2

Basic Audio - VOL 1

Basic Audio - VOL 3

Basic Engineering Mathematics

Basic mathematics for economists

Basic Training In Mathematics

Basic Concepts Of Mathematics

Basic Mathematics: Teach Yourself

Basic Mathematics for Economists

Schaum's Basic Mathematics

Basic mathematics for economists

Basic Engineering Mathematics

Basic Engineering Mathematics

Basic Mathematics for Economists

Comprehensive Mathematics Practice: Book 4

Comprehensive Mathematics for Computer Scientists

Comprehensive Mathematics Practice: Book 5

A Comprehensive Dictionary of Mathematics

Comprehensive Mathematics Practice: Book 3

Comprehensive mathematics for computer scientists

A Comprehensive Dictionary of Mathematics

Comprehensive Mathematics Practice: Book 1

Comprehensive Basic Mathematics Vol 2

Basic Audio - VOL 2

Comprehensive Mathematics Practice: Book 2

Comprehensive mathematics for computer scientists 2

Dictionary of Basic Fuseki, Vol. 2

Dictionary of Basic Joseki, Vol. 2

Let's Learn Japanese Basic I Vol. 2

Basic Jazz Conception for Saxophone Vol 2 (Volume 2)

Wolf Prize in Mathematics, vol.2

SIAM Journal on Financial Mathematics Vol 2

Basic Audio - VOL 1

Basic Audio - VOL 3

Basic Engineering Mathematics

Basic mathematics for economists

Basic Training In Mathematics

Basic Concepts Of Mathematics

Basic Mathematics: Teach Yourself

Basic Mathematics for Economists

Schaum's Basic Mathematics

Basic mathematics for economists

Basic Engineering Mathematics

Basic Engineering Mathematics

Basic Mathematics for Economists

Comprehensive Mathematics Practice: Book 4

Comprehensive Mathematics for Computer Scientists

Comprehensive Mathematics Practice: Book 5

A Comprehensive Dictionary of Mathematics

Comprehensive Mathematics Practice: Book 3

Comprehensive mathematics for computer scientists

A Comprehensive Dictionary of Mathematics

Comprehensive Mathematics Practice: Book 1

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