textbook for class xi€¦ · banashankari iii stage bengaluru 560 085 phone : 080-26725740...

MATHEMATICSTextbook for Class XI

2019-20

First EditionFebruary 2006 Phalguna 1927

ReprintedOctober 2006 Kartika 1928

November 2007 Kartika 1929

December 2008 Pausa 1930

December 2009 Agrahayana 1931

January 2011 Pausa 1932

February 2012 Magha 1933


November 2013 Kartika 1935


May 2016 Vaishakha 1938


December 2017 Agrahayana 1939

January 2019 Pausa 1940

PD 450T BS

© National Council of EducationalResearch and Training, 2006

` 210.00

ISBN 81-7450-486-9

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2019-20

Foreword

The National Curriculum Framework (NCF), 2005, recommends that children’s life

at school must be linked to their life outside the school. This principle marks a

departure from the legacy of bookish learning which continues to shape our system

and causes a gap between the school, home and community. The syllabi and textbooks

developed on the basis of NCF signify an attempt to implement this basic idea. They

also attempt to discourage rote learning and the maintenance of sharp boundaries

between different subject areas. We hope these measures will take us significantly

further in the direction of a child-centred system of education outlined in the National

Policy on Education (1986).

The success of this effort depends on the steps that school principals and

teachers will take to encourage children to reflect on their own learning and to

pursue imaginative activities and questions. We must recognise that given space,

time and freedom, children generate new knowledge by engaging with the information

passed on to them by adults. Treating the prescribed textbook as the sole basis of

examination is one of the key reasons why other resources and sites of learning are

ignored. Inculcating creativity and initiative is possible if we perceive and treat

children as participants in learning, not as receivers of a fixed body of knowledge.

These aims imply considerable change in school routines and mode of

functioning. Flexibility in the daily time-table is as necessary as rigour in implementing

the annual calendar so that the required number of teaching days are actually devoted

to teaching. The methods used for teaching and evaluation will also determine how

effective this textbook proves for making children’s life at school a happy experience,

rather than a source of stress or boredom. Syllabus designers have tried to address

the problem of curricular burden by restructuring and reorienting knowledge at

different stages with greater consideration for child psychology and the time available

for teaching. The textbook attempts to enhance this endeavour by giving higher

priority and space to opportunities for contemplation and wondering, discussion in

small groups, and activities requiring hands-on experience.

The National Council of Educational Research and Training (NCERT) appreciates

the hard work done by the Textbook Development Committee responsible for this

2019-20

book. We wish to thank the Chairperson of the advisory group in Science and

Mathematics, Professor J.V. Narlikar and the Chief Advisor for this book

Professor P.K. Jain for guiding the work of this committee. Several teachers

contributed to the development of this textbook; we are grateful to their principals

for making this possible. We are indebted to the institutions and organisations which

have generously permitted us to draw upon their resources, material and personnel.

We are especially grateful to the members of the National Monitoring Committee,

appointed by the Department of Secondary and Higher Education, Ministry of Human

Resource Development under the Chairpersonship of Professor Mrinal Miri and

Professor G.P. Deshpande, for their valuable time and contribution. As an organisation

committed to the systemic reform and continuous improvement in the quality of its

products, NCERT welcomes comments and suggestions which will enable us to

undertake further revision and refinement.

Director

New Delhi National Council of Educational

20 December 2005 Research and Training

2019-20

Textbook Development Committee

CHAIRPERSON, ADVISORY GROUP IN SCIENCE AND MATHEMATICS

J.V. Narlikar, Emeritus Professor, Chairman, Advisory Committee Inter University

Centre for Astronomy & Astrophysics (IUCCA), Ganeshkhind, Pune University, Pune

CHIEF ADVISOR

P.K. Jain, Professor, Department of Mathematics, University of Delhi, Delhi

CHIEF COORDINATOR

Hukum Singh, Professor, DESM, NCERT, New Delhi

MEMBERS

A.K. Rajput, Associate Professor, RIE Bhopal, M.P.

A.K. Wazalwar, Associate Professor, DESM NCERT, New Delhi

B.S.P. Raju, Professor, RIE Mysore, Karnataka

C.R. Pradeep, Assistant Professor, Department of Mathematics, Indian Institute of Science,

Bangalore, Karnataka.

Pradeepto Hore, Sr. Maths Master, Sarla Birla Academy Bangalore, Karnataka.

S.B. Tripathy, Lecturer, Rajkiya Pratibha Vikas Vidyalaya, Surajmal Vihar, Delhi.

S.K.S. Gautam, Professor, DESM, NCERT, New Delhi

Sanjay Kumar Sinha, P.G.T., Sanskriti School Chanakyapuri, New Delhi.

Sanjay Mudgal, Lecturer, CIET, New Delhi

Sneha Titus, Maths Teacher, Aditi Mallya School Yelaharika, Bangalore, Karnataka

Sujatha Verma, Reader in Mathematics, IGNOU, New Delhi.

Uaday Singh, Lecturer, DESM, NCERT, New Delhi.

MEMBER-COORDINATOR

V.P. Singh, Associate Professor, DESM, NCERT, New Delhi

2019-20

Acknowledgements

The Council gratefully acknowledges the valuable contributions of the following

participants of the Textbook Review Workshop: P. Bhaskar Kumar, P.G.T., Jawahar

Navodaya Vidyalaya, Ananthpur, (A.P.); Vinayak Bujade, Lecturer, Vidarbha Buniyadi

Junior College, Sakkardara Chowk Nagpur, Maharashtra; Vandita Kalra, Lecturer,

Sarvodaya Kanya Vidyalaya Vikashpuri District Centre, New Delhi; P.L. Sachdeva

Deptt. of Mathematics, Indian Institute of Science, Bangalore, Karnataka; P.K.Tiwari

Assistant Commissioner (Retd.), Kendriya Vidyalaya Sangathan; Jagdish Saran,

Department of Statistics, University of Delhi; Quddus Khan, Lecturer, Shibli National

P.G. College Azamgarh (U.P.); Sumat Kumar Jain, Lecturer, K.L. Jain Inter College

Sasni Hathras (U.P.); R.P. Gihare, Lecturer (BRC), Janpad Shiksha Kendra Chicholi

Distt. Betul (M.P.); Sangeeta Arora, P.G.T., A.P.J. School Saket, New Delhi; P.N.

Malhotra, ADE (Sc.), Directorate of Education, Delhi; D.R. Sharma, P.G.T., J.N.V.

Mungespur, Delhi; Saroj, P.G.T. Government Girls Sr. Secondary School, No. 1,

Roop Nagar, Delhi, Manoj Kumar Thakur, P.G.T., D.A.V. Public School, Rajender

Nagar, Sahibabad, Ghaziabad (U.P.) and R.P. Maurya, Reader, DESM, NCERT,

New Delhi.

Acknowledgements are due to Professor M. Chandra, Head, Department of

Education in Science and Mathematics for her support.

The Council acknowledges the efforts of the Computer Incharge, Deepak Kapoor;

Rakesh Kumar, Kamlesh Rao and Sajjad Haider Ansari, D.T.P. Operators; Kushal Pal

Singh Yadav, Copy Editor and Proof Readers, Mukhtar Hussain and Kanwar Singh.

The contribution of APC–Office, administration of DESM and Publication

Department is also duly acknowledged.

2019-20

Contents

Foreword iii

1. Sets 1

1.1 Introduction 1

1.2 Sets and their Representations 1

1.3 The Empty Set 5

1.4 Finite and Infinite Sets 6

1.5 Equal Sets 7

1.6 Subsets 9

1.7 Power Set 12

1.8 Universal Set 12

1.9 Venn Diagrams 13

1.10 Operations on Sets 14

1.11 Complement of a Set 18

1.12 Practical Problems on Union and Intersection of Two Sets 21

2. Relations and Functions 30

2.1 Introduction 30

2.2 Cartesian Product of Sets 30

2.3 Relations 34

2.4 Functions 36

3. Trigonometric Functions 49

3.1 Introduction 49

3.2 Angles 49

3.3 Trigonometric Functions 55

3.4 Trigonometric Functions of Sum and Difference of Two Angles 63

3.5 Trigonometric Equations 74

4. Principle of Mathematical Induction 86

4.1 Introduction 86

4.2 Motivation 87

4.3 The Principle of Mathematical Induction 88

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viii

5. Complex Numbers and Quadratic Equations 97

5.1 Introduction 97

5.2 Complex Numbers 97

5.3 Algebra of Complex Numbers 98

5.4 The Modulus and the Conjugate of a Complex Number 102

5.5 Argand Plane and Polar Representation 104

5.6 Quadratic Equations 108

6. Linear Inequalities 116

6.1 Introduction 116

6.2 Inequalities 116

6.3 Algebraic Solutions of Linear Inequalities in One Variable

and their Graphical Representation 118

6.4 Graphical Solution of Linear Inequalities in Two Variables 123

6.5 Solution of System of Linear Inequalities in Two Variables 127

7. Permutations and Combinations 134


7.2 Fundamental Principle of Counting 134

7.3 Permutations 138

7.4 Combinations 148

8. Binomial Theorem 160


8.2 Binomial Theorem for Positive Integral Indices 160

8.3 General and Middle Terms 167

9. Sequences and Series 177


9.2 Sequences 177

9.3 Series 179

9.4 Arithmetic Progression (A.P.) 181

9.5 Geometric Progression (G.P.) 186

9.6 Relationship Between A.M. and G.M. 191

9.7 Sum to n terms of Special Series 194

10. Straight Lines 203


10.2 Slope of a Line 204

10.3 Various Forms of the Equation of a Line 212

10.4 General Equation of a Line 220

10.5 Distance of a Point From a Line 225

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ix

11. Conic Sections 236


11.2 Sections of a Cone 236

11.3 Circle 239

11.4 Parabola 242

11.5 Ellipse 247

11.6 Hyperbola 255

12. Introduction to Three Dimensional Geometry 268


12.2 Coordinate Axes and Coordinate Planes in

Three Dimensional Space 269

12.3 Coordinates of a Point in Space 269

12.4 Distance between Two Points 271

12.5 Section Formula 273

13. Limits and Derivatives 281


13.2 Intuitive Idea of Derivatives 281

13.3 Limits 284

13.4 Limits of Trigonometric Functions 298

13.5 Derivatives 303

14. Mathematical Reasoning 321


14.2 Statements 321

14.3 New Statements from Old 324

14.4 Special Words/Phrases 329

14.5 Implications 335

14.6 Validating Statements 339

15. Statistics 347


15.2 Measures of Dispersion 349

15.3 Range 349

15.4 Mean Deviation 349

15.5 Variance and Standard Deviation 361

15.6 Analysis of Frequency Distributions 372

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x

16. Probability 383


16.2 Random Experiments 384

16.3 Event 387

16.4 Axiomatic Approach to Probability 394

Appendix 1: Infinite Series 412

A.1.1 Introduction 412

A.1.2 Binomial Theorem for any Index 412

A.1.3 Infinite Geometric Series 414

A.1.4 Exponential Series 416

A.1.5 Logarithmic Series 419

Appendix 2: Mathematical Modelling 421

A.2.1 Introduction 421

A.2.2 Preliminaries 421

A.2.3 What is Mathematical Modelling 425

Answers 433

Supplementary Material 466

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Chapter 1

SETS

Georg Cantor

(1845-1918)

vIn these days of conflict between ancient and modern studies; there

must surely be something to be said for a study which did not

begin with Pythagoras and will not end with Einstein; but

is the oldest and the youngest. — G.H. HARDY v

1.1 Introduction

The concept of set serves as a fundamental part of the

present day mathematics. Today this concept is being used

in almost every branch of mathematics. Sets are used to

define the concepts of relations and functions. The study of

geometry, sequences, probability, etc. requires the knowledge

of sets.

The theory of sets was developed by German

mathematician Georg Cantor (1845-1918). He first

encountered sets while working on “problems on trigonometric

series”. In this Chapter, we discuss some basic definitions

and operations involving sets.

1.2 Sets and their Representations

In everyday life, we often speak of collections of objects of a particular kind, such as,

a pack of cards, a crowd of people, a cricket team, etc. In mathematics also, we come

across collections, for example, of natural numbers, points, prime numbers, etc. More

specially, we examine the following collections:

(i) Odd natural numbers less than 10, i.e., 1, 3, 5, 7, 9

(ii) The rivers of India

(iii) The vowels in the English alphabet, namely, a, e, i, o, u

(iv) Various kinds of triangles

(v) Prime factors of 210, namely, 2,3,5 and 7

(vi) The solution of the equation: x2 – 5x + 6 = 0, viz, 2 and 3.

We note that each of the above example is a well-defined collection of objects in

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2 MATHEMATICS

the sense that we can definitely decide whether a given particular object belongs to a

given collection or not. For example, we can say that the river Nile does not belong to

the collection of rivers of India. On the other hand, the river Ganga does belong to this

colleciton.

We give below a few more examples of sets used particularly in mathematics, viz.

N : the set of all natural numbers

Z : the set of all integers

Q : the set of all rational numbers

R : the set of real numbers

Z+ : the set of positive integers

Q+ : the set of positive rational numbers, and

R+ : the set of positive real numbers.

The symbols for the special sets given above will be referred to throughout

this text.

Again the collection of five most renowned mathematicians of the world is not

well-defined, because the criterion for determining a mathematician as most renowned

may vary from person to person. Thus, it is not a well-defined collection.

We shall say that a set is a well-defined collection of objects.

The following points may be noted :

(i) Objects, elements and members of a set are synonymous terms.

(ii) Sets are usually denoted by capital letters A, B, C, X, Y, Z, etc.

(iii) The elements of a set are represented by small letters a, b, c, x, y, z, etc.

If a is an element of a set A, we say that “ a belongs to A” the Greek symbol ∈(epsilon) is used to denote the phrase ‘belongs to’. Thus, we write a ∈ A. If ‘b’ is not

an element of a set A, we write b ∉ A and read “b does not belong to A”.

Thus, in the set V of vowels in the English alphabet, a ∈ V but b ∉ V. In the set

P of prime factors of 30, 3 ∈ P but 15 ∉ P.

There are two methods of representing a set :

(i) Roster or tabular form

(ii) Set-builder form.

(i) In roster form, all the elements of a set are listed, the elements are being separated

by commas and are enclosed within braces { }. For example, the set of all even

positive integers less than 7 is described in roster form as {2, 4, 6}. Some more

examples of representing a set in roster form are given below :

(a) The set of all natural numbers which divide 42 is {1, 2, 3, 6, 7, 14, 21, 42}.

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SETS 3

ANote In roster form, the order in which the elements are listed is immaterial.

Thus, the above set can also be represented as {1, 3, 7, 21, 2, 6, 14, 42}.

(b) The set of all vowels in the English alphabet is {a, e, i, o, u}.

(c) The set of odd natural numbers is represented by {1, 3, 5, . . .}. The dots

tell us that the list of odd numbers continue indefinitely.

ANote It may be noted that while writing the set in roster form an element is not

generally repeated, i.e., all the elements are taken as distinct. For example, the set

of letters forming the word ‘SCHOOL’ is { S, C, H, O, L} or {H, O, L, C, S}. Here,

the order of listing elements has no relevance.

(ii) In set-builder form, all the elements of a set possess a single common property

which is not possessed by any element outside the set. For example, in the set

{a, e, i, o, u}, all the elements possess a common property, namely, each of them

is a vowel in the English alphabet, and no other letter possess this property. Denoting

this set by V, we write

V = {x : x is a vowel in English alphabet}

It may be observed that we describe the element of the set by using a symbol x

(any other symbol like the letters y, z, etc. could be used) which is followed by a colon

“ : ”. After the sign of colon, we write the characteristic property possessed by the

elements of the set and then enclose the whole description within braces. The above

description of the set V is read as “the set of all x such that x is a vowel of the English

alphabet”. In this description the braces stand for “the set of all”, the colon stands for

“such that”. For example, the set

A = {x : x is a natural number and 3 < x < 10} is read as “the set of all x such that

x is a natural number and x lies between 3 and 10.” Hence, the numbers 4, 5, 6,

7, 8 and 9 are the elements of the set A.

If we denote the sets described in (a), (b) and (c) above in roster form by A, B,

C, respectively, then A, B, C can also be represented in set-builder form as follows:

A= {x : x is a natural number which divides 42}

B= {y : y is a vowel in the English alphabet}

C= {z : z is an odd natural number}

Example 1 Write the solution set of the equation x2 + x – 2 = 0 in roster form.

Solution The given equation can be written as

(x – 1) (x + 2) = 0, i. e., x = 1, – 2

Therefore, the solution set of the given equation can be written in roster form as {1, – 2}.

Example 2 Write the set {x : x is a positive integer and x2 < 40} in the roster form.

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4 MATHEMATICS

Solution The required numbers are 1, 2, 3, 4, 5, 6. So, the given set in the roster form

is {1, 2, 3, 4, 5, 6}.

Example 3 Write the set A = {1, 4, 9, 16, 25, . . . }in set-builder form.

Solution We may write the set A as

A = {x : x is the square of a natural number}

Alternatively, we can write

A = {x : x = n2, where n ∈ N}

Example 4 Write the set 1 2 3 4 5 6

{ }2 3 4 5 6 7

, , , , , in the set-builder form.

Solution We see that each member in the given set has the numerator one less than

the denominator. Also, the numerator begin from 1 and do not exceed 6. Hence, in the

set-builder form the given set is

where is a natural number and 1 61

nx : x , n n

n

= ≤ ≤ +

Example 5 Match each of the set on the left described in the roster form with the

same set on the right described in the set-builder form :

(i) {P, R, I, N, C, A, L} (a) { x : x is a positive integer and is a divisor of 18}

(ii) { 0 } (b) { x : x is an integer and x2 – 9 = 0}

(iii) {1, 2, 3, 6, 9, 18} (c) {x : x is an integer and x + 1= 1}

(iv) {3, –3} (d) {x : x is a letter of the word PRINCIPAL}

Solution Since in (d), there are 9 letters in the word PRINCIPAL and two letters P and I

are repeated, so (i) matches (d). Similarly, (ii) matches (c) as x + 1 = 1 implies

x = 0. Also, 1, 2 ,3, 6, 9, 18 are all divisors of 18 and so (iii) matches (a). Finally, x2 – 9 = 0

implies x = 3, –3 and so (iv) matches (b).

EXERCISE 1.1

1. Which of the following are sets ? Justify your answer.

(i) The collection of all the months of a year beginning with the letter J.

(ii) The collection of ten most talented writers of India.

(iii) A team of eleven best-cricket batsmen of the world.

(iv) The collection of all boys in your class.

(v) The collection of all natural numbers less than 100.

(vi) A collection of novels written by the writer Munshi Prem Chand.

(vii) The collection of all even integers.

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SETS 5

(viii) The collection of questions in this Chapter.

(ix) A collection of most dangerous animals of the world.

2. Let A = {1, 2, 3, 4, 5, 6}. Insert the appropriate symbol ∈ or ∉ in the blank

spaces:

(i) 5. . .A (ii) 8 . . . A (iii) 0. . .A

(iv) 4. . . A (v) 2. . .A (vi) 10. . .A

3. Write the following sets in roster form:

(i) A = {x : x is an integer and –3 ≤ x < 7}

(ii) B = {x : x is a natural number less than 6}

(iii) C = {x : x is a two-digit natural number such that the sum of its digits is 8}

(iv) D = {x : x is a prime number which is divisor of 60}

(v) E = The set of all letters in the word TRIGONOMETRY

(vi) F = The set of all letters in the word BETTER

4. Write the following sets in the set-builder form :

(i) (3, 6, 9, 12} (ii) {2,4,8,16,32} (iii) {5, 25, 125, 625}

(iv) {2, 4, 6, . . .} (v) {1,4,9, . . .,100}

5. List all the elements of the following sets :

(i) A = {x : x is an odd natural number}

(ii) B = {x : x is an integer, 1

2– < x <

9

2}

(iii) C = {x : x is an integer, x2 ≤ 4}

(iv) D = {x : x is a letter in the word “LOYAL”}

(v) E = {x : x is a month of a year not having 31 days}

(vi) F = {x : x is a consonant in the English alphabet which precedes k }.

6. Match each of the set on the left in the roster form with the same set on the right

described in set-builder form:

(i) {1, 2, 3, 6} (a) {x : x is a prime number and a divisor of 6}

(ii) {2, 3} (b) {x : x is an odd natural number less than 10}

(iii) {M,A,T,H,E,I,C,S} (c) {x : x is natural number and divisor of 6}

(iv) {1, 3, 5, 7, 9} (d) {x : x is a letter of the word MATHEMATICS}.

1.3 The Empty Set

Consider the set

A = { x : x is a student of Class XI presently studying in a school }

We can go to the school and count the number of students presently studying in

Class XI in the school. Thus, the set A contains a finite number of elements.

We now write another set B as follows:

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6 MATHEMATICS

B = { x : x is a student presently studying in both Classes X and XI }

We observe that a student cannot study simultaneously in both Classes X and XI.

Thus, the set B contains no element at all.

Definition 1 A set which does not contain any element is called the empty set or the

null set or the void set.

According to this definition, B is an empty set while A is not an empty set. The

empty set is denoted by the symbol φ or { }.

We give below a few examples of empty sets.

(i) Let A = {x : 1 < x < 2, x is a natural number}. Then A is the empty set,

because there is no natural number between 1 and 2.

(ii) B = {x : x2 – 2 = 0 and x is rational number}. Then B is the empty set because

the equation x2 – 2 = 0 is not satisfied by any rational value of x.

(iii) C = {x : x is an even prime number greater than 2}.Then C is the empty set,

because 2 is the only even prime number.

(iv) D = { x : x2 = 4, x is odd }. Then D is the empty set, because the equation

x2 = 4 is not satisfied by any odd value of x.

1.4 Finite and Infinite Sets

Let A = {1, 2, 3, 4, 5}, B = {a, b, c, d, e, g}

and C = { men living presently in different parts of the world}

We observe that A contains 5 elements and B contains 6 elements. How many elements

does C contain? As it is, we do not know the number of elements in C, but it is some

natural number which may be quite a big number. By number of elements of a set S,

we mean the number of distinct elements of the set and we denote it by n (S). If n (S)

is a natural number, then S is non-empty finite set.

Consider the set of natural numbers. We see that the number of elements of this

set is not finite since there are infinite number of natural numbers. We say that the set

of natural numbers is an infinite set. The sets A, B and C given above are finite sets

and n(A) = 5, n(B) = 6 and n(C) = some finite number.

Definition 2 A set which is empty or consists of a definite number of elements is

called finite otherwise, the set is called infinite.

Consider some examples :

(i) Let W be the set of the days of the week. Then W is finite.

(ii) Let S be the set of solutions of the equation x2 –16 = 0. Then S is finite.

(iii) Let G be the set of points on a line. Then G is infinite.

When we represent a set in the roster form, we write all the elements of the set

within braces { }. It is not possible to write all the elements of an infinite set within

braces { } because the numbers of elements of such a set is not finite. So, we represent

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SETS 7

some infinite set in the roster form by writing a few elements which clearly indicate the

structure of the set followed ( or preceded ) by three dots.

For example, {1, 2, 3 . . .} is the set of natural numbers, {1, 3, 5, 7, . . .} is the set

of odd natural numbers, {. . .,–3, –2, –1, 0,1, 2 ,3, . . .} is the set of integers. All these

sets are infinite.

ANote All infinite sets cannot be described in the roster form. For example, the

set of real numbers cannot be described in this form, because the elements of this

set do not follow any particular pattern.

Example 6 State which of the following sets are finite or infinite :

(i) {x : x ∈ N and (x – 1) (x –2) = 0}

(ii) {x : x ∈ N and x2 = 4}

(iii) {x : x ∈ N and 2x –1 = 0}

(iv) {x : x ∈ N and x is prime}

(v) {x : x ∈ N and x is odd}

Solution (i) Given set = {1, 2}. Hence, it is finite.

(ii) Given set = {2}. Hence, it is finite.

(iii) Given set = φ. Hence, it is finite.

(iv) The given set is the set of all prime numbers and since set of prime

numbers is infinite. Hence the given set is infinite

(v) Since there are infinite number of odd numbers, hence, the given set is

infinite.

1.5 Equal Sets

Given two sets A and B, if every element of A is also an element of B and if every

element of B is also an element of A, then the sets A and B are said to be equal.

Clearly, the two sets have exactly the same elements.

Definition 3 Two sets A and B are said to be equal if they have exactly the same

elements and we write A = B. Otherwise, the sets are said to be unequal and we write

A ≠ B.

We consider the following examples :

(i) Let A = {1, 2, 3, 4} and B = {3, 1, 4, 2}. Then A = B.

(ii) Let A be the set of prime numbers less than 6 and P the set of prime factors

of 30. Then A and P are equal, since 2, 3 and 5 are the only prime factors of

30 and also these are less than 6.

ANote A set does not change if one or more elements of the set are repeated.

For example, the sets A = {1, 2, 3} and B = {2, 2, 1, 3, 3} are equal, since each

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element of A is in B and vice-versa. That is why we generally do not repeat any

element in describing a set.

Example 7 Find the pairs of equal sets, if any, give reasons:

A = {0}, B = {x : x > 15 and x < 5},

C = {x : x – 5 = 0 }, D = {x: x2 = 25},

E = {x : x is an integral positive root of the equation x2 – 2x –15 = 0}.

Solution Since 0 ∈ A and 0 does not belong to any of the sets B, C, D and E, it

follows that, A ≠ B, A ≠ C, A ≠ D, A ≠ E.

Since B = φ but none of the other sets are empty. Therefore B ≠ C, B ≠ D

and B ≠ E. Also C = {5} but –5 ∈ D, hence C ≠ D.

Since E = {5}, C = E. Further, D = {–5, 5} and E = {5}, we find that, D ≠ E.

Thus, the only pair of equal sets is C and E.

Example 8 Which of the following pairs of sets are equal? Justify your answer.

(i) X, the set of letters in “ALLOY” and B, the set of letters in “LOYAL”.

(ii) A = {n : n ∈ Z and n2 ≤ 4} and B = {x : x ∈ R and x2 – 3x + 2 = 0}.

Solution (i) We have, X = {A, L, L, O, Y}, B = {L, O, Y, A, L}. Then X and B are

equal sets as repetition of elements in a set do not change a set. Thus,

X = {A, L, O, Y} = B

(ii) A = {–2, –1, 0, 1, 2}, B = {1, 2}. Since 0 ∈ A and 0 ∉ B, A and B are not equal sets.

EXERCISE 1.2

1. Which of the following are examples of the null set

(i) Set of odd natural numbers divisible by 2

(ii) Set of even prime numbers

(iii) { x : x is a natural numbers, x < 5 and x > 7 }

(iv) { y : y is a point common to any two parallel lines}

2. Which of the following sets are finite or infinite

(i) The set of months of a year

(ii) {1, 2, 3, . . .}

(iii) {1, 2, 3, . . .99, 100}

(iv) The set of positive integers greater than 100

(v) The set of prime numbers less than 99

3. State whether each of the following set is finite or infinite:

(i) The set of lines which are parallel to the x-axis

(ii) The set of letters in the English alphabet

(iii) The set of numbers which are multiple of 5

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(iv) The set of animals living on the earth

(v) The set of circles passing through the origin (0,0)

4. In the following, state whether A = B or not:

(i) A = { a, b, c, d } B = { d, c, b, a }

(ii) A = { 4, 8, 12, 16 } B = { 8, 4, 16, 18}

(iii) A = {2, 4, 6, 8, 10} B = { x : x is positive even integer and x ≤ 10}

(iv) A = { x : x is a multiple of 10}, B = { 10, 15, 20, 25, 30, . . . }

5. Are the following pair of sets equal ? Give reasons.

(i) A = {2, 3}, B = {x : x is solution of x2 + 5x + 6 = 0}

(ii) A = { x : x is a letter in the word FOLLOW}

B = { y : y is a letter in the word WOLF}

6. From the sets given below, select equal sets :

A = { 2, 4, 8, 12}, B = { 1, 2, 3, 4}, C = { 4, 8, 12, 14}, D = { 3, 1, 4, 2}

E = {–1, 1}, F = { 0, a}, G = {1, –1}, H = { 0, 1}

1.6 SubsetsConsider the sets : X = set of all students in your school, Y = set of all students in your

class.

We note that every element of Y is also an element of X; we say that Y is a subset

of X. The fact that Y is subset of X is expressed in symbols as Y ⊂ X. The symbol ⊂stands for ‘is a subset of’ or ‘is contained in’.

Definition 4 A set A is said to be a subset of a set B if every element of A is also an

element of B.

In other words, A ⊂ B if whenever a ∈ A, then a ∈ B. It is often convenient to

use the symbol “⇒” which means implies. Using this symbol, we can write the definiton

of subset as follows:

A ⊂ B if a ∈ A ⇒ a ∈ B

We read the above statement as “A is a subset of B if a is an element of A

implies that a is also an element of B”. If A is not a subset of B, we write A ⊄ B.

We may note that for A to be a subset of B, all that is needed is that every

element of A is in B. It is possible that every element of B may or may not be in A. If

it so happens that every element of B is also in A, then we shall also have B ⊂ A. In this

case, A and B are the same sets so that we have A ⊂ B and B ⊂ A ⇔ A = B, where

“⇔” is a symbol for two way implications, and is usually read as if and only if (briefly

written as “iff”).

It follows from the above definition that every set A is a subset of itself, i.e.,

A ⊂ A. Since the empty set φ has no elements, we agree to say that φ is a subset of

every set. We now consider some examples :

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(i) The set Q of rational numbers is a subset of the set R of real numbes, and

we write Q ⊂ R.

(ii) If A is the set of all divisors of 56 and B the set of all prime divisors of 56,

then B is a subset of A and we write B ⊂ A.

(iii) Let A = {1, 3, 5} and B = {x : x is an odd natural number less than 6}. Then

A ⊂ B and B ⊂ A and hence A = B.

(iv) Let A = { a, e, i, o, u} and B = { a, b, c, d}. Then A is not a subset of B,

also B is not a subset of A.

Let A and B be two sets. If A ⊂ B and A ≠ B , then A is called a proper subset

of B and B is called superset of A. For example,

A = {1, 2, 3} is a proper subset of B = {1, 2, 3, 4}.

If a set A has only one element, we call it a singleton set. Thus,{ a } is a

singleton set.

Example 9 Consider the sets

φ, A = { 1, 3 }, B = {1, 5, 9}, C = {1, 3, 5, 7, 9}.

Insert the symbol ⊂ or ⊄ between each of the following pair of sets:

(i) φ . . . B (ii) A . . . B (iii) A . . . C (iv) B . . . C

Solution (i) φ ⊂ B as φ is a subset of every set.

(ii) A ⊄ B as 3 ∈ A and 3 ∉ B

(iii) A ⊂ C as 1, 3 ∈ A also belongs to C

(iv) B ⊂ C as each element of B is also an element of C.

Example 10 Let A = { a, e, i, o, u} and B = { a, b, c, d}. Is A a subset of B ? No.

(Why?). Is B a subset of A? No. (Why?)

Example 11 Let A, B and C be three sets. If A ∈ B and B ⊂ C, is it true that

A ⊂ C?. If not, give an example.

Solution No. Let A = {1}, B = {{1}, 2} and C = {{1}, 2, 3}. Here A ∈ B as A = {1}

and B ⊂ C. But A ⊄ C as 1 ∈ A and 1 ∉ C.

Note that an element of a set can never be a subset of itself.

1.6.1 Subsets of set of real numbers

As noted in Section 1.6, there are many important subsets of R. We give below the

names of some of these subsets.

The set of natural numbers N = {1, 2, 3, 4, 5, . . .}

The set of integers Z = {. . ., –3, –2, –1, 0, 1, 2, 3, . . .}

The set of rational numbers Q = { x : x = p

q , p, q ∈ Z and q ≠ 0}

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which is read “ Q is the set of all numbers x such that x equals the quotient p

q, where

p and q are integers and q is not zero”. Members of Q include –5 (which can be

expressed as 5

1– ) ,

7

5,

13

2 (which can be expressed as

7

2) and

11

3– .

The set of irrational numbers, denoted by T, is composed of all other real numbers.

Thus T = {x : x ∈ R and x ∉ Q}, i.e., all real numbers that are not rational.

Members of T include 2 , 5 and π .

Some of the obvious relations among these subsets are:

N ⊂ Z ⊂ Q, Q ⊂ R, T ⊂ R, N ⊄ T.

1.6.2 Intervals as subsets of R Let a, b ∈ R and a < b. Then the set of real numbers

{ y : a < y < b} is called an open interval and is denoted by (a, b). All the points

between a and b belong to the open interval (a, b) but a, b themselves do not belong to

this interval.

The interval which contains the end points also is called closed interval and is

denoted by [ a, b ]. Thus

[ a, b ] = {x : a ≤ x ≤ b}

We can also have intervals closed at one end and open at the other, i.e.,

[ a, b ) = {x : a ≤ x < b} is an open interval from a to b, including a but excluding b.

( a, b ] = { x : a < x ≤ b } is an open interval from a to b including b but excluding a.

These notations provide an alternative way of designating the subsets of set of

real numbers. For example , if A = (–3, 5) and B = [–7, 9], then A ⊂ B. The set [ 0, ∞)

defines the set of non-negative real numbers, while set ( – ∞, 0 ) defines the set of

negative real numbers. The set ( – ∞, ∞ ) describes the set of real numbers in relation

to a line extending from – ∞ to ∞.

On real number line, various types of intervals described above as subsets of R,

are shown in the Fig 1.1.

Here, we note that an interval contains infinitely many points.

For example, the set {x : x ∈ R, –5 < x ≤ 7}, written in set-builder form, can be

written in the form of interval as (–5, 7] and the interval [–3, 5) can be written in set-

builder form as {x : –3 ≤ x < 5}.

Fig 1.1

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12 MATHEMATICS

The number (b – a) is called the length of any of the intervals (a, b), [a, b],

[a, b) or (a, b].

1.7 Power Set

Consider the set {1, 2}. Let us write down all the subsets of the set {1, 2}. We

know that φ is a subset of every set . So, φ is a subset of {1, 2}. We see that {1}

and { 2 }are also subsets of {1, 2}. Also, we know that every set is a subset of

itself. So, { 1, 2 } is a subset of {1, 2}. Thus, the set { 1, 2 } has, in all, four

subsets, viz. φ, { 1 }, { 2 } and { 1, 2 }. The set of all these subsets is called the

power set of { 1, 2 }.

Definition 5 The collection of all subsets of a set A is called the power set of A. It is

denoted by P(A). In P(A), every element is a set.

Thus, as in above, if A = { 1, 2 }, then

P( A ) = { φ,{ 1 }, { 2 }, { 1,2 }}

Also, note that n [ P (A) ] = 4 = 22

In general, if A is a set with n(A) = m, then it can be shown that

n [ P(A)] = 2m.

1.8 Universal Set

Usually, in a particular context, we have to deal with the elements and subsets of a

basic set which is relevant to that particular context. For example, while studying the

system of numbers, we are interested in the set of natural numbers and its subsets such

as the set of all prime numbers, the set of all even numbers, and so forth. This basic set

is called the “Universal Set”. The universal set is usually denoted by U, and all its

subsets by the letters A, B, C, etc.

For example, for the set of all integers, the universal set can be the set of rational

numbers or, for that matter, the set R of real numbers. For another example, in human

population studies, the universal set consists of all the people in the world.

EXERCISE 1.3

1. Make correct statements by filling in the symbols ⊂ or ⊄ in the blank spaces :

(i) { 2, 3, 4 } . . . { 1, 2, 3, 4,5 } (ii) { a, b, c } . . . { b, c, d }

(iii) {x : x is a student of Class XI of your school}. . .{x : x student of your school}

(iv) {x : x is a circle in the plane} . . .{x : x is a circle in the same plane with

radius 1 unit}

(v) {x : x is a triangle in a plane} . . . {x : x is a rectangle in the plane}

(vi) {x : x is an equilateral triangle in a plane} . . . {x : x is a triangle in the same plane}

(vii) {x : x is an even natural number} . . . {x : x is an integer}

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Fig 1.2

2. Examine whether the following statements are true or false:

(i) { a, b } ⊄ { b, c, a }

(ii) { a, e } ⊂ { x : x is a vowel in the English alphabet}

(iii) { 1, 2, 3 } ⊂ { 1, 3, 5 }

(iv) { a } ⊂ { a, b, c }

(v) { a } ∈ { a, b, c }

(vi) { x : x is an even natural number less than 6} ⊂ { x : x is a natural number

which divides 36}

3. Let A = { 1, 2, { 3, 4 }, 5 }. Which of the following statements are incorrect and why?

(i) {3, 4} ⊂ A (ii) {3, 4} ∈ A (iii) {{3, 4}} ⊂ A(iv) 1 ∈ A (v) 1 ⊂ A (vi) {1, 2, 5} ⊂ A(vii) {1, 2, 5} ∈ A (viii) {1, 2, 3} ⊂ A (ix) φ ∈ A

(x) φ ⊂ A (xi) {φ} ⊂ A

4. Write down all the subsets of the following sets

(i) {a} (ii) {a, b} (iii) {1, 2, 3} (iv) φ5. How many elements has P(A), if A = φ?

6. Write the following as intervals :

(i) {x : x ∈ R, – 4 < x ≤ 6} (ii) {x : x ∈ R, – 12 < x < –10}

(iii) {x : x ∈ R, 0 ≤ x < 7} (iv) {x : x ∈ R, 3 ≤ x ≤ 4}

7. Write the following intervals in set-builder form :

(i) (– 3, 0) (ii) [6 , 12] (iii) (6, 12] (iv) [–23, 5)

8. What universal set(s) would you propose for each of the following :

(i) The set of right triangles. (ii) The set of isosceles triangles.

9. Given the sets A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}, which of the

following may be considered as universal set (s) for all the three sets A, B and C

(i) {0, 1, 2, 3, 4, 5, 6}

(ii) φ(iii) {0,1,2,3,4,5,6,7,8,9,10}

(iv) {1,2,3,4,5,6,7,8}

1.9 Venn Diagrams

Most of the relationships between sets can be

represented by means of diagrams which are known

as Venn diagrams. Venn diagrams are named after

the English logician, John Venn (1834-1883). These

diagrams consist of rectangles and closed curves

usually circles. The universal set is represented

usually by a rectangle and its subsets by circles.

In Venn diagrams, the elements of the sets

are written in their respective circles (Figs 1.2 and 1.3)

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Fig 1.3

Illustration 1 In Fig 1.2, U = {1,2,3, ..., 10} is the

universal set of which

A = {2,4,6,8,10} is a subset.

Illustration 2 In Fig 1.3, U = {1,2,3, ..., 10} is the

universal set of which

A = {2,4,6,8,10} and B = {4, 6} are subsets,

and also B ⊂ A.

The reader will see an extensive use of the

Venn diagrams when we discuss the union, intersection and difference of sets.

1.10 Operations on Sets

In earlier classes, we have learnt how to perform the operations of addition, subtraction,

multiplication and division on numbers. Each one of these operations was performed

on a pair of numbers to get another number. For example, when we perform the

operation of addition on the pair of numbers 5 and 13, we get the number 18. Again,

performing the operation of multiplication on the pair of numbers 5 and 13, we get 65.

Similarly, there are some operations which when performed on two sets give rise to

another set. We will now define certain operations on sets and examine their properties.

Henceforth, we will refer all our sets as subsets of some universal set.

1.10.1 Union of sets Let A and B be any two sets. The union of A and B is the set

which consists of all the elements of A and all the elements of B, the common elements

being taken only once. The symbol ‘∪’ is used to denote the union. Symbolically, we

write A ∪ B and usually read as ‘A union B’.

Example 12 Let A = { 2, 4, 6, 8} and B = { 6, 8, 10, 12}. Find A ∪ B.

Solution We have A ∪ B = { 2, 4, 6, 8, 10, 12}

Note that the common elements 6 and 8 have been taken only once while writing

A ∪ B.

Example 13 Let A = { a, e, i, o, u } and B = { a, i, u }. Show that A ∪ B = A

Solution We have, A ∪ B = { a, e, i, o, u } = A.

This example illustrates that union of sets A and its subset B is the set A

itself, i.e., if B ⊂ A, then A ∪ B = A.

Example 14 Let X = {Ram, Geeta, Akbar} be the set of students of Class XI, who are

in school hockey team. Let Y = {Geeta, David, Ashok} be the set of students from

Class XI who are in the school football team. Find X ∪ Y and interpret the set.

Solution We have, X ∪ Y = {Ram, Geeta, Akbar, David, Ashok}. This is the set of

students from Class XI who are in the hockey team or the football team or both.

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Thus, we can define the union of two sets as follows:

Definition 6 The union of two sets A and B is the set C which consists of all those

elements which are either in A or in B (including

those which are in both). In symbols, we write.

A ∪ B = { x : x ∈A or x ∈B }

The union of two sets can be represented by a

Venn diagram as shown in Fig 1.4.

The shaded portion in Fig 1.4 represents A ∪ B.

Some Properties of the Operation of Union

(i) A ∪ B = B ∪ A (Commutative law)

(ii) ( A ∪ B ) ∪ C = A ∪ ( B ∪ C)

(Associative law )

(iii) A ∪ φ = A (Law of identity element, φ is the identity of ∪)

(iv) A ∪ A = A (Idempotent law)

(v) U ∪ A = U (Law of U)

1.10.2 Intersection of sets The intersection of sets A and B is the set of all elements

which are common to both A and B. The symbol ‘∩’ is used to denote the intersection.

The intersection of two sets A and B is the set of all those elements which belong to

both A and B. Symbolically, we write A ∩ B = {x : x ∈ A and x ∈ B}.

Example 15 Consider the sets A and B of Example 12. Find A ∩ B.

Solution We see that 6, 8 are the only elements which are common to both A and B.

Hence A ∩ B = { 6, 8 }.

Example 16 Consider the sets X and Y of Example 14. Find X ∩ Y.

Solution We see that element ‘Geeta’ is the only element common to both. Hence,

X ∩ Y = {Geeta}.

Example 17 Let A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and B = { 2, 3, 5, 7 }. Find A ∩ B and

hence show that A ∩ B = B.

Solution We have A ∩ B = { 2, 3, 5, 7 } = B. We

note that B ⊂ A and that A ∩ B = B.

Definition 7 The intersection of two sets A and B

is the set of all those elements which belong to both

A and B. Symbolically, we write

A ∩ B = {x : x ∈ A and x ∈ B}

The shaded portion in Fig 1.5 indicates the

intersection of A and B.

Fig 1.4

Fig 1.5

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If A and B are two sets such that A ∩ B = φ, then

A and B are called disjoint sets.

For example, let A = { 2, 4, 6, 8 } and

B = { 1, 3, 5, 7 }. Then A and B are disjoint sets,

because there are no elements which are common to

A and B. The disjoint sets can be represented by

means of Venn diagram as shown in the Fig 1.6

In the above diagram, A and B are disjoint sets.

Some Properties of Operation of Intersection

(i) A ∩ B = B ∩ A (Commutative law).

(ii) ( A ∩ B ) ∩ C = A ∩ ( B ∩ C ) (Associative law).

(iii) φ ∩ A = φ, U ∩ A = A (Law of φ and U).

(iv) A ∩ A = A (Idempotent law)

(v) A ∩ ( B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C ) (Distributive law ) i. e.,

∩ distributes over ∪This can be seen easily from the following Venn diagrams [Figs 1.7 (i) to (v)].

A B

U

Fig 1.6

(i) (iii)

(ii) (iv)

(v)Figs 1.7 (i) to (v)

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Fig 1.8

Fig 1.9

1.10.3 Difference of sets The difference of the sets A and B in this order is the set

of elements which belong to A but not to B. Symbolically, we write A – B and read as

“ A minus B”.

Example 18 Let A = { 1, 2, 3, 4, 5, 6}, B = { 2, 4, 6, 8 }. Find A – B and B – A.

Solution We have, A – B = { 1, 3, 5 }, since the elements 1, 3, 5 belong to A but

not to B and B – A = { 8 }, since the element 8 belongs to B and not to A.

We note that A – B ≠ B – A.

Example 19 Let V = { a, e, i, o, u } and

B = { a, i, k, u}. Find V – B and B – V

Solution We have, V – B = { e, o }, since the elements

e, o belong to V but not to B and B – V = { k }, since

the element k belongs to B but not to V.

We note that V – B ≠ B – V. Using the set-

builder notation, we can rewrite the definition of

difference as

A – B = { x : x ∈ A and x ∉ B }

The difference of two sets A and B can be

represented by Venn diagram as shown in Fig 1.8.

The shaded portion represents the difference of

the two sets A and B.

Remark The sets A – B, A ∩ B and B – A are

mutually disjoint sets, i.e., the intersection of any of

these two sets is the null set as shown in Fig 1.9.

EXERCISE 1.4

1. Find the union of each of the following pairs of sets :

(i) X = {1, 3, 5} Y = {1, 2, 3}

(ii) A = [ a, e, i, o, u} B = {a, b, c}

(iii) A = {x : x is a natural number and multiple of 3}

B = {x : x is a natural number less than 6}

(iv) A = {x : x is a natural number and 1 < x ≤ 6 }

B = {x : x is a natural number and 6 < x < 10 }

(v) A = {1, 2, 3}, B = φ2. Let A = { a, b }, B = {a, b, c}. Is A ⊂ B ? What is A ∪ B ?

3. If A and B are two sets such that A ⊂ B, then what is A ∪ B ?

4. If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8 }and D = { 7, 8, 9, 10 }; find

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(i) A ∪ B (ii) A ∪ C (iii) B ∪ C (iv) B ∪ D

(v) A ∪ B ∪ C (vi) A ∪ B ∪ D (vii) B ∪ C ∪ D

5. Find the intersection of each pair of sets of question 1 above.

6. If A = { 3, 5, 7, 9, 11 }, B = {7, 9, 11, 13}, C = {11, 13, 15}and D = {15, 17}; find

(i) A ∩ B (ii) B ∩ C (iii) A ∩ C ∩ D

(iv) A ∩ C (v) B ∩ D (vi) A ∩ (B ∪ C)

(vii) A ∩ D (viii) A ∩ (B ∪ D) (ix) ( A ∩ B ) ∩ ( B ∪ C )

(x) ( A ∪ D) ∩ ( B ∪ C)

7. If A = {x : x is a natural number }, B = {x : x is an even natural number}

C = {x : x is an odd natural number}andD = {x : x is a prime number }, find

(i) A ∩ B (ii) A ∩ C (iii) A ∩ D

(iv) B ∩ C (v) B ∩ D (vi) C ∩ D

8. Which of the following pairs of sets are disjoint

(i) {1, 2, 3, 4} and {x : x is a natural number and 4 ≤ x ≤ 6 }

(ii) { a, e, i, o, u } and { c, d, e, f }

(iii) {x : x is an even integer } and {x : x is an odd integer}

9. If A = {3, 6, 9, 12, 15, 18, 21}, B = { 4, 8, 12, 16, 20 },

C = { 2, 4, 6, 8, 10, 12, 14, 16 }, D = {5, 10, 15, 20 }; find

(i) A – B (ii) A – C (iii) A – D (iv) B – A

(v) C – A (vi) D – A (vii) B – C (viii) B – D

(ix) C – B (x) D – B (xi) C – D (xii) D – C

10. If X= { a, b, c, d } and Y = { f, b, d, g}, find

(i) X – Y (ii) Y – X (iii) X ∩ Y

11. If R is the set of real numbers and Q is the set of rational numbers, then what is

R – Q?

12. State whether each of the following statement is true or false. Justify your answer.

(i) { 2, 3, 4, 5 } and { 3, 6} are disjoint sets.

(ii) { a, e, i, o, u } and { a, b, c, d }are disjoint sets.

(iii) { 2, 6, 10, 14 } and { 3, 7, 11, 15} are disjoint sets.

(iv) { 2, 6, 10 } and { 3, 7, 11} are disjoint sets.

1.11 Complement of a Set

Let U be the universal set which consists of all prime numbers and A be the subset of

U which consists of all those prime numbers that are not divisors of 42. Thus,

A = {x : x ∈ U and x is not a divisor of 42 }. We see that 2 ∈ U but 2 ∉ A, because

2 is divisor of 42. Similarly, 3 ∈ U but 3 ∉ A, and 7 ∈ U but 7 ∉ A. Now 2, 3 and 7 are

the only elements of U which do not belong to A. The set of these three prime numbers,

i.e., the set {2, 3, 7} is called the Complement of A with respect to U, and is denoted by

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A′. So we have A′ = {2, 3, 7}. Thus, we see that

A′ = {x : x ∈ U and x ∉ A }. This leads to the following definition.

Definition 8 Let U be the universal set and A a subset of U. Then the complement of

A is the set of all elements of U which are not the elements of A. Symbolically, we

write A′ to denote the complement of A with respect to U. Thus,

A′ = {x : x ∈ U and x ∉ A }. Obviously A′ = U – A

We note that the complement of a set A can be looked upon, alternatively, as the

difference between a universal set U and the set A.

Example 20 Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {1, 3, 5, 7, 9}. Find A′.

Solution We note that 2, 4, 6, 8, 10 are the only elements of U which do not belong to

A. Hence A′ = { 2, 4, 6, 8,10 }.

Example 21 Let U be universal set of all the students of Class XI of a coeducational

school and A be the set of all girls in Class XI. Find A′.

Solution Since A is the set of all girls, A′ is clearly the set of all boys in the class.

ANote If A is a subset of the universal set U, then its complement A′ is also a

subset of U.

Again in Example 20 above, we have A′ = { 2, 4, 6, 8, 10 }

Hence (A′ )′ = {x : x ∈ U and x ∉ A′}= {1, 3, 5, 7, 9} = A

It is clear from the definition of the complement that for any subset of the universal

set U, we have ( A′ )′ = A

Now, we want to find the results for ( A ∪ B )′ and A′ ∩ B′ in the followng

example.

Example 22 Let U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5}.

Find A′, B′ , A′ ∩ B′, A ∪ B and hence show that ( A ∪ B )′ = A′ ∩ B′.

Solution Clearly A′ = {1, 4, 5, 6}, B′ = { 1, 2, 6 }. Hence A′ ∩ B′ = { 1, 6 }

Also A ∪ B = { 2, 3, 4, 5 }, so that (A ∪ B )′ = { 1, 6 }

( A ∪ B )′ = { 1, 6 } = A′ ∩ B′It can be shown that the above result is true in general. If A and B are any two

subsets of the universal set U, then

( A ∪ B )′ = A′ ∩ B′. Similarly, ( A ∩ B )′ = A′ ∪ B′ . These two results are stated

in words as follows :

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20 MATHEMATICS

The complement of the union of two sets is

the intersection of their complements and the

complement of the intersection of two sets is the

union of their complements. These are called De

Morgan’s laws. These are named after the

mathematician De Morgan.

The complement A′ of a set A can be represented

by a Venn diagram as shown in Fig 1.10.

The shaded portion represents the complement of the set A.

Some Properties of Complement Sets

1. Complement laws: (i) A ∪ A′ = U (ii) A ∩ A′ = φ

2. De Morgan’s law: (i) (A ∪ B)´ = A′ ∩ B′ (ii) (A ∩ B )′ = A′ ∪ B′

3. Law of double complementation : (A′ )′ = A

4. Laws of empty set and universal set φ′ = U and U′ = φ.

These laws can be verified by using Venn diagrams.

EXERCISE 1.5

1. Let U = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = { 1, 2, 3, 4}, B = { 2, 4, 6, 8 } and

C = { 3, 4, 5, 6 }. Find (i) A′ (ii) B′ (iii) (A ∪ C)′ (iv) (A ∪ B)′ (v) (A′)′(vi) (B – C)′

2. If U = { a, b, c, d, e, f, g, h}, find the complements of the following sets :

(i) A = {a, b, c} (ii) B = {d, e, f, g}

(iii) C = {a, c, e, g} (iv) D = { f, g, h, a}

3. Taking the set of natural numbers as the universal set, write down the complements

of the following sets:

(i) {x : x is an even natural number} (ii) { x : x is an odd natural number }

(iii) {x : x is a positive multiple of 3} (iv) { x : x is a prime number }

(v) {x : x is a natural number divisible by 3 and 5}

(vi) { x : x is a perfect square } (vii) { x : x is a perfect cube}

(viii) { x : x + 5 = 8 } (ix) { x : 2x + 5 = 9}

(x) { x : x ≥ 7 } (xi) { x : x ∈ N and 2x + 1 > 10 }

4. If U = {1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = {2, 4, 6, 8} and B = { 2, 3, 5, 7}. Verify that

(i) (A ∪ B)′ = A′ ∩ B′ (ii) (A ∩ B)′ = A′ ∪ B′5. Draw appropriate Venn diagram for each of the following :

(i) (A ∪ B)′, (ii) A′ ∩ B′, (iii) (A ∩ B)′, (iv) A′ ∪ B′6. Let U be the set of all triangles in a plane. If A is the set of all triangles with at

least one angle different from 60°, what is A′?

Fig 1.10

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SETS 21

7. Fill in the blanks to make each of the following a true statement :

(i) A ∪ A′ = . . . (ii) φ′ ∩ A = . . .

(iii) A ∩ A′ = . . . (iv) U′ ∩ A = . . .

1.12 Practical Problems on Union and

Intersection of Two Sets

In earlier Section, we have learnt union, intersection

and difference of two sets. In this Section, we will

go through some practical problems related to our

daily life.The formulae derived in this Section will

also be used in subsequent Chapter on Probability

(Chapter 16).

Let A and B be finite sets. If A ∩ B = φ, then

(i) n ( A ∪ B ) = n ( A ) + n ( B ) ... (1)

The elements in A ∪ B are either in A or in B but not in both as A ∩ B = φ. So, (1)

follows immediately.

In general, if A and B are finite sets, then

(ii) n ( A ∪ B ) = n ( A ) + n ( B ) – n ( A ∩ B ) ... (2)

Note that the sets A – B, A ∩ B and B – A are disjoint and their union is A ∪ B(Fig 1.11). Therefore

n ( A ∪ B) = n ( A – B) + n ( A ∩ B ) + n ( B – A )

= n ( A – B) + n ( A ∩ B ) + n ( B – A ) + n ( A ∩ B ) – n ( A ∩ B)

= n ( A ) + n ( B ) – n ( A ∩ B), which verifies (2)

(iii) If A, B and C are finite sets, then

n ( A ∪ B ∪ C ) = n ( A ) + n ( B ) + n ( C ) – n ( A ∩ B ) – n ( B ∩ C)

– n ( A ∩ C ) + n ( A ∩ B ∩ C ) ... (3)

In fact, we have

n ( A ∪ B ∪ C ) = n (A) + n ( B ∪ C ) – n [ A ∩ ( B ∪ C ) ] [ by (2) ]

= n (A) + n ( B ) + n ( C ) – n ( B ∩ C ) – n [ A ∩ ( B ∪ C ) ] [ by (2) ]

Since A ∩ ( B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C ), we get

n [ A ∩ ( B ∪ C ) ] = n ( A ∩ B ) + n ( A ∩ C ) – n [ ( A ∩ B ) ∩ (A ∩ C)]

= n ( A ∩ B ) + n ( A ∩ C ) – n (A ∩ B ∩ C)

Therefore

n ( A ∪ B ∪ C ) = n (A) + n ( B ) + n ( C ) – n ( A ∩ B ) – n ( B ∩ C)

– n ( A ∩ C ) + n ( A ∩ B ∩ C )

This proves (3).

Example 23 If X and Y are two sets such that X ∪ Y has 50 elements, X has

28 elements and Y has 32 elements, how many elements does X ∩ Y have ?

Fig 1.11

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22 MATHEMATICS

Solution Given that

n ( X ∪ Y ) = 50, n ( X ) = 28, n ( Y ) = 32,

n (X ∩ Y) = ?

By using the formula

n ( X ∪ Y ) = n ( X ) + n ( Y ) – n ( X ∩ Y ),

we find that

n ( X ∩ Y ) = n ( X ) + n ( Y ) – n ( X ∪ Y )

= 28 + 32 – 50 = 10

Alternatively, suppose n ( X ∩ Y ) = k, then

n ( X – Y ) = 28 – k , n ( Y – X ) = 32 – k (by Venn diagram in Fig 1.12 )

This gives 50 = n ( X ∪ Y ) = n (X – Y) + n (X ∩ Y) + n ( Y – X)

= ( 28 – k ) + k + (32 – k )

Hence k = 10.

Example 24 In a school there are 20 teachers who teach mathematics or physics. Of

these, 12 teach mathematics and 4 teach both physics and mathematics. How many

teach physics ?

Solution Let M denote the set of teachers who teach mathematics and P denote the

set of teachers who teach physics. In the statement of the problem, the word ‘or’ gives

us a clue of union and the word ‘and’ gives us a clue of intersection. We, therefore,

have

n ( M ∪ P ) = 20 , n ( M ) = 12 and n ( M ∩ P ) = 4

We wish to determine n ( P ).

Using the result

n ( M ∪ P ) = n ( M ) + n ( P ) – n ( M ∩ P ),

we obtain

20 = 12 + n ( P ) – 4

Thus n ( P ) = 12

Hence 12 teachers teach physics.

Example 25 In a class of 35 students, 24 like to play cricket and 16 like to play

football. Also, each student likes to play at least one of the two games. How many

students like to play both cricket and football ?

Solution Let X be the set of students who like to play cricket and Y be the set of

students who like to play football. Then X ∪ Y is the set of students who like to play

at least one game, and X ∩ Y is the set of students who like to play both games.

Given n ( X) = 24, n ( Y ) = 16, n ( X ∪ Y ) = 35, n (X ∩ Y) = ?

Using the formula n ( X ∪ Y ) = n ( X ) + n ( Y ) – n ( X ∩ Y ), we get

35 = 24 + 16 – n (X ∩ Y)

Fig 1.12

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SETS 23

Thus, n (X ∩ Y) = 5

i.e., 5 students like to play both games.

Example 26 In a survey of 400 students in a school, 100 were listed as taking apple

juice, 150 as taking orange juice and 75 were listed as taking both apple as well as

orange juice. Find how many students were taking neither apple juice nor orange juice.

Solution Let U denote the set of surveyed students and A denote the set of students

taking apple juice and B denote the set of students taking orange juice. Then

n (U) = 400, n (A) = 100, n (B) = 150 and n (A ∩ B) = 75.

Now n (A′ ∩ B′) = n (A ∪ B)′= n (U) – n (A ∪ B)

= n (U) – n (A) – n (B) + n (A ∩ B)

= 400 – 100 – 150 + 75 = 225

Hence 225 students were taking neither apple juice nor orange juice.

Example 27 There are 200 individuals with a skin disorder, 120 had been exposed to

the chemical C1, 50 to chemical C

2, and 30 to both the chemicals C

1 and C

2. Find the

number of individuals exposed to

(i) Chemical C1 but not chemical C

2(ii) Chemical C

2 but not chemical C

1

(iii) Chemical C1 or chemical C

2

Solution Let U denote the universal set consisting of individuals suffering from the

skin disorder, A denote the set of individuals exposed to the chemical C1 and B denote

the set of individuals exposed to the chemical C2.

Here n ( U) = 200, n ( A ) = 120, n ( B ) = 50 and n ( A ∩ B ) = 30

(i) From the Venn diagram given in Fig 1.13, we have

A = ( A – B ) ∪ ( A ∩ B ).

n (A) = n( A – B ) + n( A ∩ B ) (Since A – B) and A ∩ B are disjoint.)

or n ( A – B ) = n ( A ) – n ( A ∩ B ) = 120 –30 = 90

Hence, the number of individuals exposed to

chemical C1 but not to chemical C

2 is 90.

(ii) From the Fig 1.13, we have

B = ( B – A) ∪ ( A ∩ B).

and so, n (B) = n (B – A) + n ( A ∩ B)

(Since B – A and A ∩B are disjoint.)

or n ( B – A ) = n ( B ) – n ( A ∩ B )

= 50 – 30 = 20 Fig 1.13

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24 MATHEMATICS

Thus, the number of individuals exposed to chemical C2 and not to chemical C

1 is 20.

(iii) The number of individuals exposed either to chemical C1 or to chemical C

2, i.e.,

n ( A ∪ B ) = n ( A ) + n ( B ) – n ( A ∩ B )

= 120 + 50 – 30 = 140.

EXERCISE 1.6

1. If X and Y are two sets such that n ( X ) = 17, n ( Y ) = 23 and n ( X ∪ Y ) = 38,

find n ( X ∩ Y ).

2. If X and Y are two sets such that X ∪ Y has 18 elements, X has 8 elements and

Y has 15 elements ; how many elements does X ∩ Y have?

3. In a group of 400 people, 250 can speak Hindi and 200 can speak English. How

many people can speak both Hindi and English?

4. If S and T are two sets such that S has 21 elements, T has 32 elements, and S ∩ T

has 11 elements, how many elements does S ∪ T have?

5. If X and Y are two sets such that X has 40 elements, X ∪ Y has 60 elements and

X ∩ Y has 10 elements, how many elements does Y have?

6. In a group of 70 people, 37 like coffee, 52 like tea and each person likes at least

one of the two drinks. How many people like both coffee and tea?

7. In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many

like tennis only and not cricket? How many like tennis?

8. In a committee, 50 people speak French, 20 speak Spanish and 10 speak both

Spanish and French. How many speak at least one of these two languages?

Miscellaneous Examples

Example 28 Show that the set of letters needed to spell “ CATARACT ” and the

set of letters needed to spell “ TRACT” are equal.

Solution Let X be the set of letters in “CATARACT”. Then

X = { C, A, T, R }

Let Y be the set of letters in “ TRACT”. Then

Y = { T, R, A, C, T } = { T, R, A, C }

Since every element in X is in Y and every element in Y is in X. It follows that X = Y.

Example 29 List all the subsets of the set { –1, 0, 1 }.

Solution Let A = { –1, 0, 1 }. The subset of A having no element is the empty

set φ. The subsets of A having one element are { –1 }, { 0 }, { 1 }. The subsets of

A having two elements are {–1, 0}, {–1, 1} ,{0, 1}. The subset of A having three

elements of A is A itself. So, all the subsets of A are φ, {–1}, {0}, {1}, {–1, 0}, {–1, 1},

{0, 1} and {–1, 0, 1}.

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SETS 25

Example 30 Show that A ∪ B = A ∩ B implies A = B

Solution Let a ∈ A. Then a ∈ A ∪ B. Since A ∪ B = A ∩ B , a ∈ A ∩ B. So a ∈ B.

Therefore, A ⊂ B. Similarly, if b ∈ B, then b ∈ A ∪ B. Since

A ∪ B = A ∩ B, b ∈ A ∩ B. So, b ∈ A. Therefore, B ⊂ A. Thus, A = B

Example 31 For any sets A and B, show that

P ( A ∩ B ) = P ( A ) ∩ P ( B ).

Solution Let X ∈ P ( A ∩ B ). Then X ⊂ A ∩ B. So, X ⊂ A and X ⊂ B. Therefore,

X ∈ P ( A ) and X ∈ P ( B ) which implies X ∈ P ( A ) ∩ P ( B). This gives P ( A ∩ B )

⊂ P ( A ) ∩ P ( B ). Let Y ∈ P ( A ) ∩ P ( B ). Then Y ∈ P ( A) and Y ∈ P ( B ). So,

Y ⊂ A and Y ⊂ B. Therefore, Y ⊂ A ∩ B, which implies Y ∈ P ( A ∩ B ). This gives

P ( A ) ∩ P ( B ) ⊂ P ( A ∩ B)

Hence P ( A ∩ B ) = P ( A ) ∩ P ( B ).

Example 32 A market research group conducted a survey of 1000 consumers and

reported that 720 consumers like product A and 450 consumers like product B, what is

the least number that must have liked both products?

Solution Let U be the set of consumers questioned, S be the set of consumers who

liked the product A and T be the set of consumers who like the product B. Given that

n ( U ) = 1000, n ( S ) = 720, n ( T ) = 450

So n ( S ∪ T ) = n ( S ) + n ( T ) – n ( S ∩ T )

= 720 + 450 – n (S ∩ T) = 1170 – n ( S ∩ T )

Therefore, n ( S ∪ T ) is maximum when n ( S ∩ T ) is least. But S ∪ T ⊂ U implies

n ( S ∪ T ) ≤ n ( U ) = 1000. So, maximum values of n ( S ∪ T ) is 1000. Thus, the least

value of n ( S ∩ T ) is 170. Hence, the least number of consumers who liked both products

is 170.

Example 33 Out of 500 car owners investigated, 400 owned car A and 200 owned

car B, 50 owned both A and B cars. Is this data correct?

Solution Let U be the set of car owners investigated, M be the set of persons who

owned car A and S be the set of persons who owned car B.

Given that n ( U ) = 500, n (M ) = 400, n ( S ) = 200 and n ( S ∩ M ) = 50.

Then n ( S ∪ M ) = n ( S ) + n ( M ) – n ( S ∩ M ) = 200 + 400 – 50 = 550

But S ∪ M ⊂ U implies n ( S ∪ M ) ≤ n ( U ).

This is a contradiction. So, the given data is incorrect.

Example 34 A college awarded 38 medals in football, 15 in basketball and 20 in

cricket. If these medals went to a total of 58 men and only three men got medals in all

the three sports, how many received medals in exactly two of the three sports ?

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26 MATHEMATICS

Solution Let F, B and C denote the set of men who

received medals in football, basketball and cricket,

respectively.

Then n ( F ) = 38, n ( B ) = 15, n ( C ) = 20

n (F ∪ B ∪ C ) = 58 and n (F ∩ B ∩ C ) = 3

Therefore, n (F ∪ B ∪ C ) = n ( F ) + n ( B )

+ n ( C ) – n (F ∩ B ) – n (F ∩ C ) – n (B ∩ C ) +

n ( F ∩ B ∩ C ),

gives n ( F ∩ B ) + n ( F ∩ C ) + n ( B ∩ C ) = 18

Consider the Venn diagram as given in Fig 1.14

Here, a denotes the number of men who got medals in football and basketball only, b

denotes the number of men who got medals in football and cricket only, c denotes the

number of men who got medals in basket ball and cricket only and d denotes the

number of men who got medal in all the three. Thus, d = n ( F ∩ B ∩ C ) = 3 and

a + d + b + d + c + d = 18

Therefore a + b + c = 9,

which is the number of people who got medals in exactly two of the three sports.

Miscellaneous Exercise on Chapter 1

1. Decide, among the following sets, which sets are subsets of one and another:

A = { x : x ∈ R and x satisfy x2 – 8x + 12 = 0 },

B = { 2, 4, 6 }, C = { 2, 4, 6, 8, . . . }, D = { 6 }.

2. In each of the following, determine whether the statement is true or false. If it is

true, prove it. If it is false, give an example.

(i) If x ∈ A and A ∈ B , then x ∈ B

(ii) If A ⊂ B and B ∈ C , then A ∈ C

(iii) If A ⊂ B and B ⊂ C , then A ⊂ C

(iv) If A ⊄ B and B ⊄ C , then A ⊄ C

(v) If x ∈ A and A ⊄ B , then x ∈ B

(vi) If A ⊂ B and x ∉ B , then x ∉ A

3. Let A, B, and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. Show

that B = C.

4. Show that the following four conditions are equivalent :

(i) A ⊂ B(ii) A – B = φ (iii) A ∪ B = B (iv) A ∩ B = A

5. Show that if A ⊂ B, then C – B ⊂ C – A.

6. Assume that P ( A ) = P ( B ). Show that A = B

7. Is it true that for any sets A and B, P ( A ) ∪ P ( B ) = P ( A ∪ B )? Justify your

answer.

Fig 1.14

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SETS 27

8. Show that for any sets A and B,

A = ( A ∩ B ) ∪ ( A – B ) and A ∪ ( B – A ) = ( A ∪ B )

9. Using properties of sets, show that

(i) A ∪ ( A ∩ B ) = A (ii) A ∩ ( A ∪ B ) = A.

10. Show that A ∩ B = A ∩ C need not imply B = C.

11. Let A and B be sets. If A ∩ X = B ∩ X = φ and A ∪ X = B ∪ X for some set

X, show that A = B.

(Hints A = A ∩ ( A ∪ X ) , B = B ∩ ( B ∪ X ) and use Distributive law )

12. Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty

sets and A ∩ B ∩ C = φ.

13. In a survey of 600 students in a school, 150 students were found to be taking tea

and 225 taking coffee, 100 were taking both tea and coffee. Find how many

students were taking neither tea nor coffee?

14. In a group of students, 100 students know Hindi, 50 know English and 25 know

both. Each of the students knows either Hindi or English. How many students

are there in the group?

15. In a survey of 60 people, it was found that 25 people read newspaper H, 26 read

newspaper T, 26 read newspaper I, 9 read both H and I, 11 read both H and T,

8 read both T and I, 3 read all three newspapers. Find:

(i) the number of people who read at least one of the newspapers.

(ii) the number of people who read exactly one newspaper.

16. In a survey it was found that 21 people liked product A, 26 liked product B and

29 liked product C. If 14 people liked products A and B, 12 people liked products

C and A, 14 people liked products B and C and 8 liked all the three products.

Find how many liked product C only.

Summary

This chapter deals with some basic definitions and operations involving sets. These

are summarised below:

®A set is a well-defined collection of objects.

®A set which does not contain any element is called empty set.

®A set which consists of a definite number of elements is called finite set,

otherwise, the set is called infinite set.

®Two sets A and B are said to be equal if they have exactly the same elements.

®A set A is said to be subset of a set B, if every element of A is also an element

of B. Intervals are subsets of R.

®A power set of a set A is collection of all subsets of A. It is denoted by P(A).

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28 MATHEMATICS

®The union of two sets A and B is the set of all those elements which are either

in A or in B.

®The intersection of two sets A and B is the set of all elements which are

common. The difference of two sets A and B in this order is the set of elements

which belong to A but not to B.

®The complement of a subset A of universal set U is the set of all elements of U

which are not the elements of A.

®For any two sets A and B, (A ∪ B)′ = A′ ∩ B′ and ( A ∩ B )′ = A′ ∪ B′® If A and B are finite sets such that A ∩ B = φ, then

n (A ∪ B) = n (A) + n (B).

If A ∩ B ≠ φ, then

n (A ∪ B) = n (A) + n (B) – n (A ∩ B)

Historical Note

The modern theory of sets is considered to have been originated largely by the

German mathematician Georg Cantor (1845-1918). His papers on set theory

appeared sometimes during 1874 to 1897. His study of set theory came when he

was studying trigonometric series of the form a1 sin x + a

2 sin 2x + a

3 sin 3x + ...

He published in a paper in 1874 that the set of real numbers could not be put into

one-to-one correspondence wih the integers. From 1879 onwards, he publishd

several papers showing various properties of abstract sets.

Cantor’s work was well received by another famous mathematician Richard

Dedekind (1831-1916). But Kronecker (1810-1893) castigated him for regarding

infinite set the same way as finite sets. Another German mathematician Gottlob

Frege, at the turn of the century, presented the set theory as principles of logic.

Till then the entire set theory was based on the assumption of the existence of the

set of all sets. It was the famous Englih Philosopher Bertand Russell (1872-

1970 ) who showed in 1902 that the assumption of existence of a set of all sets

leads to a contradiction. This led to the famous Russell’s Paradox. Paul R.Halmos

writes about it in his book ‘Naïve Set Theory’ that “nothing contains everything”.

The Russell’s Paradox was not the only one which arose in set theory.

Many paradoxes were produced later by several mathematicians and logicians.

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SETS 29

As a consequence of all these paradoxes, the first axiomatisation of set theory

was published in 1908 by Ernst Zermelo. Another one was proposed by Abraham

Fraenkel in 1922. John Von Neumann in 1925 introduced explicitly the axiom of

regularity. Later in 1937 Paul Bernays gave a set of more satisfactory

axiomatisation. A modification of these axioms was done by Kurt Gödel in his

monograph in 1940. This was known as Von Neumann-Bernays (VNB) or Gödel-

Bernays (GB) set theory.

Despite all these difficulties, Cantor’s set theory is used in present day

mathematics. In fact, these days most of the concepts and results in mathematics

are expressed in the set theoretic language.

—vvvvv—

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vMathematics is the indispensable instrument of

all physical research. – BERTHELOT v

2.1 Introduction

Much of mathematics is about finding a pattern – a

recognisable link between quantities that change. In our

daily life, we come across many patterns that characterise

relations such as brother and sister, father and son, teacher

and student. In mathematics also, we come across many

relations such as number m is less than number n, line l is

parallel to line m, set A is a subset of set B. In all these, we

notice that a relation involves pairs of objects in certain

order. In this Chapter, we will learn how to link pairs of

objects from two sets and then introduce relations between

the two objects in the pair. Finally, we will learn about

special relations which will qualify to be functions. The

concept of function is very important in mathematics since it captures the idea of a

mathematically precise correspondence between one quantity with the other.

2.2 Cartesian Products of Sets

Suppose A is a set of 2 colours and B is a set of 3 objects, i.e.,

A = {red, blue}and B = {b, c, s},

where b, c and s represent a particular bag, coat and shirt, respectively.

How many pairs of coloured objects can be made from these two sets?

Proceeding in a very orderly manner, we can see that there will be 6

distinct pairs as given below:

(red, b), (red, c), (red, s), (blue, b), (blue, c), (blue, s).

Thus, we get 6 distinct objects (Fig 2.1).

Let us recall from our earlier classes that an ordered pair of elements

taken from any two sets P and Q is a pair of elements written in smallFig 2.1

Chapter 2

RELATIONS AND FUNCTIONS

G . W. Leibnitz

(1646–1716)

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RELATIONS AND FUNCTIONS 31

brackets and grouped together in a particular order, i.e., (p,q), p ∈ P and q ∈ Q . This

leads to the following definition:

Definition 1 Given two non-empty sets P and Q. The cartesian product P × Q is the

set of all ordered pairs of elements from P and Q, i.e.,

P × Q = { (p,q) : p ∈ P, q ∈ Q }

If either P or Q is the null set, then P × Q will also be empty set, i.e., P × Q = φFrom the illustration given above we note that

A × B = {(red,b), (red,c), (red,s), (blue,b), (blue,c), (blue,s)}.

Again, consider the two sets:

A = {DL, MP, KA}, where DL, MP, KA represent Delhi,

Madhya Pradesh and Karnataka, respectively and B = {01,02,

03}representing codes for the licence plates of vehicles issued

by DL, MP and KA .

If the three states, Delhi, Madhya Pradesh and Karnataka

were making codes for the licence plates of vehicles, with the

restriction that the code begins with an element from set A,

which are the pairs available from these sets and how many such

pairs will there be (Fig 2.2)?

The available pairs are:(DL,01), (DL,02), (DL,03), (MP,01), (MP,02), (MP,03),

(KA,01), (KA,02), (KA,03) and the product of set A and set B is given by

A × B = {(DL,01), (DL,02), (DL,03), (MP,01), (MP,02), (MP,03), (KA,01), (KA,02),

(KA,03)}.

It can easily be seen that there will be 9 such pairs in the Cartesian product, since

there are 3 elements in each of the sets A and B. This gives us 9 possible codes. Also

note that the order in which these elements are paired is crucial. For example, the code

(DL, 01) will not be the same as the code (01, DL).

As a final illustration, consider the two sets A= {a1, a

2} and

B = {b1, b

2, b

3, b

4} (Fig 2.3).

A × B = {( a1, b

1), (a

1, b

2), (a

1, b

3), (a

1, b

4), (a

2, b

1), (a

2, b

2),

(a2, b

3), (a

2, b

4)}.

The 8 ordered pairs thus formed can represent the position of points in

the plane if A and B are subsets of the set of real numbers and it is

obvious that the point in the position (a1, b

2) will be distinct from the point

in the position (b2, a

1).

Remarks

(i) Two ordered pairs are equal, if and only if the corresponding first elements

are equal and the second elements are also equal.

DL MP KA

03

02

01

Fig 2.2

Fig 2.3

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32 MATHEMATICS

(ii) If there are p elements in A and q elements in B, then there will be pq

elements in A × B, i.e., if n(A) = p and n(B) = q, then n(A × B) = pq.

(iii) If A and B are non-empty sets and either A or B is an infinite set, then so is

A × B.

(iv) A × A × A = {(a, b, c) : a, b, c ∈ A}. Here (a, b, c) is called an ordered

triplet.

Example 1 If (x + 1, y – 2) = (3,1), find the values of x and y.

Solution Since the ordered pairs are equal, the corresponding elements are equal.

Therefore x + 1 = 3 and y – 2 = 1.

Solving we get x = 2 and y = 3.

Example 2 If P = {a, b, c} and Q = {r}, form the sets P × Q and Q × P.

Are these two products equal?

Solution By the definition of the cartesian product,

P × Q = {(a, r), (b, r), (c, r)} and Q × P = {(r, a), (r, b), (r, c)}

Since, by the definition of equality of ordered pairs, the pair (a, r) is not equal to the pair

(r, a), we conclude that P × Q ≠ Q × P.

However, the number of elements in each set will be the same.

Example 3 Let A = {1,2,3}, B = {3,4} and C = {4,5,6}. Find

(i) A × (B ∩ C) (ii) (A × B) ∩ (A × C)

(iii) A × (B ∪ C) (iv) (A × B) ∪ (A × C)

Solution (i) By the definition of the intersection of two sets, (B ∩ C) = {4}.

Therefore, A × (B ∩ C) = {(1,4), (2,4), (3,4)}.

(ii) Now (A × B) = {(1,3), (1,4), (2,3), (2,4), (3,3), (3,4)}

and (A × C) = {(1,4), (1,5), (1,6), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6)}

Therefore, (A × B) ∩ (A × C) = {(1, 4), (2, 4), (3, 4)}.

(iii) Since, (B ∪ C) = {3, 4, 5, 6}, we have

A × (B ∪ C) = {(1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6), (3,3),

(3,4), (3,5), (3,6)}.

(iv) Using the sets A × B and A × C from part (ii) above, we obtain

(A × B) ∪ (A × C) = {(1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6),

(3,3), (3,4), (3,5), (3,6)}.

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Example 4 If P = {1, 2}, form the set P × P × P.

Solution We have, P × P × P = {(1,1,1), (1,1,2), (1,2,1), (1,2,2), (2,1,1), (2,1,2), (2,2,1),

(2,2,2)}.

Example 5 If R is the set of all real numbers, what do the cartesian products R × R

and R × R × R represent?

Solution The Cartesian product R × R represents the set R × R={(x, y) : x, y ∈ R}

which represents the coordinates of all the points in two dimensional space and the

cartesian product R × R × R represents the set R × R × R ={(x, y, z) : x, y, z ∈ R}

which represents the coordinates of all the points in three-dimensional space.

Example 6 If A × B ={(p, q),(p, r), (m, q), (m, r)}, find A and B.

Solution A = set of first elements = {p, m}

B = set of second elements = {q, r}.

EXERCISE 2.1

1. If 2 5 1

13 3 3 3

x,y – ,

+ =

, find the values of x and y.

2. If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of

elements in (A×B).

3. If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.

4. State whether each of the following statements are true or false. If the statement

is false, rewrite the given statement correctly.

(i) If P = {m, n} and Q = { n, m}, then P × Q = {(m, n),(n, m)}.

(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered

pairs (x, y) such that x ∈ A and y ∈ B.

(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ φ) = φ.

5. If A = {–1, 1}, find A × A × A.

6. If A × B = {(a, x),(a , y), (b, x), (b, y)}. Find A and B.

7. Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that

(i) A × (B ∩ C) = (A × B) ∩ (A × C). (ii) A × C is a subset of B × D.

8. Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have?

List them.

9. Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1)

are in A × B, find A and B, where x, y and z are distinct elements.

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34 MATHEMATICS

10. The Cartesian product A × A has 9 elements among which are found (–1, 0) and

(0,1). Find the set A and the remaining elements of A × A.

2.3 Relations

Consider the two sets P = {a, b, c} and Q = {Ali, Bhanu, Binoy, Chandra, Divya}.

The cartesian product of

P and Q has 15 ordered pairs which

can be listed as P × Q = {(a, Ali),

(a,Bhanu), (a, Binoy), ..., (c, Divya)}.

We can now obtain a subset of

P × Q by introducing a relation R

between the first element x and the

second element y of each ordered pair

(x, y) as

R= { (x,y): x is the first letter of the name y, x ∈ P, y ∈ Q}.

Then R = {(a, Ali), (b, Bhanu), (b, Binoy), (c, Chandra)}

A visual representation of this relation R (called an arrow diagram) is shown

in Fig 2.4.

Definition 2 A relation R from a non-empty set A to a non-empty set B is a subset of

the cartesian product A × B. The subset is derived by describing a relationship between

the first element and the second element of the ordered pairs in A × B. The second

element is called the image of the first element.

Definition 3 The set of all first elements of the ordered pairs in a relation R from a set

A to a set B is called the domain of the relation R.

Definition 4 The set of all second elements in a relation R from a set A to a set B is

called the range of the relation R. The whole set B is called the codomain of the

relation R. Note that range ⊂ codomain.

Remarks (i) A relation may be represented algebraically either by the Roster

method or by the Set-builder method.

(ii) An arrow diagram is a visual representation of a relation.

Example 7 Let A = {1, 2, 3, 4, 5, 6}. Define a relation R from A to A by

R = {(x, y) : y = x + 1 }

(i) Depict this relation using an arrow diagram.

(ii) Write down the domain, codomain and range of R.

Solution (i) By the definition of the relation,

R = {(1,2), (2,3), (3,4), (4,5), (5,6)}.

Fig 2.4

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The corresponding arrow diagram is

shown in Fig 2.5.

(ii) We can see that the

domain ={1, 2, 3, 4, 5,}

Similarly, the range = {2, 3, 4, 5, 6}

and the codomain = {1, 2, 3, 4, 5, 6}.

Example 8 The Fig 2.6 shows a relation

between the sets P and Q. Write this relation (i) in set-builder form, (ii) in roster form.

What is its domain and range?

Solution It is obvious that the relation R is

“x is the square of y”.

(i) In set-builder form, R = {(x, y): x

is the square of y, x ∈ P, y ∈ Q}

(ii) In roster form, R = {(9, 3),

(9, –3), (4, 2), (4, –2), (25, 5), (25, –5)}

The domain of this relation is {4, 9, 25}.

The range of this relation is {– 2, 2, –3, 3, –5, 5}.

Note that the element 1 is not related to any element in set P.

The set Q is the codomain of this relation.

ANote The total number of relations that can be defined from a set A to a set B

is the number of possible subsets of A × B. If n(A ) = p and n(B) = q, then

n (A × B) = pq and the total number of relations is 2pq.

Example 9 Let A = {1, 2} and B = {3, 4}. Find the number of relations from A to B.

Solution We have,

A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}.

Since n (A×B ) = 4, the number of subsets of A×B is 24. Therefore, the number of

relations from A into B will be 24.

Remark A relation R from A to A is also stated as a relation on A.

EXERCISE 2.2

1. Let A = {1, 2, 3,...,14}. Define a relation R from A to A by

R = {(x, y) : 3x – y = 0, where x, y ∈ A}. Write down its domain, codomain and

range.

Fig 2.5

Fig 2.6

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36 MATHEMATICS

2. Define a relation R on the set N of natural numbers by R = {(x, y) : y = x + 5,

x is a natural number less than 4; x, y ∈N}. Depict this relationship using roster

form. Write down the domain and the range.

3. A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by

R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}. Write R in

roster form.

4. The Fig2.7 shows a relationship

between the sets P and Q. Write this

relation

(i) in set-builder form (ii) roster form.

What is its domain and range?

5. Let A = {1, 2, 3, 4, 6}. Let R be the

relation on A defined by

{(a, b): a , b ∈A, b is exactly divisible by a}.

(i) Write R in roster form

(ii) Find the domain of R

(iii) Find the range of R.

6. Determine the domain and range of the relation R defined by

R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}.

7. Write the relation R = {(x, x3) : x is a prime number less than 10} in roster form.

8. Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.

9. Let R be the relation on Z defined by R = {(a,b): a, b ∈ Z, a – b is an integer}.

Find the domain and range of R.

2.4 Functions

In this Section, we study a special type of relation called function. It is one of the most

important concepts in mathematics. We can, visualise a function as a rule, which produces

new elements out of some given elements. There are many terms such as ‘map’ or

‘mapping’ used to denote a function.

Definition 5 A relation f from a set A to a set B is said to be a function if every

element of set A has one and only one image in set B.

In other words, a function f is a relation from a non-empty set A to a non-empty

set B such that the domain of f is A and no two distinct ordered pairs in f have the

same first element.

If f is a function from A to B and (a, b) ∈ f, then f (a) = b, where b is called the

image of a under f and a is called the preimage of b under f.

Fig 2.7

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The function f from A to B is denoted by f: A à B.

Looking at the previous examples, we can easily see that the relation in Example 7 is

not a function because the element 6 has no image.

Again, the relation in Example 8 is not a function because the elements in the

domain are connected to more than one images. Similarly, the relation in Example 9 is

also not a function. (Why?) In the examples given below, we will see many more

relations some of which are functions and others are not.

Example 10 Let N be the set of natural numbers and the relation R be defined on

N such that R = {(x, y) : y = 2x, x, y ∈ N}.

What is the domain, codomain and range of R? Is this relation a function?

Solution The domain of R is the set of natural numbers N. The codomain is also N.

The range is the set of even natural numbers.

Since every natural number n has one and only one image, this relation is a

function.

Example 11 Examine each of the following relations given below and state in each

case, giving reasons whether it is a function or not?

(i) R = {(2,1),(3,1), (4,2)}, (ii) R = {(2,2),(2,4),(3,3), (4,4)}

(iii) R = {(1,2),(2,3),(3,4), (4,5), (5,6), (6,7)}

Solution (i) Since 2, 3, 4 are the elements of domain of R having their unique images,

this relation R is a function.

(ii) Since the same first element 2 corresponds to two different images 2

and 4, this relation is not a function.

(iii) Since every element has one and only one image, this relation is a

function.

Definition 6 A function which has either R or one of its subsets as its range is called

a real valued function. Further, if its domain is also either R or a subset of R, it is

called a real function.

Example 12 Let N be the set of natural numbers. Define a real valued function

f : Nà N by f (x) = 2x + 1. Using this definition, complete the table given below.

x 1 2 3 4 5 6 7

y f (1) = ... f (2) = ... f (3) = ... f (4) = ... f (5) = ... f (6) = ... f (7) = ...

Solution The completed table is given by

x 1 2 3 4 5 6 7

y f (1) = 3 f (2) = 5 f (3) = 7 f (4) = 9 f (5) = 11 f (6) = 13 f (7) =15

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38 MATHEMATICS

2.4.1 Some functions and their graphs

(i) Identity function Let R be the set of real numbers. Define the real valued

function f : R → R by y = f(x) = x for each x ∈ R. Such a function is called the

identity function. Here the domain and range of f are R. The graph is a straight line as

shown in Fig 2.8. It passes through the origin.

Fig 2.9

Fig 2.8

(ii) Constant function Define the function f: R → R by y = f (x) = c, x ∈ R where

c is a constant and each x ∈ R. Here domain of f is R and its range is {c}.

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The graph is a line parallel to x-axis. For example, if f(x)=3 for each x∈R, then its

graph will be a line as shown in the Fig 2.9.

(iii) Polynomial function A function f : R → R is said to be polynomial function if

for each x in R, y = f (x) = a0 + a

1x + a

2x2 + ...+ a

n xn, where n is a non-negative

integer and a0, a

1, a

2,...,a

n∈R.

The functions defined by f(x) = x3 – x2 + 2, and g(x) = x4 + 2 x are some examples

of polynomial functions, whereas the function h defined by h(x) =

2

3x + 2x is not a

polynomial function.(Why?)

Example 13 Define the function f: R → R by y = f(x) = x2, x ∈ R. Complete the

Table given below by using this definition. What is the domain and range of this function?

Draw the graph of f.

x – 4 –3 –2 –1 0 1 2 3 4

y = f(x) = x2

Solution The completed Table is given below:

x – 4 –3 –2 –1 0 1 2 3 4

y = f (x) = x2 16 9 4 1 0 1 4 9 16

Domain of f = {x : x∈R}. Range of f = {x2

: x ∈ R}. The graph of f is given

by Fig 2.10

Fig 2.10

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40 MATHEMATICS

Example 14 Draw the graph of the function f :R → R defined by f (x) = x3, x∈R.

Solution We have

f(0) = 0, f(1) = 1, f(–1) = –1, f(2) = 8, f(–2) = –8, f(3) = 27; f(–3) = –27, etc.

Therefore, f = {(x,x3): x∈R}.

The graph of f is given in Fig 2.11.

Fig 2.11

(iv) Rational functions are functions of the type ( )

( )

f x

g x, where f(x) and g(x) are

polynomial functions of x defined in a domain, where g(x) ≠ 0.

Example 15 Define the real valued function f : R – {0} → R defined by 1

( ) =f xx

,

x ∈ R –{0}. Complete the Table given below using this definition. What is the domain

and range of this function?

x –2 –1.5 –1 –0.5 0.25 0.5 1 1.5 2

y = 1

x... ... ... ... ... ... ... ... ...

Solution The completed Table is given by

x –2 –1.5 –1 –0.5 0.25 0.5 1 1.5 2

y = 1

x– 0.5 – 0.67 –1 – 2 4 2 1 0.67 0.5

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The domain is all real numbers except 0 and its range is also all real numbers

except 0. The graph of f is given in Fig 2.12.

Fig 2.13

(v) The Modulus function The function

f: R→R defined by f(x) = |x| for each

x ∈R is called modulus function. For each

non-negative value of x, f(x) is equal to x.

But for negative values of x, the value of

f(x) is the negative of the value of x, i.e.,

0( )

0

x,xf x

x,x

≥= − <

The graph of the modulus function is given

in Fig 2.13.

(vi) Signum function The function

f:R→R defined by

1 if 0

( ) 0 if 0

1 if 0

, x

f x , x

, x

>= =− <

Fig 2.12

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42 MATHEMATICS

is called the signum function. The domain of the signum function is R and the range is

the set {–1, 0, 1}. The graph of the signum function is given by the Fig 2.14.

Fig 2.14

(vii) Greatest integer function

The function f: R → R defined

by f(x) = [x], x ∈R assumes the

value of the greatest integer, less

than or equal to x. Such a function

is called the greatest integer

function.

From the definition of [x], we

can see that

[x] = –1 for –1 ≤ x < 0

[x] = 0 for 0 ≤ x < 1

[x] = 1 for 1 ≤ x < 2

[x] = 2 for 2 ≤ x < 3 and

so on.

The graph of the function is

shown in Fig 2.15.

2.4.2 Algebra of real functions In this Section, we shall learn how to add two real

functions, subtract a real function from another, multiply a real function by a scalar

(here by a scalar we mean a real number), multiply two real functions and divide one

real function by another.

(i) Addition of two real functions Let f : X → R and g : X → R be any two real

functions, where X ⊂ R. Then, we define (f + g): X → R by

(f + g) (x) = f (x) + g (x), for all x ∈ X.

Fig 2.15

f(x) =x

x , x x‘ 0and0for = 0

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(ii) Subtraction of a real function from another Let f : X → R and g: X → R be

any two real functions, where X ⊂ R. Then, we define (f – g) : X→R by

(f–g) (x) = f(x) –g(x), for all x ∈ X.

(iii) Multiplication by a scalar Let f : X→R be a real valued function and α be a

scalar. Here by scalar, we mean a real number. Then the product α f is a function from

X to R defined by (α f ) (x) = α f (x), x ∈X.

(iv) Multiplication of two real functions The product (or multiplication) of two real

functions f:X→R and g:X→R is a function fg:X→R defined by

(fg) (x) = f(x) g(x), for all x ∈ X.

This is also called pointwise multiplication.

(v) Quotient of two real functions Let f and g be two real functions defined from

X→R, where X ⊂ R. The quotient of f by g denoted by f

g is a function defined by ,

( )

( )( )

f f xx

g g x

=

, provided g(x) ≠ 0, x ∈ X

Example 16 Let f(x) = x2and g(x) = 2x + 1 be two real functions.Find

(f + g) (x), (f –g) (x), (fg) (x), ( )f

xg

.

Solution We have,

(f + g) (x) = x2

+ 2x + 1, (f –g) (x) = x2

– 2x – 1,

(fg) (x) = x2

(2x + 1) = 2x3

+ x2

, ( )f

xg

=

2

2 1

x

x + , x ≠ 1

2−

Example 17 Let f(x) = x and g(x) = x be two functions defined over the set of non-

negative real numbers. Find (f + g) (x), (f – g) (x), (fg) (x) and f

g

(x).

Solution We have

(f + g) (x) = x + x, (f – g) (x) = x – x ,

(fg) x =

3

2x( x ) x= and ( )f

xg

1

2 0–x

x , xx

= = ≠

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44 MATHEMATICS

EXERCISE 2.3

1. Which of the following relations are functions? Give reasons. If it is a function,

determine its domain and range.

(i) {(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)}

(ii) {(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)}

(iii) {(1,3), (1,5), (2,5)}.

2. Find the domain and range of the following real functions:

(i) f(x) = – x (ii) f(x) = 29 x− .

3. A function f is defined by f(x) = 2x –5. Write down the values of

(i) f (0), (ii) f (7), (iii) f (–3).

4. The function ‘t’ which maps temperature in degree Celsius into temperature in

degree Fahrenheit is defined by t(C) = 9C

5 + 32.

Find (i) t(0) (ii) t(28) (iii) t(–10) (iv) The value of C, when t(C) = 212.

5. Find the range of each of the following functions.

(i) f (x) = 2 – 3x, x ∈ R, x > 0.

(ii) f (x) = x2 + 2, x is a real number.

(iii) f (x) = x, x is a real number.


Example 18 Let R be the set of real numbers.

Define the real function

f: R→R by f(x) = x + 10

and sketch the graph of this function.

Solution Here f(0) = 10, f(1) = 11, f(2) = 12, ...,

f(10) = 20, etc., and

f(–1) = 9, f(–2) = 8, ..., f(–10) = 0 and so on.

Therefore, shape of the graph of the given

function assumes the form as shown in Fig 2.16.

Remark The function f defined by f(x) = mx + c ,

x ∈ R, is called linear function, where m and c are

constants. Above function is an example of a linear

function.Fig 2.16

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Example 19 Let R be a relation from Q to Q defined by R = {(a,b): a,b ∈ Q and

a – b ∈ Z}. Show that

(i) (a,a) ∈ R for all a ∈ Q(ii) (a,b) ∈ R implies that (b, a) ∈ R(iii) (a,b) ∈ R and (b,c) ∈ R implies that (a,c) ∈R

Solution (i) Since, a – a = 0 ∈ Z, if follows that (a, a) ∈ R.

(ii) (a,b) ∈ R implies that a – b ∈ Z. So, b – a ∈ Z. Therefore,

(b, a) ∈ R

(iii) (a, b) and (b, c) ∈ R implies that a – b ∈ Z. b – c ∈ Z. So,

a – c = (a – b) + (b – c) ∈ Z. Therefore, (a,c) ∈ R

Example 20 Let f = {(1,1), (2,3), (0, –1), (–1, –3)} be a linear function from Z into Z.

Find f(x).

Solution Since f is a linear function, f (x) = mx + c. Also, since (1, 1), (0, – 1) ∈ R,

f (1) = m + c = 1 and f (0) = c = –1. This gives m = 2 and f(x) = 2x – 1.

Example 21 Find the domain of the function

2

2

3 5( )

5 4

x xf x

x x

+ +=

− +

Solution Since x2 –5x + 4 = (x – 4) (x –1), the function f is defined for all real numbers

except at x = 4 and x = 1. Hence the domain of f is R – {1, 4}.

Example 22 The function f is defined by

f (x) =

1 0

1 0

1 0

x, x

, x

x , x

− < = + >

Draw the graph of f (x).

Solution Here, f(x) = 1 – x, x < 0, this gives

f(– 4) = 1 – (– 4) = 5;

f(– 3) =1 – (– 3) = 4,

f(– 2) = 1 – (– 2) = 3

f(–1) = 1 – (–1) = 2; etc,

and f(1) = 2, f (2) = 3, f (3) = 4

f(4) = 5 and so on for f(x) = x + 1, x > 0.

Thus, the graph of f is as shown in Fig 2.17 Fig 2.17

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46 MATHEMATICS


1. The relation f is defined by

20 3

( ) =3 3 10

x , xf x

x, x

≤ ≤

≤ ≤

The relation g is defined by

2, 0 2

( )3 , 2 10

x xg x

x x

≤ ≤=≤ ≤

Show that f is a function and g is not a function.

2. If f (x) = x2

, find (1 1) (1)

(1 1 1)

f . – f

. –.

3. Find the domain of the function f (x)

2

2

2 1

8 12

x x

x – x

+ +=

+.

4. Find the domain and the range of the real function f defined by f (x) = ( 1)x − .

5. Find the domain and the range of the real function f defined by f (x) = –1x .

6. Let

2

2, :1

xf x x

x

= ∈ +

R be a function from R into R. Determine the range

of f.

7. Let f, g : R→R be defined, respectively by f(x) = x + 1, g(x) = 2x – 3. Find

f + g, f – g and f

g.

8. Let f = {(1,1), (2,3), (0,–1), (–1, –3)} be a function from Z to Z defined by

f(x) = ax + b, for some integers a, b. Determine a, b.

9. Let R be a relation from N to N defined by R = {(a, b) : a, b ∈N and a = b2}. Are

the following true?

(i) (a,a) ∈ R, for all a ∈ N (ii) (a,b) ∈ R, implies (b,a) ∈ R

(iii) (a,b) ∈ R, (b,c) ∈ R implies (a,c) ∈ R.

Justify your answer in each case.

10. Let A ={1,2,3,4}, B = {1,5,9,11,15,16} and f = {(1,5), (2,9), (3,1), (4,5), (2,11)}

Are the following true?

(i) f is a relation from A to B (ii) f is a function from A to B.

Justify your answer in each case.

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11. Let f be the subset of Z × Z defined by f = {(ab, a + b) : a, b ∈ Z}. Is f a

function from Z to Z? Justify your answer.

12. Let A = {9,10,11,12,13} and let f : A→N be defined by f (n) = the highest prime

factor of n. Find the range of f.

Summary

In this Chapter, we studied about relations and functions.The main features of

this Chapter are as follows:

® Ordered pair A pair of elements grouped together in a particular order.

® Cartesian product A × B of two sets A and B is given by

A × B = {(a, b): a ∈ A, b ∈ B}

In particular R × R = {(x, y): x, y ∈ R}

and R × R × R = (x, y, z): x, y, z ∈ R}

® If (a, b) = (x, y), then a = x and b = y.

® If n(A) = p and n(B) = q, then n(A × B) = pq.

® A × φ = φ

® In general, A × B ≠ B × A.

® Relation A relation R from a set A to a set B is a subset of the cartesian

product A × B obtained by describing a relationship between the first element

x and the second element y of the ordered pairs in A × B.

® The image of an element x under a relation R is given by y, where (x, y) ∈ R,

® The domain of R is the set of all first elements of the ordered pairs in a

relation R.

® The range of the relation R is the set of all second elements of the ordered

pairs in a relation R.

® Function A function f from a set A to a set B is a specific type of relation for

which every element x of set A has one and only one image y in set B.

We write f: A→B, where f(x) = y.

® A is the domain and B is the codomain of f.

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48 MATHEMATICS

® The range of the function is the set of images.

® A real function has the set of real numbers or one of its subsets both as its

domain and as its range.

® Algebra of functions For functions f : X → R and g : X → R, we have

(f + g) (x) = f (x) + g(x), x ∈ X

(f – g) (x) = f (x) – g(x), x ∈ X(f.g) (x) = f (x) .g (x), x ∈ X

(kf) (x) = k ( f (x) ), x ∈ X, where k is a real number.

( )f

xg

= ( )

( )

f x

g x, x ∈ X, g(x) ≠ 0

Historical Note

The word FUNCTION first appears in a Latin manuscript “Methodus

tangentium inversa, seu de fuctionibus” written by Gottfried Wilhelm Leibnitz

(1646-1716) in 1673; Leibnitz used the word in the non-analytical sense. He

considered a function in terms of “mathematical job” – the “employee” being

just a curve.

On July 5, 1698, Johan Bernoulli, in a letter to Leibnitz, for the first time

deliberately assigned a specialised use of the term function in the analytical

sense. At the end of that month, Leibnitz replied showing his approval.

Function is found in English in 1779 in Chambers’ Cyclopaedia: “The

term function is used in algebra, for an analytical expression any way compounded

of a variable quantity, and of numbers, or constant quantities”.

— vvvvv —

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vA mathematician knows how to solve a problem,

he can not solve it. – MILNE v

3.1 Introduction

The word ‘trigonometry’ is derived from the Greek words

‘trigon’ and ‘metron’ and it means ‘measuring the sides of

a triangle’. The subject was originally developed to solve

geometric problems involving triangles. It was studied by

sea captains for navigation, surveyor to map out the new

lands, by engineers and others. Currently, trigonometry is

used in many areas such as the science of seismology,

designing electric circuits, describing the state of an atom,

predicting the heights of tides in the ocean, analysing a

musical tone and in many other areas.

In earlier classes, we have studied the trigonometric

ratios of acute angles as the ratio of the sides of a right

angled triangle. We have also studied the trigonometric identities and application of

trigonometric ratios in solving the problems related to heights and distances. In this

Chapter, we will generalise the concept of trigonometric ratios to trigonometric functions

and study their properties.

3.2 Angles

Angle is a measure of rotation of a given ray about its initial point. The original ray is

Chapter 3

TRIGONOMETRIC FUNCTIONS

Arya Bhatt

(476-550)

Fig 3.1

Vertex

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50 MATHEMATICS

called the initial side and the final position of the ray after rotation is called the

terminal side of the angle. The point of rotation is called the vertex. If the direction of

rotation is anticlockwise, the angle is said to be positive and if the direction of rotation

is clockwise, then the angle is negative (Fig 3.1).

The measure of an angle is the amount of

rotation performed to get the terminal side from

the initial side. There are several units for

measuring angles. The definition of an angle

suggests a unit, viz. one complete revolution from the position of the initial side as

indicated in Fig 3.2.

This is often convenient for large angles. For example, we can say that a rapidly

spinning wheel is making an angle of say 15 revolution per second. We shall describe

two other units of measurement of an angle which are most commonly used, viz.

degree measure and radian measure.

3.2.1 Degree measure If a rotation from the initial side to terminal side is

th1

360

of

a revolution, the angle is said to have a measure of one degree, written as 1°. A degree is

divided into 60 minutes, and a minute is divided into 60 seconds . One sixtieth of a degree is

called a minute, written as 1′, and one sixtieth of a minute is called a second, written as 1″.

Thus, 1° = 60′, 1′ = 60″Some of the angles whose measures are 360°,180°, 270°, 420°, – 30°, – 420° are

shown in Fig 3.3.

Fig 3.2

Fig 3.3

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TRIGONOMETRIC FUNCTIONS 51

3.2.2 Radian measure There is another unit for measurement of an angle, called

the radian measure. Angle subtended at the centre by an arc of length 1 unit in a

unit circle (circle of radius 1 unit) is said to have a measure of 1 radian. In the Fig

3.4(i) to (iv), OA is the initial side and OB is the terminal side. The figures show the

angles whose measures are 1 radian, –1 radian, 11

2 radian and –1

1

2 radian.

(i) (ii)(iii)

Fig 3.4 (i) to (iv)

(iv)

We know that the circumference of a circle of radius 1 unit is 2π. Thus, one

complete revolution of the initial side subtends an angle of 2π radian.

More generally, in a circle of radius r, an arc of length r will subtend an angle of

1 radian. It is well-known that equal arcs of a circle subtend equal angle at the centre.

Since in a circle of radius r, an arc of length r subtends an angle whose measure is 1

radian, an arc of length l will subtend an angle whose measure is l

r radian. Thus, if in

a circle of radius r, an arc of length l subtends an angle θ radian at the centre, we have

θ = l

r or l = r θ.

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52 MATHEMATICS

3.2.3 Relation between radian and real numbers

Consider the unit circle with centre O. Let A be any point

on the circle. Consider OA as initial side of an angle.

Then the length of an arc of the circle will give the radian

measure of the angle which the arc will subtend at the

centre of the circle. Consider the line PAQ which is

tangent to the circle at A. Let the point A represent the

real number zero, AP represents positive real number and

AQ represents negative real numbers (Fig 3.5). If we

rope the line AP in the anticlockwise direction along the

circle, and AQ in the clockwise direction, then every real

number will correspond to a radian measure and

conversely. Thus, radian measures and real numbers can

be considered as one and the same.

3.2.4 Relation between degree and radian Since a circle subtends at the centre

an angle whose radian measure is 2π and its degree measure is 360°, it follows that

2π radian = 360° or π radian = 180°

The above relation enables us to express a radian measure in terms of degree

measure and a degree measure in terms of radian measure. Using approximate value

of π as 22

7, we have

1 radian = 180

π

°= 57° 16′ approximately.

Also 1° = π

180 radian = 0.01746 radian approximately.

The relation between degree measures and radian measure of some common angles

are given in the following table:

AO

1

P

1

2

−1

−2

Q

0

Fig 3.5

Degree 30° 45° 60° 90° 180° 270° 360°

Radianπ

6

π

4

π

3

π

2π

3π

22π

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Notational Convention

Since angles are measured either in degrees or in radians, we adopt the convention

that whenever we write angle θ°, we mean the angle whose degree measure is θ and

whenever we write angle β, we mean the angle whose radian measure is β.

Note that when an angle is expressed in radians, the word ‘radian’ is frequently

omitted. Thus, π

π 180 and 454

= ° = ° are written with the understanding that π and π

4

are radian measures. Thus, we can say that

Radian measure = π

180 × Degree measure

Degree measure = 180

π× Radian measure

Example 1 Convert 40° 20′ into radian measure.

Solution We know that 180° = π radian.

Hence 40° 20′ = 40 1

3 degree =

π

180×

121

3 radian =

121π

540 radian.

Therefore 40° 20′ = 121π

540 radian.

Example 2 Convert 6 radians into degree measure.

Solution We know that π radian = 180°.

Hence 6 radians = 180

π× 6 degree =

1080 7

22

×degree

= 3437

11degree = 343° +

7 60

11

× minute [as 1° = 60′]

= 343° + 38′ + 2

11 minute [as 1′ = 60″]

= 343° + 38′ + 10.9″ = 343°38′ 11″ approximately.

Hence 6 radians = 343° 38′ 11″ approximately.

Example 3 Find the radius of the circle in which a central angle of 60° intercepts an

arc of length 37.4 cm (use 22

π7

= ).

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54 MATHEMATICS

Solution Here l = 37.4 cm and θ = 60° = 60π π

radian =180 3

Hence, by r =θ

l, we have

r =37.4×3 37.4×3×7

=π 22

= 35.7 cm

Example 4 The minute hand of a watch is 1.5 cm long. How far does its tip move in

40 minutes? (Use π = 3.14).

Solution In 60 minutes, the minute hand of a watch completes one revolution. Therefore,

in 40 minutes, the minute hand turns through 2

3 of a revolution. Therefore,

2θ = × 360°

3

or 4π

3 radian. Hence, the required distance travelled is given by

l = r θ = 1.5 × 4π

3cm = 2π cm = 2 × 3.14 cm = 6.28 cm.

Example 5 If the arcs of the same lengths in two circles subtend angles 65°and 110°

at the centre, find the ratio of their radii.

Solution Let r1 and r

2 be the radii of the two circles. Given that

θ1 = 65° = π

65180

× = 13π

36 radian

and θ2 = 110° = π

110180

× = 22π

36radian

Let l be the length of each of the arc. Then l = r1θ1 = r

2θ2, which gives

13π

36 × r

1 =

22π

36 × r

2 , i.e.,

1

2

r

r=

22

13

Hence r1 : r

2 = 22 : 13.

EXERCISE 3.1

1. Find the radian measures corresponding to the following degree measures:

(i) 25° (ii) – 47°30′ (iii) 240° (iv) 520°

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2 . Find the degree measures corresponding to the following radian measures

(Use 22

π7

= ).

(i)11

16(ii) – 4 (iii)

5π

3(iv)

7π

6

3. A wheel makes 360 revolutions in one minute. Through how many radians does

it turn in one second?

4. Find the degree measure of the angle subtended at the centre of a circle of

radius 100 cm by an arc of length 22 cm (Use 22

π7

= ).

5. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of

minor arc of the chord.

6. If in two circles, arcs of the same length subtend angles 60° and 75° at the

centre, find the ratio of their radii.

7. Find the angle in radian through which a pendulum swings if its length is 75 cm

and th e tip describes an arc of length

(i) 10 cm (ii) 15 cm (iii) 21 cm

3.3 Trigonometric Functions

In earlier classes, we have studied trigonometric ratios for acute angles as the ratio of

sides of a right angled triangle. We will now extend the definition of trigonometric

ratios to any angle in terms of radian measure and study them as trigonometric functions.

Consider a unit circle with centre

at origin of the coordinate axes. Let

P (a, b) be any point on the circle with

angle AOP = x radian, i.e., length of arc

AP = x (Fig 3.6).

We define cos x = a and sin x = b

Since ∆OMP is a right triangle, we have

OM2 + MP2 = OP2 or a2 + b2 = 1

Thus, for every point on the unit circle,

we have

a2 + b2 = 1 or cos2 x + sin2 x = 1

Since one complete revolution

subtends an angle of 2π radian at the

centre of the circle, ∠AOB = π

2, Fig 3.6

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56 MATHEMATICS

∠AOC = π and ∠AOD = 3π

2. All angles which are integral multiples of

π

2 are called

quadrantal angles. The coordinates of the points A, B, C and D are, respectively,

(1, 0), (0, 1), (–1, 0) and (0, –1). Therefore, for quadrantal angles, we have

cos 0° = 1 sin 0° = 0,

cos π

2= 0 sin

π

2= 1

cosπ = − 1 sinπ = 0

cos 3π

2= 0 sin

3π

2= –1

cos 2π = 1 sin 2π = 0

Now, if we take one complete revolution from the point P, we again come back to

same point P. Thus, we also observe that if x increases (or decreases) by any integral

multiple of 2π, the values of sine and cosine functions do not change. Thus,

sin (2nπ + x) = sin x , n ∈ Z , cos (2nπ + x) = cosx , n ∈ Z

Further, sin x = 0, if x = 0, ± π, ± 2π , ± 3π, ..., i.e., when x is an integral multiple of π

and cos x = 0, if x = ± π

2, ±

3π

2 , ±

5π

2, ... i.e., cos x vanishes when x is an odd

multiple of π

2. Thus

sin x = 0 implies x = nπ, π, π, π, π, where n is any integer

cos x = 0 implies x = (2n + 1) π

2, where n is any integer.

We now define other trigonometric functions in terms of sine and cosine functions:

cosec x = 1

sin x, x ≠ nπ, where n is any integer.

sec x = 1

cos x, x ≠ (2n + 1)

π


tan x = sin

cos

x

x, x ≠ (2n +1)

π


cot x = cos

sin

x

x, x ≠ n π, where n is any integer.

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not

defined

not

defined

We have shown that for all real x, sin2 x + cos2 x = 1

It follows that

1 + tan2 x = sec2 x (why?)

1 + cot2 x = cosec2 x (why?)

In earlier classes, we have discussed the values of trigonometric ratios for 0°,

30°, 45°, 60° and 90°. The values of trigonometric functions for these angles are same

as that of trigonometric ratios studied in earlier classes. Thus, we have the following

table:

0°π

6

π

4

π

3

π

2π

3π

22π

sin 01

2

1

2

3

21 0 – 1 0

cos 13

2

1

2

1

20 – 1 0 1

tan 01

3 1 3 0 0

The values of cosec x, sec x and cot x

are the reciprocal of the values of sin x,

cos x and tan x, respectively.

3.3.1 Sign of trigonometric functions

Let P (a, b) be a point on the unit circle

with centre at the origin such that

∠AOP = x. If ∠AOQ = – x, then the

coordinates of the point Q will be (a, –b)

(Fig 3.7). Therefore

cos (– x) = cos x

and sin (– x) = – sin x

Since for every point P (a, b) on

the unit circle, – 1 ≤ a ≤ 1 and Fig 3.7

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58 MATHEMATICS

– 1 ≤ b ≤ 1, we have – 1 ≤ cos x ≤ 1 and –1 ≤ sin x ≤ 1 for all x. We have learnt in

previous classes that in the first quadrant (0 < x < π

2) a and b are both positive, in the

second quadrant (π

2 < x <π) a is negative and b is positive, in the third quadrant

(π < x < 3π

2) a and b are both negative and in the fourth quadrant (

3π

2 < x < 2π) a is

positive and b is negative. Therefore, sin x is positive for 0 < x < π, and negative for

π < x < 2π. Similarly, cos x is positive for 0 < x < π

2, negative for

π

2 < x <

3π

2 and also

positive for 3π

2< x < 2π. Likewise, we can find the signs of other trigonometric

functions in different quadrants. In fact, we have the following table.

I II III IV

sin x + + – –

cos x + – – +

tan x + – + –

cosec x + + – –

sec x + – – +

cot x + – + –

3.3.2 Domain and range of trigonometric functions From the definition of sine

and cosine functions, we observe that they are defined for all real numbers. Further,

we observe that for each real number x,

– 1 ≤ sin x ≤ 1 and – 1 ≤ cos x ≤ 1

Thus, domain of y = sin x and y = cos x is the set of all real numbers and range

is the interval [–1, 1], i.e., – 1 ≤ y ≤ 1.

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Since cosec x = 1

sin x, the domain of y = cosec x is the set { x : x ∈ R and

x ≠ n π, n ∈ Z} and range is the set {y : y ∈ R, y ≥ 1 or y ≤ – 1}. Similarly, the domain

of y = sec x is the set {x : x ∈ R and x ≠ (2n + 1) π

2, n ∈ Z} and range is the set

{y : y ∈ R, y ≤ – 1or y ≥ 1}. The domain of y = tan x is the set {x : x ∈ R and

x ≠ (2n + 1) π

2, n ∈ Z} and range is the set of all real numbers. The domain of

y = cot x is the set {x : x ∈ R and x ≠ n π, n ∈ Z} and the range is the set of all real

numbers.

We further observe that in the first quadrant, as x increases from 0 to π

2, sin x

increases from 0 to 1, as x increases from π

2 to π, sin x decreases from 1 to 0. In the

third quadrant, as x increases from π to3π

2, sin x decreases from 0 to –1and finally, in

the fourth quadrant, sin x increases from –1 to 0 as x increases from 3π

2 to 2π.

Similarly, we can discuss the behaviour of other trigonometric functions. In fact, we

have the following table:

Remark In the above table, the statement tan x increases from 0 to ∞ (infinity) for

0 < x < π

2 simply means that tan x increases as x increases for 0 < x <

π

2 and

I quadrant II quadrant III quadrant IV quadrant

sin increases from 0 to 1 decreases from 1 to 0 decreases from 0 to –1 increases from –1 to 0

cos decreases from 1 to 0 decreases from 0 to – 1 increases from –1 to 0 increases from 0 to 1

tan increases from 0 to ∞ increases from –∞to 0 increases from 0 to ∞ increases from –∞to 0

cot decreases from ∞ to 0 decreases from 0 to–∞ decreases from ∞ to 0 decreases from 0to –∞

sec increases from 1 to ∞ increases from –∞to–1 decreases from –1to–∞ decreases from ∞ to 1

cosec decreases from ∞ to 1 increases from 1 to ∞ increases from –∞to–1 decreases from–1to–∞

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60 MATHEMATICS

Fig 3.10 Fig 3.11

Fig 3.8

Fig 3.9

assumes arbitraily large positive values as x approaches to π

2. Similarly, to say that

cosec x decreases from –1 to – ∞ (minus infinity) in the fourth quadrant means that

cosec x decreases for x ∈ (3π

2, 2π) and assumes arbitrarily large negative values as

x approaches to 2π. The symbols ∞ and – ∞ simply specify certain types of behaviour

of functions and variables.

We have already seen that values of sin x and cos x repeats after an interval of

2π. Hence, values of cosec x and sec x will also repeat after an interval of 2π. We

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shall see in the next section that tan (π + x) = tan x. Hence, values of tan x will repeat

after an interval of π. Since cot x is reciprocal of tan x, its values will also repeat after

an interval of π. Using this knowledge and behaviour of trigonometic functions, we can

sketch the graph of these functions. The graph of these functions are given above:

Example 6 If cos x = – 3

5, x lies in the third quadrant, find the values of other five

trigonometric functions.

Solution Since cos x = 3

5− , we have sec x =

5

3−

Now sin2 x + cos2 x = 1, i.e., sin2 x = 1 – cos2 x

or sin2 x = 1 – 9

25 =

16

25

Hence sin x = ± 4

5

Since x lies in third quadrant, sin x is negative. Therefore

sin x = – 4

5

which also gives

cosec x = – 5

4

Fig 3.12 Fig 3.13

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62 MATHEMATICS

Further, we have

tan x = sin

cos

x

x =

4

3 and cot x =

cos

sin

x

x =

3

4.

Example 7 If cot x = – 5

12, x lies in second quadrant, find the values of other five

trigonometric functions.

Solution Since cot x = – 5

12, we have tan x = –

12

5

Now sec2 x = 1 + tan2 x = 1 + 144

25 =

169

25

Hence sec x = ± 13

5

Since x lies in second quadrant, sec x will be negative. Therefore

sec x = – 13

5,

which also gives

5

cos13

x = −

Further, we have

sin x = tan x cos x = (– 12

5) × (–

5

13) =

12

13

and cosec x =1

sin x=

13

12.

Example 8 Find the value of sin 31π

3.

Solution We know that values of sin x repeats after an interval of 2π. Therefore

sin 31π

3 = sin (10π +

π

3) = sin

π

3 =

3

2.

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Example 9 Find the value of cos (–1710°).

Solution We know that values of cos x repeats after an interval of 2π or 360°.

Therefore, cos (–1710°) = cos (–1710° + 5 × 360°)

= cos (–1710° + 1800°) = cos 90° = 0.

EXERCISE 3.2

Find the values of other five trigonometric functions in Exercises 1 to 5.

1. cos x = – 1

2, x lies in third quadrant.

2. sin x = 3

5, x lies in second quadrant.

3. cot x = 4

3, x lies in third quadrant.

4. sec x = 13

5, x lies in fourth quadrant.

5. tan x = – 5

12, x lies in second quadrant.

Find the values of the trigonometric functions in Exercises 6 to 10.

6. sin 765° 7. cosec (– 1410°)

8. tan 19π

39. sin (–

11π

3)

10. cot (– 15π

4)

3.4 Trigonometric Functions of Sum and Difference of Two Angles

In this Section, we shall derive expressions for trigonometric functions of the sum and

difference of two numbers (angles) and related expressions. The basic results in this

connection are called trigonometric identities. We have seen that

1. sin (– x) = – sin x

2. cos (– x) = cos x

We shall now prove some more results:

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64 MATHEMATICS

3. cos (x + y) = cos x cos y – sin x sin y

Consider the unit circle with centre at the origin. Let x be the angle P4OP

1and y be

the angle P1OP

2. Then (x + y) is the angle P

4OP

2. Also let (– y) be the angle P

4OP

3.

Therefore, P1, P

2, P

3 and P

4 will have the coordinates P

1(cos x, sin x),

P2 [cos (x + y), sin (x + y)], P

3 [cos (– y), sin (– y)] and P

4 (1, 0) (Fig 3.14).

Consider the triangles P1OP

3 and P

2OP

4. They are congruent (Why?). Therefore,

P1P

3 and P

2P

4 are equal. By using distance formula, we get

P1P

32 = [cos x – cos (– y)]2 + [sin x – sin(–y]2

= (cos x – cos y)2 + (sin x + sin y)2

= cos2 x + cos2 y – 2 cos x cos y + sin2 x + sin2 y + 2sin x sin y

= 2 – 2 (cos x cos y – sin x sin y) (Why?)

Also, P2P

4

2 = [1 – cos (x + y)] 2 + [0 – sin (x + y)]2

= 1 – 2cos (x + y) + cos2 (x + y) + sin2 (x + y)

= 2 – 2 cos (x + y)

Fig 3.14

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Since P1P

3= P

2P

4, we have P

1P

3

2 = P2P

4

2.

Therefore, 2 –2 (cos x cos y – sin x sin y) = 2 – 2 cos (x + y).

Hence cos (x + y) = cos x cos y – sin x sin y

4 . cos (x – y) = cos x cos y + sin x sin y

Replacing y by – y in identity 3, we get

cos (x + (– y)) = cos x cos (– y) – sin x sin (– y)

or cos (x – y) = cos x cos y + sin x sin y

5. cos ( xπ

–2

) = sin x

If we replace x by π

2 and y by x in Identity (4), we get

cos (π

2x− ) = cos

π

2 cos x + sin

π

2 sin x = sin x.

6. sin ( xπ

–2

) = cos x

Using the Identity 5, we have

sin (π

2x− ) = cos

π π

2 2x

− − = cos x.

7. sin (x + y) = sin x cos y + cos x sin y

We know that

sin (x + y) = cos π

( )2

x y − +

= cos π

( )2

x y − −

= cos (π

2x− ) cos y + sin

π( )

2x− sin y

= sin x cos y + cos x sin y

8. sin (x – y) = sin x cos y – cos x sin y

If we replace y by –y, in the Identity 7, we get the result.

9. By taking suitable values of x and y in the identities 3, 4, 7 and 8, we get the

following results:

cos xπ

( + )2

= – sin x sin xπ

( + )2

= cos x

cos (πππππ – x) = – cos x sin (πππππ – x) = sin x

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66 MATHEMATICS

cos (πππππ + x) = – cos x sin (πππππ + x) = – sin x

cos (2πππππ – x) = cos x sin (2πππππ – x) = – sin x

Similar results for tan x, cot x, sec x and cosec x can be obtained from the results of sin

x and cos x.

10. If none of the angles x, y and (x + y) is an odd multiple of π

2, then

tan (x + y) = x y

x y

tan + tan

1 – tan tan

Since none of the x, y and (x + y) is an odd multiple of π

2, it follows that cos x,

cos y and cos (x + y) are non-zero. Now

tan (x + y) = sin( )

cos( )

x y

x y

++

= sin cos cos sin

cos cos sin sin

x y x y

x y x y

+−

.

Dividing numerator and denominator by cos x cos y, we have

tan (x + y) =

yx

yx

yx

yx

yx

yx

yx

yx

coscos

sinsin

coscos

coscos

coscos

sincos

coscos

cossin

−

+

= tan tan

1 – tan tan

x y

x y

+

11. tan ( x – y) = x y

x y

tan – tan

1 + tan tan

If we replace y by – y in Identity 10, we get

tan (x – y) = tan [x + (– y)]

= tan tan ( )

1 tan tan ( )

x y

x y

+ −− −

= tan tan

1 tan tan

x y

x y

−+

12. If none of the angles x, y and (x + y) is a multiple of πππππ, then

cot ( x + y) = x y

y x

cot cot – 1

cot +cot

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Since, none of the x, y and (x + y) is multiple of π, we find that sin x sin y and

sin (x + y) are non-zero. Now,

cot ( x + y) =cos ( ) cos cos – sin sin

sin ( ) sin cos cos sin

x y x y x y

x y x y x y

+=

+ +

Dividing numerator and denominator by sin x sin y, we have

cot (x + y) =cot cot –1

cot cot

x y

y x+

13. cot (x – y) =x y

y x

cot cot +1

cot – cot if none of angles x, y and x–y is a multiple of π

If we replace y by –y in identity 12, we get the result

14. cos 2x = cos2x – sin2 x = 2 cos2 x – 1 = 1 – 2 sin2 x = x

x

2

2

1 – tan

1 + tan

We know that

cos (x + y) = cos x cos y – sin x sin y

Replacing y by x, we get

cos 2x = cos2x – sin2 x

= cos2 x – (1 – cos2 x) = 2 cos2x – 1

Again, cos 2x = cos2 x – sin2 x

= 1 – sin2 x – sin2 x = 1 – 2 sin2 x.

We have cos 2x = cos2 x – sin 2 x =

2 2

2 2

cos sin

cos sin

x x

x x

−+

Dividing numerator and denominator by cos2 x, we get

cos 2x =

2

2

1 – tan

1+ tan

x

x,

ππ

2≠ +x n , where n is an integer

15. sin 2x = 2 sinx cos x = x

x2

2tan

1 + tan

ππ


We have

sin (x + y) = sin x cos y + cos x sin y

Replacing y by x, we get sin 2x = 2 sin x cos x.

Again sin 2x = 2 2

2sin cos

cos sin

x x

x x+

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68 MATHEMATICS

Dividing each term by cos2 x, we get

sin 2x = 2

2tan

1 tan

x

x+

16. tan 2x = x

x2

2tan

1 – tan if

π2 π


We know that

tan (x + y) = tan tan

1 tan tan

x y

– x y

+

Replacing y by x , we get 2

2 tantan 2

1 tan

xx

x=

−17. sin 3x = 3 sin x – 4 sin3 x

We have,

sin 3x = sin (2x + x)

= sin 2x cos x + cos 2x sin x

= 2 sin x cos x cos x + (1 – 2sin2 x) sin x

= 2 sin x (1 – sin2 x) + sin x – 2 sin3

x

= 2 sin x – 2 sin3 x + sin x – 2 sin3

x

= 3 sin x – 4 sin3 x

18. cos 3x= 4 cos3 x – 3 cos x

We have,

cos 3x = cos (2x +x)

= cos 2x cos x – sin 2x sin x

= (2cos2 x – 1) cos x – 2sin x cos x sin x

= (2cos2 x – 1) cos x – 2cos x (1 – cos2

x)

= 2cos3 x – cos x – 2cos x + 2 cos3

x

= 4cos3 x – 3cos x.

19. =x x

xx

3

2

3 tan – tantan3

1– 3tan if

π3 π


We have tan 3x =tan (2x + x)

=tan 2 tan

1 tan 2 tan

x x

– x x

+ 2

2

2tantan

1 tan

2tan tan1

1 tan

xx

– x

x . x–

– x

+=

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3 3

2 2 2

2tan tan tan 3 tan tan

1 tan 2tan 1 3tan

x x – x x – x

– x – x – x

+= =

20. (i) cos x + cos y = x y x y+ –

2cos cos2 2

(ii) cos x – cos y = – x y x y+ –

2sin sin2 2

(iii) sin x + sin y = x y x y+ –

2sin cos2 2

(iv) sin x – sin y = x y x y+ –

2cos sin2 2

We know that

cos (x + y) = cos x cos y – sin x sin y ... (1)

and cos (x – y) = cos x cos y + sin x sin y ... (2)

Adding and subtracting (1) and (2), we get

cos (x + y) + cos(x – y) = 2 cos x cos y ... (3)

and cos (x + y) – cos (x – y) = – 2 sin x sin y ... (4)

Further sin (x + y) = sin x cos y + cos x sin y ... (5)

and sin (x – y) = sin x cos y – cos x sin y ... (6)

Adding and subtracting (5) and (6), we get

sin (x + y) + sin (x – y) = 2 sin x cos y ... (7)

sin (x + y) – sin (x – y) = 2cos x sin y ... (8)

Let x + y = θ and x – y = φ. Therefore

θ θand

2 2x y

+φ −φ = =

Substituting the values of x and y in (3), (4), (7) and (8), we get

cos θ + cos φ = 2 cos θ θ

cos2 2

+φ −φ

cos θ – cos φ = – 2 sin θ θ

sin2

–+ φ φ 2

sin θ + sin φ = 2 sin θ θ

cos2 2

+φ −φ

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70 MATHEMATICS

sin θ – sin φ = 2 cos θ θ

sin2 2

+φ −φ

Since θ and φ can take any real values, we can replace θ by x and φ by y.

Thus, we get

cos x + cos y = 2 cos cos2 2

x y x y+ −; cos x – cos y = – 2 sin sin

2 2

x y x y+ −,

sin x + sin y = 2 sin cos2 2

x y x y+ −; sin x – sin y = 2 cos sin

2 2

x y x y+ −.

Remark As a part of identities given in 20, we can prove the following results:

21. (i) 2 cos x cos y = cos (x + y) + cos (x – y)

(ii) –2 sin x sin y = cos (x + y) – cos (x – y)

(iii) 2 sin x cos y = sin (x + y) + sin (x – y)

(iv) 2 cos x sin y = sin (x + y) – sin (x – y).

Example 10 Prove that

53sin sec 4sin cot 1

6 3 6 4

π π π π− =

Solution We have

L.H.S. =5

3sin sec 4sin cot6 3 6 4

π π π π−

= 3 × 1

2 × 2 – 4 sin

6

π π−

× 1 = 3 – 4 sin 6

π

= 3 – 4 × 1

2 = 1 = R.H.S.

Example 11 Find the value of sin 15°.

Solution We have

sin 15° = sin (45° – 30°)

= sin 45° cos 30° – cos 45° sin 30°

= 1 3 1 1 3 1

2 22 2 2 2

–× − × = .

Example 12 Find the value of tan 13

12

π.

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Solution We have

tan 13

12

π= tan

12

π π +

= tan tan12 4 6

π π π = −

=

tan tan4 6

1 tan tan4 6

π π−

π π+

=

11

3 132 3

1 3 113

−−

= = −++


sin ( ) tan tan

sin ( ) tan tan

x y x y

x y x y

+ +=

− − .

Solution We have

L.H.S.sin ( ) sin cos cos sin

sin ( ) sin cos cos sin

x y x y x y

x y x y x y

+ += =

− −

Dividing the numerator and denominator by cos x cos y, we get

sin ( ) tan tan

sin ( ) tan tan

x y x y

x y x y

+ +=

− − .

Example 14 Show that

tan 3 x tan 2 x tan x = tan 3x – tan 2 x – tan x

Solution We know that 3x = 2x + x

Therefore, tan 3x = tan (2x + x)

ortan 2 tan

tan31– tan 2 tan

x xx

x x

+=

or tan 3x – tan 3x tan 2x tan x = tan 2x + tan x

or tan 3x – tan 2x – tan x = tan 3x tan 2x tan x

or tan 3x tan 2x tan x = tan 3x – tan 2x – tan x.


cos cos 2 cos4 4

x x xπ π + + − =

Solution Using the Identity 20(i), we have

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72 MATHEMATICS

L.H.S. cos cos4 4

x xπ π = + + −

( )4 4 4 42cos cos

2 2

x x x – xπ π π π + + − + −

=

= 2 cos 4

π cos x = 2 ×

1

2 cos x = 2 cos x = R.H.S.

Example 16 Prove that cos 7 cos 5

cotsin 7 – sin 5

x xx

x x

+=

Solution Using the Identities 20 (i) and 20 (iv), we get

L.H.S. =

7 5 7 52cos cos

2 27 5 7 5

2cos sin2 2

x x x x

x x x x

+ −

+ − = cos

sincot

x

xx= = R.H.S.

Example 17 Prove that sin 5 2sin 3 sin

tancos5 cos

x x xx

x x

− += =

−

Solution We have

L.H.S. sin 5 2sin3 sin

cos5 cos

x x x

x x

− +=

−

sin 5 sin 2sin 3

cos5 cos

x x x

x x

+ −=

−

2sin3 cos 2 2sin3

– 2sin3 sin 2

x x x

x x

−=

sin 3 (cos2 1)

sin 3 sin 2

x x–

x x

−=

21 cos 2 2sin

sin 2 2sin cos

x x

x x x

−= = = tan x = R.H.S.

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EXERCISE 3.3

Prove that:

1. sin2 π

6+ cos2

3

π– tan2

1–

4 2

π= 2. 2sin2

6

π+ cosec2

27 3cos

6 3 2

π π=

3.2 25

cot cosec 3tan 66 6 6

π π π+ + = 4.

2 2 232sin 2cos 2sec 10

4 4 3

π π π+ + =

5. Find the value of:

(i) sin 75° (ii) tan 15°

Prove the following:

6. cos cos sin sin sin ( )4 4 4 4

x y x y x yπ π π π − − − − − = +

7.

2π

tan1 tan4

π 1 tantan

4

xx

xx

+ + = − −

8.2cos ( ) cos ( )

cot

sin ( ) cos2

x xx

x x

π + −=

π π − +

9.3π 3π

cos cos (2π ) cot cot (2π ) 12 2

x x x x + + − + + =

10. sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x

11.3 3

cos cos 2 sin4 4

x x xπ π + − − = −

12. sin2 6x – sin2 4x = sin 2x sin 10x 13. cos2 2x – cos2 6x = sin 4x sin 8x

14. sin2 x + 2 sin 4x + sin 6x = 4 cos2 x sin 4x

15. cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)

16.cos cos

sin sin

sin

cos

9 5

17 3

2

10

x x

x x

x

x

−−

= − 17. sin sin

cos costan

5 3

5 34

x x

x xx

++

=

18.sin sin

cos costan

x y

x y

x y−

+=

−

219.

sin sin

cos costan

x x

x xx

++

=3

32

20.sin sin

sin cossin

x x

x xx

−

−=

322 2 21.

cos cos cos

sin sin sincot

4 3 2

4 3 23

x x x

x x xx

+ +

+ +=

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74 MATHEMATICS

22. cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1

23.2

2 4

4tan (1 tan )tan 4

1 6 tan tan

x xx

x x

−=

− +24. cos 4x = 1 – 8sin2 x cos2 x

25. cos 6x = 32 cos6 x – 48cos4 x + 18 cos2 x – 1

3.5 Trigonometric Equations

Equations involving trigonometric functions of a variable are called trigonometric

equations. In this Section, we shall find the solutions of such equations. We have

already learnt that the values of sin x and cos x repeat after an interval of 2π and the

values of tan x repeat after an interval of π. The solutions of a trigonometric equation

for which 0 ≤ x < 2π are called principal solutions. The expression involving integer

‘n’ which gives all solutions of a trigonometric equation is called the general solution.

We shall use ‘Z’ to denote the set of integers.

The following examples will be helpful in solving trigonometric equations:

Example 18 Find the principal solutions of the equation sin x =3

2.

Solution We know that, π 3

sin3 2

= and 2π π π 3

sin sin π sin3 3 3 2

= − = =

.

Therefore, principal solutions are π

3x = and

2π

3.

Example 19 Find the principal solutions of the equation tan x = −1

3.

Solution We know that,π 1

tan6 3

= . Thus, π π 1

tan π – = – tan = –6 6 3

andπ π 1

tan 2π tan6 6 3

− = − = −

Thus5π 11π 1

tan tan6 6 3

= = − .

Therefore, principal solutions are 5π

6 and

11π

6.

We will now find the general solutions of trigonometric equations. We have already

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seen that:

sin x =0 gives x = nπ, where n ∈ Z

cos x =0 gives x = (2n + 1)π

2, where n ∈ Z.

We shall now prove the following results:

Theorem 1 For any real numbers x and y,

sin x = sin y implies x = nπ + (–1)n y, where n ∈ Z

Proof If sin x = sin y, then

sin x – sin y = 0 or 2cos x y x y+ −

2 2sin = 0

which gives cos x y+

2 = 0 or sin

x y−2

= 0

Thereforex y+

2 = (2n + 1)

π2

or x y−

2= nπ, where n ∈ Z

i.e. x = (2n + 1) π – y or x = 2nπ + y, where n∈Z

Hence x = (2n + 1)π + (–1)2n + 1 y or x = 2nπ +(–1)2n y, where n ∈ Z.

Combining these two results, we get

x = nπ + (–1)n y, where n ∈ Z.

Theorem 2 For any real numbers x and y, cos x = cos y, implies x = 2nπ ± y,

where n ∈ Z

Proof If cos x = cos y, then

cos x – cos y = 0 i.e., –2 sin x y+

2 sin

x y−2

= 0

Thus sin x y+

2 = 0 or sin

x y−

2 = 0

Thereforex y+

2 = nπ or

x y−

2 = nπ, where n ∈ Z

i.e. x = 2nπ – y or x = 2nπ + y, where n ∈ Z

Hence x = 2nπ ± y, where n ∈ Z

Theorem 3 Prove that if x and y are not odd mulitple of π

2, then

tan x = tan y implies x = nπ + y, where n ∈ Z

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76 MATHEMATICS

Proof If tan x = tan y, then tan x – tan y = 0

orsin cos cos sin

cos cos

x y x y

x y

−= 0

which gives sin (x – y) = 0 (Why?)

Therefore x – y = nπ, i.e., x = nπ + y, where n ∈ Z

Example 20 Find the solution of sin x = – 3

2.

Solution We have sin x = – 3

2 =

π π 4πsin sin π sin

3 3 3

− = + =

Hence sin x = 4π

sin3

, which gives

4π

π ( 1)3

nx n= + − , where n ∈ Z.

A Note4π

3 is one such value of x for which

3sin

2x = − . One may take any

other value of x for which sin x = −3

2. The solutions obtained will be the same

although these may apparently look different.

Example 21 Solve cos x =1

2.

Solution We have, 1 π

cos cos2 3

x = =

Thereforeπ

2 π3

x n= ± , where n ∈ Z.

Example 22 Solve π

tan 2 cot3

x x = − +

.

Solution We have, π

tan 2 cot3

x x = − +

= π π

tan2 3

x + +

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or5π

tan2 tan6

x x = +

Therefore5π

2 π6

x n x= + + , where n∈Z

or5π

π6

x n= + , where n∈Z.

Example 23 Solve sin 2x – sin 4x + sin 6x = 0.

Solution The equation can be written as

sin sin sin6 2 4 0x x x+ − =or 2 4 2 4 0sin cos sinx x x− =i.e. sin ( cos )4 2 2 1 0x x − =

Therefore sin 4x = 0 or 1

cos 22

x =

i.e.π

sin4 0 or cos 2 cos3

x x= =

Henceπ

4 π or 2 2 π3

x n x n= = ± , where n∈Z

i.e.π π

or π4 6

nx x n= = ± , where n∈Z.

Example 24 Solve 2 cos2 x + 3 sin x = 0

Solution The equation can be written as

( )2 1 3 02− + =sin sinx x

or 2 3 2 02

sin sinx x− − =

or (2sin 1) (sin 2) 0x x+ − =

Hence sin x = 1

2− or sin x = 2

But sin x = 2 is not possible (Why?)

Therefore sin x = 1

2− =

7πsin

6.

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78 MATHEMATICS

Hence, the solution is given by

7ππ ( 1)

6

nx n= + − , where n ∈ Z.

EXERCISE 3.4

Find the principal and general solutions of the following equations:

1. tan x = 3 2. sec x = 2

3. cot x = − 3 4. cosec x = – 2

Find the general solution for each of the following equations:

5. cos 4 x = cos 2 x 6. cos 3x + cos x – cos 2x = 0

7. sin 2x + cosx = 0 8. sec2 2x = 1– tan 2x

9. sin x + sin 3x + sin 5x = 0


Example 25 If sin x = 3

5, cos y = −

12

13, where x and y both lie in second quadrant,

find the value of sin (x + y).

Solution We know that

sin (x + y) = sin x cos y + cos x sin y ... (1)

Now cos2 x = 1 – sin2 x = 1 – 9

25 =

16

25

Therefore cos x = ±4

5.

Since x lies in second quadrant, cos x is negative.

Hence cos x = −4

5

Now sin2y = 1 – cos2y = 1 – 144

169

25

169=

i.e. sin y = ±5

13.

Since y lies in second quadrant, hence sin y is positive. Therefore, sin y = 5

13. Substituting

the values of sin x, sin y, cos x and cos y in (1), we get

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sin( )x y+ = × −

+ −

×

3

5

12

13

4

5

5

13=

36 20 56

65 65 65− − = − .


9 5cos 2 cos cos 3 cos sin 5 sin

2 2 2

x x xx x x− = .

Solution We have

L.H.S. = 1 9

2cos 2 cos 2cos cos 32 2 2

x xx x

−

= 1 9 9

cos 2 cos 2 cos 3 cos 32 2 2 2 2

x x x xx x x x

+ + − − + − −

= 1

2

5

2

3

2

15

2

3

2cos cos cos cos

x x x x+ − −

=

1

2

5

2

15

2cos cos

x x−

=

5 15 5 15

1 2 2 2 22sin sin2 2 2

x x x x + − −

= − −

=sin sin sin sin5

5

25

5

2x

xx

x = R.H.S.

Example 27 Find the value of tan π

8.

Solution Let π

8x = . Then

π2

4x = .

Now tantan

tan2

2

12

xx

x=

−

or 2

π2tan

π 8tanπ4

1 tan8

=−

Let y = tan π

8. Then 1 = 2

1

2

y

y

−

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80 MATHEMATICS

or y2 + 2y – 1 = 0

Therefore y = − ±

= − ±2 2 2

21 2

Sinceπ

8 lies in the first quadrant, y = tan

π

8 is positve. Hence

πtan 2 1

8= − .

Example 28 If 3 3π

tan = , π < <4 2

x x , find the value of sinx

2, cos

x

2 and tan

x

2.

Solution Since 3π

π2

x< < , cos x is negative.

Alsoπ 3π

2 2 4

x< < .

Therefore, sin x

2 is positive and cos

x

2 is negative.

Now sec2 x = 1 + tan2 x = 19

16

25

16+ =

Therefore cos2 x = 16

25 or cos x =

4

5– (Why?)

Now 22

2sin

x = 1 – cos x = 1

4

5

9

5+ = .

Therefore sin2 x

2 =

9

10

or sinx

2 =

3

10(Why?)

Again 2cos2 x

2 = 1+ cos x = 1

4

5

1

5− =

Therefore cos2 x

2 =

1

10

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or cos x

2 = −

1

10(Why?)

Hence tan x

2 =

sin

cos

x

x2

2

3

10

10

1= ×

−

= – 3.

Example 29 Prove that cos2 x + cos2 2π π 3cos

3 3 2x x

+ + − =

.

Solution We have

L.H.S. =

2π 2π1 cos 2 1 cos 2

1 cos 2 3 3

2 2 2

x xx

+ + + − + + + .

= 1 2π 2π

3 cos 2 cos 2 cos 22 3 3

x x x + + + + −

= 1 2π

3 cos 2 2cos 2 cos2 3

x x + +

= 1 π

3 cos 2 2cos 2 cos π2 3

x x + + −

= 1 π

3 cos 2 2cos 2 cos2 3

x x + −

= [ ]1 33 cos 2 cos 2

2 2x x+ − = = R.H.S.


Prove that:

1. 013

π5cos

13

π3cos

13

π9cos

13

πcos2 =++

2. (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0

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82 MATHEMATICS

3. (cos x + cos y)2 + (sin x – sin y)2 = 4 cos2 2

x y+

4. (cos x – cos y)2 + (sin x – sin y)2 = 4 sin2

2

x y−

5. sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x

6.6.6.6.6. (sin7 sin5 ) (sin 9 sin3 )tan 6

(cos7 cos5 ) (cos9 cos3 )

x x x xx

x x x x

+ + +=

+ + +

7. sin 3x + sin 2x – sin x = 4sin x cos x

2 cos

3

2

x

Find sin x

2, cos

x

2 and tan

x

2 in each of the following :

8. tan x = −4

3, x in quadrant II 9. cos x = −

1

3, x in quadrant III

10. sin x = 4

1, x in quadrant II

Summary

® If in a circle of radius r, an arc of length l subtends an angle of θ radians, then

l = r θ

®Radian measure = π

180× Degree measure

®Degree measure = 180

π× Radian measure

®cos2 x + sin2 x = 1

®1 + tan2 x = sec2 x

®1 + cot2 x = cosec2 x

®cos (2nπ + x) = cos x

® sin (2nπ + x) = sin x

® sin (– x) = – sin x

®cos (– x) = cos x

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®cos (x + y) = cos x cos y – sin x sin y

®cos (x – y) = cos x cos y + sin x sin y

®cos (π

2x− ) = sin x

® sin (π

2x− ) = cos x

® sin (x + y) = sin x cos y + cos x sin y

® sin (x – y) = sin x cos y – cos x sin y

®cos π

+2

x

= – sin x sin π

+2

x

= cos x

cos (π – x) = – cos x sin (π – x) = sin x

cos (π + x) = – cos x sin (π + x) = – sin x

cos (2π – x) = cos x sin (2π – x) = – sin x

® If none of the angles x, y and (x ± y) is an odd multiple of π

2, then

tan (x + y) = tan tan

tan tan

x y

x y

+−1

® tan (x – y) = tan tan

tan tan

x y

x y

−+1

® If none of the angles x, y and (x ± y) is a multiple of π, then

cot (x + y) = cot cot 1

cot cot

x y

y x

−+

®cot (x – y) = cot cot 1

cot cot

x y

y x−+

®cos 2x = cos2 x – sin2 x = 2cos2 x – 1 = 1 – 2 sin2 x

2

2

1 tan

1 tan

– x

x=

+

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84 MATHEMATICS

® sin 2x = 2 sinx cos x 2

2 tan

1 tan

x

x=

+

® tan 2x = 2

2tan

1 tan

x

x−

® sin 3x = 3sinx – 4sin3 x

®cos 3x = 4cos3 x – 3cosx

® tan 3x =

3

2

3tan tan

1 3tan

x x

x

−−

® (i) cos x + cos y = 2cos cos2 2

x y x y+ −

(ii) cos x – cos y = – 2sin sin2 2

x y x y+ −

(iii) sin x + sin y = 2 sin cos2 2

x y x y+ −

(iv) sin x – sin y = 2cos sin2 2

x y x y+ −

® (i) 2cos x cos y = cos (x + y) + cos ( x – y)

(ii) – 2sin x sin y = cos (x + y) – cos (x – y)

(iii) 2sin x cos y = sin (x + y) + sin (x – y)

(iv) 2 cos x sin y = sin (x + y) – sin (x – y).

® sin x = 0 gives x = nπ, where n ∈ Z.

®cos x = 0 gives x = (2n + 1) π

2, where n ∈ Z.

® sin x = sin y implies x = nπ + (– 1)n y, where n ∈ Z.

®cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

® tan x = tan y implies x = nπ + y, where n ∈ Z.

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Historical Note

The study of trigonometry was first started in India. The ancient Indian

Mathematicians, Aryabhatta (476), Brahmagupta (598), Bhaskara I (600) and

Bhaskara II (1114) got important results. All this knowledge first went from

India to middle-east and from there to Europe. The Greeks had also started the

study of trigonometry but their approach was so clumsy that when the Indian

approach became known, it was immediately adopted throughout the world.

In India, the predecessor of the modern trigonometric functions, known as

the sine of an angle, and the introduction of the sine function represents the main

contribution of the siddhantas (Sanskrit astronomical works) to the history of

mathematics.

Bhaskara I (about 600) gave formulae to find the values of sine functions

for angles more than 90°. A sixteenth century Malayalam work Yuktibhasa

(period) contains a proof for the expansion of sin (A + B). Exact expression for

sines or cosines of 18°, 36° , 54°, 72° , etc., are given by

Bhaskara II.

The symbols sin–1 x, cos–1 x, etc., for arc sin x, arc cos x, etc., were

suggested by the astronomer Sir John F.W. Hersehel (1813) The names of Thales

(about 600 B.C.) is invariably associated with height and distance problems. He

is credited with the determination of the height of a great pyramid in Egypt by

measuring shadows of the pyramid and an auxiliary staff (or gnomon) of known

height, and comparing the ratios:

H

S

h

s= = tan (sun’s altitude)

Thales is also said to have calculated the distance of a ship at sea through

the proportionality of sides of similar triangles. Problems on height and distance

using the similarity property are also found in ancient Indian works.

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vAnalysis and natural philosophy owe their most important discoveries to

this fruitful means, which is called induction. Newton was indebted

to it for his theorem of the binomial and the principle of

universal gravity. – LAPLACE v

4.1 Introduction

One key basis for mathematical thinking is deductive rea-

soning. An informal, and example of deductive reasoning,

borrowed from the study of logic, is an argument expressed

in three statements:

(a) Socrates is a man.

(b) All men are mortal, therefore,

(c) Socrates is mortal.

If statements (a) and (b) are true, then the truth of (c) is

established. To make this simple mathematical example,

we could write:

(i) Eight is divisible by two.

(ii) Any number divisible by two is an even number,

therefore,

(iii) Eight is an even number.

Thus, deduction in a nutshell is given a statement to be proven, often called a

conjecture or a theorem in mathematics, valid deductive steps are derived and a

proof may or may not be established, i.e., deduction is the application of a general

case to a particular case.

In contrast to deduction, inductive reasoning depends on working with each case,

and developing a conjecture by observing incidences till we have observed each and

every case. It is frequently used in mathematics and is a key aspect of scientific

reasoning, where collecting and analysing data is the norm. Thus, in simple language,

we can say the word induction means the generalisation from particular cases or facts.

Chapter 4

PRINCIPLE OFMATHEMATICAL INDUCTION

G . Peano

(1858-1932)

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PRINCIPLE OF MATHEMATICAL INDUCTION 87

In algebra or in other discipline of mathematics, there are certain results or state-

ments that are formulated in terms of n, where n is a positive integer. To prove such

statements the well-suited principle that is used–based on the specific technique, is

known as the principle of mathematical induction.

4.2 Motivation

In mathematics, we use a form of complete induction called mathematical induction.

To understand the basic principles of mathematical induction, suppose a set of thin

rectangular tiles are placed as shown in Fig 4.1.

Fig 4.1

When the first tile is pushed in the indicated direction, all the tiles will fall. To be

absolutely sure that all the tiles will fall, it is sufficient to know that

(a) The first tile falls, and

(b) In the event that any tile falls its successor necessarily falls.

This is the underlying principle of mathematical induction.

We know, the set of natural numbers N is a special ordered subset of the real

numbers. In fact, N is the smallest subset of R with the following property:

A set S is said to be an inductive set if 1∈ S and x + 1 ∈ S whenever x ∈ S. Since

N is the smallest subset of R which is an inductive set, it follows that any subset of R

that is an inductive set must contain N.

Illustration

Suppose we wish to find the formula for the sum of positive integers 1, 2, 3,...,n, that is,

a formula which will give the value of 1 + 2 + 3 when n = 3, the value 1 + 2 + 3 + 4,

when n = 4 and so on and suppose that in some manner we are led to believe that the

formula 1 + 2 + 3+...+ n = ( 1)

2

n n + is the correct one.

How can this formula actually be proved? We can, of course, verify the statement

for as many positive integral values of n as we like, but this process will not prove the

formula for all values of n. What is needed is some kind of chain reaction which will

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88 MATHEMATICS

have the effect that once the formula is proved for a particular positive integer the

formula will automatically follow for the next positive integer and the next indefinitely.

Such a reaction may be considered as produced by the method of mathematical induction.

4.3 The Principle of Mathematical Induction

Suppose there is a given statement P(n) involving the natural number n such that

(i) The statement is true for n = 1, i.e., P(1) is true, and

(ii) If the statement is true for n = k (where k is some positive integer), then

the statement is also true for n = k + 1, i.e., truth of P(k) implies the

truth of P (k + 1).

Then, P(n) is true for all natural numbers n.

Property (i) is simply a statement of fact. There may be situations when a

statement is true for all n ≥ 4. In this case, step 1 will start from n = 4 and we shall

verify the result for n = 4, i.e., P(4).

Property (ii) is a conditional property. It does not assert that the given statement

is true for n = k, but only that if it is true for n = k, then it is also true for n = k +1. So,

to prove that the property holds , only prove that conditional proposition:

If the statement is true for n = k, then it is also true for n = k + 1.

This is sometimes referred to as the inductive step. The assumption that the given

statement is true for n = k in this inductive step is called the inductive hypothesis.

For example, frequently in mathematics, a formula will be discovered that appears

to fit a pattern like

1 = 12 =1

4 = 22 = 1 + 3

9 = 32 = 1 + 3 + 5

16 = 42 = 1 + 3 + 5 + 7, etc.

It is worth to be noted that the sum of the first two odd natural numbers is the

square of second natural number, sum of the first three odd natural numbers is the

square of third natural number and so on.Thus, from this pattern it appears that

1 + 3 + 5 + 7 + ... + (2n – 1) = n2 , i.e,

the sum of the first n odd natural numbers is the square of n.

Let us write

P(n): 1 + 3 + 5 + 7 + ... + (2n – 1) = n2.

We wish to prove that P(n) is true for all n.

The first step in a proof that uses mathematical induction is to prove that

P (1) is true. This step is called the basic step. Obviously

1 = 12, i.e., P(1) is true.

The next step is called the inductive step. Here, we suppose that P (k) is true for some

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positive integer k and we need to prove that P (k + 1) is true. Since P (k) is true, we

have

1 + 3 + 5 + 7 + ... + (2k – 1) = k2 ... (1)

Consider

1 + 3 + 5 + 7 + ... + (2k – 1) + {2(k +1) – 1} ... (2)

= k2 + (2k + 1) = (k + 1)2 [Using (1)]

Therefore, P (k + 1) is true and the inductive proof is now completed.

Hence P(n) is true for all natural numbers n.

Example 1 For all n ≥ 1, prove that

12 + 22 + 32 + 42 +…+ n2 = ( 1)(2 1)

6

n n n+ +.

Solution Let the given statement be P(n), i.e.,

P(n) : 12 + 22 + 32 + 42 +…+ n2 = ( 1)(2 1)

6

n n n+ +

For n = 1, P(1): 1 = 1(1 1)(2 1 1)

6

+ × +=

1 2 31

6

× ×= which is true.

Assume that P(k) is true for some positive integer k, i.e.,

12 + 22 + 32 + 42 +…+ k2 = ( 1)(2 1)

6

k k k+ +... (1)

We shall now prove that P(k + 1) is also true. Now, we have

(12 +22 +32 +42 +…+k2 ) + (k + 1) 2

=2( 1)(2 1)

( 1)6

k k kk

+ ++ + [Using (1)]

=

2( 1) (2 1) 6( 1)

6

k k k k+ + + +

=

2( 1) (2 7 6)

6

k k k+ + +

=( 1)( 1 1){2( 1) 1}

6

k k k+ + + + +

Thus P(k + 1) is true, whenever P (k) is true.

Hence, from the principle of mathematical induction, the statement P(n) is true

for all natural numbers n.

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90 MATHEMATICS

Example 2 Prove that 2n > n for all positive integers n.

Solution Let P(n): 2n > n

When n =1, 21 >1. Hence P(1) is true.

Assume that P(k) is true for any positive integer k, i.e.,

2k > k ... (1)

We shall now prove that P(k +1) is true whenever P(k) is true.

Multiplying both sides of (1) by 2, we get

2. 2k

> 2k

i.e., 2 k + 1 > 2k = k + k > k + 1

Therefore, P(k + 1) is true when P(k) is true. Hence, by principle of mathematical

induction, P(n) is true for every positive integer n.

Example 3 For all n ≥ 1, prove that

1 1 1 1...

1.2 2.3 3.4 ( 1) 1

n

n n n+ + + + =

+ + .

Solution We can write

P(n): 1 1 1 1

...1.2 2.3 3.4 ( 1) 1

n

n n n+ + + + =

+ +

We note that P(1):1 1 1

1.2 2 1 1= =

+, which is true. Thus, P(n) is true for n = 1.

Assume that P(k) is true for some natural number k,

i.e.,1 1 1 1

...1.2 2.3 3.4 ( 1) 1

k

k k k+ + + + =

+ + ... (1)

We need to prove that P(k + 1) is true whenever P(k) is true. We have

1 1 1 1 1...

1.2 2.3 3.4 ( 1) ( 1) ( 2)k k k k+ + + + +

+ + +

= 1 1 1 1 1

...1.2 2.3 3.4 ( 1) ( 1)( 2)k k k k

+ + + + + + + +

=1

1 ( 1)( 2)

k

k k k+

+ + + [Using (1)]

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= ( 2) 1

( 1)( 2)

k k

k k

+ ++ +

=

2( 2 1)

( 1) ( 2)

k k

k k

+ ++ + =

( )( ) ( )

21

1 2

k

k k

++ +

= ( )1 1

2 1 1

k k

k k

+ +=

+ + +

Thus P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical

induction, P(n) is true for all natural numbers.

Example 4 For every positive integer n, prove that 7n – 3n is divisible by 4.

Solution We can write

P(n) : 7n – 3n is divisible by 4.

We note that

P(1): 71 – 31 = 4 which is divisible by 4. Thus P(n) is true for n = 1

Let P(k) be true for some natural number k,

i.e., P(k) : 7k – 3k is divisible by 4.

We can write 7k – 3k = 4d, where d ∈ N.

Now, we wish to prove that P(k + 1) is true whenever P(k) is true.

Now 7(k + 1) – 3(k + 1) = 7(k + 1) – 7.3k + 7.3k – 3(k + 1)

= 7(7k – 3k) + (7 – 3)3k = 7(4d) + (7 – 3)3k

= 7(4d) + 4.3k = 4(7d + 3k)

From the last line, we see that 7(k + 1) – 3(k + 1) is divisible by 4. Thus, P(k + 1) is true

when P(k) is true. Therefore, by principle of mathematical induction the statement is

true for every positive integer n.

Example 5 Prove that (1 + x)n ≥ (1 + nx), for all natural number n, where x > – 1.

Solution Let P(n) be the given statement,

i.e., P(n): (1 + x)n ≥ (1 + nx), for x > – 1.

We note that P(n) is true when n = 1, since ( 1+x) ≥ (1 + x) for x > –1

Assume that

P(k): (1 + x)k ≥ (1 + kx), x > – 1 is true. ... (1)

We want to prove that P(k + 1) is true for x > –1 whenever P(k) is true. ... (2)

Consider the identity

(1 + x)k + 1 = (1 + x)k (1 + x)

Given that x > –1, so (1+x) > 0.

Therefore , by using (1 + x)k ≥ (1 + kx), we have

(1 + x) k + 1 ≥ (1 + kx)(1 + x)

i.e. (1 + x)k + 1 ≥ (1 + x + kx + kx2). ... (3)

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92 MATHEMATICS

Here k is a natural number and x2 ≥ 0 so that kx2 ≥ 0. Therefore

(1 + x + kx + kx2) ≥ (1 + x + kx),

and so we obtain

(1 + x)k + 1 ≥ (1 + x + kx)

i.e. (1 + x)k + 1 ≥ [1 + (1 + k)x]

Thus, the statement in (2) is established. Hence, by the principle of mathematical

induction, P(n) is true for all natural numbers.


2.7n + 3.5n – 5 is divisible by 24, for all n ∈ N.

Solution Let the statement P(n) be defined as

P(n) : 2.7n + 3.5n – 5 is divisible by 24.

We note that P(n) is true for n = 1, since 2.7 + 3.5 – 5 = 24, which is divisible by 24.

Assume that P(k) is true

i.e. 2.7k + 3.5k – 5 = 24q, when q ∈ N ... (1)

Now, we wish to prove that P(k + 1) is true whenever P(k) is true.

We have

2.7k+1 + 3.5k+1 – 5 = 2.7k . 71 + 3.5k . 51 – 5

= 7 [2.7k + 3.5k – 5 – 3.5k + 5] + 3.5k . 5 – 5

= 7 [24q – 3.5k + 5] + 15.5k –5

= 7 × 24q – 21.5k + 35 + 15.5k – 5

= 7 × 24q – 6.5k + 30

= 7 × 24q – 6 (5k – 5)

= 7 × 24q – 6 (4p) [(5k – 5) is a multiple of 4 (why?)]

= 7 × 24q – 24p

= 24 (7q – p)

= 24 × r; r = 7q – p, is some natural number. ... (2)

The expression on the R.H.S. of (1) is divisible by 24. Thus P(k + 1) is true whenever

P(k) is true.

Hence, by principle of mathematical induction, P(n) is true for all n ∈ N.

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12 + 22 + ... + n2 >

3

3

n, n ∈ N

Solution Let P(n) be the given statement.

i.e., P(n) : 12 + 22 + ... + n2 >

3

3

n, n ∈ N

We note that P(n) is true for n = 1 since

32 1

13

>

Assume that P(k) is true

i.e. P(k) : 12 + 22 + ... + k2 >

3

3

k...(1)

We shall now prove that P(k + 1) is true whenever P(k) is true.

We have 12 + 22 + 32 + ... + k2 + (k + 1)2

( ) ( ) ( )3

2 22 2 21 2 ... 1 1

3

kk k k= + + + + + > + + [by (1)]

= 1

3 [k3 + 3k2 + 6k + 3]

= 1

3 [(k + 1)3 + 3k + 2] >

1

3 (k + 1)3

Therefore, P(k + 1) is also true whenever P(k) is true. Hence, by mathematical induction

P(n) is true for all n ∈ N.

Example 8 Prove the rule of exponents (ab)n = anbn

by using principle of mathematical induction for every natural number.

Solution Let P(n) be the given statement

i.e. P(n) : (ab)n = anbn.

We note that P(n) is true for n = 1 since (ab)1 = a1b1.

Let P(k) be true, i.e.,

(ab)k = akbk ... (1)

We shall now prove that P(k + 1) is true whenever P(k) is true.

Now, we have

(ab)k + 1 = (ab)k (ab)

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94 MATHEMATICS

= (ak bk) (ab) [by (1)]

= (ak . a1) (bk . b1) = ak+1 . bk+1

Therefore, P(k + 1) is also true whenever P(k) is true. Hence, by principle of math-

ematical induction, P(n) is true for all n ∈ N.

EXERCISE 4.1

Prove the following by using the principle of mathematical induction for all n ∈ N:

1. 1 + 3 + 32 + ... + 3n – 1 = (3 1)

2

n −.

2. 13 + 23 + 33 + … +n3 =

2( 1)

2

n n +

.

3.1 1 1 2

1(1 2) (1 2 3) (1 2 3 ) ( 1)

n...

...n n+ + + + =

+ + + + + + + .

4. 1.2.3 + 2.3.4 +…+ n(n+1) (n+2) = ( 1)( 2)( 3)

4

n n n n+ + +.

5. 1.3 + 2.32 + 3.33 +…+ n.3n =

1(2 1)3 3

4

nn

+− +.

6. 1.2 + 2.3 + 3.4 +…+ n.(n+1) = ( 1)( 2)

3

n n n+ +

.

7. 1.3 + 3.5 + 5.7 +…+ (2n–1) (2n+1) =

2(4 6 1)

3

n n n+ −.

8. 1.2 + 2.22 + 3.23 + ...+n.2n = (n–1) 2n + 1 + 2.

9.1 1 1 1 1

... 12 4 8 2 2

n n+ + + + = − .

10.1 1 1 1

...2.5 5.8 8.11 (3 1)(3 2) (6 4)

n

n n n+ + + + =

− + + .

11.1 1 1 1 ( 3)

...1.2.3 2.3.4 3.4.5 ( 1)( 2) 4( 1)( 2)

n n

n n n n n

++ + + + =

+ + + + .

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12. a + ar + ar2 +…+ arn-1 = ( 1)

1

na r

r

−−

.

13.2

2

3 5 7 (2 1)1 1 1 ... 1 ( 1)

1 4 9

nn

n

+ + + + + = +

.

14.1 1 1 1

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

nn

+ + + + = +

.

15. 12 + 32 + 52 + …+ (2n–1)2 = (2 1)(2 1)

3

n n n− +.

16.1 1 1 1

...1.4 4.7 7.10 (3 2)(3 1) (3 1)

n

n n n+ + + + =

− + + .

17.1 1 1 1

...3.5 5.7 7.9 (2 1)(2 3) 3(2 3)

n

n n n+ + + + =

+ + + .

18. 1 + 2 + 3 +…+ n < 1

8(2n + 1)2.

19. n (n + 1) (n + 5) is a multiple of 3.

20. 102n – 1 + 1 is divisible by 11.

21. x2n – y2n is divisible by x + y.

22. 32n+2 – 8n – 9 is divisible by 8.

23. 41n – 14n is a multiple of 27.

24. (2n + 7) < (n + 3)2.

Summary

®One key basis for mathematical thinking is deductive reasoning. In contrast to

deduction, inductive reasoning depends on working with different cases and

developing a conjecture by observing incidences till we have observed each

and every case. Thus, in simple language we can say the word ‘induction’

means the generalisation from particular cases or facts.

®The principle of mathematical induction is one such tool which can be used to

prove a wide variety of mathematical statements. Each such statement is

assumed as P(n) associated with positive integer n, for which the correctness

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96 MATHEMATICS

for the case n = 1 is examined. Then assuming the truth of P(k) for some

positive integer k, the truth of P (k+1) is established.

Historical Note

Unlike other concepts and methods, proof by mathematical induction is not

the invention of a particular individual at a fixed moment. It is said that the principle

of mathematical induction was known by the Pythagoreans.

The French mathematician Blaise Pascal is credited with the origin of the

principle of mathematical induction.

The name induction was used by the English mathematician John Wallis.

Later the principle was employed to provide a proof of the binomial theorem.

De Morgan contributed many accomplishments in the field of mathematics

on many different subjects. He was the first person to define and name

“mathematical induction” and developed De Morgan’s rule to determine the

convergence of a mathematical series.

G. Peano undertook the task of deducing the properties of natural numbers

from a set of explicitly stated assumptions, now known as Peano’s axioms.The

principle of mathematical induction is a restatement of one of the Peano’s axioms.

— vvvvv —

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