This document provides information about the textbook "Mathematics Textbook for Class XI 2022-23". It lists the edition number and reprint details. It mentions the price, copyright and publishing details. It also provides details about the National Council of Educational Research and Training (NCERT) including contact information of various offices. Finally, it lists the names of the textbook development committee members and acknowledges their contributions.

1. MATHEMATICS
Textbook for Class XI
2022-23

2. First Edition
February 2006 Phalguna 1927
Reprinted
October 2006 Kartika 1928
November 2007 Kartika 1929
December 2008 Pausa 1930
December 2009 Agrahayana 1931
January 2011 Pausa 1932
February 2012 Magha 1933
December 2012 Pausa 1934
November 2013 Kartika 1935
December 2014 Pausa 1936
May 2016 Vaishakha 1938
December 2016 Pausa 1938
December 2017 Agrahayana 1939
January 2019 Pausa 1940
August 2019 Shravana 1941
January 2021 Pausa 1942
November 2021 Agrahayana 1943
PD 270T RSP
© National Council of Educational
Research and Training, 2006
` 210.00
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ISBN 81-7450-486-9
11076 – MATHEMATICS
Textbook for Class XI
2022-23

3. 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
2022-23

4. 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
2022-23

5. 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 BirlaAcademy Bangalore, Karnataka.
S.B. Tripathy, Lecturer, Rajkiya Pratibha VikasVidyalaya, 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 SchoolYelaharika, 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
2022-23

6. 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 HaiderAnsari, 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.
2022-23

7. 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
2022-23

8. 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
7.1 Introduction 134
7.2 Fundamental Principle of Counting 134
7.3 Permutations 138
7.4 Combinations 148
8. Binomial Theorem 160
8.1 Introduction 160
8.2 Binomial Theorem for Positive Integral Indices 160
8.3 General and Middle Terms 167
9. Sequences and Series 177
9.1 Introduction 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
10.1 Introduction 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
2022-23

9. ix
11. Conic Sections 236
11.1 Introduction 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
12.1 Introduction 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
13.1 Introduction 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
14.1 Introduction 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
15.1 Introduction 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
2022-23

10. x
16. Probability 383
16.1 Introduction 383
16.2 Random Experiments 384
16.3 Event 387
16.4 AxiomaticApproach 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
2022-23

11. Chapter 1
SETS
Georg Cantor
(1845-1918)
In 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 
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
encounteredsetswhileworkingon“problemsontrigonometric
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
2022-23

12. 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}.
2022-23

13. SETS 3
Note 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.
Note 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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14. 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 6
1
n
x : 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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15. 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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16. 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 thatAcontains 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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17. 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.
Note 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. ThenA and P are equal, since 2, 3 and 5 are the only prime factors of
30 and also these are less than 6.
Note 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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18. 8 MATHEMATICS
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 ∈Aand 0 ∉ B, Aand 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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19. SETS 9
(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 Subsets
Consider 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 thatY 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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20. 10 MATHEMATICS
(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) LetA = {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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21. SETS 11
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
,
1
3
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  .
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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22. 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:xisanequilateraltriangleinaplane}... {x:xisatriangleinthesameplane}
(vii) {x : x is an even natural number} . . . {x : x is an integer}
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23. SETS 13
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) φ
5. 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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24. 14 MATHEMATICS
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 ofAand 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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25. SETS 15
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 representsA∪ 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 bothAand 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 }. FindA ∩ 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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26. 16 MATHEMATICS
If A and B are two sets such that A ∩ B = φ, then
A and B are called disjoint sets.
For example, letA= { 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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27. SETS 17
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 Abut not to B. Symbolically, we writeA – 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 Wehave,V–B={ e,o},sincetheelements
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 = φ
2. 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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28. 18 MATHEMATICS
(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. IfA = { 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 toA. The set of these three prime numbers,
i.e., the set {2, 3, 7} is called the Complement ofAwith respect to U, and is denoted by
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29. SETS 19
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.
Note 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
Itisclearfromthedefinitionofthecomplementthatforanysubsetoftheuniversal
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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30. 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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31. 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 inAor 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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32. 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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33. 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 C2
, and 30 to both the chemicals C1
and C2
. Find the
number of individuals exposed to
(i) Chemical C1
but not chemical C2
(ii) Chemical C2
but not chemical C1
(iii) Chemical C1
or chemical C2
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 C2
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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34. 24 MATHEMATICS
Thus, the number of individuals exposed to chemical C2
and not to chemical C1
is 20.
(iii) The number of individuals exposed either to chemical C1
or to chemical C2
, 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 AisAitself. So, all the subsets ofA are φ, {–1}, {0}, {1}, {–1, 0}, {–1, 1},
{0, 1} and {–1, 0, 1}.
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35. 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 ⊂ Aand 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 productAand 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
is170.
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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36. 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 thatA ∪ 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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37. 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 productsAand 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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38. 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 setsAand B in this order is the set of elements
which belong to A but not to B.
The complement of a subsetA 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 + a2
sin 2x + a3
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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39. 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 byAbraham
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.
—



—
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40. Mathematics is the indispensable instrument of
all physical research. – BERTHELOT 
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, setAis 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 small
Fig 2.1
Chapter 2
RELATIONSAND FUNCTIONS
G. W. Leibnitz
(1646–1716)
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41. 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 = φ
From 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
, a2
} and
B = {b1
, b2
, b3
, b4
} (Fig 2.3).
A × B = {( a1
, b1
), (a1
, b2
), (a1
, b3
), (a1
, b4
), (a2
, b1
), (a2
, b2
),
(a2
, b3
), (a2
, b4
)}.
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
, b2
) will be distinct from the point
in the position (b2
, a1
).
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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42. 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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43. RELATIONS AND FUNCTIONS 33
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
1
3 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. LetA = {1, 2} and B = {3, 4}. WriteA × 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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44. 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
2022-23

45. RELATIONS AND FUNCTIONS 35
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.
Note 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
2022-23

46. 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
2022-23

47. RELATIONS AND FUNCTIONS 37
The function f from A to B is denoted by f: A  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 : N 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
2022-23

48. 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}.
2022-23

49. RELATIONS AND FUNCTIONS 39
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
+ a1
x + a2
x2
+ ...+ an
xn
, where n is a non-negative
integer and a0
, a1
, a2
,...,an
∈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
3
x + 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 = {x
2
: x ∈ R}. The graph of f is given
by Fig 2.10
Fig 2.10
2022-23

50. 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 x
x
,
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
2022-23

51. RELATIONS AND FUNCTIONS 41
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,x
f 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
2022-23

52. 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
2022-23

53. RELATIONS AND FUNCTIONS 43
(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 x
x
g g x
 
=
 
 
, provided g(x) ≠ 0, x ∈ X
Example 16 Let f(x) = x
2
and g(x) = 2x + 1 be two real functions.Find
(f + g) (x), (f –g) (x), (fg) (x), ( )
f
x
g
 
 
 
.
Solution We have,
(f + g) (x) = x
2
+ 2x + 1, (f –g) (x) = x
2
– 2x – 1,
(fg) (x) = x
2
(2x + 1) = 2x
3
+ x
2
, ( )
f
x
g
 
 
 
=
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
2
x( x ) x
= and ( )
f
x
g
 
 
 
1
2 0
–
x
x , x
x
= = ≠
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54. 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) = 2
9 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.
Miscellaneous Examples
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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55. RELATIONS AND FUNCTIONS 45
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 x
f x
x x
+ +
=
− +
Solution Since x
2
–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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56. 46 MATHEMATICS
Miscellaneous Exercise on Chapter 2
1. The relation f is defined by
2
0 3
( ) =
3 3 10
x , x
f x
x, x
 ≤ ≤


≤ ≤


The relation g is defined by
2
, 0 2
( )
3 , 2 10
x x
g x
x x
 ≤ ≤

=
≤ ≤


Show that f is a function and g is not a function.
2. If f (x) = x
2
, 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) = –1
x .
6. Let
2
2
, :
1
x
f 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 = b
2
}.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. LetA={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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57. RELATIONS AND FUNCTIONS 47
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
productA × 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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58. 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
x
g
 
 
 
=
( )
( )
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”.
— 



 —
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59. 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
TRIGONOMETRICFUNCTIONS
Arya Bhatt
(476-550)
Fig 3.1
Vertex
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