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SENIOR
SECONDARY SCHOOL
MATHEMATICS
FOR CLASS 11
[In accordance with the latest CBSE syllabus]
R S Aggarwal, MSc, PhD
Preface
It gives me great pleasure in presenting the new edition of this book. In this
edition, the modifi cations have been dictated by the changes in the CBSE
syllabus. The structure and the methods used in the previous editions, which
have been appreciated by teachers using the book in classroom conditions,
remain unchanged.
In this edition certain topics that are no longer a part of the CBSE syllabus
have been retained so that students do not face difficulty in competitive
examinations, etc. These topics include conditional identities involving the
angles of a triangle, infinite geometric series and exponential series.
The main consideration in writing the book was to present the considerable
requirements of the syllabus in as simple a manner as possible. Special
attention has been paid to the gradation of problems. This will help students
gain confidence in problem-solving.
One problem faced by students is the lack of a comprehensive and
carefully selected set of solved problems in textbooks of this kind. I have given
due weightage to this aspect. Each set of solved examples is followed by a
comprehensive exercise section in which students will get enough questions
for practice. Hints have been given to the more difficult questions. Students
should take their help as a last resort.
I have received many suggestions and letters of appreciation from teachers
all over the country. I thank them all for contributing in the improvement of
the book and for their encouragement. I hope they will like this edition as well.
And as always, I would like to hear their views on the book.
R S Aggarwal
(iii)
Mathematics Syllabus
For Class 11
UNIT I. Sets and Functions
1. Sets 20 Periods
Sets and their representations. Empty set. Finite and Infi nite sets.
Equal sets. Subsets. Subsets of a set of real numbers especially intervals
(with notation). Power set. Universal set. Venn diagrams. Union and
Intersection of sets. Difference of sets. Complement of a set. Properties
of complement.
2. Relations and Functions 20 Periods
Ordered pairs, Cartesian product of sets. Number of elements in the
cartesian product of two fi nite sets. Cartesian product of the set of reals
with itself (up to R × R × R). Defi nition of relation, pictorial diagrams,
domain, co-domain and range of a relation. Function as a special type of
relation. Pictorial representation of a function, domain, co-domain and
range of a function. Real valued functions, domain and range of these
functions, constant, identity, polynomial, rational, modulus, signum;
exponential, logarithmic and greatest integer functions with their
graphs. Sum, difference, product and quotient of functions.
3. Trigonometric Functions 20 Periods
Positive and negative angles. Measuring angles in radians and in
degrees and conversion from one measure to another. Defi nition of
trigonometric functions with the help of unit circle. Truth of the identity
,sin cosx x 12 2 for all x. Signs of trigonometric functions. Domain and
range of trigonometric functions and their graphs. Expressing ( )sin x y!
and ( )cos x y! in terms of , , ,sin sin cosx y x ,cos y and their simple
applications. Deducing identities like the following:
, ,( ) ( )tan tan tan
tan tan
cot cot cot
cot cot
x y x y
x y
x y y x
x y
1
1
!
"
!
!
!
"
( ) ( ),sin sin sin cos2 2
1
2
1
! ! "
( ) ( ),cos cos cos cos2 2
1
2
1
( ) ( ) .cos cos sin sin2 2
1
2
1
Identities related to , , , ,sin cos tan sin cosx x x x x2 2 2 3 3 and .tan x3
General solution of trigonometric equations of the type ,sin siny a
cos cosy a and .tan tany a
(v)
UNIT II. Algebra
1. Principle of Mathematical Induction 10 Periods
Process of the proof by induction, motivating the application of the
method by looking at natural numbers as the least inductive subset
of real numbers. The principle of mathematical induction and simple
applications.
2. Complex Numbers and Quadratic Equations 15 Periods
Need for complex numbers, especially ,1 to be motivated by inability
to solve some of the quadratic equations. Algebraic properties of
complex numbers. Argand plane and polar representation of complex
numbers. Statement of Fundamental Theorem of Algebra, solution
of quadratic equations (with real coeffi cients) in the complex number
system. Square root of a complex number.
3. Linear Inequalities 15 Periods
Linear inequalities. Algebraic solutions of linear inequalities in one
variable and their representation on the number line. Graphical
representation of linear inequalities in two variables. Graphical method
of fi nding a solution of system of linear inequalities in two variables.
4. Permutations and Combinations 10 Periods
Fundamental principle of counting. Factorial ( !) .n n Permutations
and combinations, derivation of formulae for Pn r and C
n
r and their
connections, simple applications.
5. Binomial Theorem 10 Periods
History, statement and proof of the binomial theorem for positive
integral indices. Pascal’s triangle, general and middle term in binomial
expansion, simple applications.
6. Sequence and Series 10 Periods
Sequence and Series. Arithmetic progression (AP), Arithmetic mean
(AM), Geometric progression (GP), general terms of a GP, sum of fi rst
n terms of a GP, infi nite GP and its sum, geometric mean (GM), relation
between AM and GM. Formulae for the following special sums
, .andk k k
k
n
k
n
k
n
1 1
2
1
3
/ / /
UNIT III. Coordinate Geometry
1. Straight Lines 10 Periods
Brief recall of two dimensional geometry from earlier classes. Shifting
of origin. Slope of a line and angle between two lines. Various forms of
equations of a line: parallel to axis, point–slope form, slope–intercept
form, two-point form, intercept form and normal form. General
(vi)
equation of a line. Equation of family of lines passing through the point
of intersection of two lines. Distance of a point from a line.
2. Conic Sections 20 Periods
Sections of a cone: circle, ellipse, parabola, hyperbola, a point, a straight
line and a pair of intersecting lines as a degenerated case of a conic
section. Standard equations and simple properties of parabola, ellipse
and hyperbola. Standard equation of a circle.
3. Introduction to Three-dimensional Geometry 10 Periods
Coordinate axes and coordinate planes in three dimensions. Coordinates
of a point. Distance between two points and section formula.
UNIT IV. Calculus
1. Limits and Derivatives 30 Periods
Derivative introduced as rate of change both as that of distance function
and geometrically.
Intutive idea of limit. Limits of polynomials and rational functions;
trigonometric, exponential and logarithmic functions. Defi nition of
derivative, relate it to slope of tangent of the curve, derivative of sum,
difference, product and quotient of functions. Derivatives of polynomial
and trigonometric functions.
UNIT V. Mathematical Reasoning
1. Mathematical Reasoning 10 Periods
Mathematically acceptable statements. Connecting words/phrases—
consolidating the understanding of “if and only if (necessary and
suffi cient) condition”, “implies”, “and/or”, “implied by”, “and”, “or”,
“there exists” and their use through variety of examples related to real
life and mathematics. Validating the statements involving the connecting
words, difference between contradiction, converse and contrapositive.
UNIT VI. Statistics and Probability
1. Statistics 15 Periods
Measures of dispersion: Range, mean deviation, variance and standard
deviation of ungrouped/grouped data. Analysis of frequency
distributions with equal means but different variances.
2. Probability 15 Periods
Random experiments: outcomes, samplespaces (set representation).
Events: occurrence of events, ‘not’, ‘and’ and ‘or’ events, exhaustive
events, mutually exclusive events, Axiomatic (set theoretic) probability,
connections with other theories studied in earlier classes. Probability of
an event, probability of ‘not’, ‘and’ and ‘or’ events.
(vii)
Contents
Set and Functions
1. Sets 1
2. Relations 50
3. Functions 78
Algebra
4. Principle of Mathematical Induction 131
5. Complex Numbers and Quadratic Equations 150
6. Linear Inequations (In one variable) 217
7. Linear Inequations (In two variables) 245
8. Permutations 264
9. Combinations 322
10. Binomial Theorem 350
11. Arithmetic Progression 383
12. Geometrical Progression 433
13. Some Special Series 495
Trigonometry
14. Measurement of Angles 509
15. Trigonometric, or Circular, Functions 519
16. Conditional Identities Involving the
Angles of a Triangle 587
17. Trigonometric Equations 596
18. Solution of Triangles 614
19. Graphs of Trigonometric Functions 634
Coordinate Geometry
20. Straight Lines 641
21. Circle 719
22. Parabola 733
23. Ellipse 743
(ix)
24. Hyperbola 760
25. Applications of Conic Sections 775
26. Three-Dimensional Geometry 783
Calculus
27. Limits 795
28. Differentiation 838
Mathematical Reasoning
29. Mathematical Reasoning 894
Statistics and Probability
30. Statistics 918
31. Probability 946
Logarithm 980
Appendix Mathematical Modelling 985
Objective Questions
Complex Numbers 988
Permutations and Combinations 1001
Binomial Theorem 1013
Sequences and Series (AP, GP, HP) 1026
Trigonometry 1053
Coordinate Geometry 1112
(x)
Sets 1
1
Sets
INTRODUCTION
In our mathematical language, everything in this universe, whether living or non-
living, is called an object.
A given collection of objects is said to be well defined, if we can definitely say
whether a given particular object belongs to the collection or not.
SET A well-defi ned collection of objects is called a set.
The objects in a set are called its members or elements or points.
We denote sets by capital letters A, B, C, X, Y, Z, etc.
If a is an element of a set A, we write, ,a Ad which means that a belongs to A
or that a is an element of A.
If a does not belong to A, we write, .a Az
ILLUSTRATIONS
(i) The collection of all vowels in the English alphabet contains fi ve
elements, namely a, e, i, o, u.
So, this collection is well defi ned and therefore, it is a set.
(ii) The collection of all odd natural numbers less than 10 contains the
numbers 1, 3, 5, 7, 9.
So, this collection is well defi ned and therefore, it is a set.
(iii) The collection of all prime numbers less than 20 contains the numbers
2, 3, 5, 7, 11, 13, 17, 19.
So, this collection is well defi ned and therefore, it is a set.
(iv) All possible roots of the quadratic equation x x 6 02 are –2 and 3.
So, the collection of all possible roots of x x 6 02 is well defi ned
and therefore, it is a set.
(v) The collection of all rivers of India, is clearly well defi ned and
therefore, it is a set.
Clearly, river Ganga belongs to this set while river Nile does not
belong to it.
(vi) The collection of fi ve most talented writers of India is not a set, since
no rule has been given for deciding whether a given writer is talented
or not.
1
2 Senior Secondary School Mathematics for Class 11
(vii) The collection of most dangerous animals of the world is not a set,
since no rule has been given for deciding whether a given animal is
dangerous or not.
(viii) The collection of fi ve most renowned mathematicians of the world
is not a set, since there is no criterion for deciding whether a
mathematician is renowned or not.
(ix) The collection of all beautiful girls of India is not a set, since the term
‘beautiful‘ is vague and it is not well defi ned.
Similarly, ‘rich persons‘, ‘honest persons‘, ‘good players‘, ‘old people‘,
‘young men‘, etc., do not form sets.
However, ‘blind persons‘, ‘dumb persons‘, ‘illiterate persons‘, ‘retired
persons‘, etc., form sets.
HOW TO DESCRIBE OR SPECIFY A SET?
There are two methods of describing a set.
I. ROSTER FORM, OR TABULATION METHOD Under this method, we list all the members
of the set within braces { } and separate them by commas.
Note that the order in which the elements are listed, is immaterial.
EXAMPLES Write each of the following sets in the roster form:
(i) A = set of all factors of 24.
(ii) B = set of all prime numbers between 50 and 70.
(iii) C = set of all integers between 2
3 and $2
11
(iv) D = set of all consonants in the English alphabet which precede k.
(v) E = set of all letters in the word ‘TRIGONOMETRY‘.
(vi) F = set of all months having 30 days.
SOLUTION (i) All factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24.
A = {1, 2, 3, 4, 6, 8, 12, 24}.
(ii) All prime numbers between 50 and 70 are 53, 59, 61, 67.
B = {53, 59, 61, 67}.
(iii) All integers between and2
3
2
11 are –1, 0, 1, 2, 3, 4, 5.
C = {–1, 0, 1, 2, 3, 4, 5}.
(iv) All consonants preceding k are b, c, d, f, g, h, j.
D = {b, c, d, f, g, h, j}.
(v) It may be noted here that the repeated letters are taken only
once each.
E = {T, R, I, G, O, N, M, E, Y}.
(vi) We know that the months having 30 days are April, June,
September, November.
F = {April, June, September, November}.
Sets 3
NOTE We denote the sets of all natural numbers, all integers, all rational numbers
and all real numbers by N, Z, Q and R respectively.
II. SET-BUILDER FORM Under this method of describing a set, we list the property or
properties satisfi ed by all the elements of the set.
We write, {x : x has properties P}.
We read it as, ‘the set of all those x such that each x satisfies properties P’.
ILLUSTRATIONS
EXAMPLE 1 Write the set A = {1, 2, 3, 4, 5, 6, 7} in the set-builder form.
SOLUTION Clearly, A = set of all natural numbers less than 8.
Thus, in the set-builder form, we write it as
{ : } .andA x x N x 8 d
EXAMPLE 2 Write the set B = {1, 2, 4, 7, 14, 28} in the set-builder form.
SOLUTION Clearly, B = set of all factors of 28.
Thus, in the set-builder form, we write it as
{ :B x x N d and x is a factor of 28}.
EXAMPLE 3 Write the set C = {2, 4, 8, 16, 32} in the set-builder form.
SOLUTION Clearly, { , , , , } .C 2 2 2 2 21 2 3 4 5
Thus, in the set-builder form, we write it as
{ : ,C x x 2n where n Nd and } .n1 5# #
EXAMPLE 4 Write the set D = {–6, –4, –2, 0, 2, 4, 6} in the set-builder form.
SOLUTION Clearly, D = set of even integers from –6 to 6.
Thus, in the set-builder form, we write it as
{ : ,D x x n2 where n Zd and } .n3 3# #
EXAMPLE 5 Write the set E = {3, 6, 9, 12, 15, 18} in the set-builder form.
SOLUTION Clearly, { , , , , , } .E 3 1 3 2 3 3 3 4 3 5 3 6# # # # # #
Thus, in the set-builder form, we write it as
{ : , } .where andE x x n n N n3 1 6d # #
EXAMPLE 6 Write the set , , , , , , ,F 2
1
3
2
4
3
5
4
6
5
7
6
8
7
9
8 & 0 in the set-builder form.
SOLUTION Clearly, we have
,: ( ) ·where andF x x n
n n N n1 1 8# #
d& 0
EXAMPLE 7 Write the set G = {1, 3, 5, 7, 9, 11, …} in the set-builder form.
SOLUTION Clearly, G = set of all odd natural numbers.
Thus, in the set-builder form, we write it as
{ : } .and is oddG x x N x d
4 Senior Secondary School Mathematics for Class 11
EXAMPLE 8 Write the set H = {1, 4, 9, 16, 25, 36, …} in the set-builder form.
SOLUTION Clearly, H is the set of the squares of all natural numbers.
So, in the set-builder form, we write it as
{ : , } .whereH x x n n N2 d
EXAMPLE 9 Match each of the sets on the left in the roster form with the same set on
the right given in set-builder form:
(i) {23, 29} (a) { : , }x x n N and n3 1 5n # # d
(ii) {B, E, T, R} (b) { : , }x x n n N and n2 63 # # d
(iii) {3, 9, 27, 81, 243} (c) { : , }x x is prime x20 30
(iv) {8, 27, 64, 125, 216} (d) { :x x is a letter of the word ‘BETTER‘}SOLUTION (i) {23, 29} = set of prime numbers between 20 and 30
= {x : x is prime, 20 < x < 30}.
(i) (c).
(ii) {B, E, T, R} = set of letters in the word ‘BETTER‘}
= {x : x is a letter in the word ‘BETTER‘}.
(ii) (d).
(iii) { , , , , } { , , , , }3 9 27 81 243 3 3 3 3 31 2 3 4 5
{ : , } .andx x n N n3 1 5n d # #
(iii) (a).
(iv) { , , , , } { , , , , }8 27 64 125 216 2 3 4 5 63 3 3 3 3
{ : , , } .x x n n N n2 63 d # #
(iv) (b).
EXERCISE 1A
1. Which of the following are sets? Justify your answer.
(i) The collection of all whole numbers less than 10.
(ii) The collection of good hockey players in India.
(iii) The collection of all questions in this chapter.
(iv) The collection of all diffi cult chapters in this book.
(v) A collection of Hindi novels written by Munshi Prem Chand.
(vi) A team of 11 best cricket players of India.
(vii) The collection of all the months of the year whose names begin with the
letter M.
(viii) The collection of all interesting books.
(ix) The collection of all short boys of your class.
(x) The collection of all those students of your class whose ages exceed
15 years.
(xi) The collection of all rich persons of Kolkata.
Sets 5
(xii) The collection of all persons of Kolkata whose assessed annual
incomes exceed (say) ` 20 lakh in the fi nancial year 2016–17.
(xiii) The collection of all interesting dramas written by Shakespeare.
2. Let A be the set of all even whole numbers less than 10.
(a) Write A in roster form.
(b) Fill in the blanks with the approximate symbol or :
(i) 0 ..... A (ii) 10 ..... A (iii) 3 ..... A (iv) 6 ..... A
3. Write the following sets in roster form:
(i) A = {x : x is a natural number, } .x30 36#
(ii) B = {x : x is an integer and } .x4 6
(iii) C = {x : x is a two-digit number such that the sum of its digits is 9}.
(iv) D = {x : x is an integer, } .x 92#
(v) E = {x : x is a prime number, which is a divisor of 42}.
(vi) F = {x : x is a letter in the word ‘MATHEMATICS’}.
(vii) G = {x : x is a prime number and 80 < x < 100}.
(viii) H = {x : x is a perfect square and x < 50}.
(ix) { : } .andJ x x R x x 12 02 d
(x) { : ,K x x N x d is a multiple of 5 and } .x 4002
4. List all the elements of each of the sets given below:
(i) { : , } .andA x x n n N n2 5d #
(ii) { : , } .andB x x n n W n2 1 5d
(iii) ,: ·andC x x n n N n
1 6 d& 0
(iv) { : , } .andD x x n n N n2 52 # # d
(v) { : } .andE x x Z x x2 d
(vi) : ·andF x x Z x2
1
2
13 d& 0
(vii) ,: ( ) ·andG x x n n N n2 1
1 1 5d # # ' 1
(viii) { : ,| | } .H x x Z x 2d #
5. Write each of the sets given below in set-builder form:
(i) , , , , ,,A 1 4
1
9
1
16
1
25
1
36
1
49
1 & 0 (ii) , , , , , ,B 2
1
5
2
10
3
17
4
26
5
37
6
50
7 & 0
(iii) { , , , , , , }C 53 59 61 67 71 73 79 (iv) { , }D 1 1
(v) { , , , , , …, }E 14 21 28 35 42 98
6. Match each of the sets on the left described in roster form with the same set
on the right described in the set-builder form:
(i) {–5, 5} (a) { : }andx x Z x 162d
(ii) {1, 2, 3, 6, 9, 18} (b) { : }andx x N x x2 d
6 Senior Secondary School Mathematics for Class 11
(iii) {–3, –2, –1, 0, 1, 2, 3} (c) { : }andx x Z x 252d
(iv) {P, R, I, N, C, A, L} (d) { : }and is a factor ofx x N x 18d
(v) {1} (e) {x : x is a letter in the word ‘PRINCIPAL‘}
ANSWERS (EXERCISE 1A)
1. (i), (iii), (v), (vii), (x), (xii)
2. (a) { , , , , }A 0 2 4 6 8 (b) (i) (ii) (iii) (iv)
3. (i) { , , , , , }A 30 31 32 33 34 35 (ii) { , , , , , , , , }B 3 2 1 0 1 2 3 4 5
(iii) { , , , , , , , , }C 18 81 27 72 36 63 45 54 90
(iv) { , , , , , , }D 3 2 1 0 1 2 3 (v) { , , }E 2 3 7
(vi) , , , , , , ,{ }M A T H E I C SF (vii) { , , }G 83 89 97
(viii) { , , , , , , }H 1 4 9 16 25 36 49 (ix) { , }J 4 3
(x) { , , }K 5 10 15
4. (i) { , , , , }A 2 4 6 8 10 (ii) { , , , , }B 1 3 5 7 9
(iii) , , ,,C 1 2
1
3
1
4
1
5
1 & 0 (iv) { , , , }D 4 9 16 25
(v) { , }E 0 1 (vi) { , , , , , , }F 0 1 2 3 4 5 6
(vii) , , ,,G 1 3
1
5
1
7
1
9
1 & 0 (viii) { , , , , }H 2 1 0 1 2
5. (i) ,{ : }andA x x
n
n N n1 1 72 # # d
(ii) ,:
( )
andB x x
n
n n N n
1
1 72 # # d' 1
(iii) { : , }is primeC x x x50 80
(iv) { : , }D x x Z x 12 d
(v) { : , , }E x x n n N n7 2 14# # d
6. (i) (c), (ii) (d), (iii) (a), (iv) (e), (v) (b)
HINTS TO SOME SELECTED QUESTIONS
1. (i) Clearly, all whole numbers less than 10 are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
(ii) Clearly, there is no specifi c criterion to decide whether a given hockey player of
India, is good or not.
So, the given collection is not a set.
(iii) The collection of all questions in this chapter is a set, because, if a question is given
then we can easily decide whether it is a question of the chapter or not.
(iv) Clearly, the term ‘diffi cult‘ is vague. So, the given collection is not well defi ned and
therefore, it is not a set.
(v) Suppose we are given a collection of Hindi novels written by Munshi Prem
Chand. Now, if we take any Hindi novel then we can clearly decide whether it
belongs to our collection or not.
So, the given collection is a set.
(vi) The term ‘best‘ is vague. So, the given collection is not a set.
Sets 7
(vii) The given collection has defi nite members, namely March and May. So, this
collection is a set.
(viii) The term ‘interesting‘ is vague. So, the given collection is not a set.
(ix) The term ‘short’ is vague. So, the given collection is not a set.
(x) Clearly, it contains defi nite members. So, it is a set.
(xi) The term ‘rich‘ is vague. So, the given collection is not a set.
(xii) The given collection is clearly well defi ned. So, it is a set.
(xiii) The term ‘interesting‘ is vague. So, the given collection is not a set.
SOME TERMS RELATED TO SETS
EMPTY SET A set containing no element at all is called the empty set, or the null set,
or the void set, denoted by , or { }.
A set which has at least one element is called a nonempty set.
Examples of empty sets
(i) { : } ,andx x N x2 3 d since there is no natural number lying
between 2 and 3.
(ii) { : , } ,is a numberx x x x! since there is no number which is not
equal to itself.
(iii) { : , } ,andx x N x x5 7 d since there is no natural number which
is less than 5 and greater than 7.
(iv) { : } ,andx x R x 12d since there is no real number whose square
is –1.
(v) { : } ,is rational andx x x 2 02 since there is no rational number
whose square is 2.
(vi) { :x x is an even prime number greater than 2} = , since there is no
prime number which is even and greater than 2.
(vii) { :x x is a point common to two parallel lines} = , since there is no
point common to two parallel lines.
SINGLETON SET A set containing exactly one element is called a singleton set.
Examples of singleton sets
(i) {0} is a singleton set whose only element is 0.
(ii) {15} is a singleton set whose only element is 15.
(iii) {–8} is a singleton set whose only element is –8.
(iv) { : } { },andx x N x 4 22 d since 2 is the only natural number whose
square is 4.
However, { : } { , },andx x Z x 4 2 22 d which is not a singleton set.
(v) Consider the set { : } .andx x R x 1 03 d
Now, ( ) ( )x x x x1 0 1 1 03 2&
orx x1 2
1 1 4
2
1 3
·&
! !
8 Senior Secondary School Mathematics for Class 11
Thus, the given equation has one real root, namely .x 1
{ : } { },andx x R x 1 0 13d which is a singleton set.
FINITE AND INFINITE SETS
An empty set or a nonempty set in which the process of counting of elements surely
comes to an end is called a fi nite set. A set which is not fi nite is called an infi nite set.
The number of distinct elements contained in a finite set A is denoted
by ( ) .n A
Examples of fi nite sets
(i) Let { , , , , , } .A 2 4 6 8 10 12
Then, A isclearly a fi nite set and ( ) .n A 6
(ii) Let B set of all letters in the English alphabet.
Then, ( )n B 26 and therefore, B is fi nite.
(iii) Let { : } .andC x x Z x 36 02 d
Then, { , },C 6 6 which is clearly a fi nite set and ( ) .n C 2
(iv) The set of all persons on earth is a fi nite set.
(v) The set of all animals on earth is a fi nite set.
Examples of infi nite sets
(i) The set of all points on the arc of a circle is an infi nite set.
(ii) The set of all points on a line segment is an infi nite set.
(iii) The set of all circles passing through a given point is an infi nite set.
(iv) The set of all straight lines parallel to a given line, say the x-axis, is an
infi nite set.
(v) The set of all positive integral multiples of 5 is an infi nite set.
Let Z be the set of all positive integers.
Then, { : } { , , , , , …}x x Z5 5 10 15 20 25d is an infi nite set.
(vi) Each of the sets N, Z, Q and R is an infi nite set.
NOTE All infi nite sets cannot be described in roster form.
For example, the set R of all real numbers cannot be described in this
form, since the elements of this set do not follow a particular pattern.
EQUAL SETS Two nonempty 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 unequal and we write, .A B!
REMARKS (i) The elements of a set may be listed in any order.
Thus, {1, 2, 3} = {2, 1, 3} = {3, 2, 1}.
(ii) The repetition of elements in a set has no meaning.
Thus, {1, 1, 2, 2, 3} = {1, 2, 3}.
Sets 9
Some examples of equal sets
EXAMPLE 1 Let A = set of letters in the word ‘follow’,
and B = set of letters in the word ‘wolf ‘.
Show that A = B.
SOLUTION Clearly, we have
A = {f, o, l, w} and B = {w, o, l, f}.
Clearly, A and B have exactly same elements.
.A B
EXAMPLE 2 Let { , , , }A p q r s and { , , , } .B q r p s Are A and B equal?
SOLUTION Since A and B have exactly the same elements, so .A B
EXAMPLE 3 Show that , { }0 and 0 are all different.
SOLUTION We know that is a set containing no element at all.
And, {0} is a set containing one element, namely 0.
Also, 0 is a number, not a set.
Hence, , { }0 and 0 are all different.
EXAMPLE 4 Let { : , }A x x N x 9 02 d and { : , } .B x x Z x 9 02 d
Show that .A B!
SOLUTION ( ) ( ) .orx x x x x9 0 3 3 0 3 32 & &
{ : , } { }A x x N x 9 0 32 d [ ]N3a z
and { : , } { , } .B x x Z x 9 0 3 32 d
Hence, .A B!
EQUIVALENT SETS Two fi nite sets A and B are said to be equivalent, if ( ) ( ) .n A n B
Equal sets are always equivalent. But, equivalent sets need not be equal.
EXAMPLE 1 Let { , , }A 1 3 5 and { , , } .B 2 4 6
Then, ( ) ( ) .n A n B 3
So, A and B are equivalent.
Clearly, .A B!
Hence, A and B are equivalent sets but not equal.
EXAMPLE 2 Show that {0} and are not equivalent sets.
SOLUTION Let { }A 0 and .B
Then, clearly ( ) ( ) .andn A n B1 0
( ) ( )n A n B! and hence A and B are not equivalent sets.
EXERCISE 1B
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.
10 Senior Secondary School Mathematics for Class 11
(iii) { : , }A x x N x1 2 # d
(iv) { : , }B x x N x2 3 4 d
(v) { : , }is primeC x x x90 96
(vi) { : , }D x x N x 1 02 d
(vii) { : , }E x x W x 3 3# d
(viii) { : , }F x x Q x1 2 d
(ix) { }G 0
2. Which of the following are examples of the singleton set?
(i) { : , }x x Z x 42 d (ii) { : , }x x Z x 5 0 d
(iii) { : ,| | }x x Z x 1d (iv) { : , }x x N x 162 d
(v) {x : x is an even prime number}
3. Which of the following are pairs of equal sets?
(i) A = set of letters in the word, ‘ALLOY’
B = set of letters in the word, ‘LOYAL’
(ii) C = set of letters in the word, ‘CATARACT’
D = set of letters in the word ‘TRACT’
(iii) { : , } { : , }andE x x Z x F x x Z x4 42 2# d d
(iv) { , }G 1 1 and { : , }H x x Z x 1 02 d
(v) { , } { : , ( ) }andJ K x x Z x x2 3 5 6 02 d
4. Which of the following are pairs of equivalent sets?
(i) { , , } { , , }andA B2 1 0 1 2 3
(ii) { : , } { : , }andC x x N x D x x W x3 3 d d
(iii) { , , , , } { , , , , }a e i o u and p q r s tE F
5. State whether the given set is fi nite or infi nite:
(i) A = set of all triangles in a plane
(ii) B = set of all points on the circumference of a circle
(iii) C = set of all lines parallel to the y-axis
(iv) D = set of all leaves on a tree
(v) E = set of all positive integers greater than 500
(vi) { : ]F x R x0 1 d
(vii) { : }G x Z x 1 d
(viii) { : }H x Z x15 15 d
(ix) { : }and is primeJ x x N x d
(x) { : }and is oddK x x N x d
(xi) L = set of all circles passing through the origin (0, 0)
6. Rewrite the following statements using set notation:
(i) a is an element of set A.
(ii) b is not an element of A.
(iii) A is an empty set and B is a nonempty set.
Sets 11
(iv) Number of elements in A is 6.
(v) 0 is a whole number but not a natural number.
ANSWERS (EXERCISE 1B)
1. (i), (iv), (v), (vi) 2. (ii), (iv), (v)
3. (i) A = B (ii) C = D (iv) G = H
4. (i) A and B are equivalent sets (iii) E and F are equivalent sets
5. (i) infi nite (ii) infi nite (iii) infi nite (iv) fi nite (v) infi nite
(vi) infi nite (vii) infi nite (viii) fi nite (ix) infi nite (x) infi nite
(xi) infi nite
6. (i) a Ad (ii) b Az (iii) andA B! (iv) ( )n A 6
(v) butW N0 0d z
SUBSETS
SUBSET A set A is said to be a subset of set B if every element of A is also an element
of B, and we write, .A B3
SUPERSET If ,A B3 then B is called a superset of A, and we write, .B A4
PROPER SUBSET If A B3 and A B! then A is called a proper subset of B and we
write, .A B1
REMARK If there exists even a single element in A which is not in B, then A is not
a subset of B, and we write, .A B1Y
Examples of subsets
EXAMPLE 1 Let { , , }A 2 3 5 and { , , , , } .B 2 3 5 7 9
Then, every element of A is an element of B.
A B3 but .A B!
Hence, A is a proper subset of set B, i.e., .A B1
EXAMPLE 2 Let { , }A 1 2 and { , , } .B 2 3 5
Then, A1d but .B1z
.A BM
Again, B3d but .A3z
.B AM
Thus, A BM and .B AM
EXAMPLE 3 Clearly, .N W Z Q R1 1 1 1
But, W0d and .N0z
.W NM
12 Senior Secondary School Mathematics for Class 11
EXAMPLE 4 Let { , { , }, } .A 1 2 3 4
Then, which of the following statements is true?
(i) { , } A2 3 d (ii) { , } A2 3 1
Rectify the wrong statement.
SOLUTION Clearly, A is a set containing three elements, namely 1, {2, 3} and 4.
(i) { , } A2 3 d is a true statement.
(ii) { , } A2 3 1 is wrong.
On rectifying this statement, we get {{ , }} A2 3 1 as true
statement.
SOME RESULTS ON SUBSETS
THEOREM 1 Every set is a subset of itself.
PROOF Let A be any set.
Then, each element of A is in A.
.A A3
Hence, every set is a subset of itself.
THEOREM 2 The empty set is a subset of every set.
PROOF Let A be any set and be the empty set.
Since contains no element at all, so there is no element of which is
not contained in A.
Hence, .A1
THEOREM 3 The total number of subsets of a set containing n elements is .2n
PROOF Let A be a fi nite set containing n elements. Then,
number of subsets of A each containing no element .C1 n 0
number of subsets of A each containing 1 element .Cn 1
number of subsets of A each containing 2 elements .Cn 2
……… ……… ……… ……… ………
……… ……… ……… ……… ………
number of subsets of A each containing n elements .Cn n
total number of subsets of A
( )C C C C…n n n n n0 1 2
( )1 1 2n n [using binomial theorem].
UNIVERSAL SET If there are some sets under consideration then there happens to be a
set which is a superset of each one of the given sets. Such a set is known as the universal
set for those sets. We shall denote a universal set by U.
EXAMPLE 1 Let { , , }, { , , , }A b1 2 3 2 3 4 5 and { , } .C 6 7
If we consider the set { , , , , , , }U 1 23 4 5 6 7 then clearly, U is a
superset of each of the given sets.
Hence, U is the universal set.
EXAMPLE 2 When we discuss sets of lines, triangles or circles in two-
dimensional geometry, the plane in which these lines, triangles or
circles lie, is the universal set.
Sets 13
SUBSETS OF THE SET R OF ALL REAL NUMBERS
(i) { , , , , , }N 1 2 3 4 5 … is the set of all natural numbers.
(ii) { , , , , , , }W 0 1 2 3 4 5 … is the set of all whole numbers.
(iii) { , , , , , , , , , , }Z 4 3 2 1 0 1 2 3 4… … is the set of all integers.
{ , , , , , }Z 1 2 3 4 5 … is the set of all positive integers.
{ , , , , }Z 1 2 3 4 … is the set of all negative integers.
Sometimes, we denote the set of all integers by I.
(iv) ,: , ,where andQ x x q
p
p q Z q 0! d' 1 is the set of all rational
numbers.
The set of all positive rational numbers is denoted by Q.
(v) { : }andT x x R x Q d z is the set of all irrational numbers.
INTERVALS AS SUBSETS OF R
Let ,a b Rd and .a b Then, we defi ne:
(i) Closed Interval [ , ] { : } .a b x R a x b# # d
(ii) Open Interval (a, b) or ]a, b[ { : } .x R a x b d
(iii) Right Half Open Interval [a, b) or [a, b[ { : } .x R a x b# d
(iv) Left Half Open Interval (a, b] or ]a, b] { : } .x R a x b # d
On the real line, we represent these intervals as shown below:
a b
[a, b]
a b
a b a b
(a, b)
[a, b) (a, b]
LENGTH OF AN INTERVAL The length of each of the intervals [a, b], (a, b) [a, b) and
(a, b] is (b – a).
Examples on intervals
(i) [ , ] { : }x R x2 3 2 3# # d (ii) ( , ) { : }x R x2 3 2 3 d
(iii) [ , ) { : }x R x2 3 2 3# d (iv) ( , ] { : }x R x2 3 2 3d #
POWER SET The set of all subsets of a given set A is called the power set of A, denoted
by ( ) .P A
If ( )n A m then [ ( )] .n P A 2m
SOLVED EXAMPLES ON SUBSETS, POWER SET AND INTERVALS
EXAMPLE 1 Write down all possible subsets of { } .A 4
SOLUTION All possible subsets of A are , { } .4
( ) { , { }} .P A 4
Here, ( )n A 1 and [ ( )] .n P A 2 21
14 Senior Secondary School Mathematics for Class 11
EXAMPLE 2 Write down all possible subsets of { , } .A 2 3
SOLUTION All possible subsets of A are
, { }, { }, { , } .2 3 2 3
( ) { , { }, { }, { , }}P A 2 3 2 3
Thus, ( )n A 2 and { ( )} .n P A 4 22
EXAMPLE 3 Write down all possible subsets of { , , } .A 1 0 1
SOLUTION All possible subsets of A are
, { }, { }, { }, { , }, { , }, { , },1 0 1 1 0 0 1 1 1 and { , , } .1 0 1
( ) { , { }, { }, { }, { , }, { , }, { , }, { , , }} .P A 1 0 1 1 0 0 1 1 1 1 0 1
Thus, ( )n A 3 and { ( )} .n P A 8 23
EXAMPLE 4 Write down all possible subsets of { , { , }} .A 1 2 3
SOLUTION Here, A contains two elements, namely 1 and {2, 3}.
Let { , } ,B2 3 then { , } .A B1
( ) { , { }, { }, { , }}P A B B1 1
& ( ) { , { }, {{ , }}, { , { , }}} .P A 1 2 3 1 2 3
EXAMPLE 5 Write down all possible subsets of .
SOLUTION has only one subset, namely .
( ) { } .P
EXAMPLE 6 Write each of the following subsets of R as an interval:
(i) { : , }A x x R x3 5 # d (ii) { : , }B x x R x5 1 d
(iii) { : , }C x x R x2 0# d (iv) { : , }D x x R x1 4# # d
Find the length of each of the above intervals.
SOLUTION We have
(i) { : , } ( , ] .A x x R x3 5 3 5 # d Length ( ) ( ) .A 5 3 8
(ii) { : , } ( , ) .B x x R x5 1 5 1 d Length ( ) ( ) .B 1 5 4
(iii) { : , } [ , ) .C x x R x2 0 2 0# d Length ( ) ( ) .C 0 2 2
(iv) { : , } [ , ] .D x x R x1 4 1 4# # d Length ( ) ( ) .D 4 1 5
EXAMPLE 7 Write each of the following intervals in the set-builder form:
(i) ( , )A 2 5 (ii) [ , ]B 4 7 (iii) [ , )C 8 0 (iv) ( , ]D 5 9
SOLUTION We have
(i) ( , ) { : , }A x x R x2 5 2 5 d .
(ii) [ , ] { : , }B x x R x4 7 4 7# # d .
(iii) [ , ) { : , }C x x R x8 0 8 0# d .
(iv) ( , ] { : , }D x x R x5 9 5 9 # d .
EQUAL SETS Two sets A and B are said to be equal, if every element of A is in B and
every element of B is in A, and we write, A = B.
REMARKS (i) The elements of a set may be listed in any order.
Thus, { , , } { , , } { , , },1 2 3 3 1 2 2 3 1 etc.
(ii) The repetition of elements in a set is meaningless.
Thus, [ , , } { , , , , } .2 4 6 2 2 4 6 6
Sets 15
THEOREM 4 Let A and B be two sets. Then, prove that A B A B+ 3 and .B A3
PROOF Let .A B
Then, by defi nition of equal sets, every element of A is in B and every
element of B is in A.
.andA B B A3 3
Thus, ( ) ( ) .andA B A B B A& 3 3
Again, let .andA B B A3 3
Then, by the defi nition of a subset, it follows that every element of A
is in B and every element of B is in A.
Consequently, .A B
Thus, ( ) .andA B B A A B&3 3
Hence, ( ) .andA B A B B A+ 3 3
EXAMPLE 8 Let { , { }, { , }, } .A 1 2 3 4 5 Which of the following are incorrect statements?
Rectify each:
(i) A2d (ii) { } A2 1 (iii) { , } A1 2 1
(iv) { , } A3 4 1 (v) { , } A1 5 1 (vi) { } A1
(vii) A11 (viii) { , , , } A1 2 3 4 1
SOLUTION Clearly, A contains four elements, namely 1, {2}, {3, 4} and 5.
(i) A2d is incorrect. The correct statement would be { } .A2 d
(ii) { } A2 1 is incorrect. The correct statement is {{ }} .A2 1
(iii) Clearly, A2z and therefore, { , } A1 2 1 is incorrect.
The correct statement would be { , { }} .A1 2 1
(iv) Clearly, {3, 4} is an element of A.
So, { , } A3 4 1 is incorrect and {{ , }} A3 4 1 is correct.
(v) Since 1 and 5 are both elements of A, so { , } A1 5 1 is correct.
(vi) Since ,Az so { } A1 is incorrect while A1 is correct.
(vii) Since ,A1d so A11 is incorrect and therefore, { } A1 1 is
correct.
(viii) Since A2z and ,A3z so { , , , } A1 2 3 4 1 is incorrect.
The correct statement would be { , { }, { , }} .A1 2 3 4 1
EXERCISE 1C
1. State in each case whether orA B A B1 1Y .
(i) { , , , }, { , , , , }A B0 1 2 3 1 2 3 4 5
(ii) , { }A B 0
(iii) { , , }, { , , }A B1 2 3 1 2 4
(iv) { : , }, { : , }A x x Z x B x x N x1 12 2 d d
(v) { :A x x is an even natural number}, { :B x x is an integer}
(vi) { :A x x is an integer}, { :B x x is a rational number}
16 Senior Secondary School Mathematics for Class 11
(vii) { :A x x is a real number}, { :B x x is a complex number}
(viii) { :A x x is an isosceles triangle in a plane},
{ :B x x is an equilateral triangle in the same plane}
(ix) { :A x x is a square in a plane},
{ :B x x is a rectangle in the same plane}
(x) { :A x x is a triangle in a plane},
{ :B x x is a rectangle in the same plane}
(xi) { :A x x is an even natural number less than 8},
{ :B x x is a natural number which divides 32}
2. Examine whether the following statements are true or false:
(i) { , } { , , }a b b c a1Y (ii) { } { , , }a a b cd
(iii) { } { , , }a b c1
(iv) { , } { :a e x x1 is a vowel in the English alphabet}
(v) { : , }x x W x 5 5 d (vi) {{ }, }a a bd
(vii) { } {{ }, }a a b1 (viii) { , } { , { , }}b c a b c1
(ix) { , , , } { , }a a b b a b (x) { , , , , , , }a b a b a b … is an infi nite set.
(xi) If A set of all circles of unit radius in a plane and B set of all circles
in the same plane then .A B1
3. If { } {{ }, }andA B1 1 2 then show that .A B1Y
Hint A1d but .B1z
4. Write down all subsets of each of the following sets:
(i) { }A a (ii) { , }B a b (iii) { , }C 2 3
(iv) { , , }D 1 0 1 (v) E (vi) { , { }}F 2 3
(vii) { , , { , }}G 3 4 5 6
5. Express each of the following sets as an interval:
(i) { : , }A x x R x4 0 d (ii) { : , }B x x R x0 3# d
(iii) { : , }C x x R x2 6 # d (iv) { : , }D x x R x5 2# # d
(v) { : , }E x x R x3 2# d (vi) { : , }F x x R x2 0# d
6. Write each of the following intervals in the set-builder form:
(i) ( , )A 2 3 (ii) [ , ]B 4 10 (iii) [ , )C 1 8
(iv) ( , ]D 4 9 (v) [ , )E 10 0 (vi) ( , ]F 0 5
7. If { , { , }, },A 3 45 6 fi nd which of the following statements are true.
(i) { , } A4 5 3 (ii) { , } A4 5 d (iii) {{ , }} A4 5 3
(iv) A4d (v) { } A3 3 (vi) { } A3
(vii) A3 (viii) { , , } A3 4 5 3 (ix) { , } A3 6 3
8. If { , , },A a b c fi nd ( )P A and { ( )} .n P A
9. If { , { , }},A 1 2 3 fi nd ( ) { ( )} .andP A n P A
10. If A then fi nd { ( )} .n P A
11. If { , , }, { , , } { , , , , }andA B C1 3 5 2 4 6 0 2 4 6 8 then fi nd the universal set.
12. Prove that , .andA B B C C A A C&3 3 3
Sets 17
13. For any set A, prove that .A A+3
14. State whether the given statement is true or false:
(i) If andA B x Bz1 then .x Az
(ii) If A3 then .A
(iii) If A, B and C are three sets such that andA B B C1d then .A C1
Hint Let { }, {{ }, } {{ }, , } .andA a B a b C a b c
Then, { }a Bd and .B C1 But, { } .a C1Y
(iv) If A, B and C are three sets such that A B1 and B Cd then A Cd .
Hint Let { }, { , } {{ , }, } .andA a B a b C a b c
Then, .andA B B C1 d But, .A Cz
(v) If A, B and C are three sets such that .and thenA B B C A C1 1 1Y Y Y
Hint Let { }, { , } { , } .andA a B b c C a c
Then, . .and ButA B B C A C1 1 1Y Y
(vi) If A and B are sets such that x Ad and A Bd then .x Bd
Hint Let { }, {{ }, } .A x B x y
Then, .andx A A Bd d But, .x Bz
ANSWERS (EXERCISE 1C)
1. (i) A B1Y (ii) A Bf (iii) A B1Y (iv) A B1Y (v) A Bf
(vi) A Bf (vii) A Bf (viii) A B1Y (ix) A Bf (x) A B1Y
(xi) A B1Y
2. (i) False (ii) False (iii) False (iv) True (v) False
(vi) False (vii) False (viii) False (ix) True (x) False
(xi) True
4. (i) , { }a (ii) , { }, { }, { , }a b a b (iii) , { }, { }, { , }2 3 2 3
(iv) , { }, { }, { }, { , }, { , }, { , }, { , , }1 0 1 1 0 0 1 1 1 1 0 1
(v) (vi) , { }, {{ }}, { , { }}2 3 2 3
(vii) , { }, { }, {{ , }}, { , }, { , { , }}, { , { , }}, { , , { , }}3 4 5 6 3 4 3 5 6 4 5 6 3 4 5 6
5. (i) ( , )A 4 0 (ii) [ , )B 0 3 (iii) ( , ]C 2 6
(iv) [ , ]D 5 2 (v) [ , )E 3 2 (vi) [ , )F 2 0
6. (i) { : , }A x x R x2 3 d (ii) { : , }B x x R x4 10# # d
(iii) { : , }C x x R x1 8# d (iv) { : , }D x x R x4 9 # d
(v) { : , }E x x R x10 0# d (vi) { : , }F x x R x0 5 # d
7. (i) False (ii) True (iii) True (iv) False (v) True
(vi) False (vii) True (viii) False (ix) True
8. ( ) { , { }, { }, { }, { , }, { , }, { , }, { , , }}P A a b c a b b c a c a b c and { ( )}n P A 2 83
9. ( ) { , { }, {{ , }}, { , { , }}} { ( )}andP A n P A1 2 3 1 2 3 2 42
10. ( ) { } { ( )}P A n P A 1& 11. { , , , , , , , }U 0 1 2 3 4 5 6 8
14. (i) True (ii) True (iii) False (iv) False (v) False
(vi) False
18 Senior Secondary School Mathematics for Class 11
HINTS TO SOME SELECTED QUESTIONS
1. (iv) { , } { } .andA B1 1 1 So, .A B1Y
(v) { , , , , …} {…, , , , , , , , , , , …] .andA B2 4 6 8 1 0 1 2 3 4 5 6 7 8 So, .A Bf
(vi) Every integer is a rational number.
(vii) We may write a real number x as .x i0
So, every real number is a complex number.
(ix) Every square is a rectangle.
(xi) { , , } { , , , , , } .andA B2 4 6 1 2 4 8 16 32 So, .A B1Y
2. (ii) { , , }a a b cd is true. So, { } { , , }a a b cd is false.
(iii) { , , }a b c1 is true and { } { , , }a b c1 is false.
(iv) { , } { , , , , }a e a e i o u1 is true.
(v) { : , } { } .x x W x 5 5 0 d
(vi) {{ }, }a b has two elements, namely {a} and b.
(vii) { } {{ }, } .a a bd So, { } {{ }, }a a b1 is false.
(viii) { , } { , { , }} .b c a b cd So, { , } { , { , }}b c a b c1 is false.
(ix) Repetition of elements in a set has no meaning.
(x) Given set { , },a b which is fi nite.
9. Let { , } .A a1 Then, ( ) { , { }, { }, { , }},P A a a1 1 where { , } .a 2 3
( ) { , { }, {{ , }}, { , { , }}} .P A 1 2 3 1 2 3
OPERATIONS ON SETS
UNION OF SETS The union of two sets A and B, denoted by ,A B, is the set of all those
elements which are either in A or in B or in both A and B.
Thus, { : }orA B x x A x B, d d
.orx A B x A x B+, d dd
EXAMPLE 1 If { , , , }A 3 4 5 6 and { , , , },B 4 6 8 10 fi nd .A B,
SOLUTION Clearly, { , , , , , } .A B 3 4 5 6 8 10,
EXAMPLE 2 Let { :A x x is a prime number less than 10} and { : ,B x x N x d is a
factor of 12}. Find .A B,
SOLUTION We have
{ , , , } { , , , , , } .andA B2 3 5 7 1 2 3 4 6 12
{ , , , } { , , , , , } { , , , , , , , }A B 2 3 5 7 1 2 3 4 6 12 1 2 3 4 5 6 7 12, , .
EXAMPLE 3 Let { :A x x is a positive integer} and let { :B x x is a negative integer}.
Find .A B,
SOLUTION Clearly, { :A B x x, is an integer and } .x 0!
EXAMPLE 4 If { : , } { : , }A x x n n Z and B x x n n Z2 1 2 dd then fi nd .A B,
SOLUTION We have
{ :A B x x, is an odd integer } { :x x, is an even integer}
{ :x x is an integer} = Z.
Sets 19
REMARK The union of n sets , , , ,A A A A… n1 2 3 is denoted by
( … ) .A A A A An
i
n
i1 2 3
1
, , , , ,
INTERSECTION OF SETS The intersection of two sets A and B, denoted by ,A B+ is the
set of all elements which are common to both A and B.
Thus, { : } .andA B x x A x B+ d d
;andx A B x A x B++d d d
.orx A B x A x B&+z z z
EXAMPLE 5 Let { , , , , } { , , , , , } .A and B1 3 5 7 9 2 3 5 7 11 13 Find .A B+
SOLUTION We have
{ , , , , } { , , , , , } { , , } .A B 1 3 5 7 9 2 3 5 7 11 13 3 5 7+ +
EXAMPLE 6 If { : ,A x x N x d is a factor of 12} and { : ,B x x N x d is a factor of 18},
fi nd .A B+
SOLUTION We have
{ : ,A x x N x d is a factor of 12} = {1, 2, 3, 4, 6, 12},
{ : ,B x x N x d is a factor of 18} = {1, 2, 3, 6, 9, 18}.
{ , , , , , } { , , , , , } { , , , } .A B 1 2 3 4 6 12 1 2 3 6 9 18 1 2 3 6+ +
EXAMPLE 7 If { : , }A x x n n Z3 d and { : , }B x x n n Z4 d then fi nd .A B+
SOLUTION We have
{ : andA x x Z x d is a multiple of 3}
and { : andB x x Z x d is a multiple of 4}.
{ : andA B x x Z x+ d is a multiple of both 3 and 4}
{ : andx x Z x d is a multiple of 12}
{ : , } .x x n n Z12 d
Hence, { : , } .A B x x n n Z12+ d
EXAMPLE 8 If ( , ) [ , ),A and B2 4 3 5 fi nd .A B+
SOLUTION We have
( , ) { : , }A x x R x2 4 2 4 d
[ , ) { : , }B x x R x3 5 3 5# d
0 1 2 3 4 5 xx�
( )[ )
Clearly, { : , } [ , ) .A B x x R x3 4 3 4+ # d
REMARK The intersection of n sets , , , ,A A A A… n1 2 3 is denoted by
( … ) .A A A A An
i
n
i1 2 3
1
+ + + +
+
DISJOINT SETS Two sets A and B are said to be disjoint if .A B+
INTERSECTING SETS Two sets A and B are said to be intersecting if .A B+ !
20 Senior Secondary School Mathematics for Class 11
EXAMPLE 9 If { , , , , }, { , , , } { , , , , },A B and C1 3 5 7 9 2 4 6 8 2 3 5 7 11 fi nd ( )A B+
and ( ) .A C+ What do you conclude?
SOLUTION We have
{ , , , , } { , , , }A B 1 3 5 7 9 2 4 6 8+ +
and { , , , , } { , , , , } { , , } .A C 1 3 5 7 9 2 3 5 7 11 3 5 7+ + !
Thus, A and B are disjoint sets while A and C are intersecting sets.
EXAMPLE 10 Give examples of three sets A, B, C such that ( ) , ( ) ,A B B C+ +! !
( )A C+ ! and ( ) .A B C+ +
SOLUTION Consider the sets { , }, { , , } { , , } .andA B C1 2 2 3 4 1 3 5
Then, ( ) { , } { , , } { } ;A B 1 2 2 3 4 2+ + !
( ) { , , } { , , } { } ;B C 2 3 4 1 3 5 3+ + !
( ) { , } { , , } { } ;A C 1 2 1 3 5 1+ + !
( ) { , } { , , } { , , } .A B C 1 2 2 3 4 1 3 5+ + + +
Thus, , ;A B B C A C+ + +! ! ! and .A B C+ +
EXAMPLE 11 Give an example of three sets A, B, C such that A B A C+ + but .B C!
SOLUTION Consider the sets { , , }, { , } { , , } .andA B C1 2 3 3 4 3 5 7
Then, { , , } { , } { } .A B 1 2 3 3 4 3+ +
And, { , , } { , , } { } .A C 1 2 3 3 5 7 3+ +
Thus, ,A B A C+ + and clearly, .B C!
DIFFERENCE OF SETS For any sets A and B, their difference ( )A B is defi ned as
( ) { : } .A B x x A and x B d zThus, ( ) .x A B x A and x B&d d z
EXAMPLE 12 If { : ,A x x N x d is a factor of 6} and { :B x N x d is a factor of 8} then
fi nd (i) ,A B, (ii) ,A B+ (iii) ,A B (iv) .B A
SOLUTION We have
{ : ,A x x N x d is a factor of 6} { , , , }1 2 3 6
and { : ,B x x N x d is a factor of 8} { , , , } .1 2 4 8
(i) { , , , } { , , , } { , , , , , }A B 1 2 3 6 1 2 4 8 1 2 3 4 6 8, , .
(ii) { , , , } { , , , } { , } .A B 1 2 3 6 1 2 4 8 1 2+ +
(iii) { , , , } { , , , } { , } .A B 1 2 3 6 1 2 4 8 3 6
(iv) { , , , } { , , , } { , } .B A 1 2 4 8 1 2 3 6 4 8
SYMMETRIC DIFFERENCE OF TWO SETS
The symmetric difference of two sets A and B, denoted by A BT , is defi ned as
( ) ( ) .A B A B B A,T
EXAMPLE 13 Let { , , , }A a b c d and { , , , } .B b d f g Find .A BT
SOLUTION We have
( ) { , , , } { , , , } { , } .A B a b c d b d f g a c
Sets 21
( ) { , , , } { , , , } { , } .B A b d f g a b c d f g
( ) ( ) { , } { , } { , , , } .A B A B B A a c f g a c f g, ,T
COMPLEMENT OF A SET
Let U be the universal set and let .A Uf Then, the complement of A, denoted by Al or
( ),U A is defi nied as
{ : } .A x U x A d zl
Clearly, .x A x A+d zl
EXAMPLE 14 If { , , , , , , , } { , , , },U and A1 2 3 4 5 6 7 8 2 4 6 8 fi nd (i) Al (ii) ( ) .All
SOLUTION We have
(i) A U A l
{ , , , , , , , } { , , , } { , , , } .1 2 3 4 5 6 7 8 2 4 6 8 1 3 5 7
(ii) ( )A U A ll l
{ , , , , , , , } { , , , } { , , , } .A1 2 3 4 5 6 7 8 1 3 5 7 2 4 6 8
EXAMPLE 15 Let N be the universal set.
(i) If { :A x x N d and x is odd}, fi nd .Al
(ii) If { : ,B x x N x d is divisible by 3 and 5}, fi nd .Bl
SOLUTION We have
(i) { :A x x Ndl and x is not odd} { :x x N d and x is even}.
(ii) { : ,B x x N x dl is not divisible by 3 or x is not divisible by 5}.
SOME RESULTS ON COMPLEMENTATION
EXAMPLE 16 If ,A U1 prove that:
(i) U l (ii) U l (iii) ( )A All (iv) A A U, l (v) A A+ l
SOLUTION We have
(i) { : } .U x U x U d zl
(ii) { : } .x U x U d zl
(iii) ( ) { : } [ : } .A x U x A x U x A A d d dzll l
(iv) { : }A A x U x A A, , d dl l
{ : }orx U x A x A d d d l
{ : } .orx U x A x A U d d z
(v) { : }A A x U x A A+ + d dl l
{ : }andx U x A x A d d d l
{ : } .andx U x A x A d d z
EXERCISE 1D
1. If { , , , , , }, { , , , }A a b c d e f B c e g h and { , , , },C a e m n fi nd:
(i) A B, (ii) B C, (iii) A C,
(iv) B C+ (v) C A+ (vi) A B+
22 Senior Secondary School Mathematics for Class 11
2. If { , , , , }, { , , , , }, { , , , , }A B C1 2 3 4 5 4 5 6 7 8 7 8 9 10 11 and
{ , , , , },D 10 11 12 13 14 fi nd:
(i) A B, (ii) B C, (iii) A C,
(iv) B D, (v) ( )A B C, , (vi) ( )A B C, +
(vii) ( )A B D+ , (viii) ( ) ( )A B B C+ , + (ix) ( ) ( )A C C D, + ,
3. If { , , , , }, { , , , }, { , , }A B C3 5 7 9 11 7 9 11 13 11 13 15 and { , },D 15 17 fi nd:
(i) A B+ (ii) A C+ (iii) B C+
(iv) B D+ (v) ( )B C D+ , (vi) ( )A B C+ ,
4. If { : }, { :A x x N B x x N d d and x is even}, { :C x x Nd and x is odd} and
{ :D x x N d and x is prime} then fi nd:
(i) A B+ (ii) A C+ (iii) A D+
(iv) B C+ (v) B D+ (vi) C D+
5. If { : }, {( ) :andA x x N x B x x N2 1 4 2# d d and }x2 5# and
{ :C x x N d and },x4 8 fi nd:
(i) A B+ (ii) A B, (iii) ( )A B C, +
6. If { , , , , , }A 2 4 6 8 10 12 and { , , , , , , },B 3 4 5 6 7 8 10 fi nd:
(i) ( )A B (ii) ( )B A (iii) ( ) ( )A B B A,
7. If { , , , , }, { , , , } { , , , },andA a b c d e B a c e g C b e f g fi nd:
(i) ( )A B C+ (ii) ( )A B C, (iii) ( )A B C+
8. If ,: :and andA x x N x and B x x N x
1 8 2
1 4 # d d& &0 0 fi nd:
(i) A B, (ii) A B+ (iii) A B (iv) B A
9. If R is the set of all real numbers and Q is the set of all rational numbers
then what is the set ( )R Q ?
10. If { , , , , } ,andA B2 3 5 7 11 fi nd:
(i) A B, (ii) A B+
11. If A and B are two sets such that A B3 then fi nd:
(i) A B, (ii) A B+ (iii) A B
12. Which of the following sets are pairs of disjoint sets? Justify your answer:
(i) { , , , } { , , , }andA B3 4 5 6 2 5 7 9
(ii) { , , , , } { , , , }andC D1 2 3 4 5 6 7 9 11
(iii) { : ,E x x N x d is even and }x 8
{ : , }andF x x n n N n3 4 d
(iv) { : ,G x x N x d is even} and { : ,H x x N x d is prime}
(v) { : ,J x x N x d is even} and { : ,K x x N x d is odd}
13. If { , , , , , , , , }, { , , , }, { , , , }U A B1 2 3 4 5 6 7 8 9 1 2 3 4 2 4 6 8 and { , , , },C 1 4 5 6
fi nd:
(i) Al (ii) Bl (iii) Cl (iv) ( )Bll
(v) ( )A B, l (vi) ( )A C+ l (vii) ( )B C l
Sets 23
14. If { , , , , }, { , , } { , , , }U a b c d e A a b c and B b c d e then verify that:
(i) ( ) ( )A B A B, +l l l (ii) ( ) ( )A B A B+ ,l l l
15. If U is the universal set and A Uf then fi ll in the blanks:
(i) ……A A, l (ii) ……A A+ l
(iii) ……A+ l (iv) ……U A+ l
ANSWERS (EXERCISE 1D)
1. (i) {a, b, c, d, e, f, g, h} (ii) {a, c, e, g, h, m, n} (iii) {a, b, c, d, e, f, m, n}
(iv) {e} (v) {a, e} (vi) {c, e}
2. (i) {1, 2, 3, 4, 5, 6, 7, 8} (ii) {4, 5, 6, 7, 8, 9, 10, 11}
(iii) {1, 2, 3, 4, 5, 7, 8, 9, 10, 11} (iv) {4, 5, 6, 7, 8, 10, 11, 12, 13, 14}
(v) {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} (vi) {7, 8}
(vii) {4, 5, 10, 11, 12, 13, 14} (viii) {4, 5, 7, 8}
(ix) {7, 8, 9, 10, 11}
3. (i) {7, 9} (ii) {11} (iii) {11, 13} (iv)
(v) {11, 13} (vi) {7, 9, 11}
4. (i) B (ii) C (iii) D (iv)
(v) {2} (vi) { }D 2
5. (i) {4, 6} (ii) {2, 4, 5, 6} (iii) {5, 6}
6. (i) {2, 12} (ii) {3, 5, 7} (iii) {2, 3, 5, 7, 12}
7. (i) {a, c} (ii) {d} (iii) {a, b, c, d}
8. (i) , , , , , ,,1 2
1
3
1
4
1
5
1
6
1
7
1
8
1
& 0 (ii) , ,2
1
4
1
6
1
& 0 (iii) , ,,1 3
1
5
1
7
1
& 0 (iv) 8
1
& 0
9. ( ) { : ,R Q x x R x d is irrational}
10. (i) 2, 3, 5, 7, 11} (ii) 11. (i) B (ii) A (iii)
12. (ii) C and D, since C D+ (v) J and K, since J K+
13. (i) {5, 6, 7, 8, 9} (ii) {1, 3, 5, 7, 9} (iii) {2, 3, 7, 8, 9} (iv) {2, 4, 6, 8}
(v) {5, 7, 9} (vi) {2, 3, 5, 6, 7, 8, 9} (vii) {1, 3, 4, 5, 6, 7, 9}
15. (i) U (ii) (iii) A (iv)
LAWS OF OPERATIONS ON SETS
THEOREM 1 (Idempotent laws) For any set A, prove that:
(i) A A A, (ii) A A A+
PROOF We have
(i) { : } { : } .orA A x x A x A x x A A, d d d
(ii) { : } { : } .andA A x x A x A x x A A+ d d d
THEOREM 2 (Identity laws) For any set A, prove that:
(i) A A, (ii) ,A U A+ where U is the universal set
24 Senior Secondary School Mathematics for Class 11
PROOF We have
(i) { : } { : }orA x x A x x x A A, d d d [ ]xa z
(ii) { : } { : }andA U x x A x U x x A A+ d dd [ ]A Ua f
NOTE and U are the identity elements for union and intersection of sets
respectively.
THEOREM 3 (Commutative laws) For any two sets A and B, prove that:
I. A B B A, , [Commutative law for union of sets]
II. A B B A+ + [Commutative law for intersection of sets]
PROOF I. Let x be an arbitrary element of .A B, Then,
orx A B x A x B&,d d d
orx B x A& d d
x B A& ,d .
.A B B A, ,3 … (i)
Again, let y be an arbitrary element of .B A, Then,
ory B A y B y A&,d d d
ory A y B& d d
y A B& ,d .
.B A A B, ,3 … (ii)
From (i) and (ii), we get .A B B A, ,
II. Let x be an arbitrary element of .A B+ Then,
andx A B x A x B&+d d d
andx B x A& d d
.x B A& +d
.A B B A+ +3 … (iii)
Again, let y be an arbitray element of .B A+ Then,
andy B A y B y A&+d d d
andy A y B& d d
y A B& +d .
.B A A B+ +3 … (iv)
From (iii) and (iv), we get .A B B A+ +
THEOREM 4 (Associative laws) For any sets A, B, C, prove that:
I. ( ) ( )A B C A B C, , , , [Associative law for union of sets]
II. ( ) ( )A B C A B C+ + + + [Associative law for intersection of sets]
PROOF I. Let x be an arbitrary element of ( ) .A B C, , Then,
( ) ( ) orxA B C x A B x C&, , ,d d d
( )or orx A x B x C& d d d
( )or orx A x B x C& d d d
( )orx A x B C& ,d d
( ) .x A B C& , ,d
Sets 25
( ) ( ) .A B C A B C, , , ,3 … (i)
Again, let y be an arbitrary element of ( ) .A B C, , Then,
( ) ( )ory A B C y A y B C&, , ,d d d
( )or ory A y B y C& d d d
( )or ory A y B y C& d d d
( ) ory A B y C& ,d d
( ) .y A B C& , ,d
( ) ( ) .A B C A B C, , , ,3 … (ii)
From (i) and (ii), we get ( ) ( ) .A B C A B C, , , ,
II. Let x be an arbitrary element of ( ) .A B C+ + Then,
( ) ( ) andx A B C x A B x C&+ + +dd d
( )and andx A x B x C& d d d
( )and andx A x B x C& d d d
( )andx A x B C& +d d
( )x A B C& + +d .
( ) ( ) .A B C A B C+ + + +3 … (iii)
Again, let y be an arbitrary element of ( ) .A B C+ + Then,
( ) ( )andy A B C y A y B C&+ + +d d d
( )and andy A y B y C& d d d
( )and andy A y B y C& d d d
( ) andy A B y C& +d d
( )y A B C& + +d
( ) ( ) .A B C A B C+ + + +3 … (iv)
From (iii) and (iv), we get ( ) ( ) .A B C A B C+ + + +
THEOREM 5 (Distributive laws) For any three sets A, B, C prove that:
I. ( ) ( ) ( )A B C A B A C, + , + ,
[Distributive law of union over intersection]
II. ( ) ( ) ( )A B C A B A C+ , + , +
[Distributive law of intersection over union]
PROOF I. Let x be an arbitrary element of ( ) .A B C, + Then,
( ) ( )orx A B C x A x B C&, + +d d d
( )or andx A x B x C& d d d
( ) ( ))or and orx A x B x A x C& d dd d
[ ‘or‘ distributes ‘and‘]
( ) ( )andx A B x A C& , ,dd .
( ) ( ) .x A B A C& , + ,d
( ) ( ) ( ) .A B C A B A C, + , + ,3 … (i)
Again, let y be an arbitrary element of ( ) ( ) .A B A C, + , Then,
26 Senior Secondary School Mathematics for Class 11
( ) ( ) ( ) ( )andy A B A C y A B y A C&, + , , ,d d d
( ) ( )or and ory A y B y A y C& dd d d
( )or andy A y B y C& d d d
[ ‘or‘ distributes ‘and‘]
( )ory A y B C& +d d
( )y A B C& , +d .
( ) ( ) ( ) .A B A C A B C, + , , +3 … (ii)
From (i) and (ii), we get ( ) ( ) ( ) .A B C A B A C, + , + ,
II. Let x be an arbitrary element of ( ) .A B C+ , Then,
( ) ( )andx A B C x A x B C&+ , ,d d d
( )and orx A x B x C& d d d
( ) ( )and or andx A x B x A x C& d d d d
[ ‘and‘ distributes ‘or‘]
( ) ( )orx A B x A C& + +d d
( ) ( )x A B A C& d + , + .
( ) ( ) ( ) .A B C A B A C+ , + , +3 … (iii)
Again, let y be an arbitrary element of ( ) ( ) .A B A Ck j k Then,
( ) ( ) ( ) ( )ory A B A C y A B y A C&+ , + + +d d d
( ) ( )and or andy A y B y A y C& d d d d
( )and ory A y B y C& d d d
[ ‘and‘ distributes ‘or‘]
( )andy A y B C& ,d d
( )y A B C& d + , .
( ) ( ) ( )A B A C A B C+ , + + ,3 . … (iv)
From (iii) and (iv), we get ( ) ( ) ( ) .A B C A B A C+ , + , +
THEOREM 6 (De Morgan’s laws) For any two sets A and B, prove that:
I. ( ) ( )A B A B, +l l l II. ( ) ( )A B A B+ ,l l l
PROOF I. Let x be an arbitrary element of ( ) .A B, l Then,
( )x A B x A B&, ,d zl
andx A x B& z z [note this point]
andx A x B& d dl l
( )x A B& d +l l .
( ) ( ) .A B A B, +3l l l … (i)
Again, let y be an arbitrary element of ( ) .A B+l l Then,
( ) andy A B y A y B&+d d dl l l l
andy A y B& z z
( )y A B& ,z [note this point]
( )y A B& ,d l.
Sets 27
( ) ( ) .A B A B+ ,3l l l … (ii)
From (i) and (ii), we get ( ) ( ) .A B A B, +l l l
II. Let x be an arbitrary element of ( ) .A B+ l Then,
( ) ( )x A B x A B&+ +d zl
orx A x B& z z [note this point]
orx A x B& d dl l
( )x A B& ,d l l
( ) ( ) .A B A B+ ,3l l l … (iii)
Again, let y be an arbitrary element of ( ) .A B,l l Then,
( ) ory A B y A y B&,d d dl l l l
ory A y B& z z
( )y A B& +z [note this point]
( )y A B& +d l.
( ) ( ) .A B A B, +3l l l … (iv)
From (iii) and (iv), we get ( ) ( ) .A B A B+ ,l l l
SOME MORE RESULTS
THEOREM 7 For any two sets A and B, prove that:
I. A B B A&3 3l l II. A B A B+ l
III. ( )A B B+ IV. ( ) ( ) ( ) ( )A B B A A B A B, , +
V. ( )A B A A B+ +
PROOF I. Let A B3 be given and let x be an arbitrary element of .Bl Then,
x B x B&d zl
x A& z [ ]A Ba 3
.x A& d l
.B A3l l
Hence, .A B B A&3 3l l
II. Let x be an arbitrary element of ( ) .A B Then,
( ) andx A B x A x B&d d z
andx A x B& d d l
.x A B& +d l
( ) .A B A B+3 l … (i)
Again, let y be an arbitrary element of ( ) .A B+ l Then,
( ) andx A B x A x B&+d d dl l
andx A x B& d z
( )x A B& d .
( ) ( ) .A B A B+ 3 l … (ii)
Hence, from (i) and (ii), we get ( ) ( ) .A B A B+ l
III. If possible, let ( )A B B+ ! and let ( ) .x A B B+d Then,
( ) ( ) andx A B B x A B x B&+ d d d
28 Senior Secondary School Mathematics for Class 11
( )and andx A x B x B& d dz
( ) .and andx A x B x B& d dz
But, x Bz and x Bd can never hold simultaneously.
Thus, we arrive at a contradiction.
Since the contradiction arises by assuming that ( )A B B+ !
and hence ( ) .A B B+
IV. Let x be an arbitrary element of ( ) ( ) .A B B A, Then,
( ) ( ) ( ) ( )orx A B B A x A B x B A&, d d d
( ) ( )and or andx A x B x B x A& d dz z
( ) ( )or and orx A x B x A x B& d d z z
[note it]
( ) ( )andx A B x A B& , +d z
{( ) ( )} .x A B A B& , +d
( ) ( ) {( ) ( )} .A B B A A B A B, , +3 … (iii)
Again, let y be an arbitrary element of ( ) ( ) .A B A B, + Then,
( ) ( ) ( ) ( )andy A B A B y A B y A B&, + , +d d z
( ) ( )or and ory A y B y A y B& d d z z
[note it]
( ) ( )and or andy A y B y B y A& d dz z
[note it]
( ) ( )ory A B y B A& d d
( ) ( ) .y A B B A& , d
( ) ( ) ( ) ( ) .A B A B A B B A, + ,3 … (iv)
From (iii) and (iv), we get ( ) ( ) ( ) ( ) .A B B A A B A B, , +
V. Let ( )A B A be given and we have to show that .A B+
If possible, let A B+ ! and let .x A B+d Then,
andx A B x A x B&+d d d
( ) andx A B x B& d d [ ( )A A Ba (given)]
( )and andx A x B x B& d dz
( ) .and andx A x B x B& d dz
But, x Bz and x Bd both can never hold simultaneously.
Thus, we arrive at a contradiction.
Since the contradiction arises by assuming that ,A B+ ! so
.A B+
Thus, ( ) .A B A A B& +
Again, let ( )A B+ and we have to show that ( ) .A B A
Now, ( ) andx A B x A x B&d d z
x A& d (surely).
( ) .A B A3 … (v)
Sets 29
Again, y A y B&d z [ ]A Ba +
andy A y B& d z
( ) .y A B& d
( ) .A A B3 … (vi)
Thus, from (v) and (vi), we get ( ) .A B A
( ) .A B A B A&+
Hence, ( ) ( ) .A B A A B+ +
THEOREM 8 If ( ) ( )A B A B, + then prove that .A B
PROOF Let ( ) ( )A B A B, + be given.
Let x be an arbitrary element of A. Then,
x A x A B& ,d d [ ]A A Ba ,3
x A B& +d [ ]A B A Ba , +
andx A x B& d d
x B& d (surely).
.A B3 … (i)
Again, let .y Bd Then,
y B y A B& ,d d [ ]B A Ba ,3
y A B& +d [ ]A B A Ba , +
andy A y B& d d
y A& d (surely).
.B A3 … (ii)
Thus, from (i) and (ii), we get .A B
THEOREM 9 If A B3 then for any set C, prove that ( ) ( ) .C B C A3
PROOF Let A B3 be given.
Let ( ) .x C Bd Then,
( ) andx C B x C x B&d d z
andx C x A& d z [ ]A Ba 3
( ) .x C A& d
( ) ( ) .C B C A3
Hence, ( ) ( ) .A B C B C A&3 3
THEOREM 10 For any sets A and B, prove that:
(i) ( )A A B A, + (ii) ( )A A B A+ ,
PROOF (i) Since ( ) ,A B A+ 3 we have ( )A A B A, + [ ] .X Y X Y Y&a ,3
(ii)Since ( ),A A B,3 we have ( )A A B A+ , [ ] .X Y X Y X&a +3
THEOREM 11 For any sets A and B, prove that:
(i) ( ) ( )A B A B A+ , (ii) ( ) ( )A B A A B, ,
30 Senior Secondary School Mathematics for Class 11
PROOF (i) We have
( ) ( ) ( ) ( )A B A B A B A B+ , + , + l [ ( ) ( )]A B A Ba + l
( )A B B+ , l [by distributive law]
A U+ [ ]B B Ua , l
.A
Hence, ( ) ( ) .A B A B A+ ,
(ii) We have
( ) ( )A B A A B A, , + l [ ( ) ( )]B A B Aa + l
( ) ( )A B A A, + , l [by distributive law]
( )A B U, + [ ]A A Ua , l
( ) .A B,
Hence, ( ) ( ) .A B A A B, ,
THEOREM 12 If A B+ l then prove that A A B+ and hence show that .A B3
PROOF Let A B+ l be given. Then,
( ),A A U+ where U is the universal set
( )A B B+ , l [ ]B B Ua , l
( ) ( )A B A B+ , + l
( )A B+ , [ ]A Ba + l
( ) .A B+
Hence, ( ) .A A B+
Further, let A A B+ and let .x Ad Then,
x A x A B& +d d [ ]A A Ba +
andx A x B& d d
x B& d (surely).
.A B3
THEOREM 13 If A, B and C be the sets such that A B A C, , and A B A C+ +
then prove that .B C
PROOF Let A B A C, , and A B A C+ + be given. Then,
A B A C, ,
( ) ( )A B B A C B, + , + and ( ) ( )A B C A C C, + , +
( ) ( )B A B C B+ , + and ( ) ( )A C B C C+ , +
[ ( ) ( )]B A B C A Canda , ,3 3
( ) ( ) ( ) ( )andB A B B C A B B C C+ , + + , + [ ]A C A Ba + +
.B C
Hence, .B C
THEOREM 14 For any sets A, B and C, prove that:
(i) ( ) ( ) ( )A B C A B A C, +
(ii) ( ) ( ) ( )A B C A B A C+ ,
(iii) ( ) ( ) ( )A B C A C B C, ,
(iv) ( ) ( ) ( )A B C A C B C+ +
Sets 31
PROOF We have
(i) ( ) ( )A B C A B C, + , l [ ]X Y X Ya + l
( )A B C+ + l l [by De Morgan’s law]
( ) ( )A B A C+ + + l l
( ) ( ) .A B A C+
( ) ( ) ( ) .A B C A B A C, +
(ii) ( ) ( )A B C A B C+ + + l [ ]X Y X Ya + l
( )A B C+ , l l [by De Morgan‘s law]
( ) ( )A B A C+ , + l l [by distributive law]
( ) ( )A B A C, [ ]X Y X Ya + l .
( ) ( ) ( ) .A B C A B A C+ ,
(iii) ( ) ( )A B C A B C, , + l [ ]X Y X Ya + l
( ) ( )A C B C+ , + l l [by distributive law]
( ) ( ) .A C B C, [ ]X Y X Ya + l
( ) ( ) ( ) .A B C A C B C, ,
(iv) ( ) ( )A B C A B C+ + + l [ ]X Y X Ya + l
( ) ( )A C B C+ + + l l [note]
( ) ( ) .A C B C+ [ ]X Y X Ya + l
( ) ( ) ( ) .A B C A C B C+ +
THEOREM 15 Let A and B be sets. If A X B X+ + and A X B X, , for some
set X, show that .A B
PROOF A X B X, , [given]
( ) ( )A A X A B X+ , + ,
( ) ( ) ( ) ( )A A A X A B A X+ , + + , + [by distributive law]
( )A A B, + , [ ]A Xa +
( )A A B+
A B3 … (i) [ ]A B A A B&a + 3 .
Again, A X B X, ,
( ) ( )B A X B B X+ , + ,
( ) ( ) ( ) ( )B A B X B B B X+ , + + , + [by distributive law]
( )A B B+ , , [ ]andB X B A A Ba + + +
A B B+
B A3 … (ii) [ ]A B B B A&a + 3 .
From (i) and (ii), we get .A B
THEOREM 16 Show that the following four conditions are equivalent:
(i) A B1 (ii) A B (iii) A B B, (iv) A B A+
PROOF In order to prove the required result, we will show that:
(i) (ii) (iii) (iv) (i).
32 Senior Secondary School Mathematics for Class 11
Now, (i) (ii):
Let A B1 be given.
Then, there is no element of A which is not in B.
{ : }andA B x x A x B d z
[ there is no element of A which is not in B].
Hence, A B A B&1 and therefore, (i) & (ii).
(ii) & (iii):
Let A B be given. Then,
A B & every element of A is in B
A B& 1
.A B B& ,
Thus, A B A B B& , and therefore, (ii) & (iii).
(iii) & (iv):
Let A B B, be given. Then,
.A B B A B A B A& &, +1
Thus, A B B A B A&, + and therefore, (iii) (iv).
(iv) (i):
Let A B A+ be given. Then,
x A x A B& +d d [ ]A A Ba +
andx A x B& d d
x B& d (surely).
.A B3
Thus, A B A A B&+ 3 and therefore, (iv) (i).
(i) (ii) (iii) (iv) (i).
Hence, the given four conditions are equivalent.
THEOREM 17 For any sets A and B, prove that
( ) ( ) ( ) .P A B P A P B+ +
PROOF Let ( ) .X P A B+d Then,
( )X P A B X A B&+ +3d
andX A X B& 3 3
( ) ( )andX P A X P B& d d
( ) ( ) .X P A P B& +d
( ) ( ) ( )P A B P A P B+ +3 . … (i)
Again, let ( ) ( ) .Y P A P B+d Then,
( ) ( ) ( ) ( )andY P A P B Y P A Y P B&+d d d
andY A Y B& 3 3
Y A B& +3
( ) .Y P A B& +d
Sets 33
( ) ( ) ( ) .P A P B P A B+ +3 … (ii)
From (i) and (ii), we get ( ) ( ) ( ) .P A B P A P B+ +
THEOREM 18 For any two sets A and B, prove that
( ) ( ) ( ) .P A P B P A B, ,1
But, ( )P A B, is not necessarily a subset of ( ) ( ) .P A P B,
PROOF Let X be an arbitrary element of ( ) ( ) .P A P B, Then,
( ) ( ) ( ) ( )orX P A P B X P A X P B&,d d d
orX A X B& 1 1
( )X A B& ,1
( ) .X P A B& ,d
( ) ( ) ( ) .P A P B P A B, ,1
However, ( ) ( ) ( )P A B P A P B, ,1 is not always true.
For example, let { }A 1 and { } .B 2 Then, { , } .A B 1 2,
( ) { , { }}, ( ) { , { }}P A P B1 2
and ( ) { , { }, { }, { , }} .P A B 1 2 1 2,
Also, ( ) ( ) { , { }, { }} .P A P B 1 2,
( ) ( ) ( ) .P A B P A P B, ,1Y
Hence, in general, ( ) ( ) ( ) .P A B P A P B, ,!
THEOREM 19 If ( ) ( ),P A P B prove that .A B
PROOF Let ( ) ( ) .P A P B Then,
( )A A A P A&3 d
( )A P B& d [ ( ) ( )]P A P Ba
.A B& 3 … (i)
Again, ( )B B B P B& d3
( )B P A& d [ ( ) ( )]P B P Aa
.B A& 3 … (ii)
From (i) and (ii), we get .andA B B A3 3
Hence, A = B.
EXERCISE 1E
1. If { , , , , }, { , , , } { , , , },andA a b c d e B a c e g C b e f g verify that:
(i) A B B A, , (ii) A C C A, , (iii) B C C B, ,
(iv) A B B A+ + (v) B C C B+ + (vi) A C C A+ +
(vii) ( ) ( )A B C A B C, , , , (viii) ( ) ( )A B C A B C+ + + +
2. If { , , , , }, { , , , } { , , , },andA a b c d e B a c e g C b e f g verify that:
(i) ( ) ( ) ( )A B C A B A C+ + + (ii) ( ) ( ) ( )A B C A B A C+ ,
3. If { : , }, { :A x x N x B x x7# d is prime, }x 8 and { : ,C x x N x d is odd
and },x 10 verify that:
(i) ( ) ( ) ( )A B C A B A C, + , + , (ii) ( ) ( ) ( )A B C A B A C+ , + , +
34 Senior Secondary School Mathematics for Class 11
4. If { , , , , , , , , }, { , , , }U A1 2 3 4 5 6 7 8 9 2 4 6 8 and { , , , },B 2 3 5 7 verify that:
(i) ( ) ( )A B A B, +l l l (ii) ( ) ( )A B A B+ ,l l l
5. Let { , , },A a b c { , , , }B b c d e and { , , , }C c d e f be subsets of
{ , , , , , } .U a b c d e f Then, verify that:
(i) ( )A All (ii) ( ) ( )A B A B, +l l l (iii) ( ) ( )A B A B+ ,l l l
6. Given an example of three sets A, B, C such that , ,A B B C+ +! !
A C+ ! and .A B C+ +
7. For any sets A and B, prove that:
(i) ( )A B B+ (ii) ( )A B A A B, ,
(iii) ( ) ( )A B A B A, + (iv) ( )A B B A B,
(v) ( )A A B A B+
8. For any sets A and B, prove that:
(i) A B A B&+ 1l (ii) A B U A B&, 1l
HINTS TO SOME SELECTED QUESTIONS
6. Take { , }, { , } { , } .andA B C1 2 2 3 1 3
7. (i) ( ) ( ) ( ) .A B B A B B A B B A+ + + + + + l l
(ii) ( ) ( ) ( ) ( ) ( ) ( ) .A B A A B A A B A A A B U A B, , + , + , , + , l l
(iii) ( ) ( ) ( ) ( ) ( ) .A B A B A B A B A B B A U A, + + , + + , + l l
(iv) ( ) ( ) ( ) ( ) ( ) .A B B A B B A B B B A B A B A B, , + + , + + , + l l l l l
(v) ( ) ( ) ( ) ( ) ( ) ( )A A B A A B A A B A A A B A B+ + + + , + , + , + l l l l l l
( ) ( ) .A B A B+ l
8. (i) .A B A B A B& &+ 1 l
(ii) ( ) ( ) ( )A B U A A B A U A A A B A& &, + , + + , + l l l
( ) .A B A A B A A B& & &, + + 1
VENN DIAGRAMS
In order to express therelationship among sets in perspective, we represent
them pictorially by means of diagrams, called Venn diagrams.
In these diagrams, the universal set is represented by a rectangular region
and its subsets by circles inside the rectangle. We represent disjoint sets by
disjoint circles and intersecting sets by intersecting circles.
VENN DIAGRAMS IN DIFFERENT SITUATIONS
CASE 1 When the universal set and its subset are given
Let U be the universal set and let .A U3
We draw a circle inside a rectangle.
The rectangular region represents U and the circular region
represents A.
Sets 35
EXAMPLE Let { , , , , , , }U 1 2 3 4 5 6 7 and
{ , , , } .A 1 3 5 7
Then, we draw the Venn diagram,
as shown in the given fi gure.
Clearly, { , , } .A 2 4 6l
CASE 2 When two intersecting subsets of U are
given
For representing two intersecting
subsets A and B of U, we draw two
intersecting circles within the rectangle.
The common region of these circles
represents .A B+
Excluding the region of B from that of A shows ( ) .A B
Excluding the region of A from that of B shows ( ) .B A
EXAMPLE Let { , , , , , , , }U 1 2 3 4 5 6 7 8 be the
universal set, and let
{ , , , }A 1 3 4 5 and { , , , }B 2 4 5 6 be
its subsets.
Then, { , } .A B 4 5+
We draw the Venn diagram, as
shown in the given fi gure.
Clearly, ( ) { , }A B 1 3 and ( ) { , } .B A 2 6
CASE 3 When two disjoint subsets of a set be given
In order to represent two disjoint subsets A and B of the universal set
U, we draw two disjoint circles within a rectangle.
EXAMPLE Let { , , , , , , }U 1 2 3 4 5 6 7 be the
universal set, and let { , , }A 1 3 5
and { , }B 2 4 be two of its disjoint
subsets.
Clearly, .A B+
So, we may draw the Venn diagram,
as shown in the adjoining fi gure.
Clearly, , ( ) { , , }A B A B A1 3 5+ and ( ) { , } .B A B2 4
{ , , , }A 2 4 6 7l and { , , , , } .B 1 3 5 6 7l
CASE 4 When B A U3 3
In this case, we draw two concentric circles within a rectangular
region.
The inner circle represents B and the outer circle represents A.
U
A
A – B
B – A
A B
�
B
U
A B
7
8
1
3
2
6
4
5
U
A B
7
1
3 2
6
45
U
2
3
4
6
A
1
7
5
36 Senior Secondary School Mathematics for Class 11
EXAMPLE Let { , , , , , , , }U 1 2 3 4 5 6 7 8 be the
universal set, and let { , , , }A 1 3 5 7
and { , }B 3 7 be its subsets.
Then, clearly .B A3
Now, we may draw the Venn
diagram, as shown in the given
fi gure.
{ , }, { , , , },A B B A B A3 7 1 3 5 7+ ,
( ) { , }, ( ) ,A B B A1 5
{ , , , }A 2 4 6 8l and { , , , , , } .B 1 5 2 4 6 8l
IMPORTANT RESULTS FROM VENN DIAGRAMS
Let A and B be two intersecting subsets of U.
In counting the elements of ( ),A B, the elements
of A B+ are counted twice, once in counting the
elements of A and second time in counting the
elements of B.
( ) ( ) ( ) ( ) .n A B n A n B n A B, +
If ,A B+ then ( )n A B 0+ and therefore, in this case, we have
( ) ( ) ( ) .n A B n A n B,
From the Venn diagram, it is also clear that
(i) ( ) ( ) ( )n A B n A B n A+
(ii) ( ) ( ) ( )n B A n A B n B+
(iii) ( ) ( ) ( ) ( ) .n A B n A B n B A n A B+ ,
EXERCISE 1F
1. Let { , , , , }, { , , , }A a b c e f B c d e g and { , , , }C b c f g be subsets of the set
{ , , , , , , , } .U a b c d e f g h
Draw Venn diagrams to represent the following sets:
(i) A B+ (ii) ( )A B C, + (iii) A B
(iv) B A (v) ( )A B C+ (vi) ( ) ( )B C C B,
2. Let { , , , , }, { , , , }A B2 4 6 8 10 4 8 12 16 and { , , , } .C 6 12 18 24
Using Venn diagrams, verify that:
(i) ( ) ( )A B C A B C, , , , (ii) ( ) ( )A B C A B C+ + + +
3. Let { , , , , }, { , , , , }A a e i o u B a d e o v and { , , , } .C e o t m
Using Venn diagrams, verify the following:
(i) ( ) ( ) ( )A B C A B A C, + , + , (ii) ( ) ( ) ( ) .A B C A B A C+ , + , +
4. Let .A B U1 1 Exhibit it in a Venn diagram.
5. Let { , , , , , }A 2 3 5 7 11 13 and { , , , , }B 5 7 9 11 15 be subsets of
{ , , , , , , , } .U 2 3 5 7 9 11 13 15
A B
A B
�
A – B B – A
U
2
3, 7
4
6
A
1
B
8
5
Sets 37
Using Venn diagrams, verify that:
(i) ( ) ( )A B A B, +l l l (ii) ( ) ( )A B A B+ ,l l l
6. Using Venn diagrams, show that ( ), ( )A B A B+ and ( )B A are disjoint
sets, taking { , , , , , }A 2 4 6 8 10 12 and { , , , , } .B 3 6 9 12 15
SOME RESULTS DERIVED FROM VENN DIAGRAMS
RESULT 1 When A and B are disjoint sets, then we have
( ) ( ) ( ) .n A B n A n B,
RESULT 2 For any sets A and B, prove that
( ) ( ) ( ) ( ) .n A B n A n B n A B, +
PROOF From the given Venn diagram, it is clear
that the sets ( ), ( )A B A B+ and ( )B A
are disjoint and their union is ( ) .A B,
( ) ( ) ( ) ( )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+ + +
[adding and subtracting ( )]n A B+
( ) ( ) ( )n A n B n A B+
[ ( ) ( ) ( )n A B n A B n Aa + ( ) ( ) ( )]and n B A n A B n B+
Hence, ( ) ( ) ( ) ( ) .n A B n A n B n A B, +
Corollary 1 Prove that ( ) ( ) ( ) .n A B n A B n A+
PROOF Clearly, ( )A B and ( )A B+ are disjoint sets and their union is A.
( ) ( ) ( ) .n A B n A B n A+
Corollary 2 Prove that ( ) ( ) ( ) .n B A n A B n B+
PROOF Clearly, ( )B A and ( )A B+ are disjoint sets whose union is B.
( ) ( ) ( ) .n B A n A B n B+
RESULT 3 For any sets A, B, C prove that
( ) [ ( ) ( ) ( ) ( )]n A B C n A n B n C n A B C, , + +
[ ( ) ( ) ( )] .n A B n B C n A C+ + +
PROOF We have
( ) [( ) ]n A B C n A B C, , , ,
( ) ( ) [( ) ]n A B n C n A B C, , +
{ ( ) ( ) ( )} ( ) [( ) ( )n A n B n A B n C n A C B C+ + , +
{ ( ) ( ) ( ) ( )} { ( ) ( )n A n B n C n A B n A C n B C+ + +
( )}n A C B C+ + +
� � �
� �
A B
A B
A B
�
A – B B – A
38 Senior Secondary School Mathematics for Class 11
( ) ( ) ( ) ( ) ( ) ( )n A n B n C n A B n A C n B C+ + +
( )n A B C+ +
{ ( ) ( ) ( ) ( )}n A n B n C n A B C+ +
{ ( ) ( ) ( )} .n A B n B C n A C+ + +
Hence, the result follows.
SUMMARY
For any sets A, B, C we have:
(i) ( ) ( ) ( ) ( ) .n A B n A n B n A B, +
(ii) If ,A B+ then ( ) ( ) ( ) .n A B n A n B,
(iii) ( ) ( ) ( ) .n A B n A B n A+
(iv) ( ) ( ) ( ) .n B A n A B n B+
(v) ( ) { ( ) ( ) ( ) ( )}n A B C n A n B n C n A B C, , + +
{ ( ) ( ) ( )} .n A B n B C n A C+ + +
SOLVED EXAMPLES
EXAMPLE 1 If A and B are two sets such that ( ) , ( )n A n B27 35 and ( ) ,n A B 50,
fi nd ( ) .n A B+
SOLUTION We know that
( ) ( ) ( ) ( )n A B n A n B n A B, +
( ) ( ) ( ) ( ) ( ) .n A B n A n B n A B 27 35 50 12+ ,
Hence, ( ) .n A B 12+
EXAMPLE 2 If A and B are two sets containing 3 and 6 elements respectively, what can
be the maximum number of elements in A B, ?
Find also the minimum number of elements in .A B,
SOLUTION We know that
( ) ( ) ( ) ( ) .n A B n A n B n A B, + ... (i)
CASE 1 From (i), it is clear that ( )n A B, will be maximum when ( ) .n A B 0+
In that case, ( ) ( ) ( ) ( ) .n A B n A n B 3 6 9,
maximum number of elements in ( ) .A B 9,
CASE 2 From (i), it is clear that ( )n A B, will be minimum when ( )n A B+ is
maximum, i.e., when ( ) .n A B 3+
In this case, ( ) ( ) ( ) ( ) ( ) .n A B n A n B n A B 3 6 3 6, +
minimum number of elements in .A B 6,
EXAMPLE 3 A survey shows that 73% of the Indians like apples, whereas 65% like
oranges. What percentage of Indians like both apples and oranges?
SOLUTION Let A set of Indians who like apples
and B set of Indians who like oranges.
A B
A B
�
A – B B – A
Sets 39
Then, ( ) , ( ) ( ) .andn A n B n A B73 65 100,
( ) ( ) ( ) ( ) ( ) .n A B n A n B n A B 73 65 100 38+ ,
Hence, 38% of the Indians like both apples and oranges.EXAMPLE 4 In a survey of 425 students in a school, it was found that 115 drink apple
juice, 160 drink orange juice and 80 drink both apple as well as orange
juice. How many drink neither apple juice nor orange juice?
SOLUTION Let U set of all students surveyed;
A set of all students who drink apple juice
and B set of all students who drink orange juice.
Then, ( ) , ( ) , ( ) ( ) .andn U n A n B n A B425 115 160 80+
( ) ( ) ( ) ( ) ( ) .n A B n A n B n A B 115 160 80 195, +
Set of students who drink neither apple juice nor orange juice
( ) ( )A B A B+ , l l l
{( ) } ( ) ( ) ( ) .n A B n U n A B 425 195 230, , l
Hence, 230 students drink neither apple juice nor orange juice.
EXAMPLE 5 In a group of 850 persons, 600 can speak Hindi and 340 can speak Tamil.
Find (i) how many can speak both Hindi and Tamil,
(ii) how many can speak Hindi only,
(iii) how many can speak Tamil only.
SOLUTION Let A set of persons who can speak Hindi
and B set of persons who can speak Tamil.
( ) , ( ) ( ) .andn A n B n A B600 340 850,
(i) Set of persons who can speak both Hindi and Tamil ( ) .A B+
Now, ( ) ( ) ( ) ( )n A B n A n B n A B+ ,
( ) .600 340 850 90
Thus, 90 persons can speak both
Hindi and Tamil.
(ii) Set of persons who can speak Hindi
only ( ) .A B
Now, ( ) ( ) ( )n A B n A B n A+
( ) ( ) ( )n A B n A n A B+
( ) .600 90 510
Thus, 510 persons can speak Hindi only.
(iii) Set of persons who can speak Tamil only ( ) .B A
Now, ( ) ( ) ( )n B A n A B n B+
( ) ( ) ( ) ( ) .n B A n B n A B 340 90 250+
Hence, 250 persons can speak Tamil only.
EXAMPLE 6 A market research group conducted a survey of 1000 consumers and
reported that 745 consumers like product A and 430 consumers like
product B. What is the least number that must have liked both products?
A B
A B
�
A – B B – A
40 Senior Secondary School Mathematics for Class 11
SOLUTION Let P and Q be the sets of consumers who like product A and
product B respectively.
Then, ( )n P 745 and ( ) .n Q 430
Now, ( ) ( ) ( ) ( )n P Q n P n Q n P Q, +
( ) ( )n P Q n P Q745 430, +
( ) ( ) .n P Q n P Q1175, +
Clearly, ( )n P Q+ is least when ( )n P Q, is maximum and therefore,
( ) .n P Q 1000,
So, ( ) ( ) ( ) .n P Q n P Q1000 1175 1175 1000 175&+ +
Hence, the least number of consumers liking both the products
is 175.
EXAMPLE 7 Out of 600 car owners investigated, 500 owned car A; 200 owned car B
and 50 owned both A and B cars. Verify whether the given data is correct
or not.
SOLUTION Let P and Q be the sets of those who own car A and car B
respectively. Then,
( ) , ( ) ( ) .andn P n Q n P Q500 200 50+
Now, ( ) ( ) ( ) ( )n P Q n P n Q n P Q, +
( ) ( ) .n P Q 500 200 50 650,
This is a contradiction, since the maximum value of ( )n P Q, is 600.
Hence, the given data is incorrect.
EXAMPLE 8 In a group of 52 persons, 16 drink tea but not coffee and 33 drink tea.
Find (i) how many drink tea and coffee both;
(ii) how many drink coffee but not tea.
SOLUTION Let A set of persons who drink tea
and B set of persons who drink coffee.
Then, ( )A B set of persons who drink
tea but not coffee.
And, ( )B A set of persons who drink
coffee but not tea.
Given: ( ) , ( )n A B n A B52 16, and ( ) .n A 33
(i) Set of persons who drink tea and coffee both ( ) .A B+
Now, ( ) ( ) ( )n A B n A B n A+
( ) ( ) ( ) ( ) .n A B n A n A B 33 16 17+
Thus, 17 persons drink tea and coffee both.
(ii) ( ) ( ) ( ) ( )n A B n A n B n A B, +
( ) ( ) ( ) ( ) ( ) .n B n A B n A B n A 52 17 33 36, +
Now, ( ) ( ) ( ) .n B A n A B n B+
( ) ( ) ( ) ( ) .n B A n B n A B 36 17 19+
Thus, number of persons who drink coffee but not tea = 19.
A B
A B
�
A – B B – A
Sets 41
EXAMPLE 9 A school awarded 58 medals in three sports, namely 38 in football; 15 in
basketball and 20 in cricket. If 3 students got medals in all the three sports,
how many received medals in exactly two sports?
SOLUTION Let A, B and C denote the sets of students who won medals in
football, basketball and cricket respectively.
Then, ( ) , ( ) , ( ) , ( )n A n B n C n A B C38 15 20 3+ +
and ( ) .n A B C 58, ,
We know that
( ) ( ) ( ) ( ) ( ) ( )n A B C n A n B n C n A B n B C, , + +
( ) ( )n A C n A B C+ + +
( ) ( ) ( )n A B n B C n A C+ + +
( ) ( ) ( ) ( ) ( )n A n B n C n A B C n A B C+ + , ,
{( ) } ( ) .38 15 20 3 58 76 58 18
Let a, b, c and d denote respectively
the number of students who won
medals in football and basketball
both; basketball and cricket both;
football and cricket both and all
the 3 sports.
Then, ( ) ( ) ( )n A B n B C n A C+ + +
= 18
( ) ( ) ( )a d b d c d 18
( )a b c d3 18
( )a b c a b c3 3 18 9&# [ ]d 3a .
Hence, 9 students received medals in exactly two sports.
AN EASY APPROACH
EXAMPLE 10 In a survey it is found that 21 people like product A, 26 people like product
B and 29 like product C. If 14 people like products A and B; 15 people like
products B and C; 12 people like products C and A; and 8 people like all
the three products, fi nd
(i) how many people are surveyed in all;
(ii) how many like product C only.
SOLUTION Let A, B, C denote respectively the
sets of people who like product A, B
and C respectively, as shown in the
given fi gure.
Let us denote the number of elements
contained in bounded regions by a,
b, c, d, e, f, g as shown in the given
fi gure.
c
a
b
d
A B
C
A B
C
a b
c
d
e
f
g
42 Senior Secondary School Mathematics for Class 11
Then, we have
,a b c d 21
,b c e f 26
,c d f g 29
, ,b c c f c d14 15 12
and .c 8
On solving these equations, we get
, , , , , , .c d f b g e a8 4 7 6 10 5 3
(i) Total number of surveyed people ( ) .a b c d e f g 43
(ii) Number of persons who like product C only .g 10
EXAMPLE 11 In a survey of 25 students, it was found that 12 have taken physics,
11 have taken chemistry and 15 have taken mathematics; 4 have taken
physics and chemistry; 9 have taken physics and mathematics; 5 have
taken chemistry and mathematics while 3 have taken all the three subjects.
Find the number of students who have taken
(i) physics only;
(ii) chemistry only;
(iii) mathematics only;
(iv) physics and chemistry but not mathematics;
(v) physics and mathematics but not chemistry;
(vi) only one of the subjects;
(vii) at least one of the three subjects;
(viii) none of the three subjects.
SOLUTION Let P, C and M be the sets of students who have taken physics,
chemistry and mathematics respectively.
Let a, b, c, d, e, f and g denote the number of students in the respective
regions, as shown in the adjoining Venn diagram.
As per data given, we have
,
,
,
,
,
,
.
a b c d
b c e f
c d f g
b c
c d
c f
c
12
11
15
4
9
5
3
Z
[
\
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
From these equations, we get
, , , .c f d b3 2 6 1
Now, ;c d f g g g15 3 6 2 15 4& &
;b c e f e e11 1 3 2 11 5& &
c
a b
d
P C
M
e
f
g
Sets 43
;a b c d a a12 1 3 6 12 2& &
, , , , , .anda b c d e f g2 1 3 6 5 2 4
So, we have:
(i) Number of students who offered physics only .a 2
(ii) Number of students who offered chemistry only .e 5
(iii) Number of students who offered mathematics only .g 4
(iv) Number of students who offered physics and chemistry but
not mathematics .b 1
(v) Number of students who offered physics and mathematics but
not chemistry .d 6
(vi) Number of students who offered only one of the given subjects
( ) ( ) .a e g 2 5 4 11
(vii) Number of students who offered atleast one of the given
subjects ( ) ( ) .a b c d e f g 2 1 3 6 5 2 4 23
(viii) Number of students who offered none of the three given
subjects ( ) .25 23 2
EXERCISE 1G
1. If A and B are two sets such that ( ) , ( ) ( ) ,andn A n B n A B37 26 51, fi nd
( ) .n A B+
2. If P and Q are two sets such that ( ) , ( ) ( ) ,andn P Q n P Q n P75 17 49, +
fi nd ( ) .n Q
3. If A and B are two sets such that ( ) , ( )n A n B24 22 and ( ) ,n A B 8+ fi nd:
(i) ( )n A B, (ii) ( )n A B (iii) ( )n B A
4. If A and B are two sets such that ( ) , ( )n A B n B A24 19 and
( ) ,n A B 11+ fi nd:
(i) ( )n A (ii) ( )n B (iii) ( )n A B,
5. In a committee, 50 people speak Hindi, 20 speak English and 10 speak both
Hindi and English. How many speak at least one of these two languages?
6. In a group of 50 persons, 30 like tea, 25 like coffee and 16 like both. How
many like
(i) either tea or coffee? (ii) neither tea nor coffee?
7. There are 200 individuals with a skin disorder, 120 had been exposed to the
chemical ,C 501 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
8. In a class of a certain school, 50 students offered mathematics, 42 offered
biology and 24 offered both the subjects. Find the number of students offering
44 Senior Secondary School Mathematics for Class 11
(i) mathematics only,
(ii) biology only,
(iii) any of the two subjects.
9. In an examination, 56% of the candidates failed in English and 48% failed
in science. If 18% failed in both English and science, fi nd the percentage of
those who passed in both the subjects.
10. In a group of 65 people, 40 like cricket and 10 like both cricket and tennis.
How many like tennis only and not cricket? How many like tennis?
11. A school awarded 42 medals in hockey, 18 in basketball and 23 in cricket.
If these medals were bagged by a total of 65 students and only 4 students
got medals in all the three sports, how many students received medals in
exactly two of the three sports?
12. 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, and 3 read all the 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.
13. In a survey of 100 students, the number of students studying the various
languages is found as: English only 18; English but not Hindi 23; English
and Sanskrit 8; Sanskrit and Hindi 8; English 26; Sanskrit 48 and no
language 24. Find
(i) how many students are studying Hindi,
(ii) how many students are studying English and Hindi both.
14. In a town of 10,000 families, it was found that 40% of the families buy
newspaper A, 20% buy newspaper B, 10% buy newspaper C, 5% buy A and
B; 3% buy B and C and 4% buy A and C. If 2% buy all the three newspapers,
fi nd the number of families which buy
(i) A only,
(ii) B only,
(iii) none of A, B and C.
15. A class has 175 students. The following description gives the number of
students studying one or more of the subjects in this class: mathematics
100, physics 70, chemistry 46; mathematics and physics 30; mathematics
and chemistry 28; physics and chemistry 23; mathematics, physics and
chemistry 18. Find
(i) how many students are enrolled in mathematics alone, physics alone
and chemistry alone,
(ii) the number of students who have not offered any of these subjects.
ANSWERS (EXERCISE 1G)
1. 12 2. 43 3. (i) 38 (ii) 16 (iii) 14 4. (i) 35 (ii) 30 (iii) 54
5. 60 6. (i) 39 (ii) 11 7. (i) 90 (ii) 20 (iii) 140 8. (i) 26 (ii) 18 (iii) 68
Sets 45
9. 14% 10. 25, 35 11. 22 12. (i) 52 (ii) 30
13. (i) 18 (ii) 3 14. (i) 3300 (ii) 1400 (iii) 4000 15. (i) 60, 35, 13 (ii) 32
HINTS TO SOME SELECTED QUESTIONS
1. ( ) ( ) ( ) ( )n A B n A n B n A B+ , .
2. ( ) ( ) ( ) ( )n P Q n P n Q n P Q, + .
3. (i) ( ) ( ) ( ) ( ) .n A B n A n B n A B, +
(ii) ( ) ( ) ( ) .n A B n A B n A+
(iii) ( ) ( ) ( ) .n B A n A B n B+
4. (i) ( ) ( ) ( )n A B n A B n A+
( ) ( ) ( )n A n A B n A B+
( ) ( ) .n A 24 11 35
(ii) ( ) ( ) ( )n B A n A B n B+
( ) ( ) .n B 19 11 30
(iii) ( ) ( ) ( ) ( ) .n A B n A n B n A B, +
5. ( ) ( ) ( ) ( ) ( ) .n A B n A n B n A B 50 20 10 60, +
6. ( ) , ( ) ( ) .andn A n B n A B30 25 16+
(i) ( ) ( ) ( ) ( )n A B n A n B n A B 39, + .
(ii) ( ) {( ) } ( ) ( ) ( ) .n A B n A B n U n A B 50 39 11+ , , l l l
7. ( ) , ( ) , ( )n U n C n C200 120 501 2 and ( ) .n C C 301 2+
(i) ( ) ( ) ( )n C n C C n C C1 1 2 1 2+
( ) ( ) ( ) ( ) .n C C n C n C C 120 30 901 2 1 1 2+
(ii) ( ) ( ) ( )n C n C C n C C2 2 1 1 2+
( ) ( ) ( ) ( ) .n C C n C n C C 50 30 202 1 2 1 2+
(iii) ( ) ( ) ( ) ( )n C C n C n C n C C1 2 1 2 1 2, +
( ) .120 50 30 140
8. ( ) , ( ) ( ) .andn A n B n A B50 42 24+
(i) ( ) ( ) ( )n A n A B n A B+
( ) ( ) ( ) ( ) .n A B n A n A B 50 24 26+
(ii) ( ) ( ) ( )n B n B A n A B+
( ) ( ) ( ) ( ) .n B A n B n A B 42 24 18+
(iii) ( ) ( ) ( ) ( ) ( ) .n A B n A n B n A B 50 42 24 68, +
9. Failed in English only ( ) .56 18 38
Failed in science only ( ) .48 18 30
Failed in both English and science = 18.
Failed in one or both of the subjects ( ) .38 30 18 86
Passed in both the subjects ( ) .100 86 14
A B
A B
�
A – B B – A
16 148
A B
A B
�
A – B B – A
24 1911
C – C1 2
�
C C1 2 C – C2 1
C1 C2
A B
A B
�
A – B B – A
46 Senior Secondary School Mathematics for Class 11
10. Let A set of people who like cricket.
B set of people who like tennis.
( ) , ( ) .n A n A B40 10+
( ) ( )n A B 40 10 30
( ) ( ) .n B A 65 30 10 25
Number of people who like tennis only = 25.
Number of people who like tennis ( ) .25 10 35
11. Let A, B, C denote the sets of students who bagged medals in hockey, basketball and
cricket respectively.
Then, ( ) , ( ) , ( ) , ( ) ( ) .andn A n B n C n A B C n A B C42 18 23 65 4, , + + Then,
( ) { ( ) ( ) ( ) ( )} { ( ) ( ) ( )}n A B C n A n B n C n A B C n A B n B C n A C, , + + + + +
( ) ( ) .x x65 42 18 23 4 87 65 22+
12. , , ,a b c d b c e f c d f g25 26 26
, , .andc d b c c f c9 11 8 3
, , , , , .andf b d c g e a5 8 6 3 12 10 8
(i) Number of people who read at least one of the papers
( ) .a b c d e f g 52
(ii) Number of people who read exactly one newspaper
( ) ( ) .a e g 8 10 12 30
13. We have
, , , ,a a d c d c f18 23 8 8
, ,a b c d c d f g26 48
.a b c d e f g 100 24 76
, , , .anda d c f b18 5 3 5 0
( )g 48 3 5 5 35
and ( ) .e 76 18 0 3 5 5 35 76 66 10
(i) Number of students studying Hindi ( )b c e f
( ) .0 3 10 5 18
(ii) Number of students studying English and Hindi both ( ) ( ) .b c 0 3 3
14. ( ) ( % ) ,ofn A 40 10000 100
40 10000 4000# a k
( ) ( % ) , ( ) ( % ) ,of ofn B n C20 10000 2000 10 10000 1000
( ) ( % ) , ( ) ( % ) ,of ofn A B n B C5 10000 500 3 10000 300+ +
( ) ( % ) , ( ) ( % )of ofn A C n A B C4 10000 400 2 10000 200+ + + ,
and ( ) .n U 10000
,
,
,
,
,
,
.
a b c d
b c e f
c d f g
b c
c f
c d
c
4000
2000
1000
500
300
400
200
Z
[
\
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
H T
I
a
b
c
d
e
f
g
E H
S
a
b
c
d
e
f
g
A B
C
a b
c
d
e
f
g
A B
30 10 25
Sets 47
On solving these equations,we get
, , , , , , .c d f b g e a200 200 100 300 500 1400 3300
(i) ,a 3300 (ii) e 1400
(iii) ( ) ( ) .10000 3300 300 200 200 1400 100 500 10000 6000 4000
EXERCISE 1H
Very-Short-Answer Questions
1. If a set A has n elements then fi nd the number of elements in its power set
( ) .P A
2. If A then write ( ) .P A
3. If ( ) ( ) ,andn A n B3 5 fi nd:
(i) the maximum number of elements in ,A B,
(ii) the minimum number of elements in .A B,
4. If A and B are two sets such that ( ) , ( )n A n B8 11 and ( )n A B 14, then
fi nd ( ) .n A B+
5. If A and B are two sets such that ( ) , ( )n A n B23 37 and ( )n A B 8 then
fi nd ( ) .n A B,
Hint ( ) ( ) ( ) ( ) ( ) .n A n A B n A B n A B 23 8 15&+ +
6. If A and B are two sets such that ( ) , ( )n A n B54 39 and ( )n B A 13 then
fi nd ( ) .n A B,
Hint ( ) ( ) ( ) ( ) ( ) .n B n B A n A B n A B 39 13 26&+ +
7. If ,A B1 prove that .B A1l l
8. If ,A B1 show that ( ) .B A l l
Hint .A B B A B A& &1 1 l l l l
9. Let { : , }A x x n n N6 d and { : , },B x x n n N9 d fi nd .A B+
10. If { , , },A 5 6 7 fi nd ( ) .P A
11. If { , { }},A 2 2 fi nd ( ) .P A
12. Prove that ( ) .A A B+ , l
13. Find the symmetric difference ,A BT when { , , }A 1 2 3 and { , , } .B 3 4 5
14. Prove that .A B A B+ l
15. If { : , }A x x R x 5 d and { : , },B x x R x 4 d fi nd .A B+
Hint ( , ) ( , ) .andA B5 43 3 So, ( , ) .A B 4 5+
ANSWERS (EXERCISE 1H)
1. 2n 2. {} 3. (i) 8 (ii) 5 4. 5 5. 45 6. 67
9. { : , }A B x x n n N18+ d
10. ( ) { , { }, { }, { }, { , }, { , }, { , }, { , , }}P A 5 6 7 5 6 5 7 6 7 5 6 7
11. ( ) { , { }, {{ }}, { , { }}}P A 2 2 2 2 13. {1, 2, 4, 5} 15. (4, 5)
48 Senior Secondary School Mathematics for Class 11
SUMMARY OF KEY FACTS
1. A well-defi ned collection of objects is called a set.
2. The objects in a set are called its elements, or members, or points.
3. Usually, we denote sets by capital letters A, B, C, X, Y, Z, etc.
4. A set is usually described in tabular form or set-builder form.
5. In tabulation method, we make a list of all the objects of the set and put
them within braces { }. In set-builder form, we write {x : x satisfi es properties P}
which means, the set of all those x such that each x satisfi es properties P.
6. A set having no element at all is called a null set, or a void set and it is
denoted by .
7. A set having a single element is called a singleton set. For example, {3}.
8. A set having fi nite number of elements is called a fi nite set, otherwise it is
called an infi nite set.
The number of elements in a fi nite set A is denoted by ( ) .n A
9. Two sets A and B having exactly the same elements are known as equal sets
and we write, .A B
10. A set A is called a subset of a set B, if every element of A is in B and we write,
.A B3
11. If A is a subset of set B and A B! then A is called a proper subset of set B and
we write, .A Bf
12. The total number of subsets of a set A containing n elements is .2n
13. The collection of all subsets of a set A is called the power set of A, to be
denoted by ( ) .P A
14. Let a and b be real numbers such that a b then
(i) closed interval [ , ] { : , }a b x x R a x b# # d ,
(ii) open interval ( , ) { : , }a b x x R a x b d ,
(iii) right half open interval [ , ) { : , }a b x x R a x b# d ,
(iv) left half open interval ( , ] { : , }a b x x R a x b # d .
15. The union of two sets A and B, denoted by A B, , is the set of all those
elements which are either in A or in B or in both A and B.
{ : } .orA B x x A x B, d d
16. The intersection of two sets A and B, denoted by A B+ , is the set of all those
elements which are common to both A and B.
{ : } .andA B x x A x B+ d d
17. The difference between two sets A and B, denoted by ( )A B , is defi ned as
( ) { : } .andA B x x A x B d z
Similarly, ( ) { : } .andB A x x B x A d z
18. The symmetric difference between the sets A and B, denoted by A BT , is
defi ned as ( ) ( ) .A B A B B A,T
19. Let A be a subset of the universal set U. Then the complement of A, denoted
by Al, or Ac, or U A , is defi ned as { : } .andA x x U x A d zl
Sets 49
20. Various laws of operations on sets
(i) andA A A A A A, + [Idempotent laws]
(ii) andA A A U A, + [Identity law]
(iii) andA B B A A B B A, , + + [Commutative laws]
(iv) ( ) ( ) ( ) ( )andA B C A B C A B C A B C, , , , + + + +
[Associative laws]
(v) I. ( ) ( ) ( )A B C A B A C, + , + , [Distributive law]
II. ( ) ( ) ( )A B C A B A C+ , + , + [Distributive law]
(vi) I. ( ) ( )A B A B, +l l l [De Morgan‘s laws]
II. ( ) ( )A B A B+ ,l l l [De Morgan’s laws]
21. For any two sets A and B, we have
(i) ( ) ( ) ( ) ( )n A B n A n B n A B, +
(ii) ( ) ( ) ( )n A B n A B n A+
(iii) ( ) ( ) ( )n B A n A B n B+
(iv) ( ) ( ) ( )n A B n A B n B AT
22. For any three sets A, B and C, we have
( ) ( ) ( ) ( ) { ( ) ( ) ( )}n A B C n A n B n C n A B n B C n A C, , + + +
( ) .n A B C+ +
A B
A B
�
A – B B – A
50 Senior Secondary School Mathematics for Class 11
2
Relations
ORDERED PAIR Two numbers a and b listed in a specifi c order and enclosed in
parentheses form an ordered pair ( , ) .a b
In the ordered pair ( , ),a b we call a as first member (or first component) and b
as second member (or second component).
By interchanging the positions of the components, the ordered pair is
changed.
Thus, ( , ) ( , ) .a b b a!
EXAMPLE In coordinate geometry, the
position of a point in a plane is
determined by an ordered pair.
In the given fi gure, the ordered
pairs (2, 3) and (3, 2) represent
two different points A and B
respectively.
Thus, ( , ) ( , ) .2 3 3 2!
EQUALITY OF TWO ORDERED PAIRS
We have ( , ) ( , ) .anda b c d a c b d+
EXAMPLE 1 Find a and b, when ( , ) ( , ) .a b1 5 2 3
SOLUTION Using the defi nition of equality of two ordered pairs, we have
( , ) ( , ) anda b a b1 5 2 3 1 2 5 3&
.anda b3 2&
Hence, a 3 and .b 2
EXAMPLE 2 Find a and b, when ( , ) ( , ) .a b a b2 11 1 3
SOLUTION Using the defi nition of equality of two ordered pairs, we have
( , ) ( , )a b a b2 11 1 3
a b2 1 ... (i)
a b3 11 . ... (ii)
On solving (i) and (ii), we get a 2 and .b 3
Hence, a 2 and .b 3
Y
XO 1
A(2, 3)
B(3, 2)
2 3
3
2
1
)
50
Relations 51
EXAMPLE 3 If , ,,x y3 1 3
2
3
5
3
1 a ak k fi nd the values of x and y.
SOLUTION Since the given two ordered pairs are equal, we have
x x x x3 1 3
5
3 3
5 1 3
2
3 3
2 2& & & a k
and .y y y3
2
3
1
3
1
3
2
3
3 1 1& & a k
Hence, .andx y2 1
EXAMPLE 4 Express {( , ) : ,x y x y 252 2 where , }x y Wd as a set of ordered pairs.
SOLUTION It is easy to verify that each of the following ordered pairs of whole
numbers satisfi es the given relation :x y 252 2
( , ), ( , ), ( , ) ( , ) .and5 0 0 5 3 4 4 3
Hence, the set of required ordered pairs is
{( , ), ( , ), ( , ), ( , )} .5 0 0 5 3 4 4 3
CARTESIAN PRODUCT OF TWO SETS
Let A and B be two nonempty sets. Then, the Cartesian product of A and B is the set
denoted by ( ),A B# consisting of all ordered pairs (a, b) such that a Ad and .b Bd
{( , ) : } .andA B a b a A b B# d d
If A or ,B we define .A B#
REMARKS (i) If ( )n A p and ( )n B q then ( )n A B pq# and ( ) .n B A pq#
(ii) If at least one of A and B is infi nite then ( )A B# is infi nite and ( )B A#
is infi nite.
SOLVED EXAMPLES
EXAMPLE 1 If { , , } { , }A and B1 3 5 2 3 then fi nd:
(i) A B# (ii) B A# (iii) ( ) ( )A B B A# + #
SOLUTION We have
(i) { , , } { , }A B 1 3 5 2 3# #
{( , ), ( , ), ( , ), ( , ), ( , ), ( , )} .1 2 1 3 3 2 3 3 5 2 5 3
(ii) { , } { , , }B A 2 3 1 3 5# #
{( , ), ( , ), ( , ), ( , ), ( , ), ( , )} .21 2 3 2 5 3 1 3 3 3 5
(iii) ( ) ( ) {( , )} .A B B A 3 3# + #
EXAMPLE 2 If { , , }, { , } { , , }A B and C1 2 3 3 4 4 5 6 then fi nd:
(i) ( )A B C# + (ii) ( ) ( )A B A C# + #
(iii) ( )A B C# , (iv) ( ) ( )A B A C# , #
SOLUTION We have
(i) { , } { , , } { } .B C 3 4 4 5 6 4+ +
( ) { , , } { } {( , ), ( , ), ( , )} .A B C 1 2 3 4 1 4 2 4 3 4# + #
52 Senior Secondary School Mathematics for Class 11
(ii) ( ) { , , } { , }A B 1 2 3 3 4# #
{( , ), ( , ), ( , ), ( , ), ( , ), ( , )} .1 3 1 4 2 3 2 4 3 3 3 4
( ) { , , } { , , }A C 1 2 3 4 5 6# #
{( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ),1 4 1 5 1 6 2 4 2 5 2 6 3 4
( , ), ( , )}3 5 3 6
( ) ( ) {( , ), ( , ), ( , )} .A B A C 1 4 2 4 3 4# + #
(iii) { , } { , , } { , , , } .B C 3 4 4 5 6 3 4 5 6, ,
( ) { , , } { , , , }A B C 1 2 3 3 4 5 6# , #
{( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ),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) Also, from (ii), we get
( ) ( ) {( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ),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
EXAMPLE 3 Let { : }, { : }A x N x x B x W x5 6 0 0 22 # d d and
{ : }.C x N x 3 d Verify that
(i) ( ) ( ) ( )A B C A B A C# , # , #
(ii) ( ) ( ) ( )A B C A B A C# + # + #
SOLUTION We have
{ : } { : ( ) ( ) } { , };A x N x x x N x x5 6 0 2 3 0 2 32 d d
{ : } { , } { : } { , } .andB x W x C x N x0 2 0 1 3 1 2 # d d
{ , }, { , } { , } .andA B C2 3 0 1 1 2
(i) ( ) { , } ( , } { , , } .B C 0 1 1 2 0 1 2, ,
( ) { , } { , , }A B C 2 3 0 1 2# , #
{( , ), ( , ), ( , ), ( , ), ( , ), ( , )} .2 0 2 1 2 2 3 0 3 1 3 2
Now, ( ) { , } { , }A B 2 3 0 1# #
{( , ), ( , ), ( , ), ( , )}2 0 2 1 3 0 3 1
and ( ) { , } { , }A C 2 3 1 2# #
{( , ), ( , ), ( , ), ( , )} .2 1 2 2 3 1 3 2
( ) ( ) {( , ), ( , ), ( , ), ( , ), ( , ), ( , )} .A B A C 2 0 2 1 2 2 3 0 3 1 3 2# , #
Hence, ( ) ( ) ( ) .A B C A B A C# , # , #
(ii) ( ) { , } { , } { } .B C 0 1 1 2 1+ +
( ) { , } { } {( , ), ( , )} .A B C 2 3 1 2 1 3 1# + #
And,
( ) ( ) {( , ), ( , ), ( , ), ( , )} {( , ), ( , ),A B A C 2 0 2 1 3 0 3 1 2 1 2 2# + # +
( , ), ( , )}3 1 3 2
{( , ), ( , )} .2 1 3 1
Hence, ( ) ( ) ( ) .A B C A B A C# + # + #
Relations 53
EXAMPLE 4 If ( ) {( , ), ( , ), ( , ), ( , )},A B 3 2 3 4 5 2 5 4# fi nd A and B.
SOLUTION Clearly, we have
A set of all fi rst coordinates of the elements of ( )A B#
{ , } .3 5
B set of all second coordinates of the elements of ( )A B#
{ , } .2 4
Thus, { , } { , } .andA B3 5 2 4
EXAMPLE 5 A and B are two sets given in such a way that ( )A B# contains 6 elements.
If three elments of ( )A B# be (1, 3), (2, 5) and (3, 3), fi nd its remaining
elements.
SOLUTION Since (1, 3), (2, 5) and (3, 3) are in ( ),A B# it follows that 1, 2, 3 are
elements of A and 3, 5 are elements of B.
( ) { , , } { , }A B 1 2 3 3 5# #
{( , ), ( , ), ( , ), ( , ), ( , ), ( , )} .1 3 1 5 2 3 2 5 3 3 3 5
Hence, the remaining elements of ( )A B# are (1, 5), (2, 3) and (3, 5).
EXAMPLE 6 If { , },A a b fi nd ( ) .A A#
SOLUTION We have
( ) { , } { , }A A a b a b# #
{( , ), ( , ), ( , ), ( , )} .a a a b b a b b
EXAMPLE 7 Let R be the set of all real numbers. What does ( )R R# represent?
SOLUTION We have ( ) {( , ) : , } .R R x y x y R# d
Thus, ( )R R# represents the set of all coordinates of points in two-
dimensional space.
EXAMPLE 8 If ( )A A# has 9 elements two of which are ( , )1 0 and (0, 1), fi nd the set A
and the remaining elements of ( )A A# .
SOLUTION Clearly, –1, 0 and 1 are elements of A.
{ , , }A 1 0 1
{ , , } { , , }A A 1 0 1 1 0 1# #
{( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ),1 1 1 0 1 1 0 1 0 0 0 1 1 1
( , ), ( , )} .1 0 1 1
Hence, the remaining elements of ( )A A# are
(–1, –1), (–1, 1), (0, –1), (0, 0), (1, –1), (1, 0), (1, 1).
GRAPHICAL REPRESENTATION OF A # B AND B # A
Let X OXl and YOYl be the x-axis and y-axis respectively, drawn on a graph
paper.
Let { , , }A 1 2 4 and { , } .B 1 3 Then,
{( , ), ( , ), ( , ), ( , ), ( , ), ( , )} .A B 1 1 1 3 2 1 2 3 4 1 4 3#
54 Senior Secondary School Mathematics for Class 11
These points may be plotted on the graph paper, as shown below:
This is the graphical representation of .A B#
For ,B A# we take { , }B 1 3 along the x-axis and { , , }A 1 2 4 along the
y-axis and plot the points ( , ), ( , ), ( , ), ( , ), ( , ), ( , )1 1 1 2 1 4 3 1 3 2 3 4 on the graph
paper.
Arrow diagram of A # B
Let { , , } { , } .andA B1 2 4 1 3 Then,
{( , ), ( , ), ( , ), ( , ), ( , ), ( , )} .A B 1 1 1 3 2 1 2 3 4 1 4 3#
Then, A B# may be represented by arrow diagram, as shown below:
Arrow diagram of B # A
Again, {( , ), ( , ), ( , ), ( , ), ( , ), ( , )} .B A 1 1 1 2 1 4 3 1 3 2 3 4#
B
–1
2
34
A
1
O XX�
Y
�Y
42
1
3(–1, 3)
(–1, 1)
(2, 3)
(2, 1) (4, 1)
(4, 3)
A B
–1
A B#
Relations 55
We may exhibit B A# by arrow diagram, as shown below:
ORDERED TRIPLET Three numbers a, b and c listed in a specifi c order and
enclosed in parentheses form an ordered triplet ( , , ) .a b c
Thus, ( , , ) ( , , ) ( , , ),1 2 3 2 1 3 3 2 1! ! etc.
For any nonempty set A, we define:
.( ) {( , , ) : }, ,A A A a b c a b c A# # d
EXAMPLE 9 If { , },A 1 2 fi nd ( ) .A A A# #
SOLUTION { , } { , } {( , ), ( , ), ( , ), ( , )} .A A 1 2 1 2 1 1 1 2 2 1 2 2# #
( )A A A A A A# # # #
{( , ), ( , ), ( , ), ( , )} { , }1 1 1 2 2 1 2 2 1 2#
{( , , ), ( , , ), ( , , ), ( , , ), ( , , ),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 10 Let R be the set of all real numbers. What does ( )R R R# # represent?
SOLUTION ( ) {( , , ) : , , } .R R R a b c a b c R# # d
Thus, ( )R R R# # represents the set of all points in three-
dimensional space.
REMARK For any nonempty sets A, B and C, we always have
( ) ( ) .A B C A B C A B C# # # # # #
EXAMPLE 11 Let { , }, { , } { , } .A B and C1 2 3 4 4 5
Verify that ( ) ( )A B C A B C# # # # and hence fi nd .A B C# #
SOLUTION We have
{ , } { , } {( , ), ( , ), ( , ), ( , )}A B 1 2 3 4 1 3 1 4 2 3 2 4# #
( ) {( , ), ( , ), ( , ), ( , )} { , }A B C 1 3 1 4 2 3 2 4 4 5# # #
{( , , ), ( , , ), ( , , ), ( , , ), ( , , ),1 3 4 1 3 5 1 4 4 1 4 5 2 3 4
( , , ), ( , , ), ( , , )} .2 3 5 2 4 4 2 4 5
Again, { , } { , } {( , ), ( , ), ( , ), ( , )}B C 3 4 4 5 3 4 3 5 4 4 4 5# #
( ) { , } {( , ), ( , ), ( , ), ( , )}A B C 1 2 3 4 3 5 4 4 4 5# # #
{( , , ), ( , , ), ( , , ), ( , , ), ( , , ),1 3 4 1 3 5 1 4 4 1 4 5 2 3 4
( , , ), ( , , ), ( , , )} .2 3 5 2 4 4 2 4 5
( ) ( ) .A B C A B C A B C# # # # # #
Hence, ( ) {( , , ), ( , , ), ( , , ), ( , , ), ( , , ),A B C 1 3 4 1 3 5 1 4 4 1 4 5 2 3 4# #
( , , ), ( , , ), ( , , )} .2 3 5 2 4 4 2 4 5
B
–1
2
3 4
A
1
B A#
56 Senior Secondary School Mathematics for Class 11
EXERCISE 2A
1. Find the values of a and b, when:
(i) ( , ) ( , )a b3 2 5 1 (ii) ( , ) ( , )a b b2 3 4 5
(iii) ,,a b3 1 3
1
3
5
3
2 b bl l (iv) ( , ) ( , )a b b a2 2 1 1 2
2. If { , }A 0 1 and { , , },B 1 2 3 show that .A B B A# #!
3. If { , }P a b and { , , },Q x y z show that .P Q Q P# #!
4. If { , , } { , },andA B2 3 5 5 7 fi nd:
(i) A B# (ii) B A# (iii) A A# (iv) B B#
5. If { : }A x N x 3# d and { , },B x W x 2 d fi nd ( ) ( ) .andA B B A# # Is
( ) ( )A B B A# # ?
6. If { , , }, { , } { , },andA B C1 3 5 3 4 2 3 verify that:
(i) ( ) ( ) ( )A B C A B A C# , # , #
(ii) ( ) ( ) ( )A B C A B A C# + # + #
7. Let { : }, { : } { , } .andA x W x B x N x C2 1 4 3 5 # d d Verify that:
(i) ( ) ( ) ( )A B C A B A C# , # , #
(ii) ( ) ( ) ( )AB C A B A C# + # + #
Hint { , }, { , , } { , } .andA B C0 1 2 3 4 3 5
8. If {( , ), ( , ), ( , ), ( , ), ( , ), ( , )},A B 2 3 2 4 0 3 0 4 3 3 3 4# fi nd A and B.
9. Let { , } { , } .andA B2 3 4 5 Find ( ) .A B# How many subsets will ( )A B#
have?
10. Let {( , ) : } .A B a b b a3 2# If ( , )x 5 and (2, y) belong to ,A B# fi nd the
values of x and y.
11. Let A and B be two sets such that ( ) ( ) .andn A n B3 2
If a b c! ! and ( , ), ( , ), ( , )a b c0 1 0 are in ,A B# fi nd A and B.
12. Let { , } { , , } .andA B2 2 0 3 5 Find:
(i) A B# (ii) B A# (iii) A A# (iv) B B#
Represent each of the above (a) graphically and (b) by arrow diagram.
13. If { , },A 5 7 fi nd (i) A A# and (ii) A A A# # .
14. Let { , }, { , } { , } .andA B C3 1 1 3 3 5 Find:
(i) A B# (ii) ( )A B C# # (iii) B C# (iv) ( )A B C# #
ANSWERS (EXERCISE 2A)
1. (i) ,a b2 3 (ii) ,a b5 1 (iii) ,a b2 1 (iv) ,a b3 2
4. (i) {( , ), ( , ), ( , ), ( , ), ( , ), ( , )}A B 2 5 2 7 3 5 3 7 5 5 5 7#
(ii) {( , ), ( , ), ( , ), ( , ), ( , ), ( , )}B A 5 2 5 3 5 5 7 2 7 3 7 5#
(iii) {( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , )}A A 2 2 2 3 2 5 3 2 3 3 3 5 5 2 5 3 5 5#
(iv) {( , ), ( , ), ( , ), ( , )}B B 5 5 5 7 7 5 7 7#
Relations 57
5. {( , ), ( , ), ( , ), ( , ), ( , ), ( , )};A B 1 0 1 1 2 0 2 1 3 0 3 1#
{( , ), ( , ), ( , ), ( , ), ( , ), ( , )};B A 0 1 0 2 0 3 1 1 1 2 1 3#
No
8. { , , } { , }andA B2 0 3 3 4
9. {( , ), ( , ), ( , ), ( , )};A B 2 4 2 5 3 4 3 5# Number of subsets of ( )A B 2 164#
10. ,x y1 4 11. { , , } { , }andA a b c B 0 1
12. (i) {( , ), ( , ), ( , ), ( , ), ( , ), ( , )}A B 2 0 2 3 2 5 2 0 2 3 2 5#
(ii) {( , ), ( , ), ( , ), ( , ), ( , ), ( , )}B A 0 2 0 2 3 2 3 2 5 2 5 2#
(iii) {( , ), ( , ), ( , ), ( , )}A A 2 2 2 2 2 2 2 2#
(iv) {( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , )}B B 0 0 0 3 0 5 3 0 3 3 3 5 5 0 5 3 5 5#
13. (i) {( , ), ( , ), ( , ), ( , )}A A 5 5 5 7 7 5 7 7#
(ii) {( , , ), ( , , ), ( , , ), ( , , ), ( , , ), ( , , ),A A A 5 5 5 5 7 5 7 5 5 7 7 5 5 5 7 5 7 7# #
( , , ), ( , , )}7 5 7 7 7 7
14. (i) {( , ), ( , ), ( , ), ( , )}A B 3 1 3 3 1 1 1 3#
(ii) ( ) {( , , ), ( , , ), ( , , ), ( , , ), ( , , ),A B C 3 1 3 3 3 3 1 1 3 1 3 3 3 1 5# #
( , , ), ( , , ), ( , , )}3 3 5 1 1 5 1 3 5
(iii) {( , ), ( , ), ( , ), ( , )}B C 1 3 1 5 3 3 3 5#
(iv) ( ) {( , , ), ( , , ), ( , , ), ( , , ), ( , , ),A B C 3 1 3 3 1 5 3 3 3 3 3 5 1 1 3# #
( , , ), ( , , ), ( , , )}1 1 5 1 3 3 1 3 5
SOME USEFUL RESULTS
THEOREM 1 If A B3 then prove that A C B C# #3 for any set C.
PROOF Let A B3 and let (a, c) be an arbitrary element of .A C# Then,
( , ) anda c A C a A c C&#d d d
anda B c C& d d [ ]A Ba 3
( , ) .a c B C& #d
Thus, every element of A C# is contained in .B C#
.A C B C# #3
Hence, A B A C B C& # #3 3 for any set C.
THEOREM 2 If A B and C D3 3 then prove that .A C B D# #3
PROOF Let .andA B C D3 3 Then, we have to show that .A C B D# #3
Let (a, c) be an arbitrary element of .A C# Then,
( , ) anda c A C a A c C&#d d d
anda B c D& d d [ ]andA B C Da 3 3
( , ) .a c B D& #d
Thus, every element of A C# is contained in .B D#
Hence, .A C B D# #3
58 Senior Secondary School Mathematics for Class 11
THEOREM 3 If A and B are any two nonempty sets, prove that .A B B A A B+# #
PROOF Let us fi rst assume that .A B Then,
andA B B B B A B B# # # # [ ]A Ba .
.A B B A# #
Thus, .A B A B B A& # #
Conversely, let A B B A# # and we have to show that .A B
Let a be an arbitrary element of A. Then,
( , )a A a b A B& #d d for some b Bd
( , )a b B A& #d [ ]A B B Aa # #
anda B b A& d d
a B& d (surely).
.A B3
Again, let b be an arbitrary element of B. Then,
( , )b B b a B A& #d d for some a Ad
( , )b a A B& #d [ ]B A A Ba # #
andb A a B& d d
b A& d (surely).
.B A3
Thus, .andA B B A A B&3 3
.A B B A A B&# #
Hence, .A B B A A B+# #
THEOREM 4 If ,A B3 prove that ( ) ( ) .A A A B B A# # + #3
PROOF Let A B3 and let (a, a) be an arbitrary element of .A A#
Then, ( , ) anda a A A a A a A&#d d d
anda A a B& d d [ ]A Ba 3
( , ) ( , )anda A a B a B a A& d d d d
( , ) ( , )anda b A B a b B A& # #d d
( , ) ( ) ( ) .a b A B B A& # + #d
( ) ( ) .A A A B B A# # + #3
Hence, ( ) ( ) .A B A A A B B A& # # + #3 3
THEOREM 5 For any sets A, B and C, prove that
( ) ( ) ( ) .A B C A B A C# , # , #
PROOF Let (a, b) be an arbitrary element of ( ) .A B C# , Then,
( , ) ( ) anda b A B C a A b B C&# , ,d d d
( )and ora A b B b C& d d d
( ) ( )and or anda A b B a A b C& d d d d
[ ( ) ( ) ( )]and or and or andp q r p q p ra /
( , ) ( , )ora b A B a b A C& # #d d
( , ) ( ) ( )a b A B A C& # , #d .
Relations 59
( ) ( ) ( ) .A B C A B A C# , # , #3 … (i)
Again, let ( , )x y be an arbitrary element of ( ) ( ) .A B A C# , # Then,
( , ) ( ) ( ) ( , ) ( , )orx y A B A C x y A B x y A C&# , # # #d d d
( ) ( )and or andx A y B x A y C& d d d d
( )and orx A y B y C& d d d
[by distributive law]
( )andx A y B C& ,d d
( , ) ( )x y A B C& # ,d .
( ) ( ) ( ) .A B A C A B C# , # # ,3 … (ii)
Hence, from (i) and (ii), we get
( ) ( ) ( ) .A B C A B A C# , # , #
THEOREM 6 For any sets A, B and C, prove that
( ) ( ) ( ) .A B C A B A C# + # + #
PROOF Let (a, b) be an arbitrary element of ( ) .A B C# k Then,
( , ) ( ) ( )anda b A B C a A b B C&# + +d d d
( )and anda A b B b C& d d d
( ) ( )and and anda A b B a A b C& d d d d
( , ) ( , )anda b A B a b A C& # #d d
( , ) ( ) ( )a b A B A C& # + #d .
( ) ( ) ( ) .A B C A B A C# + # + #3 … (i)
Again, let (x, y) be an arbitrary element of ( ) ( ) .A B A C# + # Then,
( , ) ( ) ( ) ( , ) ( , )andx y A B A C x y A B x y A C&# + # # #d d d
( ) ( )and and andx A y B x A y C& d d d d
( )and andx A y B y C& d d d
( )andx A y B C& +d d
( , ) ( )x y A B C& # +d .
( ) ( ) ( ) .A B A C A B C# + # # +3 … (ii)
Hence, from (i) and (ii), we get
( ) ( ) ( ) .A B C A B A C# + # + #
THEOREM 7 For any sets A, B and C, prove that
( ) ( ) ( ) .A B C A B A C# # #
PROOF Let (a, b) be an arbitrary element of ( ) .A B C# Then,
( , ) ( ) ( )anda b A B C a A b B C&# d d d
( )and anda A b B b C& d d z
( ) ( )and and anda A b B a A b C& dd d z
( , ) ( , )anda b A B a b A C& # #d z
( , ) ( ) ( )a b A B A C& # #d .
( ) ( ) ( ) .A B C A B A C# # #3 … (i)
60 Senior Secondary School Mathematics for Class 11
Again, let (x, y) be an arbitrary element of ( ) ( ) .A B A C# # Then,
( , ) ( ) ( ) ( , ) ( , )andx y A B A C x y A B x y A C&# # # #d d z
( ) ( )and and andx A y B x A y C& d d d z
( )and andx A y B y C& d d z
( )andx A y B C& d d
( , ) ( )x y A B C& # d .
( ) ( ) ( ) .A B A C A B C# # #3 … (ii)
From (i) and (ii), we get
( ) ( ) ( ) .A B C A B A C# # #
THEOREM 8 For any sets A, B, C and D, prove that
( ) ( ) ( ) ( ) .A B C D A C B D# + # + # +
PROOF Let (a, b) be an arbitrary element of ( ) ( ) .A B C D# + # Then,
( , ) ( ) ( ) ( , ) ( ) ( , ) ( )anda b A B C D a b A B a b C D&# + # # #d d d
( ) ( )and and anda A b B a C b D& d d d d
( ) ( )and and anda A a C b B b D& d d d d
( ) ( )anda A C b B D& + +d d
( , ) ( ) ( ) .a b A C B D& + # +d … (i)
Again, let (x, y) be an arbitrary element of ( ) ( ) .A C B D+ # + Then,
( , ) ( ) ( ) andx y A C B D x A C y B D&+ # + + +d d d
( ) ( )and and andx A x C y B y D& d d d d
( ) ( )and and andx A y B x C y D& dd d d
( , ) ( , )andx y A B x y C D& # #d d
( , ) ( ) ( ) .x y A B C D& d # + #
( ) ( ) ( ) ( ) .A C B D A B C D+ # + # + #3 … (ii)
From (i) and (ii), we get
( ) ( ) ( ) ( ) .A B C D A C B D# + # + # +THEOREM 9 For any sets A and B, prove that
( ) ( ) ( ) ( )A B B A A B B A# + # + # + .
SOLUTION Let ( , )a b be an arbitrary element of ( ) ( ) .A B B A# + # Then,
( , ) ( ) ( ) ( , ) ( , )anda b A B B A a b A B a b B A&# + # # #d d d
( ) ( )and and anda A b B a B b A& d d d d
( ) ( )and and anda A a B b B b A& d d d d
( ) ( )anda A B b B A& + +d d
( , ) ( ) ( ) .a b A B B A& + # +d
( ) ( ) ( ) ( ) .A B B A A B B A# + # + # +3 … (i)
Again, let ( , )x y be an arbitrary element of ( ) ( ) .A B B A+ # + Then,
( , ) ( ) ( ) andx y A B B A x A B y B A&+ # + + +d d d
( ) ( )and and andx A x B y B y A& d d d d
Relations 61
( ) ( )and and andx A y B x B y A& d d d d
( , ) ( , )andx y A B x y B A& # d #d
( , ) ( ) ( ) .x y A B B A& # + #d
( ) ( ) ( ) ( ) .A B B A A B B A+ # + # + #3 … (ii)
Thus, from (i) and (ii), we get
( ) ( ) ( ) ( ) .A B B A A B B A# + # + # +
AN IMPORTANT RESULT If A and B are two nonempty sets having n elements in
common then ( )A B# and ( )B A# have n2 elements in common.
EXAMPLE Let A and B be two nonempty sets such that ( ) , ( )n A n B5 6 and
( ) .n A B 3+
Find (i) ( ),n A B# (ii) ( )n B A# and (iii) {( ) ( )} .n A B B A# + #
SOLUTION (i) ( ) ( ) ( ) ( ) .n A B n A n B 5 6 30# # #
(ii) ( ) ( ) ( ) ( ) .n B A n B n A 6 5 30# # #
(iii) Given: ( ) .n A B 3+
A and B have 3 elements in common.
So, ( ) ( )andA B B A# # have 3 92 elements in common.
Hence, {( ) ( )} .n A B B A 9# + #
THEOREM 10 If A, B and C be three nonempty sets given in such a way that
A B A C# # then prove that .B C
PROOF Let A B A C# # and we have to prove that .B C
Let .b Bd Then,
( , )b B a b A B& #d d for every a Ad
( , )a b A C& #d for every a Ad [ ]A B A Ca # #
.b C& d
.B C3 … (i)
Again, let .c Cd Then,
( , )c C a c A C& #d d for every a Ad
( , )a c A B& #d for every a Ad [ ]A C A Ba # #
.c B& d
.C B3 … (ii)
From (i) and (ii), we get .B C
Hence, .A B A C B C&# #
EXERCISE 2B
1. For any sets A, B and C, prove that:
(i) ( ) ( ) ( )A B C A B A C# , # , #
(ii) ( ) ( ) ( )A B C A B A C# + # + #
(iii) ( ) ( ) ( )A B C A B A C# # #
62 Senior Secondary School Mathematics for Class 11
2. For any sets A and B, prove that
( ) ( ) ( ) ( ) .A B B A A B B A# + # + # +
3. If A and B are nonempty sets, prove that
.A B B A A B+# #
4. (i) If ,A B3 prove that A C B C# #3 for any set C.
(ii) If A B3 and C D3 then prove that .A C B D# #3
5. If A B C D# #3 and A B# !, prove that .andA C B D3 3
6. If A and B be two sets such that ( ) , ( )n A n B3 4 and ( )n A B 2+ then fi nd:
(i) ( )n A B# (ii) ( )n B A# (iii) {( ) ( }n A B B A# + #
7. For any two sets A and B, show that A B# and B A# have an element in
common if and only if A and B have an element in common.
8. Let { , }A 1 2 and { , } .B 2 3 Then, write down all possible subsets of .A B#
9. Let { , , , }, { , , }A a b c d B c d e and { , , , }C d e f g . Then verify each of the
following identities:
(i) ( ) ( ) ( )A B C A B A C# + # + # (ii) ( ) ( ) ( )A B C A B A C# # #
(iii) ( ) ( ) ( ) ( )A B B A A B A B# + # + # +
ANSWERS (EXERCISE 2B)
6. (i) ( )n A B 12# (ii) ( )n B A 12# (iii) {( ) ( )}n A B B A 4# + #
8. , {( , )}, {( , )}, {( , )}, {( , )}, {( , ), ( , )}, {( , ), ( , )}, {( , ), ( , )},1 2 1 3 2 2 2 3 1 2 1 3 1 2 2 2 1 2 2 3
{( , ), ( , )}, {( , ), ( , )}, {( , ), ( , )}, {( , ), ( , ), ( , )},1 3 2 2 1 3 2 3 2 2 2 3 1 2 1 3 2 2
{( , ), ( , ), ( , )}, {( , ), ( , ), ( , )}, {( , ), ( , ), ( , )},1 2 1 3 2 3 1 3 2 2 2 3 1 2 2 2 2 3
{( , ), ( , ), ( , ), ( , )}1 2 1 3 2 2 2 3
HINTS TO SOME SELECTED QUESTIONS
6. (i) We know that ( ) ( ) ( )n A B n A n B# # (ii) ( ) ( ) ( ) .n B A n B n A# #
(iii) Since A and B have 2 elements in common, so ( )A B# and ( )B A# will have 2 42
elements in common.
7. We know that ( ) ( ) ( ) ( ) ( ) ( ) .A B B A A B B A A B A B# + # + # + + # +
Now, ( ) {( ) ( )}n A B n A B A B1 1 12++ + # +
{( ) ( )}n A B B A 1+ # + # [ ( ) ( ) ( ) ( )]A B A B A B B Aa + # + # + # .
8. {( , ), ( , ), ( , ), ( , )} .A B 1 2 1 3 2 2 2 3# So, it will have 2 164 subsets.
RELATIONS
RELATION Let A and B be two nonempty sets. Then, a relation R from A to B is a
subset of ( ) .A B#
Thus, R is a relation from A to ( ) .B R A B+ #3
If ( , )a b Rd then we say that ‘a is related to b‘ and we write, .a R b
If ( , )a b Rz then ‘a is not related to b‘ and we write, .a R b
Relations 63
DOMAIN, RANGE AND CO-DOMAIN OF A RELATION
Let R be a relation from A to B. Then, ( ) .R A B#3
(i) The set of all fi rst coordinates of elements of R is called the domain of R, written
as dom (R).
(ii) The set of all second coordinates of elements of R is called the range of R, denoted
by range (R).
(iii) The set B is called the co-domain of R.
Dom ( ) { : ( , ) }R a a b R d and Range ( ) { : ( , ) } .R b a b R d
Total Number of Relations from A to B
Let ( )n A p and ( ) .n B q Then, ( ) .n A B pq#
We know that every subset of A B# is a relation from A to B.
Total number of subsets of A B# is .2pq
total number of relations from A to .B 2pq
REPRESENTATION OF A RELATION
Let A and B be two given sets. Then, a relation R A B#3 can be represented in
any of the forms, given below.
(i) ROSTER FORM
In this form, R is given as a set of ordered pairs.
EXAMPLE 1 Let { , , , , } { , , , } .A and B2 1 0 1 2 0 1 4 9
Let {( , ), ( , ), ( , ), ( , ), ( , )}R 2 4 1 1 0 0 1 1 2 4 .
(i) Show that R is a relation from A to B.
(ii) Find dom (R), range (R) and co-domain of R.
SOLUTION (i) Since ,R A Bf # so R is a relation from A to B.
Note that , , , .andR R R R R2 4 1 1 0 0 1 1 2 4
(ii) Dom ( )R set of fi rst coordinates of elements of R
{ , , , , } .2 1 0 1 2
Range ( )R set of second coordinates of elements of R
{ , , } .0 1 4
Co-domain of { , , , } .R B0 1 4 9
(ii) SET-BUILDER FORM
Under this method, for every ( , ) ,a b Rd a general relation is being given between a
and b.
Using this relation, all the elements of R can be obtained.
EXAMPLE 2 Let { , , , }A 1 2 3 5 and { , , } .B 4 6 9
Defi ne a relation from A to B, given by
{( , ) : , ( )R a b a A b B and a b d d is odd}.
(i) Write R in roster form.
(ii) Find dom (R) and range (R).
64 Senior Secondary School Mathematics for Class 11
SOLUTION (i) Clearly, we have
{( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , )}R 1 4 1 6 2 9 3 4 3 6 5 4 5 6 .
(ii) Dom ( )R set of fi rst coordinates of elements of R
{ , , , }1 2 3 5 .
Range ( )R set of second coordinates of elements of R
{ , , } .4 6 9
EXAMPLE 3 Let {( , ) :R x x x3 is a prime number less than 10}.
(i) Write R in roster form.
(ii) Find dom (R) and range (R).
SOLUTION Prime numbers less than 10 are 2, 3, 5, 7.
(i) {( , ), ( , ), ( , ), ( , )}R 2 2 3 3 5 5 7 73 3 3 3
{( , ), ( , ), ( , ), ( , )} .2 8 3 27 5 125 7 343
(ii) Dom ( ) { , , , } .R 2 3 5 7
Range ( ) { , , , } .R 8 27 125 343
EXAMPLE 4 Let {( , ) :R x y x and y are integers and } .xy 4
(i) Write R in roster form.
(ii) Find dom (R) and range (R).
SOLUTION Clearly, we have
(i) {( , ), ( , ), ( , ), ( , ), ( , ), ( , )} .R 4 1 2 2 1 4 1 4 2 2 4 1
(ii) Dom ( ) { , , , , , } .R 4 2 1 1 2 4
Range ( ) { , , , , , } .R 4 2 1 1 2 4
(iii) ARROW DIAGRAM
Let R be a relation from A to B. First, we draw two bounded fi gures to represent
A and B respectively. We mark the elements of A and B in these fi gures. For
each ( , ) ,a b Rd we draw an arrow from a to b. This gives us the required arrow
diagram.
EXAMPLE 5 Let { , , , , } .A 1 2 3 4 5 Defi ne a relation R from A to A by
{( , ) : } .R x y y x2 3
(i) Depict R using an arrow diagram.
(ii) Find dom (R) and range (R).
SOLUTION ( )x y A1 2 3 1& z
( ) ,x y2 4 3 1&
( ) ,x y3 6 3 3&
( ) ,x y4 8 3 5&
( ) .x y A5 10 3 7& z
{( , ), ( , ), ( , )} .R 2 1 3 3 4 5
Relations 65
(i) We may depict it by arrow diagram, as shown below:
(ii) We have, dom (R) = {2, 3, 4} and range (R) = {1, 3, 5}.
EXAMPLE 6 Let { , , , , }A 1 2 3 4 5 and { , , } .B 1 4 5
Let R be a relation ‘is less than‘ from A to B.
(i) List the elements of R.
(ii) Find the domain, co-domain and range of R.
(iii) Depict the above relation by an arrow diagram.
SOLUTION Here, { , , , , } { , , } .andA B1 2 3 4 5 1 4 5
(i) {( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , )} .R 1 4 1 5 2 4 2 5 3 4 3 5 4 5
(ii) Dom ( ) { , , , },R 1 2 3 4 range ( ) { , }R 4 5
and co-domain ( ) { , , } .R 1 4 5
(iii) We may represent the above relation by an arrow diagram,
shown below.
REMARK Let A and B be two nonempty sets. Then, every subset of A B# is a
relation from A to B.
Since ,A B#1 so is also a relation from A to B, called an empty or
void relation.
EXAMPLE 7 Let { , , }A x y z and { , } .B 1 2 Find the number of all possible relations
that can be defi ned from A to B.
SOLUTION Here ( ) ( ) .andn A n B3 2 So, ( ) ( ) .n A B 3 2 6# #
Since a set containing n elements has 2n subsets.
( )A B# will have 2 646 subsets.
But, every subset of A B# is a relation from A to B.
Hence, there are in all 64 relations from A to B.
1
2
3
A
1
4
4
5
5
B
2
A
1
4
A
3
2
1
4
5
3
5
66 Senior Secondary School Mathematics for Class 11
EXERCISE 2C
1. Let A and B be two nonempty sets.
(i) What do you mean by a relation from A to B?
(ii) What do you mean by the domain and range of a relation?
2. Find the domain and range of each of the relations given below:
(i) {( , ), ( , ), ( , ), ( , ), ( , )}R 1 1 1 1 2 4 2 4 3 9
(ii) , : ,intis an egerR x x x x
1 0 5 b l( 2
(iii) {( , ) : , }andR x y x y x y N2 8 d
(iv) {( , ) : | |, | | }andR x y y x x Z x1 3# d
3. Let { , , , }A 1 3 5 7 and { , , , } .B 2 4 6 8
Let {( , ) : ,R x y x A y B d d and } .x y
(i) Write R in roster form.
(ii) Find dom (R) and range (R).
(iii) Depict R by an arrow diagram.
4. Let { , , }A 2 4 5 and { , , , , , } .B 1 2 3 4 6 8
Let {( , ) ,R x y x A y B d d and x divides y}.
(i) Write R in roster form.
(ii) Find dom (R) and range (R).
5. Let { , , , }A 2 3 4 5 and { , , , } .B 3 6 7 10
Let {( , ) : ,R x y x A y B d d and x is relatively prime to y}.
(i) Write R in roster form.
(ii) Find dom (R) and range (R).
6. Let { , , , }A 1 2 3 5 and { , , } .B 4 6 9
Let {( , ) : , ,R x y x A y B d d and ( )x y is odd}.
Write R in roster form.
7. Let {( , ) : ,R x y x y x N3 12 d and } .y Nd
(i) Write R in roster form.
(ii) Find dom (R) and range (R).
8. Let { , , , , , } .A 1 2 3 4 5 6
Defi ne a relation R from A to A by {( , ) : } .R x y y x 1
(i) Write R in roster form.
(ii) Find dom (R) and range (R).
(iii) What is its co-domain?
(iv) Depict R by using arrow diagram.
9. Let {( , ) : { , , , , , }} .R x x x5 0 1 2 3 4 5 d
(i) Write R in roster form.
(ii) Find dom (R) and range (R).
Relations 67
10. Let { , , , , }A 1 2 3 4 6 and let {( , ) : ,R a b a b A d and a divides b}.
(i) Write R in roster form.
(ii) Find dom (R) and range (R).
11. Defi ne a relation R from Z to Z, given by
{( , ) : , ( )andR a b a b Z a bd is an integer}.
Find dom (R) and range (R).
Hint The difference of two integers is always an integer.
12. Let {( , ) : , } .andR x y x y Z x y 42 2# d
(i) Write R in roster form.
(ii) Find dom (R) and range (R).
13. Let { , }A 2 3 and { , } .B 3 5
(i) Find ( )A B# and ( ) .n A B#
(ii) How many relations can be defi ned from A to B?
14. Let { , }A 3 5 and { , } .B 7 9 Let {( , ) : ,R a b a A b B d d and ( )a b is odd}.
Show that R is an empty relation from A to B.
Hint The difference of two odd numbers cannot be odd.
ANSWERS (EXERCISE 2C)
2. (i) dom ( ) { , , , , }R 2 1 1 2 3 and range ( ) { , , }R 1 4 9
(ii) dom ( ) { , , , }R 1 2 3 4 and range ( ) , , ,R 1 2
1
3
1
4
1 ' 1
(iii) dom ( ) { , , }R 2 4 6 and range ( ) { , , }R 3 2 1
(iv) dom ( ) { , , , , , , }R 3 2 1 0 1 2 3 and range ( ) { , , , , }R 0 1 2 3 4
3. (i) {( , ), ( , ), ( , ), ( , ), ( , ), ( , )}R 3 2 5 2 5 4 7 2 7 4 7 6
(ii) dom ( ) { , , }R 3 5 7 and range ( ) { , , }R 2 4 6
(iii)
4. (i) {( , ), ( , ), ( , ), ( , ), ( , ), ( , )}R 2 2 2 4 2 6 2 8 4 4 4 8
(ii) dom ( ) { , }R 2 4 and range ( ) { , , , }R 2 4 6 8
5. (i) {( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , )}R 2 3 2 7 3 7 3 10 4 3 4 7 5 3 5 6 5 7
(ii) dom ( ) { , , , }R 2 3 4 5 and range ( ) { , , , }R 3 6 7 10
6. {( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , )}R 1 4 1 6 2 9 3 4 3 6 5 4 5 6
7. (i) {( , ), ( , ), ( , )}R 3 3 6 2 9 1
(ii) dom ( ) { , , }R 3 6 9 and range ( ) { , , }R 3 2 1
3
A
1
7
B
5
2
4
8
6
68 Senior Secondary School Mathematics for Class 11
8. (i) {( , ), ( , ), ( , ), ( , ), ( , )}R 1 2 2 3 3 4 4 5 5 6
(ii) dom ( ) { , , , , },R 1 2 3 4 5 range ( ) { , , , , }R 2 3 4 5 6
(iii) co-domain of { , , , , , }R 1 2 3 4 5 6
(iv)
9. (i) {( , ), ( , ), ( , ), ( , ), ( , ), , )}R 0 5 1 6 2 7 3 8 4 9 5 10
(ii) dom ( ) { , , , , , }R 0 1 2 3 4 5 and range ( ) { , , , , , }R 5 6 7 8 9 10
10. (i) {( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ),R 1 1 1 2 1 3 1 4 1 6 2 2 2 4 2 6 3 3 3 6
( , ), ( , )}4 4 6 6
(ii) dom ( ) { , , , , }R 1 2 3 4 6 and range ( ) { , , , , }R 1 2 3 4 6
11. dom ( )R Z and range ( )R Z
12. (i) {( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ),R 2 0 0 2 2 0 0 2 1 0 0 1 1 0 0 1 1 1
( , ), ( , ), ( , )}1 1 1 1 1 1
(ii) dom ( ) { , , , , }R 2 1 0 1 2 and range ( ) { , , , , }R 2 1 0 1 2
13. (i) {( , ), ( , ), ( , ), ( , )}A B 2 3 2 5 3 3 3 5# and ( )n A B 4#
(ii) number of relations from A to B 2 164
BINARY RELATION ON A SET
Let A be a nonempty set. Then, every subset of ( )A A# is called a binary relation or
simply a relation on A.
REMARKS (i) Since A A#1 so is relation on A, called the empty or void
relation on A.
(ii) Since A A A A# #3 so ( )A A# is a relation on A, called the
universal relation on A.
(iii) Let {( , ) : } .I a a a AA d Then, clearly, ( )I A AA #3 and therefore, it
is a relation on A, called the identity relation on A.
SOLVED EXAMPLES
EXAMPLE 1 Let { , , }A 1 2 3 and {( , ), ( , ), ( , ), ( , )} .R 1 2 2 2 3 1 3 2 Show that R is a
binary relation on A. Find its domain and range.
SOLUTION {( , ), ( , ), ( , ), ( , )} .R 1 2 2 2 3 1 3 2
Clearly, R A A#1 and so R is a relation on A.
Dom ( )R set of fi rst coordinates of elements of { , , } .R 1 2 3
2
6
A
1
4
5
B
3
2
6
1
4
5
3
Relations 69
Range ( )R set of second coordinates of elements of { , } .R 1 2
Hence, dom ( ) { , , }R 1 2 3 and range ( ) { , } .R 1 2
EXAMPLE 2 Let N be the set of all natural numbers. Let {( , ) : ,R a b a b N d and
} .a b2 10 Show that R is a binary relation on N. Find its domain,
range and co-domain.
SOLUTION Here {( , ) : ,R a b a b N d and } .a b2 10
Now, ( ) .a b b a2 10 10 2&
( ), ( ), ( ), ( ) .a b a b a b a b1 8 2 6 3 4 4 2& & & &
{( , ), ( , ), ( , ), ( , )} .R 1 8 2 6 3 4 4 2
Since ,R N N#1 so R is a binary relation on N.
Dom (R) = set of 1st coordinates of elements of R
{ , , , } .1 2 3 4
Range (R) = set of 2nd coordinates of elements of R
{ , , , } .8 6 4 2
Co-domain of .R N
INVERSE RELATION
Let R be a binary relation on a set A. Then, the inverse of R, denoted by R 1 is a binary
relation on A, defi ned by {( , ) : ( , ) } .R b a a b R1 d
Clearly, ( , ) ( , ) .a b R b a R 1+ d d
Also, dom ( )R range ( )R 1 and range ( )R dom ( )R 1 .
EXAMPLE 3 Let A be the set of fi rst ten natural numbers.Let R be a binary relation on
A, defi ned by
{( , ) : , } .R a b a b A and a b2 10 d
Express R and R 1 as sets of ordered pairs.
Show that (i) dom (R) = range ( )R 1 (ii) range (R) = dom ( ) .R 1
SOLUTION ·
( )
a b b
a
2 10 2
10
&
Now, ( ), ( ), ( ),a b a b a b2 4 4 3 6 2& & &
( ) .a b8 1&
{( , ), ( , ), ( , ), ( , )}R 2 4 4 3 6 2 8 1
{( , ), ( , ), ( , ), ( , )} .R 4 2 3 4 2 6 1 81
(i) dom ( ) { , , , }R 2 4 6 8 range ( ) .R 1
(ii) range ( ) { , , , }R 4 3 2 1 dom ( ) .R 1
VARIOUS TYPES OF RELATIONS
Let A be a nonempty set. Then, a relation R on A is said to be:
(i) refl exive, if ( , )a a Rd for all ,a Ad i.e., a R a for all ;a Ad
(ii) symmetric, if ( , ) ( , )a b R b a R&d d for all , ,a b Ad
i.e., a R b b R a& for all , ;a b Ad
70 Senior Secondary School Mathematics for Class 11
(iii) transitive, if ( , )a b Rd and ( , ) ( , )b c R a c R& dd for all , , .a b c Ad
i.e., ,a R b b R c a R c& for all , , .a b c Ad
EQUIVALENCE RELATION
A relation which is refl exive, symmetric and transitive is called an equivalence relation.
EXAMPLE 4 Let R be a relation on the set Q of all rationals defi ned by {( , ) : ,R a b a b Q d
and } .a b Z d Show that R is an equivalence relation.
SOLUTION Given: {( , ) : ,R a b a b Q d and } .a b Z d
(i) Let .a Qd Then, .a a Z0 d
( , )a a Rd for all .a Qd
So, R is refl exive.
(ii) ( , ) ( ) ,a b R a b Z& d d i.e., ( )a b is an integer
( )a b& is an integer
( )b a& is an integer
( , ) .b a R& d
Thus, ( , ) ( , ) .a b R b a R&d d
R is symmetric.
(iii) ( , ) ( , )anda b R b c Rd d
& ( )a b is an integer and ( )b c is an integer
& {( ) ( )}a b b c is an integer
& ( )a c is an integer
& ( , ) .a c Rd
Thus, ( , ) ( , ) ( , ) .anda b R b c R a c R&d d d
R is transitive.
Thus, R is refl exive, symmetric and transitive.
So, R is an equivalence relation.
EXAMPLE 5 Let m be a given fi xed positive integer.
Let {( , ) : ,R a b a b Z d and ( )a b is divisible by m}.
Show that R is an equivalence relation on Z.
SOLUTION {( , ) : ,R a b a b Z d and ( )a b is divisible by m}.
(i) Let .a Zd Then,
,a a 0 which is divisible by m.
( , )a a Rd for all .a Zd
So, R is refl exive.
(ii) Let ( , ) .a b Rd Then,
( , ) ( )a b R a b& d is divisible by m
( )a b& is divisible by m
( )b a& is divisible by m
( , ) .b a R& d
Relations 71
Thus, ( , ) ( , ) .a b R b a R&d d
So, R is symmetric.
(iii) Let ( , )a b Rd and ( , ) .b c Rd Then,
( , )a b Rd and ( , ) .b c Rd
& ( )a b is divisible by m and ( )b c is divisible by m
& {( ) ( )}a b b c is divisible by m
& ( )a c is divisible by m
& ( , ) .a c Rd
̀ ( , )a b Rd and ( , ) ( , ) .b c R a c R&d d
So, R is transitive.
Thus, R is refl exive, symmetric and transitive.
Hence, R is an equivalence relation on Z.
EXAMPLE 6 Show that the relation ‘is parallel to’ on the set S of all straight lines in a
plane is an equivalence relation.
SOLUTION Let S be the set of all straight lines in a plane.
Then, the relation, ‘is parallel to’ on S is
(i) refl exive, since every line is parallel to itself,
i.e., ||L L for all L in S;
(ii) symmetric, since || ||L M M L& for all , ;L M Sd
(iii) transitive, since for all L, M, N in S, we have
||L M and || || .M N L N&
Thus, the given relation is refl exive, symmetric and transitive.
Hence, it is an equivalence relation on S.
EXAMPLE 7 Show that the relation ‘is congruent to’ on the set of all triangles in a
plane is an equivalence relation.
SOLUTION Let S be the set of all triangles in a plane.
Then, the congruence relation on S is
(i) refl exive, since 3b3 for every ;S3d
(ii) symmetric, since 1 2 2 1&3 b3 3 b3 for all , S1 23 3 d
(iii) transitive, since 1 23 b3 and 2 3 1 3&3 b3 3 b3
for all , , .S1 2 33 3 3 d
Hence, the given relation is an equivalence relation.
EXAMPLE 8 Let {( , ) : ,R a b a b N d and } .a b2 Show that R satisfi es none of
refl exivity, symmetry and transitivity.
SOLUTION (i) R is not refl exive, since 2 22! and therefore ( , ) .R2 2 z
(ii) Since ,4 22 so ( , ) .R4 2 d
But, .2 42! So, ( , ) .R2 4 z
Thus, ( , ) R4 2 d but ( , ) .R2 4 z
R is not symmetric.
72 Senior Secondary School Mathematics for Class 11
(iii) Since ,16 42 so ( , ) .R16 4 d
Also, ,4 22 so ( , ) .R4 2 d
But, ( , ) .R16 2 16 22 &! z
Thus, ( , ) R16 4 d and ( , ) .R4 2 d But ( , ) .R16 2 z
R is not transitive.
Hence, R satisfi es none of refl exivity, symmetry and
transitivity.
EXAMPLE 9 Let S be the set of all real numbers and let R be a binary relation on S
defi ned by ( , )a b R ab1 0+ d for all , .a b Sd Show that R is refl exive
as well as symmetric. Give an example to show that R is not transitive.
SOLUTION (i) Let a be an arbitrary real number.
Then, ( ) ( , )a a a R1 02 & d for every .a Sd
R is refl exive.
(ii) Let ( , ) .a b Rd Then,
( , )a b R ab1 0& d
ba1 0&
( , ) .b a R& d
( , ) ( , ) .a b R b a R&d d
So, R is symmetric.
(iii) In order to show that R is not transitive, consider the real
numbers ,3
2 1
b l and 2.
Now, , ,R3
2 1
db l since .1 3
2 1 1 3
2
3
1 0#
b bl l( 2
And, ( , ) ,R1 2 d since { ( )} .1 1 2 3 0#
But, , ,R3
2 2
zb l since .1 3
2 2 1 3
4
3
1 0#
b bl l( 2
Thus, , , ( , )R R3
2 1 1 2
d db l and , .R3
2 2
zb l
This shows that R is not transitive.
EXAMPLE 10 Let R be a relation on ,N N# defi ned by
( , ) ( , )a b R c d a d b c+ for all ( , ), ( , ) .a b c d N N#d
Show that R is an equivalence relation.
SOLUTION Here R is a relation on ,N N# defi ned by
( , ) ( , )a b R c d a d b c+ for all ( , ), ( , ) .a b c d N N#d
We shall show that R satisfi es the following properties.
(i) Refl exivity:
We know that a b b a for all , .a b Nd
( , ) ( , )a b R a b for all ( , ) ( ) .a b N N#d
So, R is refl exive.
Relations 73
(ii) Symmetry:
Let ( , ) ( , ) .a b R c d Then,
( , ) ( , )a b R c d a d b c&
c b d a&
( , ) ( , )c d R a b& .
( , ) ( , ) ( , ) ( , )a b R c d c d R a b& for all ( , ), ( , ) .a b c d N N#d
This shows that R is symmetric.
(iii) Transitivity:
Let ( , ) ( , )a b R c d and ( , ) ( , ) .c d R e f Then,
( , ) ( , ) ( , ) ( , )anda b R c d c d R e f
& anda d b c c f d e
& a d c f b c d e
& a f b e
& ( , ) ( , ) .a b R e f
Thus, ( , ) ( , ) ( , ) ( , ) ( , ) ( , ) .anda b R c d c d R e f a b R e f&
This shows that R is transitive.
R is refl exive, symmetric and transitive.
Hence, R is an equivalence relation on .N N#
EXERCISE 2D
1. What do you mean by a binary relation on a set A?
Defi ne the domain and range of a relation on A.
2. Let { , , }A 2 3 5 and {( , ), ( , ), ( , ), ( , )} .R 2 3 2 5 3 3 3 5
Show that R is a binary relation on A. Find its domain and range.
3. Let { , , , , , , , , }A 0 1 2 3 4 5 6 7 8 and let {( , ) : , } .andR a b a b A a b2 3 12 d
Express R as a set of ordered pairs. Show that R is a binary relation on A.
Find its domain and range.
4. If R is a binary relation on a set A, defi ne R 1 on A.
Let {( , ) : ,R a b a b W d and },a b3 2 15 where W is the set of whole
numbers.
Express R and R 1 as sets of ordered pairs.
Show that (i) dom (R) = range ( )R 1 (ii) range (R) = dom ( ) .R 1
5. What is an equivalence relation?
Show that the relation of ‘similarity‘ on the set S of all triangles in a plane is
an equivalence relation.
6. Let {( , ) : , ( )andR a b a b Z a b d is even}.
Then, show that R is an equivalence relation on Z.
7. Let { , , } {( , ) : , .and andA R a b a b A a b1 2 3 52 2 # d
Write R as set of ordered pairs.
74 Senior Secondary School Mathematics for Class 11
Mention whether R is (i) refl exive(ii) symmetric (iii) transitive.
Give reason in each case.
8. Let {( , ) : ,R a b a b Z d and } .b a2 4 If ( , )a R2 d and ( , ) .b R4 2 d Then,
write the values of a and b.
9. Let R be a relation on Z, defi ned by ( , ) .x y R x y 92 2+ d Then, write R
as set of ordered pairs. What is its domain?
10. Let A be the set of fi rst fi ve natural numbers and let R be a relation on A,
defi ned by ( , ) .x y R x y+ #d
Express R and R 1 as sets of ordered pairs.
Find: dom ( )R 1 and range (R).
11. Let {( , ) : , } .andR x y x y Z x y 252 2 d
Express R and R 1 as sets of ordered pairs. Show that .R R 1
12. Find R 1 , when
(i) {( , ), ( , ), ( , ), ( , ), ( , )}R 1 2 1 3 2 3 3 2 4 5 ,
(ii) {( , ) : , , }R x y x y N x y2 8 d .
13. Let { , } .A a b List all relations on A and fi nd their number.
Hint {( , ), ( , ), ( , ), ( , )}A A a a a b b a b b# and every subset of A A# is a relation on A.
So, their number .2 164
14. Let {( , ) : , } .andR a b a b N a b d
Show that R is a binary relation on N, which is neither refl exive nor
symmetric. Show that R is transitive.
Hint Since ,R N N#1 so it is a binary relation on N.
ANSWERS (EXERCISE 2D)
2. dom ( ) { , }R 2 3 and range ( ) { , }R 3 5
3. {( , ), ( , ), ( , )},R 0 4 3 2 6 0 dom ( ) { , , }R 0 3 6 and range ( ) { , , }R 0 2 4
4. {( , ), ( , ), ( , )}, {( , ), ( , ), ( , )}R R1 6 3 3 5 0 6 1 3 3 0 51
7. {( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , )}R 1 1 2 2 3 3 2 1 1 2 3 2 2 3 .
R is refl exive and symmetric but not transitive.
8. ,a b1 4
9. {( , ), ( , ), ( , ), ( , ),R 3 0 0 3 3 0 0 3 dom ( ) { , , }R 3 0 3
10. {( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ),R 1 1 1 2 1 3 1 4 1 5 2 2 2 3 2 4 2 5 3 3 3 4
( , ), ( , ), ( , ), ( , )}3 5 4 4 4 5 5 5
{( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ),R 1 1 2 1 3 1 4 1 5 1 2 2 3 2 4 2 5 2 3 3 4 31
( , ), ( , ), ( , ), ( , )}5 3 4 4 5 4 5 5
dom ( ) { , , , , } ( )rangeR R1 2 3 4 51
11. {( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , )}R 5 0 0 5 5 0 0 5 3 4 4 3 3 4 4 3 3 4 4 3
{( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ),R 0 5 5 0 0 5 5 0 4 3 3 4 4 3 3 4 4 31
( , )}3 4
Relations 75
12. (i) {( , ), ( , ), ( , ), ( , ), ( , )}R 2 1 3 1 3 2 2 3 5 41 (ii) {( , ), ( , ), ( , )}R 3 2 2 4 1 61
13. 16
EXERCISE 2E
Very-Short-Answer Questions
1. Let A and B be two sets such that ( ) , ( )n A n B5 3 and ( ) .n A B 2+
(i) ( )n A B, (ii) ( )n A B# (iii) {( ) ( )}n A B B A# + #
Hint (i) ( ) ( ) ( ) ( ) .n A B n A n B n A B, +
(ii) ( ) ( ) ( )n A B n A n B# $
(iii) If ( )n A B m+ then {( ) ( )} .n A B B A 2m# #+
2. Find a and b when ( , ) ( , ) .a b a b2 13 7 2 3
3. If { , },A 1 2 fi nd .A A A# #
4. If { , , } { , }andA B2 3 4 4 5 , draw an arrow diagram to represent ( ) .A B#
5. If { , }, { , }A B3 4 4 5 and { , },C 5 6 fi nd ( ) .A B C# #
6. If ,A B3 prove that .A C B C# #
7. Prove that .A B B A A B&# #
8. If { }A 5 and { , },B 5 6 write down all possible subsets of .A B#
9. Let {( , ) :R x x x2 is a prime number less than 10}.
(i) Write R in roster form.
(ii) Find dom (R) and range (R).
10. Let { , , }A 1 2 3 and { } .B 4
How many relations can be defi ned from A to B?
11. Let { , , , }A 3 4 5 6 and {( , ) : , } .andR a b a b A a b d
Write R in roster form.
Find: dom (R) and range (R).
Write R 1 in roster form.
12. Let {( , ) : , , } .R a b a b N a b d
Show that R is a binary relation which is neither refl exive, nor symmetric.
Show that R is transitive.
ANSWERS (EXERCISE 2E)
1. (i) 6 (ii) 15 (iii) 4 2. ,a b5 1
3. {( , , ), ( , , ), ( , , ), ( , , ), ( , , ), ( , , ), ( , , ), ( , , )}1 1 1 1 1 2 1 2 1 1 2 2 2 1 1 2 1 2 2 2 1 2 2 2
4.
B
2
3
4
A
54
A B#
76 Senior Secondary School Mathematics for Class 11
5. {( , , ), ( , , ), ( , , ), ( , , ), ( , , ), ( , , ), ( , , ), ( , , )}3 4 5 3 4 6 3 5 5 3 5 6 4 4 5 4 4 6 4 5 5 4 5 6
8. , {( , )}, {( , )}, {( , ), ( , )}5 5 5 6 5 5 5 6
9. (i) {( , ), ( , ), ( , ), ( , )}R 2 4 3 9 5 25 7 49
(ii) dom ( ) { , , , }R 2 3 5 7 and range ( ) { , , , }R 4 9 25 49
10. 8
11. {( , ), ( , ), ( , ), ( , ), ( , ), ( , )R 4 3 5 3 5 4 6 3 6 4 6 5
(i) dom ( ) { , , }R 4 5 6 and range ( ) { , , }R 3 4 5
{( , ), ( , ), ( , ), ( , ), ( , ), ( , )}R 3 4 3 5 4 5 3 6 4 6 5 61
SUMMARY OF KEY FACTS
1. Two numbers a and b listed in a specifi c order and enclosed in parentheses
form an ordered pair ( , ) .a b
Here a is the fi rst component and b is the second component.
In general, ( , ) ( , ) .a b b a!
2. ( , ) ( , ) .anda b c d a c b d+
3. Let A and B be two nonempty sets. Then, the Cartesian product of A and B
is defi ned as
{( , ) : } .andA B a b a A b B# d d
If orA B then .A B#
4. If ( ) ( )andn A p n B q then ( ) .n A B pq#
5. Three numbers a, b, c listed in a specifi c order and enclosed in parentheses
form an ordered triplet ( , , ) .a b c
( , , ) ( , , ) ( , , ),a b c b a c c a b! ! etc.
6. For any nonempty sets A, B, C, we have
( ) ( ),A B C A B C# # # # each denoted by .A B C# #
7. For any sets A, B and C, we have:
(i) ( ) ( ) ( )A B C A B A C# , # , #
(ii) ( ) ( ) ( )A B C A B A C# + # + #
(iii) ( ) ( ) ( )A B C A B A C# # #
(iv) ( ) ( ) ( ) ( ) ( ) ( )A B B A A B B A A B A B# + # + # + + # +
(v) A B A C B C&# #
(vi) ( ) ( ) ( )A B A A A B B A& # # + #1 1
(vii) ( ) ( )A B A C B C& # #1 1
(viii) ( ) ( )andA B C D A C B D& # #1 1 1
(ix) A B B A A B+# #
8. Let A and B be two nonempty sets and let .R A B#3
Then, R is called a relation from A to B.
If ( , ) ,a b Rd we say that ‘a is related to b‘ and we write, a R b.
Relations 77
If ( , ) ,a b Rz we say that 'a is not related to b‘ and we write, .a R bY
Dom ( ) { : ( , ) },R a a b R d range ( ) { : ( , ) } .R b a b R d
9. We defi ne, {( , ) : ( , ) }R b a a b R1 d ,
dom (R) = range ( )R 1 and range (R) = dom ( )R 1 .
10. Let A be a nonempty set. Then, every subset of A A# is called a binary
relation on A.
11. Let A be a nonempty set and R be a binary relation on A.
Then, R is said to be:
(i) refl exive, if ( , )a a Rd for all ;a Ad
i.e., a R a for all .a Ad
(ii) symmetric, if ( , ) ( , )a b R b a R&d d for all , ;a b Ad
i.e., a R b b R a& for all , .a b Ad
(iii) transitive, if ( , )a b Rd and ( , ) ( , )b c R a c R&d d for all , , ;a b c Ad
i.e., a R b and b R c a R c& for all , , .a b c Rd
12. A relation R on A, which is refl exive, symmetric and transitive, is called an
equivalence relation on A.
78 Senior Secondary School Mathematics for Class 11
3
Functions
FUNCTION Let X and Y be two nonempty sets. Then, a relation f from X to Y is called
a function, if every element in X has a unique image in Y, and we write, : .f X Y"
Thus, a relation f from X to Y is a function, if dom ( )f X and no two
distinct ordered pairs in f have the same first coordinate.
If ( , ) ,x y fd we write, ( ) .f x y
Here, y is called the image of x under f and x is called the pre-image of y.
If :f X Y" then dom ( )f X and range ( ) .f Y3
Also, Y is called the co-domain of f.
IMAGE OF A SUBSET
Let :f X Y" and let .A Xf
Then, the image of A under f is defined as
( ) { ( ) : } .f A f x x A d
Clearly, ( ) .f A Y3
EXAMPLE 1 Let { , , , } { , , , , } .X and Y1 2 3 4 1 4 9 16 25
Let {( , ) : , } .f x y x X y Y and y x2 d d
(i) Show that f is a function from X to Y. Find its domain and range.
(ii) Draw a pictorial representation of the above function.
(iii) If { , , },A 2 3 4 fi nd ( ) .f A
SOLUTION (i) We have, {( , ) : , } .andf x y x X y Y y x2 d d
Giving different values to x from the set X and getting the
corresponding value of ,y x2 weget
{( , ), ( , ), ( , ), ( , )} .f 1 1 2 4 3 9 4 16
Clearly, every element in X has a unique image in Y.
Hence, f is a function from X to Y.
Dom ( ) { , , , } .f X1 2 3 4
Range ( ) { , , , } .f Y1 4 9 16 1
Clearly, Y25d does not have its pre-image in X.
(ii) A pictorial representation of the above mapping f is given
below.
78
Functions 79
(iii) Now, let { , , } .A 2 3 4 Then,
( ) , ( ) ( ) .andf f f2 2 4 3 3 9 4 4 162 2 2
( ) { ( ) : } { , , } .f A f x x A 4 9 16 d
RELATION AS A PARTICULAR CASE OF A FUNCTION
A relation f from X to Y is a function if dom ( )f X and no two distinct ordered
pairs in f have the same fi rst coordinate.
If ( , ) ,x y fd we write, ( ) .f x y
EXAMPLE 2 Let { , , , } { , , , , , } .X and Y2 3 4 5 7 9 11 13 15 17
Defi ne a relation f from X to Y by:
{( , ) : , } .f x y x X y Y and y x2 3 d d
(i) Write f in roster form.
(ii) Find dom ( )f and range ( ) .f
(iii) Show that f is a function from X to Y.
SOLUTION Here { , , , }X 2 3 4 5 and .Y x2 3
Now, ( ) ,x y2 2 2 3 7& #
( ) ,x y3 2 3 3 9& #
( ) ,x y4 2 4 3 11& #
( ) .x y5 2 5 3 13& #
(i) {( , ), ( , ), ( , ), ( , )} .f 2 7 3 9 4 11 5 13
(ii) Clearly, dom ( ) { , , , }f 2 3 4 5 and range ( ) { , , , } .f Y7 9 11 13 1
(iii) It is clear that no two distinct ordered pairs in f have the same
fi rst coordinate.
̀ f is a function from X to Y.
EXAMPLE 3 Which of the following relations are functions? Give reasons. In case of a
function, fi nd its domain and range.
(i) {( , ), ( , ), ( , ), ( , )}f 1 3 1 5 2 3 2 5
(ii) {( , ), ( , ), ( , ), ( , )}g 2 1 5 1 8 1 11 1
(iii) {( , ), ( , ), ( , ), ( , ), ( , ), ( , )}h 2 1 4 2 6 3 8 4 10 5 12 6
SOLUTION (i) {( , ), ( , ), ( , ), ( , )} .f 1 3 1 5 2 3 2 5
Here, one element, namely 1 has two images 3 and 5 under f.
f is not a function.
2
X
1
4
Y
3
4
1
16
25
9
f
80 Senior Secondary School Mathematics for Class 11
(ii) {( , ), ( , ), ( , ), ( , )} .g 2 1 5 1 8 1 11 1
Clearly, no two distinct ordered pairs in g have the same fi rst
coordinate. So, g is a function.
dom ( ) { , , , }g 2 5 8 11 and range { } .g 1
(iii) {( , ), ( , ), ( , ), ( , ), ( , ), ( , )} .h 2 1 4 2 6 3 8 4 10 5 12 6
Clearly, no two distinct ordered pairs in h have the same fi rst
coordinate. So, h is a function.
dom ( ) { , , , , , }h 2 4 6 8 10 12 and range ( ) { , , , , , } .h 1 2 3 4 5 6
EXAMPLE 4 Let { , , , , }A 2 1 0 1 2 and : : ( ) .f A Z f x x x2 32"
Find (i) range (f ) (ii) pre-images of 6, –3 and 5.
SOLUTION We have
( ) ( ) ( ) ( ) ;f 2 2 2 2 3 4 4 3 52 #
( ) ( ) ( ) ;f 1 1 2 1 3 02 #
( ) ;f 0 3
( ) ( )f 1 1 2 3 4 and ( ) ( ) .f 2 2 2 2 3 32 #
{( , ), ( , ), ( , ), ( , ), ( , )} .f 2 5 1 0 0 3 1 4 2 3
(i) Range ( ) { , , , } .f 5 0 3 4
(ii) ( )f x x x6 2 3 62&
( ) .x x x A2 9 0 2
2 4 36
1 102& &
!
! z
6 has no pre-image in A.
( )f x x x3 2 3 32&
( ) .orx x x x x x2 0 2 0 0 22& & &
Clearly, , .A0 2d
So, the pre-images of –3 are 0 and 2.
( )f x x x5 2 3 52&
x x x x x2 8 0 4 2 8 02 2& &
( ) ( ) ( )( )x x x x x4 2 4 0 4 2 0& &
.orx x2 4&
Now, A2d but .A4z
So, the pre-image of 5 is –2.
EQUAL FUNCTIONS
Two functions f and g are said to be equal if
(i) ( ) ( )dom f dom g (ii) co-domain of f co-domain of g
(iii) ( ) ( )f x g x for every x in their common domain.
EXAMPLE 5 Let { , }A 1 2 and { , }B 3 6 and f and g be functions from A to B, defi ned
by ( )f x x3 and ( ) .g x x 22 Show that .f g
Functions 81
SOLUTION Clearly, we have
( ) ( ) { , } .dom domf g 1 2
Co-domain of f co-domain of { , } .g 3 6
Also, ( ) ( ) , ( ) ( ) ;f f1 3 1 3 2 3 2 6# #
( ) ( ) , ( ) ( ) .g g1 1 2 3 2 2 2 62 2
( ) ( ) ( ) ( ) .andf g f g1 1 2 2
Thus, ( ) ( )f x g x for all .x Ad
Hence, .f g
EXAMPLE 6 Let : : ( )f Z Z f x x2" and : : ( ) | |g Z Z g x x 2" for all .x Zd Show
that .f g
SOLUTION We have
( ) ( )dom domf g Z
and co-domain of f co-domain of .g Z
Also, for all ,x Zd we have
( ) ( ) | | .andf x x g x x x2 2 2
( ) ( )f x g x for all .x Zd
Hence, .f g
EXAMPLE 7 Let : : ( )f R R f x x 2" and : { } : ( )g R R g x x
x2 2
4 ·
2
"
Show that .f g! Re-defi ne f and g such that .f g
SOLUTION We have, ( )dom f R and ( ) { } .dom g R 2
Since ( ) ( ),dom domf g! so we have .f g!
For every real number ,x 2! we have
( ) ( )
( )( )
( )g x x
x
x
x x
x2
4
2
2 2
2
2
[ ( ) ]x 2 0a ! .
Thus, ( ) ( )f x g x for all { } .x R 2d
f g only when, they are re-defi ned as under:
: { } : ( )f R R f x x2 2" and : { } : ( )g R R g x x
x2 2
4 ·
2
"
EXAMPLE 8 Let {( , ), ( , ), ( , ), ( , )}f 1 3 0 1 1 1 2 3 be a function, described by the
formula, ( ) .f x x Then, fi nd the values of .and Also, fi nd the
formula.
SOLUTION Here ( ) .f x x … (i)
Also, ( ) , ( ) , ( ) ( ) .andf f f f1 3 0 1 1 1 2 3 [given]
Putting x 1 and ( )f 1 3 in (i), we get
.3 3& … (ii)
Putting ( )andx f0 0 1 in (i), we get
.0 1 1&# … (iii)
Putting 1 from (iii) in (ii), we get .2
.and2 1
Hence, ( )f x x2 1 is the required formula.
82 Senior Secondary School Mathematics for Class 11
EXAMPLE 9 Let : : ( ) .f R R f x x 32"
Find the pre-images of each of the following under f:
(i) 19 (ii) 28 (iii) 2
SOLUTION Given: ( ) .f x x 32
(i) Let x be the pre-image of 19. Then,
( ) .f x x x x19 3 19 16 42 2& & & !
4 and –4 are the pre-images of 19.
(ii) Let x be the pre-image of 28. Then,
( ) .f x x x x28 3 28 25 52 2& & & !
5 and –5 are the pre-images of 28.
(iii) Let x be the pre-image of 2. Then,
( ) .f x x x2 3 2 12 2& &
But, no real value of x satisfi es the equation, .x 12
2 does not have any pre-image under f.
INVERSE OF A FUNCTION
Let :f X Y" and let .y Yd
Then, we define, ( ) { : ( ) }f y x X f x y1 d set of pre-images of y.
f 1 is called the inverse of f.
REMARK In above example 9, we have
(i) { } { , }f 19 4 41 (ii) { } { , }f 28 5 51 (iii) { }f 21 .
EXAMPLE 10 Let : : ( ) .f R R f x x 12"
Find (i) { }f 41 (ii) { }f 101 (iii) { , }f 5 171 .
SOLUTION It is given that ( ) .f x x 12
(i) Let ( ) .f x41 Then,
( ) .f x x x4 1 4 52 2& &
But, there is no real value of x whose square is –5.
{ } .f 41
(ii) Let ( ) .f x101 Then,
( ) .f x x x x10 1 10 9 32 2& & & !
{ } { , } .f 10 3 31
(iii) Let ( ) .f x51 Then,
( ) .f x x x x5 1 5 4 22 2& & & !
{ } { , }f 5 2 21 .
Let ( ) .f x171 Then,
( ) .f x x x x17 1 17 16 42 2& & & !
{ } { , } .f 17 4 41
Hence, { , } { , , , } .f 5 17 2 2 4 41
Functions 83
EXAMPLE 11 Let : ,f R R" defi ned by
( ) ,
, .
f x if x Q
if x Q
1
1
d
z
Z
[
\
]
]]
]
]]
Find (i) f 2
1
b l (ii) ( . )f 0 34 (iii) ( )f 2 (iv) ( )f
(v) range ( )f (vi) { }f 11 (vii) { }f 11
SOLUTION Since each one of 2
1 and 0.34 is rational, we have
(i) f 2
1 1b l and (ii) ( . ) .f 0 34 1
Since each one of 2 and is irrational, we have
(iii) ( )f 2 1 and (iv) ( ) .f 1
(v) range ( ) { ( ) : }f f x x R d
{ ( ) : } { ( ) : }f x x Q f x x R Q, d d { , } .1 1
(vi) { } { : ( ) } .f x f x Q1 11
(vii) { } { : ( ) } ( ) .f x f x R Q1 11
EXAMPLE 12 Let : ,f R R" defi ned by
( )
,
,
, .
f x
x x
x
x x
3 2 0
1 0
4 1 0
Z
[
\
]
]
]
]]
]
]
]
]
Find (i) ( )f 2 (ii) ( )f 2 (iii) ( )f 0(iv) ( . ) .f 3 5
SOLUTION Clearly, we have
(i) ( ) ( )f 2 4 2 1 9# [ ]2 0a .
(ii) ( ) { ( ) }f 2 3 2 2 8# [ ]2 0a .
(iii) ( ) .f 0 1
(iv) ( . ) ( . )f 3 5 4 3 5 1 15# [ . ]3 5 0a .
EXAMPLE 13 Consider the relation
( ) ,
, .
f x x x
x x
0 2
3 2 10
2 # #
# #
Z
[
\
]
]]
]
]]
Show that f is not a function.
SOLUTION We have, dom ( ) [ , ] .f R0 10 1
Now, ( )f 2 2 42 and also, ( ) ( ) .f 2 3 2 6#
As such, a single element, namely 2 has two distinct images,
namely 4 and 6.
f is not a function.
EXAMPLE 14 Consider the relation
( ) ,
, .
g x x x
x x
0 3
3 3 10
2 # #
# #
Z
[
\
]
]]
]
]]
Show that g is a function.
84 Senior Secondary School Mathematics for Class 11
SOLUTION We have, dom ( ) [ , ] .g R0 10 1
It is easy to verify that every element in [0, 10] is associated with a
unique real number under g.
In particular ( )g 3 3 92 and also ( ) ,g 3 3 3 9# i.e., 3 has a
unique image, namely 9.
g is a function from [0, 10] into R.
EXERCISE 3A
1. Defi ne a function as a set of ordered pairs.
2. Defi ne a function as a correspondence between two sets.
3. What is the fundamental difference between a relation and a function? Is
every relation a function?
4. Let { , , , }, { , , , , , }X Y1 2 3 4 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 X to Y (ii) f is a function from X to Y. Justify your
answer in each case.
5. Let { , , , , }X 1 0 3 7 9 and : : ( ) .f X R f x x 13"
Express the function f as set of ordered pairs.
6. Let { , , , }A 1 0 1 2 and { , , , } .B 2 3 4 5 Find which of the following are
functions from A to B. Give reason.
(i) {( , ), ( , ), ( , ), ( , )}f 1 2 1 3 0 4 1 5
(ii) {( , ), ( , ), ( , )}g 0 2 1 3 2 4
(iii) {( , ), ( , ), ( , ), ( , )}h 1 2 0 3 1 4 2 5
7. Let { , }A 1 2 and { , , } .B 2 4 6 Let {( , ) : ,f x y x A y B d d and } .y x2 1
Write f as a set of ordered pairs.
Show that f is a relation but not a function from A to B.
8. Let { , , }A 0 1 2 and { , , , } .B 3 5 7 9 Let {( , ) : , } .andf x y x A y B y x2 3 d d
Write f as a set of ordered pairs.
Show that f is a function from A to B.
Find dom ( )f and range ( ) .f
9. Let { , , , } { , , , , } .andA B2 3 5 7 3 5 9 13 15
Let {( , ) : , } .andf x y x A y B y x2 1 d d
Write f in roster form. Show that f is a function from A to B.
Find the domain and range of f.
10. Let {( , ), ( , ), ( , ), ( , ), ( , ), ( , )} .g 1 2 2 5 3 8 4 10 5 12 6 12
Is g a function? If yes, fi nd its domain and range.
If no, give reason.
11. Let {( , ), ( , ), ( , ), ( , ), ( , )}f 0 5 1 2 2 1 3 4 4 7 be a linear function from Z into Z.
Write an expression for f.
Hint Let ( ) .f x ax b
Functions 85
12. If ( ) ,f x x3 fi nd the value of ( )
{ ( ) ( )}f f
5 1
5 1
·
13. If ( ) ,f x x2 fi nd the value of ( . )
{ ( . ) ( )}f f
1 1 1
1 1 1
·
14. Let { , , , , , }A 12 13 14 15 16 17 and : : ( )f A Z f x" highest prime factor of x.
Find range ( ) .f
15. Let R be the set of all positive real numbers.
Let : : ( ) .logf R R f x xe"
Find (i) range ( )f (ii) { : ( ) } .andx x R f x 2 d
(iii) Find out whether ( ) ( ) ( )f xy f x f y for all , .x y Rd
16. Let : : ( ) .f R R f x 2x"
Find (i) range ( )f (ii) { : ( ) }x f x 1 .
(iii) Find out whether ( ) ( ) · ( )f x y f x f y for all , .x y Rd
17. Let : : ( )f R R f x x2" and : : ( ) ,g C C g x x2" where C is the set of all
complex numbers.
Show that .f g!
18. f, g and h are three functions defi ned from R to R as follows:
(i) ( )f x x2 (ii) ( )g x x 12 (iii) ( ) sinh x x
Then, fi nd the range of each function.
19. Let : : ( ) .f R R f x x 12" Find (i) { }f 101 (ii) { } .f 31
20. The function ( )F x x5
9 32 is the formula to convert x C to Fahrenheit
units. Find
(i) ( ),F 0 (ii) ( ),F 10 (iii) the value of x when ( ) .f x 212
Interpret the result in each case.
Hints (i) % %( ) .C Ff 0 5
9 0
32 32 0 32&
# ' 1
(ii) % %( )
( )
( ) .C FF 10 5
9 10
32 14 10 14&
#
( 2
(iii) % %( ) .F CF x x x212 5
9
32 212 100 212 100& & &
ANSWERS (EXERCISE 3A)
4. (i) Yes (ii) No 5. {( , ), ( , ), ( , ), ( , ), ( , )}f 1 0 0 1 3 28 7 344 9 730
6. (i) No, since one element, namely –1, has two different images.
(ii) No, since dom ( )g A! .
(iii) Yes, since each element in A has a unique image in B.
7. {( , ), ( , ), ( , )}f 1 4 1 6 2 6
8. {( , ), ( , ), ( , )},f 0 3 1 5 2 7 dom ( ) { , , },f 0 1 2 range ( ) { , , }f 3 5 7
9. {( , ), ( , ), ( , ), ( , )},f 2 3 3 5 5 9 7 13 dom ( ) { , , , },f 2 3 5 7 range ( ) { , , , }f 3 5 9 13
86 Senior Secondary School Mathematics for Class 11
10. Yes, dom ( ) { , , , , , }f 1 2 3 4 5 6 and range ( ) { , , , , }f 2 5 8 10 12
11. ( )f x x3 5 12. 31 13. 2.1 14. range ( ) { , , , , , }f 3 13 7 5 2 17
15. (i) R (ii) { }e 2 (iii) Yes 16. (i) R (ii) {0} (iii) Yes
18. (i) { : }R x R x 0$ d (ii) { : }x R x 1$d (iii) { : }x R x1 1# #d
19. (i) {–3, 3} (ii) 20. (i) 32 (ii) 14 (iii) x 100
HINTS TO SOME SELECTED QUESTIONS
4. (i) Since ,f X Y#1 so f is a relation from X to Y.
(ii) Dom ( )f X and range ( ) { , , , } .f Y1 5 9 11 1
But, two different ordered pairs namely (2, 9) and (2, 11) have the same fi rst
coordinate. So, f is not a function.
6. (i) No, since two different ordered pairs (–1, 2) and (–1, 3) have the same fi rst
coordinate.
(ii) No, since dom ( ) { , , } .g A0 1 2 !
(iii) Yes, dom ( )h A and no two different ordered pairs have the same fi rst coordinate.
11. ( ) .f a b b0 5 0 5 5& &#
( ) .f a b a a1 2 1 2 5 2 3& & &#
12.
( )
{ ( ) ( )} {( ) } ( )
.
f f
5 1
5 1
4
5 1
4
125 1
4
124 31
3 3
13.
( . )
{ ( . ) ( )}
( . )
{( . ) }
( . ) . .
f f
1 1 1
1 1 1
1 1 1
1 1 1
1 1 1 2 1
2 2
14. Highest prime factors of 12, 13, 14, 15, 16, 17 are 3, 13, 7, 5, 2, 17 respectively.
15. (i) For every ,x Rd we have .log x Re So, range ( ) .f R
(ii) ( ) .logf x x x e2 2e
2& & So, { : ( ) } { } .andx x R f x e2 2 d
(iii) ( ) ( ) ( ) ( ) .log log logf xy xy x y f x f ye e e
16. (i) ( )f x 2 0x for every .x Rd
If ,x Rd there exists log x2 such that ( ) .logf x x2
log x
2
2
range ( ) .f R
(ii) ( ) .f x x1 2 1 2 0x 0& & So, { : ( ) } { } .x f x 1 0
(iii) ( ) ( ) ( ) .f x y f x f y2 2 2x y x y# $
17. Dom ( )f R and dom ( ) ( ) ( )dom domg C f g& !
18. (i) Clearly, ( )f x x 02$ for all .x Rd
(ii) ( )f x x 1 12 $ for all .x Rd
(iii) [ , ] .sin x 1 1d
19. (i) x x x1 10 9 32 2& & ! .
(ii) x x1 3 42 2& and there is no real number whose square is –4.
REAL FUNCTIONS
REAL-VALUED FUNCTIONS A function which has either R or one of its subsets as its
range, is called a real-valued function.
Functions 87
REAL FUNCTIONS A function :f X Y" is called a real function, if x R3 and .y R3
In general, real functions are described as general expressions or formulae,
without mentioning their domains and co-domains. Following are some
examples of real functions.
SOLVED EXAMPLES
EXAMPLE 1 If ( ) ,f x x x3 5 103 2 fi nd ( ) .f x 1
SOLUTION We have, ( ) .f x x x3 5 103 2 … (i)
Replacing x by ( )x 1 in (i), we get
( ) ( ) ( )f x x x1 3 1 5 1 103 2
{ ( )} ( )x x x x x3 1 3 1 5 2 1 103 2
( ) ( )x x x x x3 3 3 1 5 2 1 103 2 2
.x x x3 14 19 23 2
( ) .f x x x x1 3 14 19 23 2
EXAMPLE 2 If ,( )f x x x
1 show that ·{ ( )} ( )f x f x f x3
13 3 b l
SOLUTION We have, ·( )f x x x
1 … (i)
On cubing both sides of (i), we get
{ ( )}f x x
x
x x x x
1 3 1 13 3 3 # # # b l
x
x x
x1 3 13 3 d bn l
( )f x f x3
13 b l f x x
x
x x
1 1
1
1 1
a ·b bl l
R
T
S
S
S
S
S
SS
V
X
W
W
W
W
W
WW
Z
[
\
]
]]
]
]]
_
`
a
b
bb
b
bb
Hence, ·{ ( )} ( )f x f x f x3
13 3 b l
EXAMPLE 3 If ,( )f x x
x x1
1 1!
then show that ,{ ( )}f f x x
1 where .x 0!
SOLUTION We have, ,( )f x x
x
1
1
where .x 1!
{ ( )}f f x f x
x
x
x
x
x
1
1
1
1 1
1
1 1
b l
'
'
1
1
·( )
{( ) ( )}
{( ) ( )}
( )
x
x x
x x
x
x x1
1 1
1 1
1
2
2 1
#
Hence, ,{ ( )}f f x x
1 where .x 0!
EXAMPLE 4 If ( )y f x
bx a
ax b
and a b2 2! then prove that ( ) .x f y
88 Senior Secondary School Mathematics for Class 11
SOLUTION We have, ·( )y f x
bx a
ax b
( ) { ( )}f y f f x f ax a
ax b
b
bx a
ax b a
a
bx a
ax b b
b
d
d
l
n
n
)
)
3
3
( )
{( ) ( )}
{( ) ( )}
( )
bx a
a x ab b x ab
abx b abx a
bx a2 2
2 2#
( )
( )
( )
( )
.
a b
a x b x
a b
a b x
x2 2
2 2
2 2
2 2
Hence, ( ) .x f y
EXAMPLE 5 If ( )f x x
x
1
1
then prove that ·( ) ( )
( )
f x
f x
f x
2 3
3 1
SOLUTION We have, ·( )f x x
x
1
1
( )
( )
( )
( )f x
f x
x
x
x
x
3
3 1
1
1
3
3 1
1 1
b l
( )
( ) ( )
( ) ( )
( )
( ) .
x
x x
x x
x
x
x f x1
3 3 1
1 3 3
1
2 1
2 1 2#
Hence, ·( ) ( ) )
( )
f x
f x
f x
2 3
3 1
EXAMPLE 6 Let the functions f and g be defi ned by
,
, .
x
x when x
k when x
3
9 3
3
2
!
( ) ( ) ( )f x x and g x3
Z
[
\
]
]
]
]]
]
]
]
]
Find the value of k such that ( ) ( )f x g x for all .x Rd
SOLUTION We have, ( ) ( )f x g x for all .x Rd
( ) ( )f g3 3
& ( ) k3 3 [ ( ) ( ) ( ) ]andf g k3 3 3 3a
& .k 6
Hence, .k 6
EXAMPLE 7 If ,( )af x bf x x
1 1 b l where a b! and x 0! , fi nd ( ) .f x
SOLUTION We have, ·( )af x bf x x
1 1 b l … (i)
Replacing x by x
1 in (i), we get
( ) .af x bf x x
1 b l … (ii)
Functions 89
Adding (i) and (ii), we get
( ) ( )a f x f x b f x f x x x
1 1 1 b b bl l l( (2 2
& ( ) ( )a b f x f x x x
1 1 b bl l( 2
& ·( ) ( )f x f x a b x x
1 1 1 b bl l … (iii)
On subtracting (ii) from (i), we get
( ) ( )a f x f x b f x f x x x
1 1 1 b b bl l l( (2 2
& ( ) ( )a b f x f x x x
1 1 b bl l( 2
& ·( ) ( )f x f x a b x x
1 1 1 b bl l( 2 … (iv)
On adding (iii) and (iv), we get
( ) ( ) ( ) ( ) ( )f x a b
x
a b
x
a b x a b x
2 1 1 ( (2 2
( )
( ) ( )
( )
( ) ( )
a b
a b x a b x
a b x
a b a b
2 2 2 2
·
( ) ( )a b x
bx d
a b x
a bx2 2 22 2
2
2 2
a k
Hence, ·( )
( )
f x
a b x
a bx12 2
a k
EXERCISE 3B
1. If ( )f x x x3 42 and ( ) ( ),f x f x2 1 fi nd the values of x.
2. If ( )f x x
x
1
1
then show that (i) ( )f x f x
1 b l (ii) ·
( )f x f x
1 1 b l
3. If ( )f x x
x
13
3
then show that ( ) .f x f x
1 0 b l
4. If ( )f x x
x
1
1
then show that { ( )} .f f x x
5. If ( ) ( )f x x2 1
1 and x 2
1
!
then prove that ,{ ( )}f f x x
x
2 3
2 1
when it is
given that ·x 2
3
!
6. If ( ) ( )f x x1
1 then show that [ { ( )}] .f f f x x
7. If ( )
( )
f x
x
x
1
2
2
then show that ( ) .tan sinf 2
8. If ,( )y f x x
x
5 3
3 1
prove that ( ) .x f y
90 Senior Secondary School Mathematics for Class 11
ANSWERS (EXERCISE 3B)
1. orx x1 3
2
HINTS TO SOME SELECTED QUESTIONS
1. ( ) ( ) ( ) ( )f x f x x x x x2 1 3 4 2 1 3 2 1 42 2&
( ) ( ) .x x x x x x x x3 4 4 2 2 3 2 0 1 3 2 02 2 2& & &
5. ·( ) ( ) { ( )}
( )
f x
x
f f x
x
x
x
2 1
1
2 2 1
1 1
1
2 3
2 1
&
#
( 2
6. ( ) ( ) { ( )}
( )
f x
x
f f x
x
x
x
x
x
1
1
1 1
1
1 1 1
&
( 2
[replace x by
( )]x1
1
[ { ( )}]
( )
( )
( )
( )
.f f f x
x
x
x
x x x
1
1
1
1 1
1 1
1
& #
( 2
7. ( )
( )
· .tan
tan
tan
sec
tan
cos
sin cos sin cos sinf
1
2 2 2 2 22 2
2#
8. ( ) .f y y
y
x
x
x
x
x x
x x x x5 3
3 1
5 5 3
3 1 3
3 5 3
3 1 1
15 5 15 9
9 3 5 3
14
14
c
c
m
m
PROBLEMS BASED ON DOMAINS AND RANGES OF REAL FUNCTIONS
SOLVED EXAMPLES
EXAMPLE 1 Find the domain and the range of the real function, ·( )f x x 3
1
SOLUTION We have, ·( ) ( )f x x 3
1
Clearly, ( )f x is defi ned for all real values of x except that at which
,x 3 0 i.e., .x 3
dom ( ) { } .f R 3
Let ( ) .y f x Then,
y x x y3
1 3 1&
·x y
1 3& d n … (i)
It is clear from (i) that x assumes real values for all real values of y
except .y 0
range ( ) { } .f R 0
Hence, dom ( ) { }f R 3 and range ( ) { } .f R 0
Functions 91
EXAMPLE 2 Find the domain and the range of the real function, ·( )f x x
x
5
3
SOLUTION We have, ·( )f x x
x
5
3
Clearly, ( )f x is defi ned for all real values of x except that at which
,x 5 0 i.e., .x 5
dom ( ) { } .f R 5
Let ( ) .y f x Then,
y x
x xy y x5
3 5 3&
·( )x y y x y
y
1 5 3 1
5 3
& &
… (i)
It is clear from (i) that x is not defi ned when ,y 1 0 i.e., when
.y 1
range ( ) { } .f R 1
Hence, dom ( ) { }f R 5 and range ( ) { } .f R 1
EXAMPLE 3 Find the domain and the range of the real function, ·( )f x
x
x
1
1
2
2
SOLUTION We have, ·( )f x
x
x
1
1
2
2
Clearly, ( )f x is defi ned for all real values of x except those for which
,x 1 02 i.e., .x 1!
dom ( ) { , } .f R 1 1
Let ( ) .y f x Then,
( ) ( )y
x
x x y y x x y y
1
1 1 1 12
2
2 2 2& &
·x y
y
x y
y
1
1
1
1
2& & !
… (i)
It is clear from (i) that x is not defi ned when y 1 0 or when
.y
y
1
1
0
Now, .y y1 0 1& … (ii)
And ( ) ( )and or andy
y
y y y y1
1
0 1 0 1 0 1 0 1 0 &
( ) ( )and or andy y y y1 1 1 1 &
.y1 1 & … (iii)
[ andy y1 1 a is not possible]
Thus, x is not defi ned when .y1 1 # [using (ii) and (iii)]
range ( ) ( , ] .f R 1 1
Hence, dom ( ) { , }f R 1 1 and range ( ) ( , ] .f R 1 1
92 Senior Secondary School Mathematics for Class 11
EXAMPLE 4 Find the domain of the real-valued function:
·( )f x
x x
x x
5 4
1
2
2
SOLUTION We have, ·( )f x
x x
x x
5 4
1
2
2
Clearly, ( )f x is defi ned for all real values of x except those at which
.x x5 4 02
But, ( )( ) .orx x x x x x5 4 0 1 4 0 1 42 & &
dom ( ) { , } .f R 1 4
EXAMPLE 5 Find the domain of each of the following real functions:
(i) ( )f x x 3 (ii) ( )g x x4 2 (iii) ·( )h x
x1
1
SOLUTION (i) We have, ( ) .f x x 3
Clearly, ( )f x is defi ned for all real values of x for which
,x 3 0$ i.e., .x 3$
dom ( ) [ , ) .f 3 3
(ii) We have, ( ) .g x x4 2
Clearly, ( )g x is defi ned for all real values of x for which
( ) .x4 02 $
But, ( ) ( )x x4 0 4 02 2&$ #
( ) ( )x x x4 0 2 2 02& &# #
[ , ] .x x2 2 2 2& &# # d
dom ( ) [ , ] .g 2 2
(iii) We have, ·( )h x
x1
1
Clearly, ( )h x is defi ned for all real values of x for which
( ) .x1 0
But, ( , ) .x x x x1 0 1 1 1 & & & 3 d
dom ( ) ( , ) .h 13
EXAMPLE 6 Find the domain of the function, ·( )f x x
x
3
1
1
2
SOLUTION We have, ·( )f x x
x
3
1
1
2
Clearly, ( )f x is defi ned for all real values of x for which
andx x3 0 1 02$
& ( )( )andx x x3 0 1 1 0#
& ( )and orx x x3 1 1 #
& ( ) ( )and or andx x x x3 1 3 1 # #
Functions 93
& ( ) ( )orx x1 1 3 #
& ( , ) ( , ]x 1 1 3,3 d .
dom( ) ( , ) ( , ] .f 1 1 3,3
EXAMPLE 7 Find the range of each of the following functions:
(i) ( ) ,f x x x R and x2 3 0 d (ii) ( ) , .g x x x R22 d
SOLUTION (i) We have, ( ) ,f x x2 3 where .andx R x 0d
Now, x x x0 3 0 3 0 & &
x x3 2 0 2 2 3 2 & &
( ) ( ) ( , ) .f x f x2 2& & 3d
Hence, range ( ) ( , ) .f 23
(ii) We have, ( ) , .g x x x R22 d
Now, x R x x0 2 0 22 2& &$ $ d
( )x g x2 2 22& &$ $
( ) [ , ) .g x 2& 3d
Hence, range ( ) [ , ) .g 2 3
EXAMPLE 8 Find the domain and the range of the function, ·( )f x x
x
3
2
SOLUTION We have, ·( )f x x
x
3
2
Clearly, ( )f x is defi ned for all real values of x for which ,x 3 0!
i.e., .x 3!
dom ( ) { } .f R 3
Let ( ) .y f x Then,
y x
x xy y x3
2 3 2&
( )x y y1 3 2&
·x y
y
1
3 2
&
… (i)
It follows from (i) that x assumes real values for all y except that for
which ,y 1 0 i.e., .y 1
range ( ) { } .f R 1
Hence, dom ( ) { }f R 3 and range ( ) { } .f R 1
EXAMPLE 9 Find the domain and the range of the real function, ( ) .f x x 3
SOLUTION We have, ( ) .f x x 3
Clearly, ( )f x is defi ned for all real values of x for which ,x 3 0$
i.e., .x 3$
dom ( ) [ , ) .f 3 3
Also, ( ) .x f x x3 3 0&$ $
range ( ) [ , ) .f 0 3
Hence, dom ( ) [ , )f 3 3 and range ( ) [ , ) .f 0 3
94 Senior Secondary School Mathematics for Class 11
EXAMPLE 10 Find the domain and the range of each of the functions given below.
(i) ( ) | |f x x 1 (ii) ( ) | |g x x
SOLUTION (i) We have, ( ) | |.f x x 1
Clearly, ( )f x is defi ned for all .x Rd So, dom ( ) .f R
For all ,x Rd we have
| | ( ) .x f x1 0 0&$ $
range ( ) [ , ) .f 0 3
Hence, dom ( )f R and range ( ) [ , ) .f 0 3
(ii) We have, ( ) | |.g x x
Clearly, ( )g x is defi ned for all .x Rd So, dom ( ) .g R
For all ,x Rd we have
| | | | ( ) .x x g x0 0 0& &$ # #
range ( ) ( , ] .g 03
Hence, dom ( )g R and range ( ) ( , ] .g 03
EXAMPLE 11 Find the domain and the range of the real function, ·( )f x
x 2
1
SOLUTION We have, ·( )f x
x 2
1
Clearly, ( )f x is defi ned for all real values of x for which ,x 2 0
i.e., .x 2
dom ( ) ( , ) .f 2 3
For any real value of ,x 2 we have
x x x2 2 0 2 0 & &
( ) .
x
f x
2
1 0 0 & &
range ( ) ( , ) .f 0 3
Hence, dom ( ) ( , )f 2 3 and range ( ) ( , ) .f 0 3
EXAMPLE 12 Find the domain and the range of the real function, ( ) .f x x9 2
SOLUTION We have, ( ) .f x x9 2
Clearly, ( )f x is defi ned for all real values of x for which ( ) .x9 02 $
And, ( )( )x x x x9 0 9 0 3 3 02 2& &$ # #
[ , ] .x x3 3 3 3& &# # d
dom ( ) [ , ] .f 3 3
Let ( ) .y f x Then,
.y x y x x y x y9 9 9 92 2 2 2 2 2& & &
Clearly, x will take real values only when .y9 02$
Now, ( )( )y y y y9 0 9 0 3 3 02 2& &$ # #
[ , ]y y3 3 3 3& &# # d
Functions 95
[ , ]y 0 3& d { y x9 02a $ for all [ , ]}x 3 3d .
range ( ) [ , ] .f 0 3
Hence, dom ( ) [ , ]f 3 3 and range ( ) [ , ] .f 0 3
EXAMPLE 13 Find the domain and the range of the real function, ·( )
( )
f x
x
x
1 2
SOLUTION We have, ·( )
( )
f x
x
x
1 2
Clearly, ( )f x is defi ned for all .x Rd So, dom ( ) .f R
Let ( ) .y f x Then,
( )
y
x
x x y x y
1
02
2&
·x y
y
2
1 1 4 2
&
!
… (i)
It is clear from (i) that x will take real values, when
( ) andy y1 4 0 02 !$
& ( ) andy y4 1 0 02 !#
& ( )( ) andy y y2 1 2 1 0 0!#
& andy y y2
1
2
1 0 0!# b bl l
& andy y2
1
2
1 0!# #
& , { } .y 2
1
2
1 0
d; E
Also, .x y0 0&
range , ·( )f 2
1
2
1 ; E
Hence, dom ( )f R and range , ·( )f 2
1
2
1 ; E
EXAMPLE 14 Find the domain and the range of the real function, ·( )
( )
f x
x2
3
2
SOLUTION We have, ·( )
( )
f x
x2
3
2
Clearly, ( )f x is defi ned for all real values of x except those for which
,x2 02 i.e., .x 2!
dom ( ) { , } .f R 2 2
Let ( ) .y f x Then,
( )
y
x
y x y x y y
2
3 2 3 2 32
2 2& &
·x y
y
x y
y2 3 2 3
2& & !
… (i)
96 Senior Secondary School Mathematics for Class 11
It is clear from (i) that x will take real values only when .y
y2 3
0$
Now, ( ) ( )and or andy
y
y y y y
2 3
0 2 3 0 0 2 3 0 0 +$ # $
and or andy y y y2
3 0 2
3 0 + # $b bl l
( ) ory y0 2
3+ $b l
,( , ) ory y0 2
3
+ 3 3d d l;
, ·( , )y 0 2
3
+ ,3 3d l;
range , ·( ) ( , )f 0 2
3
,3 3 l;
Hence, dom ( ) { , }f R 2 2 and range , ·( ) ( , )f 0 2
3
,3 3 l;
EXAMPLE 15 Find the domain and the range of the real function , :f x
x
x x R
12
2
d
e o
Z
[
\
]
]]
]
]]
_
`
a
b
bb
b
bb
from R into R.
SOLUTION We have, ,( ) .f x
x
x x R
12
2
d
Clearly, x 1 02 ! for any real value of x.
( )f x is defi ned for all real values of x.
dom ( ) .f R
Let ( ) .y f x Then,
( )
( )y
x
x x y y x x y y
1
12
2
2 2 2& &
·( )x y
y
x
y
y
1 1
2& & ! … (i)
It is clear from (i) that x will take real values only when ( ) .y
y
1 0$
Now, ( ) ( ) ( )and or andy
y
y y y y1 0 0 1 0 0 1 0 +$ # $
( ) ( )and or andy y y y0 1 0 1 + # $
( )andy y0 1+ $
[ andy y0 1a # is not possible]
[ , ) .y 0 1+ d
range ( ) [ , ) .f 0 1
Hence, dom ( )f R and range ( ) [ , ) .f 0 1
EXAMPLE 16 Find the domain and the range of the function
·, : andf x
x
x R x
1
1 12 !! dd n) 3
Functions 97
SOLUTION We have, ,( )
( )
.f x
x
x R
1
1
2
d
Clearly, ( )f x is defi ned for all real values of x for which ( ) .x1 02 !
Now, ( ) ( )( ) .orx x x x x1 0 1 1 0 1 12 & &
Thus, ( )f x is defi ned for all values of x Rd except .1!
dom ( ) { , } .f R 1 1
Let ( ) .y f x Then,
y
x
y x y x y y
1
1 1 12
2 2& &
·x y
y
x y
y1 1
2& & !
… (i)
It is clear from (i) that x will take real values only when .y
y 1
0$
Now, ( ) ( )and or andy
y
y y y y
1
0 1 0 0 1 0 0 +$ # $
( ) ( )and or andy y y y1 0 1 0 + # $
( ) ( )ory y0 1+ $
( , ) [ , ) .ory 0 1+ 3 3d
range ( ) ( , ) [ , ) .f 0 1,3 3
Hence, dom ( ) { , }f R 1 1 and range ( ) ( , ) [ , ) .f 0 1,3 3
EXAMPLE 17 Find the domain and the range of the real function, ·( )f x x
x
5
252
SOLUTION We have, ·( )f x x
x
5
252
Clearly, ( )f x is defi ned for all real values of x for which ,x 5 0!
i.e., .x 5!
dom ( ) { } .f R 5
Let ( ) .y f x Then,
,y x
x y x5
25 5
2
&
when x 5 0!
,y x 5& when x 5!
.y y5 5 10& &! !
Then, y can be assigned any real value except 10.
range ( ) { } .f R 10
Hence, dom ( ) { }f R 5 and range ( ) { } .f R 10
EXAMPLE 18 Find the domain and the range of the real function, ·( )f x x
x
3
3
SOLUTION We have, ·( )f x x
x
3
3
98 Senior Secondary School Mathematics for Class 11
Clearly, ( )f x is defi ned for all real values of x for which ,x 3 0!
i.e., .x 3!
dom ( ) { } .f R 3
Let ( ) .y f x Then,
,y x
x y3
3 1&
when x 3 0!
,y 1& when .x 3!
range ( ) { } .f 1
Hence, dom ( ) { }f R 3 and range ( ) { } .f 1
EXAMPLE 19 Find the domain and the range of the real function, ·( ) ( )
| |
f x
x
x
3
3
SOLUTION We have, ·( ) ( )
| |
f x
x
x
3
3
Clearly, ( )f x is defi ned for all real values of x for which ,x 3 0!
i.e., .x 3!
dom ( ) { } .f R 3
Now, when ,x 3! we have
( ) , { | | ( )}
, { | | ( )}
when
when
f x x x x
x x x
1 3 0 3 3
1 3 0 3 3
a
a
Z
[
\
]
]]
]
]]
range ( ) { , } .f 1 1
Hence, dom ( ) { }f R 3 and range ( ) { , } .f 1 1
EXAMPLE 20 Find the domain of the real function, ·( )
| |
f x
x x
1
SOLUTION We have, ·( )
| |
f x
x x
1
Now, | | ,
,
when
when
x x x
x x
0
0
$
Z
[
\
]
]]
]
]]
& | | ,
,
when
when
x x x x x
x x x
0
0
$
Z
[
\
]
]]
]
]]
& | | ,
,
when
when
x x x x
x
2 0
0 0
$
Z
[
\
]
]]
]
]]
& | | ,x x 0 when x 0
& ( )
| |
f x
x x
1
assumes real values only when | |x x 0 and
this happens only when .x 0
dom ( ) ( , ) .f 0 3
EXAMPLE 21 Show that ( )
| |
f x
x x
1
is not defi ned for any .x Rd How will you
defi ne dom ( )f and range ( )?f
Functions 99
SOLUTION We have, ·( )
| |
f x
x x
1
Now, | | ,
,
when
when
x x x
x x
0
0
$
Z
[
\
]
]]
]
]]
& | | ,
,
when
when
x x x x x
x x x
0
0
$
Z
[
\
]
]]
]
]]
& | | ,
,
when
when
x x x
x x
0 0
2 0
$
Z
[
\
]
]]
]
]]
& | |x x 0# for all x Rd
&
| |x x
1
is not defi ned for any .x Rd
dom ( )f and range ( ) .f
GREATEST INTEGER FUNCTION (STEP FUNCTION)
The function : : ( ) [ ],f R R f x x" where [ ]x denotes the greatest integer less than or
equal to x, is called the Greatest Integer Function.
EXAMPLES [3.65] = 3, [3.02] = 3, [5] = 5, [–2.6] = –3, [–4] = –4.
NOTE [ ] [ ] .for and forx x x Z x x x R Z0 0 1 d d
EXAMPLE 22 Find the domain and the range of the real function, ·( )
[ ]
f x
x x
1
SOLUTION We have, ·( )
[ ]
f x
x x
1
We know that
[ ]
[ ] ,
[ ] .
for all
when
for all
x x x
x x x
x x x
0 0
0 0
0 0
Z
[
\
]
]
]
]]
]
]
]
]]
( )
[ ]
f x
x x
1
is defi ned only when [ ]x x 0 and this
happens only when .x 0
dom ( ) ( , ) .f 0 3
Let ( ) .y f x Then,
·
[ ]
[ ]y
x x
x x y
1 1
&
… (i)
Now, [ ] [ ] .x x x x x y y0 0 0
1 0 0 & & & &
range ( ) ( , ) .f 0 3
Hence, dom ( ) ( , )f 0 3 and range ( ) ( , ) .f 0 3
EXAMPLE 23 Find the domain and the range of the real function, ·( )
[ ]
f x
x x
1
100 Senior Secondary School Mathematics for Class 11
SOLUTION We have, ·( )
[ ]
f x
x x
1
We know that
[ ]x x0 1# for all x Rd
and [ ]x x 0 for all .x Zd
[ ]x x0 1 for all x R Zd
& ( )
[ ]
f x
x x
1
exists for all x R Zd
& dom ( ) .f R Z
Also, [ ]x x0 1 for all x R Zd
& [ ]x x0 1 for all x R Zd
&
[ ]x x
1 1 3
for all x R Zd
& ( )f x1 3 for all x R Zd
& range ( ) ( , ) .f 1 3
Hence, dom ( )f R Z and range ( ) ( , ) .f 1 3
EXERCISE 3C
1. Find the domain of each of the following real functions:
(i) ( )f x
x
x
9
3 5
2
(ii) ( )f x
x x
x
2
2 3
2
(iii) ( )f x
x x
x x
8 12
2 1
2
2
(iv) ( )f x
x
x
1
8
2
3
Find the domain and the range of each of the following real functions:
2. ( )f x x
1 3. ( ) ( )f x x 5
1 4. ( )f x x
x
2
3
5. ( )f x x
x
2
3 2
6. ( )f x x
x
4
162
7. ( )f x
x2 3
1
8. ( )f x
cx d
ax b
9. ( )f x x3 5 10. ( )f x x
x
3
5
11. ( )f x
x 1
1
2
12. ( ) | |f x x1 2 13. ( )
| |
f x x
x
4
4
14. ( )f x x
x
3
92
15. ( ) sinf x x2 3
1
ANSWERS (EXERCISE 3C)
1. (i) { , }R 3 3 (ii) { , }R 1 2 (iii) { , }R 2 6 (iv) { , }R 1 1
2. dom ( ) { },f R 0 range ( ) { }f R 0
3. dom ( ) { },f R 5 range ( ) { }f R 0
Functions 101
4. dom ( ) { },f R 2 range ( ) { }f R 1
5. dom ( ) { },f R 2 range ( ) { }f R 3
6. dom ( ) { },f R 4 range ( ) { }f R 8
7. dom ,( ) ,f R 2
3
3 b E range ( ) { }f R 0
8. dom ,( )f R c
d ' 1 range ( )f R c
a % /
9. dom , ,( )f 3
5
3 l; range ( ) [ , )f R 0 3
10. dom ( ) ( , ],f 3 5 range ( ) [ , )f 0 3
11. dom ( ) ( , ) ( , ),f 1 1,3 3 range ( ) { }f R 0
12. dom ( ) ,f R range ( ) ( , ]f 13
13. dom ( ) { },f R 4 range ( ) { , }f 1 1
14. dom ( ) { },f R 3 range ( ) { }f R 6
15. dom ( ) ,f R range ,( )f 3
1 1 ; E
HINTS TO SOME SELECTED QUESTIONS
2. ( )f x x
1 is not defi ned when .x 0 So, dom ( ) { } .f R 0
Let ( ) .y f x Then, ·y x x y
1 1
&
x is not defined when .y 0 So, range ( ) { } .f R 0
3. ( ) ( )f x x 5
1 is not defi ned when .x 5 So, dom ( ) { } .f R 5
Let ( ) .y f x Then, ·( )y x x y x y5
1 5 1 1 5& & d n
x is not defined when .y 0 So, range ( ) { } .f R 0
4. ( )f x x
x
2
3
is not defi ned when .x 2 So, dom ( ) { } .f R 2
Let ( ) .y f x Then,
y x
x y xy x xy x y2
3 2 3 2 3& &
·( )x y y x y
y
1 2 3 1
2 3
& &
x is not defined when .y 1 So, range ( ) { } .f R 1
5. ( )f x x
x
2
3 2
is not defi ned when .x 2 So, dom ( ) { } .f R 2
Let ( ) .y f x Then,
y x
x xy y x x xy y2
3 2 2 3 2 3 2 2& &
·( )x y y x y
y
3 2 2 3
2 2
& &
… (i)
It follows from (i) that x is not defined when .y 3
range ( ) { } .f R 3
102 Senior Secondary School Mathematics for Class 11
6. ( )f x x
x
4
162
is not defi ned when .x 4 So, dom ( ) { } .f R 4
Let ( ) .y f x Then,
( ),y x
x x4
16 4
2
when x 4!
& ( ) .y y4 4 8&! ! So, range ( ) { } .f R 8
7. ( )f x
x2 3
1
is not defi ned when ,x2 3 0# i.e., when ·x 2
3
#
dom ·( ) ,f R 2
3
3 c E
Let ( ) .y f x Then,
·
( ) ( )y x
y
x
x
y
x
y2 3
1
2 3
1 2 3 1 2
1 1 32 2 2& & &
e o
x is not defined when .y 0 So, range ( ) { } .f R 0
8. ( )f x
cx d
ax b
is not defi ned when ·cx d x c
d0 &
dom ·( )f R c
d ' 1
Let ( ) .y f x Then,
y
cx d
ax b cxy dy ax b cxy ax dy b& &
·( )x cy a dy b x cy a
dy b
& &
Clearly, x is not defined when ·cy a y c
a0 &
range ·( )f R c
a % /
9. ( )f x x3 5 is defi ned only when ,x3 5 0$ i.e., when ,x3 5 0$ i.e., when ·x 3
5
$
dom ·( ) ,f 3
5
3 m;
( ) ( ) [ , ) .rangex f x f R3
5 0 0& & 3$ $
10. ( )f x x
x
3
5
is defi ned only when .x
x
3
5 0$
Now, .x
x x x3
5 0 5 0 5& &
And, ( ) ( )and or andx
x x x x x3
5 0 5 0 3 0 5 0 3 0 &
( ) ( )and or andx x x x5 3 5 3 &
( ) ( )and or andx x x x3 5 5 3 &
x3 5 & [ andx x5 3 a cannot hold].
Thus, ( )f x is defined when .x3 5 #
dom ( ) ( , ] .f 3 5
Clearly, ( ) .f x x
x
3
5 0$
Let ( ) .y f x Then,
Functions 103
y x
x y x
x y xy x3
5
3
5 3 52 2 2& &
·( ) ( )x xy y x y y x
y
y
3 5 1 3 5
1
3 52 2 2 2
2
2
& & &
… (i)
It follows from (i) that x is defined for each real value of y.
range ( ) [ , )f 0 3 [ ( ) ]f x 0a $ .
11. ( )f x
x 1
1
2
is defi ned only when .x 1 02
Now, ( ) ( ) ( ) ( )orx x x x x1 0 1 1 0 1 1 2 & &
( , ) ( , )orx x1 1& 3 3 d d .
dom ( ) ( , ) ( , ) .f 1 1,3 3
Let ( ) .y f x Then,
·
( )
y
x
y
x
x
y
x
y y
y
1
1
1
1 1 1 1 1
1
2
2
2
2
2 2 2
2
& & &
Clearly, x is not defined when .y 0
range ( ) { } .f R 0
12. ( ) | |f x x1 2 is defi ned for all .x Rd So, dom ( ) .f R
Now, | | | | | |andx x x0 2 2 0 2 &3 3# $
| | | |andx x2 0 2 & 3#
| | | |andx x1 2 1 1 2 & 3#
| | ( ) ( , ] .x f x1 2 1 1& &3 3# d
range ( ) ( , ] .f 13
13. ( )
| |
f x x
x
4
4
is not defi ned at .x 4 So, dom ( ) { } .f R 4
Now, when ,x 4! we have
, { | | }
, { | | ( )} .
when
when
x x x
x x x
1 4 0 4 4
1 4 0 4 4
a
a
( )f x
Z
[
\
]
]]
]
]]
range ( ) [ , ] .f 1 1
14. ( )f x x
x
3
92
is defi ned for all real values of x for which .x 3!
dom ( ) { } .f R 3
Let ( ) .y f x Then,
, wheny x
x y x x3
9 3 3
2
& !
.y y3 3 6& &! !
range ( ) { } .f R 6
15. Clearly, sin x1 3 1# # for all x Rd
& ( ) ( )sin x2 1 2 3 2 1# # for all x Rd
& sin x1 2 3 3# # for all x Rd
& sin x2 3 0! for any x Rd .
( ) ( )sinf x x2 3
1 is defi ned for all .x Rd
104 Senior SecondarySchool Mathematics for Class 11
We have, sin sinandx x2 3 1 2 3 3$ #
.sin sin sinandx x x2 3
1 1 2 3
1
3
1
3
1
2 3
1 1&# $ # #
range ·( ) ,f 3
1 1 ; E
GRAPHS OF SOME STANDARD REAL FUNCTIONS
1. IDENTITY FUNCTION
The function : : ( )f R R f x x" for all x Rd is called an identity function on R.
Dom ( )f R and range ( ) .f R
EXAMPLE 1 Draw the graph of the identity function : : ( )f R R f x x" for all .x Rd
SOLUTION Here ( )f x x for all .x Rd
We may prepare its table as under.
x –2 –1 0 1 2 3
( )f x x –2 –1 0 1 2 3
On a graph paper, we draw X OXl and YOYl as the x-axis and the
y-axis respectively.
We take 10 small divisions = 1 unit.
Now, on this graph paper, we plot the points
( , ), ( , ), ( , ), ( , ), ( , ) ( , ) .andA B O C D E2 2 1 1 0 0 1 1 2 2 3 3
Join these points successively to obtain the straight line ABOCDE
as the required graph line, given below.
Graph of the identity function, ( )f x x
X
Y
D
C
X� O
B
A
�Y
–3 –2
–1
1 2 3–1
1
2
E
3
–2
Functions 105
2. CONSTANT FUNCTION
Let c be a fi xed real number.
Then, the function : : ( )f R R f x c" for all x Rd is called the constant function.
For the constant function ( )f x c for all ,x Rd we have
dom ( )f R and range ( ) { } .f c
EXAMPLE 2 Draw the graph of each of the following constant functions:
(i) ( )f x 2 for all x Rd
(ii) ( )f x 0 for all x Rd
(iii) ( )f x 2 for all x Rd
SOLUTION (i) Let ( )f x 2 for all .x Rd
Then, dom ( )f R and range ( ) { } .f 2
We may prepare the table given below.
x –2 –1 0 1 2 3
( )f x 2 2 2 2 2 2 2
On a graph paper, we draw X OXl and YOYl as the x-axis and
the y-axis respectively.
We take the scale: 10 small divisions = 1 unit.
On this graph paper, we plot the points A(–2, 2), B(–1, 2),
C(0, 2), D(1, 2), E(2, 2) and F(3, 2). Join A, B, C, D, E and F
successively to obtain the required graph line ABCDEF, whose
equation is .y 2
Graph of the function, ( )f x 2
106 Senior Secondary School Mathematics for Class 11
(ii) Let ( )f x 0 for all .x Rd Then,
dom ( )f R and range ( ) { } .f 0
Now, we have:
x –2 –1 0 1 2
( )f x 0 0 0 0 0 0
On a graph paper, we draw the horizontal line X OXl as the
x-axis and the vertical line YOYl as the y-axis.
We take the scale: 10 small divisions = 1 unit.
Now, on this graph paper, we plot the points A(–2, 0), B(–1, 0),
O(0, 0), C(1, 0) and D(2, 0) and join them successively to get the
graph line ABOCD, whose equation is .y 0
Graph of the function, ( )f x 0
(iii) Let ( )f x 2 for all .x Rd Then,
dom ( )f R and range ( ) .f 2
Now, we have:
x –2 –1 0 1 2 3
( )f x 2 –2 –2 –2 –2 –2 –2
On a graph paper, we draw the horizontal line X OXl as the
x-axis and the vertical line YOYl as the y-axis.
We take the scale: 10 small divisions = 1 unit.
On this graph paper, we plot the points A(–2, –2), B(–1, –2),
C(0, –2), D(1, –2), E(2, –2) and F(3, –2) and join these points
successively to get the required graph line ABCDE, as shown
below. Its equation is .y 2
Functions 107
Graph of the function, ( )f x 2
3. THE MODULUS FUNCTION
The function f, defi ned by ( ) | | ,
,
when
when
f x x x x
x x
0
0
$
Z
[
\
]
]]
]
]]
is called the modulus function.
We know that | |x 0$ for all .x Rd
dom ( )f R and range ( ) [ , ) .f 0 3
EXAMPLE 3 Draw the graph of the modulus function, defi ned by
: : ( ) | | ,
, .
when
when
f R R f x x x x
x x
0
0
" $
Z
[
\
]
]]
]
]]
SOLUTION We have
x –3 –2 –1 0 1 2 3 4
( ) | |f x x 3 2 1 0 1 2 3 4
On a graph paper draw the horizontal line X OXl as the x-axis and
the vertical line YOYl as the y-axis.
Take the scale: 10 small divisions = 1 unit.
On this graph paper, plot the points A(–3, 3), B(–2, 2), C(–1, 1),
O(0, 0), D(1, 1), E(2, 2), F(3, 3) and G(4, 4).
Join these points successively to obtain the graph lines ABCO and
ODEFG, as shown below.
108 Senior Secondary School Mathematics for Class 11
Graph of ( ) | |f x x
4. THE GREATEST INTEGER FUNCTION, OR STEP FUNCTION
For any real number x, the symbol [ ]x denotes the greatest integer less than or equal
to x.
EXAMPLES (i) [6.85] = 6 (ii) [8] = 8 (iii) [0.536] = 0 (iv) [–5.8] = –6
(v) [–5] = –5
The function : : ( ) [ ]f R R f x x" for all x Rd is called the greatest integer
function, or step function.
Clearly, dom ( )f R and range ( ) .f Z
EXAMPLE 4 Draw the graph of the greatest integer function:
: : ( ) [ ]f R R f x x" for all .x Rd
SOLUTION Using the defi nition of [x], we have
[ ] forx x2 2 1#
[ ] forx x1 1 0#
[ ] forx x0 0 1#
[ ] forx x1 1 2#
[ ] forx x2 2 3#
[ ] .forx x3 3 4#
Taking the scale: 10 small divisions = 1 unit, the graph of the
function ( ) [ ]f x x may be drawn as shown below.
Functions 109
Graph of greatest integer function [x]
5. THE SMALLEST INTEGER FUNCTION
For any real number x, the smallest integer greater than or equal to x is denoted by .x_ i
For example: . , , . , . ,3 8 4 6 6 0 76 1 6 2 6 _ _ _ _i i i i etc.
The function : : ( )f R R f x x" _ i for all x Rd is called the smallest integer
function.
Clearly, dom ( ) ,f R range ( ) .f Z
EXAMPLE 5 Draw the graph of the smallest integer function : : ( )f R R f x x" _ i for
all .x Rd
SOLUTION Using the defi nition of ,x_ i we have
x ( )f x x _ i
x3 2 # –2
x2 1 # –1
x1 0 # 0
x0 1 # 1
x1 2 # 2
x2 3 # 3
SCALE: 10 small divisions = 1 unit.
110 Senior Secondary School Mathematics for Class 11
Now, the graph may be drawn as shown below.
Graph of smallest integer function x_ i
6. THE FRACTIONAL PART FUNCTION
Let : : ( ) [ ] { } .f R R f x x x x"
Here ( ) { }f x x denotes the fractional part of x.
For example: { . } . , { } , { } , { . } . , { . } . ,2 35 0 35 6 0 7 0 0 65 0 35 3 75 0 25 etc.
Clearly, dom ( )f R and range ( ) [ , ) .f 0 1
EXAMPLE 6 Draw the graph of the fractional part function:
: : ( ) [ ] { }f R R f x x x x"
SOLUTION We have, ( ) [ ] { } .f x x x x
( ) , ( . ) . , ( . ) . , , ( . ) . , ( ) .f f f f f3 0 3 1 0 1 3 2 0 2 3 9 0 9 4 0…
( ) , ( . ) . , ( . ) . , , ( . ) . , ( ) .f f f f f2 0 2 1 0 1 2 2 0 2 2 9 0 9 3 0…
( ) , ( . ) . , ( . ) . , …, ( . ) . , ( ) .f f f f f1 0 1 1 0 1 1 2 0 2 1 9 0 9 2 0
( ) , ( . ) . , ( . ) . , , ( . ) . , ( ) .f f f f f0 0 0 1 0 1 0 2 0 2 0 9 0 9 1 0…
( ) , ( . ) . , ( . ) . , …, ( . ) . , ( ) .f f f f f1 0 0 9 0 1 0 8 0 2 0 1 0 9 0 0
( ) , ( . ) . , ( . ) . , …, ( . ) . , ( ) .f f f f f2 0 1 9 0 1 1 8 0 2 1 1 0 9 1 0
( ) , ( . ) . , ( . ) . , , ( . ) . , ( ) .f f f f f3 0 2 9 0 1 2 8 0 2 2 1 0 9 2 0…
Functions 111
Plotting these points on a graph paper, we may get the graph as
shown below.
Graph of fractional part function, ( ) { }f x x
7. THE SIGNUM FUNCTION
The function | |
,
,
when
when
x
x x
x
0
0 0
!
: : ( )f R R f x"
Z
[
\
]
]
]]
]
]
]]
is called the signum function.
The above function may be simplified as under:
( )
,
,
, .
if
if
if
f x
x
x
x
1 0
0 0
1 0
Z
[
\
]
]
]
]]
]
]
]
]]
Clearly, dom ( )f R and range ( ) { , , } .f 1 0 1
EXAMPLE 7 Draw the graph of the signum function, : ,f R R" defi ned by
( ) | |
,
,
when
when
f x x
x x
x
0
0 0
!
Z
[
\
]
]
]]
]
]
]]
or ( )
,
,
,
if
if
if
f x
x
x
x
1 0
0 0
1 0
Z
[
\
]
]
]
]]
]
]
]
]]
SOLUTION Clearly, we have
( )x f x0 1 &
( )x f x0 0&
( ) .x f x0 1 &
112 Senior Secondary School Mathematics for Class 11
We may now draw the graph as shown below.
Graph of signum function
8. POLYNOMIAL FUNCTION
A function of the form ( ) ,f x a a x a x a x… n
n
0 1 2
2 where n is a non-negative
integer and , , , , ,a a a a R… n0 1 2 d is called a polynomial function.
EXAMPLES (i) ( )f x x x2 53 2 is a polynomial function.
(ii) ( )g x x x4 22 is a polynomial function.
(iii) ( )h x x x2 3 5 is not a polynomial function.
EXAMPLE 8 Let : : ( )f R R f x x2" for all .x Rd
Find its domain and range.
Also, draw its graph.
SOLUTION Here, : : ( )f R R f x x2" for all .x Rd
Dom ( )f R and range ( ) { : } [ , ) .f x R x 0 0 3$ d
Now, we have:
x –2 –1.5 –1 0 1 1.5 2
( )f x x2 4 2.25 1 0 1 2.25 4
On a graph paper, we draw X OXl and YOYl as the x-axis and the
y-axis respectively.
Now, we plot the points A(–2, 4), B(–1.5, 2.25), C(–1, 1), O(0, 0),
D(1, 1), E(1.5, 2.25) and F(2, 4).
Join these points freehand successively to obtain the required
curve.
Functions 113
Scale: 10 small divisions = 1 unit.
Graph of ( )f x x2
EXAMPLE 9 Let : : ( )f R R f x x3" for all .x Rd
Find its domain and range.
Also, draw its graph.
SOLUTION Let : : ( )f R R f x x3" for all .x Rd
Then,
dom ( )f R and range ( ) .f R
Now, we have:
x –2 –1.5 –1 0 1 1.5 2
( )f x x3 –8 –3.375 –1 0 1 3.375 8
On a graph paper, we draw X OXl
and YOYl as the x-axis and the y-axis
respectively.
We take the scale as
5 small divisions = 1 unit.
Now, we plot the points A(–2, –8),
B(–1.5, –3.375), C(–1, –1), O(0, 0),
D(1, 1), E(1.5, 3.375) and F(2, 8).
We join these points freehand
successively to obtain the required
curve.
114 Senior Secondary School Mathematics for Class 11
9. RATIONAL FUNCTION
The functions of the type ,( )
( )
g x
f x
where ( )f x and ( )g x are polynomial functions of
x, defi ned in a domain, where ( ) .g x 0!
EXAMPLE 10 Let : ( { }) : ( )f R R f x x0
1" for all values of { } .x R 0d
Find its domain and range. Also, draw its graph.
SOLUTION Let : ( { }) : ( )f R R f x x0
1" $ Then,
dom ( ) { }f R 0 and range ( ) { } .f R 0
We have:
x –4 –2 –1 –0.5 –0.25 0.25 0.5 1 2 4
( )f x x
1 –0.25 –0.5 –1 –2 –4 4 2 1 0.5 0.25
On a graph paper, let X OXl and YOYl be the x-axis and the y-axis
respectively.
We take the scale: 10 small divisions = 1 unit.
Now, we plot the points A(–4, –0.25), B(–2, –0.5), C(–1, –1),
D(–0.5, –2), E(–0.25, –4) and F(0.25, 4), G(0.5, 2), H(1, 1), J(2, 0.5) and
K(4, 0.25).
Graph of the function, ( )f x x
1
Functions 115
We join A, B, C, D and E freehand successively to obtain a curve.
Also, we join F, G, H, J and K freehand successively to obtain
another curve, as shown above.
EXERCISE 3D
1. Consider the real function : : ( )f R R f x x 5" for all .x Rd
Find its domain and range.
Draw the graph of this function.
2. Consider the function : ,f R R" defi ned by
( )
,
,
, .
when
when
when
f x
x x
x
x x
1 0
1 0
1 0
Z
[
\
]
]
]
]]
]
]
]
]]
Write its domain and range.
Also, draw the graph of ( ) .f x
3. Find the domain and the range of the square root function,
: { } : ( )f R R f x x0 ", for all non-negative real numbers.
Also, draw its graph.
4. Find the domain and the range of the cube root function,
: : ( )f R R f x x /1 3" for all .x Rd
Also, draw its graph.
HINTS TO SOME SELECTED QUESTIONS
1. Clearly, dom ( )f R and range ( ) .f R
116 Senior Secondary School Mathematics for Class 11
Now, .y x 5
Putting ,x 0 we get .y 5
Putting ,y 0 we get .x 5
On the graph paper, plot the points A(–5, 0) and B(0, 5).
Join AB to get the required graph.
2. .x y x0 1 &
.x y1 1&
x y x1 1 & .
(i) For ,x 0 we have y x1 .
x –3 –2 –1
y 4 3 2
(ii) For ,x 0 we have .y 1
x 0
[ ]x y0 1&a
y 1
(iii) For ,x 0 we have .y x 1
x 1 2 3
y 2 3 4
On a graph paper, draw horizontal line X OXl and the vertical line .YOYl Now, plot
the points ( , ), ( , ), ( , ), ( , ), ( , ), ( , ) ( , ) .andA B C D E F G3 4 2 3 1 2 0 1 1 2 2 3 3 4
Join ABCD and DEFG to get the required graph.
Y
�Y
X� X
A
B
C
O
D
E
F
G
x
+
y
=
1
–x
+
y
=
1
Functions 117
3. Let : { } : ( ) .f R R f x x0 ", Then,
dom ( ) [ , )f 0 3 and range ( ) [ , ) .f 0 3
We have
, , . . , , . . .and0 0 1 1 2 25 1 5 4 2 6 25 2 5 9 3
On a graph paper, we plot the points O(0, 0), A(1, 1), B(2.25, 1.5), C(4, 2), D(6.25, 2.5),
and E(9, 3). Join these points freehand successively to get the required curve.
Graph of the curve, ( )f x x
4. Let : : ( )f R R f x x /1 3" for all .x Rd
Clearly, dom ( )f R and range ( ) .f R
We have
, . , . , , , , . , . ,8 2 5 1 7 3 1 4 1 1 0 0 1 1 3 1 4 5 1 7 3 3 3 3 3 3 3 3
and .8 2 3
On a graph paper, take the horizontal line X OXl as the x-axis and the vertical line
YOYl as the y-axis.
Take the scale: 5 small divisions = 1 unit.
Now, plot the points A(8, 2), B(5, 1.7), C(3, 1.4), D(1, 1), O(0, 0), E(–1, –1), F(–3, –1.4),
G(–5, –1.7) and H(–8, –2).
Join ABCDOEFGH freehand to get the required curve.
Remember: . . .and3 1 4 5 1 7 3 3
118 Senior Secondary School Mathematics for Class 11
OPERATIONS ON FUNCTIONS
(i) SUM OF TWO REAL FUNCTIONS
Let :f D R1" and : ,g D R2" where D R13 and .D R23 Then,
( ) : ( ) : ( ) ( ) ( ) ( )f g D D R f g x f x g x1 2 "+ for all ( )x D D1 2+d .
(ii) DIFFERENCE OF TWO REAL FUNCTIONS
Let :f D R1" and : ,g D R2" where D R13 and .D R23 Then,
( ) : ( ) : ( ) ( ) ( ) ( )f g D D R f g x f x g x1 2 "+ for all ( )x D D1 2+d .
(iii) MULTIPLICATION OF TWO REAL FUNCTIONS
Let :f D R1" and : ,g D R2" where D R13 and .D R23 Then,
( ) : ( ) : ( ) ( ) ( ) ( )fg D D R fg x f x g x1 2 "+ $ for all ( )x D D1 2+d .
NOTE This is known as pointwise multiplication of functions.
(iv) QUOTIENT OF TWO REAL FUNCTIONS
Let :f D R1" and : ,g D R2" where D R13 and .D R23
Let ( ) { : ( ) } .D D D x g x 01 2+ Then,
: : ( ) ( )
( )
g
f
D R g
f
x
g x
f x
" e eo o for all .x Dd
(v) RECIPROCAL OF A FUNCTION
Let :f D R1" and let { : ( ) } .D D x f x 01
Then, the reciprocal of f is the function ,
f
1 defined by
: : ( ) ( )f D R f x f x
1 1 1" e o for all .x Dd
(vi) SCALAR MULTIPLE OF A FUNCTION
Let :f D R" and let be a scalar (a real number). Then,
( ) : : ( ) ( ) ( )f D R f x f x" $ for all .x Dd
SOLVED EXAMPLES
EXAMPLE 1 Let : : ( )f R R f x x2" and : : ( ) .g R R g x x2 1"
Find (i) ( ) ( )f g x (ii) ( ) ( )f g x (iii) ( ) ( )fg x (iv) ( )g
f
xe o
SOLUTION Here, dom ( )f R and dom ( ) .g R
dom ( )f + dom ( ) ( ) .g R R R+
(i) ( ) :f g R R" is given by
( ) ( ) ( ) ( ) ( ) ( ) .f g x f x g x x x x2 1 12 2
(ii) ( ) :f g R R" is given by
( ) ( ) ( ) ( ) ( ) ( ) .f g x f x g x x x x x2 1 2 12 2
Functions 119
(iii) ( ) :fg R R" is given by
( ) ( ) ( ) ( ) ( ) ( ) .fg x f x g x x x x x2 1 2· ·2 3 2
(iv) ·{ : ( ) } { : }x g x x x0 2 1 0 2
1 ' 1
dom ·g
f
R R R2
1
2
1
+
e o ' '1 1
The function :g
f
R R2
1 "
' 1 is given by
·( ) ( )
( )
,g
f
x
g x
f x
x
x x2 1 2
12
!
e o
EXAMPLE 2 Let ( )f x x and ( )g x x be two functions defi ned over the set of non-
negative real numbers. Find:
(i) ( ) ( )f g x (ii) ( ) ( )f g x (iii) ( ) ( )fg x (iv) ( )g
f
x
SOLUTION Here : [ , ) : ( )f R f x x0 "3 and : [ , ) : ( ) .g R g x x0 "3
dom ( ) [ , )f 0 3 and dom ( ) [ , ) .g 0 3
So, dom ( )f + dom ( ) [ , ) [ , ) [ , ) .g 0 0 0+3 3 3
(i) ( ) : [ , )f g R0 "3 is given by
( ) ( ) ( ) ( ) ( ) .f g x f x g x x x
(ii) ( ) : [ , )f g R0 "3 is given by
( ) ( ) ( ) ( ) ( ) .f g x f x g x x x
(iii) ( ) : [ , )fg R0 "3 is given by
( ) ( ) ( ) ( ) ( ) .fg x f x g x x x x· /3 2#
(iv) { : ( ) } { } .x g x 0 0
dom ( ) ( ) { : ( ) }dom domg
f
f g x g x 0+ e o
[ , ) [ , ) { } ( , ) .0 0 0 0+3 3 3
So, : ( , )g
f
R0 "3 is given by
,( ) ( )
( )
.g
f
x
g x
f x
x
x
x
x1 0! e o
EXAMPLE 3 Let f and g be real functions, defi ned by ( ) ( )f x x 4
1 and ( ) ( ) .g x x 4
3
Find (i) ( ) ( )f g x (ii) () ( )f g x (iii) ( ) ( )fg x (iv) ( )g
f
xe o (v) ( )
f
x1e o
SOLUTION Clearly, ( ) ( )f x x 4
1 is defi ned for all real values of x except that at
which ,x 4 0 i.e., .x 4
dom ( ) { } .f R 4
120 Senior Secondary School Mathematics for Class 11
And, ( ) ( )g x x 4 3 is defi ned for all .x Rd So, dom ( ) .g R
dom ( )f + dom ( ) { } { } .g R R R4 4+
(i) ( ) : { }f g R R4 " is given by
·( ) ( ) ( ) ( ) ( ) ( ) ( )
( )
f g x f x g x
x
x
x
x
4
1 4 4
1 43
4
(ii) ( ) : { }f g R R4 " is given by
·( ) ( ) ( ) ( ) ( ) ( ) ( )
( )
f g x f x g x
x
x
x
x
4
1 4 4
1 43
4
(iii) ( ) : { }fg R R4 " is given by
( ) ( ) ( ) ( ) ( ) ( ) ( ) .fg x f x g x x x x4
1 4 4· 3 2#
(iv) { : ( ) } { : ( ) } { : } { } .x g x x x x x0 4 0 4 0 43
dom ( ) ( ) { : ( ) } { } .dom domg
f
f g x g x R0 4+ e o
: { }g
f
R R4 " e o is given by
,( ) ( )
( )
( )
( )
( )
.g
f
x
g x
f x
x
x
x
x
4
4
1
4
1 43 4 !
e o
(v) Clearly, ( )f x 0! for any { } .x R 4 d
: { }
f
R R1 4 " is given by
( ) ( )
( )
( ) .
f
x
f x
x
x1 1
4
1
1 4
e o
EXAMPLE 4 Let f and g be real functions defi ned by ( ) ( ) .f x x and g x x1 1
Find (i) ( ) ( )f g x (ii) ( ) ( )f g x (iii) ( ) ( )fg x (iv) ( )g
f
xe o .
SOLUTION Clearly, ( )f x x 1 is defi ned for all real values of x for which
,x 1 0$ i.e., .x 1$ So, dom ( ) [ , ) .f 1 3
Also, ( )g x x 1 is defi ned for all real values of x for which
,x 1 0$ i.e., .x 1$ So, dom ( ) [ , ) .g 1 3
dom ( ) ( ) { , ) [ , ) [ , ) .domf g 1 1 1+ +3 3 3
(i) ( ) : [ , )f g R1 "3 is given by
( ) ( ) ( ) ( ) ( ) .f g x f x g x x x1 1
(ii) ( ) : [ , )f g R1 "3 is given by
( ) ( ) ( ) ( ) ( ) .f g x f x g x x x1 1
(iii) ( ) : [ , )fg R1 "3 is given by
( ) ( ) ( ) ( ) .fg x f x g x x x x1 1 1· 2#
Functions 121
(iv) { : ( ) } { : } { : } { } .x g x x x x x0 1 0 1 0 1
dom ( ) ( ) { : ( ) }domf g x g x 0+
[ , ) [ , ) { } [ , ) .1 1 1 1+3 3 3
[ , )g
f
R1" "3 is given by
·( ) ( )
( )
g
f
x
g x
f x
x
x
1
1
e o
EXAMPLE 5 Let f and g be real functions, defi ned by ( ) ( ) .f x x and g x x2 4 2
Find (i) ( ) ( )f g x (ii) ( ) ( )f g x (iii) ( ) ( )fg x (iv) ( ) ( )ff x
(v) ( ) ( )gg x (vi) ( )g
f
xe o .
SOLUTION Clearly, ( )f x x 2 is defi ned for all x Rd such that
,x 2 0$ i.e., .x 2$
dom ( ) [ , ) .f 2 3
Again, ( )g x x4 2 is defi ned for all x Rd such that
.x4 02$
But, ( ) ( ) [ , ]x x x x x4 0 4 0 2 2 0 2 22 2& & &$ # # d .
dom ( ) [ , ] .g 2 2
dom ( ) ( ) [ , ) [ , ] [ , ] .domf g 2 2 2 2 2+ +3
(i) ( ) : [ , ]f g R2 2 " is given by
( ) ( ) ( ) ( ) .f g x f x g x x x2 4 2
(ii) ( ) : [ , ]f g R2 2 " is given by
( ) ( ) ( ) ( ) .f g x f x g x x x2 4 2
(iii) ( ) : [ , ]fg R2 2 " is given by
( ) ( ) ( ) ( ) ( ) ( )fg x f x g x x x2 4· 2
( ) ( ) ( ) ( ) .x x x x2 2 2 22
(iv) ( ) : [ , ]ff R2 2 " is given by
( ) ( ) ( ) ( ) ( ) ( ) ( ) .ff x f x f x x x x2 2 2·
(v) ( ) : [ , ]gg R2 2 " is given by
( ) ( ) ( ) ( ) ( ) ( ) ( ) .gg x g x g x x x x4 4 4· 2 2 2
(vi) { : ( ) } { : ] { : ( ) ( ) } { , } .x g x x x x x x0 4 0 2 2 0 2 22
dom ( ) { } { : ( ) }dom domg
f
f g x g x 0+ e o
[ , ] { , } ( , ) .2 2 2 2 2 2
: ( , )g
f
R2 2 " is given by
122 Senior Secondary School Mathematics for Class 11
·( ) ( )
( )
( ) ( ) ( )g
f
x
g x
f x
x
x
x x
x
x4
2
2 2
2
2
1
2
e o
EXAMPLE 6 Let f be the exponential function and g be the logarithmic function. Then,
fi nd:
(i) ( ) ( )f g 1 (ii) ( ) ( )fg 1 (iii) ( ) ( )f4 1 (iv) ( ) ( )g3 1
SOLUTION Let : : ( )f R R f x ex" and : : ( ) .logg R R g x xe"
Then, dom ( ) ( ) .domf g R R R+ +
(i) ( ) :f g R R" is given by
( ) ( ) ( ) ( ) ( ) .logf g x f x g x e xx e
( ) ( ) ( ) ( ) .logf g e e e1 1 0e
1
(ii) ( ) :fg R R" is given by
( ) ( ) ( ) ( ) ( )logfg x f x g x e xx e$ .
( ) ( ) ( ) ( ) .logfg e e1 1 0 0e
1 #
(iii) ( ) :f R R4 " is given by
( ) ( ) ( ) .f x f x e4 4 4 x#
( ) ( ) ( ) .f e e4 1 4 41#
(iv) ( ) :g R R3 " is given by
( ) ( ) ( ) ( )logg x g x x3 3 3 e# #
( ) ( ) ( ) ( ) ( ) .logg g3 1 3 1 3 1 3 0 0e# # #
EXAMPLE 7 If ( ) ( ) ( ) [ ]logf x x and g x x1e then fi nd:
(i) ( ) ( )f g x (ii) ( ) ( )fg x (iii) ( )g
f
xe o (iv) ( ) .
f
g
xe o
Also fi nd ·( ) ( ), ( ) ( ), ( ),f g fg g
f
f
g
1 0 1 2
1 e e bo o l
SOLUTION Clearly, ( )log x1e is defi ned only when ,x1 0 i.e., .x 1
dom ( ) ( , ) .f 13
Also, dom ( ) .g R
dom ( ) ( ) ( , ) ( , ) .domf g R1 1+ +3 3
(i) ( ) : ( , )f g R1 "3 is given by
( ) ( ) ( ) ( ) ( ) [ ]logf g x f x g x x x1e
(ii) ( ) : ( , )fg R1 "3 is given by
( ) ( ) ( ) ( ) { ( )} [ ] .logfg x f x g x x x1e# #
(iii) { : ( ) } { : [ ] } [ , ) .x g x x x0 0 0 1
dom ( ) ( ) { : ( ) }dom domg
f
f g x g x 0+ e o
( , ) [ , ) ( , ) .R1 0 1 0+3 3
Functions 123
: ( , )g
f
R0 "3 is given by
·( ) ( )
( )
[ ]
( )log
g
f
x
g x
f x
x
x1e
e o
(iv) { : ( ) } { : ( ) } { }logx f x x x0 1 0 0e .
( ) ( ) { : ( ) }dom dom dom
f
g
g f x f x 0+ e o
( , ) { } ( , ) ( , )R 1 0 0 0 1+ ,3 3 .
: ( , ) ( , )
f
g
R0 0 1 ",3 is given by
·( ) ( )
( )
( )
[ ]
logf
g
x
f x
g x
x
x
1e
e o
Now, we have:
( ) ( ) ( ) ( ) [ ] ( ) ( ) .log logf g f g1 1 1 1 1 1 2 1e e
( ) ( ) ( ) ( ) ( ) [ ] ( ) ( ) .log logfg f g0 0 0 1 0 0 1 0 0 0 0e e# # # #
·( ) ( )
( )
( )
[ ]
log logg
f
g
f
1 1
1
1 1
1
2
1
e e
e o
[ . ]
.
log logf
g
f
g
2
1
2
1
2
1
1 2
1
2
1
2
1
0 5
0
e e
e b
b
b
b b
o l
l
l
l l
; E
EXAMPLE 8 Find the sum and the difference of the identity function and the modulus
function.
SOLUTION Let : : ( )f R R f x x" be the identity function.
And, let : : ( ) | |g R R g x x" be the modulus function.
Then, dom ( )f R and dom ( ) .g R
dom ( ) ( ) .domf g R R R+ +
(i) dom ( ) ( ) ( ) .dom domf g f g R+
Now, ( ) :f g R R" is given by
( ) ( ) ( ) ( )f g x f x g x
,
,
when
when
x x
x x
0
0
$
| |x x x
Z
[
\
]
]]
]
]]
,
,
when
when
x x x
x x x
0
0
$
Z
[
\
]
]]
]
]]
,
, .
when
when
x x
x
2 0
0 0
$
Z
[
\
]
]]
]
]]
Hence,
,
, .
when
when
x x
x
2 0
0 0
$
( ) ( )f g x
Z
[
\
]
]]
]
]]
(ii) dom ( ) ( ) ( ) .dom domf g f g R+
( ) ( ) ( ) ( )f g x f x g x
,
,
when
when
x x
x x
0
0
$
| |x x x
Z
[
\
]
]]
]
]]
124 Senior Secondary School Mathematics for Class 11
,
,
when
when
x x x
x x x
0
0
$
Z
[
\
]
]]
]
]]
,
,
when
when
x
x x
0 0
2 0
$
Z
[
\
]
]]
]
]]
,
, .
when
when
x
x x
0 0
2 0
$
( ) ( )f g x
Z
[
\
]
]]
]
]]
EXAMPLE 9 Find the sum and the difference of the identity function and the reciprocal
function.
SOLUTION Let : : ( )f R R f x x" and : { } : ( )g R R g x x0
1" be the identity
function and the reciprocal function respectively.
Then, ( ) ( ) { } { } .dom domf g R R R0 0+ +
·( ) : { } : ( ) ( ) ( ) ( )f g R R f g x f x g x x x0
1" b l
Hence, ( ) ( )f g x x x
1 b l for all { } .x R 0d
And, ·( ) : { } : ( ) ( ) ( ) ( )f g R R f g x f x g x x x0
1" b l
Hence, ( ) ( )f g x x x
1 b l for all { } .x R 0d
EXAMPLE 10 Find the product of the identity functionby the modulus function.
SOLUTION Let : : ( )f R R f x x" and : : ( ) | |g R R g x x" be the identity
function and the modulus function respectively.
Then, dom ( ) ( ) ( ) .dom domfg f g R R R+ +
( ) : : ( ) ( ) ( ) · ( ) .fg R R fg x f x g x"
Now, ( ) ( ) ( ) ( ) | |fg x f x g x x x· ·
,
,
,
, .
when
when
when
when
x x
x x
x x
x x
0
0
0
0
2
2
$ $
·x
Z
[
\
]
]]
]
]]
Z
[
\
]
]]
]
]]
Hence,
,
, .
when
when
x x
x x
0
0
2
2
$
( ) ( )fg x
Z
[
\
]
]]
]
]]
EXAMPLE 11 Find the product of the identity function by the reciprocal function.
SOLUTION Let : : ( )f R R f x x" and : { } : ( )g R R g x x0
1" be the identity
function and the reciprocal function respectively.
Then, dom ( ) ( ) ( ) { } { } .dom domfg f g R R R0 0+ +
( ) : { } : ( ) ( ) ( ) ( ) .fg R R fg x f x g x x x0
1 1·" # b l
Hence, ( ) ( )fg x 1 for all { } .x R 0d
EXAMPLE 12 Find the quotient of the identity function by the modulus function.
SOLUTION Let : : ( )f R R f x x" and : : ( ) | |g R R g x x" be the identity
function and the modulus function respectively.
Functions 125
Now, dom ( ) ( ) { : ( ) }dom domg
f
f g x g x 0+ e o
and { : ( ) } { :| | } { }x g x x x0 0 0 .
[ { }] { } { } .dom g
f
R R R0 0 0+ e o
So,
( )
( )
| |
,
, .
when
wheng
f
g
f
g x
f x
x
x x
x
1 0
1 0
: { } : ( )R R x0 " e o
Z
[
\
]
]]
]
]]
Hence, ( ) ,
, .
when
when
g
f
x x
x
1 0
1 0
e o
Z
[
\
]
]]
]
]]
EXAMPLE 13 Find the quotient of the identity function by the reciprocal function.
SOLUTION Let : : ( )f R R f x" and : { } : ( )g R R g x x0
1" be the identity
function and the reciprocal function respectively.
Now, dom ( ) ( ) { : ( ) }dom domg
f
f g x g x 0+ e o
and { : ( ) } : .x g x x x0
1 0 ' 1
dom [ { }] { } .g
f
R R R0 0+ e o
So, : { } : ( ) ( )
( )
.g
f
R R g
f
x
g x
f x
x
x x0 1
2" e o
Hence, ( )g
f
x x2e o for all { } .x R 0d
EXERCISE 3E
1. Let : : ( )f R R f x x 1" and : : ( ) .g R R g x x2 3"
Find (i) ( ) ( )f g x (ii) ( ) ( )f g x (iii) ( ) ( )fg x (iv) ( )g
f
xe o .
2. Let : : ( ) : : ( ) .andf R R f x x g R R g x x x2 5 2" "
Find (i) ( ) ( )f g x (ii) ( ) ( )f g x (iii) ( ) ( )fg x (iv) ( )g
f
xe o .
3. Let : : ( )f R R f x x 13" and : : ( ) ( ) .g R R g x x 1"
Find (i) ( ) ( )f g x (ii) ( ) ( )f g x (iii) ( )
f
x1e o (iv) ( )g
f
xe o .
4. Let : : ( ) ,f R R f x c
x" where c is a constant.
Find (i) ( ) ( )cf x (ii) ( ) ( )c f x2 (iii) ( )c f x
1
b l .
5. Let : [ , ) : ( )f R f x x2 2"3 and : [ , ) : ( ) .g R g x x2 2"3
Find (i) ( ) ( )f g x (ii) ( ) ( )f g x (iii) ( ) ( )fg x .
126 Senior Secondary School Mathematics for Class 11
ANSWERS (EXERCISE 3E)
1. (i) x3 2 (ii) x 4 (iii) x x2 32 (iv) x
x
2 3
1
2. (i) x x3 52 (ii) x x 52 (iii) x x x2 7 53 2 (iv)
x x
x2 5
2
3. (i) x x 23 (ii) x x3 (iii)
x 1
1
3
(iv) x x 12
4. (i) x (ii) cx (iii)
c
x
2
5. (i) x x2 2 for all [ , )x 2 3d (ii) x x2 2 for all [ , )x 2 3d
(iii) x 42 for all [ , )x 2 3d
HINTS TO SOME SELECTED QUESTIONS
1. (iv) dom ( ) ( ) { : ( ) }dom domg
f
f g x g x 0+ e o
·R R R2
3
2
3
+ ' '1 1
2. (iv) dom ( ) ( ) { : ( ) }dom domg
f
f g x g x 0+ e o
{ , } { , } .R R R0 1 0 1+
3. (iii) dom ( ) { : ( ) } { } .dom
f
f x f x R1 0 1 e o
(iv) dom ( ) ( ) { : ( ) }dom domg
f
f g x g x 0+ e o
{ } { } .R R R1 1+
EXERCISE 3F
Very-Short-Answer Questions
1. Find the set of values for which the function ( )f x x1 3 and ( )g x x2 12
are equal.
2. Find the set of values for which the functions ( )f x x 3 and ( )g x x3 12
are equal.
3. Let { , , , }X 1 0 2 5 and : : ( ) .f X R f x x 13" Then, write f as a set of
ordered pairs.
4. Let { , , , , }A 2 1 0 1 2 and : : ( ) .f A Z f x x x2 32" Find ( ) .f A
5. Let : : ( ) .f R R f x x2"
Determine (i) range (f ) (ii) { : ( ) }x f x 4
6. Let : : ( ) .f R R f x x 12" Find { } .f 101
7. Let : : ( ) .logf R R f x xe"
Find { : ( ) } .x f x 2
Functions 127
8. Let { , , , , }A 6 10 11 15 21 and let : : ( )f A N f n" is the highest prime factor
of n. Find range (f ).
9. Find the range of the function ( ) .sinf x x
10. Find the range of the function ( ) | |.f x x
11. Write the domain and the range of the function, ( ) [ ] .f x x x
12. If ( )f x x
x
5
5
then fi nd dom ( )f and range ( )f .
13. Let {( , ), ( , ), ( , ), ( , ), ( , ), ( , )}f 1 6 2 5 4 3 5 2 8 1 10 3
and {( , ), ( , ), ( , ), ( , ), ( , ), ( , )} .g 2 0 3 2 5 6 7 10 8 12 10 16
Find (i) dom ( )f g (ii) dom ·g
f
e o
14. If ,( )f x x
x 1 fi nd the value of ·f f x
1
b l( 2
15. If ,( )f x x
kx
1
where x 1! and { ( )}f f x x for x 1! then fi nd the value
of k.
16. Find the range of the function, ·( )
| |
f x
x
x
17. Find the domain of the function, ( ) | | .logf x x
18. If f x x x x
1 12
2
b dl n for all { }x R 0d then write an expression for ( ) .f x
19. Write the domain and the range of the function, ·( )f x
bx a
ax b
20. Write the domain and the range of the function, ( ) .f x x 1
21. Write the domain and the range of the function, ( ) | | .f x x
ANSWERS (EXERCISE 3F)
1. ,2 2
1' 1 2. ,3
4 1' 1 3. {( , ), ( , ), ( , ), ( , )}f 1 0 0 1 2 9 5 126
4. ( ) { , , , }f A 5 0 3 4 5. (i) [ , )0 3 (ii) {–2, 2} 6. {–3, 3}
7. { }e 2 8. {3, 5, 7, 11} 9. [–1, 1] 10. [ , )0 3
11. dom ( ) ,f R range ( ) [ , )f 0 1 12. dom ( ) { },f R 5 range ( ) { }f 1
13. (i) {2, 5, 8, 10} (ii) {5, 8, 10} 14. x
x
1 15. k 1
16. {1, –1} 17. { }R 0 18. ( ) ,f x x x R22 d
19. dom ,( )f R
b
a ' 1 range ( )f R
b
a ' 1 20. dom ( ) [ , ),f 1 3 range ( )f R
21. dom ( ) ,f R range ( ) ( , ]f 03
HINTS TO SOME SELECTED QUESTIONS
1. ( ) ( ) .x x x x x x2 1 1 3 2 3 2 0 2 2 1 02 2& &
128 Senior Secondary School Mathematics for Class 11
2. ( ) ( ) .x x x x x x3 1 3 3 4 0 3 4 1 02 2& &
5. (i) ( ) .f x x 02$ So, range ( ) [ , ) .f 0 3
(ii) ( ) .f x x x4 4 22& & !
6. ( ) .f x x x x10 1 10 9 32 2& & & !
7. ( ) .logf x x x e2 2e
2& &
8. ( ) , ( ) , ( ) , ( ) , ( ) .f f f f f6 3 10 5 11 11 15 5 21 7
range ( ) { , , , } .f 3 5 7 11
10. ( ) | | .f x x 0$ So, range ( ) [ , ) .f 0 3
11. [ ]x x 0 for x Zd and [ ]x x 0 for .x R Zd
[ ]x x 0$ for .x Rd So, dom ( ) .f R
Also, [ ] [ ] .x x x x0 1 0 1 &# # So, range ( ) [ , ) .f 0 1
12. ( )f x is not defi ned when .x 5 So, dom ( ) { } .f R 5
( ) ( )
( )
.f x
x
x
5
5
1
So, range ( ) { } .f 1
13. dom ( ) ( ) ( ) { , , , }dom domf g f g 2 5 8 10+ .
dom ( ) ( ) { : ( ) } { , , , } { } { , , }dom domg
f
f g x g x 0 2 5 8 10 2 5 8 10+ e o .
14. ·( )
( )
f x
x
x x f f x x
x
x
x
x
x1
1
1 1
1 1 1
1 1
1 1&
c cm m( 2
15. ·( ( ))f f x f x
kx
x
kx
k x
kx
kx x
k x
1
1 1
1
1
2#
d n So, .kx x
k x x
1
2
( )
( ) ( )
k kx x k kx x k
x x x x x
1 1 0 2
4 1
2
22 2
2
& &
! !
& ork x k k1 1 1& [ ka is a constant].
18. ( ) , .f x x x x f x x x R
1 1 2 2
2
2& dc cm m
SUMMARY OF KEY FACTS
1. (i) Function or Mapping
Let X and Y be two nonempty sets and let f X Y#3 be a relation from
X to Y such that
(i) dom ( )f X and
(ii) no two different ordered pairs in f have the same fi rst coordinate.
Then, f is called a function or a mapping from X to Y and we write,
: .f X Y"
(ii) Dom ( ) { : ( , ) } ,f x x y f X d range ( ) { : ( , ) }f y x y f Y3 d and
co-domain( ) .f Y
Functions 129
(iii) If ( , ) ,x y fd we write, ( ) .f x y
Here, y is called the image of x under f, while x is called the pre-image
of y.
(iv) Two functions f and g having same domain X are said to be equal only
when ( ) ( )f x g x for all .x Xd
2. (i) Function or Mapping
Let X and Y be two nonempty sets. Then, a rule f which associates to
each ,x Xd a unique element ( )f x of Y, is called a function from X to Y.
We write, : .f X Y" Dom ( ) ,f X range ( ) { ( ) : }f f x x X Y3 d and
co-domain ( ) .f Y
(ii) A mapping :f X Y" is one-one if ( ) ( ) .f x f x x x1 2 1 2&
(iii) Real Functions
Let R be the set of all real numbers. Then, a function :f R R" is called
a real function.
3. (i) Constant Function
Let k be a fi xed real number. Then, a function : : ( )f R R f x k" for all
x Rd is called a constant function. Dom ( )f R and range ( ) { } .f k
(ii) Identity Function
The function : : ( )I R R I x x" for all x Rd is called the identity
function. Dom (I) = Range ( ) .I R
(iii) Modulus Function
The function : : ( ) | |f R R f x x" for all x Rd is called the modulus
function. Dom ( )f R and range ( ) { : } [ , ) .andf x x R x 0 0 3$ d
(iv) Greatest Integer Function, or Step Function
The function : : ( ) [ ],f R R f x x" where [ ]x denotes the greatest integer
less than or equal to x, is called the greatest integer function.
Dom ( ) ,f R range ( ) .f Z
EXAMPLES [3.85] = 3, [0.94] = 0, [5.0] = 5, [–7.35] = –8 and [–6.0] = –6, etc.
(v) Smallest Integer Function
The function : : ( ) ,f R R f x x" _ i where x_ i denotes the smallest
integer greater than or equal to x. Dom ( )f R and range ( ) .f Z
EXAMPLES [6.8] = 7, [5] = 5, [0.85] = 1, [– 4.5] = – 4, etc.
(vi) Fractional Part Function
The function : : ( ) { },f R R f x x" where { }x denotes the fractional part
or decimal part of x. Note that { } [ ] .x x x
Dom ( )f R and range ( ) [ , ) .f 0 1
EXAMPLES {6.83} = 0.83, {–3.65} = 0.35, {8} = 0, {–3} = 0, etc.
(vii) Signum Function
The function ,: : ( )
| |
,
when
when
f R R f x x
x
x
x
0
0 0
" !
Z
[
\
]
]
]]
]
]
]]
is called the signum function.
130 Senior Secondary School Mathematics for Class 11
Dom ( )f R and range ( ) { , , } .f 1 0 1
(viii) Exponential Function
Let a be a real number such that a 0 and .a 1!
Then, : : ( )f R R f x ax" for all x Rd is called the exponential function.
Dom ( )f R and range ( ) ( , ) .f 0 3
In particular, : : ( )f R R f x ex" for all .x Rd
(ix) Logarithmic Function
The function : : ( ) logf R R f x xe"
for all x Rd is called the
logarithmic function. Dom ( ) ( , )f 0 3 and range ( ) .f R
(x) Reciprocal Function
The function : { } : ( )f R R f x x0
1" for all { }x R 0d is called the
reciprocal function. Dom ( ) { }f R 0 and range ( ) { } .f R 0
(xi) Square Root Function
The function : { } : ( )f R R f x x0 ", for all { }x R 0,d is called the
square root function. Dom ( ) { } [ , ),f R 0 0, 3 range ( ) [ , ) .f 0 3
(xii) Cube Root Function
The function : : ( )f R R f x x /1 3" for all x Rd is called the cube root
function. Dom ( )f R and range ( ) .f R
(xiii) Rational Function
The function : : ( ) ( )
( )
f R R f x
q x
p x
" where ( )p x and ( )q x are polynomials
and ( ) ,q x 0! is called a rational function. Dom ( ) { : ( ) } .f R x q x 0
4. OPERATIONS ON REAL FUNCTIONS
Let f and g be real functions having domains D1 and D2 respectively. Then,
(i) ( ) : : ( ) ( ) ( ) ( ) .f g D D R f g x f x g x1 2"+
(ii) ( ) : : ( ) ( ) ( ) ( ) .f g D D R f g x f x g x1 2"+
(iii) ( ) : : ( ) ( ) ( ) ( )fg D D R fg x f x g x·1 2"+
(iv) ( ) : : ( ) ( ) ( ) .f D R f x f x·1"
(v) ,: ( ) { : ( ) } : ( ) ( )
( )
( ) .g
f
D D x g x R g
f
x
g x
f x
g x0 01 2 "+ ! e o
Principle of Mathematical Induction 131
4
Principle of Mathematical Induction
PRINCIPLE OF MATHEMATICAL INDUCTION
Let ( )P n be a statement involving the natural number n such that
(i) ( )P 1 is true and (ii) ( )P k 1 is true, whenever ( )P k is true
then ( )P n is true for all .n Nd
SOLVED EXAMPLES
EXAMPLE 1 Using the principle of mathematical induction, prove that
( )n n1 3 5 7 2 1… 2 for all .n Nd
SOLUTION Let the given statement be ( ) .P n Then,
( ) : … ( ) .P n n n1 3 5 7 2 1 2
Putting n 1 in the given statement, we get
LHS = 1 and RHS .1 12
LHS = RHS.
Thus, ( )P 1 is true.
Let ( )P k be true. Then,
( ) : … ( ) .P k k k1 3 5 7 2 1 2 … (i)
Now, ( ) { ( ) }k k1 3 5 7 2 1 2 1 1…
{ ( )} ( )k k1 3 5 7 2 1 2 1…
( )k k2 12 [using (i)]
( ) .k 1 2
( ) : …( ) { ( ) } ( ) .P k k k k1 1 3 5 7 2 1 2 1 1 1 2
This shows that ( )P k 1 is true, whenever ( )P k is true.
Thus, ( )P 1 is true and ( )P k 1 is true, whenever ( )P k is true.
Hence, by the principle of mathematical induction, we have
( )n n1 3 5 7 2 1… 2 for all .n Nd
EXAMPLE 2 Using the principle of mathematical induction, prove that
… ( ) ( )n n n1 4 7 10 3 2 2
1 3 1 for all .n Nd
SOLUTION Let the given statement be ( ) .P n Then,
( ) : … ( ) ( ) .P n n n n1 4 7 10 3 2 2
1 3 1
131
132 Senior Secondary School Mathematics for Class 11
Putting n 1 in the given statement, we get
LHS = 1 and RHS ( ) .2
1 1 3 1 1 1# # #
LHS = RHS.
Thus, P(1) is true.
Let ( )P k be true. Then,
( ) : … ( ) ( ) .P k k k k1 4 7 10 3 2 2
1 3 1 … (i)
Now, ( ) { ( ) }k k1 4 7 3 2 3 1 2…
{ ( )} ( )k k1 4 7 3 2 3 1…
( ) ( )k k k2
1 3 1 3 1 [using (i)]
( ) ( )k k k k k2
1 3 6 2 2
1 3 5 22 2
( ) { ( ) ( )}k k k k k k2
1 3 3 2 2 2
1 3 1 2 1·2
( ) ( ) ( ) { ( ) } .k k k k2
1 1 3 2 2
1 1 3 1 1
( ) : … { ( ) } ( ) { ( ) } .P k k k k1 1 4 7 3 1 2 2
1 1 3 1 1
This shows that ( )P k 1 is true, whenever ( )P k is true.
Thus, ( )P 1 is true and ( )P k 1 is true, whenever ( )P k is true.
Hence, by the principle of mathematical induction, we have
( ) ( )n n n1 4 7 3 2 2
1 3 1… for all .n Nd
EXAMPLE 3 Using the principle of mathematical induction, prove that
( ) ( )n n n n1 2 3 6
1 1 2 1…2 2 2 2 for all .n Nd
SOLUTION Let the given statement be ( ) .P n Then,
( ): … ( )( ) .P n n n n n1 2 3 6
1 1 2 12 2 2 2
Putting n 1 in the given statement, we get
LHS 1 12 and RHS ( ) .6
1 1 2 2 1 1 1# # # #
LHS = RHS.
Thus, ( )P 1 is true.
Let ( )P k be true. Then,
( ): … ( )( ) .P k k k k k1 2 3 6
1 1 2 12 2 2 2 … (i)
Now, ( )k k1 2 3 1…2 2 2 2 2
{ } ( )k k1 2 3 1…2 2 2 2 2
( )( ) ( )k k k k6
1 1 2 1 1 2 [using (i)]
Principle of Mathematical Induction 133
( ) { ( ) ( )}k k k k6
1 1 2 1 6 1·
( ) ( ) ( ) {( ) ( )}k k k k k k k6
1 1 2 7 6 6
1 1 2 4 3 62 2
( )( )( ) ( )( )[ ( ) ] .k k k k k k6
1 1 2 2 3 6
1 1 1 1 2 1 1
( ): … ( )P k k1 1 2 3 12 2 2 2
( )( )[ ( ) ] .k k k6
1 1 1 1 2 1 1
This shows that ( )P k 1 is true, whenever ( )P k is true.
Thus, ( )P 1 is true and ( )P k 1 is true, whenever ( )P k is true.
Hence, by the principle of mathematical induction, we have
( )( )n n n n1 2 3 6
1 1 2 1…2 2 2 2 for all .n Nd
EXAMPLE 4 Using the principle of mathematical induction, prove that
…
( )
n
n n
1 2 3 2
13 3 3 3
2
( 2 for all .n Nd
SOLUTION Let the given statement be ( ) .P n Then,
·( ): …
( )
P n n
n n
1 2 3 2
13 3 3 3
2
( 2
Putting n 1 in the given statement, we get
LHS 1 13 and RHS .2
1 2 1 1
2
2# b l
LHS = RHS.
Thus, ()P 1 is true.
Let ( )P k be true for some .k Nd Then,
·( ): …
( )
P k k
k k
1 2 3 2
13 3 3 3
2 2
) 3 … (i)
Now, … ( )k k1 2 3 13 3 3 3 3
{ } ( )k k1 2 3 1…3 3 3 3 3
( )
( )
k k
k2
1
1
2
3
( 2 [using (i)]
( ) ( ) ( )k k k k k k1 4 1 1 4
4 42 2 2 2 ( (2 2
( ) ( ) ( ) {( ) }k k k k
4
1 2
2
1 1 12 2 2
( 2
·( ): … ( )
( ) {( ) }
P k k
k k
1 1 2 3 1 2
1 1 13 3 3 3
2
< F
This shows that ( )P k 1 is true, whenever ( )P k is true.
134 Senior Secondary School Mathematics for Class 11
( )P 1 is true and ( )P k 1 is true, whenever ( )P k is true.
Hence, by the principle of mathematical induction, we have
( )
n
n n
1 2 3 2
1
…3 3 3 3
2
( 2 for all .n Nd
EXAMPLE 5 Using the principle of mathematical induction, prove that
· · · … ( ) ( )( ) .n n n n n1 2 2 3 3 4 1 3
1 1 2
SOLUTION Let the given statement be ( ) .P n Then,
( ): · · · … ( ) ( )( ) .P n n n n n n1 2 2 3 3 4 1 3
1 1 2
When ,n 1 LHS ·1 2 2 and RHS ( ) .3
1 1 2 1 2 2# # #
LHS = RHS.
Thus, the given statement is true for ,n 1 i.e., ( )P 1 is true.
Let ( )P k be true. Then,
( ): · · · … ( ) ( )( ) .P k k k k k k1 2 2 3 3 4 1 3
1 1 2 … (i)
Now, · · · … ( ) ( )( )k k k k1 2 2 3 3 4 1 1 2
{ · · · … ( ) ( )( )k k k k1 2 2 3 3 4 1 1 2
( )( ) ( )( )k k k k k3
1 1 2 1 2 [using (i)]
· [ ( )( ) ( )( )] ( )( )( ) .k k k k k k k k3
1 1 2 3 1 2 3
1 1 2 3
( ) : · · · … ( )( )P k k k1 1 2 2 3 3 4 1 2
( )( )( ) .k k k3
1 1 2 3
This shows that ( )P k 1 is true, whenever ( )P k is true.
( )P 1 is true and ( )P k 1 is true, whenever ( )P k is true.
Hence, by the principle of mathematical induction, we have
· · · … ( ) ( )( )n n n n n1 2 2 3 3 4 1 3
1 1 2 for all .n Nd
EXAMPLE 6 Using the principle of mathematical induction, prove that
· · · … ( )( ) ( ) .n n n n n1 3 3 5 5 7 2 1 2 1 3
1 4 6 12
SOLUTION Let the given statement be ( ) .P n Then,
( ): · · · … ( )( ) ( ) .P n n n n n n1 3 3 5 5 7 2 1 2 1 3
1 4 6 12
When ,n 1 we have
LHS ·1 3 3 and RHS ( ) .3
1 1 4 1 6 1 1 3
1 1 9 32# # # # # #
LHS = RHS.
Principle of Mathematical Induction 135
Thus, P(1) is true.
Let ( )P k be true. Then,
( ): · · · … ( )( ) ( ) .P k k k k k k1 3 3 5 5 7 2 1 2 1 3
1 4 6 12 … (i)
Now, · · · … ( )( ) { ( ) }{ ( ) }k k k k1 3 3 5 5 7 2 1 2 1 2 1 1 2 1 1
{ · · · … ( )( )} ( )( )k k k k1 3 3 5 5 7 2 1 2 1 2 1 2 3
( ) ( )( )k k k k k3
1 4 6 1 2 1 2 32 [using (i)]
[( ) ( )] ( )k k k k k k k k3
1 4 6 3 4 8 3 3
1 4 18 23 93 2 2 3 2
( )( ) ( )[ ( ) ( ) ] .k k k k k k3
1 1 4 14 9 3
1 1 4 1 6 1 12 2
( ): · · · … { ( ) } { ( ) }P k k k1 1 3 3 5 5 7 2 1 1 2 1 1
( ) { ( ) ( ) } .k k k3
1 1 4 1 6 1 12
Thus, ( )P k 1 is true, whenever ( )P k is true.
( )P 1 is true and ( )P k 1 is true, whenever ( )P k is true.
Hence, by the principle of mathematical induction, we have
· · · … ( )( ) ( )n n n n n1 3 3 5 5 7 2 1 2 1 3
1 4 6 12 for all .n Nd
EXAMPLE 7 Using the principle of mathematical induction, prove that
· · · · … ( )( ) ( )( )( )n n n n n n n1 2 3 2 3 4 1 2 4
1 1 2 3
for all .n Nd
SOLUTION Let the given statement be ( ) .P n Then,
( ): · · · · … ( )( ) ( )( )( ) .P n n n n n n n n1 2 3 2 3 4 1 2 4
1 1 2 3
When ,n 1 we have
LHS · ·1 2 3 6 and RHS .4
1 1 2 3 4 6# # # #
LHS = RHS.
Thus, the given statement is true for ,n 1 i.e., P(1) is true.
Let ( )P k be true. Then,
( ): · · · · … ( )( )P k k k k1 2 3 2 3 4 1 2
( )( )( ) .k k k k4
1 1 2 3 … (i)
Now, · · · · … ( )( ) ( )( )( )k k k k k k1 2 3 2 3 4 1 2 1 2 3
{ · · · · … ( )( )} ( )( )( )k k k k k k1 2 3 2 3 4 1 2 1 2 3
( )( )( )
( )( )( )
k k k k
k k k
4
1 1 2 3 4
4 1 2 3
[using (i)]
( )( )( )( ) .k k k k4
1 1 2 3 4
136 Senior Secondary School Mathematics for Class 11
( ): · · · · … ( )( )( )P k k k k1 1 2 3 2 3 4 1 2 3
( )( )( )( ) .k k k k4
1 1 2 3 4
Thus, ( )P k 1 is true, whenever ( )P k is true.
P(1) is true and ( )P k 1 is true, whenever ( )P k is true.
Hence, by the principle of mathematical induction, we have
· · · · … ( )( ) ( )( )( )n n n n n n n1 2 3 2 3 4 1 2 4
1 1 2 3
for all values of .n Nd
EXAMPLE 8 Using the principle of mathematical induction, prove that
·… ( ) ( )n n n
n
1 2
1
2 3
1
3 4
1
1
1
1· · ·
SOLUTION Let the given statement be ( ) .P n Then,
( ): … ( ) ( ) ·P n n n n
n
1 2
1
2 3
1
3 4
1
1
1
1· · ·
Putting n 1 in the given statement, we get
LHS ·1 2
1
2
1 and RHS ·( )1 1
1
2
1
LHS = RHS.
Thus, ( )P 1 is true.
Let ( )P k be true. Then,
( ): · · · … ( ) ( ) ·P k k k k
k
1 2
1
2 3
1
3 4
1
1
1
1
… (i)
Now, · · · … ( ) ( )( )k k k k1 2
1
2 3
1
3 4
1
1
1
1 2
1
· · · ( ) ( )( )k k k k1 2
1
2 3
1
3 4
1
1
1
1 2
1… ( 2
( ) ( )( )k
k
k k1 1 2
1 [using (i)]
·( )( )
( )
( )( )
( )
( )
( )
k k
k k
k k
k
k
k
1 2
2 1
1 2
1
2
12
·( ): ( )( ) ( )
( )
P k
k k k
k
1 1 2
1
2 3
1
3 4
1
1 2
1
2
1
· · ·
Thus, ( )P k 1 is true, whenever ( )P k is true.
P(1) is true and ( )P k 1 is true, whenever ( )P k is true.
Hence, by the principle of mathematical induction, we have
· · · … ( )n n n
n
1 2
1
2 3
1
3 4
1
1
1
1
for all .n Nd
EXAMPLE 9 Using the principle of mathematical induction, prove that
·· · · … ( )( ) ( )n n n
n
3 5
1
5 7
1
7 9
1
2 1 2 3
1
3 2 3
Principle of Mathematical Induction 137
SOLUTION Let the given statement be ( ) .P n Then,
·( ): · · · … ( )( ) ( )P n n n n
n
3 5
1
5 7
1
7 9
1
2 1 2 3
1
3 2 3
Putting n 1 in the given statement, we get
LHS ·3 5
1
15
1 and RHS ( ) ·3 2 1 3
1
15
1
#
LHS = RHS.
Thus, P(1) is true.
Let ( )P k be true. Then,
·( ): … ( )( ) ( )P k k k k
k
3 5
1
5 7
1
7 9
1
2 1 2 3
1
3 2 3· · ·
… (i)
Now, · · … ( )( ) { ( ) } { ( ) }k k k k3 5
1
5 7
1
2 1 2 3
1
2 1 1 2 1 3
1
· · ( )( ) ( )( )k k k k3 5
1
5 7
1
2 1 2 3
1
2 3 2 5
1… ( 2
( ) ( )( )k
k
k k3 2 3 2 3 2 5
1 [using (i)]
( )( )
( )
( )( )
( )
( )( )
( )( )
k k
k k
k k
k k
k k
k k
3 2 3 2 5
2 5 3
3 2 3 2 5
2 5 3
3 2 3 2 5
1 2 32
·( )
( )
{ ( ) }
( )
k
k
k
k
3 2 5
1
3 2 1 3
1
( ): · · · … { ( ) } { ( ) }P k k k1 3 5
1
5 7
1
7 9
1
2 1 1 2 1 3
1
·{ ( ) }
( )
k
k
3 2 1 3
1
This shows that ( )P k 1 is true, whenever ( )P k is true.
P(1) is true and ( )P k 1 is true, whenever ( )P k is true.
Hence, by the principle of mathematical induction, we have
· · · … ( )( ) ( )n n n
n
3 5
1
5 7
1
7 9
1
2 1 2 3
1
3 2 3
for all values of .n Nd
EXAMPLE 10 Using the principle of mathematical induction, prove that
· · · · … ( )( ) ( )( )
( )
n n n n n
n n
1 2 3
1
2 3 4
1
1 2
1
4 1 2
3
for all .n Nd
SOLUTION Let the given statement be ( ) .P n Then,
·( ): · · · · … ( )( ) ( )( )
( )
P n
n n n n n
n n
1 2 3
1
2 3 4
1
1 2
1
4 1 2
3
Putting n 1 in the given statement, we get
LHS · ·1 2 3
1
6
1 and RHS ·( ) ( )
( )
4 1 1 1 2
1 1 3
4 2 3
1 4
6
1
#
#
# #
#
138 Senior Secondary School Mathematics for Class 11
LHS = RHS.
Thus,the given statement is true for ,n 1 i.e., P(1) is true.
Let ( )P k be true. Then,
·( ): · · · · … ( )( ) ( )( )
( )
P k
k k k k k
k k
1 2 3
1
2 3 4
1
1 2
1
4 1 2
3
… (i)
Now, · · · · … ( )( ) ( )( )( )k k k k k k1 2 3
1
2 3 4
1
1 2
1
1 2 3
1
· · · · ( )( ) ( )( )( )k k k k k k1 2 3
1
2 3 4
1
1 2
1
1 2 3
1… ( 2
( )( )
( )
( )( )( )k k
k k
k k k4 1 2
3
1 2 3
1
) 3 [using (i)]
( )( )( )
( )
( )( )( )
( )
k k k
k k
k k k
k k k
4 1 2 3
3 4
4 1 2 3
6 9 42 3 2
·( )( )( )
( )( )( )
( )( )
( )( )
k k k
k k k
k k
k k
4 1 2 3
1 1 4
4 2 3
1 4
( ): · · · · … ( )( )( )P k k k k1 1 2 3
1
2 3 4
1
1 2 3
1
·( )( )
( )( )
k k
k k
4 2 3
1 4
This shows that ( )P k 1 is true, whenever ( )P k is true.
P(1) is true and ( )P k 1 is true, whenever ( )P k is true.
Hence, by the principle of mathematical induction, we have
· · · · · · … ( )( ) ( )( )
( )
n n n n n
n n
1 2 3
1
2 3 4
1
3 4 5
1
1 2
1
4 1 2
3
for all values of .n Nd
EXAMPLE 11 Using the principle of mathematical induction, prove that
· · · … ·
( )
n
n
1 3 2 3 3 3 3 4
2 1 3 3n
n
2 3
1
for all .n Nd
SOLUTION Let the given statement be ( ) .P n Then,
·( ): · · · … ·
( )
P n n
n
1 3 2 3 3 3 3 4
2 1 3 3n
n
2 3
1
Putting n 1 in the given statement, we get
LHS ·1 3 3 and RHS
( )
.4
2 1 1 3 3
4
1 9 3
4
12 3
1 1# #
LHS = RHS.
Thus, P(1) is true.
Let ( )P k be true. Then,
·( ): · · · … ·
( )
P k k
k
1 3 2 3 3 3 3 4
2 1 3 3k
k
2 3
1
… (i)
· · · … · ( )k k1 3 2 3 3 3 3 1 3( )k k2 3 1
Principle of Mathematical Induction 139
{ · · · … · } ( )k k1 3 2 3 3 3 3 1 3( )k k2 3 1
( )
( )
k
k4
2 1 3 3
1 3
k
k
1
1
[using (i)]
( ) ( ) ( )k k k k
4
2 1 3 3 4 1 3
4
2 1 4 4 3 3k k k1 1 1
·
( ) { ( ) }k k
4
6 3 3 3
4
2 1 1 3 3( )k k1 1 1
( ): · · · … ( ) ·P k k1 1 3 2 3 3 3 1 3k2 3 1
·
{ ( ) }k
4
2 1 1 3 3( )k 1 1
This shows that ( )P k 1 is true, whenever ( )P k is true.
Thus, P(1) is true and ( )P k 1 is true, whenever ( )P k is true.
Hence, by the principle of mathematical induction, we have
· · · … ·
( )
n
n
1 3 2 3 3 3 3 4
2 1 3 3n
n
2 3
1
for all values of .n Nd
EXAMPLE 12 Using the principle of mathematical induction, prove that
( )n n1 2
1 1 3
1 1 4
1 1 1
1
1
1… b b b bl l l l for all .n Nd
SOLUTION Let the given statement be ( ) .P n Then,
·( ): … ( )P n n n1 2
1 1 3
1 1 4
1 1 1
1
1
1 b b b bl l l l
When ,n 1 LHS 1 2
1
2
1 b l and RHS ·( )1 1
1
2
1
LHS = RHS.
Thus, P(1) is true.
Let ( )P k be true. Then,
·( ): … ( )P k k k1 2
1 1 3
1 1 4
1 1 1
1
1
1 b b b bl l l l … (i)
Now, ·
k k
1 2
1 1 3
1 1 4
1 1 1
1 1 2
1… b b b b bl l l l l( 2
· ·( ) ( )
( )
( ) ( )
( )
( )k k
k
k k
k
k1
1
2
2 1
1
1
2
1
2
1
= G [using (i)].
·( ): … ( )P k k k1 1 2
1 1 3
1 1 4
1 1 2
1
2
1 b b b bl l l l
This shows that ( )P k 1 is true, whenever ( )P k is true.
Thus, P(1) is true and ( )P k 1 is true, whenever ( )P k is true.
Hence, by the principle of mathematical induction, we have
( )n n1 2
1 1 3
1 1 4
1 1 1
1
1
1… b b b bl l l l for all .n Nd
140 Senior Secondary School Mathematics for Class 11
EXAMPLE 13 Using the principle of mathematical induction, prove that
( )
( )
a ar ar ar
r
a r
1
1
… n
n
2 1
for r 1 and all .n Nd
SOLUTION Let the given statement be ( ) .P n Then,
·( ): … ( )
( )
P n a ar ar ar
r
a r
1
1n
n
2 1
When ,n 1 we have
LHS a and RHS ( )
( )
.
r
a r
a1
11
LHS = RHS.
Thus, P(1) is true.
Let ( )P k be true. Then,
·( ): … ( )
( )
P k a ar ar ar
r
a r
1
1k
k
2 1
… (i)
Now, ( … ) ( )
( )
a ar ar ar ar
r
a r
ar1
1k k
k
k2 1
[using (i)]
·( )
( )
r
a r
1
1k 1
·( ): … ( )
( )
P k a ar ar ar ar
r
a r
1 1
1k k
k
2 1
1
This shows that ( )P k 1 is true, whenever ( )P k is true.
P(1) is true and ( )P k 1 is true, whenever ( )P k is true.
Hence, by the principle of mathematical induction, we have
( )
( )
a ar ar ar
r
a r
1
1
… n
n
2 1
for r 1 and all .n Nd
EXAMPLE 14 Let a and b be arbitrary real numbers. Using the principle of mathematical
induction, prove that
( )ab a bn n n for all .n Nd
SOLUTION Let the given statement be ( ) .P n Then,
( ): ( ) .P n ab a bn n n
When ,n 1 we have
LHS ( )ab ab1 and RHS .a b ab1 1
LHS = RHS.
Thus, the given statement is true for ,n 1 i.e., P(1) is true.
Let ( )P k be true. Then,
( ): ( ) .P k ab a bk k k … (i)
Now, ( ) ( ) ( )ab ab abk k1 ( )( )a b abk k [using (i)]
( ) ( )a a b b· ·k k [by commutativity and associativity of
multiplication on real numbers]
( ) .a b·k k1 1
Principle of Mathematical Induction 141
( ): ( ) ( · ) .P k ab a b1 k k k1 1 1
This shows that ( )P k 1 is true, whenever ( )P k is true.
P(1) is true and ( )P k 1 is true, whenever ( )P k is true.
Hence, by the principle of mathematical induction, we have
( )ab a bn n n for all .x Nd
EXAMPLE 15 Using the principle of mathematical induction, prove that ( )n n2 is even
for all .n Nd
SOLUTION Let ( ): ( )P n n n2 is even.
For ,n 1 the given expression becomes ( ) ,1 1 22 which is even.
So, the given statement is true for ,n 1 i.e., ( )P 1 is true.
Let ( )P k be true. Then,
( ): ( )P k k k2 is even
& ( )k k m22 for some natural number m. … (i)
Now, ( ) ( )k k k k1 1 3 22 2 ( ) ( )k k k2 12
( )m k2 2 1 [using (i)]
[ ( )],m k2 1 which is clearly even.
( ): ( ) ( )P k k k1 1 12 is even.
Thus, ( )P k 1 is true, whenever ( )P k is true.
P(1) is true and ( )P k 1 is true, whenever ( )P k is true.
Hence, by the principle of induction, it follows that ( )n n2 is even
for all .n Nd
EXAMPLE 16 Using the principle of mathematical induction, prove that ( )( )n n n1 5
is a multiple of 3 for all .n Nd
SOLUTION Let ( ): ( )( )P n n n n1 5 is a multiple of 3.
For ,n 1 the given expression becomes ( ) ,1 2 6 12# # which is a
multiple of 3.
So, the given statement is true for ,n 1 i.e., P(1) is true.
Let ( )P k be true. Then,
( ): ( )( )P k k k k1 5 is a multiple of 3
& ( )( )k k k m1 5 3 for some natural number m. ... (i)
Now, ( )( )( )k k k1 2 6
( )( ) ( )( )k k k k k1 2 6 1 2
( )( ) ( )( )k k k k k1 5 3 6 1 2
( )( ) ( ) ( )( )k k k k k k k1 5 3 1 6 1 2
( )( )m k k k3 3 1 2 4
( )( )m k k3 3 1 4
{ ( )( )},m k k3 1 4 which is a multiple of 3.
( ): ( )( )( )P k k k k1 1 2 6 is a multiple of 3.
142 Senior Secondary School Mathematics for Class 11
Thus, ( )P k 1 is true, whenever ( )P k is true.
( )P 1 is true and ( )P k 1 is true, whenever ( )P k is true.
Hence, by the principle of mathematical induction, it follows that
( )( )n n n1 5 is a multiple of 3 for all .n Nd
EXAMPLE 17 Using the principle of mathematical induction, prove that ( )7 3n n is
divisible by 4 for all .n Nd
SOLUTION Let ( ): ( )P n 7 3n n is divisible by 4.
For ,n 1 the given expression becomes ( ) ,7 3 41 1 which is
divisible by 4.
So, the given statement is true for ,n 1 i.e., P(1) is true.
Let ( )P k be true. Then,
( ): ( )P k 7 3k k is divisible by 4.
& ( ) m7 3 4k k for some natural number m. … (i)
Now, { }7 3( ) ( )k k1 1
· ·7 7 3 7 3 3( ) ( )k k k k1 1 [subtracting and adding · ]7 3k
( ) ( )7 7 3 3 7 3k k k
·( )m7 4 4 3k# [using (i)]
( ),m4 7 3k which is clearly divisible by 4.
( ): { }P k 1 7 3( ) ( )k k1 1 is divisible by 4.
Thus, ( )P k 1 is true, whenever ( )P k is true.
P(1) is true and ( )P k 1 is true, whenever ( )P k is true.
Hence, by the principle of mathematical induction, it follows that
( )7 3n n is divisible by 4 for all values of .n Nd
EXAMPLE 18 Using the principle of mathematical induction, prove that ( )10 1n2 1 is
divisible by 11 for all .n Nd
SOLUTION Let ( ): ( )P n 10 1n2 1 is divisible by 11.
For ,n 1 the given expression becomes { } ,10 1 11( )2 1 1 # which
is divisible by 11.
So, the given statement is true for ,n 1 i.e., P(1) is true.
Let ( )P k be true. Then,
( ): ( )P k 10 1k2 1 is divisible by 11
& ( ) ,m10 1 11k2 1 for some natural number m. … (i)
Now, ·{ } { }10 1 10 10 1( ) ( )k k2 1 1 2 2 1
{ }100 10 1 99k2 1#
( )m100 11 99# [using (i)]
( ),m11 100 9# which is divisible by 11.
( ): { }P k 1 10 1( )k2 1 1 is divisible by 11.
Thus, ( )P k 1 is true, whenever ( )P k is true.
Principle of Mathematical Induction 143
P(1) is true and ( )P k 1 is true, whenever ( )P k is true.
Hence, by the principle of mathematical induction, it follows that
{ }10 1n2 1 is divisble by 11 for all .n Nd
EXAMPLE 19 Using the principle of mathematical induction, prove that ( · · )2 7 3 5 5n n
is divisible by 24 for all .n Nd
SOLUTION Let ( ): ( · · )P n 2 7 3 5 5n n is divisible by 24.
For ,n 1 the given expression becomes ( · · ) ,2 7 3 5 5 241 1
which is clearly divisible by 24.
So, the given statement is true for ,n 1 i.e., P(1) is true.
Let ( )P k be true. Then,
( ): ( · · )P k 2 7 3 5 5k k is divisible by 24
& ( · · ) ,m2 7 3 5 5 24k k for some .m Nd … (i)
Now, ( · · )2 7 3 5 5k k1 1
( · · · · )2 7 7 3 5 5 5k k
( · · ) ·7 2 7 3 5 5 6 5 30k k k
( ) ( )m7 24 6 5 5k# [using (i)]
( ) ( )m24 7 6 5 5 1k 1# # #
( )m p24 7 5 24# # ( ) ( ), . .,
( )
is divisible by i e
p
5 1 5 1 4
5 1 4
k
k
1
1&
a
R
T
S
S
S
S
S
SS
V
X
W
W
W
W
W
WW
( ),m p24 7 5# which is clearly divisible by 24.
( ): ( · · )P k 1 2 7 3 5 5k k1 1 is divisible by 24.
Thus, ( )P k 1 is true, whenever ( )P k is true.
P(1) is true and ( )P k 1 is true, whenever ( )P k is true.
Hence, by the principle of mathematical induction, it follows that
( · · )2 7 3 5 5n n is divisible by 24 for all .n Nd
EXAMPLE 20 Using the principle of mathematical induction, prove that ( )x yn n is
divisible by ( )x y for all .n Nd
SOLUTION Let the given statement be ( ) .P n Then,
( ): ( )P n x yn n is divisible by ( ) .x y
When ,n 1 the given statement becomes: ( )x y1 1 is divisible by
( ),x y which is clearly true.
P(1) is true.
Let ( )P k be true. Then,
( ): ( )P k x yk k is divisible by ( ) .x y … (i)
Now, ( )x yk k1 1
{ }x x y x y yk k k k1 1 [on adding and subtracting ]x yk
( ) ( ),x x y y x yk k k which is divisible by ( )x y [using (i)].
144 Senior Secondary School Mathematics for Class 11
Thus, ( )P k 1 is true, whenever ( )P k is true.
P(1) is true and ( )P k 1 is true, whenever ( )P k is true.
Hence, by the principle of mathematical induction, it follows that
( )x yn n is divisible by ( )x y for all .n Nd
EXAMPLE 21 Using the principle of mathematical induction, prove that
( ) ( )x nx1 1n$ for all ,n Nd where .x 1
SOLUTION Let the given statement be ( ) .P n Then,
( ): ( ) ( ),P n x nx1 1n$ where .x 1
When ,n 1 we have
LHS ( ) ( )x x1 11 and RHS ( ) ( ) .x x1 1 1#
LHS RHS$ is true.
Thus, P(1) is true.
Let ( )P k be true. Then,
( ): ( ) ( ),P k x kx1 1k$ where .x 1 … (i)
Now, ( ) ( ) ( )x x x1 1 1k k1
( )( )kx x1 1$ [using (i)]
( )k x kx1 1 2
( )k x1 1$ [ ]kx 02a $ .
( ): ( ) ( ) .P k x k x1 1 1 1k 1$
Thus, ( )P k 1 is true, whenever ( )P k is true.
P(1) is true and ( )P k 1 is true, whenever ( )P k is true.
Hence, by the principle of mathematical induction, we have
( ) ( ),x nx1 1n$ where x 1 for all .n Nd
EXAMPLE 22 Using the principle of mathematical induction, prove that n 2 n for all
.n Nd
SOLUTION Let the given statement be ( ) .P n Then,
( ): .P n n 2 n
When ,n 1 we get LHS = 1 and RHS .2 21
Clearly, 1 < 2.
( )P n is true for ,n 1 i.e., P(1) is true.
Let ( )P k be true. Then,
( ): .P k k 2 k
Now, k k2 2 2 k k 1&
( )k k 2 k 1&
( ) ( )k k k1 2 k 1& # [ ]k1a #
( )k 1 2 k 1& .
( ): ( ) .P k k1 1 2 k 1
Thus, ( )P k 1 is true, whenever ( )P k is true.
Principle of Mathematical Induction 145
P(1) is true and ( )P k 1 is true, whenever ( )P k is true.
Hence, by the principle of mathematical induction, it follows that
n 2 n for all .n Nd
EXAMPLE 23 Using the principle of mathematical induction, prove that
( ) ( )n n2 7 3 2 for all values of .n Nd
SOLUTION Let the given statement be ( ) .P n Then,
( ): ( ) ( ) .P n n n2 7 3 2
When ,n 1 we have
LHS ( )2 1 7 9# and RHS ( ) .1 3 4 162 2
LHS < RHS [ ]9 16a .
Thus, ( )P n is true for ,n 1 i.e., P(1) is true.
Let ( )P k be true. Then,
( ): ( ) ( ) .P k k k2 7 3 2 … (i)
Now, ( ) ( )k k2 1 7 2 7 2
( ) ( )k k k3 2 6 11 2 2 [using (i)]
( ) ( ) ( ) .k k k k8 16 4 1 3 2 2 2
( ): ( ) ( ) .P k k k1 2 1 7 1 3 2
Thus, ( )P k 1 is true, whenever ( )P k is true.
P(1) is true and ( )P k 1 is true, whenever ( )P k is true.
Hence, by the principle of mathematical induction, it follows that
( ) ( )n n2 7 3 2 for all .n Nd
EXAMPLE 24 Using the principle of mathematical induction, prove that
( )n n1 2 3…
2 2 2
3
for all values of .n Nd
SOLUTION Let ·( ): ( … )P n n n1 2 3
2 2 2
3
When ,n 1 LHS 1 12 and RHS ·3
1
3
13
Since ,1 3
1 it follows that P(1) is true.
Let ( )P k be true. Then,
·( ): ( … )P k k k1 2 3
2 2 2
3
… (i)
Now, ( )k k1 2 1…2 2 2 2
{ } ( )k k1 2 1…2 2 2 2
( )k k3 1
3
2 [using (i)]
· ·{ ( ) } { }k k k k k3
1 3 1 3
1 3 6 33 2 3 2
146 Senior Secondary School Mathematics for Class 11
·[{( ( )} ( )] [( ) ( )]k k k k k k3
1 1 3 1 3 2 3
1 1 3 23 3
( ) .k3
1 1 3
( ): … ( ) ( ) .P k k k k1 1 2 1 3
1 12 2 2 2 3
Thus, ( )P k 1 is true, whenever ( )P k is true.
P(1) is true and ( )P k 1 is true, whenever ( )P k is true.
Hence, by the principle of mathematical induction, it follows that
( )n n1 2 3 3…
2 2 2 2
3
for all .n Nd
EXAMPLE 25 Using the principle of mathematical induction, prove that
( ) ( )n n1 2 3 8
1 2 1… 2 for all values of .n Nd
SOLUTION Let the given statement be ( ) .P n Then,
( ): ( … ) ( ) .P n n n1 2 3 8
1 2 1 2
When ,n 1 we have
LHS = 1 and RHS ·( )8
1 2 1 1 8
92#
Clearly, ·1 8
9
P(1) is true.
Let ( )P k be true. Then,
( ): ( … ) ( ) .P k k k1 2 3 8
1 2 1 2 … (i)
Now, ( )k k1 2 3 1…
{ } ( )k k1 2 3 1…
( ) ( )
( ) ( )
k k
k k
8
1 2 1 1 8
2 1 8 1
2
2
[using (i)]
·
( ) ( ) { ( ) }k k k k
8
4 12 9
8
2 3
8
2 1 12 2 2
·{ … ( )}
{ ( ) }
k k
k
1 2 3 1 8
2 1 1
2
This shows that ( )P k 1 is true.
Thus, ( )P k 1 is true, whenever ( )P k is true.
P(1) is true and ( )P k 1 is true, whenever ( )P k is true.
Hence, by the principle of mathematical induction, it follows that
( ) ( )n n1 2 3 8
1 2 1… 2 for all values of .n Nd
Principle ofMathematical Induction 147
EXERCISE 4
Using the principle of mathematical induction, prove each of the following for
all :n Nd
1. ( ) .n n n1 2 3 4 2
1 1…
2. ( ) .n n n2 4 6 8 2 1…
3. ( ) .1 3 3 3 3 2
1 3 1… n n2 3 1
4. … ( ).2 6 18 2 3 3 1· n n1
5. ·…2
1
4
1
8
1
2
1 1
2
1
n n d n
6. ·… ( )
( )( )
n
n n n
1 3 5 7 2 1 3
2 1 2 12 2 2 2 2
7. · · · … · ( ) .n n1 2 2 2 3 2 2 1 2 2n n2 3 1
8. · · · … · ( ) .3 2 3 2 3 2 3 2 5
12 6 1n n n2 2 3 3 4 1
9. ·( ) ( ) … ( ) ( )n n
n1 1 2
1
1 2 3
1
1 2 3
1
1
2
…
10. ·· · · … ( )( ) ( )n n n
n
2 5
1
5 8
1
8 11
1
3 1 3 2
1
6 4
11. ·… ( )( ) ( )n n n
n
1 4
1
4 7
1
7 10
1
3 2 3 1
1
3 1· · ·
12. ·… ( )( ) ( )n n n
n
1 3
1
3 5
1
5 7
1
2 1 2 1
1
2 1· · ·
13. ·… ( )( ) ( )n n n
n
2 5
1
5 8
1
8 11
1
3 1 3 2
1
6 4· · ·
14.
( )
( ) .
n
n
n1 1
3 1 4
5 1 9
7 1
2 1
1… 2
2
b b bl l l ) 3
15. ( ) .n n1 1
1 1 2
1 1 3
1 1 1 1… b b b bl l l l
16. ( )( )n n n1 2 is a multiple of 6.
17. ( )x yn n2 2 is divisible by ( ) .x y
18. ( )x 1n2 is divisible by ( ),x 1 where .x 1!
19. {( ) ( ) }41 14n n is divisible by 27.
20. ( )n4 15 1n is divisible by 9.
21. ( )n3 8 9n2 2 is divisible by 8.
22. ( )2 1n3 is a multiple of 7.
23. 3 2n n$
148 Senior Secondary School Mathematics for Class 11
HINTS TO SOME SELECTED QUESTIONS
6. ·( ): … ( )
( )( )
P k k
k k k
1 3 5 2 1 3
2 1 2 12 2 2 2
Now, { … ( ) } { ( ) }k k1 3 5 2 1 2 1 12 2 2 2 2
·
( ) ( )
( ) { ( ) ( ) ( ) }
k k k
k k k k k3
2 1 2 1
2 1 3
1 2 1 2 1 3 2 12 2
( ) { ( ) ( )} ( ) ( )k k k k k k k3
1 2 1 2 1 3 2 1 3
1 2 1 2 5 32
( ) ( ) ( ) .k k k3
1 1 2 1 2 3
7. ( ): … ( ) .P k k k1 2 2 2 3 2 2 1 2 2· · · · ·k k2 3 1 … (i)
Now, { · · · … · } ( ) ·k k1 2 2 2 3 2 2 1 2k k2 3 1
{( ) · } ( ) · · · .k k k k1 2 2 1 2 2 2 2 2 2k k k k1 1 1 2
8. ( ): · · … · ( ) .P k 3 2 3 2 3 2 5
12 6 1k k k2 2 3 1
Now, ( · · … · ) ·3 2 3 2 3 2 3 2k k k k2 2 3 1 1 2
·( ) { ( ) }5
12 6 1 3 2 2 5
1 12 6 1 10 6· ·k k k k k1 1 1# ' 1
· { ( ) } ( ) .5
1 2 6 6 10 6 5
12 6 1k k k1 1 1#
14. ( ): ( ) …
( )
( ) .P k
k
k
k1 3 1 4
5 1 9
7 1
2 1
12
2
c cm m
Z
[
\
]
]]
]
]]
_
`
a
b
bb
b
bb
Now, ( )
( )
( )
k
k
k
k
1 3 1 4
5 1 2 1 1
1
2 1 1
… 2 2
c em o
Z
[
\
]
]]
]
]]
_
`
a
b
bb
b
bb
( )
( )
( ) ( )
( ) ( ) .k
k
k k
k k k1
1
1 2 3
4 4 22 2
2
2 2#
15. ( ): ( ) … ( ) .P k
k
k1 1 1 2
1 1 3
1 1 1 1 c c dm m n
Now, ( )
k k
1 1 1 2
1 1 3
1 1 1 1
1
1… c c d dm m n n) 3
·( ) ( )
( )
( )
( ) .k
k
k
k
k
k1 1
1
1 1
1
2
2
d n
16. Let ( ) ( )k k k p1 2 6 … (i). Then,
( ) ( ) ( ) ( ) ( ) ( ) ( )k k k k k k k k1 2 3 1 2 3 1 2 ( ) ( ) ( ) ( )k k k k k1 2 3 1 2
( ) ( )p k k6 3 1 2 [using (i)]
( ),p q p q6 6 6 where ( ) ( )k k q1 2 2
{ ( ) ( )andk k1 2a being consecutive integers, one
of them must be even and so their product is even}.
17. Let
( )
( )
x y
x y
p
n n2 2
… (i). Then,
{ } ( )x y x x y y( ) ( )n n n n2 1 2 1 2 2 2 2$ $ x x x y x y y yn n n n2 2 2 2 2 2 2 2$ $
Principle of Mathematical Induction 149
( ) ( )x x y y x yn n n2 2 2 2 2 2 ( ) ( )x p x y y x yn2 2 2 2 [using (i)]
( ) [ ],x y x p y n2 2 which is divisible by ( ) .x y
18. Let
( )
( )
.
x
x
p1
1n2
… (i). Then,
{ } ( ) ( ) ( )x x x x x x x x1 1 1 1·( )n n n2 1 2 2 2 2 2 2 2
( ) ( )x p x x1 1·2 2 [using (i)]
( ) ( ),x px x1 12 which is divisible by ( ) .x 1
19. Let
( ) ( )
.p27
41 14k k
… (i). Then,
{( ) ( ) } ( ) ( ) ( ) ( )41 14 41 41 14 41 14 14· ·k k k k k k1 1 1 1
( ) ( ) {( ) ( ) } ( ) p41 41 14 14 41 14 27 41 14 27k k k k# #
[( ) ] .27 41 14k#
20. Let ( )k p4 15 1 9k … (i). Then,
{ ( ) } { } ( )k k k k4 15 1 1 4 15 14 4 4 60 4 45 18·( ) ( )k k k1 1
( ) ( ) ( ) ( )k k p k4 4 15 1 9 5 2 4 9 9 5 2k # [using (i)]
( ),p k9 4 5 2# which is divisible by 9.
21. Let ( )k p3 8 9 8k2 2 … (i). Then,
{ ( ) } { } ( · ) ( )k k k k3 8 1 9 3 3 8 17 9 3 72 81 64 64·( ) ( ) ( )k k k2 1 2 2 2 2 2 2
{ } ( )k k9 3 8 9 64 1· ( )k2 2
( ) ( )p k9 8 64 1# [using (i)]
[ ( )],p k8 9 8 1# which is divisible by 8.
22. Let ( ) p2 1 7k3 … (i). Then,
{ } ( ) ( )2 1 2 2 1 8 2 8 7· ·( )k k k3 1 3 3 3 { ( ) } ( )p8 2 1 7 8 7 7k3 # [using (i)]
( ) ( ),p p56 7 7 8 1 which is divisible by 7.
23. Clearly, .3 21 1$ So, the result is true for .n 1
Let it be true for .n k Then,
3 2k k$ and .3 2 3 3 2 2 3 2· · k k k k1 1& &$ $
So, whenever the result is true for k, then it is also true for ( ) .k 1
SUMMARY OF KEY FACTS
PRINCIPLE OF MATHEMATICAL INDUCTION
Let ( )P n be a statement involving the natural number n such that
(i) P(1) is true and (ii) ( )P k 1 is true, whenever ( )P k is true then ( )P n is true for
all .n Nd
150 Senior Secondary School Mathematics for Class 11
5
Complex Numbers and
Quadratic Equations
IMAGINARY NUMBERS
If the square of a given number is negative then such a number is called an imaginary
number.
For example, , ,1 2 etc., are imaginary numbers.
We denote 1 by the Greek letter iota ‘’, which is transliterated as ‘i ’.
Thus, ,i i4 2 9 3 and ,i5 5 etc.
POWERS OF i We have
, , , ( )i i i i i i i i i1 1 10 1 2 3 2# #
and ( ) ( ) .I i i 1 1 14 2 2# #
Thus, we have
, , , ,i i i i i i i1 1 10 1 2 3 4 .
Let us consider ,in where n is a positive integer and .n 4
On dividing n by 4, let the quotient be m and the remainder be r. Then,
,n m r4 where r0 4# .
( )i i i i i i in m r m r m r r4 4 4# # . [ ]i 14a
EXAMPLES (i) ( )i i i i i 198 4 24 2 4 24 2 2# # . [ ]i 14a
(ii) i
i i i
i
i
i1 1
1
1 1( )
98
98 98 2
2
98 2
2
#
. [ ]andi i1 1
100 2a
SOLVED EXAMPLES
EXAMPLE 1 Evaluate:
(i) i23 (ii) i998 (iii) i 998 (iv) i 71
(v) ( )1 91 (vi) ( )i i37 61# (vii) i 1
SOLUTION We have
(i) ( ) .i i i i i i( )23 4 5 3 4 5 3 3# # [ ]i 14a
(ii) ( ) ( ) .i i i i i i1 1998 4 249 2 4 249 2 2 2# # # [ ]i 14a
150
Complex Numbers and Quadratic Equations 151
(iii) .i
i i
i
i
i1
1
1 1998 998 2
2
1000
2
#
[ ( ) ]i i 11000 4 250a
(iv)
( )
.i
i i
i
i
i
i
i i i1 1
71
71 72 4 18#
[ ]i 14a
(v) ( ) ( ) ( ) .i i i i i i1 191 91 4 22 3 4 22 3# # # [ ]i i3a
(vi) ( ) .i i i i i i137 4 9 1 4 9# # #
( )
.i
i i
i
i
i
i
i i i1 1
61
61 3
3
64
3
4 16#
( ) ( ) .i i i i 037 61
(vii) .i i i i
i
i
i i i1 1 1
1
3
3
4#
EXAMPLE 2 Prove that:
(i) i i i i 0n n n n1 2 3
(ii) i i i i 0107 112 117 122
(iii) ( )i i1 1
1 164
4
# b l .
SOLUTION We have
(i) i i i in n n n1 2 3
( )i i i i1n 2 3
( ) ( ) .i i i i1 1 0 0n n# [ ]andi i i12 3a
(ii) i i i i107 112 117 122
( ) ( )i i i i i i i i i i i1 1107 5 10 15 107 4 8 2 12 3# # #
( )i i i i1107 2 3 [ , ]andi i i1 1 14 8 12a
( ) ( ) .i i i i1 1 0 0107 107# [ , ]i i i12 3a
(iii)( )i i1 1
14 4# b l
( ) ( ) ( )i i i
i i i1 1 1 1 14
4
4 4# # b l [ ]i 12a
{( ) ( )} ( ) { ( )} .i i i1 1 1 1 1 2 164 2 4 4 4
EXAMPLE 3 Show that .i i
1 423
29 2
b l( 2
SOLUTION We have
( ) ( ) .i i i i i i1( )23 4 5 3 4 5 3# # # [ ]andi i i14 3a
i i i i
i
i
i i i1 1 1 1
29
29 29 3
3
32
3
#
b l . [ ]andi i i 13 32a
( ) ( ) ( ) .i i i i i i
1 2 4 4 1 423
29 2
2 2 2 # b l( 2
152 Senior Secondary School Mathematics for Class 11
EXAMPLE 4 Simplify:
(i) ( )i i2 6
1 b l (ii) ( ) ( )i i i3 6
1 3 b l (iii) 4 4 5 9 3 16
SOLUTION We have
(i) ·( ) ( )i i i2 6
1 2 6
1
3
1 1 3
12# # #
b bl l
(ii) ( ) ( ) ( )i i i i i3 6
1 3 216
13 2 3 b bl l
( ) ( ) ( )i3 1 216
1
# #
; E [ ]i i3a
.i i3 216
1
72
1
# # b l
(iii) 4 4 5 9 3 16
( ) ( ) ( )i i i4 2 5 3 3 4# # # [ , , ]i i i4 2 9 3 16 4a
( ) ( ) .i i i i8 15 12 11
EXAMPLE 5 Show that ( ) ,i1 n4 3 where n is a positive integer.
SOLUTION We have
( ) ( )i1 n n4 3 4 3 [ ]i1a
( ) ( )i in4 3#
{( ) } ( ) ( ) .i i i i1n4 3# #
[ ( ) ( ) ]andi i i i14 3a
Hence, ( ) .i1 n4 3
EXAMPLE 6 Show that the sum ( )i i i1 … n2 4 2 is 0 when n is odd and 1 when n
is even.
SOLUTION Let .S i i i1 … n2 4 2
This is clearly a GP having ( )n 1 terms with a 1 and .r i 12
( )
( )
( )
{ ( ) }
S
r
a r
i
i
1
1
1
1 1n n1
2
2 1#
( )
{ ( ) } { ( ) }
1 1
1 1
2
1 1n n1 1
( ) ,
( ) , .
when is odd
when is even
n
n
2
1 1 1 0
2
1 1 1 1
Z
[
\
]
]
]
]]
]
]
]
]]
AN IMPORTANT RESULT
For any two real numbers a and b, the result a b ab# is true only
when at least one of the given numbers is either zero or positive.
Thus, ( ) ( )2 3 2 3 6# # is wrong.
In fact, ( ) ( ) .i i i2 3 2 3 6 62# #
Complex Numbers and Quadratic Equations 153
EXAMPLE 7 Explain the fallacy:
( ) ( ) ( ) .i i1 1 1 1 1 1 1# # #
SOLUTION We know that for any real numbers a and ,b a b ab# is true
only when at least one of a and b is either 0 or positive.
( ) ( ) .1 1 1 1# #!
EXAMPLE 8 Evaluate:
(i) 25 49# (ii) 36 16#
SOLUTION We have
(i) ( ) ( ) ( ) ( ) .i i i25 49 5 7 35 35 1 352# # # #
(ii) ( ) .i i36 16 6 4 24# #
EXAMPLE 9 Evaluate .16 3 25 36 625
SOLUTION We have
16 3 25 36 625
( ) ( ) .i i i i i i i i4 3 5 6 25 4 15 6 25 0#
EXERCISE 5A
1. Evaluate: (i) i19 (ii) i62 (iii) i373
2. Evaluate: (i) ( )1 192 (ii) ( )1 93 (iii) ( )1 30
3. Evaluate: (i) i 50 (ii) i 9 (iii) i 131
4. Evaluate: (i) i
i
141
71
d n (ii) i
i
153
53
d n
Prove that:
5. i i i1 02 4 6 . 6. i i i i i6 5 2 6 750 33 15 48 .
7. i i i i
1 1 1 1 02 3 4 . 8. ( )i i i1
10 20 30 is a real number.
9. i i i
1 221
46 2
b l( 2 . 10. ( )i i i
1 2 118
25 3
b l( 2 .
11. ( )i i1 1
1 2n
n
n b l for all values of .n Nd
12. 16 3 25 36 625 0 .
13. ( )i i i i i1 1…2 4 6 8 20 .
14. i i i i i253 72 93 102 .
15. ( ) ( ),i i i n N1
n
n n
1
13
1
/
d .
ANSWERS (EXERCISE 5A)
1. (i) i (ii) –1 (iii) i 2. (i) 1 (ii) i (iii) –1
3. (i) –1 (ii) i (iii) i 4. (i) 2i (ii) 0
154 Senior Secondary School Mathematics for Class 11
HINTS TO SOME SELECTED QUESTIONS
7. , ,i i i
i i i
i i i i
i i1 1 1 1
1 1 1 13
3
3
2 3 3# # and .i
1 14
( ) .i i i i
i i1 1 1 1 1 1 02 3 4 ) 3
8. ( ) { }i i i i i i1 110 20 30 4 2 2 4 5 4 7 2 # # #
{ ( ) ( ) ( ) }i i i i i1 4 2 2 4 5 4 7 2# #
( ) ( ) .i i1 1 1 1 1 1 02 2
11. .i i i
i
i
i i i1 1 3
3
4
3
3#
( ) ( ) ( )i i i i1 1
1 1 1n
n n n c m i i
1
a ; E
{( ) ( )} ( )i i i1 1 1 2n n n2 . [ ]i 12a
13. This is a GP in which ,a r i1 12 and .n 11
( )
( )
{ ( )}
{ ( ) }
.S
r
a r
1
1
1 1
1 1 1
2
2 1
n 11#
14. ( ) ( ) ( ) ( )i i i i i i i i i i i53 72 93 102 4 13 4 18 4 23 4 25 2# # #
( ) ( ) ( ) .i i i i i1 1 1 1 1 1 1 2# # #
15. ·( ) ( ) ( )i i i i i i1 1
n
n n
n
n
n
n
1
13
1
1
13
1
13
/ / /
( ) { }i i i i i1 …2 3 13$
( ) ( )
( )
,i
r
a r
1 1
1 n
where , anda i r i n 13
· ·( ) ( )
( )
( ) ( )
( )
( ) .i
i
i i
i i
i
i
i1 1
1
1 1
1
1
13
[ ( ) ]i i i i i113 4 3a # #
COMPLEX NUMBERS
COMPLEX NUMBERS The numbers of the form ( ),a ib where a and b are real numbers
and ,i 1 are known as complex numbers. The set of all complex numbers is
denoted by C.
{( ) : , } .C a ib a b R d
EXAMPLES Each of the numbers ( ), ( ) andi i i5 8 3 2 3
2
7
5 b l is a complex
number.
For a complex number, ( ),z a ib we have
a real part of z, written as Re(z)
and b imaginary part of z, written as Im(z).
Complex Numbers and Quadratic Equations 155
EXAMPLES (i) If ( )z i5 9 then Re( )z 5 and Im( ) .z 9
(ii) If z i3
2 3 b l then Re( )z 3
2 and Im( ) .z 3
(iii) If z i7 8
5 b l then Re( )z 7 and Im ·( )z 8
5
PURELY REAL AND PURELY IMAGINARY NUMBERS
A complex number z is said to be
(i) purely real, if Im( ) ,z 0
(ii) purely imaginary, if Re( ) .z 0
Thus, each of the numbers , ,2 7 3 is purely real.
And, each of the numbers , ( ),i i i2 3 2
3
b l is purely imaginary.
CONJUGATE OF A COMPLEX NUMBER
Conjugate of a complex number ( )z a ib is defi ned as, ( ) .z a ib
EXAMPLES (i) ( ) ( )i i3 8 3 8 (ii) ( ) ( )i i6 2 6 2 (iii) .3 3
MODULUS OF A COMPLEX NUMBER
Modulus of a complex number ( ),z a ib denoted by | |,z is defi ned as | | .z a b2 2
EXAMPLES (i) If ( )z i2 3 then | | .z 2 3 132 2
(ii) If ( )z i5 4 then | | ( ) ( ) .z 5 4 412 2
EQUALITY OF COMPLEX NUMBERS If z a ib1 1 1 and z a ib2 2 2 then
.andz z a a b b1 2 1 2 1 2+
EXAMPLE If ( ) ( ),y x y i i2 3 5 2 fi nd the values of x and y.
SOLUTION Equating the real and imaginary parts, we get
( ) ( )y x y i i2 3 5 2
+ andy x y2 5 3 2
+ andy x2
5 3 2
5 2
+ ·andy x2
5
6
1
Hence, ·andx y6
1
2
5
SUM AND DIFFERENCE OF COMPLEX NUMBERS
If ( ) ( )z a ib and z a ib1 1 1 2 2 2 then we defi ne:
(i) ( ) ( )z z a a i b b1 2 1 2 1 2
(ii) ( ) ( ) .z z a a i b b1 2 1 2 1 2
156 Senior Secondary School Mathematics for Class 11
EXAMPLES (i) If ( ) ( )andz i z i3 5 5 21 2 then
{ ( )} { } ( )z z i i3 5 5 2 2 71 2
and { ( )} ( ) ( ) .z z i i3 5 5 2 8 31 2
(ii) If ( ) ( )andz i z i6 2 3 51 2 then
{( ) ( )} {( ) ( )} ( )z z i i6 3 2 5 9 71 2
and { ( )} {( ) ( )} ( ) .z z i i6 3 2 5 3 31 2
REMARKS (i) ( ) ( ) ( ) ( ) ( ) ( ) .Re Re Re Im Im Imandz z z z z z z z1 2 1 2 1 2 1 2
(ii) ( ) ( ) ( ) ( ) ( ) ( ) .Re Re Re Im Im Imandz z z z z z z z1 2 1 2 1 2 1 2
PROPERTIES OF ADDITION OF COMPLEX NUMBERS
(i) CLOSURE PROPERTY
The sum of two complex numbers is always a complex number.
Let ( ) ( )andz a ib z a ib1 1 1 2 2 2 be any two complex numbers.
Then, ( ) ( )z z a ib a ib1 2 1 1 2 2
( ) ( ),a a i b b1 2 1 2 which is a complex number.
Thus, if z1 and z2 are any two complex numbers then ( )z z1 2 is also a
complex number.
(ii) COMMUTATIVE LAW
For any two complex numbers z1 and ,z2 prove that
.z z z z1 2 2 1
PROOF Let ( ) ( ),andz a ib z a ib1 1 1 2 2 2 where , , ,a a b b1 2 1 2 arereal numbers.
( ) ( )z z a ib a ib1 2 1 1 2 2
( ) ( )a a i b b1 2 1 2
( ) ( )a a i b b2 1 2 1 [by commutative law of addition in R]
( ) ( ) .a ib a ib z z2 2 1 1 2 1
Thus, z z z z1 2 2 1 for all , .z z C1 2d
Hence, addition of complex numbers is commutative.
(iii) ASSOCIATIVE LAW
For any complex numbers ,z z1 2 and ,z3 prove that
( ) ( ) .z z z z z z1 2 3 1 2 3
PROOF Let ( ), ( ) ( ),andz a ib z a ib z a ib1 1 1 2 2 2 3 3 3 where , ,a a a1 2 3 and
, ,b b b1 2 3 are all real numbers.
( ) {( ) ( )} ( )z z z a ib a ib a ib1 2 3 1 1 2 2 3 3
{( ) ( )} ( )a a i b b a ib1 2 1 2 3 3
{( ) } {( ) }a a a i b b b1 2 3 1 2 3
{ ( )} { ( }a a a i b b b1 2 3 1 2 3
[by associative law of addition in R]
( ) {( ) ( )}a ib a a i b b1 1 2 3 2 3
Complex Numbers and Quadratic Equations 157
( ) {( ) ( )}a ib a ib a ib1 1 2 2 3 3
( ) .z z z1 2 3
( ) ( )z z z z z z1 2 3 1 2 3 for all , , .z z z C1 2 3d
Hence, addition of complex numbers is associative.
(iv) EXISTENCE OF ADDITIVE IDENTITY
For any complex number z, prove that
.z z z0 0
PROOF Let ( )z a ib and we may write, ( ) .i0 0 0 Then,
( ) ( ) ( ) ( ) ( )z a ib i a i b a ib0 0 0 0 0
and ( ) ( ) ( ) ( ) ( ) .z i a ib a i b a ib0 0 0 0 0
z z z0 0 for all values of .x Cd
Thus, 0 is the additive identity for complex numbers.
(v) EXISTENCE OF ADDITIVE INVERSE
For every complex number z, prove that
( ) ( ) .z z z z 0
PROOF Let ( ) .z a ib Then, ( ) ( ) ( ) .z a i b
( ) ( ) {( ) ( )}z z a ib a i b
{ ( )} { ( )} ( ) .a a i b b i0 0
Similarly, ( ) ( ) .z z i0 0 0
Hence, ( ) ( ) .z z z z 0
Thus, every complex number z has ( )z as its additive inverse.
MULTIPLICATION OF COMPLEX NUMBERS
Let ( ) ( ) .andz a ib z a ib1 1 1 2 2 2 Then, we defi ne
( ) ( ) ( ) ( ) .z z a ib a ib a a b b i a b b a1 2 1 1 2 2 1 2 1 2 1 2 1 2
{ ( ) ( ) ( ) ( )}Re Re Im Imz z z z z z· ·1 2 1 2 1 2
{ ( ) ( ) ( ) ( )} .Re Im Re Imi z z z z· ·1 2 2 1
EXAMPLES (i) Let ( ) ( ) .andz i z i3 2 5 41 2 Then,
{( ) ( )} {( ) ( )}z z i3 5 2 4 3 4 2 51 2 # # # #
( ) ( ) ( ) .i i15 8 12 10 7 22
(ii) Let ( ) ( ) .andz i z i2 3 7 51 2 Then,
{( ) ( )} {( ) ( ) }z z i2 7 3 5 2 5 3 71 2 # # # #
( ) ( ) ( ) .i i14 15 10 21 1 31
PROPERTIES OF MULTIPLICATION OF COMPLEX NUMBERS
(i) CLOSURE PROPERTY The product of two complex numbers is always a complex
number.
PROOF Let ( ) ( )andz a ib z a ib1 1 1 2 2 2 be any two complex numbers.
Then, , ,anda a b b1 2 1 2 are real numbers.
158 Senior Secondary School Mathematics for Class 11
( ) ( )z z a ib a ib1 2 1 1 2 2
( ) ( ),a a b b i a b b a1 2 1 2 1 2 1 2 which is a complex number.
Thus, the product of two complex numbers is always a complex
number.
(ii) COMMUTATIVE LAW
For any two complex numbers ,andz z1 2 prove that
.z z z z1 2 2 1
PROOF Let ( ) ( ) .andz a ib z a ib1 1 1 2 2 2 Then,
( ) ( ) ( ) ( )z z a ib a ib a a b b i a b b a1 2 1 1 2 2 1 2 1 2 1 2 1 2
( ) ( )a a b b i b a a b2 1 2 1 2 1 2 1
[by commutativity of multiplication on R]
( ) ( ) .a ib a ib z z2 2 1 1 2 1
z z z z1 2 2 1 for all , .z z C1 2d
Hence, multiplication of complex numbers is commutative.
(iii) ASSOCIATIVE LAW
For any three complex numbers ,z z1 2 and ,z3 prove that
( ) ( ) .z z z z z z1 2 3 1 2 3
PROOF Let ( ), ( ) ( )andz a ib z a ib z a ib1 1 1 2 2 2 3 3 3 be any three complex
numbers. Then , ,a a a1 2 3 and , ,b b b1 2 3 are all real numbers.
( ) {( ) ( )} ( )z z z a ib a ib a ib1 2 3 1 1 2 2 3 3
{( ) ( )} ( )a a b b i a b b a a ib1 2 1 2 1 2 1 2 3 3
{( ) ( ) }a a b b a a b b a b1 2 1 2 3 1 2 1 2 3
{( ) ( ) }i a a b b b a b b a a1 2 1 2 3 1 2 1 2 3
{ ( ) ( )}a a a b b b a b b a1 2 3 2 3 1 2 3 2 3
{ ( ) ( )}i a a b a b b a a b b1 2 3 3 2 1 2 3 2 3
( ) {( ) ( )}a ib a a b b i a b a b1 1 2 3 2 3 2 3 3 2
( ) {( ) ( )} ( ) .a ib a ib a ib z z z1 1 2 2 3 3 1 2 3
Thus, ( ) ( )z z z z z z1 2 3 1 2 3 for all , , .z z z C1 2 3d
Hence, multiplication of complex numbers is associative.
(iv) EXISTENCE OF MULTIPLICATIVE IDENTITY
The complex number ( )i1 0 is the multiplicative identity in C.
Let ( ) .z a ib Then,
( ) ( ) {( ) ( )} {( ) ( )} ( ) .z a ib i a b i a b a ib z1 1 0 1 0 0 1# # # # #
Similarly, .z z1#
Thus, ( ) ( )z z z1 1# # for all .z Cd
Hence, the complex number ( )i1 1 0 is the multiplicative identity.
Complex Numbers and Quadratic Equations 159
(v) EXISTENCE OF MULTIPLICATIVE INVERSE
Let ( ) .z a ib Then,
·( ) ( ) ( )
( )
( )
( )
z z a ib a ib a ib
a ib
a b
a ib1 1 11
2 2#
Clearly, .z z z z 11 1# #
Thus, every ( )z a ib has its multiplicative inverse, given by
·
( )
( )
| |
z z a b
a ib
z
z11
2 2 2
:
| |
Remember z
z
z1
2
R
T
S
S
S
S
SS
V
X
W
W
W
W
WW
| | .zz z 2
THINGS TO REMEMBER
(i) ( ) ( ) | | ( ) .andz a ib z a ib z a b2 2 2&
(ii) ·( )
| | ( )
( )
z a ib z
z
z
a b
a ib1
2 2 2&
(vi) DISTRIBUTIVE LAWS
For complex numbers , , ,z z z1 2 3 prove that
( ) .z z z z z z z·1 2 3 1 2 1 3
( ) · .z z z z z z z1 2 3 1 3 2 3
PROOF Let ( ), ( ) ( )andz a ib z a ib z a ib1 1 1 2 2 2 3 3 3 be any three complex
numbers. Then, , ,a a a1 2 3 and , ,b b b1 2 3 are real numbers.
( )z z z·1 2 3
( ) [( ) ( )]a ib a ib a ib1 1 2 2 3 3
( ) [( ) ( )]a ib a a i b b1 1 2 3 2 3
{ ( ) ( )} { ( ) ( )}a a a b b b i a b b b a a1 2 3 1 2 3 1 2 3 1 2 3
{( ) ( )} {( ) ( )}a a b b i a b b a a a b b i a b b a1 2 1 2 1 2 1 2 1 3 1 3 1 3 1 3
.z z z z1 2 1 3
Thus, ( )z z z z z z z1 2 3 1 2 1 3 for all , , .z z z C1 2 3d
Similarly, we can prove that
( ) .z z z z z z z1 2 3 1 3 2 3
DIVISION OF TWO COMPLEX NUMBERS
Let z1 and z2 be complex numbers such that .z 02!
Then, · .z
z
z z z z
1
2
1
1
2
1 2
1
EXAMPLE Find ,z
z
2
1 when ( )z i6 31 and ( ) .z i32
SOLUTION We have .z
z
z z
2
1
1 2
1
160 Senior Secondary School Mathematics for Class 11
Now,
| |
z
z
z
2
2
2
2 1 ·
| |
( )
{ ( ) }
( ) ( )
i
i i i
3
3
3 1
3
10
3
2 2 2
· ·( )
( ) ( ) ( )
z
z
z z i
i i i
6 3 10
3
10
6 3 3
2
1
1 2
1
·
( ) ( )i i i
10
15 15
10
15 1
2
3 1
SOME IDENTITIES ON COMPLEX NUMBERS
THEOREM 1 For any complex numbers ,z and z1 2 prove that:
(i) ( )z z z z z z21 2
2
1
2
2
2
1 2
(ii) ( )z z z z z z21 2
2
1
2
2
2
1 2
(iii) ( ) ( ) ( )z z z z z z1
2
2
2
1 2 1 2
(iv) ( ) ( )z z z z z z z z31 2
3
1
3
2
3
1 2 1 2
(v) ( ) ( )z z z z z z z z31 2
3
1
3
2
3
1 2 1 2
PROOF We have
(i) ( ) ( ) ( )z z z z z z1 2
2
1 2 1 2
( ) ( )z z z z z z1 2 1 1 2 2 [by distributive law]
z z z z z z1
2
2 1 1 2 2
2 [by distributive law]
z z z z21
2
1 2 2
2 [ ]z z z z2 1 1 2a
z z z z21
2
2
2
1 2 .
( ) .z z z z z z21 2
2
1
2
2
2
1 2
(ii) ( ) ( ) ( )z z z z z z1 2
2
1 2 1 2
( ) ( )z z z z z z1 2 1 1 2 2 [by distributive law]
z z z z z z1
2
2 1 1 2 2
2 [by distributive law]
z z z z21
2
1 2 2
2 [ ]z z z z2 1 1 2a
.z z z z21
2
2
2
1 2
( ) .z z z z z z21 2
2
1
2
2
2
1 2
(iii) RHS ( ) ( )z z z z1 2 1 2
( ) ( )z z z z z z1 2 1 1 2 2 [by distributivelaw]
z z z z z z1
2
2 1 1 2 2
2 [by distributive law]
( )z z1
2
2
2 LHS. [ ]z z z z2 1 1 2a
( ) ( ) ( ) .z z z z z z1
2
2
2
1 2 1 2
Complex Numbers and Quadratic Equations 161
(iv) ( ) ( ) ( )z z z z z z·1 2
3
1 2
2
1 2
( ) ( )z z z z z z21
2
2
2
1 2 1 2 [expanding ( ) ]z z1 2
2
( ) ( )z z z z z z z z z z2 21
2
2
2
1 2 1 1
2
2
2
1 2 2
[by distributive law]
z z z z z z z z z z2 21
3
2
2
1 1
2
2 1
2
2 2
3
1 2
2
z z z z z z3 31
3
2
3
1
2
2 1 2
2 ( ) .z z z z z z31
3
2
3
1 2 1 2
( ) ( ) .z z z z z z z z31 2
3
1
3
2
3
1 2 1 2
(v) LHS ( )z z1 2
3
( ) ( )z z z z1 2
2
1 2 ( ) ( )z z z z z z21
2
1 2 2
2
1 2
( ) ( )z z z z z z z z z z2 21
2
1 2 2
2
1 1
2
1 2 2
2
2
z z z z z z z z z z2 21
3
1
2
2 2
2
1 1
2
2 1 2
2
2
3
.RHSz z z z z z3 31
3
1
2
2 1 2
2
2
3
This may be written as ( ) ( ) .z z z z z z z z31 2
3
1
3
2
3
1 2 1 2
SOLVED EXAMPLES
EXAMPLE 1 Simplify:
(i) ( ) ( )i i i3 6 6 6 6 (ii) ( ) ( )i i1 3 6
(iii) i i3
1
3
2 4 2
3 b bl l (iv) i i i5
1
5
7 6 5
1
5
4 b b bl l l( 2
SOLUTION We have
(i) ( ) ( ) .i i i i i i i i3 6 6 6 6 18 18 6 6 18 24 6 12 242
(ii) ( ) ( ) .i i i i i1 3 6 1 3 6 4 7
(iii) i i i i3
1
3
2 4 2
3
3
1
3
2 4 2
3 b bl l
.i i3
1 4 3
2
2
3
3
11
6
13 b bl l
(iv) i i i5
1
5
7 6 5
1
5
4 b b bl l l( 2
i i i i i5
1
5
7 6 5
1
5
4
5
1 6 5
7
5
1
5
4 b b b b bl l l l l( 2
·i i i i i5
29
5
6
5
4
5
29
5
4
5
6 5 5
1 b b bl l l
EXAMPLE 2 Express ( ) ( )i3 2 2 3 in the form ( ) .a ib
SOLUTION We have
( ) ( )i3 2 2 3 ( ) ( )i i3 2 2 3
i i i6 3 2 6 2 2
( ) ( ) .i6 2 3 1 2 2
162 Senior Secondary School Mathematics for Class 11
EXAMPLE 3 Express each of the following in the form ( ):a ib
(i) ( ) ( )3 5 3 5 (ii) ( ) ( )2 3 3 2 3
(iii) ( )i2 3 2 (iv) ( )i5 7 2
SOLUTION We have
(i) ( ) ( ) ( ) ( ) { ( ) }i i i3 5 3 5 3 5 3 5 3 52 2
( ) ( ) .i i9 5 9 5 14 02
[ ( ) ( ) ( )]z z z z z z1 2 1 2 1
2
2
2a
(ii) ( ) ( ) ( ) ( )i i2 3 3 2 3 2 3 3 2 3
( )i i i6 4 3 3 3 6 2
.i i7 3 0 7 3
(iii) ( ) ( ) ( ) ( )i i i2 3 2 3 2 2 32 2 2 # #
( ) ( ) ( ) .i i i i4 9 12 4 9 12 5 122
(iv) ( ) ( ) ( )i i i5 7 5 7 2 5 72 2 2 # #
( ) ( )i i i25 49 14 5 25 49 14 52
( ) .i24 14 5
EXAMPLE 4 Prove that ( ) .i1 44
SOLUTION We have
( ) ( ) ( ) ( ) ( )i i i i i i i1 1 1 1 2 1 24 2 2 2 2# #
( ) ( ) ( ) .i i i2 2 4 4 1 42 # [ ]i 12a
EXAMPLE 5 Express ( )i2 3 3 in the form ( ) .a ib
SOLUTION We know that ( ) ( ) .z z z z z z z z31 2
3
1
3
2
3
1 2 1 2
( ) ( ) ( )i i i i2 3 2 3 3 2 3 2 33 3 3 # # #
i i i8 27 36 543 2
( )i i8 27 36 54 [ ]andi i i 13 2a
( ) .i46 9
Hence, ( ) ( ) .i i2 3 46 93
EXAMPLE 6 Express ( )i3 5 3 in the form ( ) .a ib
SOLUTION We know that ( ) ( ) .z z z z z z z z31 2
3
1
3
2
3
1 2 1 2
( ) ( ) ( )i i i i3 5 3 5 3 3 5 3 53 3 3 # # #
i i i27 125 135 2253 2
( )i i27 125 135 225 [ ]andi i i 13 2a
( ) .i198 10
Complex Numbers and Quadratic Equations 163
EXAMPLE 7 If ( ),z 2 3 fi nd ( ), ( ),Re Imz z z and | |.z
SOLUTION The given number may be written as
.z i2 3
( ) , ( ) ,Re Imz z z2 3 ( ) ( )i i2 3 2 3
and | | {( ) ( ) } ( ) | | .z z2 3 2 3 5 52 2 2 &
EXAMPLE 8 Write down the modulus of:
(i) 4 3 (ii) i2 5 (iii) i (iv) ( )i1 3 2
SOLUTION We know that, if ( )z a ib then | | .z a b2 2
(i) Let .z 4 3 Then, .z i4 3
| | {( ) ( ) } ( ) | | .z z4 3 16 3 19 192 2 2 &
(ii) Let .z i2 5 Then,
| | { ( ) } ( ) | | .z z2 5 4 25 29 292 2 2 &
(iii) Let .z i0 Then,
| | { ( ) } ( ) | | .z z0 1 0 1 1 1 12 2 2 &
(iv) Let ( ) .z i1 3 2 Then,
( ) ( ) ( ) ( ) .z i i i1 3 2 1 3 8 62 2 # #
| | {( ) ( ) } ( ) | | .z z8 6 64 36 100 100 102 2 2 &
EXAMPLE 9 Write down the conjugate of each of the following:
(i) ( )5 1 (ii) ( )6 3 (iii) i3 (iv) ( )i4 5 2
SOLUTION We know that the conjugate of ( )z a ib is given as ( ) .z a ib
(i) Let ( ) .z i5 1 5 Then,
z ( ) ( ) .i i5 5
(ii) Let .z i6 3 6 3 Then,
z ( ) ( ) .i i6 3 6 3
(iii) Let .z i i i03 Then,
z ( ) ( ) .i i i0 0
(iv) Let ( ) ( ) ( )z i i i4 5 4 5 2 4 52 2 2 # #
( ) ( )i i16 25 40 9 40 .
z ( ) ( ) .i i9 40 9 40
EXAMPLE 10 Find the multiplicative inverse of each of the following:
(i) i5 3 (ii) i4 3 (iii) ( )i3 1 2 (iv) i
SOLUTION We know that the multiplicative inverse of the complex number z
is given by ·
||
z
z
z1
2
164 Senior Secondary School Mathematics for Class 11
(i) Let ( ) .z i5 3 Then,
( )z i5 3 and | | ( ) ( ) .z 5 3 5 9 142 2 2
·
| |
( )
z
z
z i i14
5 3
14
5
14
31
2
e o
(ii) Let .z i4 3 Then,
z ( ) ( ) | | {( ) ( ) } ( ) .andi i z4 3 4 3 4 3 16 9 252 2 2
·
| |
( )
z
z
z i i25
4 3
25
4
25
31
2
b l
(iii) Let ( ) ( ) ( ) ( ) .z i i i i i3 1 9 1 6 9 1 6 8 62 2
z ( ) ( )i i8 6 8 6
and | | {( ) ( ) } ( ) .z 8 6 64 36 1002 2 2
Hence, ·
| |
( )
z
z
z i i i100
8 6
100
8
100
6
25
2
50
31
2
b bl l
(iv) Let ( ) .z i0 Then,
z ( ) | | ( ) .andi i z0 0 1 12 2 2
| |
.z
z
z i i1
1
2
EXAMPLE 11 Express each of the following in the form ( ):a ib
(i) ( )i
i
1 (ii) ( )i1 3
1 (iii)
i
i
1 2
5 2
SOLUTION We have
(i) ( ) ( ) ( )
( )
( )
( )
i
i
i
i
i
i
i
i i
1 1 1
1
1
1
2 2#
·( )
( ) ( ) ( )i i i i
i1 1 2
1
2
1
2
1
2
1
2
b l
(ii) ( )
( ) ( )
( )
i
i i
i
1 3
1 3
1
1 3
1 31 #
( ) ( )
( )
( ) ( )
( )
i i
i
i
i
1 3 1 3
1 3
1 3
1 3
2 2
·
( )
( ) ( )
i
i i
i
1 3
1 3
4
1 3
4
1
4
3
2
e o
(iii)
( )
( )
( )
( )
( )
( ) ( )
i
i
i
i
i
i
i
i i
1 2
5 2
1 2
5 2
1 2
1 2
1 2
5 2 1 2
2#
( )
( ) .
i
i3
3 6 2
1 2 2
Complex Numbers and Quadratic Equations 165
EXAMPLE 12 Reduce i i i
i
1 2
1
1
3
1 3
3 2
b dl n to the form ( ) .a ib
SOLUTION We have
i i i
i
1 2
1
1
3
1 3
3 2
b dl n
( ) ( )
( ) ( )
( )
( )
( )
( )
i i
i i
i
i
i
i
i
i
1 2 1
1 3 6
1 3
3 2
3
4 5
1 3
3 2
#
d n) 3
( ) ( )
( ) ( )
( ) ( )
( ) ( ) ( )
i i
i i
i i
i
i
i
i
i
3 1 3
4 5 3 2
3 3 9
12 10 15 8
10
22 7
#
·
( )
i
i i i i i
10
7 22
10
7 22
10
7
10
22
10
7
5
11
2
2
b bl l
EXAMPLE 13 Reduce i
i
i
i
1
1
1
1
b l to the form ( )a ib and hence fi nd its modulus.
SOLUTION Let ( ) ( )
( ) ( )
( )
( )
.z i
i
i
i
i i
i i
i
i i i1
1
1
1
1 1
1 1
1
4 1
2
4 2
2 2
2
# #
b l
Thus, | | .z i z0 2 0 2 4 22 2&
Hence, z i0 2 and | | .z 2
EXAMPLE 14 If ( )i
i a ib1
1
then show that ( ) .a b 12 2
SOLUTION We have
( )
( )
aib i
i
i
i
i
i
i
i
1
1
1
1
1
1
1
1
2
#
( )i
i
2
1
2
1
2
1
e o
& | | .a ib
2
1
2
1
2
1
2
1 12
2 2
e e bo o l
& ( ) .a b 12 2
Hence, ( ) .a b 12 2
EXAMPLE 15 Find the least positive integer m for which .i
i
1
1 1
m
b l
SOLUTION We have
( )
( )
( )
( )
( )
( ) ( )
.i
i
i
i
i
i
i
i i i i i1
1
1
1
1
1
1
1
2
1 2
2
2
2
2 2
#
b l
.i
i i1
1 1 1
m
m&
b l
And, we know that 4 is the least positive integer such that i 14 and
therefore, .m 4
166 Senior Secondary School Mathematics for Class 11
EXAMPLE 16 Separate
2 1
3 1
J
L
K
K
K
KK
N
P
O
O
O
OO
into real and imaginary parts and hence fi nd its
modulus.
SOLUTION Let ( )
( )
( )
( )
( )
( )
z i
i
i
i
i
i
i
i
2 1
3 1
2
3
2
3
2
3
2
2
#
d
J
L
K
K
K
KK
n
N
P
O
O
O
OO
( ) ( )
( ) ( )
( )
( )
{ ( )}
( )
( )
i i
i i
i
i i i i i2 2
3 2
4
6 5
4 1
5 5
5
5 1
12
2
.
| | .z 1 1 22 2
Hence, ( ) | | .andz i z1 2
EXAMPLE 17 Reduce
i i
i i
5 12 5 12
5 12 5 12
Z
[
\
]
]]
]
]]
_
`
a
b
bb
b
bb
to the form ( )a ib and hence fi nd its
conjugate.
SOLUTION We have
z
i i
i i
5 12 5 12
5 12 5 12
( )
( )
( )
( )
i i
i i
i i
i i
5 12 5 12
5 12 5 12
5 12 5 12
5 12 5 12
#
( ) ( )
( ) ( ) ( ) ( ) ( )
i i
i i
i
i i i
5 12 5 12
5 12 5 12
24
5 12 5 12 2 5 122 2 2
( )
i i i24
10 2 25 144
24
10 2 169
24
10 2 13#
.i i i i
i
i
i i24
36
2
3
2
3
2
3
2
3
2#
Hence, z i z0 2
3 and b l ·i i0 2
3 0 2
3 b bl l
EXAMPLE 18 If ( ) ( )x iy a ib/1 3 then prove that:
(i) ( )a
x
b
y
a b4 2 2 (ii) ( )a
x
b
y
a b2 2 2
SOLUTION We have
( ) ( )x iy a ib/1 3
& ( ) ( )x iy a ib 3 [on cubing both sides]
& ( ) ( )x iy a i b iab a ib33 3 3
a ib a bi ab3 33 3 2 2 ( ) ( )a ab i a b b3 33 2 2 3
& andx a ab y a b b3 33 2 2 3
[on equating real and imaginary parts separately]
Complex Numbers and Quadratic Equations 167
& ( ) ( )anda
x a b
b
y
a b3 32 2 2 2
& ( ) ( ) .anda
x
b
y
a b a
x
b
y
a b4 22 2 2 2 d dn n
Hence, (i) ( )a
x
b
y
a b4 2 2 and (ii) ( ) .a
x
b
y
a b2 2 2
EXAMPLE 19 Let z be a complex number such that z 1! and | | .z 1 Then, prove that
z
z
1
1
b l is purely imaginary. What will be your conclusion when z 1 ?
SOLUTION Let ( )z x iy be the given complex number such that z 1! and
| | .z 1 Then,
| | | | .z z x y x y1 1 1 02 2 2 2 2& & & … (i)
( )
( )
( )
( )
( )
( )
z
z
z
z
x iy
x iy
x iy
x iy
x iy
x iy
1
1
1
1
1
1
1
1
1
1
#
b l
{( ) }
{( ) } {( ) }
x y
x iy x iy
1
1 1
2 2
{( ) }
( ) {( ) ( ) }
x y
x y i x y x y
1
1 1 1
2 2
2 2
{( ) }
( ) ( )
x y
x y y i
1
1 2
2 2
2 2
( )x y
yi
1
2
2 2
Z
[
\
]
]]
]
]]
_
`
a
b
bb
b
bb
[using (i)],
which is purely imaginary.
Particular Case When .z 1
In this case, z x iy1 1&
( )x iy1 0&
, , .x y x y1 0 0 1 0& &
( )
( )
( )
( )
( )
( )
.z
z
x iy
x iy
x iy
x iy
i
i
1
1
1
1
1
1
1 1 0
1 1 0
2
0 0
#
#
Thus, z z
z1 1
1
&
is purely real.
EXAMPLE 20 Find the real value of for which sin
sin
i
i
1 2
3 2
d n is purely real.
SOLUTION We have
( )
( )
( )
( )
sin
sin
sin
sin
sin
sin
i
i
i
i
i
i
1 2
3 2
1 2
3 2
1 2
1 2
#
d n
( )
( ) ( )
sin
sin sin
i
i i
1 4
3 2 1 2
2 2
168 Senior Secondary School Mathematics for Class 11
( )
( ) ( )
sin
sin sin sini
1 4
3 4 6 2
2
2
·
( )
( ) ( )
sin
sin sini
1 4
3 4 8
2
2
Now, sin
sin
i
i
1 2
3 2
d n will be purely real only when
( )
.
sin
sin
1 4
8 02
This happens only when , .sin sin n n N8 0 0+ + d
Hence, the required value of is ,n where .n Nd
EXAMPLE 21 Show that | |i1 2x x has no nonzero integral solution.
SOLUTION | | ( )i1 2 2 2x x x x& [ | | ( ) ]i1 1 1 22 2a
2 2
2
2 1/ /
x x
x
x
2
2& &
.x x2 1 2 2 0 0
/x 2 0& & &
Thus, x 0 is the only solution of the given equation.
Hence, the given equation has no nonzero integral solution.
EXAMPLE 22 Solve for x and y:
(i) ( ) ( )x iy y x i3 7 2 5 5
(ii) ( ) ( )x iy i i2 3 4
SOLUTION (i) The given equation is
( ) ( )x iy y x i3 7 2 5 5 .
Equating the real parts and the imaginary parts of the given
equation separately, we get
andx y y x3 7 5 2 5
& andx y x y3 5 7 2 5
& andx y1 2 [on solving , ]x y x y3 5 7 2 5 .
Hence, .andx y1 2
(ii) The given equation is
( ) ( )x iy i i2 3 4
& ( ) ( )x y i y x i2 3 2 3 4
& andx y y x2 3 4 2 3 1
[equating real parts and imaginary parts separately]
& andx y x y2 3 4 3 2 1
& andx y13
5
13
14 [on solving , ]x y x y2 3 4 3 2 1 .
Hence, andx y13
5
13
14 is the required solution.
Complex Numbers and Quadratic Equations 169
EXAMPLE 23 Find the real values of x and y for which .i
x
i
y
i3
1
3
1
e o
SOLUTION We have
i
x
i
y
i3
1
3
1
& ( )
( )
( )
( )
( )
( )
( )
( )
i
x
i
i
i
y
i
i
i3
1
3
3
3
1
3
3
# #
&
( )
( ) ( )
( )
( ) ( )
i
x i
i
y i
i
9
1 3
9
1 3
2 2
&
( ) ( ) ( ) ( )x i x y i y
i10
3 1 1
10
3 1 1
& ( ) ( ) ( ) ( )x i x y i y i3 1 1 3 1 1 10
& ( ) ( )x y i y x i3 3 3 3 1 1 10
& ( ) ( )x y i y x i3 3 6 10
& ( ) andx y y x3 2 0 10
[equating real parts and the imaginary parts separately]
& andx y y x2 0 10
& andx y x y2 10
& andx y4 6 [on solving , ]x y x y2 10 .
Hence, andx y4 6 are the required values.
EXAMPLE 24 Find the complex number z for which | | ( ) .z z i1 2 1
SOLUTION Let the required complex number be ( ) .z x iy Then,
| | ( )z z i1 2 1
& |( ) | ( ) ( )x iy x iy i1 2 1
& ( ) ( ) ( )x y x i y1 2 22 2
& ( ) ( ) andx y x y1 2 2 02 2
[equating real parts and imaginary parts separately]
& ( ) ( ) ( )andy x x2 1 2 22 2
& ( )andy x x x2 2 5 22
& ( ) ( )andy x x x2 2 5 22 2
& andx x x x y2 5 4 4 22 2
& .and andx y x y2 1 2 2
1 2&
Hence, the required complex number is ·z i2
1 2 b l
170 Senior Secondary School Mathematics for Class 11
EXAMPLE 25 Solve the equation | | ( )z z i2 for complex value of z.
SOLUTION Let ( ) .z x iy Then,
| |z z i2
& | | ( )x iy x iy i2
& { } ( )x y x iy i22 2
& andx y x y2 12 2
[equating real parts and imaginary parts separately]
& ( )and andy x x y x x1 1 2 1 1 22 2 2&
& ( )and andy x x y x x x1 1 2 1 1 4 42 2 2 2&
& .and andx y x y4 3 1 4
3 1&
Hence, z i4
3 b l is the desired solution.
EXAMPLE 26 Solve the equation | |z z i2 1 0 for complex value of z.
SOLUTION Let the required complex number be .z x iy Then,
| |z z i2 1 0
& ( ) |( ) |x iy x iy i2 1 0
& { ( ) } ( )x x y y i2 1 1 02 2
& { ( ) } andx x y y2 1 0 1 02 2
[equating real parts and imaginary parts separately
on both sides]
& { ( ) ( ) } ( )andy x x1 2 1 12 2
& ( )andy x x x1 2 2 22$
& ( )andy x x x1 2 2 22 2
& ( )and andy x x xy1 4 4 0 2 0 12 2&
& .and andx y x y2 0 1 2 1&
Hence, the required complex number is ( ) .z i2
EXAMPLE 27 If ,z i3 2 prove that z z6 13 02 and hence deduce that
.z z z3 13 9 65 03 2
SOLUTION ( )z i z i z i3 2 3 2 3 42 2& &
.z z6 13 02& … (i)
Thus, .z z6 13 02
Now, ( ) ( )z z z z z z z z3 13 9 65 3 6 13 5 6 133 2 2 2
( ) ( )z3 0 5 0 0# # [using (i)].
Hence, .z z z3 13 9 65 03 2
Complex Numbers and Quadratic Equations 171
EXAMPLE 28 If ,z i5 3 fi nd the value of ( ) .z z z z9 26 14 84 3 2
SOLUTION We have
( )z i z i5 3 5 3&
( ) .z i z z5 9 10 34 02 2 2& & … (i)
Now, z z z z9 26 14 84 3 2
( ) ( ) ( )z z z z z z z z10 34 10 34 2 10 34 602 2 2 2
( ) ( ) ( )z z0 0 2 0 60 602# # # [using (i)].
Hence, the value of z z z z9 26 14 84 3 2 is –60.
EXAMPLE 29 If ,z i2 prove that ( ) .z z z i3 9 8 1 143 2
SOLUTION We have
z i z i2 2&
( ) .z i z z2 4 5 02 2 2& & … (i)
z z z3 9 83 2
( ) ( )z z z z z z4 5 7 4 5 14 272 2
( ) ( ) ( )z z z0 7 0 14 27 14 27# # [using (i)]
( ) ( ) .i i14 2 27 1 14
Hence, ( ) .z z z i3 9 8 1 143 2
EXERCISE 5B
1. Simplify each of the following and express it in the form a ib :
(i) ( ) ( )i i i2 3 4 5 6 (ii) ( ) ( )3 16 4 9
(iii) ( ) ( )i i5 6 2 (iv) ( ) ( )i i8 4 3 5
(v) ( ) ( ) ( )i i i1 1 3 42 2 (vi) ( ) ( )5 3 5 3
(vii) ( ) ( )i i3 4 2 3 (viii) ( ) ( )2 3 3 2 3
2. Simplify and express each of the following in the form ( ) :a ib
(i) ( )2 3 2 (ii) ( )i5 2 2 (iii) ( )i3 5 3
(iv) i2 3
1 3 b l (v) ( )i4 3 1 (vi) ( )2 3 1
(vii) ( )i2 2 (viii) ( )i1 2 3 (ix) ( ) ( )i i1 13 3
3. Express each of the following in the form ( ) :a ib
(i) ( )i4 3
1
(ii) ( )
( )
i
i
4 5
3 4
(iii)
( )
( )
i
i
1 2
5 2
(iv) ( )
( )
i
i
3 6
2 5
(v) ( ) ( )
( )
i i
i
4 2 1
3 4
(vi) ( ) ( )
( ) ( )
i i
i i
1 2 2
3 2 2 3
(vii) ( )
( )
i
i
2
2 3 2
(viii)
( )
( )
i
i
1
1
3
3
(ix) ( ) ( )
( )
i i
i
1 2
1 2 3
172 Senior Secondary School Mathematics for Class 11
4. Simplify and express each of the following in the form ( ) :a ib
(i) i i i
i
3 2
5
1
2
3 2
4 5
b dl n (ii) i i i
i
1 4
1
1
2
5 3
1
b dl n
5. Show that
(i) ( )
( )
( )
( )
i
i
i
i
2 3
3 2
2 3
3 2
) 3 is purely real,
(ii)
( )
( )
( )
( )
i
i
i
i
7 3
7 3
7 3
7 3
Z
[
\
]
]]
]
]]
_
`
a
b
bb
b
bb
is purely real.
6. Find the real values of for which cos
cos
i
i
1 2
1
d n is purely real.
7. If | | | |,z i z i prove that z is real.
8. Give an example of two complex numbers z1 and z2 such that z z1 2! and
| | | |.z z1 2
9. Find the conjugate of each of the following:
(i) ( )i5 2 (ii) ( )i4 3
1
(iii) ( )
( )
i
i
3
1 2
(iv) ( )
( ) ( )
i
i i
3
1 2
(v) 3 (vi) 2 (vii) 1 (viii) ( )i2 5 2
10. Find the modulus of each of the following:
(i) ( )3 5 (ii) ( )i3 4 (iii) ( )i7 24 (iv) i3
(v) ( )
( )
i
i
4 3
3 2 2
(vi) ( )
( ) ( )
i
i i
1
2 1
(vii) 5 (viii) ( ) ( )i i1 2 1
11. Find the multiplicative inverse of each of the following:
(i) ( )i1 3 (ii) ( )i2 5 (iii) ( )
( )
i
i
1
2 3
(iv) ( )
( ) ( )
i
i i
1 3
1 1 2
12. If ( ),i
i a ib1
1 100
b l fi nd the values of a and b.
13. If ,i
i
i
i x iy1
1
1
13 3
b bl l fi nd x and y.
14. If ,( )x iy
a ib
a ib
prove that .x y 12 2
15. If ,( )a ib c i
c i
where c is real, prove that a b 12 2 and ·a
b
c
c
1
2
2
16. Show that ( )i i1 1
1 2n
n
n b l for all .n Nd
17. Find the smallest positive integer n for which ( ) ( ) .i i1 1n n2 2
18. Prove that ( ) ( ) ( ) ( ) ( ) .x i x i x i x i x1 1 1 1 44
19. If ( ),cos sina i prove that .cota
a i1
1
2
b l
Complex Numbers and Quadratic Equations 173
20. If ( )z i21 and ( ),z i12 fi nd ·z z i
z z 1
1 2
1 2
21. Find the real values of x and y for which:
(i) ( ) ( )i x i y i1 1 1 3 (ii) ( ) ( ) ( )x iy i i3 2 12 5
(iii) x yi ix y4 3 (iv) ( ) ( ) ( )i y i i x1 6 22
(v) ( )
( )
( )
iy
x i
i2
3
1
(vi) ( )
( )
( )
( )
i
i x i
i
i y i
i3
1 2
3
2 3
22. Find the real values of x and y for which ( ) ( )x iy i3 5 is the conjugate of
( ) .i6 24
23. Find the real values of x and y for which the complex numbers ( )iyx3 2
and ( )x y i42 are conjugates of each other.
24. If ( ),z i2 3 prove that z z4 13 02 and hence deduce that
.z z4 3 169 03 2
25. If ( ) ( )i z i z1 1 then prove that .z i z
26. If z
z
1
1
b l is purely an imaginary number and z 1! then fi nd the value
of | |.z
27. Solve the system of equations, ( ) , | | .Re z z0 22
28. Find the complex number z for which | | .z z i1 2
ANSWERS (EXERCISE 5B)
1. (i) ( )i12 13 (ii) ( )i1 7 (iii) ( )i3 5 (iv) ( )i11 9
(v) ( )i9 22 (vi) 28 (vii) ( )i18 (viii) i7 3
2. (i) ( )i1 4 3 (ii) ( )i21 20 (iii) ( )i198 10 (iv) i3
22
27
107 b l
(v) i25
4
25
3b l (vi) i7
2
7
3 e o (vii) i25
3
25
4b l (viii) i125
11
125
2 b l
(ix) i4
3. (i) i25
4
25
3b l (ii) i41
8
41
31 b l (iii) ( )i1 2 2 (iv) i5
4
15
1 b l
(v) i4
1
4
3b l (vi) i25
63
25
16b l (vii) i5
22
5
19 b l (viii) ( )i2 0
(ix) i2
7
2
1 b l
4. (i) i13
5
13
4b l (ii) i17
4
17
5b l 6. ( ) ,n n I2 1 2
d
9. (i) ( )i5 2 (ii) i25
4
25
3b l (iii) i5
1
5
3 b l (iv) i5
3
5
4b l
(v) i3 (vi) 2 (vii) i (viii) ( )i21 20
174 Senior Secondary School Mathematics for Class 11
10. (i) 14 (ii) 5 (iii) 25 (iv) 3
(v) 5
13 (vi) 5 (vii) 5 (viii) 10
11. (i) , i4
1
4
3
e o (ii) i29
2
29
5b l (iii) i13
5
13
1b l (iv) i5
4
5
3b l
12. ,a b1 0 13. ,x y0 2 17. n 2 20. 2 2
21. (i) ,x y2 1 (ii) ,x y2 3 (iii) ,x y4 1
(iv) ( , ) ( , )orx y x y5 2 5 2 (v) ,x y7 5 (vi) ,x y3 1
22. ,x y3 3 23. ( , )x y1 4 or ( , )x y1 4
26. | |z 1 27. ( ) ( )orz i z i2 1 2 1! ! 28. i2
3 2b l
HINTS TO SOME SELECTED QUESTIONS
2. (iii) ( ) ( ) ( ) ( ) ( )i i i i i3 5 5 3 5 3 3 5 3 5 33 3 3 3 # #
( ) ( ) .i i i125 27 225 135 198 10
(iv) ·( )i i i i i2 3
1 1 2 3
1 8 27
1 2 2 3
13 3 3 3 c c cm m m( 2
(v) ·( ) ( ) ( )
( ) ( )
i
i i
i i
4 3 4 3
1
4 3
4 3
25
4 31 #
(vi) ·( )
( ) ( )
( )
i
i i
i
2 3
2 3
1
2 3
2 31 #
(vii) ·( )
( ) ( ) ( ) ( )
( )
i
i i i i i
i
2
2
1
4 4
1
3 4
1
3 4
3 42
2 2 #
(viii) ·( )
( ) { ( )} ( ) ( )
( )
i
i i i i i i
i
1 2
1 2
1
1 8 6 1 2
1
11 2
1
11 2
11 23
3 3 #
4. (i)
( ) ( )
( ) ( )
( )
( )
i i i
i
i i
i i
i
i
3 2
5
1
2
3 2
4 5
3 2 1
5 1 2 3 2
3 2
4 5
#
d dn n
( ) ( )
( ) ( )
( ) ( )
( ) ( )
i i
i i
i i
i i
1 5 3 2
1 4 5
1 5 3 2
1 4 5
·
( )
( )
( )
( ) ( )
i
i
i
i i
i13 1
9
1
1
26
10 8
13
5
13
4
#
c m
5. (i) Given expression
( ) ( )
( ) ( ) ( ) ( )
.
i i
i i i i
2 3 2 3
3 2 2 3 3 2 2 3
13
0 0
(ii) Given expression ·
( ) ( )
( ) ( )
( )
{( ) ( ) }
i i
i i i
7 3 7 3
7 3 7 3
7 3
2 7 3
10
8
5
4
2 2 2 2
6. Given expression ·
( )
( )
( )
( )
()
( ) ( )
cos
cos
cos
cos
cos
cos cos
i
i
i
i i
1 2
1
1 2
1 2
1 4
1 2 3
2
2
#
This expression will be purely real when .cos3 0
Now, ,( )cos cos n3 0 0 2 1 2+ +
where .n Id
Complex Numbers and Quadratic Equations 175
7. Let ( ) .z x iy Then,
| | | |z i z i2 2
& |( ) | |( ) |x iy i x iy i2 2
& | ( ) | | ( ) | ( ) ( )x y i x y i x y x y1 1 1 12 2 2 2 2 2&
& ( ) ( ) .y y y y1 1 0 4 0 02 2 & &
Hence, z is purely real.
8. Let ( )z i3 21 and ( ) .z i3 22 Then, .z z1 2!
But, | | | | .z z 132
10. (v)
( )
( )
( )
( )
( )
( )
( )
( )
z
i
i
i
i i
i
i
i
i i
4 3
3 2
4 3
9 4 12
4 3
5 12
4 3
4 3
25
16 63
2 2
#
& | |z 25
16
25
63
625
256
625
3969
625
42252 2 2 c c cm m m) 3
& ·| |z 625
4225
25
65
5
13
(vi)
( )
( ) ( )
( )
( )
( )
( ) ( )
( ) .z
i
i i
i
i
i
i i
i1
2 1
1
3
1
1
2
2 4
1 2#
11. Use the formula, ·
| |
z
z
z1
2
(iii) ·
( )
( )
( )
( ) ( )
z
i
i
i
i i
i1
2 3
1
1
2
5
2
5
2
1
#
c m
·| |
( )
andz z i
i
4
25
4
1
4
26
2
13
2
5
2
1
2
52
c cm m
·
| |
( ) ( )
z
z
z i i i2
5
13
2
13
5
13
5
13
11
2 #
c m
(iv) ·
( )
( ) ( )
( )
( )
( )
( ) ( )
z
i
i i
i
i
i
i i
i1 3
1 1 2
1 3
1 3
1 3
1 3
10
8 6
5
4
5
3
#
c m
( )
| | .andz i
i
z5
4
5
3
5
4 3
25
16
25
9
25
25 12
c cm m
·
| |
z
z
z i5
4
5
31
2
c m
12.
( )
( )
( )
( )
( )
( ) ( ) ( )
.
i
i
i
i
i
i i i i
i1
1
1
1
1
1
2
1
2
1 22 2
#
( ) {( ) } ( )i
i i i1
1 1 1
100
100 4 25 25
d n [ ( ) ]i i 14 4a .
Hence, .anda b1 0
13.
( )
( )
( )
( ) ( ) ( )
.i
i
i
i
i
i i i i i i1
1
1
1
1
1
2
1
2
1 2
2
2
2 2
#
d n
Similarly, .i
i i1
1
d n
Given expression ( ) ( ) .i i i i i2 2 0 23 3 3
14. ·( ) ( )
( )
( )
( )
( )
( )
x iy
a ib
a ib x iy
a ib
a ib
x iy
a ib
a ib
& &
176 Senior Secondary School Mathematics for Class 11
( ) ( )
( )
( )
( )
( )
.x iy x iy
a ib
a ib
a ib
a ib
x y1 12 2&#
15. ( ) ( )
( )
( )
( )
( )
( )
( ) ( )
( )
( )
.a ib
c i
c i
c i
c i
c
c i
c
c i ci
c
c
c
c i
1 1
2
1
1
1
2
2
2
2
2 2
2
2
2#
& | |
( ) ( )
( )
( )
( )
.a ib
c
c
c
c
c
c c
c
c
1
1
1
4
1
1 4
1
1
12 2
2 2
2 2
2
2 2
2 2 2
2 2
2 2
J
L
K
K
K
KK
N
P
O
O
O
OO
Z
[
\
]
]
]]
]
]
]]
_
`
a
b
b
bb
b
b
bb
& .a b 12 2
Now, ·
( )
( )
( ) ( )
anda
c
c
b
c
c
a
b
c
c
1
1
1
2
1
2
2
2
2 2&
16.
( ) ( ) ( )
( ) .i i
i
i
i
i
i i i
i1 1
1
1
1
12
2
#
c m
given expression ( ) ( ) {( ) ( )} ( ) .i i i i i1 1 1 1 1 2n n n n2 2#
17.
( )
( )
( )
( )
( )
( ) ( ) ( )
.
i
i
i
i
i
i i i i i i1
1
1
1
1
1
2
1
2
1 2
2
2
2 2
#
( ) ( )
( )
( )
.i i
i
i
i
i i1 1
1
1
1 1
1 1 1n n n
n n
n2 2
2
2 2
2
& & &
d n
Least value of n2 is 4 and so the least value of n is 2.
18. Given expression {( ) } {( ) } {( ) } {( ) }x i x i x i x i1 1 1 1
{( ) } {( ) } {( ) } {( ) }x i x i x x1 1 1 1 1 12 2 2 2 2 2
{( ) } {( ) }x x x x2 2 2 22 2
( ) ( ) .x x x2 4 42 2 2 4
19.
( )
( )
( )
( )
cos sin
cos sin
cos sin
cos sin
a
a
i
i
i
i
1
1
1
1
1
1
#
( )
( )
( )
( )
cos sin
cos sin sin
cos sin cos
cos sin sini i
1
1 2
1 2
1 2
2 2
2 2
2 2
2 2
·
( ) ( ) ( / )
( / ) ( / )
.cos
sin
cos
sin
sin
sin cos
coti i i i2 1
2
1 2 2
2 2 2
22
c m
20.
( ) ( )
( ) ( )
( ) ( )
( )
( )z z i
z z
i i i
i i
i i
i
i
1
2 1
2 1 1
1
4
1
1
2 1
1 2
1 2
#
& ( ) .z z i
z z 1
2 1 1 2 2
1 2
1 2 2#
21. (iv) ( ) ( ) ,y x y x i y x y x6 2 1 0 6 2 0 1 02 2 2 2&
& .andy x y x2 6 12 2
On subtracting, we get .x 5
.y y y5 1 4 22 2& & !
(v) ( ) ( ) ( )x i i iy3 1 2
& ( ) ( ) ( ) ( ) ( )x i y y i x y i y3 2 2 2 5 0&
& .andx y y2 0 5 0
Complex Numbers and Quadratic Equations 177
(vi) { ( ) } ( ) { ( ) } ( ) ( ) ( )x x i i y y i i i i i2 3 2 1 3 3 3 3
& { ( )} { ( ) } { ( )} { ( }x x x x i y y y y i i3 2 3 2 6 1 3 2 3 9 10
& ( ) ( )x y x y i i4 2 9 1 2 6 3 7 10
& ( ) ( )x y x y i i4 9 3 2 7 3 10
& .andx y x y4 9 3 2 7 13
22. ( ) ( ) ( ) ( )x iy i i i3 5 6 24 6 24
& ( ) ( )x y x y i i3 5 5 3 6 24
& .andx y x y3 5 6 5 3 24
23. ( ) ( )iyx x y i3 42 2
& ( ) ( )iyx x y i3 42 2
& ( ) ( )x y yx i3 4 02 2
& andx y yx3 0 4 02 2
& x y 32 … (i) and yx 42 … (ii).
Putting ( )x y32 from (i) in (ii), we get
( ) ( ) ( )y y y y y y3 4 3 4 0 4 1 02& & .ory y4 1&
Now, y x x1 42& & is imaginary. So, .y 1!
When ,y 4 we get ( ) .x x3 4 1 12 & !
( , ) ( , ) .orx y x y1 4 1 4
25. ( ) ( ) ( )
( )
( )
( ) ( ) ( )
.i z i z
z
z
i
i
i
i i i i
i1 1 1
1
1
1
2
1
2
1 22 2
& #
Hence, .z i z
26. Let .z x iy Then,
·
( )
( )
( )
( )
( )
( ) ( )
z
z
x iy
x iy
x iy
x iy
x y
x y x y i
1
1
1
1
1
1
1
1
2 2
2 2
#
Now, z
z
1
1
is purely imaginary | | .x y x y z1 0 1 12 2 2 2+ + +
27. Let ( ) .z x iy Then, ( ) .z x y ixy22 2 2
Now, ( ) .Re z x y0 02 2 2& … (i)
| | | | .z z x y2 4 42 2 2& & … (ii)
On solving (i) and (ii), we get
.andx y2 22 2
x 2! and .y 2!
( ) ( ) .orz i z i2 1 2 1! !
28. Let the required complex number be ( ) .z x iy Then,
| |z z i1 2
& | | ( )x iy x iy i1 2
& ( ) ( )x y x y i1 22 2
178 Senior Secondary School Mathematics for Class 11
& ( ) andx y x y1 2 02 2
& ( ) ( )andy x x2 2 12 2
& ( ) .and andy x x x y2 4 1 2
3 22 2 &
SOME RESULTS ON MODULUS AND CONJUGATE OF COMPLEX NUMBERS
THEOREM 1 For any complex numbers ,z z1 and z2 prove that:
(i) ( )z z (ii) | | { } { ( )}Re Imzz z z z2 2 2
(iii) ( )Rez z z2 (iv) · ( )Imz z i z2
(v) z z z+ is purely real (vi) z z z0 + is purely imaginary
(vii) z z z z1 2 1 2 (viii) z z z z1 2 1 2
(ix) ( ) ( )z z z z1 2 2 2 (x) ,z
z z
z 0
2
1
2
1
2! z
d n
PROOF Let ( ), ( ) ( ) .andz a ib z a ib z a ib1 1 1 2 2 2 Then,
(i) ( )z a ib z& ( ) ( )a ib a ib
( ) ( ) ( ) .z a ib a ib z&
Hence, ( ) .z z
(ii) zz ( ) ( ) ( ) ( )a ib a ib a ib a ib
( ) { ( )} { ( )} | |Re Ima b z z z2 2 2 2 2 .
{ ( )} { ( )} | | .Re Imzz z z z2 2 2
(iii) z z ( ) ( )a ib a ib
( ) ( ) ( ) .Rea ib a ib a z2 2
( ) .Rez z z2
(iv) z z ( ) ( )a ib a ib
( ) ( ) ( ) .Ima ib a ib ib i z2 2 ·
· ( ) .Imz z i z2
(v) z z + ( ) ( )a ib a ib
( ) ( )a ib a ib ib2 0+ +
( ) .Imb z0 0+ +
( )Imz z z z0+ + is purely real.
(vi) z z 0 + ( ) ( )a ib a ib 0
( ) ( )a ib a ib a a0 2 0 0+ + +
z+ is purely imaginary.
z z z0 + is purely imaginary.
Complex Numbers and Quadratic Equations 179
(vii) ( ) ( )zz a ib a ib1 2 1 1 2 2
( ) ( ) ( ) ( )a a i b b a a i b b1 2 1 2 1 2 1 2
( ) ( ) .a ib a ib z z1 1 2 2 1 2
.z z z z1 2 1 2
(viii) ( ) ( )z z a ib a ib1 2 1 1 2 2
( ) ( ) ( ) ( )a a i b b a a i b b1 2 1 2 1 2 1 2
( ) ( ) .a ib a ib z z1 1 2 2 1 2
.z z z z1 2 1 2
(ix) ( ) ( ) ( ) ( )z z a ib a ib a a b b i a b b a1 2 1 1 2 2 1 2 1 2 1 2 1 2
& ( ) ( )z z a a b b i a b b a1 2 1 2 1 2 1 2 1 2
( ) ( )a a b b i a b b a1 2 1 2 1 2 1 2
( ) ( ) .a ib a ib z z1 1 2 2 1 2
.z z z z1 2 1 2
(x) ( )
( )
( )
( )
( )
( )
z
z
a ib
a ib
a ib
a ib
a ib
a ib
2
1
2 2
1 1
2 2
1 1
2 2
2 2
#
( )
( ) ( )
( )
( ) ( )
a b
a ib a ib
a b
a a b b i b a a b
2
2
2
2
1 1 2 2
2
2
2
2
1 2 1 2 1 2 1 2
( )
( )
( )
( )
a b
a a b b
i
a b
b a a b
2
2
2
2
1 2 1 2
2
2
2
2
1 2 1 2
·
( )
( )
( )
( )
z
z
a b
a a b b
i
a b
b a a b
2
1
2
2
2
2
1 2 1 2
2
2
2
2
1 2 1 2
d n … (i)
Also, · ·( ) ( ) ( )
( )
z
z
z
z
a ib
a ib a ib
a ib1 11
1 1 1
2 2 2 2
2 2
#
2 2
Z
[
\
]
]]
]
]]
_
`
a
b
bb
b
bb
·
( )
( ) ( )
( )
( )
( )
( )
a b
a ib a ib
a b
a a b b
i
a b
b a a b
2
2
2
2
1 1 2 2
2
2
2
2
1 2 1 2
2
2
2
2
1 2 1 2
… (ii)
From (i) and (ii), we get
z
z
2
1 d n ·
z
z1
2
THEOREM 2 For all complex numbers ,z z1 and ,z2 prove that:
(i) | |zz z 2 (ii) | | | | | |z z z
(iii) | |z z0 0+ (iv) | | ( ) | |Rez z z# #
(v) | | ( ) | |Imz z z# # (vi) | | | || |z z z z1 2 1 2
(vii)
| |
| |
,z
z
z
z
z 0
2
1
2
1
2!
180 Senior Secondary School Mathematics for Class 11
PROOF Let ( ), ( ) ( ) .andz a ib z a ib z a ib1 1 1 2 2 2 Then,
(i) zz ( ) ( ) ( ) ( ) ( ) | | .a ib a ib a ib a ib a b z2 2 2
| | .z z z 2
(ii) | | | | , | | | | ( )z a ib a b z a ib a b a b2 2 2 2 2 2
and | | | ( )| |( ) ( )| ( ) ( ) .z a ib a i b a b a b2 2 2 2
| | | | | |.z z z
(iii) | | | | | |z z a ib0 0 02 2+ +
.anda b a b z0 0 0 02 2+ + +
(iv) Let ( ) .z a ib Then, | | .z a b2 2
Clearly, .a b a a b2 2 2 2# #
| | ( ) | |.Rez z z# #
(v) Let ( ) .z a ib Then, | | .z a b2 2
Clearly, a b b a b2 2 2 2# # .
| | ( ) | |.Imz z z# #
(vi) ( ) ( ) ( ) ( ) .z z a ib a ib a a b b i a b b a1 2 1 1 2 2 1 2 1 2 1 2 1 2
| | ( ) ( )z z a a b b a b b a1 2 1 2 1 2
2
1 2 1 2
2
a a b b a b b a1
2
2
2
1
2
2
2
1
2
2
2
1
2
2
2
( ) ( ) ( ) ( )a a b b a b a b a b1
2
2
2
2
2
1
2
2
2
2
2
1
2
1
2
2
2
2
2
·| | | |a b a b z z1
2
1
2
2
2
2
2
1 2 $ $. .
·| | | | | |.z z z z1 2 1 2
(vii) · ·( ) ( ) ( )
( )
z
z
z z a ib a ib a ib
a ib1 1
2
1
1
2
1 1
2 2 2 2
2 2
#
Z
[
\
]
]]
]
]]
_
`
a
b
bb
b
bb
( ) ( )
( ) ( )
( )
( ) ( )
a ib a ib
a ib a ib
a b
a a b b i b a a b
2 2 2 2
1 1 2 2
2
2
2
2
1 2 1 2 1 2 1 2
( )
( )
( )
( )
a b
a a b b
i
a b
a b a b
2
2
2
2
1 2 1 2
2
2
2
2
2 1 1 2
& z
z
a b
a a b b
a b
a b a b
2
1
2
2
2
2
1 2 1 2
2
2
2
2
2
2 1 1 2
2
Z
[
\
]
]]
]
]]
Z
[
\
]
]]
]
]]
_
`
a
b
bb
b
bb
_
`
a
b
bb
b
bb
( )
( ) ( )
a b
a a b b a b a b
2
2
2
2 2
1 2 1 2
2
2 1 1 2
2
( )a b
a a b b a b a b
2
2
2
2 2
1
2
2
2
1
2
2
2
2
2
1
2
1
2
2
2
Complex Numbers and Quadratic Equations 181
( )
( ) ( )
a b
a a b b a b
2
2
2
2 2
1
2
2
2
2
2
1
2
2
2
2
2
·
( )
( ) ( )
( )
( )
| |
| |
a b
a b a b
a b
a b
a b
a b
z
z
2
2
2
2 2
2
2
2
2
1
2
1
2
2
2
2
2
1
2
1
2
2
2
2
2
1
2
1
2
2
1
Hence, ,
| |
| |
.z
z
z
z
z 0
2
1
2
1
2!
THEOREM 3 For all complex numbers ,z and z1 2 prove that:
(i) · ·( ) ( ) ( ) ( ) ( )Re Re Re Im Imz z z z z z1 2 1 2 1 2
(ii) · ·( ) ( ) ( ) ( ) ( )Im Re Im Im Rez z z z z z1 2 1 2 1 2
PROOF Let ( ) ( ) .z a ib and z a ib1 1 1 2 2 2 Then,
( ) ( )z z a ib a ib1 2 1 1 2 2
& ( ) ( ) .z z a a b b i a b b a1 2 1 2 1 2 1 2 1 2
(i) ( ) ( )Re z z a a b b1 2 1 2 1 2
· ·( ) ( ) ( ) ( ) .Re Re Im Imz z z z1 2 1 2
(ii) ( ) ( )Im z z a b b a1 2 1 2 1 2
· ·( ) ( ) ( ) ( ) .Re Im Im Rez z z z1 2 1 2
THEOREM 4 For all , , ,a b R and z z C1 2d d prove that
| | | | ( )(| | | | ).az bz bz az a b z z1 2
2
1 2
2 2 2
1
2
2
2
PROOF We have
| | ( )( )az bz az bz az bz1 2
2
1 2 1 2
( ) ( )az bz az bz1 2 21 [ , ]a b Ra d
a z z b z z abz z abz z2 1
2
2 1 2 2 21 1
| | | | ( ) .a z b z ab z z z z2 1
2 2
2
2
1 2 2 1 … (i)
| | ( )( )bz az bz az bz az1 2
2
1 2 1 2
( ) ( )bz az bz az1 2 21 [ , ]a b Ra d
b z z a z z abz z abz z2 1
2
2 1 2 1 2 2 1
| | | | ( ) .b z a z ab z z z z2 1
2 2
2
2
1 2 2 1 … (ii)
Adding the corresponding sides of (i) and (ii), we get
| | | | ( )(| | | | ).az bz bz az a b z z1 2
2
1 2
2 2 2
1
2
2
2
SOLVED EXAMPLES
EXAMPLE 1 If ,z i
z i
5
5 1
show that z is a real number.
SOLUTION Let ( ) .z x iy Then,
| |
| |
z i
z i
z i
z i
5
5 1
5
5
1&
| |
| |
z
z
z
z
2
1
2
1
a
R
T
S
S
S
S
SS
V
X
W
W
W
W
WW
182 Senior Secondary School Mathematics for Class 11
& | | | | | | | |z i z i z i z i5 5 5 52 2&
& |( ) | |( ) |x iy i x iy i5 52 2 [ ( )]z x iya
& | ( )| | ( )|x i y x i y5 52 2
& ( ) ( )x y x y5 52 2 2 2 [ | | ( )]x iy x y2 2 2a
& ( ) ( ) .y y y y5 5 0 4 5 0 02 2 & &# #
,z x i z x0 & where x is real.
Hence, z is a real number.
EXAMPLE 2 If ( )
( )
( )
a ib
x
x i
2 12
2
then prove that ·( )
( )
( )
a b
x
x
2 1
12 2
2 2
2 2
SOLUTION We have
( )
( )
( )
( )
( )
( )
( ) ( )
a ib
x
x i
x
x i ix
x
x i x
2 1 2 1
2
2 1
1 2
2
2
2
2 2
2
2
& ·( )
( )
( )
( )
a ib
x
x
i
x
x
2 1
1
2 1
2
2
2
2
& ·| |
( )
( )
( )
a ib
x
x
i
x
x
2 1
1
2 1
22
2
2
2
2
& ( )
( )
( )
( )
a b
x
x
x
x
2 1
1
2 1
42 2
2 2
2 2
2 2
2
Z
[
\
]
]]
]
]]
_
`
a
b
bb
b
bb
·
( )
( )
( )
( )
x
x x
x
x
2 1
1 4
2 1
1
2 2
2 2 2
2 2
2 2
Hence, ·( )
( )
( )
a b
x
x
2 1
12 2
2 2
2 2
EXAMPLE 3 If ( ) ( )
( )
p iq
a i
a i
2
2
then prove that ·( )
( )
( )
p q
a
a
4 1
12 2
2
2 2
SOLUTION We have
( ) ( )
( )
( )
( )
( )
( )
p iq
a i
a i
a i
a i ai
a i
a ai
2 2
2
2
1 22 2 2 2
& | |
| |
|( ) |
{ ( ) }
( )
( )
( )
p iq
a i
a ai
a
a a
a
a
2
1 2
4 1
1 4
4 1
12
2
2 2
2 2
2 2 2
2
2 2
& ·( )
( )
( )
p q
a
a
4 1
12 2
2
2 2
Hence, ·( )
( )
( )
p q
a
a
4 1
12 2
2
2 2
Complex Numbers and Quadratic Equations 183
EXAMPLE 4 Let z be a complex number such that .z i
iz1 1
Show that z is purely
real.
SOLUTION Let .z x iy Then,
| |
| |
z i
iz
z i
iz1 1
1
1&
| |
| |
z
z
z
z
2
1
2
1
a
R
T
S
S
S
S
SS
V
X
W
W
W
W
WW
& | | | | | | | |iz z i iz z i1 1 2 2&
& | ( )| |( ) |i x iy x iy i1 2 2
& |( ) | | ( ) |y ix x y i1 12 2
& ( ) ( ) ( )y x x y1 12 2 2 2
& y y x x y y1 2 2 12 2 2 2
& .y y4 0 0&
z x y0 and hence z is purely real.
EXAMPLE 5 If z1 and z2 are complex numbers such that z
z
3
2
2
1 is purely imaginarythen
prove that .z z
z z
1
1 2
1 2
SOLUTION Let z
z
ki3
2
2
1 for some real number k. Then,
·z
z
ki z
z ki
3
2
2
3
2
1
2
1
& … (i)
| |
| |
z z
z z
z z
z z
z
z
z
z
1
1
1 2
1 2
1 2
1 2
2
1
2
1
| |
| |
z
z
z
z
2
1
2
1
a
R
T
S
S
S
S
SS
V
X
W
W
W
W
WW
| |
| |
ki
ki
ki
ki
k
k
2
3 1
2
3 1
3 2
3 2
9 4
9 4
1
2
2
[using (i)].
Hence, .z z
z z
1
1 2
1 2
EXAMPLE 6 If | | | | | | | |z z z z 1… n1 2 3 then prove that
| |.z z z z z z z z
1 1 1 1… …
n
n
1 2 3
1 2 3
SOLUTION We have
| | | | | | | |z z z z 1… n1 2 3
& | | | | | | | |z z z z 1… n1
2
2
2
3
2 2
& , , , …,z z z z z z z z1 1 1 1n1 2 3 n1 2 3
& , , , …,z z z z z z z z
1 1 1 1
n1 2 3
n1 2 3
184 Senior Secondary School Mathematics for Class 11
& | … |z z z z z z z z
1 1 1 1…
n1 2 3
n1 2 3
| … |z z z zn1 2 3
| |z z z z… n1 2 3 [ | | | |]z za .
| |.z z z z z z z z
1 1 1 1… …
n
n
1 2 3
1 2 3
EXAMPLE 7 If ,
c id
a ib x iy
prove that
(i)
c id
a ib x iy
and (ii) .
c d
a b x y2 2
2 2
2 2
SOLUTION (i) ( ) ( )
c id
a ib x iy
c id
a ib x iy&
d dn n
( )
( )
( ) ( )
( )
( ) .
c id
a ib
x iy
c id
a ib
x iy& &
(ii) We have
( )
( )
( ) ( )
( )
( )and
c id
a ib
x iy
c id
a ib
x iy
& ( )
( )
( )
( )
( ) ( )
c id
a ib
c id
a ib
x iy x iy#
& ( ) ( )
( ) ( )
( ) ( )
c id c id
a ib a ib
x iy x iy
&
( )
( )
( ) .
c d
a b
x y2 2
2 2
2 2
EXAMPLE 8 If ,( )x iy
c id
a ib
prove that ·( )x y
c d
a b2 2 2
2 2
2 2
SOLUTION We have
( )x iy
c id
a ib
c id
a ib
& ( )x iy
c id
a ib
& ( ) ( )
( ) ( )
( ) ( )
x iy x iy
c id
a ib
c id
a ib
c id c id
a ib a ib
#
& ·( ) ( )
( )
( )
x y
c d
a b
x y
c d
a b2 2
2 2
2 2
2 2 2
2 2
2 2
&
Hence, ·( )
( )
( )
x y
c d
a b2 2 2
2 2
2 2
Complex Numbers and Quadratic Equations 185
EXAMPLE 9 If ( ) ( ) ( )…( ) ( )i i i ni x iy1 1 2 1 3 1 then prove that
… ( ) ( ) .n x y2 5 10 1 2 2 2# # # #
SOLUTION ( ) ( ) ( ) ( ) ( )i i i ni x iy1 1 2 1 3 1…
& |( )( ) ( ) ( )| | |i i i ni x iy1 1 2 1 3 1… 2 2
& | | | | | | | | ( )i i i ni x y1 1 2 1 3 1…2 2 2 2 2 2# # # #
& ( ) ( ) ( ) … ( ) ( )n x y2 5 10 12 2 2 2 2 2 2# # # #
& ( ) ( ) .n x y2 5 10 1… 2 2 2# # # #
Hence, ( ) ( ) .n x y2 5 10 1… 2 2 2# # # #
EXAMPLE 10 If ( ) ( ) ( ) ( ) ( )a ib c id e if g ih A iB then show that
( ) ( ) ( ) ( ) ( ) .a b c d e f g h A B2 2 2 2 2 2 2 2 2 2
SOLUTION ( ) ( ) ( ) ( ) ( )a ib c id e if g ih A iB
& |( )( ) ( ) ( )| | |a ib c id e if g ih A iB
& · · ·| | | | | | | | | |a ib c id e if g ih A iB
& · · ·| | | | | | | | | |a ib c id e if g ih A iB2 2 2 2 2
& ( ) ( ) ( ) ( ) ( ) .a b c d e f g h A B2 2 2 2 2 2 2 2 2 2
Hence, ( ) ( ) ( ) ( ) ( ) .a b c d e f g h A B2 2 2 2 2 2 2 2 2 2
EXAMPLE 11 If and are different complex numbers such that | | ,1 show that
.
1
1
SOLUTION We have
1 1 1
2
Z
[
\
]
]]
]
]]
Z
[
\
]
]]
]
]]
_
`
a
b
bb
b
bb
_
`
a
b
bb
b
bb
[ | | ]z zz2a
·
( )
( )
( )
( )
( ) ( )
( ) ( )
1 1 1 1
( ) ( )1
( ) | | | |
| | | | ( )
1 2 2
2 2
[ | | , | | ]2 2a
| | ( )
| | ( )
1
1
12
2
[ | | ]1a .
Hence, .
1
1
186 Senior Secondary School Mathematics for Class 11
EXAMPLE 12 If | | | |z z1 12 2 then show that z is imaginary.
SOLUTION Let ( ) .z x iy Then, ( )z x y ixy22 2 2 and | | ( ) .z x y2 2 2
Now, | | | |z z1 12 2
& |( ) ( )| ( )x y i xy x y1 2 12 2 2 2
& |( ) ( )| ( )x y i xy x y1 2 12 2 2 2 2 2
& ( ) ( )x y x y x y1 4 12 2 2 2 2 2 2 2
& [ ( )] [ ( )]x y x y x y1 1 42 2 2 2 2 2 2 2
& ( ) {( ) }x y x y x y y4 1 4 4 1 02 2 2 2 2 2 2&
& .x x4 0 02 &
Hence, ,z iy0 which shows that z is imaginary.
EXAMPLE 13 If iz z z i 03 2 then show that | | .z 1
SOLUTION We have
iz z z i 03 2
& z iz iz 1 03 2 [on dividing both sides by i]
& ( ) ( )z z i i z i 02
& ( ) ( )z i z i 02
& .orz i z i2
Now, | | || | | .z i z i z 1& &
And, | | | | | | | | .z i z i z z1 1 12 2 2& & &
Hence, in either case, we have | | .z 1
EXAMPLE 14 Find all nonzero complex numbers z satisfying .z iz2
SOLUTION Let .z x iy Then, z x iy and ( ) .z x y ixy22 2 2
( )z iz x iy i x y xy22 2 2&
( ) ( ) .x xy i x y y2 02 2& … (i)
On equating real parts and imaginary parts on both sides of (i)
separately, we get
x xy2 0 … (ii) and x y y 02 2 … (iii).
From (ii), we get
( )x xy x y2 0 1 2 0&
·or orx y x y0 1 2 0 0 2
1
& &
Case I When .x 0
Putting x 0 in (iii), we get
( ) .ory y y y y y0 1 0 0 12 & &
Complex Numbers and Quadratic Equations 187
Thus, ( , ) ( , )orx y x y0 0 0 1
( ) ( ) .orz i z i0 0 0 1 ·
Case II When ·y 2
1
Putting y 2
1 in (iii), we get
x x2
1
2
1 0 4
1
2
1 02
2
2&
b bl l
& ·x x4
1
2
1
4
3
2
32 & ! b l
, ,orx y x y2
3
2
1
2
3
2
1
e eo o
·orz i z i2
3
2
1
2
3
2
1
e eo o
Hence, the required nonzero complex numbers are
·, andi i i2
3
2
1
2
3
2
1
e eo o
EXAMPLE 15 Solve the equation, ,z z2 where z is a complex number.
SOLUTION Let .z x iy Then,
( )z z x iy x iy2 2&
( ) .x y ixy x iy22 2& … (i)
Equating real parts and imaginary parts on both sides of (i)
separately, we get
x y x2 2 … (ii) and xy y2 … (iii).
From (iii), we get
·( ) orxy y y x y x2 0 2 1 0 0 2
1
& &
Case I When .y 0
Putting y 0 in (ii), we get
( ) .orx x x x x x0 1 0 0 12 & &
( , ) ( , )orx y x y0 0 1 0
Thus, ( ) ( ) .orz i z i0 0 1 0
Case II When ·x 2
1
Putting x 2
1 in (ii), we get
·y y y2
1
2
1
4
1
2
1
4
3
2
32 2 2& & !
b b bl l l
188 Senior Secondary School Mathematics for Class 11
, , ·orx y x y2
1
2
3
2
1
2
3
e eo o
Thus, ·orz i z i2
1
2
3
2
1
2
3
e eo o
Hence, , , andz i i0 1 2
1
2
3
2
1
2
3
e eo o are the required roots of
the given equation.
EXAMPLE 16 For all complex numbers ,z and z1 2 prove that
| | | | (| | | | ).z z z z z z21 2
2
1 2
2
1
2
2
2
SOLUTION We have
| | ( )( )z z z z z z1 2
2
1 2 1 2
( ) ( )z z z z1 2 21 [ ( )z z1 2a ]z z1 2
z z z z z z z z1 2 1 2 1 2 2 1
| | | |z z z z z z1
2
2
2
1 2 12 . … (i)
| | ( )( )z z z z z z1 2
2
1 2 1 2 ( ) ( )z z z z1 2 1 2
z z z z z z z z1 2 1 2 1 12 2
| | | | .z z z z z z1
2
2
2
1 2 12 … (ii)
On adding the corresponding sides of (i) and (ii), we get
| | | | (| | | | ).z z z z z z21 2
2
1 2
2
1
2
2
2
EXERCISE 5C
1. Express each of the following in the form ( )a ib and fi nd its conjugate:
(i) ( )i4 3
1
(ii) ( )i2 3
2 (iii)
( )
( )
i
i
1 2
2
2
(iv) ( )
( ) ( )
i
i i
1 3
1 1 2
(v) i
i
2
1 2 2
b l (vi) ( ) ( )
( )
i i
i
3 1 2
2
2. Express each of the following in the form ( )a iband fi nd its multiplicative
inverse:
(i) i
i
1 3
1 2
(ii)
( )
( )
i
i
2
1 7
2
(iii)
( )i1 3
4
3. If ( ) ( )x iy u iv3 then prove that ( ) .x
u
y
v x y4 2 2 b l
4. If ( ) ( )x iy a ib/1 3 then prove that ( ) .a
x
b
y
a b4 2 2 d n
5. Express ( )i1 2 3 in the form ( ) .a ib
Complex Numbers and Quadratic Equations 189
6. Find real values of x and y for which
( ) ( ) ( ) ( )x xi x iy i iy2 3 3 5 1 24 2 .
7. If | | ,z z 02 2 show that z is purely imaginary.
8. If z
z
1
1
is purely imaginary and ,z 1! show that | | .z 1
9. If z1 is a complex number other than –1 such that | |z 11 and z z
z
1
1
2
1
1
then
show that z2 is purely imaginary.
10. For all ,z Cd prove that
(i) ( ) ( )Rez z z2
1 (ii) ( ) ( )Imi z z z2
1 (iii) | |zz z 2
(iv) ( )z z is real (v) ( )z z is 0 or imaginary.
11. If ( )z i11 and ( ),z i2 42 prove that .Im z
z z
21 2
1
d n
12. If a and b are real numbers such that a b 12 2 then show that a real value
of x will satisfy the equation, ( ) .ix
ix a ib1
1
ANSWERS (EXERCISE 5C)
1. (i) ,z i z i25
4
25
3
25
4
25
3 b bl l (ii) ( ), ( )z i z i5 12 5 12
(iii) ,z i z i25
2
25
11
25
2
25
11 b bl l (iv) ,z i z i5
4
5
3
5
4
5
3 b bl l
(v) ,z i z i25
7
25
24
25
7
25
24 b bl l (vi) ,z i z i10
3
10
1
10
3
10
1 b bl l
2. (i) , ( )z i z i2
1
2
1 11
b l (ii) ( ),z i z i1 2
1
2
11 b l
(iii) ( ),z i z i1 3 4
1
4
31 e o
5. i125
11
125
2 b l 6. ( , ) ,orx y x y2 3 2 3
1 b l
HINTS TO SOME SELECTED QUESTIONS
3. ( ) ( ) ( )u iv x iy x iy ixy x iy33 3 3
& ( ) ( ) ( )u iv x xy i x y y3 33 2 2 3
& andu x x y v x y y3 33 2 2 3
& ( ) ( ) ( ) .x
u
y
v x y x y x y3 3 42 2 2 2 2 2 c m
4. ( ) ( ) ( )x iy a ib a ib iab a ib33 3 3
& ( ) ( ) ( )x iy a ab i a b b3 33 2 2 3
190 Senior Secondary School Mathematics for Class 11
& andx a ab y a b b3 33 2 2 3
& ( ) ( ) ( ) .a
x
b
y
a b a b a b3 3 42 2 2 2 2 2 d n
5. ( )
( ) { ( )}
i
i i i i
1 2
1 2
1
1 8 6 1 2
13
3 3
·
( ) ( )
( ) ( )
i i
i i
11 2
1
11 2
11 2
125
11 2
#
6. ( ) ( ) ( )x x i x y y i3 2 4 2 54 2
& ( ) ( )x x i x y y3 4 2 2 5 04 2
& .andx x x y3 4 0 2 3 5 04 2
Now, ( ) ( ) .orx x x x x x3 4 0 4 1 0 2 24 2 2 2& &
·( ) andx y x y2 3 2 3
1
& & c m
7. Let ( ) .z x iy Then,
| |z z x y ixy x y0 2 02 2 2 2 2 2&
.andx ixy x xy x0 0 0 02 2& & &
z is purely imaginary.
8. Let ( ) .z x iy Then,
·
( )
( )
( )
( )
( )
( )
( )
( )
z
z
x iy
x iy
x iy
x iy
x iy
x iy
x y
x y iy
1
1
1
1
1
1
1
1
1
1 2
2 2
2 2
#
z
z
1
1
is purely imaginary | | .x y x y z1 0 1 12 2 2 2+ + +
9. Let .z x iy1 1 1 Then, | | | | .z z x y1 1 11 1
2
1
2
1
2& &
( )
( )
( )
( )
( )
( )
z z
z
x iy
x iy
x iy
x iy
x iy
x iy
1
1
1
1
1
1
1
1
2
1
1
1 1
1 1
1 1
1 1
1 1
1 1
#
,
( )
( )
( )x y
x y iy
x y
iy
1
1 2
1
2
1
2
1
2
1
2
1
2
1
1
2
1
2
1
which is purely imaginary. [ | | ]z 11
2a
12. We have
( ) ( )
( ) ( )
( )
( )
ix
ix a ib
ix ix
ix ix
a ib
a ib
1
1
1 1 1
1 1
1
1
&
[by componendo and dividendo]
( )
( )
ix a ib
a ib
2
2
1
1
&
( )
( )
( )
( )
ix
a ib
a ib
a ib
a ib
1
1
1
1
& #
( )
( ) {( ) ( ) }
ix
a b
a b a b a b i
1
1 1 1
2 2
2 2
&
( )a b
bi
1
2
2 2
[ ]a b 12 2a
,
( )
x
a b
b
1
2
2 2&
which is purely real.
Complex Numbers and Quadratic Equations 191
POLAR REPRESENTATION OF COMPLEX NUMBERS
COMPLEX PLANE OR ARGAND PLANE
Let X OXl and YOYl be the mutually perpendicular lines, known as the x-axis
and the y-axis respectively. The complex number ( )x iy corresponds to the
ordered pair ( , )x y and it can be represented by the point ( , )P x y in the x-y plane.
The x-y plane is known as the complex plane, or the Argand plane. x-axis is called
the real axis and y-axis is called the imaginary axis.
Note that every number on the x-axis is a real number, while each on the
y-axis is an imaginary number.
The complex numbers represented geometrically in the above diagram are
( ), ( ), ( ), ( ),i i i i3 5 4 3 5 3 2 4
( ), ( ), ( ) ( ),andi i i i6 0 3 0 0 2 0
represented by the points
( , ), ( , ), ( , ), ( , ), ( , ),A B C D E3 5 4 3 5 3 2 4 6 0
( , ), ( , ) ( , )andF G H3 0 0 2 0 1 respectively.
POLAR FORM OF A COMPLEX NUMBER
Let the complex number z x iy be
represented by the point ( , )P x y in the
complex plane.
Let XOP+ and | | .OP r 0 Then,
( , )P r are called the polar coordinates
of P.
We call the origin O as pole.
Clearly, cosx r and siny r
( ) .cos sinz r i
O
A(3, 5)
E(6, 0)
Y
�Y
X� X
B(–4, 3)
D(2, –4)
C(–5, –3)
F(–3, 0)
G(0, 2)
H(0, –1)
O
Y
�Y
X� X
r
�
�P(r, )
x
y
192 Senior Secondary School Mathematics for Class 11
This is called the polar form, or trigonometric form, or modulus-amplitude
form, of z.
Here, | |r x y z2 2 is called the modulus of z.
And is called the argument, or amplitude of z, written as arg(z), or amp(z).
The value of such that # is called the principal argument of z.
METHOD FOR FINDING THE PRINCIPAL ARGUMENT OF A COMPLEX NUMBER
Case I When ( )z x iy lies on one of the axes:
I. When z is purely real.
In this case, z lies on the x-axis.
(i) If z lies on positive side of the x-axis then .0
(ii) If z lies on negative side of the x-axis then .
II. When z is purely imaginary.
In this case, z lies on the y-axis.
(i) If z lies on the y-axis and above the x-axis then ·2
(ii) If z lies on the y-axis and below the x-axis then ·2
Case II When ( )z x iy does not lie on any axes:
Step 1. Find the acute angle given by ·( )
( )
tan Re
Im
z
z
Step 2. Find the quadrant in which ( , )P x y lies.
Then, ( ) ( )arg or ampz z may be obtained as under.
(i) When z lies in quad. I;
then, ( ) .arg z&
(ii) When z lies in quad. II;
then, ( ) ( ) ( ) .arg z&
(iii) When z lies in quad. III;
then, ( ) ( )or
& ( ) ( ) ( ) .arg orz
(iv) When z lies in quad. IV;
then, ( )or 2
& ( ) ( ) .arg orz 2
Complex Numbers and Quadratic Equations 193
SUMMARY
Polar form of z x iy is ( ) .cos sinr i
(i) | | .r z x y2 2
(ii) ·
( )
( )
tan Re
Im
x
y
z
z
(iii) When , # then is the principal argument of z.
(iv) Quad. in which z lies ( )arg z
I
II
III ( )
IV ( )or 2
(v) When z is purely real;
then, z lies on the x-axis.
( ) ( ) .andx x0 0 0 & &
(vi) When z is purely imaginary;
then, z lies on the y-axis.
·andy y0 2 0 2 & &
a ak k
X� X
Y�
Y
O
= ( – )
X� X
Y�
Y
OM M
P P
X� X
Y�
Y
O X� X
Y�
Y
O
M
P
M
P
( – )
194 Senior Secondary School Mathematics for Class 11
SOLVED EXAMPLES
EXAMPLE 1 Convert each of the following complex numbers into polar form:
(i) 3 (ii) –5 (iii) i (iv) i2
SOLUTION (i) The given complex number is .z i3 0
Let its polar form be ( ) .cos sinz r i
Now, | | .r z 3 0 9 32 2
Clearly, z i3 0 is represented by the point ( , ),P 3 0 which lies
on the positive side of the x-axis.
( ) .arg z 0
Thus, .andr 3 0
Hence, the required polar form of z i3 0 is ( ) .cos sini30 0
(ii) The given complex number is .z i5 0
Let its polar form be ( ) .cos sinz r i
Now, | | ( ) .r z 5 0 25 52 2
Clearly, z i5 0 is represented by the point ( , ),P 5 0 which
lies on the negative side of the x-axis.
( ) .arg z &
Thus, .andr 5
Hence, the required polar form of z i5 0 is
( ) .cos sini5
(iii) The given complex number is .z i0
Let its polar form be ( ) .cos sinz r i
Now, | | .r z 0 1 12 2
Clearly, z i0 is represented by the point ( , ),P 0 1 which lies
on the y-axis and above the x-axis.
·( )arg z 2 2&
Thus, ·andr 1 2
Hence, the required polar form of z i0 is
,cos sini1 2 2·
a k i.e., ·cos sini2 2
a k
(iv) The given complex number is .z i0 2
Let its polar form be ( ) .cos sinz r i
Now, | | ( ) .r z 0 2 4 22 2
Clearly, ( )z i0 2 is represented by the point ( , ),P 0 2 which
lies on the y-axis and below the x-axis.
·( )arg z 2 2&
Thus, ·andr 2 2
Complex Numbers and Quadratic Equations 195
Hence, the required polar form of z i0 2 is
, ·. .,cos sin cos sini ez i i2 2 2 2 2 2
a a ak k k' 1
EXAMPLE 2 Find the modulus and argument of each of the complex numbers given below:
(i) i1 (ii) i3 (iii) i1 3
SOLUTION (i) Let .z i1 Then, | | .z 1 1 22 2
Let be the acute angle given by
·( )
( )
tan Re
Im
z
z
1
1 1 4&
Clearly, the point representing z i1 is ( , )P 1 1 , which lies in
the fi rst quadrant.
·( )arg z 4
Hence, ·| | ( )argandz z2 4
(ii) Let .z i3 Then, | | ( ) .z 3 1 4 22 2
Let be the acute angle given by
·( )
( )
tan Re
Im
z
z
3
1
3
1
6&
Now, ( )z i3 is represented by the point ( , )P 3 1 , which
lies in the second quadrant.
·( ) ( )arg z 6 6
5 a k
Hence, | |z 2 and ·( )arg z 6
5
(iii) Let .z i1 3 Then, | | ( ) ( ) .z 1 3 4 22 2
Let be the acute angle given by
·( )
( )
tan Re
Im
z
z
1
3
3 3&
Now, ( )z i1 3 is represented by the point ( , ),P 1 3
which lies in the third quadrant.
·( ) ( )arg z 3 3
2
a k
Hence, ·| | ( )argandz z2 3
2
EXAMPLE 3 Convert the complex number ( )i1 3 into polar form.
SOLUTION The given complex number is ( ) .z i1 3
Let its polar form be ( ) .cos sinz r i
Now, | | ( ) .r z 1 3 4 22 2
196 Senior Secondary School Mathematics for Class 11
Let be the acute angle, given by
·( )
( )
tan Re
Im
z
z
1
3
3 3&
Clearly, the point representing ( )z i1 3 is ( , ),P 1 3 which lies
in the fi rst quadrant.
·( )arg z 3
Thus, ·| | ( )argandr z z2 3
Hence, the required polar form of ( )z i1 3 is ·cos sini2 3 3
a k
EXAMPLE 4 Convert the complex number ( )i2 3 2 into polar form.
SOLUTION The given complex number is ( ) .z i2 3 2
Let its polar form be ( ) .cos sinz r i
Now, | | ( ) ( ) .r z 2 3 2 12 4 16 42 2
Let be the acute angle, given by
·( )
( )
tan Re
Im
z
z
2 3
2
3
1
6&
Clearly, the point representing ( )z i2 3 2 is ( , ),P 2 3 2 which
lies in the fourth quadrant.
·( )arg z 6
Thus, ·| | ( )argandr z z4 6
Hence, the required polar form of ( )z i2 3 2 is
·cos sin cos sini i4 6 6 4 6 6
a a ak k k; E
EXAMPLE 5 Convert the complex number ( )i2 2 3 into polar form.
SOLUTION The given complex number is ( ) .z i2 2 3
Let its polar form be ( ) .cos sinz r i
Now, | | ( ) ( ) .r z 2 2 3 4 12 16 42 2
Let be the acute angle, given by
·( )
( )
tan Re
Im
z
z
2
2 3
3 3&
Clearly, the point representing ( )z i2 2 3 is ( , ),P 2 2 3 which
lies in the second quadrant.
·( ) ( )arg z 3 3
2 a k
Thus, ·| | andr z 4 3
2
Hence, the required polar form is ·cos sinz i4 3
2
3
2 b l
Complex Numbers and Quadratic Equations 197
EXAMPLE 6 Convert
( )
( )
i
i
2
1 7
2
into polar form.
SOLUTION Let
( )
( )
( )
( )
( )
( )
( )
( )
z
i
i
i i
i
i
i
i
i
2
1 7
4 4
1 7
3 4
1 7
3 4
3 4
2 2 #
& ( )
( ) ( )
( ) .z
i i i i9 16
1 7 3 4
25
25 25 1
Let its polar form be ( ) .cos sinz r i
Now, | | ( ) .r z 1 1 22 2
Let be the acute angle, given by
·( )
( )
tan Re
Im
z
z
1
1 1 4&
Clearly, the point representing ( )z i1 is ( , ),P 1 1 which lies in
the second quadrant.
·( ) ( )arg z 4 4
3 a k
Thus, ·| | andr z 2 4
3
Hence, the required polar form is ·cos sinz i2 4
3
4
3 b l
EXAMPLE 7 Convert the complex number ( )i1
1
into polar form.
SOLUTION Let ·( ) ( )
( )
( )
( ) ( )
z
i i
i
i
i i
i1
1
1
1
1
1
2
1
2
1
2
1
2#
b l
Let its polar form be ( ) .cos sinz r i
Now, ·| |r z 2
1
2
1
4
1
4
1
2
1
2
12 2 b bl l
Let be the acute angle, given by
·( )
( )
( / )
( / )
tan Re
Im
z
z
1 2
1 2
1 4&
Clearly, the point representing z i2
1
2
1 b l is , ,P 2
1
2
1
b l which lies
in the fourth quadrant.
·( )arg z 4
Thus, ·| | andr z
2
1
4
Hence, the required polar form is
,cos sinz i
2
1
4 4
a ak k' 1 i.e., ·cos sini
2
1
4 4
a k
198 Senior Secondary School Mathematics for Class 11
EXAMPLE 8 Convert the complex number
( )
cos sini
i
3 3
1
a k
into polar form.
SOLUTION Let
( ) ( )
( )
( )
cos sin
z
i
i
i
i
i
i
3 3
1
2
1
2
3
1
1 3
2 1
a
e
k
o
&
( )
( )
( )
( ) ( ) ( )
z
i
i
i
i i i
1 3
2 1
1 3
1 3
2
1 1 3
#
·
( ) ( )
i2
3 1
2
3 1
) 3
Let its polar form be ( ) .cos sinz r i
Now, | |
( ) ( )
.r z r4
3 1
4
3 1
4
8 2 22 2
2 2
&
) 3
Let be the acute angle, given by
( )
( ) ( )
( ) ( )
( )
tan Re
Im
z
z
2
3 1
3 1
2
3 1
3 1
1
3
1
1
3
1
#
e
e
o
o
&
·
tan
tan tan
tan tan
tan tan
1 4 6
4 6
4 6 12
5
a
a
a
k
k
k
& ·12
5
Clearly, the point representing the given complex number is
, ,P 2
3 1
2
3 1
e o which lies in the fi rst quadrant.
·( )arg z 12
5
12
5
&
Thus, ·| | andr z 2 12
5
Hence, the required polar form is ·cos sinz i2 12
5
12
5 b l
EXAMPLE 9 Express the complex number ( )i1 in polar form.
SOLUTION The given complex number is .z i1
Let its polar form be ( ) .cos sinz r i
Now, | | ( ) ( ) .r z 1 1 22 2
Let be the acute angle, given by
·( )
( )
tan Re
Im
z
z
1
1 1 4&
Complex Numbers and Quadratic Equations 199
Clearly, the point representing the complex number z i1 is
( , ),P 1 1 which lies in the third quadrant.
·( ) ( )arg z 4 4
3
a k
Thus, | |r z 2 and ·4
3
Hence, the required polar form of ( )z i1 is given by
, ·. .,cos sin cos sini ez i i2 4
3
4
3 2 4
3
4
3 b b bl l l( 2
EXAMPLE 10 Express the complex number ( )i3 in polar form.
SOLUTION The given complex number is ( ) .z i3
Let its polar form be ( ) .cos sinz r i
Now, | | ( ) ( ) .r z 3 1 4 22 2
Let be the acute angle, given by
·( )
( )
tan Re
Im
z
z
3
1
3
1
6&
Clearly, the point representing the complex number ( )z i3 is
( , ),P 3 1 which lies in the third quadrant.
·( ) ( )arg z 6 6
5
a k
Thus, ·| | andr z 2 6
5
Hence, the polar form of ( )z i3 is given by
, ·. .,cos sin cos sini ez i i2 6
5
6
5 2 6
5
6
5 b b bl l l( 2
EXAMPLE 11 Convert the complex number i
i
1 3
1 2
into polar form.
SOLUTION Let ·( )
( )
( )
( )
( )
( ) ( )
z
i
i
i
i i i i i1 3
1 2
1 3
1 3
1 9
1 2 1 3
10
5 5
2
1
2
1
#
b l
Let its polar form be ( ) .cos sinz r i
Now, ·| |r z 2
1
2
1
2
1
2
12 2 b bl l
Let be the acute angle, given by
·( )
( )
( / )
( / )
tan Re
Im
z
z
1 2
1 2
1 4&
Clearly, the point representing z i2
1
2
1 b l is , ,P 2
1
2
1
b l which lies
in the second quadrant.
200 Senior Secondary School Mathematics for Class 11
·( ) ( )arg z 4 4
3 a k
Thus, ·| | andr z
2
1
4
3
Hence, the required polar form is ·cos sinz i
2
1
4
3
4
3 b l
EXAMPLE 12 Express the complex number sin cosi5 1 5
a k in polar form.
SOLUTION Let ·sin cosz i5 1 5
a k
Let its polar form be ( ) .cos sinz r i
Now, | | sin cos sin cos cosr z 5 1 5 5 5 1 2 5
2 2 2 2 2 2 a ak k
& ·cos sin sinr r2 1 5 4 10 2 10
2 2 &
a k
Let be the acute angle, given by
·
·
( )
( )
tan Re
Im
sin
cos
sin cos
sin
tan
z
z
5
1 5
2 10 10
2 10
10 10
2
&
Clearly, the point representing z lies in the fi rst quadrant as x 0
and .y 0
·( )arg z 10
Thus, ·sin andr 2 10 10
Hence, the required polar form is ·sin cos sini2 10 10 10
a k
EXAMPLE 13 Convert ( )cos sini4 300 300c c into Cartesian form.
SOLUTION ( ) [ ( ) ( )]cos sin cos sini i4 300 300 4 360 60 360 60 c c c c c c
[ ( ) ( )]cos sini4 60 60 c c
·[ ]cos sini i4 60 60 4 2
1
2
3
c c e o
( ) .i2 2 3
EXAMPLE 14 For any complex numbers ,z z1 and z2 prove that:
(i) ( ) ( )arg argz z
(ii) ( ) ( ) ( )arg arg argz z z z1 2 1 2
(iii) ( ) ( ) ( )arg arg argz z z z1 1 2 2
(iv) ( ) ( )arg arg argz
z
z z
2
1
1 2 d n
SOLUTION (i) Let ( ) .cos sinz r i Then, | |z r and ( ) .arg z
Now, ( )cos sinz r i
Complex Numbers and Quadratic Equations 201
& ( )cos sinz r i r
& ( ) ( )cos sin cos sinz r i r r i
{ ( ) ( )}cos sinr i
& | | ( ) ( ) .arg argandz r z z
Hence, ( ) ( ) .arg argz z
(ii) Let ( ) ( ) .cos sin cos sinandz r i z r i1 1 1 1 2 2 2 2 Then,
| | , ( ) | | , ( ) .arg argandz r z z r z1 1 1 2 2 2 2
( ) ( )cos sin cos sinz z r i r i·1 2 1 1 1 2 2 2
{( )cos cos sin sinr r1 2 1 2 1 2
( )}sin cos cos sini 1 2 1 2
{ ( ) ( )}cos sinr r i1 2 1 2 1 2
& ( ) ( ) ( ) ( ) .arg arg argz z z z1 2 1 2 1 2
REMARKS (I) Note here that | | | || |.z z r r z z1 2 1 2 1 2
(II) In general, we have
| … | | | | |…| |z z z z z z·n n1 2 1 2
and ( ) ( ) ( ) ( ) .arg arg arg argz z z z z z… …n n1 2 1 2
(iii) Let ( )cos sinz r i1 1 1 1 and ( ) .cos sinz r i2 2 2 2 Then,
z2 ( ) ( )cos sin cos sinr i r r i r2 2 2 2 2 2 2 2
& { ( ) ( )}cos sinz r i2 2 2 2 .
( ) · { ( ) ( )}cos sin cos sinz z r i r i1 1 1 1 2 2 2 2
( ) { ( ) ( )}cos sin cos sinr r i i1 2 1 1 2 2
[ { ( )} { ( )}]cos sinr r i1 2 1 2 1 2
{ ( ) ( )} .cos sinr r i1 2 1 2 1 2
Hence, ( ) ( ) ( ) ( ) .arg arg argz z z z1 1 2 1 2 2
(iv) Let ( ) ( ) .cos sin cos sinandz r i z r i1 1 1 1 2 2 2 2 Then,
| | ,| | , ( ) ( ) .arg argandz r z r z z1 1 2 2 1 1 2 2
( )
( )
( )
( )
cos sin
cos sin
cos sin
cos sin
z
z
r i
r i
i
i
2
1
2 2 2
1 1 1
2 2
2 2
#
·
( )
( )
( )
cos sin
cos cos sin sin
sin cos cos sin
r
r i
· ·
· ·
2
1
2
2
2
2
1 2 1 2
1 2 1 2
Z
[
\
]
]
]
]]
]
]
]
]]
_
`
a
b
b
b
bb
b
b
b
bb
· { ( ) ( )}cos sinr
r
i
2
1
1 2 1 2
& ( ) ( ) ( ) .arg arg argz
z
z z
2
1
1 2 1 2 d n
Hence, ( ) ( ) .arg arg argz
z
z z
2
1
1 2 d n
202 Senior Secondary School Mathematics for Class 11
EXERCISE 5D
Find the modulus and argument of each of the following complex numbers and
hence express each of them in polar form:
1. 4 2. 2 3. i 4. i2
5. i1 6. i1 7. i3 8. i1 3
9. i1 3 10. i2 2 11. i4 4 3 12. i3 2 3 2
13. i
i
1
1
14. i
i
1
1
15. i
i
1 2
1 3
16. i
i
1 2
1 3
17. i
i
2 3
5
18.
i1 3
16
19.
i
i
5 3
2 6 3
20. i
i
1
1
21. i3 22. ( )i25 3 23.
( )
cos sini
i
3 3
1
a k
24. ( )sin cosi120 120c c
ANSWERS (EXERCISE 5D)
1. , , ( )cos sini4 0 4 0 0 2. , , ( )cos sini2 2
3. ,, cos sini1 2 2 2
a ak k' 1 4. ,, cos sini2 2 2 2 2
a k
5. ,, cos sini2 4 2 4 4
a ak k' 1 6. ,, cos sini2 4
3 2 4
3
4
3 b l
7. ,, cos sini2 6 2 6 6
a k 8. ,, cos sini2 3
2 2 3
2
3
2 b l
9. ,, cos sini2 3 2 3 3
a ak k' 1 10. ,, cos sini2 2 4 2 2 4 4
a ak k' 1
11. ,, cos sini8 3
2 8 3
2
3
2 ' 1 12. ,, cos sini6 4
3 6 4
3
4
3 b l
13. ,, cos sini1 2 2 2
a k 14. ,, cos sini1 2 2 2
a ak k' 1
15. ,, cos sini2 4
3 2 4
3
4
3 b l 16. ,, cos sini2 4
3
4
3
4
3 b bl l( 2
17. ,, cos sini2 4 2 4 4
a k 18. ,, cos sini8 3
2 8 3
2
3
2 b l
19. ,, cos sini2 3 2 3 3
a k 20. ,, cos sini1 4 4 4
a k
21. ,, cos sini2 6
5 2 6
5
6
5 b bl l( 2 22. ,, cos sini1 2 2 2
a ak k' 1
23. ,, cos sini2 12
7 2 12
7
12
7 b l 24. ( )cos sini30 30c c
HINTS TO SOME SELECTED QUESTIONS
21. Let | | ( ) ( ) | | .z i r z r z3 3 1 4 22 2 2 2& &
·tan
3
1
3
1
6&
Complex Numbers and Quadratic Equations 203
The given number represents the point ( , )P 3 1 which lies in the third quadrant.
·( ) ( )arg z 6 6
5
a k
·cos sinz i2 6
5
6
5 c cm m< F
23. Given number
( )
( )
( )
( ) ( ) ( )
i
i
i
i
i
1 3
2 1
1 3
1 3
2
1 3
2
1 3
#
) 3
| |
( ) ( ) ( )
| | .z z4
1 3
4
1 3
4
2 1 3
2 22
2 2
&
·( )
( )
( )
( )
tan
tan tan
tan tan
2
1 3
2
1 3
3 1
3 1
1
3
1
1
3
1
1 4 6
4 6
e
e
o
o
·tan tan4 6 12
5 a k
The given number z is represented by the point , ·
( ) ( )
P 2
3 1
2
3 1
e o
So, it lies in the third quadrant.
·( ) ( )arg z 12
5
12
7
c m
Hence, ·cos sinz i2 12
7
12
7 c cm m( 2
QUADRATIC EQUATIONS (With Complex Roots)
FUNDAMENTAL THEOREM OF ALGEBRA
A polynomial equation of degree n has at the most n roots.
SOLVING A QUADRATIC EQUATION
Let the given equation be ,ax bx c 02 where , ,a b c Rd and .a 0!
Then, ,ax bx c a0 02 !
& ax bx c2
& ·x a
b x a
c2
& ·x a
b x a
b
a
c
a
b
2 2
2
2 2
b bl l [adding a
b
2
2
b l on both sides]
&
( )
x a
b
a
b ac
2 4
42
2
2
b l
& x a
b
a
b ac
2 2
42
!
204 Senior Secondary School Mathematics for Class 11
& x a
b
a
b ac
2 2
42
!
& ·x a
b b ac
2
42!
Hence, the roots of the equation ,ax bx c a0 02 ! are
a
b b ac
2
42
and ·a
b b ac
2
42
The expression ( )b ac42 is called the discriminant.
If ( ) ,b ac4 02 the given quadratic equation will have complex roots.
Here, we shall consider only the quadratic equations having complex roots.
SOLVED EXAMPLES
EXAMPLE 1 Solve: .x 3 02
SOLUTION We have
.x x x i3 0 3 3 32 2& & ! !
solution set { , } .i i3 3
EXAMPLE 2 Solve: .x x3 9 02
SOLUTION The given equation is .x x3 9 02
This is of the form ,ax bx c 02 where , .anda b c1 3 9
( ) ( ) () .b ac4 3 4 1 9 9 36 27 02 2 # #
So, the given equation has complex roots.
These roots are given by
a
b b ac
2
4
2 1
3 272!
#
!
[ ]b ac4 272a
·
i i
2
3 27
2
3 3 3! !
solution set ,
i i
2
3 3 3
2
3 3 3
) 3
·,i i2
3
2
3 3
2
3
2
3 3
) 3
EXAMPLE 3 Solve: .x x9 10 3 02
SOLUTION The given equation is .x x9 10 3 02
This is of the form ,ax bx c 02 where ,a b9 10 and .c 3
( ) {( ) } ( ) .b ac4 10 4 9 3 100 108 8 02 2 # #
So, the given equation has complex roots.
Complex Numbers and Quadratic Equations 205
These roots are given by
a
b b ac
2
4
2 9
10 82!
#
!
[ ( ) ]b ac4 82a
·
i i
18
10 2 2
9
5 2! !
solution set , ·,
i i
i i9
5 2
9
5 2
9
5
9
2
9
5
9
2
) )3 3
EXAMPLE 4 Solve: .x x2 2 02
SOLUTION The given equation is .x x2 2 02
This is of the form ,ax bx c 02 where , .anda b c2 1 2
( ) ( ) ( ) .b ac4 1 4 2 2 1 8 7 02 2 # #
So, the given equation has complex roots.
These roots are given by
a
b b ac
2
4
2 2
1 72! !
[ ]b ac4 72a
·
i
2 2
1 7!
solution set ,
i i
2 2
1 7
2 2
1 7
Z
[
\
]
]]
]
]]
_
`
a
b
bb
b
bb
·,i i
2 2
1
2 2
7
2 2
1
2 2
7
Z
[
\
]
]]
]
]]
_
`
a
b
bb
b
bb
EXAMPLE 5 Solve: .x x3 2 3 3 02
SOLUTION The given equation is
.x x3 2 3 3 02
This is of the form ,ax bx c 02 where ,a b3 2 and .c 3 3
( ) {( ) } ( ) .b ac4 2 4 3 3 3 2 36 34 02 2 # #
So, the given equation has complex roots.
These roots are given by
a
b b ac
2
4
2 3
2 342!
#
!
[ ]b ac4 342a
·
i
2 3
2 34!
solution set ,
i i
2 3
2 34
2 3
2 34
Z
[
\
]
]]
]
]]
_
`
a
b
bb
b
bb
·,i i
6
1
2 3
34
6
1
2 3
34
Z
[
\
]
]]
]
]]
_
`
a
b
bb
b
bb
206 Senior Secondary School Mathematics for Class 11
EXAMPLE 6 Solve: .x ix3 8 3 02
SOLUTION The given equation is
.x ix3 8 3 02
This is of the form ,ax bx c 02 where , .anda b i c3 8 3
( ) {( ) } ( ) .b ac i4 8 4 3 3 64 36 100 02 2 # #
So, the given equation has complex roots.
These roots are given by
a
b b ac i
2
4
2 3
8 1002!
#
!
[ ]b ac4 1002a
·i i i i6
8 10
3
4 5! !
Thus, the roots of the given equation are
.andi i i i i i i3
4 5
3 3
4 5
3
9 3
solution set , ·i i3 3
' 1
EXERCISE 5E
Solve:
1. x 2 02 2. x 5 02 3. x2 1 02
4. x x 1 02 5. x x 2 02 6. x x2 2 02
7. x x2 4 3 02 8. x x3 5 02 9. x x5 5 02
10. x x25 30 11 02 11. x x8 2 1 02 12. x x27 10 1 02
13. x x2 3 1 02 14. x x17 8 1 02 15. x x3 5 72
16. x x3 4 3
20 02 17. x ix3 7 6 02 18. x x21 28 10 02
19. x x13 42 20. x ix3 10 02 21. x ix2 3 2 02
ANSWERS (EXERCISE 5E)
1. { , }i i2 2 2. { , }i i5 5 3. ,i i
2 2
) 3
4. ,i i2
1
2
3
2
1
2
3 ) 3 5. ,i i2
1
2
7
2
1
2
7
) 3 6. { , }i i1 1
7. ,i i1
2
1 1
2
1 ) 3 8. ,i i2
3
2
11
2
3
2
11 ) 3
9. ,i i
2 5
1
2 5
19
2 5
1
2 5
19
Z
[
\
]
]]
]
]]
_
`
a
b
bb
b
bb
10. ,i i5
3
5
2
5
3
5
2
) 3
Complex Numbers and Quadratic Equations 207
11. ,i i8
1
8
7
8
1
8
7 ) 3 12. ,i i27
5
27
2
27
5
27
2 ) 3
13. ,i i4
3
4
5
4
3
4
5
) 3 14. ,i i17
4
17
1
17
4
17
1 ' 1
15. ,i i6
7
6
11
6
7
6
11
) 3 16. ,i i3
2
3
4
3
2
3
4 ' 1
17. ,i i3
2 3' 1 18. ,i i3
2
21
14
3
2
21
14
) 3
19. { , }i i2 3 2 3 20. { , }i i2 5 21. ,i i2
1 2' 1
HINTS TO SOME SELECTED QUESTIONS
17. , .anda b i c3 7 6
( ) {( ) } ( ) ( ) .b ac i4 7 4 3 6 49 72 121 02 2 # #
So, the given equation has complex roots, given by
·a
b b ac i i i
2
4
2 3
7 121
6
7 11
2!
#
! !
Solution set , ·,i i i i i i6
7 11
6
7 11
3
2 3
' '1 1
SQUARE ROOTS OF A COMPLEX NUMBER
To Evaluate a ib :
SOLUTION Let ( ) .a ib x iy … (i)
On squaring both sides of (i), we get
( ) ( )a ib x iy 2
& ( ) ( ) ( ) .a ib x y i xy22 2 … (ii)
On comparing real parts and imaginary parts on both sides of (ii),
we get
( ) andx y a xy b22 2
& ( ) ( )x y x y x y a b42 2 2 2 2 2 2 2 2 .
( )x y a2 2 … (iii) and ( )x y a b2 2 2 2 … (iv)
On solving (iii) and (iv), we get
·andx a b a y a b a2
1
2
12 2 2 2 2 2 $ $. .
Hence, ·a ib a b a i a b a2
1
2
12 2 2 2! < F$ $. .
REMARK Similarly, by assuming that ( ),a ib x iy we may fi nd .a ib
208 Senior Secondary School Mathematics for Class 11
SOLVED EXAMPLES
EXAMPLE 1 Evaluate .i6 8
SOLUTION Let ( ) .i x iy6 8 … (i)
On squaring both sides of (i), we get
( ) ( ) ( ) .i x iy i x y i xy6 8 6 8 22 2 2& … (ii)
On comparing real parts and imaginary parts on both sides of (ii),
we get
andx y xy6 2 82 2
& andx y xy6 42 2
& ( ) ( )x y x y x y4 6 4 16 100 102 2 2 2 2 2 2 2 #
& andx y x y6 102 2 2 2
& andx y2 16 2 42 2
& andx y8 22 2
& .andx y2 2 2! !
Since ,xy 0 so x and y are of the same sign.
( ) ( ) .and or andx y x y2 2 2 2 2 2
Hence, ( ) ( ) .ori i i6 8 2 2 2 2 2 2
EXAMPLE 2 Evaluate .i5 12
SOLUTION Let ( ) .i x iy5 12 … (i)
On squaring both sides of (i), we get
( ) ( ) ( ) .i x iy i x y i xy5 12 5 12 22 2 2& … (ii)
On comparing real parts and imaginary parts on both sides of (ii),
we get
x y xy5 2 12and2 2
& andx y xy5 62 2
& ( ) ( ) ( )x y x y x y4 5 4 36 169 132 2 2 2 2 2 2 2 #
& andx y x y5 132 2 2 2
& andx y2 8 2 182 2
& andx y4 92 2
& .andx y2 3! !
Since ,xy 0 so x and y are of the same sign.
( ) ( )and or andx y x y2 3 2 3 .
Hence, ( ) ( ) .ori i i5 12 2 3 2 3
Complex Numbers and Quadratic Equations 209
EXAMPLE 3 Evaluate .i8 15
SOLUTION Let ( ) .i x iy8 15 … (i)
On squaring both sides of (i), we get
( ) ( ) ( ) ( ) ( ) .i x iy i x y i xy8 15 8 15 22 2 2& … (ii)
On comparing real parts and imaginary parts on both sides of (ii),
we get
andx y xy8 2 152 2
& andx y xy8 2
152 2
& ( ) ( )x y x y x y4 64 225 289 172 2 2 2 2 2 2
& andx y x y8 172 2 2 2
& andx y2 25 2 92 2
& andx y2
25
2
92 2
& ·andx y
2
5
2
3
! !
Since ,xy 0 so x and y are of the same sign.
Hence, ·ori i i8 15
2
5
2
3
2
5
2
3 e eo o
EXAMPLE 4 Evaluate .i24 10
SOLUTION Let ( ) .i x iy24 10 … (i)
On squaring both sides of (i), we get
( ) ( ) ( ) .i x y i xy24 10 22 2 … (ii)
On comparing real parts and imaginary parts on both sides of (ii),
we get
andx y xy24 2 102 2
& andx y xy24 52 2
& ( ) ( ) ( )x y x y x y4 24 4 25 676 262 2 2 2 2 2 2 2 #
& andx y x y24 262 2 2 2
& andx y2 2 2 502 2
& and andx y x y1 25 1 52 2 & ! ! .
Since ,xy 0 so x and y are of the same sign.
( ) ( ) .ori i i24 10 1 5 1 5
EXAMPLE 5 Evaluate .i
SOLUTION Let ( ) .i x iy … (i)
On squaring both sides of (i), we get
( ) ( ) ( ) ( )i x iy i x y i xy22 2 2& . … (ii)
210 Senior Secondary School Mathematics for Class 11
On comparing real parts and imaginary parts on both sides of (ii),
we get
( ) .andx y xy0 2 12 2
( ) ( ) .x y x y x y4 0 1 0 1 1 12 2 2 2 2 2 2 2 2
Thus, ( )x y 02 2 … (iii) and ( )x y 12 2 … (iv).
On solving (iii) and (iv), we get ·andx y2
1
2
12 2
·andx y
2
1
2
1
! !
Since ,xy 0 so x and y are of the same sign.
, , ·orxy x y
2
1
2
1
2
1
2
1 e eo o
Hence, · · ·ori i i
2
1
2
1
2
1
2
1 e eo o
EXAMPLE 6 Evaluate .4 3 20 4 3 20
SOLUTION .andi i4 3 20 4 6 5 4 3 20 4 6 5
Let ( )i x iy4 6 5 . … (i)
On squaring both sides of (i), we get
( ) ( ) ( ) ( ) ( )i x iy i x y i xy4 6 5 4 6 5 22 2 2& . … (ii)
On comparing real parts and imaginary parts on both sides of (ii),
we get
( ) .andx y xy4 2 6 52 2
( ) ( ) ( ) ( )x y x y x y4 4 6 52 2 2 2 2 2 2 2 2
.16 180 196 14
Thus, ( )x y 42 2 … (iii) and ( )x y 142 2 … (iv).
On solving (iii) and (iv), we get .andx y9 52 2
.andx y3 5! !
Since ,xy 0 so x and y are of the same sign.
( , ) ( , ) .orx y x y3 5 3 5
( ) ( ) .ori i i4 3 20 4 6 5 3 5 3 5
Similarly, ( ) ( )ori i i4 3 20 4 6 5 3 5 3 5 .
or
{( ) ( )}
{ } { } .
i i
i i
4 3 20 4 3 20
3 5 3 5 6
3 5 3 5 6
Z
[
\
]
]
]
]]
]
]
]
]]
Hence, .or4 3 20 4 3 20 6 6
Complex Numbers and Quadratic Equations 211
EXERCISE 5F
Evaluate:
1. i5 12 2. i7 24 3. i2 2 3
4. 1 4 3 5. i 6. i4
7. 3 4 7 8. i16 30 9. i4 3
10. i15 8 11. i11 60 12. 7 30 2
13. i8 14. i1
ANSWERS (EXERCISE 5F)
1. ( ) ( )ori i3 2 3 2 2. ( ) ( )ori i3 4 3 4
3. ( ) ( )ori i1 3 1 3 4. ( ) ( )ori i2 3 2 3
5. ( ) ( )ori i
2
1 1
2
1 1 6. ( ) ( )ori i2 1 2 1
7. ( ) ( )ori i7 2 7 2 8. ( ) ( )ori i5 3 5 3
9. ( ) ( )ori i
2
1 1 3
2
1 1 3 10. ( ) ( )ori i1 4 1 4
11. ( ) ( )ori i5 6 5 6 12. ( ) ( )ori i5 3 2 5 3 2
13. ( ), ( )i i2 2 2 2
14. ori i2
2 1
2
2 1
2
2 1
2
2 1
) ) ) )3 3 3 3
HINTS TO SOME SELECTED QUESTIONS
14. Let .i x iy1 … (i)
On squaring both sides, we get
( ) ( ) ( ) ( )i x iy x y i xy1 22 2 2
( ) andx y xy1 2 12 2
& ( ) ( )x y x y x y4 1 1 22 2 2 2 2 2 2 2 2
& andx y x y1 22 2 2 2
& andx y2
2 1
2
2 12 2
& ·andx y2
2 1
2
2 1
! !
Since ,xy 0 so x and y have the same sign.
, , ·orx y x y2
2 1
2
2 1
2
2 1
2
2 1
) )3 3
·ori i i1 2
2 1
2
2 1
2
2 1
2
2 1
) )3 3
212 Senior Secondary School Mathematics for Class 11
EXERCISE 5G
Very-Short-Answer Questions
1. Evaluate ·
i
1
78
2. Evaluate ( ) .i i i i i57 70 91 101 104
3. Evaluate ·
i i i i i
i i i i i
170 168 166 164 162
180 178 176 174 172
J
L
K
K
K
KK
N
P
O
O
O
OO
4. Evaluate ( ) .i in n4 1 4 1
5. Evaluate ( ) .36 25#
6. Find the sum ( ),i i i in n n n1 2 3 where .n Nd
7. Find the sum ( … ).up to termsi i i i 4002 3 4
8. Evaluate ( ) .i i i1 10 20 30
9. Evaluate ·i
i
141
71
d n
10. Find the least positive integer n for which .i
i
1
1 1
n
b l
11. Express ( )i2 3 3 in the form ( ) .m a ib
12. Express
( ) ( )
( ) ( )
i i
i i
3 2 3 2
3 5 3 5
in the form ( ) .a ib
13. Express
1 9
3 16
J
L
K
K
K
KK
N
P
O
O
O
OO
in the form ( ) .a ib
14. Solve for x: ( ) ( ) .i x i y i1 1 1 3
15. Solve for x: .x ix5 6 02
16. Find the conjugate of ·( )i3 4
1
17. If ( ),z i1 fi nd .z 1
18. If ( ),z i5 3 fi nd .z 1
19. Prove that ( ) ( ) .arg argz z 0
20. If | |z 6 and ( ) ,arg z 4
3 fi nd z.
21. Find the principal argument of ( ) .i2
22. Write the principal argument of ( ) .i1 3 2
23. Write 9 in polar form.
24. Write i2 in polar form.
25. Write i3 in polar form.
26. Write ( )z i1 in polar form.
Complex Numbers and Quadratic Equations 213
27. Write ( )z i1 3 in polar form.
28. If | |z 2 and ,( )arg z 4
fi nd z.
ANSWERS (EXERCISE 5G)
1. 1 2. i 3. 1 4. i2 5. 30 6. 0 7. 0 8. 0 9. i2
10. n 4 11. i46 9 12. ,a b0
2
7 13. i2
3
2
1b l
14. ,x y2 1 15. orx i x i3 2 16. i25
3
25
4b l 17. i2
1
2
1b l
18. i14
5
14
3e o 20. ( )z i3 2 1 21. 2
22. 3
2
23. ( )cos sini9 24. cos sini2 2 2
a k
25. cos sini3 2 2
a ak k' 1 26. cos sini2 4 4
a ak k' 1
27. cos sini2 3
2
3
2 b l 28. ( )z i2 1
HINTS TO SOME SELECTED QUESTIONS
1. .
i i i
i1 1 178 78 2
2
# [ ( ) ]i i 180 4 20a
2. Given expression ( )i i i i i i i i i48 3 68 2 88 3 100 104# # # #
( ) ( ) .i i i i i i i i1 1 13 2 3
3. Given expression
( )
( )
.
i i i i i
i i i i i
i i i i
1
1
1( )162 8 6 4 2
172 8 6 4 2
172 162 10 8 2#
4. Given expression ( )
( )
.i i i i i i i
i
i
i i i1
1
1
2 2n n4 4 1
2
# # #
c m
5. ( ) ( ) .i i i36 25 6 5 30 302# #
6. Given sum ( ) ( ) .i i i i i i i1 1 1 0n n2 3#
7. Given sum ( ) ( )i i i i i i i i ……2 3 4 5 6 7 8
( ) ( ) .i i i i i i i1 1 0…2 3 5 2 3 [ ( ) ]i i i1 02 3a
8. Given sum ( ) .i i1 1 02 2
9. ( ) .i
i
i
i
i
i i
i i i i1 1 1 241 71 3 3 # e e eo o o
10.
( )
( )
( )
( )
( )
( ) ( )
.i
i
i
i
i
i
i
i i i i i1
1
1
1
1
1
1
1
2
1 2
2
2
2
2 2
#
d n
.i
i i i i n1
1 1 1 4
n
n n 4
& & &
d n
214 Senior Secondary School Mathematics for Class 11
11. ( ) ( ) ( ) ( )i i i i2 3 2 3 3 2 3 2 33 3 3 # #
( ) ( ) ( ) .i i i i i8 27 36 54 8 27 36 54 46 93
12. Given expression ·
( )
,
i
i
i i
i i a b
2 2
9 5
2 2
14
2
7 0
2
7
2
&#
13. Given expression
( )
( ) ( )
( ) .i
i
i
i i i i i1 3
3 4
1 3
1 3
1 9
3 4 1 3
10
15 5
2
1 3#
14. ( ) ( ) ( ) ( )i x i y i x y y x i i1 1 1 3 1 3+
, .andx y y x x y1 3 2 1&
15. x ix x ix ix5 6 0 3 2 6 02 2& ( ) ( )x x i i x i3 2 3 0&
( ) ( ) .orx i x i x i x i3 2 0 3 2& &
16. ·
( ) ( )
( )
z
i i
i i i z i3 4
1
3 4
3 4
25
3 4
25
3
25
4
25
3
25
4
&#
c cm m
17. ·
| | ( )
( )
z
z
z i i
1 1
1
2
1
2
11
2 2 2
c m
18. ·
| | ( )
( )
z
z
z i i5 9
5 3
14
5
14
31
2
e o
20. cos sin cos sinz i i6 4
3
4
3 6 4 4
c a am k k' 1
& ( )cos sinz i i i6 4 4 6 2
1
2
1
2
6 1·
a ek o
& ( ) .z i3 2 1
22. ( ) ( ) .z i i1 3 2 2 32
REVIEW OF FACTS AND FORMULAE
1. (i) Imaginary Numbers A given number is said to be imaginary, if its
square is negative, e.g., , , , ,1 2 3 4 etc.
(ii) We denote 1 by the Greek letter iota ‘’ transliterated as ‘i ’.
(iii) , , ,i i i i i i1 10 1 2 3 and .i 14
(iv) Let n be a positive integer. On dividing n by 4, let m be the quotient
and r be the remainder.
Then, ,n m r4 where .r0 4#
( ) .i i i i i·n m r m r r4 4 [ ]i 14a
(v) ( ) ( ) .i i i16 9 4 3 12 122# #
2. Complex Numbers
(i) The numbers of the form ( )z a ib are called complex numbers,
where , .a b Rd Here, ( )Re z a and ( ) .Im z b
Complex Numbers and Quadratic Equations 215
If b 0 then z is purely real.
If a 0 then z is purely imaginary.
(ii) Let ( ) ( )andz a ib z a ib1 1 1 2 2 2 then
I. ( ) ( ) ( )z z a a i b b1 2 1 2 1 2
II. ( ) ( ) ( )z z a a i b b1 2 1 2 1 2
III. ( ) ( ) .z z a a b b i a b a b1 2 1 2 1 2 1 2 2 1
(iii) The conjugate of ( )z a ib is z ( ) ( ) .a ib a ib
(iv) The modulus of ( )z a ib is | | .z a b2 2
(v) If ( )z a ib1 1 1 and ( )z a ib2 2 2 then .andz z a a b b1 2 1 2 1 2+
(vi) If ( )z a ib then ·
| | ( )
( )
z
z
z
a b
a ib1
2 2 2
3. (i) z z z z1 2 1 2 (ii) z z z z1 2 1 2(iii) z z z z1 2 1 2
(iv) z
z
2
1
d n
z
z
2
1 (v) ( ) ( )Re Imandz z z z z z2 2
(vi) | |zz z 2
4. Polar form of ( )z x iy is ( ) .cos sinz r i
Here, | |r x y z2 2 and the value of such that # is called the
principal argument or amplitude of z.
5. When z is purely real; then, it lies on the x-axis.
(i) If z lies on positive side of the x-axis then .0
(ii) If z lies on negative side of the x-axis then .
6. When z is purely imaginary; then, it lies on the y-axis.
(i) If z lies on the y-axis and above the x-axis then ·2
(ii) If z lies on the y-axis and below the x-axis then ·2
7. When ( )z x iy does not lie on any axes:
Step 1. Find for which ·( )
( )
tan Re
Im
z
z
Step 2. Find the quadrant in which ( , )P x y lies.
(i) When z lies in quadrant I;
then, ( ) .arg z&
(ii) When z lies in quadrant II;
then, ( ) .arg z&
(iii) When z lies in quadrant III;
then, ( ) ( ) .arg z&
(iv) When z lies in quadrant IV;
then, ( ) .arg z&
216 Senior Secondary School Mathematics for Class 11
8. (i) Fundamental theorem of algebra
A polynomial equation of degree n has at most n roots.
(ii) Let ,ax bx c 02 where , ,a b c Rd and .a 0!
Its roots are given by
a
b b ac
2
42
and ·a
b b ac
2
42
Here, ( )D b ac42 is called the discriminant.
If ,D 0 the given quadratic equation has complex roots.
9. ( ),a ib x iy! where ·andx a b a y a b a2
1
2
12 2 2 2 2 2 $ $. .
Similarly, by assuming that ( ),a ib x iy! we may fi nd .a ib
Linear Inequations (In one variable) 217
6
Linear Inequations
(In one variable)
Linear Inequations in One Variable
Inequalities of the form:
(i) ax b c (ii) ax b c# (iii) ax b c (iv) ax b c$ ,
where a, b, c are real numbers, a 0! and x is a variable, are called inequations in x.
Thus, each of the following is a linear inequation:
(i) x2 3 7 (ii) x3 5 10# (iii) x3 2 8 (iv) x4 3 9$
Replacement Set or Domain of the Variable
A set given to us from which the values of x are replaced in an inequation in x,
is called the replacement set.
SOLUTION SET The set of all those values of x taken from the replacement set
which satisfy the given inequation is called the solution set of the inequation.
EXAMPLE Write down the solution set of the inequation ,x 6 when the replacement
set is (i) N, (ii) W, (iii) Z.
SOLUTION (i) Solution set { : } { , , , , } .x N x 6 1 2 3 4 5 d
(ii) Solution set { : } { , , , , , } .x W x 6 0 1 2 3 4 5 d
(iii) Solution set { : } { , , , , , , , , , } .x Z x 6 5 4 3 2 1 0 1 2 3 … d
Rules for Solving an Inequation
RULE 1 Adding the same number or expression to each side of an inequation
does not change the inequality.
RULE 2 Subtracting the same number or expression from each side of an
inequation does not change the inequality.
RULE 3 Multiplying (or dividing) each side of an inequation by the same
positive number does not change the inequality.
RULE 4 Multiplying (or dividing) each side of an inequation by the same
negative number reverses the inequality.
Thus, x x3 3 & [on multiplying both sides by –1].
And, x x3 12 4&# $ [on dividing both sides by –3].
TRANSPOSITION Using Rule 1 and Rule 2, we can drop any term from one side
of an inequation and put it on the other side with the opposite sign.
This process is called transposition.
217
218 Senior Secondary School Mathematics for Class 11
Thus, .x x x3 2 2 3 1 & &
And, ·x x x x2 3 6 2 6 3 2 9 2
9
& & &$ $ $ $
SOLVED EXAMPLES
EXAMPLE 1 Solve x5 24 when (i) ,x Nd (ii) .x Zd
In each case, represent the solution set on the number line.
SOLUTION x x5 24 5
24 &
. .x 4 8&
(i) Solution set { : . }x N x 4 8 d
{ , , , } .1 2 3 4
On the number line, we may represent it as shown below.
The darkened circles indicate the natural numbers contained
in the set.
(ii) Solution set { : . }x Z x 4 8 d
{ , , , , , , , , }4 3 2 1 0 1 2 3 …
{ , , , , , , , , } .3 2 1 0 1 2 3 4…
On the number line, we may represent it as shown below.
The darkened circles show the integers contained in the set
and three dark dots above the left part of the line show that the
negative integers are continued idefi nitely.
EXAMPLE 2 Solve x x12 1 6
5 5 3# when (i) ,x Nd (ii) .x Rd
Draw the graph of the solution set in each case.
SOLUTION x x12 1 6
5 5 3#
& x x12 6
11 5 3#
& x x72 11 30 18# [multiplying both sides by 6]
& x x11 18 42# [adding 72 to both sides]
& x7 42# [adding x18 to both sides]
& x 6$ [dividing both sides by ] .7
(i) Solution set { : }x N x 6$ d
{ , , , , } .6 7 8 9 …
The graph of this set is the number line, shown below.
1 2 3 4 5 60–1–2–3
1 2 3 4 5 60–1–2–3–4
1 2 3 4 5 60–1–2 7 8
Linear Inequations (In one variable) 219
The darkened circles indicate the natural numbers contained
in the set. Three dots above the right part of the line show that
the natural numbers are continued indefi nitely.
(ii) Solution set { : } [ , [.x R x 6 6 3$ d
The graph of this set is shown below.
This graph consists of 6 and all real numbers greater than 6.
EXAMPLE 3 Solve x
x
2
4 0
and draw the graph of the solution set.
SOLUTION Note that
b
a 0 when ( )anda b0 0 or ( ) .anda b0 0
either ( )andx x4 0 2 0 or ( ) .andx x4 0 2 0
Case I When .x and x4 0 2 0
In this case, andx x4 0 2 0
& andx x4 2
& x 2
& ( , ) .x 2 3d
Case II When .x and x4 0 2 0
In this case, andx x4 0 2 0
& andx x4 2
& x 4
& ( , ) .x 43 d
solution set ( , ) ( , ) .4 2,3 3
The graph of the solution set is given below.
EXAMPLE 4 Solve x
x
4
3 0
and draw the graph of the solution set.
SOLUTION Note that
b
a 0 when ( )anda b0 0 or ( ) .anda b0 0
either ( )andx x3 0 4 0 or ( ) .andx x3 0 4 0
Case I When .x and x3 0 4 0
In this case, andx x3 0 4 0
& .andx x3 4
This is not possible, as we can never fi nd a real number
which is greater than 3 and less than .4
Case II When .x and x3 0 4 0
In this case, andx x3 0 4 0
& andx x3 4
& ( , ) .x x4 3 4 3 & d
1 2 3 4 5 60–1–2 7 8
–1–2 43210–3–4
220 Senior Secondary School Mathematics for Class 11
solution set ( , ) .4 3
The graph of the solution set is given below.
EXAMPLE 5 Solve .x
x
5
2 2
SOLUTION We have
x
x
5
2 2
& x
x
5
2 2 0
& x
x x
5
2 2 10 0
& ( )
( )
x
x
5
12
0
& x
x
5
12 0
[on multiplying both sides by ]1 .
either ( ) ( ) .and or andx x x x12 0 5 0 12 0 5 0
Case I When .x and x12 0 5 0
In this case, andx x12 0 5 0
& andx x12 5
& ,x5 12 which is impossible [ ]12 5a .
Case II When .x and x12 0 5 0
In this case, andx x12 0 5 0
& andx x12 5
& ( , ) .x x12 5 12 5 & d
solution set ( , ) .12 5
EXAMPLE 6 Solve x 2
5 3 and represent the solution set on the number line.
SOLUTION x x2
5 3 2
5 3 0 & [adding 3 to both sides]
x
x
2
5 3 6 0&
.x
x
2
11 3 0&
either ( )andx x11 3 0 2 0 or ( ) .andx x11 3 0 2 0
Case I When .x and x11 3 0 2 0
Now, andx x11 3 0 2 0
& andx x3 11 2
& andx x3
11 2
& ·x2 3
11 … (i)
–1–2 43210–3–4 5
Linear Inequations (In one variable) 221
Case II When .x and x11 3 0 2 0
Now, andx x11 3 0 2 0
& andx x3 11 2
& .andx x3
11 2
This is not possible, as we cannot fi nd a real number which is
greater than 3
11 and less than 2.
solution set ·: ,x R x2 3
11 2 3
11 d b l' 1
We can represent this set on the number line, as given below.
SOME USEFUL RESULTS
Let abe a positive real number. Then,
(i) | | ( , ) .x a a x a x a a + + d
(ii) | | [ , ] .x a a x a x a a+ +# # # d
(iii) | | .orx a x a x a +
(iv) | | .orx a x a x a+$ # $
EXAMPLE 7 Solve | | .x4 2
SOLUTION We have, | | .x a a x a +
| |x4 2
+ x2 4 2
+ andx x2 4 4 2
+ andx x2 4 2 4
+ andx x6 2
+ andx x6 2
+ .x2 6
solution set { : } ( , ) .x R x2 6 2 6 d
EXAMPLE 8 Solve ,| | .x x R3 2 2
1
# d
SOLUTION We have, | | .x a a x a+# # #
| |x x3 2 2
1
2
1 3 2 2
1
+# # #
andx x2
1 3 2 3 2 2
1
+ # #
andx x2
1 2 3 3 2
1 2+ # #
21 430 11
3
222 Senior Secondary School Mathematics for Class 11
andx x2
3 3 3 2
5
+ # #
andx x2
1
6
5
+ # #
·x2
1
6
5
# #
solution set , ·:x R x2
1
6
5
2
1
6
5
# # d ; E' 1
EXAMPLE 9 Solve | | , .x x R1 4 d
SOLUTION We have, | | .orx a x a x a +
| | orx x x1 4 1 4 1 4 +
orx x4 1 4 1 +
orx x5 3 +
( , ) ( , ) .orx x5 3+ 3 3 d d
solution set ( , ) ( , ) .5 3,3 3
EXAMPLE 10 Solve | | , .x x R3 4 9$ d
SOLUTION We have, | | .orx a x a x a+$ # $
| | orx x x3 4 9 3 4 9 3 4 9+$ # $
orx x4 9 3 4 9 3+ # $
orx x4 12 4 6+ # $
orx x3 2
3
+ $ #
orx x2
3 3+ # $
, [ , ) .orx x2
3 3+ 3 3
d db E
solution set , [ , ) .2
3 3,3 3
b E
EXAMPLE 11 Solve
| |
, .
x
x R
3
2 5 d
SOLUTION Clearly, x 3 0! and therefore, .x 3!
We have,
| |
.
x 3
2 5 … (i)
Since | |x 3 is positive, we may multiply both sides of (i) by | |.x 3
This gives
| |x2 5 3
+ | |x5
2 3
+ | |x 3 5
2
+ x5
2 3 5
2
[ | | ]x a a x a +a
Linear Inequations (In one variable) 223
+ andx x5
2 3 3 5
2
+ andx x5
2 3 5
2 3
+ ·andx x5
13
5
17
+ ·x5
13
5
17
Also, as shown above, .x 3!
solution set : { }x R x5
13
5
17 3 d' 1
( . , . ) { } ( . , ) ( , . ) .2 6 3 4 3 2 6 3 3 3 4,
EXAMPLE 12 Solve
| |
| |
, .
x
x
x R
2 2
2 1
0#
d
SOLUTION We have either .orx x2 0 2 0$
Case I When .x 2 0$
In this case, | | .andx x x2 2 2$
So, the given inequation becomes
.x
x
x
x
2 2
2 1 0 4
3 0&# #
( ) ( )and or andx x x x3 0 4 0 3 0 4 0 $ #
& ( ) ( )and or andx x x x3 4 3 4 $
& x3 4# [ andx x3 4 a is not possible]
& [ , )x 3 4d [this includes ]x 2$ .
Case II When .x 2 0
In this case, | | ( ) .andx x x x2 2 2 2
So, the given inequation becomes
x
x
x
x
2 2
2 1 0 1 0&# #
x
x 1 0& #
[multiplying num. and denom. by –1]
( ) ( )and or andx x x x1 0 0 1 0 0 # $
& ( ) ( )and or andx x x x1 0 1 0 # $
& x0 1 # [ andx x1 0a $ is not possible]
& ( , ]x 0 1d [this includes ]x 2 .
Hence, from the above two cases, we get
solution set ( , ] [ , ) .0 1 3 4,
EXAMPLE 13 Solve ( )
| |
, .
x
x x
x R2
3
1
d
224 Senior Secondary School Mathematics for Class 11
SOLUTION We have
( )
| |
x
x x
2
3
1
& ( )
| |
x
x x
2
3
1 0
& ( )
| |
x
x x x
2
3 2
0
& ( )
| |
x
x
2
3 2
0
. … (i)
Now, we have either .orx x3 0 3 0$
Case I When .x 3 0$
In this case, | | .andx x x3 3 3$
(i) becomes
.x
x
x
x
2
3 2 0 2
1 0 &
( ) ( )and or andx x x x1 0 2 0 1 0 2 0
& ( ) ( )and or andx x x x1 2 1 2
& ( ) ( )orx x1 2
& ( ) ( )orx x2 1
& ( , ) ( , )orx x2 13 3 d d
& [ , ) ( , )orx x3 2 1 3 d d [ ]x 3a $
& [ , ) ( , ) .x 3 2 1, 3 d
Case II When .x 3 0
In this case, | | ( ) .andx x x x3 3 3 3
(i) becomes
( )
( )
x
x
x
x
x
x
2
3 2 0 2
5
0 2
5 0 & &
.
( ) ( )and or andx x x x5 0 2 0 5 0 2 0
& ( ) ( )and or andx x x x5 2 5 2
& x5 2 [ andx x5 2 a is not possible]
& x5 3 [ ]x 3a
& ( , ) .x 5 3 d
solution set ( , ) [ , ) ( , )5 3 3 2 1, , 3
( , ) ( , ) .5 2 1, 3
EXAMPLE 14 Solve | | | | , .x x x R1 2 4$ d
SOLUTION Putting ,andx x1 0 2 0 we get andx x1 2 as the critical
points. These points divide the whole real line into three parts,
namely ( , ), [ , ) [ , ) .and1 1 2 23 3 So, we consider the following
three cases.
Linear Inequations (In one variable) 225
Case I When .x 1 3
In this case, .andx x1 0 2 0
| | ( ) | | ( ) .andx x x x x x1 1 1 2 2 2
Now, | | | |x x1 2 4$
& x x1 2 4$
& x x x x2 3 4 2 4 3 2 1 2
1
& & &$ $ $ #
& ·,x 2
1
3
db E
But, .x 1 3
solution set in this case ·, ( , ) ,2
1 1 2
1
+3 3 3
b bE E
Case II When .x1 2#
In this case, .andx x1 0 2 0$
| | | | ( ) .andx x x x x1 1 2 2 2
Now, | | | |x x1 2 4$
& ,x x1 2 4 1 4&$ $ which is absurd.
So, the given inequation has no solution in [1, 2).
Case III When .x2 3#
In this case, .andx x2 0 1 0$
| | | | .andx x x x2 2 1 1
Now, | | | |x x1 2 4$
& ·x x x x x1 2 4 2 3 4 2 7 2
7
& & &$ $ $ $
Also, in this case, we have .x 2$
solution set in this case , ·, [ , )2
7 2 2
7
+3 3 3 l l; ;
Hence, from all the above cases, we have
solution set , ·, 2
1
2
7
,3 3
b l;E
SOLVING SIMULTANEOUS INEQUATIONS IN ONE VARIABLE
METHOD Suppose we have to solve two inequations simultaneously. Find the
solution set of each of them. Then, the intersection of these solution
sets is the required solution set.
EXAMPLE 15 Solve the inequations , .x x R3 3 2 9# d Represent the solution set
on the real line.
SOLUTION We have
.andx x x3 3 2 9 3 3 2 3 2 9 &# #
Now, x x3 3 2 3 2 3&# # [adding 2x on both sides]
x2 6& # [adding 3 on both sides]
x 3& # [dividing both sides by 2]
( , ] .x 3& 3d
226 Senior Secondary School Mathematics for Class 11
And, x x3 2 9 2 6 & [subtracting 3 from both sides]
x 3& [dividing both sides by 2 ]
( , ) .x 3& 3d
solution set ( , ] ( , ) ( , ] .3 3 3 3+3 3
We may represent it on the number line as shown below.
EXAMPLE 16 Solve the inequations , .x x x x R2 3 2 3 5 # d Draw the graph of
the solution set.
SOLUTION We have
andx x x x x x x2 3 2 3 5 2 3 2 2 3 5 &# # .
Now, x x x2 3 2 3 2 & [adding x on both sides]
x 5& [adding 3 on both sides]
( , ) .x 5& 3d
And, x x x x2 3 5 3 5 2&# $ [ ]a b b a&a # $
x2 5 2& $ [adding x on both sides]
x2 3& $ [adding 5 on both sides]
x 2
3
& $
[dividing both sides by 2]
, ·x 2
3
& 3
d l;
solution set , , ·( , )5 2
3
2
3 5+3 3
l l; ;
We may represent it on the number line, as shown below.
EXAMPLE 17 Solve the system of inequations , .x x x x R2 1 3
7 2
d
Represent the solution set on the number line.
SOLUTION We have
.andx x x x x x x x2 1 3
7 2 2 1 3
7
3
7 2 &
Now, x x x x x x2 1 3
7 6 3 3 7 &
[multiplying both sides by 3]
x x6 3 2 7&
x4 3 7& [adding x2 on both sides]
x4 10& [adding 3 on both sides]
x 2
5& [dividing both sides by 4]
, ·x 2
5
& 3db l
–1–2 3210–3–4
–1–2 3210 4 5–3
2
6
Linear Inequations (In one variable) 227
And, x x x x3
7 2 3 7 6 &
[multiplying both sides by 3]
x2 7 6&
x2 1& [adding –7 on both sides]
x 2
1&
[dividing both sides by 2]
, ·x 2
1
& 3
db l
solution set , , , ·2
5
2
1
2
5
+3 3 3
b b bl l l
The solution set on the number line may be represented as shown
below.
EXAMPLE 18 Solve the system of inequations .x5 4
2 3 9# #
SOLUTION We have.andx x x5 4
2 3 9 5 4
2 3
4
2 3 9&# # # #
Now, x x5 4
2 3 20 2 3&# #
[multiplying both sides by 4]
x22 3& # [adding 2 on both sides]
x22 3& $ [multiplying both sides by 1 ]
x3 22& #
x 3
22
& # [dividing both sides by 3]
·,x 3
22
& 3db E
And, x x4
2 3 9 2 3 36&# #
[multiplying both sides by 4]
x3 34& #
x 3
34
& $
[dividing both sides by 3 ]
, ·x 3
34
& 3
d l;
solution set , , ·, 3
22
3
34
3
34
3
22
+3 3
b l; ;E E
EXAMPLE 19 Solve the system of inequations:
, ·x
x
x
x
4 1
6
2
1
2 1 4
1 $
SOLUTION The fi rst inequation of the given system is
x
x
4 1
6
2
1
–1 3210 45
2
228 Senior Secondary School Mathematics for Class 11
& x
x
4 1
6
2
1 0
& ( )x
x x
2 4 1
12 4 1 0
& ( )x
x
x
x
2 4 1
8 1 0 4 1
8 1 0 &
. … (i)
( ) ( )and or andx x x x8 1 0 4 1 0 8 1 0 4 1 0
& ( ) ( )and or andx x x x8 1 4 1 8 1 4 1
& and or andx x x x8
1
4
1
8
1
4
1
b bl l
& x8
1
4
1
and is not possiblex x8
1
4
1 a
; E
& , ·x 8
1
4
1
db l
The second inequation of the given system is
x
x
2 1 4
1
$
& x
x
2 1 4
1 0$
& ( )x
x x
x
x
4 2 1
4 2 1 0 2 1
2 1 0&$ $
. … (ii)
( ) ( )and or andx x x x2 1 0 2 1 0 2 1 0 2 1 0 # $
& ( ) ( )and or andx x x x2 1 2 1 2 1 2 1 # $
& and or andx x x x2
1
2
1
2
1
2
1 # $
b bl l
& orx x2
1
2
1 $
b bl l
& ,, orx x2
1
2
1
3 3
d db l l;
& ,,x 2
1
2
1
,3 3
db l l;
solution set , ,, .8
1
4
1
2
1
2
1
+ ,3 3
b bl l l;( 2
Hence, the given system of inequations has no solution.
EXAMPLE 20 Solve the system of inequations: | | , | | .x x1 5 2# $
SOLUTION The fi rst inequation is | | .x 1 5#
Using | | ,x a a x a+# # # we get
| |x x1 5 5 1 5&# # #
[ , ] .x x4 6 4 6& &# # d
The second inequation is | | .x 2$
Using | | ,orx a x a x a+$ # $ we get
Linear Inequations (In one variable) 229
| | orx x x2 2 2&$ # $
( , ] [ , ) .x 2 2& ,3 3 d
solution set for the given system is
{( , ] [ , )} [ , ] [ , ] [ , ] .2 2 4 6 4 2 2 6, + ,3 3
WORD PROBLEMS
EXAMPLE 1 Find all pairs of consecutive odd positive integers both of which are
smaller than 10 such that their sum is more than 11.
SOLUTION Let the required consecutive odd positive integers be x and .x 2
Then,
( )andx x x2 10 2 11
& andx x8 2 2 11
& andx x8 2 9
& andx x8 2
9
& . .x x2
9 8 4 5 8 &
x can take the odd integral values 5 and 7.
Hence, the required pairs of odd integers are (5, 7) and (7, 9).
EXAMPLE 2 Find all pairs of consecutive even positive integers both of which are larger
than 5 such that their sum is less than 23.
SOLUTION Let the required consecutive even positive integers be x and .x 2
Then,
( )andx x x5 2 23
& andx x5 2 21
& .andx x5 10 5
& .x5 10 5 .
x can take the even integral values 6, 8 and 10.
Hence, the required pairs of even integers are (6, 8), (8, 10) and
(10, 12).
EXAMPLE 3 The longest side of a triangle is 3 times the shortest side and the third side
is 2 cm shorter than the longest side. If the perimeter of the triangle is at
least 61 cm, fi nd the minimum length of the shortest side.
SOLUTION Let the minimum length of the shortest side be x cm.
Then, the longest side x3 cm and the third side ( )x3 2 cm.
( ) .x x x x x3 3 2 61 7 63 9& &$ $ $
Hence, the minimum length of the shortest side is 9 cm.
–2 2
–4 6
230 Senior Secondary School Mathematics for Class 11
EXAMPLE 4 The cost and revenue functions of a product are given by ( )C x x20 4000
and ( )R x x60 2000 respectively, where x is the number of items
produced and sold. How many items must be sold to realise some profi t?
SOLUTION Let the required number of items sold be x and let ( )P x be the profi t
function. Then,
profi t function = (revenue function) – (cost function)
& ( ) ( ) ( )P x x x60 2000 20 4000
& ( ) .P x x40 2000
For some profi t, we have ( ) .P x 0
x40 2000 0
& x40 2000
& .x 50
Hence, the manufacturer must sell more than 50 items to realise
some profi t.
EXAMPLE 5 A man wants to cut three lengths from a single piece of cloth of length
91 cm. The second length is to be 3 cm longer than the shortest and the
third length is to be twice as long as the shortest. What are the possible
lengths of the shortest piece of cloth if the third piece is to be at least 5 cm
longer than the second.
SOLUTION Let the length of the shortest piece be x cm. Then, second length
( )x 3 cm and third length .cmx2
( ) ( )andx x x x x3 2 91 2 3 5# $
& andx x x4 3 91 2 8# $
& andx x x4 91 3 2 8# $
& andx x4 88 8# $
& andx x22 8# $
& .x8 22# #
Hence, the length of the shortest piece is to be greater than or equal
to 8 but less than or equal to 22.
EXAMPLE 6 In drilling world’s deepest hole it was found that the temperature
( )T x in degree Celsius, x km below the earth’s surface was given by
( ) ( ),T x x30 25 3 .x3 15# # At what depth will the temperature be
between 155C and 205C?
SOLUTION We have,
( ) ( )T x x30 25 3
& ( )T x x30 25 75
& ( ) .T x x25 45 … (i)
Let x km below the earth’s surface, the temperature lies between
155C and 205C. Then,
( )T x155 205
& x155 25 45 205
& andx x155 25 45 25 45 205
Linear Inequations (In one variable) 231
& andx x155 45 25 25 205 45
& andx x200 25 25 250
& andx x8 10
& .x8 10
Clearly, x8 10 lies in the range .x3 15# #
Hence, the required depth is more than 8 km and less than 10 km.
EXAMPLE 7 A manufacturer has 460 litres of a 9% acid solution. How many litres
of a 3% acid solution must be added to it so that the acid content in the
resulting mixture be more than 5% but less than 7%?
SOLUTION Let x litres of a 3% acid solution be added to 460 litres of 9% acid
solution. Then,
total quantity of mixture ( )x460 litres.
Total acid content in ( )x460 litres of mixture
x460 100
9
100
3
# # b bl l( 2 litres x5
207
100
3 b l litres.
Now, the acid content in the resulting mixture must be more than
5% and less than 7%.
5% of ( ) %x x460 5
207
100
3 7 b l of ( )x460
& ( ) ( )x x x100
5 460 100
4140 3
100
7 460 # #
& ( ) ( )x x x5 460 4140 3 7 460
& x x x2300 5 4140 3 3220 7
& andx x x x2300 5 4140 3 4140 3 3220 7
& andx x x x5 3 4140 2300 4140 3220 7 3
& andx x2 1840 920 4
& andx x920 230
& .x230 920
Hence, the required quantity of 3% acid solution to be added must
be more than 230 litres and less than 920 litres.
EXAMPLE 8 A solution is to be kept between 40C and 45C. What is the range
of temperature in degree Fahrenheit, if the conversion formula is
?F C5
9 32
SOLUTION ( )F C C F5
9 32 9
5 32& . … (i)
C40 45
& ( )F40 9
5 32 45
& ( ) ( )andF F40 9
5 32 9
5 32 45
232 Senior Secondary School Mathematics for Class 11
& andF F40 5
9 32 32 45 5
9 # #
& andF F72 32 32 81
& andF F72 32 81 32
& andF F104 113
& .F104 113
Hence, the solution is to be kept between 104F and 113F.
EXAMPLE 9 The IQ of a person is given by the formula, ,IQ c
m 100# where m is the
mental age and c is the chronological age. If IQ80 140# # for a group of
12-year children, fi nd the range of their mental age.
SOLUTION When ,c 12 we have IQ m m12 100 3
25
# $ a k
IQ m80 140 80 3
25 140&# # # #
andm m80 3
25
3
25 140& # #
andm m25
3 80 140 25
3
& # ## #
andm m5
48
5
84
& # #
. . .m9 6 16 8& # #
Hence, the required mental age for a group of 12-year children is
9.6 years or more and 16.8 years or less.
EXERCISE 6A
1. Fill in the blanks with correct inequality sign ( , , , ) . $ #
(i) x x5 20 4…… &
(ii) x x3 9 3…… &
(iii) x x4 16 4…… &
(iv) x x6 18 3……&#
(v) x x3 2 6…… &
(vi) anda b c c
a
c
b0 …… &
(vii) p q p q3 ……&
(viii) u v u v2 ……&
Solve each of the following inequations and represent the solution set on the
number line.
2. ,x6 25# where (i) ,x Nd (ii) x Zd .
3. ,x2 5 where (i) ,x Zd (ii) x Rd .
4. ,x3 8 2 where (i) ,x Zd (ii) x Rd .
5. ,x5 2 17 where (i) ,x Zd (ii) x Rd .
Linear Inequations (In one variable) 233
6. ,x x3 4 6 where x Rd .
7. ,x x3 2 4 9$ where x Rd .
8. ,x x3
5 8
2
4 7
$
where x Rd .
9. ,x x4
5
3
4 1 1
where x Rd .
10. ( ),x x4
1
3
2 1 3
1 2$ b l where x Rd .
11. ,x x x12
2 1
3
1
4
3 1
where x Rd .
12. ,
( ) ( )x x x
4 3
5 2
5
7 3
where x Rd .
13. ,
( ) ( ) ( )x x x
3
2 1
4
3 2
5
2
$
where x Rd .
Solve:
14. ,x
x x R1
3 0
d 15. ,x
x x R4
3 0
d
16. ,x
x x R3 7
2 3 0
d 17. ,x
x x R2
7 0$
d
18. ,x x R2
3 2 d 19. ,x x R1
1 2# d
20. ,x
x x R4
5 8 2
d 21. | | ,x x R3 7 4 d
22. | | ,x x R5 2 3# d 23. ,| |x x R4 5 3
1
# d
24. ,
| |x
x R
3
1
2
1
# d 25.
| |
,x
x x
x R
2
2
d
26. ,x
x x R1
2 1 2
d 27.
| |
,x
x
x R3
3
0
d
28.
| |
| |
, { , }
x
x
x R
2
1
0 2 2$
d 29. | |
, { , }
x
x R
2
1 1 2 2$
d
30. | | | | ,x x x R1 3 d 31. ,x x4
2 1 4 !
Solve the following systems of linear inequations:
32. ,x x x1
4 3 1
6 0# # 33. x11 4 3 13# #
34. ( ),x x x x5 7 3 3 1 2
3 4 $ 35. x2 4
6 5 7
36. x x x3 2 3
4 3
37. ,x x2
7 1 3 5
3 8 11 0
38. x12 4 5
3 2 # 39. | |x1 2 3# #
234 Senior Secondary School Mathematics for Class 11
40. Find all pairs of consecutive odd positive integers, both of which are
smaller than 18 such that their sum is more than 20.
41. Find all pairs of consecutive even positive integers both of which are larger
than 8 such that their sum is less than 25.
42. A company manufactures cassettes. Its cost and revenue functions are
( )C x x26000 30 and ( )R x x43 respectively, where x is the number of
cassettes produced and sold in a week. How many cassettes must be sold
by the company to realise some profi t?
43. The water acidity in a pool is considered normal when the average pH
reading of three daily measurements is between 8.2 and 8.5. If the fi rst two
pH readings are 8.48 and 8.35, fi nd the range of the pH values for the third
reading that will result in the acidity level being normal.
44. A manufacturer has 640 litres of a 8% solution of boric acid. How many
litres of a 2% boric acid solution be added to it so that the boric acid content
in the resulting mixture will be more than 4% but less than 6%?
45. How many litres of water will have to be added to 600 litres of the 45%
solution of acid so that the resulting mixture will contain more than 25%,
but less than 30% acid content?
46. To receive grade A in a course one must obtain an average of 90 marks or
more in fi ve papers, each of 100 marks. If Tanvy scored 89, 93, 95 and 91
marks in fi rst four papers, fi nd the minimum marks that she must score in
the last paper to get grade A in the course.
ANSWERS (EXERCISE 6A)
1. (i) < (ii) < (iii) > (iv) $ (v) < (vi) > (vii) < (viii) >
2. (i) {1, 2, 3, 4}
–1–2 3210 4
(ii) {…, –3, –2, –1, 0, 1, 2, 3, 4}
3. (i) {–3, –4, –5, –6, …}
–5–6 0–4 –3 –2 –1
(ii) ( , . )2 53
0–3 –2 –1
2.5
4. (i) {–1, 0, 1, 2, 3, 4, …}
–1 3210 4
(ii) ( , )2 3
0–2 –1
Linear Inequations (In one variable) 235
5. (i) {2, 1, 0, –1, –2, …}
1 20–2 –1
(ii) ( , )33
0 321
6. ( , )5 3
1 20 43 5
7. ( , ]23
0 21
8. , 2
5
3b E
5
2
9. ( , )83
–8 –7 –6 –5 –4 –3 –2 –1 0
10. ( , . ]5 53
11. ( , )0 3
0
12. ( , )4 3
43210
13. ( , ]23
210
14. ( , )1 3 15. ( , ) ( , )4 3,3 3 16. ,, 2
3
3
7
,3 3b bl l
17. ( , ) [ , )2 7,3 3 18. ,2 2
7
b l 19. ,( , )1 2
3
,3 3 l;
20. ( , ) ( , )0 4,3 3 21. ,( , )1 3
11
,3 3 b l 22. [1, 4]
23. ,6
7
3
4
; E 24. ( , ] ( , ) [ , )5 3 3 5, ,3 3
25. ( , ) ( , )2 1,3 3 26. , ( , )4
3 1 1, 3b l 27. ( , )3 3
28. [ , ] ( , ) ( , )1 1 2 2, ,3 3 29. ( , ] [ , )2 1 1 2,
30. ( , ) ( , )2 1,3 3 31. ( , ) ( , )2 4 4 6, 32. x3
1 1# #
33. x2 4# # 34. x 23 # 35. . .x4 4 2 8
236 Senior Secondary School Mathematics for Class 11
36. x2
5 3 37. x 21 3 38. x3
80
3
10 #
39. [ , ] [ , ]1 1 3 5, 40. (11, 13), (13, 15), (15, 17)
41. (10, 12) 42. More than 2000 43. Between 7.77 and 8.67
44. More than 320 litres and less than 1280 litres
45. More than 300 litres but less than 480 litres 46. 82 marks
HINTS TO SOME SELECTED QUESTIONS
1. (ii) .x x3 9 3 & (iii) .x x4 16 4 &
(iv) x x6 18 3&# $ . (v) x x3 2 6 & .
(vi) ·anda b c c
a
c
b0 & (vii) p q p q3 & .
3. . .x x x2 5 2
5 2 5 & &
(i) All integers less than .2 5 are , , , , … .3 4 5 6
(ii) Set of all real numbers less than .2 5 are ( , . ) .2 53
4. .x x3 6 2 &
(i) All integers greater than 2 are , , , , , , .1 0 1 2 3 4 …
(ii) Set of all real numbers greater than 2 is ( , ) .2 3
7. .x x x x3 2 4 9 6 12 2& &$ $ #
solution set ( , ] .23
8. ·x x x x10 16 12 21 2 5 2
5
& &$ $ #
solution set ·, 2
5
3 c E
9. ( ) .x x x x x x15 4 4 1 12 15 16 4 12 8 8 & & &
solution set ( , ) .83
10. · ·
( )x x x x x x4
1
3
2 3
3
2 2 3 4 8 2 11 2
11
& & &$ $ $ #
solution set ( , . ] .5 53
11. ( ) ( ) ( ) .x x x x x x x2 1 4 1 3 3 1 2 3 9 3 11 0 0 & & &
solution set ( , ) .0 3
12. ( ) ( )x x x x x x15 20 5 2 12 7 3 15 100 40 84 36 &
.x x x x15 16 4 4 4 & & &
solution set ( , ) .4 3
13. ( ) ( ) ( )x x x x x x20 2 1 15 3 2 12 2 40 20 45 30 24 12&$ $
.x x17 34 2& &$ #
solution set ( , ] .23
14. ( ) ( )and or andx x x x3 0 1 0 3 0 1 0
& ( ) ( )and or andx x x x3 1 3 1
& ( , ) .x x1 3 1 3 & d [ andx x3 1$ is not possible]
Linear Inequations (In one variable) 237
15. ( ) ( )and or andx x x x3 0 4 0 3 0 4 0
& ( ) ( )and or andx x x x3 4 3 4
& ( ) ( ) ( , ) ( , ) .orx x x4 3 4 3 & ,3 3 d
16. ( ) ( )and or andx x x x2 3 0 3 7 0 2 3 0 3 7 0
& and or andx x x x2
3
3
7
2
3
3
7 c cm m
& , ·,orx x x2
3
3
7
2
3
3
7 & ,3 3dc c c cm m m m
17. ( ) ( )and or andx x x x7 0 2 0 7 0 2 0 # $
& ( ) ( )and or andx x x x7 2 7 2 # $
& ( ) ( ) ( , ) [ , ) .orx x x2 7 2 7 & ,3 3$ d
18. x x
x
x
x
2
3 2 0 2
3 2 4 0 2
7 2 0 & &
& ( ) ( )and or andx x x x7 2 0 2 0 7 2 0 2 0
& ( ) ( )and or andx x x x7 2 2 7 2 2
& and or andx x x x2
7 2 2
7 2 c cm m
& ·,x x2 2
7 2 2
7 & dc m [ andx x2
7 2 cannot hold]
19. x x
x
x
x
1
1 2 0 1
1 2 2 0 1
3 2 0& &# # #
& ( ) ( )and or andx x x x3 2 0 1 0 3 2 0 1 0 # $
& ( ) ( )and or andx x x x3 2 1 2 3 1 # $
& and or andx x x x2
3 1 2
3 1 $ #c cm m
& , ·( ) ( , )orx x x1 2
3 1 2
3 ,3 3$ dc m m;
20. x
x
x
x x
x
x
4
5 8 2 0 4
5 8 8 2 0 4
7 0 & &
& ( ) ( )and or andx x x x7 0 4 0 7 0 4 0
& ( ) ( )and or andx x x x0 4 0 4
& ( ) ( )and or andx x x x0 4 0 4
& ( ) ( ) ( , ) ( , ) .orx x x0 4 0 4 & ,3 3d
21. Using | | ,orx a x a x a & we get
| | orx x x3 7 4 3 7 4 3 7 4 &
or orx x x x3 3 3 11 1 3
11 & &
, ·( , )x 1 3
11
& ,3 3d c m
22. Using | | ,x a a x a+# # # we get
| |x x5 2 3 3 5 2 3&# # #
andx x3 5 2 5 2 3& # #
and andx x x x2 8 2 2 4 1& &# $ # $
[ , ] .x x1 4 1 4& &# # d
238 Senior Secondary School Mathematics for Class 11
23. Using | | ,x a aMais conteúdos dessa disciplina
saneamento questionario 3- Resoluções Exercícios Aula 4
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