Std. XI Sci.
Perfect Mathematics - II
Edition: July 2014
Mr. Vinodkumar J. Pandey
Dr. Sidheshwar S. Bellale
B.Sc. (Mathematics)
M.Sc., B.Ed., PhD. (Maths)
Department of Mathematics,
G. N. Khalsa College, Mumbai
Dayanand Science College, Latur. Mr. Vinod Singh
M.Sc. (Mathematics)
Mumbai University Published by
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Written according to the New Text book (2012-2013) published by the Maharashtra State
Board of Secondary and Higher Secondary Education, Pune.
Std. XI Sci.
Perfect Mathematics - II
Salient Features :
• Exhaustive coverage of entire syllabus.
• Covers answers to all textual and miscellaneous exercises.
• Precise theory for every topic.
• Neat, labelled and authentic diagrams.
• Written in systematic manner.
• Self evaluative in nature.
• Practice problems and multiple choice questions for effective
preparation.
TEID : 752
PREFACE
In the case of good books, the point is not how many of them you can get through, but rather how many can get
through to you.
“Std. XI Sci. : PERFECT MATHEMATICS - II” is a complete and thorough guide critically analysed and
extensively drafted to boost the students confidence. The book is prepared as per the Maharashtra State board syllabus
and provides answers to all textual questions. At the beginning of every chapter, topic – wise distribution of all
textual questions including practice problems has been provided for simpler understanding of different types of
questions. Neatly labelled diagrams have been provided wherever required.
Practice Problems and Multiple Choice Questions help the students to test their range of preparation and the
amount of knowledge of each topic. Important theories and formulae are the highlights of this book. The steps are
written in systematic manner for easy and effective understanding.
The journey to create a complete book is strewn with triumphs, failures and near misses. If you think we’ve
nearly missed something or want to applaud us for our triumphs, we’d love to hear from you.
Please write to us on : mail@targetpublications.org
Best of luck to all the aspirants!
Yours faithfully,
Publisher
No.
Topic Name
Page No.
1
Sets, Relations and Functions
1
2
Logarithms
39
3
Complex Numbers
67
4
Sequence and Series
119
5
Permutations and Combinations
172
6
Method of Induction and Binomial Theorem
213
7
Limits
250
8
Differentiation
302
9
Integration
329
10
Statistics (Measures of Dispersion)
345
11
Probability
389
Log Tables
Logarithms
439
Antilogarithms
441
Target Publications Pvt. Ltd.
01
Chapter 01: Sets, Relations and Functions
Sets, relations and functions
Type of Problems
Exercise
1.1
To describe sets in Roster form
Practice Problems
(Based on Exercise 1.1)
1.1
To describe sets in Set-Builder
form
Q.11 (i. to iv.)
Miscellaneous
Q.1 (i., ii., iii.)
Q.3 to Q.11
Q.13 (i. to iv.)
Q.3 to Q.10
(Based on Exercise 1.1)
Q.12 (i., ii.)
Practice Problems
(Based on Miscellaneous)
1.2
Practice Problems
(Based on Exercise 1.2)
1.2
Practice Problems
(Based on Exercise 1.2)
Miscellaneous
Practice Problems
(Based on Miscellaneous)
1.2
Practice Problems
(Based on Exercise 1.2)
Miscellaneous
Practice Problems
(Based on Miscellaneous)
1
Q.1(i., ii.)
Practice Problems
Miscellaneous
To find domain and range of a
given relation
Q.2 (i., ii., iii.), Q.12(i. to iv.)
(Based on Exercise 1.1)
1.1
Cartesian product of two Sets
Q.1 (i., ii., iii.)
Q.2 (i., ii., iii.)
(Based on Miscellaneous)
Ordered Pairs
Q.1 (i., ii., iii.)
Practice Problems
Practice Problems
Operations on Sets
Q. Nos.
Q.2, 3, 4
Q.2, 3, 4
Q.1, 2, 6, 11
Q.1, 2, 5, 9
Q.3, 4, 5
Q.3, 4
Q.5 (i., ii.)
Q.5
Q.7, 8, 9, 10
Q.6, 7, 8
Q.6, 7
Q.6, 7
1
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Std. XI Sci.: Perfect Maths - II
1.3
Practice Problems
Types of Functions
(Based on Exercise 1.3)
Miscellaneous
Practice Problems
(Based on Miscellaneous)
1.3
Practice Problems
To find values of the given
function
(Based on Exercise 1.3)
Miscellaneous
Practice Problems
(Based on Miscellaneous)
1.3
Operations on functions
Practice Problems
(Based on Exercise 1.3)
1.3
Practice Problems
Composite function
(Based on Exercise 1.3)
Miscellaneous
Practice Problems
(Based on Miscellaneous)
Inverse function
Q.9, 10, 22
Q.9, 10, 21
Q.3, 5, 6, 7
Q.3, 4, 6, 7
Q.13 to Q.17
Q.12 to Q.16
Q.11 (i. to iii.)
Q.11 (i. to iii.)
Q.12 to Q.15
Q.12 to Q.15
Q.18 to Q.21
Q.17 to Q.20
Q.16
Practice Problems
Q.16
(Based on Exercise 1.3)
Miscellaneous
(Based on Miscellaneous)
1.3
Practice Problems
(Based on Exercise 1.3)
Miscellaneous
Practice Problems
(Based on Miscellaneous)
2
Q.9, 10
1.3
Practice Problems
To find domain and range of a
given function
Q.9, 10
Q.11, 12
Q.11
Q.1, 2, 4, 8
Q.1, 2, 5, 8
Q.8, 23, 24
Q.8
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Syllabus:
Sets, Subset, Intervals, Types of sets, Power set,
Ordered pair, Cartesian products of two sets,
Definition of Relation, Domain, Co-domain and
Range of Relation, Types of Relation, Definition of
function, Types of functions, Operations on
functions, Composite function, Inverse function,
Binary operation, Real valued functions of the real
variable.
Introduction
All basic concepts of modern mathematics are based
on set theory. The concepts involving logic can be
explained more easily with the help of set theory.
It plays a crucial role in the study of relations,
functions, probability and is used extensively in
various other branches of mathematics. We shall
briefly revise and study some more concepts about
sets.
Sets
A set is a well-defined collection of objects. These
objects may be actually listed or may be specified by
a rule. A set is usually denoted by the capital letters
A, B, C, N, R, etc. Each object in a set is called an
element or a member of the set and is denoted by the
small letters a, b, c, etc.
If x is an element of set A, then we write it as x ∈ A
and read it as ‘x belongs to A’ and if y is not an
element of set A, then we write it as y ∉ A and read
it as ‘y does not belong to A’.
Example:
If A = {2, 4, 6, 8},
then 4 ∈ A, 7 ∉ A, 8 ∈ A, 10 ∉ A
The set of natural numbers, whole numbers, integers,
rational numbers and real numbers are denoted by N,
W, I, Q and R respectively.
Methods of Representation of Sets
There are two methods of representing a set which
are as follows:
i.
Roster method (Listing method):
In this method all the elements are listed or
tabulated. The elements are separated by
commas and are enclosed within two
braces(curly brackets).
3
Chapter 01: Sets, Relations and Functions
Example:
The set A of all positive even integers less
than 9 can be written as A = {2, 4, 6, 8}.
ii.
Set-Builder method:
In this method, the set is described by the
characteristic property of its elements.
In general, if all the elements of set A satisfy
some property P, then write A in set-builder
notation as A = {x/x has property P} and read
it as ‘A is the set of all x such that x has the
property P’.
Example:
Let B = {3, 4, 5, 6, 7, 8}
Using the set-builder method, B can be written
as B = {x/x ∈ N, 3 ≤ x ≤ 8}
Since B = {3, 4, 5, 6, 7, 8} can also be stated
as the set of natural numbers from 3 to 8
including 3 and 8.
Some standard sets are as follows:
N = set of all natural numbers
= {1, 2, 3,…..}
Z or I = set of all integers
= { ….−3, −2, −1, 0, 1, 2, 3 …}
Q = set of all rational numbers
⎧p
⎫
= ⎨ / p,q ∈ Z,q ≠ 0 ⎬
⎩q
⎭
Subset:
Set A is called a subset of set B, if every element of
set A is also an element of set B
i.e., if x ∈ A, then x ∈ B.
We denote this relation as A ⊆ B and read it as
‘A is a subset for B’. It s clear that
i.
Every set is a subset of itself i.e., A ⊆ A.
ii.
An empty set φ is a subset of every set.
Example:
If A = {2, 4, 6, 8} and B = {2, 4, 6, 8, 10, 12},
then A ⊆ B.
If A ⊆ B, then B is called a superset of A, denoted
by B ⊇ A.
Proper subset:
If every element of set A is an element of set B and
B contains at least one element which is not in A,
then A is said to be a proper subset of B and it is
denoted as A ⊂ B.
3
Target Publications Pvt. Ltd.
Std. XI Sci.: Perfect Maths - II
Example:
If A = {2, 3, 5, 6}, B = {1, 2, 3, 4, 5, 6, 7}
Here every element of set A i.e., 2, 3, 5, 6 is an
element of set B. But B contains elements 1, 4, 7
which are not in A. Hence in the case we say that A
is a proper subset of B and is denoted by A ⊂ B.
Set of all real numbers ≥ p
i.e., [p, ∞) = {x/x∈R, x ≥ p}
ii.
Intervals
Set of real numbers < q
i.e., (−∞, q) {x/x∈R, x < q}
q
Open interval
If p, q ∈ R and p < q, then the set {x/x∈R, p < x < q}
is called open interval and is denoted by (p, q). Here
all the numbers between p and q ∈ (p, q) except p and
q.
R
q
x
p
∴
R
p
R
Set of real numbers ≤ q
i.e., (−∞, q] = {x/x∈R, x ≤ q}
q
iii.
R
Set of all real numbers R is (−∞, ∞)
R
R = (−∞, ∞) = {x/x ∈ R, − ∞ < x < ∞}
(p, q) = {x/x∈R, p < x < q}
Closed interval
Types of sets
If p, q ∈ R and p < q, then the set {x/x∈R, p ≤ x ≤ q}
is called closed interval and is denoted by [p,q].
Here all the numbers between p and q ∈ [p, q]
including p, q.
Empty set:
A set which does not contain any element is called
an empty set and it is denoted by φ or { }. It is also
called null set or void set.
Example:
A = {x/x ∈ N, 3 < x < 4}
B = {x/x is a positive integer < 1}
p
∴
R
q
x
[p, q] = {x/x∈R, p ≤ x ≤ q}
Semi-closed interval
If p, q ∈ R and p < q, then the set {x/x∈R, p ≤ x < q}
is called semi-closed interval and is denoted by [p, q).
p
∴
R
q
x
[p, q) = {x/x∈R, p ≤ x < q}
[p, q) includes p but excludes q.
Semi-open interval
If p, q ∈ R and p < q, then the set {x/x∈R, p < x ≤ q}
is called semi-open interval and is denoted by (p, q].
p
∴
x
q
R
Remarks:
i.
Set of all real numbers > p
i.e., (p, ∞) = [x/x∈R, x > p}
4
Singleton set:
A set which contains only one element is called a
singleton set.
Example:
A = {5},
B = {3},
X = {x/x ∈ N, 1 < x < 3}
The set A = set of all integers which are neither
positive nor negative is a singleton set since A = {0}
Finite set:
A set which contains countable number of elements
is called a finite set.
(p, q] = {x/x ∈ R, p < x ≤ q}
(p, q] excludes p but includes q.
p
Note:
The set {0} and {φ} are not empty sets as they
contain one element, namely 0 and φ respectively.
R
Example:
A = {a, b, c}
B = {1, 2, 3, 4, 5}
C = {a, e, i, o, u}
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Chapter 01: Sets, Relations and Functions
Infinite set:
A set which contains uncountable number of
elements is called an infinite set.
Example:
N = {1, 2, 3, 4……}
Z = {…−3, −2, −1, 0, 1, 2, 3,…..}
Note:
i.
An empty set is a finite set.
ii.
N, W, I, Q and R are infinite sets.
Equivalent sets:
Two sets A and B are said to be equivalent, if they
contain the same number of elements and we denote
it as A ≈ B.
Example:
If A = {1, 2, 3, 4, 5}, B = {a, b, c, d, e}, then
n(A) = n(B)
∴ A and B are equivalent sets.
Note:
Equal sets are always equivalent but equivalent sets
need not be equal.
Universal set:
A non-empty set of which all the sets under
consideration are subsets, is called a universal set. It
is usually denoted by X or U.
Example:
If A = {1, 2, 3, 4}, B = {2, 8, 13, 15} and
C = {1, 2, 3, …, 50} are sets under consideration,
then the set N of all natural numbers can be taken as
the universal set.
Venn diagram:
A set is represented by any closed figure such as
circle, rectangle, triangle, etc. The diagrams
representing sets are called venn diagrams.
Example:
A
.4
i.
A = {4, 6, 9}
.6 .9
A = {a, b, c, d, e, f}
B = {b, e, f}
B⊂A
5
B .a
.b
.e
.f
Complement of a set:
Let A be a subset of a universal set X then the set of
all those elements of X which do not belong to A is
called the complement of set A and it is denoted by
A′ or Ac.
Thus, A′ = {x/x ∈ X, x ∉ A}
X
Equal sets:
Two sets A and B are said to be equal if they have
the same elements and we denote this as A = B.
From this definition it follows that “two sets A and
B are equal if and only if A ⊆ B and B ⊆ A”
Example:
If A = {1, 2, 3, 4}, B = {2, 4, 1, 3}, then A = B.
ii.
Operations on sets
.d
A
.c
A
A′
The shaded region in the above figure represents A′.
Example:
Let X = {1, 2, 3, 4, 5, 6, 7, 8, 9} be an universal set
and A = {1, 3, 5, 6, 8}.
Then A′ = {2, 4, 7, 9}
Properties:
If X is the universal set and A, B ⊆ X, then
i.
(A′)′ = A
ii.
X′ = φ
iii. φ′ = X
iv. A ∩ A′ = φ
v.
A ∪ A′ = X
vi. If A ⊆ B, then B′ ⊆ A′.
Union of sets:
If A and B are two sets, then the set of those
elements which belong to A or to B or to both A and
B is called the union of the sets A and B and is
denoted by A ∪ B.
i.e., A ∪ B = {x/x ∈ A or x ∈ B}
The shaded portion in the below venn diagram
represents A ∪ B.
X
A
B
A∪B
Example:
i.
If A = {1, 2, 3, 4}, B = {2, 4, 6, 8},
then A ∪ B = {1, 2, 3, 4, 6, 8}
ii.
If A is the set of all odd integers and B is the
set of all even integers, then A ∪ B is the set
of all integers.
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Target Publications Pvt. Ltd.
Std. XI Sci.: Perfect Maths - II
Properties:
If A, B, C are any three sets, then
i.
A∪φ=A
ii.
A∪X=X
iii. A ∪ B = B ∪ A
(Commutative law)
iv. (A ∪ B) ∪ C = A ∪ (B ∪ C)
(Associative law)
v.
A∪A=A
(Idempotent law)
vi. If A ⊂ B, then A ∪ B = B
vii. A ⊂ A ∪ B, B ⊂ A ∪ B
Intersection of sets
If A and B are two sets, then the set of those
elements which belong to both A and B i.e., which
are common to both A and B is called the
intersection of the sets A and B and is denoted by
A ∩ B.
Thus, A ∩ B = { x/x ∈ A and x ∈ B}
The shaded portion in the below venn diagram
represents A ∩ B.
X
A
B
A∩B
Example:
If A = {1, 2, 3, 4, 5}, B = {1, 3, 5, 7, 9},
then A ∩ B = {1, 3, 5}
Properties:
If A, B, C are any three sets, then
i.
A∩φ=φ
ii.
A∩X=A
iii. A ∩ B = B ∩ A
(Commutative law)
iv. (A ∩ B) ∩ C = A ∩ (B ∩ C)
(Associative law)
v.
A∩A=A
(Idempotent law)
vi. If A ⊂ B, then A ∩ B = A
vii. A ∩ B ⊂ A, A ∩ B ⊂ B
Disjoint sets:
Two sets A and B are said to be disjoint, if they have
no element in common i.e., A ∩ B = φ.
Example:
If A = {2, 4, 6} and B = {3, 5, 7}, then A ∩ B = φ
∴
A and B are disjoint sets.
The venn diagram of the disjoint sets A and B is
shown below:
6
X
A
B
A∩B=φ
Difference of sets:
If A and B are two sets then the set of all the
elements of A which are not in B is called difference
of sets A and B and is denoted by A − B.
Thus, A − B = {x/x ∈ A and x ∉ B}
Similarly, B − A = {x/x ∉ A and x ∈ B}
In the below venn diagrams
shaded region
represents A − B and B − A .
X
X
A
B
A
B
B−A
A−B
Example:
If A = {1, 2, 3, 4, 5, 6}, B = {2, 4, 6, 8}, then
A − B = {1, 3, 5} and B − A = {8}
Distributive Properties of union and intersection
If a, b, c ∈ R, then
a × (b + c) = (a × b) + (a × c)
This is known as distributive property of
multiplication over addition. In set theory, the
operation of union and intersection of sets are both
distributive over each other i.e.,
If A, B, C are any three sets, then
i.
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
ii.
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
We verify these distributive laws using Venn
diagrams shown below. The shaded portion in each
figure shows the set obtained by performing the
operation given below the figure.
X
A
X
A
i.
B
C
=
B
(A ∪ B) ∩ (A ∪ C)
A ∪ (B ∩ C)
X
X
A
A
ii.
B
C
A ∩ (B ∪ C)
C
=
B
C
(A ∩ B) ∪ (A ∩ C)
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Chapter 01: Sets, Relations and Functions
De-Morgan’s laws
Exercise 1.1
If A and B are two subsets of a universal set X, then
i.
(A ∪ B)′ = A′ ∩ B′
ii.
(A ∩ B)′ = A′ ∪ B′
We verify these laws using Venn diagrams shown
below. The shaded portion in each figure shows the
set obtained by performing the operation below the
figure:
i.
X
X
A
A
B
=
B
1.
(A ∪ B)′
ii.
A′ ∩ B′
X
X
A
B
=
(A ∩ B)′
Number of elements in a set
A
B
A′ ∪ B′
Let A be a set. Then the total number of elements in it
is denoted by n(A).
Example:
Let A = {8, 9, 10, 11, 12}
∴
n (A) = 5
The number of elements in the empty set φ is zero.
i.e., n (φ) = 0
Results:
For given sets A, B, C
i.
n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
ii.
When A and B are disjoint sets, then
n(A ∪ B) = n(A) + n(B)
iii. n(A ∩ B′) + n(A ∩ B) = n(A)
iv. n(A′ ∩ B) + n(A ∩ B) = n(B)
v.
n(A ∩ B′) + n(A ∩ B) + n(A′ ∩ B) = n(A ∪ B)
vi. n(A∪ B∪ C) = n(A) + n(B) + n(C) − n(A∩B)
− n(B ∩ C) − n(A ∩ C) + n(A ∩ B ∩ C)
Power set
The set of all subsets of set A is called the power set
of A and it is denoted by P(A).
Example:
If A = {a, b, c}, then
P(A) ={φ, {a}, {b}, {c}, {a, b}, {b, c}, {a, c},
{a, b, c}}
Note:
If A contains n elements, then the power set of A
i.e., P(A) contains 2n elements.
7
Describe the following sets in Roster form:
i.
{x/x is a letter of the word
‘MARRIAGE’}
1
9⎫
⎧
ii.
⎨ x / x isan integer and − < x < ⎬
2
2⎭
⎩
iii. {x/x = 2n, n ∈ N}
Solution:
i.
Let A = {x/x is a letter of the word
‘MARRIAGE’}
∴
A = {M, A, R, I, G, E}
1
9⎫
⎧
ii.
Let B = ⎨ x / x is an integer and − < x < ⎬
2
2⎭
⎩
∴
B = {0, 1, 2, 3, 4}
iii. Let C = {x/x = 2n, n ∈ N}
∴
C = {2, 4, 6, 8, ….}
2.
Describe the following sets in Set-Builder
form:
i.
{0}
ii.
{0, ± 1, ± 2, ± 3}
⎧1 2 3 4 5 6 7 ⎫
iii.
,
,
, ⎬
⎨ , , ,
⎩ 2 5 10 17 26 37 50 ⎭
Solution:
i.
Let A = {0}
0 is a whole number but it is not a natural
number
∴
A = {x / x ∈ W, x∉ N}
∴
Let B = {0, ± 1, ± 2, ± 3}
B is the set of elements which belongs to Z
from −3 to 3
B = {x / x ∈ Z, −3 ≤ x ≤ 3}
iii.
Let
ii.
⎧1 2 3 4 5 6 7 ⎫
C= ⎨ , , , , , , ⎬
⎩ 2 5 10 17 26 37 50 ⎭
n
⎧
⎫
C = ⎨x / x = 2
, n ∈ N, n ≤ 7 ⎬
⎩
⎭
n +1
∴
If A = {x / 6x2 + x − 15 = 0}
B = {x / 2x2 − 5x − 3 = 0}
C = {x / 2x2 − x − 3 = 0}, then find
i.
A∪B∪C
ii.
A∩B∩C
Solution:
3.
{
}
A = x / 6 x 2 + x − 15 = 0
∴
∴
2
6x + x − 15 = 0
6x2 + 10x − 9x − 15 = 0
7
Target Publications Pvt. Ltd.
∴
∴
2x(3x + 5) − 3(3x + 5) = 0
(3x + 5) (2x − 3) = 0
−5
3
∴
x=
or x =
3
2
⎧ −5 3 ⎫
∴
A =⎨ , ⎬
⎩ 3 2⎭
2
B = {x/2x − 5x − 3 = 0}
∴
2x2 − 5x − 3 = 0
∴
2x2 − 6x + x − 3 = 0
∴
2x(x − 3) + 1(x − 3) = 0
∴
(x − 3)(2x + 1) = 0
−1
x = 3 or x =
∴
2
⎧ −1 ⎫
B = ⎨ , 3⎬
⎩2 ⎭
2
C = {x/2x − x − 3 = 0}
∴
2x2 − x − 3 = 0
∴
2x2 − 3x + 2x − 3 = 0
x(2x − 3) + 1(2x − 3) = 0
∴
∴
(2x − 3) (x + 1) = 0
3
∴
x = or x = −1
2
3⎫
⎧
∴
C = ⎨ −1, ⎬
2⎭
⎩
Thus,
⎧ 5 3 ⎫ ⎧ −1 ⎫ ⎧ 3 ⎫
i.
A ∪ B ∪ C = ⎨− , ⎬ ∪ ⎨ ,3⎬ ∪ ⎨−1, ⎬
⎩ 3 2⎭ ⎩ 2 ⎭ ⎩ 2⎭
−1 3 ⎫
⎧ −5
= ⎨ , − 1, , , 3⎬
2 2 ⎭
⎩3
ii.
A∩B∩C={}
∴
If A, B, C are the sets of the letters in the
words ‘college’, ‘marriage’ and ‘luggage’
respectively, then verify that
[A − (B ∪ C)] = [(A − B) ∩ (A − C)].
Solution:
A = {c, o, l, g, e}
B = {m, a, r, i, g, e,}
C = {l, u, g, a, e}
B ∪ C = {m, a, r, i, g, e, l, u}
A − (B ∪ C) = {c, o}
A − B = {c, o, l}
A − C = {c, o}
∴
[(A − B) ∩ (A − C)] = {c, o} = A − (B ∪ C)
∴
[A − (B ∪ C)] = [(A − B) ∩ (A − C)]
Std. XI Sci.: Perfect Maths - II
5.
If A = {1, 2, 3, 4}, B = {3, 4, 5, 6},
C = {4, 5, 6, 7, 8} and universal set
X = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, then verify
the following:
i.
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
ii.
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
iii. (A ∪ B)′ = A′ ∩ B′
iv. (A ∩ B)′ = A′ ∪ B′
v.
A = (A ∩ B) ∪ (A ∩ B′)
vi. B = (A ∩ B) ∪ (A′ ∩ B)
vii. n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
Solution:
A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {4, 5, 6, 7, 8}
X ={1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
i.
(B ∩ C) = {4, 5, 6}
∴
A ∪ (B ∩ C)= {1, 2, 3, 4, 5, 6}
(A ∪ B) = {1, 2, 3, 4, 5, 6}
(A ∪ C) = {1, 2, 3, 4, 5, 6, 7, 8}
∴
(A ∪ B) ∩ (A ∪ C) = {1, 2, 3, 4, 5, 6}
∴
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
ii.
∴
∴
∴
iii.
∴
∴
∴
iv.
4.
8
∴
∴
v.
∴
∴
vi.
∴
∴
(B ∪ C) = {3, 4, 5, 6, 7, 8}
A ∩ (B ∪ C) = {3, 4}
A ∩ B = {3, 4}
A ∩ C = {4}
(A ∩ B) ∪ (A ∩ C) = {3, 4}
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
A ∪ B = {1, 2, 3, 4, 5, 6}
(A ∪ B)′ ={7, 8, 9, 10}
A′ = {5, 6, 7, 8, 9, 10},
B′ = {1, 2, 7, 8, 9, 10}
A′ ∩ B′ = {7, 8, 9, 10}
(A ∪ B)′ = (A′ ∩ B′)
A ∩ B = {3, 4}
(A ∩ B)′ = {1, 2, 5, 6, 7, 8, 9, 10}
A′ = {5, 6, 7, 8, 9, 10}
B′ = {1, 2, 7, 8, 9, 10}
A′ ∪ B′ = {1, 2, 5, 6, 7, 8, 9, 10}
(A ∩ B)′ = A′ ∪ B′
A ∩ B = {3, 4}
A ∩ B′ = {1, 2}
(A ∩ B) ∪ (A ∩ B′) = {1, 2, 3, 4}
A = (A ∩ B) ∪ (A ∩ B′)
A ∩ B = {3, 4}
A′ ∩ B = {5, 6}
(A ∩ B) ∪ (A′ ∩ B) = {3, 4, 5, 6}
B = (A ∩ B) ∪ (A′ ∩ B)
Target Publications Pvt. Ltd.
vii.
Chapter 01: Sets, Relations and Functions
A = {1, 2, 3, 4}, B = {3, 4, 5, 6},
A ∩ B = {3, 4}, A ∪ B = {1, 2, 3, 4, 5, 6}
n(A) = 4, n(B) = 4,
n(A ∩ B) = 2, n(A ∪ B) = 6
n(A) + n(B) − n(A ∩ B) = 4 + 4 − 2 = 6
n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
ii.
If A and B are subsets of the universal set X
and n(X) = 50, n(A) = 35, n(B) = 20,
n(A′ ∩ B′) = 5, find
i.
n(A ∪ B)
ii.
n(A ∩ B)
iv. n(A ∩ B′).
iii. n(A′ ∩ B)
Solution:
n(X) = 50, n(A) = 35, n(B) = 20, n(A′ ∩ B′) = 5
i.
n(A ∪ B) = n(X) − [n(A ∪ B)′]
= n(X) − n(A′ ∩ B′)
= 50 − 5
= 45
ii.
n(A ∩ B) = n(A) + n(B) − n(A ∪ B)
= 35 + 20 − 45
= 10
iii. n(A′ ∩ B) = n(B) − n(A ∩ B)
= 20 − 10
= 10
iv. n(A ∩ B′) = n(A) − n(A ∩ B) = 35 − 10 = 25
8.
∴
∴
∴
6.
7.
In a class of 200 students who appeared
certain examinations, 35 students failed in
MHT-CET, 40 in AIEEE and 40 in IIT
entrance, 20 failed in MHT-CET and
AIEEE, 17 in AIEEE and IIT entrance,
15 in MHT-CET and IIT entrance and 5
failed in all three examinations. Find how
many students
i.
did not fail in any examination.
ii.
failed in AIEEE or IIT entrance.
Solution:
Let A = set of students who failed in MHT-CET
B = set of students who failed in AIEEE
C = set of students who failed in IIT entrance
X = set of all students
∴
n(X) = 200, n(A) = 35, n(B) = 40, n(C) = 40,
n(A ∩ B)= 20, n(B ∩ C) = 17, n(A ∩ C) = 15,
n(A ∩ B ∩ C) = 5
i.
n(A ∪ B ∪ C)
= n(A) + n(B) + n(C) − n(A ∩ B) − n(B ∩ C)
− n(A ∩ C) + n(A ∩ B ∩ C)
= 35 + 40 + 40 − 20 − 17 − 15 + 5
= 68
∴
No. of students who did not fail in any exam
= n(X) − n(A ∪ B ∪ C) = 200 − 68 = 132
9
No. of students who failed in AIEEE or IIT
entrance = n(B ∪ C)
= n(B) + n(C) − n(B ∩ C)
= 40 + 40 − 17
= 63
From amongst 2000 literate individuals of a
town, 70% read Marathi newspapers, 50%
read English newspapers and 32.5% read
both Marathi and English newspapers.
Find the number of individuals who read
i.
at least one of the newspapers.
ii.
neither
Marathi
nor
English
newspaper.
iii. only one of the newspapers.
Solution:
Let M = set of individuals who read Marathi
newspapers
E = set of individuals who read English
newspapers
X = set of all literate individuals
70
∴
n(X) = 2000, n(M) =
× 2000 = 1400
100
50
n(E) =
× 2000 = 1000
100
32.5
× 2000 = 650
n(M ∩ E) =
100
n(M ∪ E) = n(M) + n(E) − n(M ∩ E)
= 1400 + 1000 − 650
= 1750
i.
No. of individuals who read at least one of the
newspapers = n(M ∪ E) = 1750.
ii.
No. of individuals who read neither Marathi
nor English newspaper
= n(M′ ∩ E′)
= n(M ∪ E)′
= n(X) − n(M ∪ E)
= 2000 − 1750
= 250
iii.
No. of individuals who read only one of the
newspaper = n(M ∩ E′) + n(M′ ∩ E)
= n(M ∪ E) – n(M ∩ E)
= 1750 – 650
= 1100
9
Target Publications Pvt. Ltd.
9.
Std. XI Sci.: Perfect Maths - II
In a hostel, 25 students take tea, 20 students
take coffee, 15 students take milk, 10
students take both tea and coffee, 8 students
take both milk and coffee. None of them
take tea and milk both and everyone takes
atleast one beverage, find the number of
students in the hostel.
Solution:
Let T = set of students who take tea
C = set of students who take coffee
M = set of students who take milk
∴
n(T) = 25, n(C) = 20, n(M) = 15,
n(T ∩ C) = 10, n(M ∩ C) = 8, n(T ∩ M) = 0,
n(T ∩ M ∩ C) = 0
∴
No. of students in the hostel
= n(T ∪ C ∪ M)
= n(T) + n(C) + n(M) − n(T ∩ C) − ( M ∩ C)
− (T ∩ M) + n(T ∩ M ∩ C)
= 25 + 20 + 15 − 10 − 8 − 0 + 0
= 42
11.
10.
There are 260 persons with a skin disorder.
If 150 had been exposed to the chemical A,
74 to the chemical B, and 36 to both
chemicals A and B, find the number of
persons exposed to
i.
Chemical A but not Chemical B
ii.
Chemical B but not Chemical A
iii. Chemical A or Chemical B.
Solution:
Let A = set of persons exposed to chemical A
B = set of persons exposed to chemical B
X = set of all persons
∴
n(X)=260, n(A)=150, n(B)= 74, n(A ∩ B)= 36
i.
No. of persons exposed to chemical A but not
to chemical B
= n(A ∩ B′)
= n(A) − n(A ∩ B)
= 150 − 36
= 114
ii.
No. of persons exposed to chemical B but not
to chemical A
= n(A′∩ B)
= n(B) − n(A ∩ B)
= 74 − 36
= 38
iii. No. of persons exposed to chemical A or
chemical B
= n(A ∪ B)
= n(A) + n(B) − n(A ∩ B)
= 150 + 74 − 36
= 188
10
If A = {1, 2, 3}, write down the set of all
possible subsets of A i.e., the power set of A.
Solution:
A = {1, 2, 3}
The power set of A is given by
P(A) = {φ,{1},{2},{3}, {1, 2}, {2, 3},{1, 3},
{1, 2, 3}}
12.
Write the following intervals in Set-Builder
form:
i.
(−3, 0)
ii.
[6, 12]
iii. (6, 12]
iv. [−23, 5).
Solution:
i.
Let A = (−3, 0)
∴
A = {x/x ∈ R and − 3 < x < 0}
ii.
∴
Let B = [6, 12]
B = {x/x ∈ R and 6 ≤ x ≤ 12}
iii.
∴
Let C = (6, 12]
C = {x/x ∈ R and 6 < x ≤ 12}
iv.
∴
Let D = [−23, 5)
D = {x/x ∈ R and −23 ≤ x < 5}
13.
Using venn diagrams represent
i.
(A ∪ B)′
ii.
A′ ∪ B′
iv. A ∩ B′
iii. A′ ∩ B
Solution:
X
i.
A
B
(A ∪ B)′
X
ii.
A
B
A′ ∪ B′
X
iii.
A
A′ ∩ B
B
Target Publications Pvt. Ltd.
Chapter 01: Sets, Relations and Functions
X
iv.
A
B
A ∩ B′
Relations
Ordered Pair:
If (a, b) is a pair of numbers then the order in which
the numbers appear is important, is called an ordered
pair. Ordered pairs (a, b) and (b, a) are different.
Two ordered pairs (a, b) and (c, d) are equal, if and
only if a = c and b = d
Also, (a, b) = (b, a) if and only if a = b
Cartesian products of two sets
Let A and B be any two non-empty sets. The set of
all ordered pairs (a, b) such that a ∈ A and b ∈ B is
called the cartesian product of A and B and is
denoted by A × B.
Thus, A × B = {(a, b)/a ∈ A, b ∈ B}
where a is called the first element and b is called the
second element of the ordered pair (a, b).
If A ≠ B, then A × B ≠ B × A
If A = φ or B = φ or both A and B are empty sets,
then A × B = φ.
Example:
If A = {1, 2, 3}, B = {a, b}, then
A × B = {(1, a), (1, b), (2, a), (2, b), (3, a), (3,b)}
Number of elements in the Cartesian products of
two finite sets
If A and B are any two finite sets with n(A) = m1
and n(B) = m2, then the number of elements in the
cartesian product of two finite sets A and B is given
by n(A × B) = m1.m2.
Cartesian product of real with itself (R3)
Ordered Triplet:
Three real numbers a, b, c listed in a specific order
and enclosed in parentheses form ordered triplet
(a, b, c).
For any non-empty set A, we define
A × A × A = {(a, b, c) / a, b, c ∈ A}
Note:
(a, b, c) ≠ (b, a, c) ≠ (c, a, b)
11
Definition of Relation
Consider the following statements:
i.
Ram is taller than Shyam.
ii.
Harshal and Ravi have shirts of same colour.
iii. 25 is the square of 5.
iv. 2 and 4 are even integers.
Here we can say that Ram is related to Shyam by the
relation “is taller than”. Harshal is related to Ravi by
the relation “have shirts of same colour”. 25 is
related to 5 by the relation “is the square of ” and 2
is related to 4 by the relation “are even integers”.
Definition:
If A and B are two non-empty sets, then any subset
of A × B is called relation from A to B and is
denoted by capital letters P, Q, R etc
Consider the following illustration:
a
b
c
A
R
l
m
n
o
p
B
Let us consider A = {a, b, c}, B = {l, m, n, o, p}
∴
A × B = {(a, l), (a, m), (a, n), (a, o), (a, p),
(b, l), (b, m), (b, n), (b, o), (b, p),
(c, l), (c, m), (c, n), (c, o), (c, p)}
In the above figure, the arrow starting from the
element ‘a’ and pointing to the elements ‘l’ and ‘n’
indicates that ‘a’ is related to ‘l’ and ‘n’. Similarly,
‘b’ is related to ‘m’ and ‘o’ and ‘c’ is related to ‘p’.
This relation is also represented by set of ordered
pairs, R = {(a, l), (a, n), (b, m), (b, o), (c, p)}
This relation R is a subset of A × B. Thus, relation
from set A to B is a subset of A × B i.e., R ⊆ A × B.
If R is a relation and (x, y) ∈ R, then it is denoted by
xRy.
y is called image of x under R and x is called
pre-image of y under R.
Domain:
The set of all first components of the ordered pairs in
a relation R is called the domain of the relation R.
i.e., domain (R) = {a/(a, b) ∈ R}
Range:
The set of all second components of the ordered
pairs in a relation R is called the range of the relation
R.
i.e., range (R) = {b/(a, b) ∈ R}
11
Target Publications Pvt. Ltd.
Std. XI Sci.: Perfect Maths - II
Co-domain:
If R is a relation from A to B, then set B is called the
co-domain of the relation R.
Example:
Let A = {−2, 1, 0, 1, 2, 3}, B = {0, 1, 2, 3, 4}
and R = {(−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4)}
Types of Relation
i.
A
−2
−1
0
1
2
3
One-One relation:
If every element of A has at most one image in
B and distinct elements in A have distinct
images in B, then a relation R from A to B is
said to be one-one.
Example:
Let A = {3, 4, 5, 6}, B = {4, 5, 6, 7, 9}
and R = {(3, 5), (4, 6), (5, 7)}
A
R
3
4
5
6
iv.
Then R is a one-one relation from A to B.
Here, domain of R = {3, 4, 5} and
range of R = {5, 6, 7}
ii.
1
2
3
4
R
B
2
3
4
5
6
7
Then R is a many-one relation from A to B.
Here, domain of R = {1, 3, 4} and
range of R = {4, 7}
iii.
Into relation:
If there exists at least one element in B which
has no pre-image in A, then a relation R from
A to B is said to be into relation.
12
0
1
2
3
4
Onto relation:
If every element of B is the image of some
element of A, then a relation R from A to B is
said to be onto relation.
Example:
Let A = {−2, −1, 1, 3, 4}, B = {1, 4, 9}
and R = {(−2, 4), (−1, 1), (1, 1), (3, 9)}
A
R
−2
−1
1
3
4
Many-one relation:
If two or more than two elements in A have
same image in B, then a relation R from A to
B is said to be many-one.
Example:
Let A = {1, 2, 3, 4}, B = {2, 3, 4, 5, 6, 7}
and R = {(1, 4), (3, 7), (4, 4)}
A
B
Then R is into relation from A to B.
Here, domain of R = {–2, –1, 0, 1, 2} and
range of R = {0, 1, 4}
B
4
5
6
7
9
R
B
1
4
9
Then R is onto relation from A to B.
Here, domain of R = {–2, –1, 1, 3} and
range of R = {1, 4, 9}
∴
Range = co-domain (B)
Various types of Relations
Let A be a non – empty set. Then a relation R on A
is said to be
i.
Reflexive:
if (a, a) ∈R V a ∈ A
i.e., if a R a V a ∈ A
ii.
Symmetric:
if (a,b) ∈R ⇒ (b,a) ∈R V a, b ∈A
i.e., if a R b ⇒ b R a V a, b∈A
iii. Transitive:
if (a,b)∈R & (b,c)∈R
⇒ (a,c)∈R V a, b, c ∈A
if a R b & b R c ⇒ a R c V a, b, c ∈ A
Target Publications Pvt. Ltd.
Chapter 01: Sets, Relations and Functions
Equivalence Relation: A relation R on a set A is
said to be an equivalence relation on A when R is
i.
Reflexive
ii.
Symmetric
iii. Transitive
4.
If P = {1, 2, 3} and Q = {4},
find sets P × Q and Q × P.
Solution:
P = {1, 2, 3}, Q = {4}
∴
P × Q = {(1, 4), (2, 4), (3, 4)}
and Q × P = {(4, 1), (4, 2), (4, 3)}
Exercise 1.2
1.
If (x − 1, y + 4) = (1, 2), find the values of x
and y.
Solution:
By the definition of equality of ordered pairs, we
have
(x − 1, y + 4) = (1, 2)
∴
x − 1 = 1 and y + 4 = 2
∴
x = 2 and y = − 2
2.
1 y
⎛
⎞ ⎛ 1 3⎞
If ⎜ x + , − 1 ⎟ = ⎜ , ⎟ , find x and y.
3 2
⎝
⎠ ⎝ 2 2⎠
5.
Let A = {1, 2, 3, 4}, B = {4, 5, 6}, C = {5, 6}
Find
i.
A × (B ∩ C)
ii.
(A × B) ∩ (A × C)
iii. A × (B ∪ C)
iv. (A × B) ∪ (A × C)
Solution:
A = {1, 2, 3, 4}, B = {4, 5, 6}, C = {5, 6}
i.
B ∩ C = {5, 6}
∴
A × (B ∩ C) = {(1, 5), (1, 6), (2, 5), (2, 6),
(3,5), (3, 6), (4, 5), (4, 6)}
ii.
Solution:
By the definition of equality of ordered pairs, we
have
1 y ⎞ ⎛1 3⎞
⎛
⎜ x + , − 1⎟ = ⎜ , ⎟
3 2 ⎠ ⎝2 2⎠
⎝
∴
x+
1
1
y
3
= and − 1 =
3
2
2
2
∴
x=
1 1
y 3
− and = + 1
2 3
2 2
∴
x=
1
and y = 5
6
3.
If A = {a, b, c}, B = {x, y},
iii.
B ∪ C = {4, 5, 6}
A × (B ∪ C)
= {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6),
(3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}
iv.
(A × B) ∪ (A × C)
= {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6),
(3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}
Express {(x, y)/ x2 + y2 = 100, where
x, y ∈ W} as a set of ordered pairs.
Solution:
{(x, y) / x2 + y2 = 100, where x, y ∈ W}
We have, x2 + y2 = 100
When x = 0, y = 10 ⇒ x2 + y2 = 02 + 102 = 100
When x = 6, y = 8 ⇒ x2 + y2 = 62 + 82 = 100
When x = 8, y = 6 ⇒ x2 + y2 = 82 + 62 = 100
When x = 10, y = 0 ⇒ x2 + y2 = 102 + 02 = 100
∴
Set of ordered pairs
= {(0, 10), (6, 8), (8, 6), (10, 0)}
6.
find A × B, B × A, A × A, B × B.
Solution:
A = {a, b, c}, B = {x, y}
A × B = {(a, x), (a, y), (b, x), (b, y), (c, x), (c, y)}
B × A = {(x, a), (x, b), (x, c), (y, a), (y, b), (y, c)}
A × A = {(a, a), (a, b), (a, c) (b, a) (b, b), (b, c),
(c, a), (c, b) (c, c)}
B × B = {(x, x), (x, y), (y, x), (y, y)}
13
∴
A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5),
(2, 6), (3, 4), (3, 5), (3, 6), (4, 4),
(4, 5), (4, 6)}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5),
(3, 6), (4, 5), (4, 6)}
(A × B) ∩ (A × C)
= {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6),
(4, 5), (4, 6)}
13
Target Publications Pvt. Ltd.
7.
Write the domain and range of the
following relations:
i.
{(a, b) / a ∈ N, a < 6 and b = 4}
ii.
{(a, b) / a, b ∈ N, a + b = 12}
iii. {(2, 4), (2, 5), (2, 6), (2, 7)}.
Solution:
i.
Let R1 = {(a, b)/ a ∈ N, a < 6 and b = 4}
Set of values of ‘a’ are domain and set of
values of ‘b’ are range
a ∈ N and a < 6
∴
a = 1, 2, 3, 4, 5 and b = 4
Domain (R1) = {1, 2, 3, 4, 5}
Range (R1) = {4}
ii.
Let R2 = {(a, b)/a, b ∈ N and a + b = 12}
Now, a, b ∈ N and a + b = 12
When a = 1, b = 11
When a = 2, b = 10
When a = 3, b = 9
When a = 4, b = 8
When a = 5, b = 7
When a = 6, b = 6
When a = 7, b = 5
When a = 8, b = 4
When a = 9, b = 3
When a = 10, b= 2
When a = 11, b = 1
∴
Domain (R2) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
Range (R2) = {11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1}
iii.
Let R3 ={(2, 4), (2, 5), (2, 6), (2, 7)}
Domain (R3)= {2}
Range (R3)= {4, 5, 6, 7}
Let A = {6, 8} and B = {1, 3, 5}.
Let R = {(a, b) / a∈A, b∈B, a − b is an even
number}.
Show that R is an empty relation from A to B.
Solution:
A = {6, 8}, B = {1, 3, 5}
R = {(a, b)/ a ∈ A, b ∈ B, a − b is an even number}
a∈A
∴
a = 6, 8
b∈B
∴
b = 1, 3, 5
When a = 6 and b = 1, a − b = 5 which is odd
When a = 6 and b = 3, a − b = 3 which is odd
When a = 6 and b = 5, a − b = 1 which is odd
When a = 8 and b = 1, a − b = 7 which is odd
When a = 8 and b = 3, a − b = 5 which is odd
When a = 8 and b = 5, a − b = 3 which is odd
Thus, no set of values of a and b gives a − b even
∴
R is an empty relation from A to B.
Std. XI Sci.: Perfect Maths - II
9.
Write the relation in the Roster form and
hence find its domain and range.
i.
R1 = {(a, a2) / a is a prime number
less than 15}
⎧⎛ 1 ⎞
⎫
ii.
R2 = ⎨⎜ a, ⎟ / 0 < a ≤ 5,a ∈ N ⎬
⎩⎝ a ⎠
⎭
Solution:
i.
R1 = {(a, a2) / a is a prime number less than
15}
∴
a = 2, 3, 5, 7, 11, 13
∴
a2 = 4, 9, 25, 49, 121, 169
∴
R1 = {(2, 4), (3, 9), (5, 25), (7, 49), (11, 121),
(13, 169)}
∴
Domain (R1)
= {a/a is a prime number less than 15}
= {2, 3, 5, 7, 11, 13}
Range (R1)
= {a2/a is a prime number less than 15}
= {4, 9, 25, 49, 121, 169}
ii.
∴
∴
∴
∴
8.
14
⎧⎛ 1 ⎞
⎫
R2 = ⎨⎜ a, ⎟ 0 < a ≤ 5,a ∈ N ⎬
⎩⎝ a ⎠
⎭
a = 1, 2, 3, 4, 5
1
1 1 1 1
= 1, , , ,
a
2 3 4 5
⎧
⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞⎫
R2 = ⎨(1,1) , ⎜ 2, ⎟ , ⎜ 3, ⎟ , ⎜ 4, ⎟ , ⎜ 5, ⎟ ⎬
⎝ 2 ⎠ ⎝ 3 ⎠ ⎝ 4 ⎠ ⎝ 5 ⎠⎭
⎩
Domain (R2) = {a/0 < a ≤ 5, a ∈ N}
= {1, 2, 3, 4, 5}
⎧1
⎫
Range (R2) = { ⎨ / 0 < a ≤ 5,a ∈ N ⎬
⎩a
⎭
⎧ 1 1 1 1⎫
= ⎨1, , , , ⎬
⎩ 2 3 4 5⎭
R = {(a, b) / b = a + 1, a ∈ Z, 0 < a < 5}.
Find the range of R.
Solution:
R = {(a, b) / b = a + 1, a ∈ Z, 0 < a < 5}
∴
a = 1, 2, 3, 4
∴
b = 2, 3, 4, 5
∴
Range (R) = {2, 3, 4, 5}
10.
11.
Write the following relations as sets of
ordered pairs:
i.
{(x, y) / y = 3x, x ∈ {1, 2, 3},
y ∈ {3, 6, 9, 12}}
ii.
{(x, y) / y > x + 1, x = 1, 2 and
y = 2, 4, 6}
iii. {(x, y) / x + y = 3, x, y ∈ (0, 1, 2, 3)}
Target Publications Pvt. Ltd.
Chapter 01: Sets, Relations and Functions
Solution:
i.
{(x, y)/y = 3x, x ∈ {1, 2, 3}, y ∈ {3, 6, 9, 12}}
Here y = 3x
When x = 1, y = 3(1) = 3
When x = 2, y = 3(2) = 6
When x = 3, y = 3(3) = 9
∴
Ordered pairs are {(1, 3), (2, 6), (3, 9)}
ii.
∴
iii.
∴
{(x, y) / y > x + 1, x = 1, 2 and y = 2, 4, 6}
Here, y > x + 1
When x = 1 and y = 2, 2 > 1 + 1
When x = 1 and y = 4, 4 > 1 + 1
When x = 1 and y = 6, 6 > 1 + 1
When x = 2 and y = 2, 2 > 2 + 1
When x = 2 and y = 4, 4 > 2 + 1
When x = 2 and y = 6, 6 > 2 + 1
Ordered pairs are {(1, 4), (1, 6), (2, 4), (2, 6)}
{(x, y) / x + y = 3, x, y ∈ (0, 1, 2, 3)}
Here, x + y = 3
When x = 0, y = 3
When x = 1, y = 2
When x = 2, y = 1
When x = 3, y = 0
Ordered pairs are {(0, 3), (1, 2), (2, 1), (3, 0)}
The set of all values of function f is {2, 4, 6, 8}.
This set is called range of the function f.
Range of function:
If f is a function from set A to set B, then the set of
all values of the function f is called the range of the
function f.
Thus the range set of the function f: A → B is
{f(x) / x ∈ A}
Note that the range set is a subset of co-domain.
This subset may be proper or improper.
f
1
1
4
2
9
3
16
25
4
36
B
A
In the figure for f: A → B
A = {1, 2, 3, 4} is domain, B = {1, 4, 9, 16, 25, 36}
is co-domain and set {1, 4, 9, 16} is the range of the
function f.
Representation of functions
i.
Functions
Definition:
A function from set A to the set B is a relation
which associates every element of a set A to unique
element of set B and is denoted by f:A →B. If f is a
function from A to B and (x, y) ∈ f, then we write it
as y = f(x)
Example:
Let A = {1, 2, 3, 4}, B = {2, 3, 4, 5, 6, 7, 8}
f
1
2
3
4
A
ii.
2
3
4
5
6
7
8
B
Let a relation from A to B be given as “twice of”
then we observe that every element x of set A is
related to one and only one element of set B. Hence
this relation is a function from set A to set B. In this
case f(1) = 2, f(2) = 4, f(3) = 6, f(4) = 8 are the
values of function f(x) = 2x at x = 1, 2, 3, 4
respectively.
15
iii.
Arrow diagram:
In this diagram, we use arrows. Arrows start
from the element of domain and point out it’s
value.
x
p
y
z
q
r
s
A
B
Ordered pair:
If f: A → B be a function , then the element is
the ordered pair (a, b) ∈ f, where a ∈ A and
b ∈ B.
Example:
f = {(a, 1), (b, 2), (c, 3), (d, 4)}
Tabular form:
If the sets A and B are finite and contain very
few elements, then a function f: A→B can be
exhibited by means of a table of
corresponding elements.
Let f = {(1, 7), (2, 9), (3, 11), (4, 13)}
We can represent the above function in tabular
form as follows:
x
f(x)
1
7
2
9
3
11
4
13
15
Target Publications Pvt. Ltd.
iv.
Std. XI Sci.: Perfect Maths - II
Y
By Formula:
This is the most usual way of exhibiting
function.
Let A = {1, 2, 3, 4, 5}, B = {5, 7, 9, 11, 13}
and f: A → B is a function represented by
arrow diagram.
5
7
9
11
3.
Constant function:
A function f defined by f(x) = k, for all x ∈ R,
where k is a constant, is called a constant
function. The graph of a constant function is a
line parallel to the X-axis, intersecting Y-axis
at (0, k).
Y
P(x, k)
f(x) = k
(0, k)
O
X
(x, 0)
Y′
For example, f(x) = 5 is a constant function.
2.
Identity function:
The function f(x) = x, where x ∈ R is called an
identity function. The graph of the identity
function is the line which bisects the first and
the third quadrants. Observe the following
table of some values of f.
x
−3
−2
−1
0
1
2
3
f(x)
−3
−2
−1
0
1
2
3
16
O 1
−1
(−1, −1)
(−2, −2) −2
2
3
X
−3
Y′
Types of functions
X′
(1, 1)
−3 −2 −1
(−3, −3)
In this case we observe that, if we take any
element x of the set A, then the element of the
set B related to x is obtained by adding 3 to
twice of x. Applying this rule we get in
general
f(x) = 2x + 3, for all x ∈ A.
This is the formula which exhibits the function f.
If we denote the value of f at x by y, then we
get y = 2x + 3, for all x ∈A.
1.
1
(0, 0)
B
A
(3, 3)
(2, 2)
2
X′
f
1
2
3
4
3
One-one function:
A function f: A → B is said to be one-one
function, if different elements in A have
different images in B.
Consider the function f: A → B such that each
element of its range set is the value of the
function at only one element of the domain
set.
f
1
1
2
4
3
9
16
4
B
A
In this case, f: A → B is one-one function.
4.
Into function:
If the function f: A → B is such that there
exists at least one element in B which is not
the image of any element in A, then f is said to
be into function. In this case, the range of a
function f is a proper subset of its co-domain.
Consider the function f: A → B represented by
the following arrow diagram.
1
2
3
4
f
1
5
7
9
13
B
A
In this case range = {1, 5, 7, 9} is a proper
subset of co-domain {1, 5, 7, 9, 13}
Hence, f: A → B is into function.
Target Publications Pvt. Ltd.
5.
Chapter 01: Sets, Relations and Functions
Onto function:
7.
If the function f: A → B is such that each
element in B is the image of some element in
A, then f is said to be a onto function. In this
case range of function f is same as its
co-domain B.
a
b
m
c
n
Hence f: A → B is onto function.
2
4
X′
8.
(1, 1)
O 1
−3 −2 −1
2
3
X
Y′
Odd function:
A function f is called an odd function,
if f(−x) = − f(x) for all x ∈ R
Let f: R → R : f (x) = x3 for all x ∈ R
Then, domain of f = R and range of f = R.
We have
x
f(x) = x3
−2
−8
−1
−1
0
0
1
1
2
8
Y
(2, 8)
8
7
6
5
4
3
2
1 (1, 1)
1
4
X′
−3 −2 −1
(−1, −1)
B
In this case, f: A → B is many-one function.
Note:
We have the following functions.
i.
one-one onto function.
ii.
one-one into function.
iii. many-one into function.
iv. many-one onto function.
1
(−1, 1)
f
9
16
2
2
The function f: A → B represented by the
following arrow diagram is such that the
co-domain B contains 1, 4 and 9 each of
which is the value of the function f at two
distinct elements of the domain set A.
17
1
1
3
If the function f : A → B is such that two or
more elements in a set A have the same image
in set B i.e. there is at least one element in B
which has more than one pre-image in A then
the function f is called many-one function.
A
0
0
4
(−2, 4)
Many-one function:
1
−1
2
−2
3
−3
−1
1
Y
l
A
B
In this case, range is equal to co-domain
= {l, m, n}
6.
−2
4
x
f(x) = x2
Consider the function f: A → B represented by
the following arrow diagram
f
Even function:
A function f is said to be an even function, if
f(−x) = f(x) for all x ∈ R
Let f: R → R : f (x) = x2 for all x ∈ R
Domain of f = R, range of f = {x/x ∈ R, x ≥ 0}
We have
(−2, −8)
O1 2 3
−1
−2
−3
−4
−5
−6
−7
−8
X
Y′
17
Target Publications Pvt. Ltd.
Std. XI Sci.: Perfect Maths - II
9.
Y
Polynomial function:
A function of the form
5
2
n
f(x) = a0 + a1x + a2x +… + anx , where n is a
4
non-negative integer and a0, a1, a2,…, an ∈ R is
3
called a polynomial function.
2
(0.5, 2)
Example:
1
(1, 1)
X′
2
f(x) = x − 2x − 3 for x ∈ R
−2 −1 0
x
f(x) = x2 − 2x − 3
5
−3
0
1
2
3 4
−4
−3
0
5
X′
(−1, 0)
11.
O
1 2 3
X
4
Modulus function:
Let f: R → R, the function f(x) = |x| such that
4
⎧ x, for x ≥ 0
|x| = ⎨
⎩ − x,for x < 0
3
is called modulus or absolute value function.
2
The graph of the absolute value function
5
(4, 5)
1
(3, 0)
−3 −2 −1 O 1
−1
2
3 4
consist of two rays having common end point
X
origin and bisecting the first and second
quadrant.
−2
Consider table of same values of f(x) = |x|
−3 (0, −3) (2, −3)
−4
x
−3
f(x) 3
(1, −4)
−2
2
Y′
The function of the type
(−2, 2)
f ( x)
, where f(x) and
g( x)
(−1, 1)
X′
g(x) are polynomial functions of x, defined in
0
0
1
1
2
2
(3, 3)
2
1
−3 −2 −1
3
3
f(x) = x
3
(−3, 3)
Rational function:
−1
1
Y
f(x) = −x
10.
(2, 0.5)
Y′
Y
(−2, 5)
(0.25, 4)
(2, 2)
B
1
2
3
X
a domain, where g (x) ≠ 0 is called a rational
function.
Y′
Let f: (R−(0)) ⎯→ R : f (x) =
1
for all values
x
of x ∈ R − (0)
12.
Signum function:
The function f: R ⎯→ R : f (x) = 1, if x > 0
= 0, if x = 0
We have
x
f(x) = 1/x
18
= −1, if x < 0
0.25
4
0.5
2
1
1
2
0.5
is called the signum function.
Domain of f = R and Range of f = {−1, 0, 1}
Target Publications Pvt. Ltd.
Chapter 01: Sets, Relations and Functions
14.
Y
y=1
X′
Exponential Function
Let f: R → R+. The function f is defined by
f(x) = ax, where a > 0, a ≠ 1 is called an
exponential function.
X
O
Y
f(x) = ax (a > 0)
y = −1
Y′
13.
X′
X
O
Step function or the greatest integer
function:
Let x be a real number, the symbol [x] is the
Y′
f(x) = bx (b < 0)
greatest integer not greater than x.
Y
For example, [4] = 4, [−1] = −1, [−3. 8] = 4
For a given real number x, [x] is unique.
Hence the function f(x) = [x] is called the step
X′
function, its graph look like steps. The graph
Y′
consist of segments parallel to X-axis
corresponding to different intervals.
For example, f(x) = 2x is an exponential
function.
The right hand end points of the line segment
do not lie on the graph. There are breaks in
15.
the graph at all values of x = n, where n is an
ay = x, x ∈ R then y is called the logarithm of x
Thus, f(x) = [x], n ≤ x < n + 1 where n is an
with base ‘a’ and we write it as y = loga x.
integer.
if
if
if
if
if
Logarithmic function:
Let a be a positive real number with a ≠ 1, if
integer.
Thus [x] = −2,
= −1,
= 0,
= 1,
= 2,
X
O
i.e. A function f : R+ → R defined by
−2 ≤ x < − 1
−1 ≤ x < 0
0≤x<1
1≤x<2
2≤x<3
f(x) = loga x is called logarithmic function.
∴
f = {(x, loga x)/x ∈ R, a > 0, a ≠ 1}
Y
y = logax
(a > 1)
Y
3
2
X′
1
X′
–3
–2
–1
0
1
2
3
O
X
X
–1
–2
–3
Y′
19
Y′
19
Target Publications Pvt. Ltd.
Std. XI Sci.: Perfect Maths - II
Operations on functions
Inverse function
Let X ⊂ R, then various operations on function are
defined as follows:
i.
Sum of functions:
Let f: X → R and g: X → R be the two
functions, then f + g: X → R is defined by
(f + g)(x) = f(x) + g(x) for all x ∈ X.
If a function f: A → B is one-one and onto function
defined by y = f(x), then the function g: B → A
defined by g(y) = x is called the inverse of f and is
denoted by f −1.
ii.
Difference of functions:
Let f: X → R and g: X → R be the two
functions, then f − g: X → R is defined by
(f − g)(x) = f(x) − g(x) for all x ∈ X.
iii.
Product of functions:
Let f: X → R and g: X → R be the two
functions, then f.g : X → R is defined by
(f.g)(x) = f(x).g(x) for all x ∈ X.
iv.
Quotient of functions:
Let f: X → R and g: X → R be the two
functions.
Let X0 = {x ∈ X/ g(x) = 0}.
f
Then :X − X0 → R is defined by
g
⎛f ⎞
f ( x)
for all x ∈ X such that
⎜ ⎟ (x) =
g( x)
⎝g⎠
g (x) ≠ 0 (i.e., for all x ∈ X − X0)
v.
Scalar multiplication of a function:
Let f: X → R be a function and k be a scalar,
then (kf): X → R is defined by
(kf)(x) = kf(x) for all x ∈ X
Composite function
If f: A → B and g: B → C are two functions then the
composite function of f and g is the function
gof: A → C given by (gof)(x) = g[f(x)], for all x ∈ A
Let z = g(y) then z = g(y) = g[f(x)] ∈ C
gof
x
f
y
g
z
Thus f −1 : B → A is defined by x = f −1(y)
We also write if y = f(x) then x = f −1 (y)
Note that if the function is not one-one nor onto,
then its inverse does not exist.
The geometrical representation of the function is
called graph of the function.
We know that a function can be expressed as a set of
ordered pairs.
Let f: A → B be a function, where A and B are non
empty subsets of R. Let (x,y) be an element of f,
where x ∈ A, y ∈ B.
Since x, y are real numbers, we can plot the point
(x, y) in a plane by choosing a suitable co-ordinate
system. On plotting all such ordered pairs from the
set representing f we get a geometrical
representation of the function. This is called graph of
the function f.
Binary operation
If f: A → A then f is called unary operation on A.
i.e. f = {(a, f(a)/a, f(a) ∈ A}
If g: A × A → A then g is called binary operation on
A.
i.e. g = {(a, b), g(a, b)/(a, b)∈ A × A and g(a, b)∈A}
Let g(a, b) = a * b such that (a, b) ∈ A × A and a * b
∈A
i.e. g(a, b) = a * b where a, b, a * b ∈ A
Where * is called the binary operation in A. * can be
+, −, × for A = R(real numbers).
÷ is a binary operation on R − {0}
If * is a binary operation in set A, then set A is
closed with respect to binary operation *.
Real valued functions of the real variable
C
A
B
This shows that every element x of the set A is
related to one and only one element z = g[f(x)] of C.
This gives rise to a function from the set A to the set
C. This function is called the composite of f and g.
Note that (fog) (x) ≠ (gof)(x)
20
A function f : A → B is called a real valued function
of real variable, if A and B both are subsets of R.
The domain of the real function f(x), is the set of
all those real numbers for which f(x) assumes real
values only.
The range of the real function, is the set of all real
values taken by f(x) at points in its domain.
Target Publications Pvt. Ltd.
Exercise 1.3
1.
iii.
Find the domain and range of the following
functions:
i.
f(x) = x2
( x − 1)( 3 − x )
ii.
f(x) =
iii.
f(x) =
iv.
x+3
f(x) =
x−3
v.
f(x) = 9 − x 2
vi.
f(x) =
∴
∴
∴
∴
∴
∴
∴
∴
f(x) =
1
x −1
x−2
.
3− x
( x − 1)( 3 − x )
For this to exist
(x − 1).(3 − x) ≥ 0
(x − 1) ≥ 0 and 3 − x ≥ 0
x ≥ 1 and 3 ≥ x
x ≥ 1 and x ≤ 3
1≤x≤3
x ∈ [1, 3]
or
x − 1 ≤ 0 and 3 − x ≤ 0
x ≤ 1 and 3 ≤ x
x ≤ 1 and x ≥ 3
Which is not possible
Domain is [1, 3] and
For range
Let y = f(x) =
∴
∴
∴
∴
∴
∴
∴
∴
∴
∴
2
∴
2
Solution
i.
f(x) = x2
Domain = set of all real numbers
∴
Range = {x / x ∈ R and x ≥ 0}
ii.
Chapter 01: Sets, Relations and Functions
( x −1) (3 − x)
y = (x − 1) (3 – x)
y2 = −x2 + 4x − 3
x2 − 4x + (3 + y2) = 0
Disc > 0 (∵ x is real)
(− 4)2 − 4 (1)(3 + y2) > 0
16 − 12 − 4y2 ≥ 0
4y2< 4
y2< 1
−1<y<1
− 1 ≤ f (x) ≤ 1
Range is [−1, 1]
∴
∴
f(x) =
1
x2 − 1
f(x) is defined, when x2−1 > 0
i.e., when x2 > 1
i.e., when x > 1 or x < −1
Domain of f = (−∞, −1) ∪ (1, ∞)
1
Let y = f(x) =
x2 − 1
1
1
= x2 − 1
∴
x2−1 = 2
y
y
x2 =
1
+1
y2
∴
x2 =
∴
1+ y 2
y
Clearly, x is not defined, if y = 0
Range of f is R − {0}
iv.
f(x) =
∴
∴
∴
∴
∴
∴
x=+
x+3
x−3
f(x) is not defined, when x – 3 = 0 i.e., when
x=3
Domain of f = R − {3}
x+3
Let y = f(x) =
x−3
xy − 3y = x + 3
xy − x = 3 + 3y
3 + 3y
x=
y −1
which is not defined, when y – 1 = 0
i.e., when y = 1
Range of f = R − {1}
∴
∴
∴
∴
f(x) = 9 − x 2
f(x) is defined, when 9 − x2 > 0
9 ≥ x2
x2 ≤ 9
x ≤ 3 and x ≥ −3
−3≤x≤3
Domain of f = [−3, 3]
Now, − 3 ≤ x ≤ 3
0 ≤ x2 ≤ 9
0 ≥ − x2 ≥ − 9
0 + 9 ≥ 9 − x2 ≥ 9 − 9
0≤9−x2≤9
∴
∴
∴
− 3 ≤ 9 − x2 ≤ 3
− 3 ≤ f(x ) ≤ 3
Range of f = [−3, 3]
v.
∴
∴
∴
∴
∴
1+ y 2
y2
21
Target Publications Pvt. Ltd.
vi.
x−2
3− x
f(x) =
x−2
>0
3− x
i.e., i. x − 2 > 0 and 3 − x > 0
or
ii. x − 2 < 0 and 3 − x < 0
Std. XI Sci.: Perfect Maths - II
iii.
= (x2 − 6x + 9) + 2
f(x) is defined when
i.
When x − 2 > 0 and 3 − x > 0
x > 2 and 3 > x
i.e., x > 2 and x < 3
i.e., x ∈ [2, 3)
When x − 2 < 0 and 3 − x < 0
x < 2 and 3 < x
x < 2 and x > 3
which is not possible
Domain of f is [2, 3)
ii.
∴
Let y =
∴
∴
∴
∴
∴
x−2
3− x
x−2
3− x
3y2 − xy2 = x − 2
3y2 + 2 = x (1+ y2)
3y2 + 2
x=
1 + y2
Which is defined for all real values
Range of f (x) is R.
y2 =
Find the range of each of the following
functions:
i.
f(x) = 3x − 4, for − 1 ≤ x ≤ 3
ii.
f(x) = 9 − 2x2, for − 5 ≤ x ≤ 3
iii. f(x) = x2 − 6x + 11, for all x ∈ R.
Solution:
i.
f(x) = 3x − 4 for −1 ≤ x ≤ 3
As – 1 ≤ x ≤ 3
∴
–3 ≤ 3x ≤ 9
∴
–3 – 4 ≤ 3x – 4 ≤ 9 – 4
∴
–7 ≤ 3x – 4 ≤ 5
∴
Range of f is [−7, 5].
ii.
f(x) = 9 – 2x2, for –5 ≤ x ≤ 3
As – 5 ≤ x ≤ 3
∴
0 ≤ x2 ≤ 25
∴
0 ≤ 2x2 ≤ 50
∴
0 ≥ –2x2 ≥ –50
∴
0 + 9 ≥ 9 – 2x2 ≥ 9 – 50
∴
9 ≥ 9 – 2x2 ≥ –41
∴
9 ≥ f(x) ≥ –41
∴
Range of f is [–41, 9]
= (x − 3)2 + 2
But (x − 3)2 ≥ 0, for all x ∈ R
∴
(x − 3)2 + 2 ≥ 0 + 2
∴
f(x) ≥ 2
∴
Range = [2, ∞)
3.
Solve
i.
if f(x) =
x3 + 1
x2 + 1
, find f(−3), f(−1)
ii.
if f(x) = (x − 1)(2x + 1), find f(1), f(2), f(−3)
iii. if f(x) = 2x2 − 3x − 1, find f(x + 2).
Solution:
i.
f(x) =
x3 + 1
x2 + 1
3
−3) + 1
(
f(−3) =
( −3)2 + 1
f(−1) =
ii.
2.
22
f(x) = x2 − 6x + 11, for all x ∈ R
iii.
∴
∴
4.
=
−27 + 1 −26
=
= −2.6
9 +1
10
( −1)3 + 1 −1 + 1 0
= =0
=
( −1)2 + 1 1 + 1 2
f(x) = (x − 1)(2x + 1)
f(1) = (1 − 1)[2(1) + 1] = 0
f(2) = (2 − 1)[2(2) + 1] = 5
f(−3) = (−3 −1)[2 (− 3) + 1] = 20
f(x) = 2x2 − 3x − 1
f(x + 2) = 2(x + 2)2 − 3(x + 2) − 1
= 2(x2 + 4x + 4) − 3x − 6 − 1
= 2x2 + 8x + 8 − 3x − 7
f(x + 2) = 2x2 + 5x + 1
Which of the following relations are
functions? Justify your answer. If it is a
function, determine its domain and range.
Also state the function by formula.
i.
{(2, 1), (4, 2), (6, 3), (8, 4), (10, 5),
(12, 6), (14, 7)}
ii.
{(2, 1), (3, 1), (5, 2)}
iii. {(2, 3), (3, 2), (2, 5), (5, 2)}
iv. {(0, 0), (1, 1), (1, −1), (4, 2), (4,−2),
(9, 3), (9, −3), (16, 4), (16, −4}}
Target Publications Pvt. Ltd.
Chapter 01: Sets, Relations and Functions
Solution:
i.
iv.
Let f = {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5),
(12, 6), (14, 7)}
0
1
2
3
4
5
6
7
2
4
6
8
10
12
14
1
4
9
16
From fig. every element of set A is associated
with unique element of set B
∴
It is a function
Domain = {2, 4, 6, 8, 10, 12, 14}
Range = {1, 2, 3, 4, 5, 6, 7}
Also, each element of domain is half of the
corresponding element of co-domain.
x
2
∴
Function is y =
ii.
Let f = {(2, 1), (3, 1), (5, 2)}
A
B
2
1
3
5
2
It is a function
Domain = {2, 3, 5} and Range = {1, 2}
iii.
Let f = {(2, 3), (3, 2), (2, 5), (5, 2)}
A
2
3
5
∴
∴
B
−4
−3
−2
−1
0
1
2
3
4
‘1’, ‘4’, ‘9’, ‘16’ have two images
It is not a function
5.
Find a, if f(x) = ax + 5 and f(1) = 8.
Solution:
f(x) = ax + 5 and f(1) = 8
∴
f(1) = a(1) + 5
∴
8=a+5
∴
a=3
If f(x) = f(3x − 1), for f(x) = x2 − 4x + 11,
find x.
Solution:
f(x) = x2 − 4x + 11
Also, f(x) = f(3x − 1)
x2 − 4x + 11 = (3x − 1)2 − 4(3x − 1) + 11
∴
∴
x2 − 4x = 9x2 − 6x + 1 − 12x + 4
∴
8x2 − 14x + 5 = 0
∴
8x2 − 4x − 10x + 5 = 0
∴
4x(2x − 1) − 5(2x − 1) = 0
∴
(2x − 1)(4x − 5) = 0
1
5
∴
x = or x =
2
4
6.
From fig. every element of set A is associated
with unique element of set B
∴
A
B
A
Let f = {(0,0), (1, 1), (1, −1), (4, 2), (4, −2),
(9, 3), (9, −3), (16, 4), (16, −4)}
B
2
3
5
Since ‘2’ has 2 images i.e. 3 and 5, therefore it
is not a function
If f(x) = x2 − 3x + 4, then find the value of x
satisfying f(x) = f(2x + 1).
Solution:
f(x) = x2 − 3x + 4
Also, f(x) = f(2x + 1)
7.
∴
∴
∴
∴
∴
∴
∴
x2 − 3x + 4 = (2x + 1)2 − 3(2x + 1) + 4
x2 − 3x + 4 = 4x2 + 4x + 1 − 6x − 3 + 4
3x2 + x − 2 = 0
3x2 + 3x − 2x − 2 = 0
3x(x + 1) − 2(x + 1) = 0
(x + 1)(3x − 2) = 0
2
x = −1 or x =
3
23
Target Publications Pvt. Ltd.
8.
Let A = {1, 2, 3, 4} and Z be the set of
integers. Define f: A → Z by f(x) = 3x + 7.
Show that f is a function from A to Z. Also
find the range of f.
Solution:
A = {1, 2, 3, 4}
f(x) = 3x + 7
when x = 1, f(1) = 3(1) + 7 = 10
when x = 2, f(2) = 3(2) + 7 = 13
when x = 3, f(3) = 3(3) + 7 = 16
when x = 4, f(4) = 3(4) + 7 = 19
∴
f = {(1, 10), (2, 13), (3, 16), (4, 19)}
It is a function because each element in A has
one and only one image in Z
∴
Range of f = {10, 13, 16, 19}
Std. XI Sci.: Perfect Maths - II
ii.
∴
∴
∴
∴
∴
Let f : Z → Z given by f(x) = x2 + 4 for all x ∈ Z
Let x1, x2 ∈ R be such that f(x1) = f(x2)
x12 + 4 = x22 + 4
x12 = x22
x1 = ± x2
f is not one-one function.
Here 0 ∈ co-domain Z, but there does not
exist x ∈ domain Z such that f(x) = 0
f is not onto function.
Find whether following functions are
one-one, onto or not:
i.
f: R→R given by f(x) = x3 + 5 for all
x∈R
ii.
f: Z→Z given by f(x) = x2 +4 for all
x∈Z
Solution:
i.
Let f: R → R given as f(x) = x3 + 5 for all
x ∈ R.
First we have to show that f is one-one
function
For this we have to show that
if f(x1) = f(x2), then x1 = x2
Here, f(x) = x3 + 5
Let f(x1) = f(x2)
x13 + 5 = x23 + 5
∴
x13 = x23
∴
∴
x1 = x2
∴
f is one-one function
Now we have to show that f is onto.
For that we have to prove that for any
y ∈ co-domain R, there exist an element
x ∈ domain R such that f(x) = y.
Let y ∈ R be such that
y = f(x)
∴
y = x3 + 5
x3 = y − 5
∴
x = 3 y − 5 ∈R
∴
Find which of the functions are one-one
onto, many-one onto, one-one into, manyone into. Justify your answer.
i.
f:R→R given as f(x) = 3x + 7 for all
x∈R
ii.
f: R→R given as f(x) = x2 for all x ∈ R
iii. f = {(1, 3), (2, 6), (3, 9), (4, 12)} defined
from A to B where A = {1, 2, 3, 4},
B = {3, 6, 9, 12, 15}.
Solution:
i.
f: R→R given as f(x) = 3x + 7 for all x ∈ R
First we have to show that f is one−one
For this we have to show that
if f(x1) = f(x2) then x1 = x2
Here, f(x) = 3x + 7
Let f(x1) = f(x2)
∴
3x1 + 7 = 3x2 + 7
∴
x1 = x2
∴
f is one-one function.
Now we have to show that f is onto function.
For that we have to prove that for any
y ∈ co-domain R, there exist an element
x ∈ domain R such that f(x) = y.
Let y ∈ R be such that
y = f(x)
y = 3x + 7
∴
∴
y − 7 = 3x
y−7
∴
x=
∈R
3
∴
for any y ∈ co-domain R, there exist an element
y−7
x=
∈ R such that f(x) = y.
3
∴
f is onto function.
∴
f(x) is one-one onto function.
∴
for any y ∈ co-domain R, there exist an
element x = 3 y − 5 ∈ domain R such f(x) = y.
ii.
∴
∴
f is onto function.
f is one-one onto function.
∴
∴
9.
24
10.
f: R→R given as f(x) = x2 for all x ∈ R
To find whether it is one-one or many-one
Let f(x1) = f(x2)
x12 = x22
x1 = ± x2
Target Publications Pvt. Ltd.
∴
∴
∴
∴
∴
∴
iii.
∴
∴
∴
f is not one-one function.
f is many one function
Now we have to show that f is onto function.
For that we have to prove that for any
y ∈ co-domain R, there exist an element
x ∈ domain R such that f(x) = y.
Let y ∈ R be such that
y = f(x)
y = x2
x = ± y ∉ R if y < 0
for any y ∈ co-domain R, there does not exist
an element x ∈domain R such that f(x) = y.
f is into function
Hence f is many-one into function.
f = {(1, 3), (2, 6), (3, 9), (4, 12)}
A = {1, 2, 3, 4}, B = {3, 6, 9, 12, 15}
f is defined from A to B
f: A → B
Here each and every element of A have their
distinct images in B
f is one-one function.
Also element 15 in the co-domain B don’t
have any pre-image in the domain A.
f is into function.
f it is one-one into function.
11.
Let f and g be two real valued functions
defined by f(x) = x + 1 and g(x) = 2x − 3.
Find
f
iii.
.
i.
f+g
ii.
f−g
g
Solution:
f(x) = x + 1 and g(x) = 2x − 3
i.
(f + g) x = f(x) + g(x) = x + 1 + 2x − 3 = 3x − 2
ii.
(f − g) x = f(x) − g(x) = x + 1 − (2x − 3) = 4 − x
⎛f ⎞
f ( x)
x +1
=
iii.
⎜ ⎟x =
g( x) 2 x − 3
⎝g⎠
12.
Find gof and fog, where
i.
f(x) = x − 2, g(x) = x2 + 3x + 1
x−2
1
ii.
f(x) = , g(x) =
.
x+2
x
Solution:
i.
f(x) = x − 2 and g(x) = x2 + 3x + 1
(gof)(x) = g[f(x)]
= g(x – 2)
= (x − 2)2 + 3(x − 2) + 1
= x2 − 4x + 4 + 3x − 6 + 1
= x2 − x − 1
Chapter 01: Sets, Relations and Functions
ii.
13.
(fog)x = f[g(x)] = f(x2 + 3x + 1)
= x2 + 3x + 1 − 2
= x2 + 3x − 1
1
x−2
f(x) =
and g(x) =
x+2
x
1
−2
1 − 2x
⎛1⎞
(gof) = g[f(x)] = g ⎜ ⎟ = x
=
1
⎝ x⎠
+ 2 1 + 2x
x
x+2
1
⎛ x−2⎞
(fog)x = f[g(x)] = f ⎜
⎟= x−2 =
x−2
⎝ x+2 ⎠
x+2
If f(x) =
2x + 3
, prove that fof is an identity
3x − 2
function.
Solution:
2x + 3
f(x) =
3x − 2
(fof)(x) = f[f(x)]
⎛ 2x + 3 ⎞
= f⎜
⎟
⎝ 3x − 2 ⎠
⎛ 2x + 3 ⎞
2⎜
⎟+3
3x − 2 ⎠
= ⎝
⎛ 2x + 3 ⎞
3⎜
⎟−2
⎝ 3x − 2 ⎠
4 x + 6 + 9 x − 6 13 x
=
=x
6x + 9 − 6x + 4
13
(fof)(x) = x
fof is an identity function.
=
∴
∴
3x + 2
x+2
and g(x) =
, prove
4x − 1
4x − 3
that (gof) (x) = (fog) (x) = x.
Solution:
3x + 2
x+2
f(x) =
, g(x) =
4x −1
4x − 3
14.
If f(x) =
3x + 2
+2
⎛ 3x + 2 ⎞
4x −1
(gof)(x) = g[f(x)] = g ⎜
⎟ = ⎛ 3x + 2 ⎞
⎝ 4x −1 ⎠
4⎜
⎟−3
⎝ 4x −1 ⎠
3x + 2 + 8 x − 2
11x
=x
=
=
12 x + 8 − 12 x + 3
11
⎛ x+2 ⎞
3⎜
⎟+2
4x − 3 ⎠
⎛ x+2 ⎞
⎝
(fog)(x) = f[g(x)] = f ⎜
⎟=
⎝ 4x − 3 ⎠ 4⎛ x + 2 ⎞ −1
⎜
⎟
⎝ 4x − 3 ⎠
3x + 6 + 8 x − 6 11x
=x
=
=
4x + 8 − 4x + 3
11
∴
(gof)(x) = (fog)(x) = x
25
Target Publications Pvt. Ltd.
15.
If f = {(2, 4), (3, 6), (4, 8), (5, 10), (6, 12)},
g = {(4, 13), (6, 19), (8, 25), (10, 31),
(12, 37)}, find (gof).
Solution:
f = {(2, 4), (3, 6), (4, 8), (5, 10), (6, 12)}
g ={(4, 13), (6, 19), (8, 25), (10, 31), (12, 37)}
Let A = { 2, 3, 4, 5, 6}, B = {4, 6, 8, 10, 12} and
C = {13, 19, 25, 31, 37}
∴
f(x) = 2x and g(x) = 3x+1
∴
(gof)(x) = g[f(x)] = 9(2x)
= 3(2x) + 1 = 6x+1
∴
(gof)(2) = 6(2) + 1 = 13
(gof)(3) = 6(3) + 1 = 19
(gof)(4) = 6(4) + 1 = 25
(gof)(5) = 6(5) + 1 = 31
(gof)(6) = 6(6) + 1 = 37
∴
(gof) = {(2, 13), (3, 19), (4, 25), (5, 31), (6, 37)}
Show that f:R → R given by f(x) = 3x − 4 is
one-one and onto. Find its inverse function.
Also find f−1 (9) and f−1 (−2).
Solution:
f: R→ R given by f(x) = 3x − 4
Let x1, x2 ∈ R be such that
f(x1) = f(x2)
∴
3x1 − 4 = 3x2 −4
∴
3x1 = 3x2
∴
x1 = x2
∴
f is one-one function.
Now we have to show that f is onto function.
Let y ∈ R be such that
y = f(x)
∴
y = 3x − 4
y+4
∴
x=
∈R
3
∴
for any y ∈ co-domain R, there exist an element
y+4
∈ R such that f(x) = y.
x=
3
∴
f is onto function.
∴
f is a one-one onto function.
∴
f−1 exists
y+4
∴
f −1(y) =
3
x+4
−1
∴
f (x) =
3
9
+
4 13
f −1(9) =
=
3
3
−2 + 4 2
=
f −1(−2) =
3
3
16.
26
Std. XI Sci.: Perfect Maths - II
Miscellaneous Exercise - 1
1.
Write down the following sets in set-builder
form:
i.
{10, 20, 30, 40, 50}
ii.
{a, e, i, o, u}
iii. {Sunday,
Monday,
Tuesday,
Wednesday,
Thursday,
Friday,
Saturday}
Solution:
i.
Let A = {10, 20, 30, 40, 50}
∴
A = {x/x = 10n, n ∈ N and n ≤ 5}
ii.
∴
Let B = {a, e, i, o, u}
B = {x/x is a vowel of alphabets}
iii.
Let C = {Sunday, Monday, Tuesday, Wednesday,
Thursday, Friday, Saturday}
C = {x/x represents days of a week}
∴
If U = {x/x ∈ N, 1 ≤ x ≤ 12}
A = {1, 4, 7, 10}, B = {2, 4, 6, 7, 11}
C = {3, 5, 8, 9, 12}
Write down the sets
ii.
B∩C
i.
A∪B
iii. A – B
iv. B ∩ C′
vi. A ∩ (B ∪ C)
v.
A∪B∪C
Solution:
U = {x / x ∈ N, 1 ≤ x ≤ 12} = {1, 2, 3, …., 12}
A = {1, 4, 7, 10}, B = {2, 4, 6, 7, 11},
C ={3, 5, 8, 9, 12}
i.
A ∪ B = {1, 2, 4, 6, 7, 10, 11}
ii.
B∩C={}=φ
iii. A − B = {1, 10}
iv. C′ = {1, 2, 4, 6, 7, 10, 11}
∴
B ∩ C′ = {2, 4, 6, 7, 11}
v.
A ∪ B ∪ C ={1, 2, 3, 4, 5, 6, 7, 8, 9,10, 11, 12}
vi. B ∪ C = {2, 3, 4, 5, 6, 7, 8, 9, 11, 12}
∴
A ∩ (B ∪ C) = {4, 7}
2.
3.
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 A = set of students who drink apple juice
B = set of students who drink orange juice
X = set of all students
∴
n(X) = 425, n(A) = 115, n(B) = 160,
n(A ∩ B) = 80
Target Publications Pvt. Ltd.
No. of students who neither drink apple juice nor
orange juice = n(A′ ∩ B′) = n(A ∪ B)′
= n(X) − n(A ∪ B)
= 425 − [n(A) + n(B) − n(A ∩ B)]
= 425 − (115 + 160 − 80)
= 230
4.
In a school there are 20 teachers who teach
mathematics or physics. Of these, 12 teach
mathematics and 4 teach both physics and
mathematics. How many teachers teach
physics?
Solution:
Let A = set of teachers who teach mathematics
B = set of teachers who teach physics
∴
n(A ∪ B) = 20, n(A) = 12, n(A ∩ B) = 4
But n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
∴
20 = 12 + n(B) − 4
∴
12 = n(B)
∴
No. of teachers who teach physics = 12
5.
i.
If A = {1, 2, 3} and B = {2, 4}, state the
elements of A × A, A × B, B × A,
B × B, (A × B) ∩ (B × A)
If A = {−1, 1}, find A × A × A.
ii.
Solution:
i.
A = {1, 2, 3} and B = {2, 4}
A × A = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2),
(2, 3), (3, 1), (3, 2), (3, 3)}
A × B = {(1, 2), (1, 4), (2, 2), (2, 4), (3, 2),
(3, 4)}
B × A = {(2, 1), (2, 2), (2, 3), (4, 1), (4, 2),
(4, 3)}
B × B = {(2, 2), (2, 4), (4, 2), (4, 4)}
(A × B) ∩ (B × A) = {(2, 2)}
ii.
∴
6.
A = {−1, 1}
A×A×A
= {(−1, −1, −1), (−1, −1, 1), (−1, 1, −1), (−1, 1, 1),
(1, −1, −1), (1, −1, 1), (1, 1, −1), (1, 1, 1)}
If A = {1, 2, 3}, B = {4, 5, 6}, which of the
following are relations from A to B:
i.
R1 = {(1, 4), (1, 5), (1, 6)}
ii.
R2 = {(1, 5), (2, 4), (3, 6)}
iii. R3 = {(1, 4), (1, 5), (3, 6), (2, 6), (3, 4)}
iv. R4 = {(4, 2), (2, 6), (5, 1), (2, 4)}.
Solution:
A = {1, 2, 3}, B = {4, 5, 6}
∴
A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5),
(2, 6), (3, 4), (3, 5), (3, 6)}
Chapter 01: Sets, Relations and Functions
i.
∴
ii.
∴
iii.
∴
iv.
∴
∴
R1 = {(1, 4), (1, 5), (1, 6)}
Since, R1 ⊆ A × B
R1 is a relation from A to B
R2 = {(1, 5), (2, 4), (3, 6)}
Since, R2 ⊆ A × B
R2 is a relation from A to B
R3 = {(1, 4), (1, 5), (3, 6), (2, 6), (3, 4)}
Since, R3 ⊆ A × B
R3 is a relation from A to B
R4 = {(4, 2), (2, 6), (5, 1), (2, 4)}
Since, (4, 2) ∈ R4, but (4, 2) ∉ A × B
R4 ⊄ A × B
R4 is not a relation from A to B
7.
Determine the domain and range of the
following relations:
i.
R = {(a, b) / a ∈ N, a < 5, b = 4}
ii.
S = {(a, b) / b = | a – 1|, a ∈ Z, | a | < 3}.
Solution:
i.
R = {(a, b) / a ∈ N, a < 5, b = 4}
∴
Domain (R) = {1, 2, 3, 4}
Range (R) = {4}
ii.
S = {(a, b) / b = |a − 1|, a ∈ Z, |a| < 3}
Since, a ∈ Z and |a| < 3
∴
a < 3 and a > −3
∴
−3 < a < 3
∴
a = −2, −1, 0, 1, 2
When a = −2, b = 3
When a = −1, b = 2
When a = 0, b = 1
When a = 1, b = 0
When a = 2, b = 1
∴
Domain (S) = {−2, −1, 0, 1, 2}
Range (S) = {3, 2, 1, 0}
8.
Which of the following relations are
functions? If it is a function, determine its
domain and range:
i.
{(2, 1), (4, 2), (6, 3), (8, 4), (10, 5),
(12, 6), (14, 7)}
ii.
{(0, 0), (1, 1), (1, −1), (4, 2), (4, −2),
(9, 3), (9, −3), (16, 4), (16, −4)}
iii. {(2, 1), (3, 1), (5, 2)}.
Solution:
i.
Refer Solution Ex. 1.3 Q. 4 - i
ii.
Refer Solution Ex. 1.3 Q. 4 – iv
iii. Refer Solution Ex. 1.3 Q. 4 - ii
27
Target Publications Pvt. Ltd.
9.
Std. XI Sci.: Perfect Maths - II
Find whether following functions are oneone or not :
i.
f: R → R defined by f (x) = x2 + 5
ii.
f: R − {3}
f (x) =
→
R
defined
by
5x + 7
for x ∈ R – {3}.
x−3
Solution:
i.
f: R → R defined by f(x) = x2 + 5
To prove that f is one-one we have to prove
that if f(x1) = f(x2) then x1 = x2
Here f(x) = x2 + 5
Let f(x1) = f(x2)
∴
x12 + 5 = x22 + 5
∴
x12 = x22
∴
x1 = ± x2
∴
f is not one-one function.
ii.
f: R − {3} → R defined by f(x) =
5x + 7
x−3
To prove that f is one-one we have to prove
that if f(x1) = f(x2) then x1 = x2
Here, f(x) =
5x + 7
x−3
Let f(x1) = f(x2)
∴
5 x1 + 7 5 x 2 + 7
=
x2 − 3
x1 − 3
∴
(5x1 + 7)(x2 − 3) = (5x2 + 7)(x1 − 3)
∴
5x1x2 − 15x1 + 7x2 − 21 = 5x1x2 − 15x2 + 7x1 − 21
∴
−15x1 + 7x2 = −15x2 + 7x1
∴
22x2 = 22x1
∴
x1 = x2
∴
f is one-one function.
28
10.
Find whether following functions are onto
or not:
i.
f: Z → Z defined by f(x) = 6x − 7 for
all x ∈ Z.
ii.
f: R → R defined by f(x) = x2 + 3 for
all x ∈ R.
Solution:
i.
f: Z → Z defined by f(x) = 6x − 7 for all x ∈ Z
We want to find whether f is a onto function
For that we have to prove that for any
y ∈ co-domain Z, there exist an element
x ∈ domain Z such that f(x) = y
Let y ∈ Z be such that
y = f(x)
∴
y = 6x − 7
y+7
∉Z
∴
6x = y + 7
∴
x=
6
∴
for any y ∈ co-domain Z, there does not exist
an element x ∈ domain Z such that f(x) = y
∴
f is not onto function.
ii.
f : R → R defined by f(x) = x2 + 3 for all x ∈ R
As x ∈ R, x2 ≥ 0
∴
f(x) ≥ 3
∴
x2 + 3 ≥ 3
∴
Value of all element of domain R is greater
than or equal to 3
This means that in co-domain for all the
elements which are less than 3 will not have
their pre-image in the domain.
∴
f is not onto function.
Let f: R→R be a function defined by
f(x) = 5x3 − 8 for all x ∈ R, show that f is
one-one and onto. Hence find f −1.
Solution:
f: R → R defined by f(x) = 5x3 − 8
First, we have to prove that f one-one function
For that we have to prove that if f(x1) = f(x2) then
x1 = x2
Here f(x) = 5x3 − 8
Let f(x1) = f(x2)
∴
5x13 − 8 = 5x23 − 8
∴
x1 = x2
∴
x13 = x23
∴
f is one-one function
Now we have to show that f is onto function.
Let y ∈ R be such that
y = f(x)
∴
y = 5x3 − 8
∴
5x3 = y + 8
y +8
∈R
∴
x= 3
5
11.
Target Publications Pvt. Ltd.
∴
∴
∴
∴
∴
for any y ∈ co-domain R, there exist an element
y +8
∈ domain R such that f(x) = y
x= 3
5
f is onto function.
f is one-one onto function
f−1 exists
y +8
x+8
f−1(y) = 3
∴
f−1(x) = 3
.
5
5
3x
+ 2,
5
x ∈ R. Show that f is one-one and onto.
Hence find f −1.
Solution:
3x
f: R → R defined by f(x) = + 2
5
First we have to prove that f is one-one function for
that we have to prove if f(x1) = f(x2) then x1 = x2
3x
Here f(x) = + 2
5
Let f(x1) = f(x2)
3 x1
3x
+2= 2 +2
∴
5
5
3x1 3x2
=
∴
5
5
∴
x1 = x2
∴
f is one-one function
Now, we have to that f is onto function
Let y ∈ R be such that
y = f(x)
3x
+2
∴
y=
5
3x
∴
y−2=
5
5( y − 2)
∴
x=
∈R
3
∴
for any y ∈ co-domain R, the exist an element
5( y − 2)
x=
∈ domain R such that f(x) = y
3
∴
f is onto function.
∴
f is one-one onto function.
∴
f−1 exists
5( y − 2)
∴
f−1(y) =
3
5( x − 2)
∴
f −1 (x) =
3
12.
A function f: R → R defined by f(x) =
Chapter 01: Sets, Relations and Functions
13.
A function f is defined as follows:
f(x) = 4x + 5, for −4 ≤ x < 0, find the values
of f(−1), f(−2), f(0), if they exist.
Solution:
f(x) = 4x + 5, for x ∈ [−4, 0)
− 1 ∈ [−4, 0)
∴
f(−1) = 4(−1) + 5 = 1
− 2 ∈ [−4, 0)
∴
f(−2) = 4(−2) + 5 = − 3
But 0 ∉ [−4, 0)
∴
f(0) does not exist.
14.
A function f is defined as follows:
f(x) = 5 − x for 0 ≤ x ≤ 4
Find the value of x such that
f(x) = 3 and f(x) = 5.
Solution:
f(x) = 5 − x for 0 ≤ x ≤ 4
When f(x) = 3,
5−x=3
∴
x=2
When f(x) = 5,
5−x=5
∴
x=0
15. If f(x) = 3x4 − 5x2 + 7, find f(x − 1).
Solution:
f(x) = 3x4 − 5x2 + 7
∴
f(x − 1) = 3(x − 1)4 − 5(x − 1)2 + 7
= 3(x4 − 4x3 + 6x2 − 4x + 1) − 5(x2 − 2x + 1) + 7
= 3x4 − 12x3 + 18x2 −12x + 3 − 5x2 + 10x − 5 + 7
= 3x4 − 12x3 + 13x2 − 2x + 5
16. If f(x) = 3x + a and f(1) = 7, find a and f(4).
Solution:
f(x) = 3x + a and f(1) = 7
∴
f(1) = 3(1) + a
∴
7=3+a
∴
a=4
∴
f(x) = 3x + 4
∴
f(4) = 3(4) + 4 = 16
If f(x) = ax2 + bx + 2 and f(1) = 3, f(4) = 42,
find a and b.
Solution:
f(x) = ax2 + bx + 2
f(1) = 3
∴
a(1)2 + b(1) + 2 = 3
∴
a+b+2=3
17.
29
Target Publications Pvt. Ltd.
∴
a=1−b
….(i)
and f(4) = 42
∴
a(4)2 + b(4) + 2 = 42
∴
16a + 4b = 40
∴
4a + b = 10
∴
4(1 − b) + b = 10
….[From (i)]
∴
4 − 4b + b = 10
∴
−3b = 6
∴
b=−2
Putting the value of b in (i), we get
a = 1− (−2) = 3
∴
a = 3, b = − 2
Find composite of f and g
i.
f = {(1, 3), (2, 4), (3, 5), (4, 6)}
g = {(3, 6), (4, 8), (5, 10), (6, 12)}
ii.
f = {(1, 1), (2, 2), (3, 3), (4, 4)}
g = {(1, 1), (2, 8), (3, 27), (4, 64)}
Solution:
i.
f = {(1, 3), (2, 4), (3, 5), (4, 6)}
g = {(3, 6), (4, 8), (5, 10), (6, 12)}
Let A = {1, 2, 3, 4}, B = {3, 4, 5, 6} and
C = {6, 8, 10, 12}
∴
f(x) = x + 2 and g(x) = 2x
∴
(gof) (x) = g[f(x)] = g(x + 2) = 2(x + 2) = 2x + 4
∴
(gof) (1) = 2(1) + 4 = 6
(gof) (2) = 2(2) + 4 = 8
(gof) (3) = 2(3) + 4 = 10
(gof) (4) = 2(4) + 4 = 12
∴
gof = {(1, 6), (2, 8), (3, 10), (4, 12)}
ii.
f = {(1,1), (2, 2), (3, 3), (4, 4)}
g = {(1, 1), (2, 8), (3, 27), (4, 64)}
Let A = {1, 2, 3, 4}, B = {1, 2, 3, 4} and
C = {1, 8, 27, 64}
∴
f(x) = x and g(x) = x3
∴
(gof) (x) = g[f(x)] = g(x) = x3
∴
(gof) (1) = (1)3 = 1
(gof) (2) = (2)3 = 8
(gof) (3) = (3)3 = 27
(gof) (4) = (4)3 = 64
gof = {(1, 1), (2, 8), (3, 27), (4, 64)}
Std. XI Sci.: Perfect Maths - II
Solution:
i.
f(x) = x2 + 5, g(x) = x − 8
fog(x) = f[g(x)]
= f(x – 8)
= (x − 8)2 + 5
= x2 − 16x + 64 + 5
= x2 − 16x + 69
gof(x) = g[f(x)]
= g(x2 + 5)
= x2 + 5 − 8 = x2 − 3
ii.
f(x) = 3x − 2, g(x) = x2
fog(x) = f[g(x)]
= f(x2)
= 3x2 − 2
gof(x) = g[f(x)] = [f(x)]2
= (3x − 2)2 = 9x2 − 12x + 4
iii.
f(x) = 256 x4, g(x) = x
fog(x) = f[g(x)]
18.
19.
Find fog and gof
i.
f(x) = x2 + 5, g(x) = x − 8
ii.
f(x) = 3x − 2, g(x) = x2
iii. f(x) = 256x4, g(x) = x .
30
( x)
4
= 256 ( x )
=f
= 256 x2
gof(x) = g[f(x)]
= g(256 x4)
= 256 x 4 = 16x2
2x + 1
2
,x≠ ,
5x − 2
5
show that (fof)(x) = x.
Solution:
2x + 1
f(x) =
5x − 2
(fof)x = f[f(x)]
⎛ 2x + 1 ⎞
= f⎜
⎟
⎝ 5x − 2 ⎠
⎛ 2x + 1 ⎞
2⎜
⎟ +1
5x − 2 ⎠
⎝
=
⎛ 2x + 1 ⎞
5⎜
⎟−2
⎝ 5x − 2 ⎠
4x + 2 + 5x − 2
=
10 x + 5 − 10 x + 4
9x
=
9
=x
20.
If f(x) =
Target Publications Pvt. Ltd.
Chapter 01: Sets, Relations and Functions
x+3
3 + 5x
, g(x) =
,
4x − 5
4x − 1
then show that (fog)(x) = x.
Solution:
x+3
3 + 5x
, g(x) =
f(x) =
4x − 5
4x −1
∴
(fog) x = f[g(x)]
3 + 5x
+3
⎛ 3 + 5x ⎞
x
−
4
1
=f ⎜
⎟=
⎝ 4 x − 1 ⎠ 4 ⎛ 3 + 5x ⎞ − 5
⎜
⎟
⎝ 4x −1 ⎠
3 + 5 x + 12 x − 3
17 x
4x −1
=
=
=x
12 + 20 x − 20 x + 5
17
4x −1
21.
Y
If f(x) =
Let f: R − {2} → R be defined by
x2 − 4
and g: R → R be defined by
f(x) =
x−2
g(x) = x + 2. Find whether f ≡ g or not.
Solution:
x2 − 4
f: R − {2} → R be defined by f(x) =
x−2
g: R → R be defined by g(x) = x + 2
x2 − 4
is not defined by when x − 2 = 0
f(x) =
x−2
i.e., x = 2
Here domain of f is R − {2}
But 2 ∉ domain of f
∴
f(2) does not exist
But 2 ∈ domain of g
∴
g(2) exist
and g(2) = 2 + 2 = 4
∴
f(x) ≠ g(x) ∀ x
∴
f≠g
22.
Let f: R → R be defined by f(x) = x + 5 for
all x ∈ R. Find its domain and range. Also
draw its graph.
Solution:
f(x) = x + 5
Domain of f is R
range of f is R
Let y = f(x)
i.e., y = x + 5
We have
6
(1, 6)
5 (0, 5)
4
3
2
1
(−5, 0)
X′
−6 −5 −4 −3 −2 −1
1
6
0
5
−5
0
−6
−1
O1 2
−1
(−6, −1)
3
4
X
−2
Y′
24. Let f: R → R be defined by f(x) = x3 for all x
∈ R. Find its domain and range. Also draw its
graph.
Solution:
f (x) = x3
Domain of f is R
Range of f is R
Let y = f(x) i.e., y = x3
We have
x
y = x3
2
8
1
1
−1
−1
0
0
−2
−8
Y
8
7
6
5
4
3
2
23.
x
y=x+5
y = f(x)
1
X′
−3 −2 −1
(−1, −1)
(−2, −8)
(2, 8)
y = x3
(1, 1)
O1 2 3
−1
X
−2
−3
−4
−5
−6
−7
−8
Y′
31
Target Publications Pvt. Ltd.
Additional Problems for Practice
Std. XI Sci.: Perfect Maths - II
8.
From amongst 800 individuals using internet,
55% use Facebook, 40% use Whatsapp and
35% use both sites. Find the number of
individuals who use
i.
at least one of the sites.
ii.
neither Facebook nor Whatsapp.
iii. only one of the sites.
9.
In a class, 50 students study Marathi, 40
students study English, 30 students study
Hindi, 20 students study both English and
Marathi, 16 students study both English and
Hindi. None of them study Marathi and Hindi
both. If every student study at least one
subject, find the number of students in the
class.
10.
If A = {1, 2, 3, 4}, write down the set of all
possible subsets of A, i.e., the power set of A.
11.
Write the following intervals in set-builder form:
i.
(−2, 0)
ii.
(2, 6]
iii. [−2, 5)
iv. [−1, 1]
12.
Using Venn diagrams represent:
i.
(A ∪ B ∪ C)′
ii.
A ∩ (B ∪ C)
Based on Exercise 1.1 1.
Describe the following sets in Roster form:
i.
{x / x is a letter of the word ‘APPLE’}
3⎫
⎧
ii.
⎨ x / x is an integer and − 2 < x < ⎬ .
2⎭
⎩
iii. {x / x = 2n − 1, n ∈ N}
2.
Describe the following sets in set-builder
form:
i.
{0, 1, 2, 3, ….}
ii.
{2, 3, 5, 7, 11, 13}
⎧ 1 1 1 1⎫
iii.
⎨1, , , , ⎬
⎩ 3 5 7 9⎭
3.
If A = {x / 2x2 + x − 6 = 0},
B = {x / x2 − 4 = 0},
C = {x / x2 − 3x − 10 = 0},
then find
i.
A∪B∪C
ii.
A∩B∩C
4.
If A, B, C are the sets of the letters in the
words ‘language’, ‘luggage’ and ‘drainage’
respectively, then verify that
[A − (B ∪ C)] = [(A − B) ∩ (A − C)].
5.
If A = {a, b, c, d}, B = {c, d, e, f}, C = {f, g, h, i}
and universal set U = {a, b, c, d, e, f, g, h, i, j},
then verify the following:
i.
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
ii.
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
iii. (A ∪ C)′ = A′ ∩ C′
iv. (B ∩ C)′ = B′ ∪ C′
v.
C = (A ∩ C) ∪ (A′ ∩ C)
vi. n (B ∪ C) = n(B) + n(C) − n(B ∩ C)
6.
7.
If A and B are subsets of the universal set U
and n (U) = 100, n (A) = 80, n (B) = 40,
n (A′ ∩ B′) = 10, then find
i.
n (A ∪ B)
ii.
n (A ∩ B)
iii. n (A′ ∩ B)
iv. n (A ∩ B′)
In a survey of 75 students of a class, 40 like
apple juice, 35 like orange juice, 30 like
pineapple juice, 15 like both apple and orange
juices, 12 like both apple and pineapple juices,
10 like both orange and pineapple juices and 4
like all three juices. Find the number of
students who
i.
did not like any juice.
ii.
like orange juice or pineapple juice.
32
Based on Exercise 1.2 1.
2.
If (x + 3, y − 1) = (4, 1), find the values of x
and y.
1
1⎞ ⎛1 5⎞
⎛
If ⎜ x + , y − ⎟ = ⎜ , ⎟ , find x and y.
4
2⎠ ⎝2 2⎠
⎝
3.
If A = {1, 2, 3}, B = {x, y}, then find A × B,
B × A, A × A and B × B.
4.
If A = {1, 2, 3}, B = {3, 4}, C = {4, 5}, then
find i.
(A × B) ∪ (A × C)
ii.
(A × B) ∩ (A × C)
5.
Express A = {(x, y) / x2 + y2 = 25, where x, y ∈ W}
as a set of ordered pairs.
6.
Write the domain and range of the following
relations:
i.
{(a, b) / a, b ∈ N, a < 3 and b < 2}
ii
{(a, b) / a, b ∈ N, a + b = 5}
7.
Let A = {1, 2, 3}, B = {4, 10} and
R = {(a, b) / a ∈ A, b ∈ B, a.b is odd}. Show
that R is an empty relation from A to B.
Target Publications Pvt. Ltd.
8.
Write the following relation in Roster form
and hence find its domain and range:
⎧⎛
⎫
1 ⎞
R = ⎨⎜ a,
⎟ a ∈ N,0 < a < 4 ⎬
⎩⎝ a + 1 ⎠
⎭
9.
Write the following relations as sets of
ordered pairs:
i.
{(x, y) / x + y = 5, x, y ∈ {1, 2, 3, 4}}
ii.
{(x, y) / x > y + 1, x = 4, 6, 8 and y = 1, 3}
Chapter 01: Sets, Relations and Functions
11.
Let f and g be two real valued functions
defined by f(x) = 3x − 1 and g(x) = x − 2. Find
i.
f+g
ii.
f−g
f
iii.
g
12.
Find gof and fog, where f(x) =
g(x) =
Based on Exercise 1.3 1.
3.
If f(x) =
14.
If f(x) =
Find the range of each of the following
functions:
i.
f (x) = x2 − 8x + 19, for all x ∈ R
ii.
f (x) = 5x + 4, −2 ≤ x ≤ 3
15.
If f = {(1, 3), (2, 4), (3, 5), (4, 6), (5, 7)},
g = {(3, 9), (4, 16), (5, 25), (6, 36), (7, 49)},
find (gof).
If f(x) = (x + 5) (3x − 1), then find f(1), f (−2).
16.
Show that f : R → R given by f(x) = 4x + 7 is
one-one onto. Find its inverse function. Also
find f −1 (19) and f −1 (−5).
f (x) =
25 − x 2
2
4.
If f(x) = 3x + 2x + 5, then find f (x + 1).
5.
Which of the following relations are
functions? If it is a function, determine its
domain and range. Also find the function by
formula (if possible).
i.
{(1, 2), (2, 3), (3, 4), (4, 5)}
ii.
{(1, 2), (1, 4), (2, 4), (3, 6)}
6.
If f(x) = ax − 10 and f (1) = 6, then find a.
7.
If f(x) = f(2x + 1) for f(x) = x2 − 5x + 7, then
find x.
8.
Let A = {1, 2, 3, 4, 5} and Z be the set of
integers. If f : A → Z is defined by f(x) = 2x + 1,
then show that f is a function from A to Z. Also
find the range of f.
9.
10.
5x + 1
, then prove that fof is an
x −5
identity function.
13.
Find the domain and range of the following
functions:
i.
f (x) = ( x − 2)(4 − x)
ii.
2.
x +1
.
x −1
x
and
2
Find whether the following functions are oneone, onto or not:
i.
f : R → R given by f(x) = x2
ii.
f : R → R given by f(x) = 2x + 3.
Find which of the following functions are oneone onto, many-one onto, one-one into, manyone into.
i.
f = {(1, 3), (2, 6), (3, 11), (4, 18)} defined
from A to B, where A = {1, 2, 3, 4} and
B = {3, 6, 11, 18, 27}
ii.
f : R → R given as f(x) = 2x2 + 7 for all
x ∈ R.
7x + 4
3x + 4
and g(x) =
, then show
5x − 3
5x − 7
that (fog) (x) = (gof) (x) = x.
Based on Miscellaneous Exercise − 1 1.
Write down the following sets in set-builder
form:
i.
{1, 4, 9, 16, 25, 36}
ii.
{0, ±2, ±4, ±6}
2.
If U = {x / x ∈ I, −2 ≤ x ≤ 10},
A = {−2, 0, 2, 4, 6, 8}, B = {1, 2, 3, 9, 10},
C = {3, 6, 9}, write down the following sets:
i.
A∪C
ii.
A∩B
iii. A′ ∩ C
iv. (A ∩ B) ∩ C′
3.
In a class of 180 students, 95 like English, 110
like Hindi and 30 like both subjects. How many
students neither like English nor like Hindi?
4.
Out of 40 players participating in Cricket and
Football, 18 play Cricket and 9 play both
Cricket and Football. How many players play
Football?
5.
If A = {1, 3, 4} and B = {3, 5}, find A × B,
B × B, B × A, (A × B) ∩ (B × A).
6.
If X = {a, b, c} and Y = {p, q}, which of the
following are relations from X to Y:
i.
R1 = {(a, p), (a, q), (b, p), (c, q)}
ii.
R2 = {(p, a), (a, p), (a, q), (b, q), (q, c)}
33
Target Publications Pvt. Ltd.
Std. XI Sci.: Perfect Maths - II
7.
Determine the domain and range of the
following relation:
R = {(a, b) / |a| < 2, b = |a + 1|, a ∈ Z}
8.
Which of the following relations are
functions? If it is a function, determine its
domain and range:
i.
{(1, 1) (2, 8) (3, 27) (4, 64)}
ii.
{(1, 1), (1, −1), (2, 2), (2, −2)}
9.
Find whether the following function is oneone or not:
f : R−{−2} → R defined by
4x + 3
for x ∈ R − {−2}
f(x) =
x+2
10.
Find whether the following functions are onto
or not:
i.
f : R → R defined by f(x) = 5x + 11 for
all x ∈ R
ii.
f : Z → Z defined by f(x) = 5x − 11 for
all x ∈ Z.
x+2
2 + 7x
and g(x) =
, then show
3x − 7
3x − 1
that (fog) (x) = x.
20.
If f(x) =
21.
Let f : R → R be defined by f(x) = x + 3 and
x2 − 9
.
g : R − {3} → R be defined by g(x) =
x−3
Find whether f ≡ g or not.
Multiple Choice Questions
1.
The set of intelligent students in a class is
(A) A null set
(B) A singleton set
(C) A finite set
(D) Not a well defined collection.
2.
Which of the following is the empty set?
(A) {x / x is a real number and x2 − 1 = 0}
(B) {x / x is a real number and x2 + 1 = 0}
(C) {x / x is a real number and x2 − 9 = 0}
(D) {x / x is a real number and x2 = x + 2}
11.
A function f : R → R defined by f(x) = 5 +
x
,
6
x ∈ R. Show that f is one-one and onto. Hence
find f −1 .
3.
If a set A has n elements, then the total
number of subsets of A is
(A) n
(B) n2
n
(C) 2
(D) 2n
12.
A function f is defined as follows:
f(x) = 3x + 7, for − 3 ≤ x ≤ 1.
Find the values of f(−2), f(1), f(2), if they
exist.
4.
13.
A function f is defined as follows:
f(x) = 3 + x for −2 < x < 2. Find the values of
x such that f(x) = 2 and f(x) = 4.
If A = {1, 2, 3}, B = {3, 4} C = {4, 5, 6}, then
A ∪ (B ∩ C) is
(A) {3}
(B) {1, 2, 3, 4}
(C) {1, 2, 4, 5C}
(D) {1, 2, 3, 4, 5, 6}
5.
If A and B are any two sets, then A ∪ (A ∩ B)
is equal to
(A) A
(B) B
(D) Bc
(C) Ac
6.
If the sets A and B are defined as
1
A = {(x, y) / y = , 0 ≠ x ∈ R}
x
B = {(x, y) / y = −x, x ∈ R}, then
(A) A ∩ B = A
(B) A ∩ B = B
(C) A ∩ B = (D) None of these
7.
Let n (U) = 700, n(A) = 200, n (B) = 300 and
n (A ∩ B) = 100, then n (Ac ∩ Bc) =
(A) 400
(B) 600
(C) 300
(D) 200
14.
If f(x) = 2x3 − 3x + 11, find f (x + 1).
15.
If f(x) = 2x + a and f(2) = 9, find a and f(3).
16.
If f(x) = ax2 + bx + 5 and f(1) = 12, f(2) = 21,
find a and b.
17.
Find composite of f and g and express it by
formula:
f = {(1, 4), (2, 5), (3, 6), (4, 7)}
g = {(4, 9), (5, 11), (6, 13), (7, 15)}
18.
If f(x) = 16x2 and g(x) =
19.
If f(x) =
x , find fog and gof.
3x + 1
3
, x ≠ , then show that
5x − 3
5
(fof)(x) = x.
34
Target Publications Pvt. Ltd.
8.
9.
10.
Chapter 01: Sets, Relations and Functions
In a city 20 percent of the population travels
by car, 50 percent travels by bus and 10
percent travels by both car and bus. Then
persons travelling by car or bus is
(A) 80 percent
(B) 40 percent
(C) 60 percent
(D) 70 percent
If A, B and C are any three sets, then
A × (B ∪ C) is equal to
(A) (A × B) ∪ (A × C)
(B) (A ∪ B) × (A ∪ C)
(C) (A × B) ∩ (A × C)
(D) None of these
11.
Which set is the subset of all given sets
(A) {1, 2, 3, 4,......} (B) {1}
(C) {0}
(D) {}.
12.
The smallest set A such that
A ∪ {1, 2} = {1, 2, 3, 5, 9} is
(A) {2, 3, 5}
(B) {3, 5, 9}
(C) {1, 2, 5, 9}
(D) None of these
13.
Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10},
A = {1, 2, 5}, B = {6, 7}, then A ∩ B′ is
(A) B′
(B) A
(C) A′
(D) B
14.
The shaded region in the given figure is
15.
A ∩ (B ∪ C)
A ∪ (B ∩ C)
A ∩ (B – C)
A – (B ∪ C)
19.
20.
21.
22.
A
C
17.
18.
If A = {2, 4, 5}, B = {7, 8, 9}, then n(A × B) is
equal to
(A) 6
(B) 9
(C) 3
(D) 0
(A)
(B)
(C)
(D)
16.
B
Of the members of three athletic teams in a
school 21 are in the cricket team, 26 are in the
hockey team and 29 are in the football team.
Among them, 14 play hockey and cricket, 15
play hockey and football, and 12 play football
and cricket. Eight play all the three games.
The total number of members in the three
athletic teams is
(A) 43
(B) 76
(C) 49
(D) None of these
23.
If A, B, C are three sets, then A ∩ (B ∪ C) is
equal to
(A) (A ∪ B) ∩ (A ∪ C)
(B) (A ∩ B) ∪ (A ∩ C)
(C) (A ∪ B) ∪ (A ∪ C)
(D) None of these
If A = {2, 3, 5}, B = {2, 5, 6}, then
(A – B) × (A ∩ B) is
(A) {(3, 2), (3, 3), (3, 5)}
(B) {(3, 2), (3, 5), (3, 6)}
(C) {(3, 2), (3, 5)}
(D) None of these
If n(A) = 4, n(B) = 3, n(A × B × C) = 24, then
n(C) =
(A) 288
(B) 1
(C) 12
(D) 2
If A = {1, 2, 3, 4}; B = {a, b} and f is a
mapping such that f : A → B, then A × B is
(A) {(a, 1), (3, b)}
(B) {(a, 2), (4, b)}
(C) {(1, a), (1, b), (2, a), (2, b), (3, a), (3, b),
(4, a), (4, b)}
(D) None of these
If A = {x, y}, then the power set of A is
(A) {xx, yy}
(B) {φ, x, y}
(C) {φ, {x}, {2y}}
(D) {φ, {x}, {y}, {x, y}}
Let X = {1, 2, 3, 4, 5} and Y = {1,3, 5, 7, 9}.
Which of the following is/are not relations
from X to Y?
(A) R1 = {(x, y)| y = 2 + x, x ∈ X, y ∈ Y}
(B) R2 = {(1, 1), (2, 1), (3, 3), (4, 3), (5, 5)}
(C) R3 = {(1, 1), (1, 3), (3, 5), (3, 7), (5, 7)}
(D) R4 = {(1, 3), (2, 5), (2, 4), (7, 9)}
Given two finite sets A and B such that
n(A) = 2, n(B) = 3. Then total number of
relations from A to B is
(A) 4
(B) 8
(C) 64
(D) None of these
Since every subset of A × B defines a relation
from A to B, number of relation from A to B is
equal to number of subsets of A × B = 26 = 64.
The relation R defined on the set of natural
numbers as {(a, b) / a differs from b by 3}, is
given by
(A) {(1, 4, (2, 5), (3, 6),.....}
(B) {(4, 1), (5, 2), (6, 3),.....}
(C) {(1, 3), (2, 6), (3, 9),..}
(D) None of these
35
Target Publications Pvt. Ltd.
24.
The relation R = {(1, 1), (2, 2), (3, 3), (1, 2),
(2, 3), (1, 3)} on set A = {1, 2, 3} is
(A) Reflexive but not symmetric
(B) Reflexive but not transitive
(C) Symmetric and Transitive
(D) Neither symmetric nor transitive
25.
The relation “less than” in the set of natural
numbers is
(A) Only symmetric
(B) Only transitive
(C) Only reflexive
(D) Equivalence relation
26.
Std. XI Sci.: Perfect Maths - II
31.
Let A = { 1,2,3,4} and B = {1,6,8,11,15} which
of the following are functions from A to B
i.
f: A → B defined by
f(1) = 1, f(2) = 6, f(3) = 8, f(4) = 8
ii.
f(1) = 1, f(2) = 6, f(3) = 15
iii. f(1) = 6, f(2) = 6, f(3) = 6, f(4) = 6
(A) (ii) & (iii)
(B) (i) & (ii)
(C) (ii)
(D) (i) &(iii)
32.
If f: R → R is defined by f(x) = x2 − 3x + 2,
then the value of f[f(5)] is
(A) 111
(B) 110
(C) 109
(D) 101
33.
The domain of the function:
1
is
−
2
x
3
(
)( x + 1)
Let X be a family of sets and R be a relation
on X defined by ‘A is disjoint from B’. Then
R is
(A) Reflexive
(B) Symmetric
(C) Anti-symmetric (D) Transitive
(A)
R − {− 1}
(C)
R {−1,
3
}
2
(B)
⎧3⎫
R− ⎨ ⎬
⎩2⎭
(D)
R
27.
Let A = {a, b, c} and B = {1, 2}. Consider a
relation R defined from set A to set B. Then R
is equal to set
(A) A
(B) B
(C) A × B
(D) B × A
34.
If f(x) = 3x − 5, then f−1(x) is
1
x+5
(A)
(B)
3x − 5
3
y+3
(C)
(D) does not exist
5
28.
A relation R is defined from {2, 3, 4, 5} to
{3, 6, 7, 10} by xRy
x is relatively prime
to y. Then domain of R is
(A) {2, 3, 5}
(B) {3, 5}
(C) {2, 3, 4}
(D) {2, 3, 4, 5}
35.
If f(x) =
29.
30.
Let R be a relation on N defined by x + 2y = 8.
The domain of R is
(A) {2, 4, 8}
(B) {2, 4, 6, 8}
(C) {2, 4, 6}
(D) {1, 2, 3, 4}
Let A = {1, 2, 3, 4} and R be a relation in A
given by
R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 1),
(3, 1), (1, 3)}.
Then R is
(A) Reflexive
(B) Symmetric
(C) Transitive
(D) Both (A) and (B)
36
(A)
(C)
3x + 2
for x ≠
4x − 3
17 x
4x
3
, then fof (x) is
4
(B) 3x
(D) x
36.
If f(x) = x2 + 5x + 7, then the value of x for
which f(x) = f(x + 1) is
(A) 3
(B) −6
(C) −3
(D) 6
37.
The domain of the function
(A)
(C)
(1, ∞)
(1,1)
x
is
2 + x2
(B) (−∞, 1)
(D) (−∞, ∞)
38.
The range of the function 4 − x 2 is
(A) [−3, 2]
(B) [0, 2]
(C) (0, 2)
(D) (−2, 2)
39.
If f(x) = x2 +
(A)
(C)
9
2
0
1
, x ≠ 0, then the value of f(−1) is
x
(B)
1
(D)
2
Target Publications Pvt. Ltd.
Chapter 01: Sets, Relations and Functions
Answers to Additional Practice Problems
Based on Exercise 1.1 1.
i.
ii.
iii.
{A, P, L, E}
{−1, 0, 1}
{1, 3, 5, 7, ….}
2.
i.
ii.
{x / x ∈ W}
{x / x is a prime number, x < 14}
1
⎧
⎫
, n ∈ N, 1≤ n ≤ 5 ⎬
⎨x / x =
2n − 1
⎩
⎭
iii.
3.
i.
3
⎧
⎫
⎨−2, , 2, 5⎬
2
⎩
⎭
6.
i.
iii.
90
10
ii.
iv.
30
50
7.
i.
3
ii.
55
8.
i.
iii.
480
200
ii.
320
9.
84
10.
P(A) = {φ, {1}, {2}, {3}, {4}, {1, 2}, {1, 3},
{1, 4}, {2, 3}, {2, 4}, {3, 4}, {1, 2, 3},
{1, 2, 4}, {1, 3, 4}, {2, 3, 4}, {1, 2, 3, 4}}
11.
i.
ii.
iii.
iv.
12.
i.
ii.
{− 2}
(−2, 0) = {x / x ∈ R, −2 < x < 0}
(2, 6] = {x / x ∈ R, 2 < x ≤ 6}
[−2, 5) = {x / x ∈ R, −2 ≤ x < 5}
[−1, 1] = {x / x ∈ R, −1 ≤ x ≤ 1}
A
B
C
X
Based on Exercise 1.2 1.
x = 1, y = 2
2.
x=
3.
A × B = {(1, x), (1, y), (2, x), (2, y), (3, x), (3, y)}
B × A = {(x, 1), (x, 2), (x, 3), (y, 1), (y, 2), (y, 3)}
A × A = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2),
(2, 3), (3, 1), (3, 2), (3, 3)}
B × B = {(x, x), (x, y), (y, x), (y, y)}
4.
i.
1
,y=3
4
{(1, 3), (1, 4), (1,5), (2, 3), (2, 4), (2, 5),
(3, 3), (3, 4), (3, 5)}
{(1, 4), (2, 4), (3, 4)}
ii.
5.
A = {(0, 5), (3, 4), (4, 3), (5, 0)}
6.
i.
ii.
8.
Domain = {1, 2}, Range = {1}
Domain = {1, 2, 3, 4}, Range = {4, 3, 2, 1}
⎧⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎫
R = ⎨⎜1, ⎟ , ⎜ 2, ⎟ , ⎜ 3, ⎟ ⎬ ,
⎩⎝ 2 ⎠ ⎝ 3 ⎠ ⎝ 4 ⎠ ⎭
⎧1 1 1 ⎫
Domain = {1, 2, 3}, Range = ⎨ , , ⎬
⎩2 3 4⎭
9.
i.
ii.
{(1, 4), (2, 3), (3, 2), (4, 1)}
{(4, 1), (6, 1), (8, 1), (6, 3), (8,3)}
Based on Exercise 1.3 1.
i.
Domain = [2, 4]
Range = [−1, 1]
Domain = [−5, 5]
Range = [−5, 5]
ii.
Range = [3, ∞)
Range = [− 6, 19]
2.
i.
ii.
3.
f(1) = 12, f(−2) = −21
4.
3x2 + 8x + 10
5.
i.
It is a function.
Domain = {1, 2, 3, 4}
Range = {2, 3, 4, 5}
f(x) = x + 1
Not a function.
(A ∪ B ∪ C)′
ii.
ii.
B
A
C
X
A ∩ (B ∪ C)
6.
16
7.
−1,
8.
f = {(1, 3), (2, 5), (3, 7), (4, 9), (5, 11)}
Range of f = {3, 5, 7, 9, 11}
4
3
37
Target Publications Pvt. Ltd.
Std. XI Sci.: Perfect Maths - II
9.
i.
ii.
Not one-one, not onto
One-one, onto
10.
i.
One-one into
ii.
Many-one into
11.
i.
4x − 3
3x − 1
x−2
ii.
2x + 1
iii.
16.
a = 1, b = 6
17.
gof = {(1, 9), (2, 11), (3, 13), (4, 15)}
gof (x) = 2x + 7
18.
fog(x) = 16x, gof(x) = 4x
21.
f≠g
Answers to Multiple Choice Questions
12.
(gof)(x) =
x +1
x+2
, (fog)(x) =
2( x − 1)
x−2
1.
(D)
2.
(B)
3.
(C)
4.
(B)
15.
(gof) = {(1, 9), (2, 16), (3, 25), (4, 36), (5, 49)}
5.
(A)
6.
(C)
7.
(C)
8.
(C)
16.
f −1(x) =
9.
(A)
10. (B)
11. (D)
12. (B)
13. (B)
14. (D)
15. (A)
16. (B)
17. (C)
18. (D)
19. (C)
20. (D)
x − 7 −1
, f (19) = 3, f −1 (−5) = −3
4
Based on Miscellaneous Exercise − 1 1.
2.
i.
ii.
{x / x = n2, n ∈ N, n ≤ 6}
{x / x = 2n, n ∈ Z, −3 ≤ n ≤ 3}
21. (D)
22. (C)
23. (B)
24. (A)
25. (B)
26. (B)
27. (C)
28. (D)
i.
ii.
iii.
iv.
{−2, 0, 2, 3, 4, 6, 8, 9}
{2}
{3, 9}
{2}
29. (C)
30. (D)
31. (D)
32. (B)
33. (C)
34. (B)
35. (D)
36. (C)
37. (D)
38. (B)
39. (C)
3.
5
4.
31
5.
A × B = {(1,3), (1,5), (3, 3), (3, 5), (4, 3), (4, 5)}
B × B = {(3, 3), (3, 5), (5, 3), (5, 5)}
B × A = {(3,1), (3,3), (3, 4), (5, 1), (5, 3), (5, 4)}
(A × B) ∩ (B × A) = {(3, 3)}
6.
i.
ii.
7.
Domain = {−1, 0, 1}
Range = {0, 1, 2}
8.
i.
ii.
R1 is a relation.
R2 is not a relation.
It is a function.
Domain = {1, 2, 3, 4}
Range = {1, 8, 27, 64}
Not a function.
9.
one-one
10.
i.
11.
f −1 ( x) = 6(x − 5)
12.
f(−2) = 1, f(1) = 10, f(2) does not exist.
13.
−1, 1
14.
2x3 + 6x2 + 3x + 10
15.
a = 5, f(3) = 11
38
Onto
ii.
Not onto