NCERT Exemplar Class 11 Maths Solutions Chapter 2 Relations and Functions

# NCERT Exemplar Class 11 Maths Solutions Chapter 2 Relations and Functions

Edited By Ravindra Pindel | Updated on Sep 12, 2022 04:42 PM IST

NCERT Exemplar Class 11 Maths solutions chapter 2 Relations and Functions deals with advanced operations related to sets learned in previous chapter sets. NCERT Exemplar Class 11 Maths chapter 2 solutions help define the link between two elements of different sets in a pair is also known as a relation. This could be understood with an easy example. Let’s say there are two different sets, one is A containing name of students in a class and set B contains the marks obtained by the students of the same class. So to find the result and progress of students, teachers link the marks, and names of the students which are defining a relation between the pairs of elements belonging to different sets.

## Question:1

Let and . Determine
I) AXB ii)BXA iii) BXB iv) AXA

Given data: and .

Now, let’s solve the problems one by one-
i)
ii)
iii)
iv)

Question:2

Given data:

Now,

Thus,

Question:3

If ,, find

I) (ii)

Given data:

Now, we can find,

&

i)
ii)

Question:4

i) Given data:

If & only if the corresponding coordinates are equal the two ordered pairs will be equal.

Thus, …… (i)

& ……. (ii)

From (i) & (ii), we get,

A = 11/3 & b = 2/3

ii) Given data:

If & only if the corresponding coordinates are equal the two ordered pairs will be equal.

Thus,

&

$b = -2$

Thus, a = 0 & b = -2

Question:5

Given data:

&

Now,

1. The ordered pairs that satisfying the given condition, are, .

2. The ordered pairs that satisfying the given condition, are, .

3. The ordered pairs that satisfying the given condition, are, .

Question:6

Given data:

, are the ordered pairs satisfying the condition .

Therefore, here,

&

Question:7

Given data:

Thus, domain of R1 will be

Domain =

&

Thus, the values of y will be-

$\left \{-3,-1,1,3,5,7,9,11,13,15,17\right \}$

Thus, domain of

& range of

Question:8

Given data:, where x & y Z

Thus, it is clear that 64 is the sum of the squares of 2 integers

Thus, for, x=0

Y will be = $\pm8$ & vice-versa.

Therefore, $R_2 = \left \{(0,8),(0,-8),(8,0),(-8,0)\right \}$

Question:9

Given data: where x is a real no.

Thus, domain of R3 will be equal to that of R & its range will be

Question:10

Is the given relation a function? Give reasons for your answer.

(i)

(ii)

(iii)

(iv)

(v) .

(i) Given data:

‘h’ is not a function since there are two images- 9 & 11 for the relation 3.

(ii)

Here, f is a function because every element of the domain has a unique image.

(iii)

‘g’ is a function because there is a unique image, ‘1/n’ for every element in the domain.

(iv)

‘S’ is a function since square of any integer is a unique no. & thus, for every element in the domain there is a unique image.

(V)

Here ‘t’ is a constant function, since we can observe that there is a constant no. 3 for every real element in the domain.

Question:11

If f and g are real functions defined by and , find each of the following
(a)

(b)

(c)

(d)

(e)

Given data: &

1. , (t is not equal to 5) =

$=( t^2 + 7 - 32)/t - 5$

$= (t^2 - 25)/ (t - 5)$

Question:12

Let f and g be real functions defined byand .
(a) For what real numbers x, ?

(b) For what real numbers x, ?

Given data: &

i) Now, for, f(x) = g(x),

Thus, , viz, the required real no.

ii) For, f(x) < g(x),

Thus,

Thus, , viz., the required real no.

Question:13

If f and g are two real valued functions defined as , $g (x) = x^2 + 1$, then find.

(i) f + g (ii) f - g (iii) fg (iv)f/g

Given data: & $g (x) = x^2 + 1$
i) $f + g = f(x) + g(x)$

ii) $f- g = f(x) - g(x)$

$= 2x - x^2$

iii) $f.g = f(x).g(x)$

iv) $f/8 = f(x)/g(x)$

Question:14

Express the following functions as set of ordered pairs and determine their range.

Given data:

We know that, here,

For x = -1,

For x = 0,

For x = 3,

For x = 9,

For x = 7,

Thus, are the ordered pairs &

Question:15

Find the values of x for which the functions

are equal.

Given data:

Now, it is given that -

Thus,

$x(3x - 4) + 1(3x - 4) = 0$

$x = 4/3$

Thus, -1 & 4/3 are the values of x.

Question:16

Is a function? Justify. If this is described by the relation, , then what values should be assigned to $\alpha \ and \ \beta$ ?

Given data:

Here, ‘g’ is a function since every element of the domain has unique image.

Now, for (1,1)

& for (2,3),

On solving (i) & (ii), we get,

$g(x) = 2x -1$

It is satisfying for other values of x hence it is a function.

Question:17

Find the domain of each of the following functions given by

i)

ii)

iii)

iv) $f(x) =\frac{x^3-x+3}{x^2-1}$

v)

i) Given data:
Now, we know that,

Now, for the real value of the domain,

But,
Thus, domain of

ii)Given data:
Now,
&
Now, is not defined so far, hence,
The domain =
iii)Given data:
For all x R, f(x) is defined
Thus, domain of f = R.
iv)Given data:$f(x) =\frac{x^3-x+3}{x^2-1}$
Here, only if, f(x) is defined,

Therefore, domain of
v)Given data:

F(x) is only defined at $2x - 8 \neq 0, x \neq 4$,

Thus, the domain = $R - \left \{4\right \}$.

Question:18

Find the range of the following functions given by

i)

ii)

iii)

iv)

i)Given data:
Let us consider that, y = f(x)
Thus, $y = 3/(2 - x^2)$
Thus,

$yx^2 = 2y - 3$
$x^2 = 2 - 3/y$
x is real, if
Thus,
Therefore,
Range of $f = (3/2, \infty)$
ii)Given data:
Now, we know that,

Thus, $\left |-x-2 \right | \geq 0$
Thus,
Therefore,
Range of
iii)Given data:

Now, we know that,

Thus,
Therefore, range of
iv)Given data:

Now, we know that,

$-1 \leq cos 2x \leq 1$

Thus,

Thus,

Therefore, range of

Question:19

Given data:

Now,

Now,

Now, since

Thus,

$f(x) = \left\{\begin{matrix}-2x,-3\leq x<-2 \\ 4,-2\leq x<2 \\ 2x,2\leq x\leq 3 \end{matrix}\right.$

Question:20

If , then show that:

i) ii)

Given data:
i)
$= \frac{1-x}{1+x}$
=
Thus,
ii)

Thus, .

Question:21

Let be two functions defined in the domain . Find

(i)

(ii)

(iii)

(iv)

Given data: are the two functions which are defined in the domain

Now, (i)

ii)

iii)

iv)

Question:22

Given data:

When, , f(x) is real

Thus,

Now, in order to find the range, we will put

Thus,

Therefore, for .

Therefore, range of

Question:23

Given data:

Now, let us put x = y in f(x),

f(y) = ay-b/cy-a

Thus, f(y) = $\frac{a^2x-ab-bcx+ab}{cax-bc-cax+a^2}$

=

Therefore, .

Question:24

Let n(A) = m, and n(B) = n. Then the total number of non-empty relations that can be defined from A to B is
(a) mn
(b) nm- 1
(c) mn - 1
(d) 2mn- 1

Given data: n(A) = m & n(B) = n

Thus, n(AxB) = n(A).n(B)

= mn

Thus, is the total no. of relations,
Therefore, opt (d) is correct.

Question:25

Given data:

Thus,

Thus,

Or we can say that $x \in [2,4)$

Therefore, opt (d) is the correct option.

Question:26

Range of is

A.
B.
C.
D.

Given data:

Now, we know that,

Thus,

$\\ -1 \leq 1-2 \cos x<0 \text { or } 0<1-2 \cos x \leq 3 \\\\ -1 \geq \frac{1}{1-2 \cos x}>-\infty \text { or } \infty>\frac{1}{1-2 \cos x} \geq \frac{1}{3} \\\\\ \frac{1}{1-2 \cos x} \in(-\infty,-1] \cup\left[\frac{1}{3}, \infty\right)$
Therefore, (c) is the correct option.

Question:27

Let , then
A.
B.
C.
D. None of these

Given data: ,

Thus,

i.e.,

Thus, (c) is the correct answer.

Question:28

Domain of (a>0) is
A. (- a, a)
B. [- a, a]
C. [0, a]
D. (- a, 0]

Let us take,

& f(x) is defined at

Thus,

Thus, domain of f(x) will be [-a,a]

Therefore, (b) is the correct answer.

Question:29

If, where a and b are integers, f(-1) = -5 and f(3) - 3, then a and b are equal to
(a) a = -3, b =-1
(b) a = 2, b =-3
(c) a = 0, b = 2
(d) a = 2, b = 3

Given data:

Now,

i.e.,

Thus, ......... (i)

Now,

i.e.,

Thus, ..... (ii)

From (i) & (ii), we get,

a = 2 & b = -3

Therefore, (b) is the correct answer.

Question:30

B.
C.
D.

Given data:

Now, we know that,

f(x) is defined when, $4-x\geq0\: \: and \: \: x^2-1>0$

Thus, $-x\geq-4 \: \: and \: \: (x-1)(x+1)>0$

Thus, $x\leq 4\: \: and\: \: x<-1/x>1$

Thus, the domain of f(x) =$(-\infty,-1)\cup(1,4]$

Therefore, opt (a) is the correct answer.

Question:31

The domain and range of the real function f defined by$f(x) = \frac{4-x}{x-4}$ is given by

Given data: $y=f(x) = \frac{4-x}{x-4}$

We know that, the domain of

Thus,

$yx+x = 4y+4$

$x(y+1) = 4y+4$

$x = 4(1+y)/1+y$

Now, if x is a real no. then,

Thus, $y \neq -1$

Thus, the range of

Thus, opt (c) is the correct answer.

Question:32

The domain and range of real function f defined by is given by

Given data:

f(x) is defined $x-1 \geq 0$

& domain of $f(x) = [1,\infty)$

Now, let

Thus,

Now, if x is real then y should R

Thus, Range of $f(x) = [0,\infty)$
Hence opt (d) is the correct answer.

Question:33

Given data:

Now, f(x) is defined by

Thus,

Thus,

Thus domain of

Hence the correct answer is opt (a).

Question:34

The domain and range of the function f given by is
A.
B.
C.
D.

Given data:
& f(x) is defined by

Thus, its domain is f(x) = R

$-|x-5| \leq 0$

Thus,

Thus, range of $f(x) =(-\infty,2]$

Therefore, opt (b) is the correct answer.

Question:35

Given data: and

Now, f(x) = g(x)

Thus,

Thus,

Thus,

Thus, its domain is $\{ -1, 4/3\}$

Therefore, (a) is the correct option.

Question:36

Given data:

Thus, domain of f is & that of g =

Now, domain of f.g =

Thus, the filler is

Question:37

Let and be two real functions. Then match the following:

 a f-g i b f+g ii c fxg iii d f/g iv

Given data: and

Now, f-g, f+g, f.g & f/g are defined in the domain $\{2,8,10\}$

$(f-g)(8) = -5$

$(f-g)(10) = -16$

Thus,

1. $(f+g) (2)= f(2) + g(2) = 9$

$(f+g)(8) = 3$

$(f+g)(10) = 10$

Thus,

1. $(f.g)(2) = f(2) . g(2) = 20$

$(f.g) (8)= -4$

$(f.g)(10) = -39$

Thus,

1. $(f/g)(2) = f(2)/g(2) = 4/52(f/g) = f(2)/g(2) = 4/5$

$(f/g)(8) = -1/4$

$(f/g)(10) = -3/13$

Thus,

Thus, the correct matches will be-

Question:38

The ordered pair (5,2) belongs to the relation

Given data:

Now, for (5,2),

Putting

Thus, (5,2) is not the ordered pair of R, hence it is false.

Question:39

If , then

Given data:

Now,

&

Thus, the statement is false.

Question:40

If , then

Given data:

Now,

Now,

Thus, the given statement is true.

Question:41

State True or False for the following statements

If are two equal ordered pairs, then.

Given data:

Now, & , thus,

Thus, the given statement is false.

Question:42

Given data:

Therefore, the statement is true.

## More About NCERT Exemplar Class 11 Maths Chapter 2

NCERT Exemplar Class 11 Maths chapter 2 solutions are very useful and will help the students in understanding different concepts thoroughly. NCERT Exemplar Class 11 Maths solutions chapter 2 pdf download offers a step-wise explanation of the questions making it easy for the students to understand the concept and the method easily.

## Main Subtopics in NCERT Exemplar Class 11 Maths Solutions Chapter 2 Relations and Functions

• 2.1 Overview
• 2.1.1 Cartesian Product of sets
• 2.1.2 Relations
• 2.1.3 Functions
• 2.1.4 Some specific types of functions
• i. Identity function
• ii. Constant function
• iii. Polynomial function
• iv. Rational function
• v. Modulus function
• vi. Signum function
• vii. Greatest integer function
• 2.1.5 Algebra of real sets
• i. Addition of two real functions
• ii. Subtraction of a real function from another
• iii. Multiplication by a scalar
• iv. Multiplication of two real functions
• v. Quotient of two real functions

2.2 Solved examples

## What will the students learn through NCERT Exemplar Class 11 Maths Solutions Chapter 2?

Through Class 11 Maths NCERT Exemplar solutions chapter 2, the students will learn briefly about different types of relations and functions along with different functions associated with them. NCERT Exemplar solutions for Class 11 Maths Chapter 2 will help students in further learning about sets and will help them to relate this concept to real life situations. With NCERT Exemplar Class 11 Maths solutions chapter 2, the students will understand the use of function and relation in daily routine. It can also be related to different marks a student scores during different semesters for college or different scores of exams in a year at the school of a particular student is seen as established through the use of relation and functions.

## NCERT Solutions for Class 11 Mathematics Chapters

 Chapter 1 Sets Chapter 3 Trigonometric Functions Chapter 4 Principle of Mathematical Induction Chapter 5 Complex Numbers and Quadratic Equations Chapter 6 Linear Inequalities Chapter 7 Permutations and Combinations Chapter 8 Binomial Theorem Chapter 9 Sequences and Series Chapter 10 Straight lines Chapter 11 Conic Sections Chapter 12 Introduction to Three Dimensional Geometry Chapter 13 Limits and Derivatives Chapter 14 Mathematical Reasoning Chapter 15 Statistics Chapter 16 Probability

Important topics to cover in NCERT Exemplar Class 11 Maths Solutions Chapter 2 Relations and Functions

• NCERT Exemplar solutions for Class 11 Maths chapter 2 defines a function which is denoted by f and defines relation f from set S to T as a function only if set S has only one image in set T.

• NCERT Exemplar Class 11 Maths chapter 2 solutions also defines various terms related to function including domain, co-domain and range and their identification indifferent sets of relations and functions along with the interpretation of different kinds of functions as identity function, constant function, polynomial function, modulus function, rational function, signum function, and greatest integer function.

• NCERT Exemplar Class 11 Maths solutions chapter 2 also covers different algebra for a real function includes addition, subtraction multiplication, and quotient with the help of given illustrations and examples. It also contains a variety of numerical for practice and illustration of different kinds of questions relating to relations and functions and operations associated with it.

Check Chapter-Wise NCERT Solutions of Book

 Chapter-1 Sets Chapter-2 Relations and Functions Chapter-3 Trigonometric Functions Chapter-4 Principle of Mathematical Induction Chapter-5 Complex Numbers and Quadratic equations Chapter-6 Linear Inequalities Chapter-7 Permutation and Combinations Chapter-8 Binomial Theorem Chapter-9 Sequences and Series Chapter-10 Straight Lines Chapter-11 Conic Section Chapter-12 Introduction to Three Dimensional Geometry Chapter-13 Limits and Derivatives Chapter-14 Mathematical Reasoning Chapter-15 Statistics Chapter-16 Probability

### NCERT Exemplar Class 11 Solutions

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### Frequently Asked Questions (FAQs)

1. How is this chapter helpful for higher education?

This chapter is one of the basic chapters of calculus maths and is helpful in solving many problems related to maths and physics during higher education and engineering.

2. Are these solutions available offline?

Yes, these NCERT Exemplar Class 11 Maths solutions chapter 2 are available offline as one can download these solutions through a download link.

3. How many questions are solved in these solutions?

Our team has solved 23 questions from three exercises along with 12 miscellaneous questions mentioned in the NCERT book.

4. Who has prepared these NCERT Exemplar class 11 maths solutions chapter 2?

These NCERT Exemplar Solutions for Class 11 Maths chapter 2 are prepared by our team of teachers of maths who have CBSE teaching experience of many years.

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