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.

JEE Main Scholarship Test Kit (Class 11): Narayana | Physics Wallah | Aakash | Unacademy

Suggested: JEE Main: high scoring chapters | Past 10 year's papers

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NCERT Exemplar Class 11 Maths Solutions Chapter 2: Excercise-1.3
Question:1
More About NCERT Exemplar Class 11 Maths Chapter 2
Main Subtopics in NCERT Exemplar Class 11 Maths Solutions Chapter 2 Relations and Functions
What will the students learn through NCERT Exemplar Class 11 Maths Solutions Chapter 2?
NCERT Solutions for Class 11 Mathematics Chapters

NCERT Exemplar Class 11 Maths Solutions Chapter 2: Excercise-1.3

Question:1

Let $A = \left \{ -1,2,3\right \}$ and $B = \left \{ 1,3\right \}$ . Determine
I) AXB ii)BXA iii) BXB iv) AXA

Answer:

Given data: $A = \left \{ -1,2,3\right \}$ and $B = \left \{ 1,3\right \}$ .

Now, let’s solve the problems one by one-
i) $A \times B = \left \{-1, 2, 3}{ \right \} X \left \{1, 3}{ \right \}= \left \{(-1, 1), (-1, 3), (2, 1), (2, 3), (3, 1), (3, 3) \right \}$
ii) $B \times A = \left \{1, 3}{ \right \} \times \left \{-1, 2, 3}{ \right \}= \left \{(1, -1), (3, -1), (1, 2), (3, 2), (1, 3), (3, 3)}{ \right \}$
iii) $B \times B = \left \{1, 3}{ \right \} \times \left \{1, 3}{ \right \} = \left \{(1, 1), (1, 3), (3, 1), (3, 3)}{ \right \}$
iv) $A \times A = \left \{-1, 2, 3}{ \right \} \times \left \{-1, 2, 3}{ \right \}= \left \{(-1, -1), (-1, 2), (-1, 3), (2,-1), (2, 2), (2, 3), (3, -1), (3, 2), (3, 3)}{ \right \}$

Question:2

If $P = \left \{ {x : x < 3, x \in N} \right \}$ , $Q = \left \{ {x : x \leq 2, x \in W} \right \}$ . Find $(P \cup Q) \times (P \cap Q)$ , where W is the set of whole numbers.
Answer:

Given data:

$P = \left \{x:x < 3, x \in N}{ \right \} \rightarrow P = \left \{1, 2}{ \right \}$

$Q = \left \{x:x \leq 2, x \in W}{ \right \} \rightarrow Q =\left \{ 0,1,2 \right \}$

Now, $(P \cup Q) = \left \{0, 1, 2}{ \right \} \: \: and\: \: (P \cap Q) = \left \{1, 2}{ \right \}$

Thus, $(P \cup Q) \times (P \cap Q) =\left \{ {0, 1, 2} \right \} \times \left \{ {1, 2} \right \} = \left \{ {(0, 1), (0, 2), (1, 1), (1, 2), (2, 1), (2, 2)} \right \}$

Question:3

If $A = \left \{ {x : x \in W, x < 2} \right \}$ , $B = \left \{ {x : x \in N, 1 < x < 5} \right \}$ , $C = \{ 3, 5\}$ find

I) $A \times (B \cap C)$ (ii) $A \times (B \cup C)$

Answer:

Given data:

$C = \left \{3, 5\right \}$

Now, we can find,

$(B \cap C) = \left \{3\right \}$ &

$(B \cup C) = \left \{2, 3, 4, 5\right \}$
i) $A \times (B \cap C) = \left \{0, 1\right \} \times \left \{3\right \} = \left \{(0, 3), (1, 3)\right \}$
ii) $A \times (B \cup C) = \left \{0, 1\right \} \times \left \{2, 3, 4, 5\right \} = \left \{(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)\right \}$

Question:4

In each of the following cases, find a and b.
(i) $(2a + b, a - b) = (8, 3)$
(ii) $(a/4 , a - 2b) = (0, 6 + b)$

Answer:

i) Given data:
$(2a + b, a - b) = (8, 3)$
If & only if the corresponding coordinates are equal the two ordered pairs will be equal.

Thus, $2a + b = 8$ …… (i)

& $a - b = 3$ ……. (ii)

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

A = 11/3 & b = 2/3

ii) Given data:

$(a/4, a - 2b) = (0, 6 + b)$

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

Thus, $a/4 = 0 \rightarrow a = 0$

& $a - 2b = 6 + b \rightarrow a � 3b = 6$

$0 - 3b = 6$

$b = -2$

Thus, a = 0 & b = -2

Question:5

Given $A = \left \{1, 2, 3, 4, 5\right \}$ , $S = \left \{(x, y) : x \in A, y \in A\right \}$ . Find the ordered pairs which satisfy the conditions given below:
(i) $x + y = 5$
(ii) $x + y < 5$
(iii) $x + y > 8$

Answer:

Given data:

$A = \left \{1, 2, 3, 4, 5\right \}$ &

$S = \left \{(x, y) : x \in A, y \in A\right \}$

Now,

The ordered pairs that satisfying the given condition, $x + y = 5$ are, $(1,4), (4,1), (2,3) \& (3,2)$ .
The ordered pairs that satisfying the given condition, $x + y < 5$ are, $(1,1), (1,2), (2,1), (1,3), (2,2), (3,1)$ .
The ordered pairs that satisfying the given condition, $x + y > 8$ are, $(4,5), (5,4), (5,5)$ .

Question:6

Given $R = \left \{(x, y) : x, y \in W, x^2 + y^2 = 25\right \}$ . Find the domain and Range of R.

Answer:

Given data:

$R = \left \{(x,y) : x, y \in W, x^2 + y^2 = 25\right \}$

$(0,5), (3,4), (5,0) \& (4,3)$ , are the ordered pairs satisfying the condition $x^2 + y^2 = 25$ .

Therefore, here,

$Domain = \left \{0,3,4,5\right \}$ &

$Range = \left \{0,3,4,5\right \}$

Question:7

If $R1 = \left \{(x, y) | y = 2x + 7, where\: \: x \in R \: \: and - 5 \leq x \leq 5\right \}$ is a relation. Then find the domain and Range of $R1$ .

Answer:

Given data: $R1 = \left \{(x, y) / y = 2x + 7 \: \: where\: \: x \in R \and -5 \leq x \leq 5\right \}$

Thus, domain of R₁ will be $\left \{x: -5 \leq x \leq 5 = [-5,5] \right \}$

Domain = $\left \{-5,-4,-3,-2,-1,0,1,2,3,4,5\right \}$

& $y = 2x + 7$

Thus, the values of y will be-

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

Thus, domain of $R_1 = [-5,5]$

& range of $R_1 = [-3,17]$

Question:8

If $R_2 = \left \{(x, y) | \: \: x \: \: and \: \: y \: \: are \: \: integers \: \: and\: \: x^2 + y^2 = 64\right \}$ is a relation. Then find $R_2$ .

Answer:

Given data: $x^2 + y^ 2 = 64$ , where x & y $\in$ 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

If $R3 = \left \{(x, |x|) |x \: \: is\: \: a\: \: real\: \: number\right \}$ is a relation. Then find domain and range of R₃.

Answer:

Given data: $R_3 = \left \{(x,|x|\right \}$ where x is a real no.

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

$R_3 = (o, \infty) ... (since,\left|x\right| = R_+]$

Question:10

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

(i) $h = \left \{(4, 6), (3, 9), (- 11, 6), (3, 11)\right \}$

(ii) $f = \left \{(x, x) | x \: \: is \: \: a \: \: real\: \: number\right \}$

(iii) $g = n, (1/n) |n \: \: is\: \: a\: \: positive\: \: integer$

(iv) $s = \left \{(n, n2) | n \: \: is\: \: a\: \: positive\: \: integer\right \}$

(v) $t = \left \{(x, 3) | x \: \: is \: \: a \: \: real\: \: number\right \}$ .

Answer:

(i) Given data: $h = \left \{(4,6),(3,9),(-11,6),(3,11)\right \}$

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

(ii) $f = \left \{(x,x)/ x \: \: is\: \: a\: \: real\: \: no.\right \}$

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

(iii) $g = \left \{(n, 1/n)/ n \: \: is\: \: a\: \: positive\: \: integer\right \}$

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

(iv)
$S = \left \{(n, n2)/ n \: \: is\: \: a\: \: positive\: \: integer.\right \}$

‘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) $T = \left \{(x,3)/ x \: \: is\: \: a\: \: real\: \: number\right \}$

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 $f (x) = x^2 + 7$ and $g (x) = 3x + 5$ , find each of the following
(a) $f (3) + g (- 5)$

(b) $f(1/2) \times g(14)$

(d) $f (t) - f (- 2)$

(e) $(f(t) - f(5))/ (t - 5), if \: \: t \neq 5$

Answer:

Given data: $f(x) = x^2 + 7$ & $g(x) = 3x + 5$

$f(3) + g(-5) = [(3)^2 + 7] + [3(-5) + 5]$

$= 16 - 10$

$= 6$

$f( 1/2 ) \times g(14) = [(1/2)^2 + 7] \times [3 \times 14 + 5]$

$= 29/4 \times 47$

$= 1363/4$

$f(-2) + g(-1) = [(-2)^2 + 7] + [(-1) + 5]$

$= 11 + 2 = 13$

$f(t) - f(-2) = (t^2 + 7) - [(-2)^2 + 7]$

$= t^2 - 4$

$f(t) - f(5)/ t - 5$ , (t is not equal to 5) = $(t^2 + 7) - (5^2 + 7)/ t - 5$

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

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

$=(t - 5) (t + 5)/(t - 5)$

$= t + 5$

Question:12

Let f and g be real functions defined by $f (x) = 2x + 1$ and $g (x) = 4x - 7$ .
(a) For what real numbers x, $f (x) = g (x)$ ?

(b) For what real numbers x, $f (x) < g (x)$ ?

Answer:

Given data: $f(x) = 2x + 1$ & $g(x) = 4x -7$

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

$2x + 1 = 4x - 7$

$-2x = -8$

Thus, $x = 4$ , viz, the required real no.

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

$2x + 1 < 4x - 7$

$-2x < -8$

Thus, $2x > 8$

Thus, $x > 4$ , viz., the required real no.

Question:13

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

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

Answer:

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

$= 2x+1+x^2+1$

$= x^2 + 2x + 2$
ii) $f- g = f(x) - g(x)$

$= (2x + 1) - (x^2 + 1)$

$= 2x - x^2$

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

$= (2x + 1) (x^2 + 1)$

$= 2x^3 + x^2 + 2x + 1$

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

$= 2x + 1/x^2 + 1$

Question:14

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

$f: X \rightarrow R, f (x) = x^3 + 1, where X = \left \{-1, 0, 3, 9, 7\right \}$

Answer:

Given data: $F: x \rightarrow R, f(x) = x^3 + 1, where, x = \left \{-1,0,3,9,7\right \}$

We know that, here, $x = \left \{-1,0,3,9,7\right \}$

For x = -1,

$f(-1)^3 + 1 = 0$

For x = 0,

$f(0) = (0)^3 + 1 = 1$

For x = 3,

$f(3) = (3)^3 + 1 = 28$

For x = 9,

$F(9) = (9)^3 + 1 = 730$

For x = 7,

$F(7) = (7)^3 + 1 = 344$

Thus, $(-1,0),(0,1),(3,28),(7,344) \& (9,730)$ are the ordered pairs &

$Range = \left \{0,1,28,344,730\right \}.$

Question:15

Find the values of x for which the functions

$f (x) = 3x^2 - 1 \: \: and \: \: g (x) = 3 + x$ are equal.

Answer:

Given data: $f (x) = 3x^2 - 1 \: \: and \: \: g (x) = 3 + x$

Now, it is given that - $f(x) = g(x),$

Thus, $3x^2 - 1 = 3 + x$

$3x^2 - x - 4 = 0$

$3x^2 - 4x + 3x - 4 = 0$

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

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

$3x - 4 = 0 \: \: or\: \: x + 1= 0$

$3x = 4 \: \: or\: \: x = -1$

$x = 4/3$

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

Question:16

Is $g = \left \{(1, 1), (2, 3), (3, 5), (4, 7)\right \}$ a function? Justify. If this is described by the relation, $g (x) = \alpha x + \beta$ , then what values should be assigned to $\alpha \ and \ \beta$ ?

Answer:

Given data:

$g = \left \{(1,1),(2,3),(3,5),(4,7)\right \}$

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

$g(x) = \alpha x + \beta ... (given)$

Now, for (1,1)

$g(1) = \alpha (1) + \beta = 1$

$\alpha + \beta = 1 ....... (i)$

& for (2,3),

$g(2) = \alpha (2) + \beta = 3$

$2\alpha + \beta = 3 ........ (ii)$

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

$\alpha = 2 \: \: and\: \: \beta = -1.$

$g(x) = 2x -1$

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) $f(x) = \frac{1}{\sqrt{1-\cos x}}$

ii) $f(x) = \frac{1}{\sqrt{x+\left | x \right |}}$

iii) $f(x) =x\left | x \right |$

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

v) $f(x) =\frac{3x}{2x-8}$

Answer:

i) Given data: $f(x) = \frac{1}{\sqrt{1-\cos x}}$
Now, we know that,
$-1 \leq cosx \leq 1$
$1 \geq -cosx \geq -1$
$1 + 1 \geq 1 - cos x \geq -1+1$
$2 \geq 1 - cos x \geq 0$
$0 \leq 1 - cos x \leq 2$
Now, for the real value of the domain,
$1 - cos x\neq0,cosx \neq 1$
But, $x \neq 0\: \: 2n\pi \: \: \forall n \in Z$
Thus, domain of $f = R - \left \{2n\pi, n \in Z\right \}$

ii)Given data: $f(x) = \frac{1}{\sqrt{x+\left | x \right |}}$
Now, $x + \left |x \right | = x+x = 2x ...... if \: \: x \geq0$
& $x + \left |x \right | = x - x = 0 ...... if \: \: x < 0$
Now, $x<0$ is not defined so far, hence,
The domain = $R^+$
iii)Given data: $f(x) =x\left | x \right |$
For all x $\in$ 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 $x^2 - 1 \neq 0$ , f(x) is defined,
$(x-1)(x+1) \neq 0,$
$Thus, x \neq 1$
$and\: \: x \neq -1$
Therefore, domain of $f = R - \left \{-1,1\right \}$
v)Given data: $f(x) =\frac{3x}{2x-8}$

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) $f\left ( x \right )= \frac{3}{2-x^2}$

ii) $f\left ( x \right )= 1-\left | x-2 \right |$

iii) $f\left ( x \right )= \left | x-3 \right |$

iv) $f\left ( x \right )= 1+3\cos 2x$

Answer:

i)Given data: $f\left ( x \right )= \frac{3}{2-x^2}$
Let us consider that, y = f(x)
Thus, $y = 3/(2 - x^2)$
Thus,
$y(2 - x^2) = 3$
$2y - yx^2 = 3$
$yx^2 = 2y - 3$
$x^2 = 2 - 3/y$
x is real, if $2y - 3 \geq0 \: \: and\: \: y \geq0$
Thus, $y \geq 3/2$
Therefore,
Range of $f = (3/2, \infty)$
ii)Given data: $f\left ( x \right )= 1-\left | x-2 \right |$
Now, we know that,
$\left |x-2 \right | = -(x-2) , if x < 2, \: \: and\: \: \left | x-2 \right |= (x-2), if \geq 2$
Thus, $\left |-x-2 \right | \geq 0$
Thus, $1 -\left | x-2 \right | \leq 1$
Therefore,
Range of $f = (-\infty, 1)$
iii)Given data: $f\left ( x \right )= \left | x-3 \right |$

Now, we know that,

$\left |x-3 \right | \geq 0$

Thus, $f(x) = 0$
Therefore, range of $f = (0,\infty)$
iv)Given data: $f\left ( x \right )= 1+3\cos 2x$

Now, we know that,

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

Thus, $-3 \leq 3 cos 2x \leq 3$

$-3 + 1 \leq 1 + 3 cos 2x \leq 3 + 1$

Thus, $-2 \leq 1 + 3 cos 3x \leq 4$

$-2 \leq f(x) \leq 4$

Therefore, range of $f = [-2, 4].$

Question:19

Redefine the function $f(x) = |x - 2| + |2 + x|, - 3 \leq x \leq 3$
Answer:

Given data: $f(x) = |x - 2| + |2 + x|, - 3 \leq x \leq 3$

Now, $\left |x-2 \right | = (x-2), x \geq 2, since\: \: \left |x-2 \right | = -(x-2), x < 2$

Now, $\left |2+x \right |= -(2+x), x < -2$

$\left |2+x \right |= (2 + x), x \geq -2$

Now, since

$f(x) = x-2+ x-3, -3\leq x\leq 3$

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

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 $f(x)=\frac{x-1}{x+1}$ , then show that:

i) $f(\frac{1}{x})=-f(x)$ ii) $f(-\frac{1}{x})=\frac{-1}{f(x)}$

Answer:

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

Question:21

Let $f(x) = \sqrt{x}\: \: and\: \: g (x) = x$ be two functions defined in the domain $R^+\cup \left \{0\right \}$ . Find

(i) $(f + g) (x)$

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

(iii) $(fg) (x)$

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

Answer:

Given data: $f(x) = \sqrt{x}\: \: and\: \: g (x) = x$ are the two functions which are defined in the domain $R^+ \cup \left \{0\right \}$

Now, (i) $(f+g)(x) = f(x) + g(x)$

$=\sqrt{x} + x$

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

$=\sqrt{x} - x$

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

$= \sqrt{x}.x$

$= x^{3/2}$

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

$= \sqrt{x}/x$

$= 1/\sqrt{x}$

Question:22

Find the domain and Range of the function $f (x) = 1/\sqrt{(x-5)}$ .
Answer:

Given data: $f (x) = 1/\sqrt{(x-5)}$

When, $x-5>0, i.e., x>5$ , f(x) is real

Thus, $domain = (5, \infty)$

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

$y = f(x) = 1/\sqrt{x-5}$

Thus, $\sqrt{x-5 }= 1/y$

$x-5 = 1/y^2$

$x = 1/y^2 + 5$

Therefore, for $x \in (5, \infty), y \in R$ .

Therefore, range of $f = R^+$

Question:23

If $f( x)= y = ax-b/ cx-a$ then prove that $f (y) = x$

Answer:

Given data: $f( x)= y = ax-b/ cx-a$

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

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

$= \frac{a\frac{\left |ax-b \right |}{\left | cx-a \right |}-b}{c\frac{\left |ax-b \right |}{\left | cx-a \right |}-a}$

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

= $\frac{a^2x-bcx}{a^2-bc}$

$= \frac{x(a^2^-bc)}{a^2-bc}$

$=x$

Therefore, $f(y) = x$ .

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) n^m- 1
(c) mn - 1
(d) 2^mn- 1

Answer:

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

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

= mn

Thus, $2^{mn-1}$ is the total no. of relations,
Therefore, opt (d) is correct.

Question:25

If $[x]^2 - 5[x] + 6 = 0$ , where [. ] denote the greatest integer function, then
(a) $x \in [3,4]$
(b) $x\in (2, 3]$
(c) $x\in [2, 3]$
(d) $x \in [2, 4)$

Answer:

$x \in [2,3]$ Given data: $[x]^2 - 5[x] + 6 = 0$

Thus, $[x]^2 - 3[x]2[x] + 6 = 0$

$[x]([x]-3) - 2([x]-3) = 0$

$([x]-3)([x]-2) = 0$

Thus, $[x] = 2,3$

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

Therefore, opt (d) is the correct option.

Question:26

Range of $f\left ( x \right )=\frac{1}{1-2\cos x}$ is

A. $\left [ \frac{1}{3},1 \right ]$
B. $\left [-1, \frac{1}{3} \right ]$
C. $\left ( -\infty ,-1 \right ]\cup \left [ -\frac{1}{3},1 \right )$
D. $\left [ -\frac{1}{3},1 \right ]$

Answer:

Given data: $f\left ( x \right )=\frac{1}{1-2\cos x}$

Now, we know that,

$-1 \geq \cos x \geq1$

Thus, $-1\leq cosx \leq 1$

$-2 \leq -2cosx \leq 2$

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

Question:27

Let $f(x)=\sqrt{(1+x^2 )}$ , then
A. $f (xy) = f (x) . f (y)$
B. $f (xy) \geq f (x) . f (y)$
C. $f (xy) \leq f (x) . f (y)$
D. None of these

Answer:

Given data: $f(x)=\sqrt{(1+x^2 )}$ ,
$f(xy) = \sqrt{(1+x^2 y^2 )}$
$f(x).f(y) = \sqrt{(1+x^2 )}. \sqrt{(1+x^2 y^2 )}$
$=\sqrt{1+x^2+y^2+ x^2 y^2 }$
Thus, $\sqrt{1+x^2 y^2 } \leq \sqrt{1+x^2+y^2+ x^2 y^2 }$

i.e., $f(xy)\leq f(x).f(y)$

Thus, (c) is the correct answer.

Question:28

Domain of $\sqrt{a^2-x^2 }$ (a>0) is
A. (- a, a)
B. [- a, a]
C. [0, a]
D. (- a, 0]

Answer:

Let us take,

$f(x) = a^2-x^2$

& f(x) is defined at $a^2-x^2\geq 0$

$x^2-a^2 \leq 0$

$x^2\leq a^2$

Thus, $-a\leq x\leq a$

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

Therefore, (b) is the correct answer.

Question:29

If $f(x)= ax+ b$ , 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

Answer:

Given data: $f(x)= ax+ b$

Now, $f(-1) = a(-1) + b$

i.e., $-5 = -a +b$

Thus, $a-b = 5$ ......... (i)

Now, $f(3) = 3a + b$

i.e., $3 = 3a +b$

Thus, $3a + b = 3$ ..... (ii)

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

a = 2 & b = -3

Therefore, (b) is the correct answer.

Question:30

The domain of the function f defined by $f(x)=\sqrt{4-x}+\frac{1}{\sqrt{x^2-1}}$ is equal to
A. $(- \infty, - 1) \cup (1, 4]$
B. $(- \infty, - 1] \cup (1, 4]$
C. $(- \infty, - 1) \cup [1, 4]$
D. $(- \infty, - 1) \cup [1, 4)$

Answer:

Given data: $f(x)=\sqrt{4-x}+\frac{1}{\sqrt{x^2-1}}$

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

1. $Domain = R, Range = \left \{-1, 1\right \}$
2. $Domain = R - \left \{1\right \}, Range = R$
3. $Domain = R - \left \{4\right \}, Range = \left \{- 1\right \}$
4. $Domain = R - \left \{- 4\right \}, Range = \left \{-1, 1\right \}$

Answer:

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

We know that, the domain of $f(x) = R-\left \{4\right \}$

Thus, $yx - 4y = 4-x$

$yx+x = 4y+4$

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

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

Now, if x is a real no. then,

$1+y \neq 0$

Thus, $y \neq -1$

Thus, the range of $f(x) = R - (-1)$

Thus, opt (c) is the correct answer.

Question:32

The domain and range of real function f defined by $f (x) =\sqrt{ x-1}$ is given by

1. $Domain = (1, \infty), Range = (0, \infty)$
2. $Domain = [1, \infty), Range = (0, \infty)$
1. $Domain = [1, \infty), Range = [0, \infty)$
2. $Domain = [1, \infty), Range = [0, \infty)$

Answer:

Given data: $f (x) =\sqrt{ x-1}$

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

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

Now, let $y = f(x) =\sqrt{ x-1}$

$y^2 = x - 1$

Thus, $x = y^2 + 1$

Now, if x is real then y should $\in$ R

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

Question:33

The domain of the function f given by $f(x) = (x^2+2x+1)/(x^2-x-6)$
A. $R - \left \{3, - 2\right \}$
B. $R - \left \{-3, 2\right \}$
C. $R - [3, - 2]$
D. $R - (3, - 2)$

Answer:

Given data: $f(x) = (x^2+2x+1)/(x^2-x-6)$

Now, f(x) is defined by $x^2-x-6=0$

Thus, $x^2-3x+2x-6\neq0$

$(x-3)(x+2) \neq 0$

Thus, $x \neq -2 or 3$

Thus domain of $f(x) = R-\left \{-2,3\right \}$

Hence the correct answer is opt (a).

Question:34

The domain and range of the function f given by $f (x) = 2 - \left |x -5 \right |$ is
A. $Domain = R+, Range = ( - \infty, 1]$
B. $Domain = R, Range = ( - \infty, 2]$
C. $Domain = R, Range = (- \infty, 2)$
D. $Domain = R+, Range = (- \infty, 2]$

Answer:

Given data: $f (x) = 2 - \left |x -5 \right |$
& f(x) is defined by $x \in R$

Thus, its domain is f(x) = R

$\left |x-5 \right | \geq 0$

$-|x-5| \leq 0$

$2-\left |x-5 \right |\leq 2$

Thus, $f(x) \leq 2$

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

Therefore, opt (b) is the correct answer.

Question:35

The domain for which the functions defined by $f (x) = 3x^2 - 1$ and $g (x) = 3 + x$ are equal is

A. $\left \{ -1,\frac{4}{3} \right \}$
B. $\left [ -1,\frac{4}{3} \right ]$
C. $\left ( -1,\frac{4}{3} \right )$
D. $\left [ -1,\frac{4}{3} \right )$

Answer:

Given data: $f (x) = 3x^2 - 1$ and $g (x) = 3 + x$

Now, f(x) = g(x)

Thus,

$3x^2 - 1 = 3 + x$

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

Thus, $x+1 = 0 or 3x-4 = 0$

Thus, $x = -1 or x = 4/3$

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

Therefore, (a) is the correct option.

Question:36

Let f and g be two real functions given by $f= \left \{(0, 1), (2,0), (3,.-4), (4,2), (5, 1)\right \}$
$g= \left \{(1,0), (2,2), (3,-1), (4,4), (5, 3)\right \}$ then the domain of $f \times g$ is given by________ .

Answer:

Given data: $f = \left \{(0,1),(2,0),(3,-4),(4,2),(5,1)\right \} \& g = \left \{(1,0),(2,2),(3,-1),(4,4),(5,3)\right \}$

Thus, domain of f is $\left \{0,2,3,4,5\right \}$ & that of g = $\left \{1,2,3,4,5\right \}$

Now, domain of f.g = $\left \{2,3,4,5\right \}$

Thus, the filler is $\left \{2,3,4,5\right \}.$

Question:37

Let $f= \left \{(2,4), (5,6), (8, -1), (10, -3)\right \}$ and $g = \left \{(2, 5), (7,1), (8,4), (10,13), (11, 5)\right \}$ be two real functions. Then match the following:

a	f-g	i	$\left \{ \left ( 2,\frac{4}{5} \right ),\left ( 8,\frac{-1}{4} \right ),\left ( 10,\frac{-3}{13} \right ) \right \}$
b	f+g	ii	$\left \{ \left ( 2,20 \right ),\left ( 8,-4 \right ),\left ( 10,-39 \right ) \right \}$
c	fxg	iii	$\left \{ \left ( 2,-1 \right ),\left ( 8,-5 \right ),\left ( 10,-16 \right ) \right \}$
d	f/g	iv	$\left \{ \left ( 2,9 \right ),\left ( 8,3 \right ),\left ( 10,10 \right ) \right \}$

Answer:

Given data: $f= \left \{(2,4), (5,6), (8, -1), (10, -3)\right \}$ and $g = \left \{(2, 5), (7,1), (8,4), (10,13), (11, 5)\right \}$
$Domain of f(x)$ is \{2,5,8,10\}, Domain of g(x)$ is \{2,7,8,10,11\}$

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

$(f-g)2 = f(2) - g(2) = -1$

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

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

Thus, $(f-g) = \left \{(2,-1),(8,-5),(10,16)\right \}$

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

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

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

Thus, $(f+g) = \left \{(2,9),(8,3),(10,10)\right \}$

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

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

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

Thus, $(f.g) = \left \{(2,20),(8,-4),(10,-39)\right \}$

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

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

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

Thus, $V = \left \{(2,4/5), (8,(-1/4), (10,-3/13)\right \}$

Thus, the correct matches will be-

$(a)\rightarrow (iii), (b)\rightarrow (iv), (c) \rightarrow (ii) \& (d) \rightarrow (i)$

Question:38

The ordered pair (5,2) belongs to the relation $R =\left \{(x,y): y = x - 5, x,y \in Z \right \}$

Answer:

Given data: $R =\left \{(x,y): y = x - 5, x,y \in Z \right \}$

Now, for (5,2),

$Y = x-5$

Putting $x = 5 \: \: and\: \: y = 5-5 = 0\neq2$

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

Question:39

If $P = \left \{1, 2\right \}$ , then $P \times P \times P = \left \{(1, 1,1), (2,2, 2), (1, 2,2), (2,1, 1)\right \}$

Answer:

Given data: $P = \left \{1, 2\right \}$

Now, $P \times P = \left \{1,2\right \} \times \left \{1,2\right \}$

$= \left \{(1,1),(1,2),(2,1),(2,2)\right \}$

& $P \times P \times P = \left \{1,2\right \} \times \left \{1,2\right \} \times \left \{1,2\right \}$

$= \left \{(1,1,1),(1,1,2),(1,2,1),(1,2,2),(2,1,1),(2,1,2),(2,2,1),(2,2,2)\right \}$

Thus, the statement is false.

Question:40

If $A= \left \{1,2, 3\right \}, 5= \left \{3,4\right \} and\: \: C= \left \{4, 5, 6\right \}$ , then $(A \times B) \cup (A \times C) = \left \{(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3,4), (3, 5), (3,6)\right \}.$

Answer:

Given data: $A= \left \{1,2, 3\right \}, 5= \left \{3,4\right \} and\: \: C= \left \{4, 5, 6\right \}$

Now, $A \times B = \left \{(1,3),(1,4),(2,3),(3,3),(3,4)\right \}$

$A \times C = \left \{(1,4),(1,5),(1,6),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6)\right \}$

Now, $(A \times B) \cup (A \times C) = \left \{(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,3),(3,4),(3,5),(3,6)\right \}.$

Thus, the given statement is true.

Question:41

State True or False for the following statements

If $(x - 2, y + 5) = \left ( -2,\frac{1}{3} \right )$ are two equal ordered pairs, then $x = 4, y=\frac{-14}{3}$ .

Answer:

Given data: $(x - 2, y + 5) = \left ( -2,\frac{1}{3} \right )$

Now, $x - 2 = -1, i.e., x = 0$ & $y +5 = 1/3$ , thus, $y = -14/3$

Thus, the given statement is false.

Question:42

If $A\times B= \left \{(a, x), (a, y), (b, x), (b, y)\right \}$ , then $M = \left \{a, b\right \},B= \left \{x, y\right \}.$

Answer:

Given data: $A = \left \{a,b\right \} , B = \left \{x,y\right \}$

$A \times B = \left \{(a,x),(a,y),(b,x),(b,y)\right \}$

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

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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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NCERT Exemplar Class 11 Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry

NCERT Exemplar Class 11 Maths Solutions Chapter 10 Straight Lines

NCERT Exemplar Class 11 Maths Solutions Chapter 6 Linear Inequalities

NCERT Exemplar Class 11 Maths Solutions Chapter 16 Probability

NCERT Exemplar Class 11 Maths Solutions Chapter 14 Mathematical Reasoning

NCERT Exemplar Class 11 Maths Solutions Chapter 7 Permutations and Combinations

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Exemplar Class 11 Maths Solutions Chapter 2 Relations and Functions

NCERT Exemplar Class 11 Maths Solutions Chapter 2: Excercise-1.3

Question:1

More About NCERT Exemplar Class 11 Maths Chapter 2

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NCERT Solutions for Class 11 Mathematics Chapters

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