NCERT Exemplar Class 11 Maths Solutions Chapter 2 Relations and Functions

NCERT Exemplar Class 11 Maths Solutions Chapter 2 Relations and Functions

Question 1

Let $A = \{ -1,2,3\}$ and $B = \{ 1,3\}$. Determine
i) $A\times B$ ii)$B\times A$ iii) $B\times B$ iv) $A\times A$

Answer:

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

Now, let’s solve the problems one by one.

i) $A \times B = \{-1, 2, 3 \} \times \{1, 3\}= \{(-1, 1), (-1, 3), (2, 1), (2, 3), (3, 1), (3, 3)\}$

ii) $B \times A = \{1, 3\} \times \{-1, 2, 3\}= \{(1, -1), (3, -1), (1, 2), (3, 2), (1, 3), (3, 3)\}$

iii) $B \times B = \{1, 3\} \times \{1, 3\} = \{(1, 1), (1, 3), (3, 1), (3, 3)\}$

iv)$A \times A =\{-1, 2, 3\} \times \{-1, 2, 3\}= \{(-1, -1), (-1, 2), (-1, 3), (2,-1), (2, 2), (2, 3), (3, -1), (3, 2), (3, 3)\}$

Question 2

If $P = \ \{ {x : x < 3, x \in N} \ \}$, $Q = \ \{ {x : x \leq 2, x \in W} \ \}$. Find $(P \cup Q) \times (P \cap Q)$, where W is the set of whole numbers.
Answer:
Given data:
$P = \{x: x\leq 3, x \in N \} \Rightarrow P = \{1, 2 \}$
$Q = \{x:x \leq 2, x \in W \} \Rightarrow Q =\{ 0,1,2 \}$

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

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

Question 3

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

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

Answer:

Given data:

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

Now, we can find,

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

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

Question 4

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

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,

$a = \frac{11}{3}$ & $b = \frac{2}{3}$

ii) Given:

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

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

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

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

$0 - 3b = 6$

$b = -2$

Thus, a = 0 & b = -2

Question 5

Given $A = \ \{1, 2, 3, 4, 5\ \}$, $S = \ \{(x, y) : x \in A, y \in A\ \}$. 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:

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

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

Now,

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

Question 6

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

Answer:

Given:

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

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

Therefore, here,

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

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

Question 7

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

Answer:

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

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

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

& $y = 2x + 7$

Thus, the values of y will be

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

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

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

Question 8

If $R_2 = \ \{(x, y) | \: \: x \: \: and \: \: y \: \: are \: \: integers \: \: and\: \: x^2 + y^2 = 64\ \}$ 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, $R_2 = \ \{(0,8),(0,-8),(8,0),(-8,0)\ \}$

Question 9

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

Answer:

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

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

$R_3 = (0, \infty) ... (since,\|x\| = R_+]$

Question 10

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

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

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

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

(iv)$s = \ \{(n, n^2) | n \: \: is\: \: a\: \: positive\: \: integer\ \}$

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

Answer:

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

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

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

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

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

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

(iv)
$S = | \{(n, n^2)|n \: \: is\: \: a\: \: positive\: \: integer.\ \}$

‘S’ is a function since the square of any integer is a unique number. & thus, for every element in the domain, there is a unique image.

(V) $T = | \{(x,3)| x \: \: is\: \: a\: \: real\: \: number\ \}$

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) $f\left(\frac{1}{2}\right) \times g(14)$

(c) $f (- 2) + g (- 1)$

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

(e) $\frac{(f(t) - f(5))}{ (t - 5)}, if \: \: t \neq 5$

Answer:

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

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

$= 16 - 10$

$= 6$

2. $f(\frac{1}{2} ) \times g(14) = [\left(\frac{1}{2}\right)^2 + 7] \times [3 \times 14 + 5]$

$= \frac{29}{4} \times 47$

$= \frac{1363}{4}$

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

$= 11 + 2 = 13$

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

$= t^2 - 4$

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

$=\frac{( t^2 + 7 - 32)}{t – 5}$

$= \frac{(t^2 - 25)}{(t - 5)}$

$=(t - 5) \frac{(t + 5)}{(t - 5)}$

$= t + 5$

Question 12

Let f and g be real functions defined by $(x) = 2x + 1$ 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)$\frac{f}{g}$

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) $\frac{f}{8} = \frac{f(x)}{g(x)}$

$= 2x + \frac{1}{x^2} + 1$

Question 14

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

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

Answer:

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

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

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 &

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

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

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

Here, ‘g’ is a function since every element of the domain has a 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) $f(x) = \frac{1}{\sqrt{x+\ | x \ |}}$

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

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,
$-1 \leq \cos x \leq 1$
$1 \geq -\cos x \geq -1$
$1 + 1 \geq 1 - \cos x \geq -1+1$
$2 \geq 1 - \cos x \geq 0$
$0 \leq 1 - \cos x \leq 2$
Now, for the real value of the domain,
$1 - \cos x\neq0,\cos x \neq 1$
But, $x \neq 0\: \: 2n\pi \: \: \forall n \in Z$
Thus, domain of $f = R - \ \{2n\pi, n \in Z\ \}$

ii)Given data: $f(x) = \frac{1}{\sqrt{x+\ | x \ |}}$
Now, $x + \ |x \ | = x+x = 2x ...... if \: \: x \geq0$
&$x + \ |x \ | = x - x = 0 ...... if \: \: x < 0$
Now, $x<0$ is not defined so far; hence,
The domain = $R^+$
iii)Given data:$f(x) =x\ | x \ |$
For all x $\in$ R, f(x) is defined
Thus, the 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 $f = R - \ \{-1,1\ \}$
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 = $R - \ \{4\ \}$.

Question 18

Find the range of the following functions given by

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

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

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

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

Answer:

i)Given data: $f\ ( x \ )= \frac{3}{2-x^2}$
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: $f\ ( x \ )= 1-\ | x-2 \ |$
Now, we know that,
$\ |x-2 \ | = -(x-2) , if x < 2, \: \: and\: \: \ | x-2 \ |= (x-2), if \geq 2$
Thus, $\ |-x-2 \ | \geq 0$
Thus, $1 -\ | x-2 \ | \leq 1$
Therefore,
Range of$f = (-\infty, 1)$
iii)Given data: $f\ ( x \ )= \ | x-3 \ |$

Now, we know that,

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

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

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:

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

$
|x-2| =
\begin{cases}
x - 2, & x \geq 2 \\
-(x - 2), & x < 2
\end{cases}
$

$
|2+x| =
\begin{cases}
-(2 + x), & x < -2 \\
(2 + x), & x \geq -2
\end{cases}
$

$
f(x) =
\begin{cases}
-(x - 2) - (2 + x), & -3 \leq x < -2 \\
-(x - 2) + (2 + x), & -2 \leq x < 2 \\
(x - 2) + (2 + x), & 2 \leq x \leq 3
\end{cases}
$

$
f(x) =
\begin{cases}
-2x, & -3 \leq x < -2 \\
4, & -2 \leq x < 2 \\
2x, & 2 \leq x \leq 3
\end{cases}
$

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) $f\left(\frac{1}{x}\right) = \frac{\frac{1}{x}-1}{\frac{1}{x}+1}$
$= \frac{1-x}{1+x}$
= $-\frac{(x-1)}{x+1} = -f(x)$
Thus, $f\left(\frac{1}{x}\right) = -f(x)$
ii) $f(-\frac 1 x) = \frac{\frac{-1}{x}-1}{\frac{-1}{x}+1}$
$= \frac{1 + x}{1 – x}$
$=\frac{ 1}{1 + x/1 - x}$
$=\frac{ -1}{ f(x)}$
Thus, $f\left(\frac{-1}{x}\right) = \frac{-1}{f(x)}$.

Question 21

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

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

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

(iii) $(fg) (x)$

(iv) $\left(\frac{f}{g}\right) (x)$

Answer:

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

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

$=\sqrt{x} + x$

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

$=\sqrt{x} - x$

iii)$(fg)(x) = f(x)\cdot g(x)$

$= \sqrt{x}.x$

$= x^{\frac{3}{2}}$

iv) $\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}$

$= \frac{\sqrt{x}}{x}$

$= \frac{1}{\sqrt{x}}$

Question 22

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

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

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

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

Now, to find the range, we will put

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

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

$x-5 = \frac{1}{y^2}$

$x = \frac{1}{y^2} + 5$

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

Therefore, the range of $ = R^+$

Question 23

If $f( x)= y = \frac{ax-b}{cx-a}$ then prove that $f (y) = x$

Answer:

Given: $f( x)= y = \frac{ax-b}{cx-a}$

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

$f(y) = \frac{ay-b}{cy-a}$

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

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

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

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

$=x$

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) $m^{n}$
(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 : $[x]^2 - 5[x] + 6 = 0$

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 $f\ ( x \ )=\frac{1}{1-2\\cos x}$ is

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

Answer:

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

Now, we know that,

$-1 \geq \\cos x \geq1$

Thus, $-1\leq cost \leq 1$

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

$\\ -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} $

$\\\\\ \frac{1}{1-2 \\cos x} \in(-\infty,-1] \cup [\frac{1}{3}, \infty)$
Therefore, (c) is the correct option.

Question 27

Let $f(x)=\sqrt{(1+x^2 )}$, then
(A) $f(x y)=f(x) \cdot f(y)$

(B) $f(x y) \geq f(x) . f(y)$

(C) $f(x y) \leq f(x) \cdot f(y)$

(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

1. 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, $x\leq 4\: \: and\: \: x<-\frac 1 x>1$

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

Answer:

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

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

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

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 (4) 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. $R - \ \{3, - 2\ \}$
B. $R - \ \{-3, 2\ \}$
C. $R - [3, - 2]$
D. $R - (3, - 2)$

Answer:

Given data: $f(x) = \frac{(x^2+2x+1)}{(x^2-x-6)}$

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,$x \neq -2 \text{ or} 3$

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

Hence, the correct answer is opt (a).

Question 34

The domain and range of the function f given by $f (x) = 2 - \ |x -5 \ |$ 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: $f (x) = 2 - \ |x -5 \ |$
& f(x) is defined by $x \in R$

Thus, its domain is f(x) = R

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

$-|x-5| \leq 0$

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

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

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 $f= \ \{(0, 1), (2,0), (3,.-4), (4,2), (5, 1)\ \}$
$g= \ \{(1,0), (2,2), (3,-1), (4,4), (5, 3)\ \}$ then the domain of $f \times g$ is given by________ .

Answer:

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

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

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

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

Question 37

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

Answer:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Thus, the correct matches will be

$(a) \rightarrow(i i i),(b) \rightarrow(i v),(c) \rightarrow(i i) \&(d) \rightarrow(i)$

Question 38

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

Answer:

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

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

Answer:

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

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

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

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

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

Thus, the statement is false.

Question 40

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

Answer:

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

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

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

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

Thus, the given statement is true.

Question 41

State True or False for the following statements

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

Answer:

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

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

Answer:

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

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

Therefore, the statement is true.

Importance of solving NCERT Exemplar Class 11 Maths Solutions Chapter 2

NCERT Exemplar Class 11 Maths Solutions Chapter 2 play a crucial role in building a strong conceptual foundation in Relations and Functions. Regular practice of these solutions helps students understand abstract concepts in a simple and practical manner.

Here are some points on why these NCERT Exemplar Class 11 Maths Solutions Chapter 2 are important.

• Provides a clear understanding of different types of relations and functions.
• Strengthens the basics of sets, which are essential for higher-level mathematics.
• Helps students relate mathematical concepts to real-life examples and situations.
• Explains how functions work as input–output models, improving conceptual clarity.
• Shows practical applications, such as analysing exam scores across semesters or years using relations.
• Enhances logical reasoning and problem-solving skills through structured questions.
• Aids in effective preparation for school exams and competitive examinations.

NCERT Solutions for Class 11 Mathematics Chapterwise

All NCERT Class 11 Maths Solutions are available in one spot on Careers360 for easy access. Click the links below to view them.

NCERT Exemplar Class 11 Mathematics Chapterwise

Careers360 offers all NCERT Class 11 Maths Exemplar Solutions in one place for students. Just click the links below to see them.

NCERT Solutions of class 11 - Subject-wise

Here are the subject-wise links for the NCERT Solutions of class 11:

• NCERT Solutions for Class 11 Maths
• NCERT Solutions for Class 11 Physics
• NCERT Solutions for Class 11 Chemistry
• NCERT Solutions for Class 11 Biology

NCERT Notes of class 11 - Subject Wise

Given below are the subject-wise NCERT Notes of class 11 :

• NCERT Notes for Class 11 Maths
• NCERT Notes for Class 11 Physics
• NCERT Notes for Class 11 Chemistry
• NCERT Notes for Class 11 Biology

NCERT Books and NCERT Syllabus

Checking the updated syllabus at the start of the academic year helps students stay prepared. Below, you’ll find the syllabus links along with useful reference books.

• NCERT Books Class 11 Maths
• NCERT Syllabus Class 11 Maths
• NCERT Books Class 11
• NCERT Syllabus Class 11

NCERT Exemplar Class 11 Solutions

Given below are the subject-wise Exemplar solutions of class 11 NCERT:

• NCERT Exemplar Class 11 Biology Solutions
• NCERT Exemplar Class 11 Chemistry Solutions
• NCERT Exemplar Class 11 Physics Solutions