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

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:

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

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 $A = \ \{1, 2, 3, 4, 5\ \}$, $S = \ \{(x, y) : x \in A, y \in A\ \}$. 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) $g = n, (1/n) |n \: \: is\: \: a\: \: positive\: \: integer$

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

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

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

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

‘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,
$-1 \leq cosx \leq 1$
$1 \geq -cosx \geq -1$
$1 + 1 \geq 1 - cos x \geq -1+1$
$2 \geq 1 - cos x \geq 0$
$0 \leq 1 - cos x \leq 2$
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 $f = R - \ \{2n\pi, n \in Z\ \}$

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) $f(-1/x) = \frac{\frac{-1}{x}-1}{\frac{-1}{x}+1}$
$= \frac{1 + x}{1 – x}$
$=\frac{ 1}{1 + x/1 - x}$
$=\frac{ -1}{ f(x)}$
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, $-1\leq cost \leq 1$

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

$\\ -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} $

$\\\\\ \frac{1}{1-2 \cos x} \in(-\infty,-1] \cup [\frac{1}{3}, \infty)$
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, $x\leq 4\: \: and\: \: x<-1/x>1$

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

Through Class 11 Maths NCERT Exemplar Solutions Chapter 2, the students will learn briefly about different types of relations and functions and their associated functions. NCERT Exemplar solutions for Class 11 Maths Chapter 2 will help students further learn about sets and will help them 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 as seen established through the use of relation and functions.

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