Solution

$A=\lbrace-1,2,3\rbrace$ and $B=\lbrace1,3\rbrace$

(i) $A \times B=\lbrace(-1,1),(-1,3),(2,1),(2,3),(3,1),(3,3) \rbrace$

(ii) $B \times A=\lbrace(1,-1),(1,2),(1,3),(3,-1),(3,2),(3,3)\rbrace$

(iii) $B \times B=\lbrace(1,1),(1,3),(3,1),(3,3)\rbrace$

(iv)$A \times A= \lbrace(-1,-1),(-1,2),(-1,3),(2,-1),(2,2),(2,3),(3,-1),(3,2),(3,3)\rbrace$

Solution

We have, $P =\lbrace x: x<3, x \in N\rbrace=\lbrace1,2\rbrace $

And $Q =\lbrace x: x \leq 2, x \in W\rbrace=\lbrace0,1,2\rbrace $

Now, $P \cup Q =\lbrace0,1,2\rbrace \text { and } P \cap Q=\lbrace1,2\rbrace $

$(P \cup Q) \times(P \cap Q) =\lbrace0,1,2\rbrace \times\lbrace1,2\rbrace $

$=\lbrace(0,1),(0,2),(1,1),(1,2),(2,1),(2,2)\rbrace$

$ \therefore \quad P \cup Q=\lbrace0,1,2\rbrace \text { and } P \cap Q=\lbrace1,2\rbrace $

Solution

We have, $A =\lbrace x: x \in W, x<2\rbrace=\lbrace0,1\rbrace $,

$B =\lbrace x: x \in N, 1<x<5\rbrace $ $ =\lbrace2,3,4\rbrace $

And $C=\lbrace3,5\rbrace$

(i) $A \times(B \cap C)$ $:$

$B \cap C=\lbrace3\rbrace$

$\therefore \quad A \times(B \cap C)=\lbrace0,1\rbrace \times\lbrace3\rbrace=\lbrace(0,3),(1,3)\rbrace$

(ii) $A \times(B \cup C)$ $:$

$\because(B \cup C)= \lbrace 2,3,4,5\rbrace$

$ \begin{aligned} \therefore \quad A \times(B \cup C) & = \lbrace 0,1\rbrace \times \lbrace 2,3,4,5 \rbrace \\ & =\lbrace(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)\rbrace \end{aligned} $

Solution

(i) We have, $(2a+b, a-b)=(8,3)$

$ \Rightarrow \quad 2 a+b=8 \text { and } a-b=3 $

[since, two ordered pairs are equal, if their corresponding first and second elements are equal]

On substituting, $b=a-3$ in $2 a+b=8$, we get

$ 2a+a-3 =8 \\ \Rightarrow 3 a-3=8 \\ \Rightarrow 3a =11 \\ \Rightarrow a=\frac{11}{3} $

$ \text {Again, substituting a} = \frac{11}{3} \ \text{in b=a-3, we get}$

$b=\frac{11}{3}-3=\frac{11-9}{3}=\frac{2}{3} $

$\therefore \ a=\frac{11}{3} \ \text{and b}=\frac{2}{3}$

(ii) We have, $\quad \begin{pmatrix} \frac{a}{4}, a-2b\end{pmatrix}=(0,6+b) $

$\Rightarrow \frac{a}{4} =0 \Rightarrow a=0 $

And $a-2 b =6+b $

$\Rightarrow 0-2 b =6+b $

$\Rightarrow -3 b =6 $

$\therefore b =-2 $

$\therefore a =0, b=-2$

Solution

We have, $A=\lbrace 1,2,3,4,5\rbrace$ and $S=\lbrace(x, y): x \in A, y \in A \rbrace$

(i) The set of ordered pairs satisfying $x+y=5$ is

$\lbrace (1,4),(2,3),(3,2),(4,1)\rbrace$.

(ii) The set of ordered pairs satisfying $x+y<5$ is

$\lbrace(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)\rbrace$.

(iii) The set of ordered pairs satisfying $x+y>8$ is

$\lbrace(4,5),(5,4),(5,5)\rbrace$.

Thinking Process

First, write the relation in Roaster form, then find the domain and range of $R$.

Solution

We have,

$ \begin{aligned} R & =\lbrace(x, y): x, y \in W, x^{2}+y^{2}=25\rbrace \\ & =\lbrace(0,5),(3,4),(4,3),(5,0)\rbrace \end{aligned} $

Domain of $R=$ Set of first element of ordered pairs in $R$

$=\lbrace 0,3,4,5 \rbrace $

Range of $R=$ Set of second element of ordered pairs in $R$

$=\lbrace 5,4,3,0 \rbrace$

Solution

We have,

$R_1 =\lbrace(x, y) \mid y=2 x+7, \text { where } x \in R \text { and }-5 \leq x \leq 5\rbrace $

$\text { Domain of } R_1 =\lbrace-5 \leq x \leq 5, x \in R\rbrace $

$ =[-5,5] $

For range of $R_1 ,$

$y =2 x+7 $

$\text { When } x=-5 \text {, then } $ $y =2(-5)+7=-3 $

$ \text { When } x=5 \text {, then } $ $y =2(5)+7=17 $

$\therefore \ $ Range of $R_1 =\lbrace-3 \leq y \leq 17, y \in R\rbrace $

$ =[-3,17]$

Solution

We have, $R_2=\lbrace (x, y) \mid x$ and $y$ are integers and $x^{2}+y^{2}=64\rbrace$

Since, 64 is the sum of squares of 0 and $\pm 8$.

When $x=0$, then $y^{2}=64 \Rightarrow y= \pm 8$

$x=8$, then $y^{2}=64-8^{2} \Rightarrow 64-64=0$

$x=-8$, then $y^{2}=64-(-8)^{2}=64-64=0$

$\therefore \quad R_2= \lbrace (0,8),(0,-8),(8,0),(-8,0)\rbrace$

Solution

We have

$R_3=\lbrace (x,|x|) \mid x \text { is real number }\rbrace $

Clearly, domain of $R_3=R$

Since, image of any real number under $R_3$ is positive real number or zero.

$ \therefore \quad \text { Range of } R_3=R^{+} \cup \lbrace 0 \rbrace \text { or }[0, \infty) $

Solution

(i) We have, $h=\lbrace(4,6),(3,9),(-11,6),(3,11)\rbrace$.

Since, 3 has two images 9 and 11. So, it is not a function.

(ii) We have, $f=\lbrace(x, x) \mid x$ is a real number $\rbrace $.

We observe that, every element in the domain has unique image. So, it is a function.

(iii) We have, $g=\lbrace (n, \frac{1}{n} )\mid n $ $\text{is a positive integer }\rbrace$.

For every $n$, it is a positive integer and $\frac{1}{n}$ is unique and distinct. Therefore, every element in the domain has unique image. So, it is a function.

(iv) We have, $s=\lbrace(n, n^{2}) \mid n$ is a positive integer $\rbrace$.

Since, the square of any positive integer is unique. So, every element in the domain has unique image. Hence, it is a function.

(v) We have, $t=\lbrace(x, 3) \mid x$ is a real number $\rbrace$.

Since, every element in the domain has the image 3 . So, it is a constant function.

Solution

Given, $f$ and $g$ are real functions defined by $f(x)=x^{2}+7$ and $g(x)=3 x+5$.

(i) $f(3)=(3)^{2}+7=9+7=16$

And $g(-5)=3(-5)+5=-15+5=-10$

$\therefore f(3)+g(-5)=16-10=6$

(ii) $f (\frac{1}{2})=(\frac{1}2)^{2}+7=\frac{1}{4}+7=\frac{29}{4}$

And $g(14)=3(14)+5=42+5=47$

$\therefore \quad f (\frac{1}{2})\times g(14)=\frac{29}{4} \times 47=\frac{1363}{4}$

(iii) $f(-2)=(-2)^{2}+7=4+7=11$

And $g(-1)=3(-1)+5=-3+5=2$

$\therefore \quad f(-2)+g(-1)=11+2=13$

(iv) $f(t)=t^{2}+7$

And $f(-2)=(-2)^{2}+7=4+7=11$

$ \therefore \quad f(t)-f(-2)=t^{2}+7-11=t^{2}-4 $

(v) $f(t)=t^{2}+7$

And $f(5)=5^{2}+7=25+7=32$

$ \therefore \quad \frac{f(t)-f(5)}{t-5}, \text { if } t \neq 5 $

$ =\frac{t^{2}+7-32}{t-5} $

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

$ =t+5 $

Solution

We have,

$ f(x)=2 x+1 \text { and } g(x)=4 x-7 $

$\text { (i) } \because \quad f(x)=g(x) $

$\Rightarrow \quad 2 x+1=4 x-7 $

$\Rightarrow \quad 2 x=8 $

$\therefore \quad x=4 $

$\text { (ii) } \because \quad f(x)<g(x) $

$\Rightarrow \quad 2 x+1<4 x-7 $

$\Rightarrow \quad 1+7 < 4x-2x$

$\Rightarrow \quad 8 < 2x$

$\therefore \quad x>4$

Solution

We have, $f(x)=2 x+1$ and $g(x)=x^{2}+1$

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

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

(ii) $(f-g)(x)=f(x)-g(x)=(2 x+1)-(x^{2}+1)$

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

(iii) $(f g)(x)=f(x) \cdot g(x)=(2 x+1)(x^{2}+1)$

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

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

Solution

We have,

$ f: X \rightarrow R, f(x)=x^{3}+1 $

Where

$X=\lbrace-1,0,3,9,7\rbrace$

When

$x=-1$, then $f(-1)=(-1)^{3}+1=-1+1=0$

$x=0$, then $f(0)=(0)^{3}+1=0+1=1$

$x=3$, then $f(3)=(3)^{3}+1=27+1=28$

$x=9$, then $f(9)=(9)^{3}+1=729+1=730$

$x=7$, then $f(7)=(7)^{3}+1=343+1=344$

$f=\lbrace(-1,0),(0,1),(3,28),(9,730),(7,344)\rbrace$

$\therefore \quad$ Range of $f=\lbrace0,1,28,730,344\rbrace$

Solution

$ f(x)=g(x) $

$ \Rightarrow \quad 3 x^{2}-1 =3+x $

$ \Rightarrow \quad 3 x^{2}-x-4 =0 $

$ \Rightarrow \quad 3 x^{2}-4 x+3 x-4 =0 $

$ \Rightarrow \quad x(3 x-4)+1(3 x-4) =0 $

$ \Rightarrow \quad (3 x-4)(x+1) =0 $

$ \therefore \quad x =-1, \frac{4}{3}$

Thinking Process

First, find the two equation by substitutions different values of $x$ and $g(x).$

Solution

We have,

$ g=\lbrace(1,1),(2,3),(3,5),(4,7)\rbrace $

Since, every element has unique image under $g$. So, $g$ is a function.

Now, $ g(x) =\alpha x+\beta $

When $x=1$, then

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

$ 1 =\alpha+\beta \quad \ldots (i)$

When $x=2$, then

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

$ \Rightarrow \quad 3=2 \alpha+\beta \quad \ldots (ii) $

On solving Eqs. (i) and (ii), we get

$ \alpha=2, \beta=-1 $

Solution

(i) We have, $f(x)=\frac{1}{\sqrt{1-\cos x}}$

$\Rightarrow \quad -1 \leq \cos x \leq 1$

$\Rightarrow \quad -1 \leq -\cos x \leq 1$

$\Rightarrow \quad -1+1 \leq 1- \cos x \leq 1$

$\Rightarrow \quad 0 \leq 1- \cos x \leq 1$

Now, $f(x)$ is defined iff

$1-\cos x \neq 0$

$\Rightarrow \quad \cos x \neq 1$

$\Rightarrow \quad x \neq 2n \pi \quad \forall n \in Z$

$\therefore $ Domain of $f= R- \lbrace 2n \pi : n \in Z \rbrace $

(ii) We have, $f(x) =\frac{1}{\sqrt{x+|x|}} $

$x+|x| =$ $\begin{cases} x-x=0, & x<0 \\ x+x=2 x, & x \geq 0 \end{cases}$

$ \because \quad x+|x|=x-x=0, x<0 $

Hence, $f(x)$ is defined, if $x>0$.

$ \therefore \quad \text { Domain of } f=R^{+} $

(iii) We have, $f(x)=x|x|$

Clearly, $f(x)$ is defined for any $x \in R$.

$\therefore$ Domain of $f=R$

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

$f(x)$ is not defined, if

$x^{2}-1 =0 $

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

$x =-1,1 $

Domain of $f =R-\lbrace-1,1\rbrace $

(v) We have, $ f(x)=\frac{3 x}{28-x} $

Clearly, $f(x)$ is defined, if

$28-x \neq 0$

$\Rightarrow \quad x \neq 28$

$\therefore \quad$ Domain of $f=R-\lbrace28\rbrace$

Thinking Process

First, find the value of $x$ in terms of $y$, where $y=f(x)$. Then, find the values of $y$ for which $x$ attain real values.

Solution

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

Let $y =f(x) $

Then, $y =\frac{3}{2-x^{2}} \Rightarrow 2-x^{2}=\frac{3}{y} $

$\Rightarrow \quad x^{2} =2-\frac{3}{y} $

$\Rightarrow \quad x=\sqrt{\frac{2 y-3}{y}}$

$x$ assums real values, if $2 y-3 \geq 0$ and $y>0 \Rightarrow y \geq \frac{3}{2}$

$ \therefore \quad \text { Range of } f=[\frac{3}{2}, \infty) $

(ii) We have, $f(x)=1-|x-2|$

We know that, $ \mid x-2 \mid \geq 0 $

$\Rightarrow -\mid x-2 \mid \leq 0$

$\Rightarrow$ $1- \mid x-2 \mid \leq 1 $

$\Rightarrow $ $f(x) \leq 1 $

$\therefore$ Range of $f=(-\infty , 1] $

(iii) We have, $f(x)=|x-3|$

We know that, $ |x-3| \geq 0 $

$ \Rightarrow f(x) \geq 0 $

$ \therefore \text { Range of } f=[0, \infty) $

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

We know that, $-1 \leq \cos 2 x \leq 1 $

$\Rightarrow \quad -3 \leq 3 \cos 2 x \leq 3$

$\Rightarrow \quad 1-3 \leq 1 + 3 \cos 2 x \leq 1+3 \\ \Rightarrow \quad -2 \leq 1+3 \cos 2 x \leq 1+3 \\ -2 \leq f(x) \leq 4 $

$\therefore \text { Range of } f =[-2,4] $

Thinking Process

First find the interval in which $|x-2|$ and $|2+x|$ is defined, then find the value of $f(x)$ in that interval.

Solution

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

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

Given that

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

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

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

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

Solution

We have, $\quad f(x)=\frac{x-1}{x+1}$

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

(ii) $f(-\frac{1}{x})=\frac{-\frac{1}{x}-1}{-\frac{1}{x}+1}=\frac{(-1-x) / x}{(-1+x) / x} \Rightarrow f(-\frac{1}{x})=\frac{-(x+1)}{x-1}$

Now, $\quad \frac{-1}{f(x)}=\frac{-1}{\frac{x-1}{x+1}}=\frac{-(x+1)}{x-1}$

$\therefore \quad f(-\frac{1}{x})=-\frac{1}{f(x)}$

Solution

We have, $f(x)=\sqrt{x}$ and $g(x)=x$ be two function defined in the domain $R^{+} \cup\lbrace0\rbrace$.

(i) $(f+g)(x)=f(x)+g(x)=\sqrt{x}+x$

(ii) $(f-g)(x)=f(x)-g(x)=\sqrt{x}-x$

(iii) $(f g)(x)=f(x) \cdot g(x)=\sqrt{x} \cdot x=x^{(\frac{3}{2})}$

(iv) $\big(\frac{f}{g}\big)x=\frac{f(x)}{g(x)}=\frac{\sqrt{x}}{x}=\frac{1}{\sqrt{x}}$

Solution

We have, $\quad f(x)=\frac{1}{\sqrt{x-5}}$

$f(x)$ is defined, if $x-5>0 \Rightarrow x>5$

$\therefore \quad$ Domain of $f=(5, \infty)$

Let $f(x)=y$

$\therefore \quad y=\frac{1}{\sqrt{x-5}} \Rightarrow \sqrt{x-5}=\frac{1}{y}$

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

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

$x \in (5, \infin )$ and $\frac{1}{y^2} > 0$

$\Rightarrow \quad y \in R^{+}$

Hence, range of $f=R^{+} $

Solution

We have, $ f(x)=y=\frac{a x-b}{c x-a} $

$ \begin{aligned} \therefore \quad f(y) & =\frac{a y-b}{c y-a}=\frac{a (\frac{a x-b}{c x-a})-b}{c (\frac{a x-b}{c x-a})-a} \\ & =\frac{a(a x-b)-b(c x-a)}{c(a x-b)-a(c x-a)}=\frac{a^{2} x-a b-b c x+a b}{a c x-b c-a c x+a^{2}} \\ & =\frac{a^{2} x-b c x}{a^{2}-b c}=\frac{x(a^{2}-b c)}{(a^{2}-b c)}=x \\ \therefore \quad f(y) & =x \end{aligned} $

Hence proved.

Thinking Process

First find the number of element in $A \times B$ and then find the number of non-empty relation by using $2^{n(A \times B)}-1$

Solution

Option (d): We have, $n(A) =m \text { and } n(B)=n $

Now, $n(A \times B) =n(A) \cdot n(B) $

$\quad n(A \times B) =m n$

$\therefore \quad $ Total number of relation from $A$ to $B=2^{n(A \times B)}-1=2^{m n}-1$

• Option (a) $ m^n $: This option represents the number of functions from set $ B $ to set $ A $, not the number of non-empty relations. A function is a specific type of relation where each element in $ A $ is related to exactly one element in $ B $. Therefore, this does not account for all possible non-empty relations.

• Option (b) $ n^m - 1 $: This option is incorrect because it represents the number of functions from $ A $ to $ B $ minus one. Similar to option (a), it only considers functions, not all possible relations.

• Option (c) $ mn - 1 $: This option is incorrect because it represents the number of non-empty pairs in the cartesian product $ A \times B $ minus one.

Thinking Process

If $a$ and $b$ are two successive positive integer and $[x]=a, b$, then $x \in [a, a+1 )$ & $x \in [b, b+1)$.

Solution

Option (d): We have,

$ [x]^{2}-5[x]+6 =0 $

Let $[x] = z$

$\Rightarrow \quad z^2-5z +6 =0$

$\Rightarrow \quad z^2-3z-2z+6=0$

$\Rightarrow \quad z(z-3)-2(z-3)=0$

$\Rightarrow \quad z=2,3$

$\therefore \quad [x]=2$ and $[x]=3$

if $[x]=2$ then $x \in [2,3)$

if $[x]=3$ then $x \in [3,4)$

Combining the intervals, we get

$x \in [2,3) \cup [3,4)$

$\Rightarrow \quad x \in [2,4)$

• Option (a) $x \in[3,4]$: This option is incorrect because this intervals suggests that $x$ can take any value from $3$ to $4$ inclusive of both endpoints. $x$ cannot be in the interval $[3,4]$ as it would imply $[x] = 4$ which does not satisfy the equation.

• Option (b) $x \in(2,3]$: This option is incorrect because if $x \in (2,3]$, then $[x]$ would be 2 or 3. While $[x] = 2$ and $[x] = 3$ are solutions to the equation, the interval $(2,3]$ excludes $x = 2$, which is a valid solution. Therefore, the correct interval should include 2, making this option incorrect.

• Option (c) $x \in[2,3]$: This option is incorrect because if $x \in [2,3]$, then $[x]$ would be 2 or 3. For $[x]=3, x \in [3, 3+1)$. Therefore, this option is incorrect.

Solution

Option (c): We have, $f(x)=\frac{1}{1-2 \cos x}$

We know that,

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

$\Rightarrow \quad -1 \leq - \cos x \leq 1$

$\Rightarrow \quad -2 \leq -2 \cos x \leq 2$

$\Rightarrow \quad 1-2 \leq 1-2 \cos x \leq 1+2$

$\Rightarrow \quad -1 \leq 1-2 \cos x \leq 3$

$\because \quad \cos x \neq \frac{1}{2}$ when $1-2 \cos x \neq 0$

$\Rightarrow \quad -1 \leq 1- 2 \cos x <0 \ $ & $ \ 0 < 1-2 \cos x \leq 3$

$\Rightarrow \quad -\infin < \frac{1}{1- 2 \cos x } \leq -1 \ $ & $ \ \frac{1}{3} \leq \frac{1}{1-2 \cos x} < \infin$

$\Rightarrow \quad - \infin < f(x) \leq 1 \ $ & $ \ \frac{1}{3} \leq f(x) < \infin$

$\therefore $ Range of $f= (-\infin , -1] \cup [\frac{1}{3}, \infin)$

• Option (a) $[\frac{1}{3}, 1]$: $[\frac{1}{3}, 1]$ is incorrect because the range of $f(x)$ does not include 1. The correct range is $(-\infin , -1] \cup [\frac{1}{3}, \infin)$, and $1$ is not within this interval.

• Option (b) $[-1, \frac{1}{3}]$: $[-1, \frac{1}{3}]$ is incorrect because $f(x)$ can take values much larger than $\frac{1}{3}$ and less than $-1$.

• Option (d) $[-\frac{1}{3}, 1]$: $[-\frac{1}{3}, 1]$ is incorrect because the range of $f(x)$ does not include $-\frac{1}{3}$. The correct range is $(-\infin , -1] \cup [\frac{1}{3}, \infin)$, and $-\frac{1}{3}$ is not within this interval.

Solution

Option (c): We have,

$ \begin{aligned} f(x) & =\sqrt{1+x^{2}} \\ f(x y) & =\sqrt{1+x^{2} y^{2}} \\ f(x) \cdot f(y) & =\sqrt{1+x^{2}} \cdot \sqrt{1+y^{2}} \\ & =\sqrt{(1+x^{2})(1+y^{2})} \\ & =\sqrt{1+x^{2}+y^{2}+x^{2} y^{2}} \end{aligned} $

$ \begin{matrix} \because & \sqrt{1+x^{2} y^{2}} \leq \sqrt{1+x^{2}+y^{2}+x^{2} y^{2}} \\ \Rightarrow & f(x y) \leq f(x) \cdot f(y) \end{matrix} $

• Option (a) $f(x y)=f(x) \cdot f(y)$: This is incorrect because $ f(x y) = \sqrt{1 + x^2 y^2} \quad \text{and} \quad f(x) \cdot f(y) = \sqrt{(1 + x^2)(1 + y^2)} = \sqrt{1 + x^2 + y^2 + x^2 y^2} $

  Since $\sqrt{1 + x^2 y^2} \neq \sqrt{1 + x^2 + y^2 + x^2 y^2}$, the equality does not hold.

• Option (b) $f(x y) \geq f(x) \cdot f(y)$: This is incorrect because $ f(x y) = \sqrt{1 + x^2 y^2} \quad \text{and} \quad f(x) \cdot f(y) = \sqrt{(1 + x^2)(1 + y^2)} = \sqrt{1 + x^2 + y^2 + x^2 y^2} $

  Since $\sqrt{1 + x^2 y^2} \leq \sqrt{1 + x^2 + y^2 + x^2 y^2}$, the inequality $f(x y) \geq f(x) \cdot f(y)$ does not hold.

Solution

Option (b): Let $ f(x)=\sqrt{a^{2}-x^{2}} $

$f(x)$ is defined, if

$ a^{2}-x^{2} \geq 0 $

$\Rightarrow \quad x^{2}-a^{2} \leq 0 $

$\Rightarrow \quad (x-a)(x+a) \leq 0$

$\Rightarrow \quad -a \leq x \leq a$

$\therefore \quad \text {Domain of }f=[-a, a]$

• Option (a): $(-a, a)$ is incorrect because it excludes the endpoints $-a$ and $a$. The function $\sqrt{a^2 - x^2}$ is defined at $x = -a$ and $x = a$ since $a^2 - (-a)^2 = 0$ and $a^2 - a^2 = 0$, making the square root of zero valid.

• Option (c): $[0, a]$ is incorrect because it excludes the negative values within the interval $[-a, 0)$. The function $\sqrt{a^2 - x^2}$ is defined for all $x$ in the interval $[-a, a]$, not just the non-negative part.

• Option (d): $(-a, 0]$ is incorrect because it excludes the positive values within the interval $(0, a]$. The function $\sqrt{a^2 - x^2}$ is defined for all $x$ in the interval $[-a, a]$, not just the non-positive part.

Solution

Option (b): We have,

$ f(x) =a x+b $

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

$ -5 =-a+b \quad \ldots (i)$

$ \text{ and,} \ f(3) =a(3)+b $

$ 3 =3 a+b \quad \ldots (ii)$

On solving Eqs. (i) and (ii), we get

$ a=2 \text { and } b=-3 $

• Option (a) $a=-3, b=-1$:

  If $a = -3$ and $b = -1$, then:

  For $f(-1) = -5$: $-3(-1) + (-1) = 3 - 1 = 2 \neq -5$

  For $f(3) = 3$: $-3(3) + (-1) = -9 - 1 = -10 \neq 3$

  Therefore, this option does not satisfy the given conditions.

• Option (c) $a=0, b=2$:

  If $a = 0$ and $b = 2$, then:

  For $f(-1) = -5$: $0(-1) + 2 = 2 \neq -5$

  For $f(3) = 3$: $0(3) + 2 = 2 \neq 3$

  Therefore, this option does not satisfy the given conditions.

• Option (d) $a=2, b=3$:

  If $a = 2$ and $b = 3$, then:

  For $f(-1) = -5$: $2(-1) + 3 = -2 + 3 = 1 \neq -5$

  For $f(3) = 3$: $2(3) + 3 = 6 + 3 = 9 \neq 3$

  Therefore, this option does not satisfy the given conditions.

Solution

Option (a): We have,

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

$f(x)$ is defined, if

$ \begin{aligned} 4-x & \geq 0 \text { or } x^{2}-1>0 \\ x-4 & \leq 0 \text { or }(x+1)(x-1)>0 \\ x & \leq 4 \text { or } x<-1 \text { and } x>1 \end{aligned} $

$ \therefore \quad \text { Domain of } f=(-\infty,-1) \cup(1,4] $

• Option (b): $(-\infty,-1] \cup(1,4]$ is incorrect because $x = -1$ is not in the domain of $f(x)$. At $x = -1$, the term $\frac{1}{\sqrt{x^2 - 1}}$ becomes undefined as the denominator becomes zero.

• Option (c): $(-\infty,-1) \cup[1,4]$ is incorrect because $x = 1$ is not in the domain of $f(x)$. At $x = 1$, the term $\frac{1}{\sqrt{x^2 - 1}}$ becomes undefined as the denominator becomes zero.

• Option (d): $(-\infty,-1) \cup[1,4)$ is incorrect because $x = 4$ is in the domain of $f(x)$. At $x = 4$, the term $\sqrt{4 - x}$ becomes zero, which is defined.

Thinking Process

A function $\frac{f(x)}{g(x)}$ is defined, if $g(x) \neq 0$.

Solution

Option (c): We have,

$ f(x)=\frac{4-x}{x-4} $

$f(x)$ is defined, if $x-4 \neq 0$ i.e., $x \neq 4$

$\therefore \quad$ Domain of $f=R-\lbrace4\rbrace$

Let $\quad f(x)=y$

$ \therefore \quad y=\frac{4-x}{x-4} \Rightarrow x y-4 y=4-x $

$\Rightarrow \quad x y+x=4+4 y $

$\Rightarrow \quad x(y+1)=4(1+y) $

$\therefore \quad x=\frac{4(1+y)}{y+1}$

$x$ assumes real values, if $y+1 \neq 0$ i.e., $y \neq-1$.

$\therefore \quad$ Range of $f=R-\lbrace-1\rbrace$

• Option (a) Domain $=R$, Range $=\lbrace-1,1\rbrace$: This is incorrect because the function $f(x)=\frac{4-x}{x-4}$ is not defined at $x=4$, so the domain cannot be all real numbers $R$. Additionally, the range is not limited to ${-1, 1}$; it includes all real numbers except $-1$.

• Option (b) Domain $=R-\lbrace1\rbrace$, Range $=R$: This is incorrect because the function $f(x)=\frac{4-x}{x-4}$ is not defined at $x=4$, not $x=1$. Therefore, the domain should exclude $4$, not $1$. Also, the range is not all real numbers $R$; it excludes $-1$.

• Option (d) Domain $=R-\lbrace-4\rbrace$, Range $=\lbrace-1,1\rbrace$: This is incorrect because the function $f(x)=\frac{4-x}{x-4}$ is not defined at $x=4$, not $x=-4$. Therefore, the domain should exclude $4$, not $-4$. Additionally, the range is not limited to ${-1, 1}$; it includes all real numbers except $-1$.

Solution

Option (d) We have $f(x)=\sqrt{x-1}$

The function $f$ is defined for

$x-1 \geq 0 \Rightarrow x \geq 1$

$\Rightarrow x \in [1 , \infin)$

$\therefore \quad $ Domain of $f = [1, \infin)$

For $x \geq 1$

$x-1 \geq 0 \Rightarrow \sqrt{x-1} \geq 0$

$f(x) \geq 0$

$\therefore $ Range of $f = [0, \infin)$

• Option (a) Domain $=(1, \infty)$, Range $=(0, \infty)$: This is incorrect because the domain should include 1, as $ f(x) = \sqrt{x-1} $ is defined for $ x = 1 $. Therefore, the domain should be $[1, \infty)$ instead of $(1, \infty)$.

• Option (b) Domain $=[1, \infty)$, Range $=(0, \infty)$: This is incorrect because the range $(0, \infin)$ incorrectly excludes $0,$ but $0$ is a value the function can take at $x=1$.

• Option (c) Domain $=(1, \infty)$, Range $=[0, \infty)$: This is incorrect because the domain should include 1, as $ f(x) = \sqrt{x-1} $ is defined for $ x = 1 $. Therefore, the domain should be $[1, \infty)$ instead of $(1, \infty)$.

Thinking Process

A function $\frac{f(x)}{g(x)}$ is defined, if $g(x) \neq 0$.

Solution

Option (a) We have, $f(x)= \frac{x^2+2x+1}{x^2-x-6}$

The function $f$ is defined if

$x^2-x-6 \neq 0$

$\Rightarrow \quad x^2-3x+2x-6 \neq 0$

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

$\Rightarrow \quad x \neq 3, -2$

$\therefore \quad $ Domain of $f=R-\lbrace -2, 3 \rbrace$

• Option (b) $\mathbf{R}-{-3,2}$: This is incorrect because this suggests that the domain excludes $x=-3 $ and $x=2$ but these value do not make the denominator zero.

• Option (c) $\mathbf{R}-[3,-2]$: This is incorrect because the interval notation implies a range of numbers, not specific points.

• Option (d) $\mathbf{R}-(3,-2)$: This is incorrect because this suggests excluding the open interval from $3 $ to $-2$. Similar to option C, this does not correctly represent the exclusion of specific points.

Solution

Option (b) We have, $f(x)= 2- \mid x-5 \mid$

Since, $\mid x-5 \mid $ is defined for all $x \in \mathbf{R}$.

$\therefore \quad f$ is defined for every $x \in \mathbf{R}$.

$\therefore \quad $ Domain of $f = \mathbf{R}$.

Now, we know that $\mid x-5 \mid \geq 0$

$\Rightarrow \quad - \mid x-5 \mid \leq 0$

$\Rightarrow \quad 2- \mid x-5 \mid \leq 2$

$\Rightarrow \quad f(x) \leq 2$

$\therefore \quad $ Range of $f= (- \infin , 2 ]$.

Option (a) Domain $=\mathbf{R}^{+}$, Range $=(-\infty, 1]$: This is incorrect because it states that the function is only defined for positive real numbers. Since $ f(x) = 2 - |x - 5| $ involves an absolute value, which is defined for all real $ x $, the domain should actually be all real numbers, $ \mathbf{R} $.

Option (c) Domain $=\mathbf{R}$, Range $=(-\infty, 2)$: This is incorrect because the range is given as $(- \infty, 2)$, suggesting that 2 is not included. However, the function reaches a maximum value of 2 at $ x = 5 $. Hence, 2 should be included in the range, making the correct range $(- \infty, 2]$.

Option (d) Domain $=\mathbf{R}^{+}$, Range $=(-\infty, 2]$: This is incorrect because it states that the function is only defined for positive real numbers. Since $ f(x) = 2 - |x - 5| $ involves an absolute value, which is defined for all real $ x $, the domain should actually be all real numbers, $ \mathbf{R} $.

Solution

Option (a) We have $f(x) = 3x^2-1$ amd $g(x)=3+x$

$f(x) = g(x)$

$\Rightarrow \quad 3x^2-1 = 3+x$

$\Rightarrow \quad 3x^2-x-4 =0$

$\Rightarrow \quad 3x^2-4x+3x-4= 0$

$\Rightarrow \quad x(3x-4)+1(3x-4)=0$

$\Rightarrow \quad x=-1, \frac{4}{3}$

$\therefore \quad $ For $x \in \lbrace -1, \frac{4}{3} \rbrace , f(x)= g(x)$

Option (b) $\left[-1, \frac{4}{3}\right]$: This is incorrect because this suggests that $ f(x) $ and $ g(x) $ are equal for all $ x $ in the closed interval $[-1, \frac{4}{3}]$. However, the functions are only equal at the specific points $ x = -1 $ and $ x = \frac{4}{3} $, not throughout the interval.

Option (c) $\left(-1, \frac{4}{3}\right)$: This is incorrect because this suggests that the functions are equal for all $ x $ in the open interval $(-1, \frac{4}{3})$. However, the functions are not equal at any point in this interval except at the endpoints.

Option (d) $\left[-1, \frac{4}{3}\right)$: This is incorrect because this suggests that the functions are equal at $ x = -1 $ and for all $ x $ less than $\frac{4}{3}$. Again, the functions are only equal at the specific points $ x = -1 $ and $ x = \frac{4}{3} $.

Thinking Process

First find the domain of $f$ and domain of $g$. Then,

$ \text { domain of } f \cdot g=\text { domain of } f \cap \text { domain of } g \text {. } $

Solution

We have, $ f=\lbrace(0,1),(2,0),(3,-4),(4,2),(5,1)\rbrace $

And $g=\lbrace(1,0),(2,2),(3,-1),(4,4),(5,3)\rbrace$

$ \text { Domain of } f=\lbrace0,2,3,4,5\rbrace $

and Domain of $g=\lbrace1,2,3,4,5\rbrace$

$\therefore$ Domain of $(f \cdot g)=$ Domain of $ f \ \cap$ Domain of $g=\lbrace2,3,4,5\rbrace$

Solution

We have, $f=\lbrace(2,4),(5,6),(8,-1),(10,-3)\rbrace $

$\text { And } \quad g=\lbrace(2,5),(7,1),(8,4),(10,13),(11,5)\rbrace $

$\text { So, } f-g, f+g, f . g, \frac{f}{g} \text { are defined in the domain as (domain of } f \cap \text { domain of } g \text { ) } $

$\text { i.e., }\lbrace2,5,8,10\rbrace \cap\lbrace2,7,8,10,11\rbrace = \lbrace2,8,10\rbrace $

$\text { (i) }(f-g)(2)=f(2)-g(2)=4-5=-1 $

$\quad (f-g)(8)=f(8)-g(8)=-1-4=-5 $

$\quad (f-g)(10)=f(10)-g(10)=-3-13=-16 $

$\therefore \quad f-g=\lbrace(2,-1),(8,-5),(10,-16)\rbrace $

(ii) $(f+g)(2)=f(2)+g(2)=4+5=9 $

$\quad (f+g)(8)=f(8)+g(8)=-1+4=3 $

$\quad (f+g)(10)=f(10)+g(10)=-3+13=10 $

$\therefore \quad f+g=\lbrace(2,9),(8,3),(10,10)\rbrace $

(iii) $(f \cdot g)(2)=f(2) \cdot g(2)=4 \times 5=20$

$ \begin{aligned} (f \cdot g)(8) & =f(8) \cdot g(8)=-1 \times 4=-4 \\ (f \cdot g)(10) & =f(10) \cdot g(10)=-3 \times 13=-39 \\ \therefore \quad f \cdot g & =\lbrace(2,20),(8,-4),(10,-39)\rbrace \end{aligned} $

(iv) $\frac{f}{g}(2)=\frac{f(2)}{g(2)}=\frac{4}{5}$

$\frac{f}{g}(8)=\frac{f(8)}{g(8)}=\frac{-1}{4}$

$\frac{f}{g}(10)=\frac{f(10)}{g(10)}=\frac{-3}{13}$

$\therefore \quad \frac{f}{g}= \lbrace (2, \frac{4}{5}), (8,-\frac{1}{4}), (10, \frac{-3}{13}) \rbrace $

Hence, the correct matches are (i) $\rightarrow$ (c), (ii) $\rightarrow$ (d), (iii) $\rightarrow$ (b), (iv) $\rightarrow$ (a).

Solution

False

We have, $\ R=\lbrace(x, y): y=x-5, x, y \in Z\rbrace$

If $\ x=5$, then $y=5-5=0$

Hence, $(5, 2)$ does not belong to $R$.

Solution

False

We have, $\ P=\lbrace1,2\rbrace$ and $n(P)=2$

$\therefore \quad n(P \times P \times P)=n(P) \times n(P) \times n(P)=2 \times 2 \times 2=8$

But given $P \times P \times P$ has 4 elements.

Thinking Process

First, we find $A \times B$ and $A \times C$, then we will find $(A \times B) \cup(A \times C)$.

Solution

True

We have, $ \ A=\lbrace1,2,3\rbrace, B=\lbrace3,4\rbrace$ and $C=\lbrace4,5,6\rbrace$

$A \times B=\lbrace(1,3),(1,4),(2,3),(2,4),(3,3),(3,4)\rbrace$

$A \times C=\lbrace(1,4),(1,5),(1,6),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6)\rbrace$

$(A \times B) \cup(A \times C)=\lbrace(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,3),(3,4),(3,5)$, $(3,6)\rbrace$

Solution

False

We have, $ \ (x-2, y+5)=(-2, \frac{1}{3})$

$\Rightarrow \quad x-2 =-2 \ \text{and} \ y+5=\frac{1}{3} $

$\Rightarrow \quad x=-2+2 \ \text{and} \ y=\frac{1}{3}-5 $

$\therefore \quad x =0 \ \text{and} \ y=\frac{-14}{3}$

Solution

True

We have, $\ A \times B=\lbrace(a, x),(a, y),(b, x),(b, y)\rbrace$

$A=$ Set of first element of ordered pairs in $A \times B=\lbrace a, b\rbrace$

$B=$ Set of second element of ordered pairs in $A \times B=\lbrace x, y\rbrace$