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,

and

$ \begin{aligned} P & =\lbrace x: x<3, x \in N\rbrace=\lbrace1,2\rbrace \\ Q & =\lbrace x: x \leq 2, x \in W\rbrace=\lbrace0,1,2\rbrace \\ 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 \end{aligned} $

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

Solution

We have,

and

$ \begin{aligned} 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 \text { and } C=\lbrace3,5\rbrace \end{aligned} $

(i) $:$

$B \cap C=\lbrace3\rbrace$

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

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

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

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

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

$ \begin{matrix} \Rightarrow & \frac{a}{4} & =0 \Rightarrow a=0 \\ \text { and } & a-2 b & =6+b \\ \Rightarrow & 0-2 b & =6+b \\ \Rightarrow & -3 b & =6 \\ \therefore & b & =-2 \\ \therefore & a & =0, b=-2 \end{matrix} $

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

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} $

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

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

$R=$ Set of second element of ordered pairs in $R$ $=\lbrace 5,4,3,0 \rbrace$,

Solution

We have,

$ \begin{aligned} 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] \\ y & =2 x+7 \\ y & =2(-5)+7=-3 \\ y & =2(5)+7=17 \\ \text { Range of } R_1 & =\lbrace-3 \leq y \leq 17, y \in R\rbrace \\ & =[-3,17] \end{aligned} $

$ \begin{aligned} & \text { When } x=-5 \text {, then } \\ & \text { When } x=5 \text {, then } \\ & \therefore \quad \text { Range of } R_1=\lbrace-3 \leq y \leq 17, y \in R\rbrace \end{aligned} $

Solution

We have, $R_2=\lbrace(x, y)\rbrace x$ 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 $y^{2}=64-(-8)^{2}=64-64=0$

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

Solution

We have

$ \begin{aligned} & R_3=\lbrace (x,|x|) \mid x \text { is real number }\rbrace \\ & R_3=R \end{aligned} $

Clearly, domain of

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

$ \therefore \quad \text { Range of } R_3=R^{+} \cup0\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.

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

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

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

(iv) We have, $s=\lbrace(x, x^{2}) \mid x.$ 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 $f(5)=5^{2}+7=25+7=32$

$ \begin{aligned} \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 \end{aligned} $

Solution

We have,

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

$ \begin{aligned} & \text { (i) } \because \quad f(x)=g(x) \\ & \Rightarrow \quad 2 x+1=4 x-7 \Rightarrow 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 2 x-4 x+1<4 x-7-4 x \\ & \Rightarrow \quad-2 x+1<-7 \\ & \Rightarrow \quad-2 x<-7-1 \\ & \Rightarrow \quad-2 x<-8 \\ & \Rightarrow \quad \frac{-2 x}{-2}>\frac{-8}{-2} \\ & \therefore \quad x>4 \end{aligned} $

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) $

$ \begin{aligned} & \Rightarrow & 3 x^{2}-1 & =3+x \\ & \Rightarrow & 3 x^{2}-x-4 & =0 \\ & \Rightarrow & 3 x^{2}-4 x+3 x-4 & =0 \\ & \Rightarrow & x(3 x-4)+1(3 x-4) & =0 \\ & \Rightarrow & (3 x-4)(x+1) & =0 \\ & \therefore & x & =-1, \frac{4}{3} \end{aligned} $

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

$ \begin{aligned} g(x) & =\alpha x+\beta \\ g(1) & =\alpha(1)+\beta \\ 1 & =\alpha+\beta \\ g(2) & =\alpha(2)+\beta \end{aligned} $

$ \begin{aligned} & \text { When } x=1 \text {, then } \\ & \Rightarrow \end{aligned} $

$ \Rightarrow \quad 3=2 \alpha+\beta $

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

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

Solution

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

So, $f(x)$ is defined, if $1-\cos x \neq 0$

$\cos x \neq 1$

$x \neq 2 n \pi-\forall n \in Z$

$\therefore \quad$ Domain of $f=R-\lbrace2 n \pi: n \in Z\rbrace$

(ii) We have,

$ \begin{aligned} f(x) & =\frac{1}{\sqrt{x+|x|}} \\ +|x| & =x-x=0, x<0 \\ & =x+x=2 x, x \geq 0 \end{aligned} $

$ \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

$ \begin{aligned} x^{2}-1 & =0 \\ (x-1)(x+1) & =0 \\ x & =-1,1 \\ \text { Domain of } f & =R-\lbrace-1,1\rbrace \end{aligned} $

(v) We have,

$ f(x)=\frac{3 x}{28-x} $

Clearly, $f(x)$ is defined, $\quad$ if $28-x \neq 0$

$\Rightarrow \quad x \neq 28$

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

Solution

(i) We have,

$ \begin{aligned} f(x) & =\frac{3}{2-x^{2}} \\ y & =f(x) \\ y & =\frac{3}{2-x^{2}} \Rightarrow 2-x^{2}=\frac{3}{y} \\ x^{2} & =2-\frac{3}{y} \Rightarrow x=\sqrt{\frac{2 y-3}{y}} \end{aligned} $

Let

Then,

$\Rightarrow$

$x$ 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 know that, $\begin{vmatrix} x-2 \end{vmatrix} \geq 0 \Rightarrow $ $-\begin{vmatrix} x-2 \end{vmatrix}$

$\Rightarrow$ 1- $\begin{vmatrix} x-2 \end{vmatrix} \leq 1 \Rightarrow $ $f(x) \leq 1 $

$\therefore$ $ \text {Range of f}=(-\infty , 1) $

(iii) We know that,

$\therefore$ $ |x-3| \geq 0 \Rightarrow f(x) \geq 0 $

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

(iv) We know that,

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

$\Rightarrow 1-3 \leq 1 + 3 \cos 2 x \leq 1+3 \\ \\ \Rightarrow-2 \leq 1+3 \cos 2 x \leq 1+3 \\ -2 \leq f(x) \leq 4 \\ $ $\therefore \text { Range of } f =[-2,4] $

Solution

Since,

$ \text { and } $

$ \begin{aligned} Since, |x-2|=-(x-2), x<2 \\ x-2, x \geq \geq 2 \\ And|2+x|=-(2+x), x<-2 \\ (2+x), x \geq-2 \\ f(x)=|x-2|+|2+x|,-3 \leq x \leq 3 \\ = \begin{cases}-(x-2)-(2+x), \quad-3 \leq x<-2 \\ -(x-2)+2+x, \quad-2 \leq x<2 \\ x-2+2+x, \quad 2 \leq x \leq 3 \end{cases} \\ =\begin{cases}-2 x, \quad-3 \leq x<-2 \\ =-2 \leq x<2 \\ 2, \quad 2 \leq x \leq 3 \end{cases} \end{aligned} $

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

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

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

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

$f(x)=y$

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

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

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

$\because$ Hence, range of $f=R^{+} \quad x \in(5, \infty) \Rightarrow y \in R^{+}$

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.

Solution

(d) We have,

$ \begin{aligned} n(A) & =m \text { and } n(B)=n \\ n(A \times B) & =n(A) \cdot n(B) \\ & =m n \end{aligned} $

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

Solution

(c) We have,

$ \begin{aligned} \Rightarrow [x]^{2}-5[x]+6 =0 \\ \Rightarrow [x]^{2}-3[x]-2[x]+6 =0 \\ \Rightarrow \quad([x]-3)([x]-2)=0 \\ \Rightarrow [x] =2,3 \\ \therefore x \in[2,3] \end{aligned} $

Solution(b) We know that,

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

$ \begin{matrix} \Rightarrow & -2 \leq-2 \cos x \leq 2 \\ \Rightarrow & 1-2 \leq 1-2 \cos x \leq 1+2 \\ \Rightarrow & -1 \leq 1-2 \cos x \leq 3 \\ \Rightarrow & -1 \leq \frac{1}{1-2 \cos x} \leq \frac{1}{3} \\ \Rightarrow & -1 \leq f(x) \leq \frac{1}{3} \\ \therefore & \text { Range of } f=-1, \frac{1}{3} \end{matrix} $

Solution

(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} $

Solution

(b) Let

$ f(x)=\sqrt{a^{2}-x^{2}} $

$f(x)$ is defined, if

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

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

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

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

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

Solution

(b) We have,

$ \begin{aligned} f(x) & =a x+b \\ f(-1) & =a(-1)+b \\ -5 & =-a+b \\ \text{ and,}f(3) & =a(3)+b \\ 3 & =3 a+b \end{aligned} $

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

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

Solution

(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] $

Solution

(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 $

$ \begin{matrix} \Rightarrow & x y+x=4+4 y \Rightarrow x(y+1)=4(1+y) \\ \therefore & x=\frac{4(1+y)}{y+1} \end{matrix} $

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

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

(a) Domain $=(1, \infty)$, Range $=(0, \infty)$

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

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

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

Thinking Process

Solution

(d) We have,

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

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

$ \begin{aligned} & \Rightarrow x \geq 1 \\ & \therefore \text { Domain of } f=[1, \infty) \\ & \text { Let } f(x)=y \\ & \therefore y=\sqrt{x-1} \\ & \Rightarrow y^{2}=x-1 \\ & \therefore x=y^{2}+1 \end{aligned} $

Solution

(a) We have,

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

$ \begin{matrix} f(x) \text { is defined, if } x^{2}-x-6 & =0 \\ \Rightarrow x^{2}-3 x+2 x-6 =0 \\ \Rightarrow x(x-3)+2(x-3) =0 \\ \Rightarrow (x-3)(x+2) =0 \\ \therefore x =-3,-2 \\ \therefore \text { Domain of } f =R-\lbrace3,-2\rbrace \end{matrix} $

Solution

(b) We have,

$ f(x)=2-|x-5| $

$f(x)$ is defined for all $x \in R$

$ \therefore $

$ \text { Domain of } f=R $

$ \text {we know that} $

$ \Rightarrow $ $ \begin{aligned} |x-5| & \geq 0 \Rightarrow-|x-5| \leq 0 \\ 2-|x-5| & \leq 2 \\ f(x) & \leq 2 \end{aligned} $

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

Solution

(a) We have, $f(x)=3 x^{2}-1$ and $g(x)=3+x$

$ f(x) =g(x) \\ \\ \Rightarrow 3x^{2}-1 =3+x \\ \\ \Rightarrow 3x^{2}-x-4 =0 \\ \\ \Rightarrow 3x^{2}-4 x+3 x-4 =0 \\ \\ \Rightarrow x(3x-4)+1(3 x-4) =0 \\ \\ \Rightarrow (3x-4)(x+1) =0 \\ \\ \therefore x =-1, \frac{4}{3} $

So, domain for which $f(x)$ and $g(x)$ are equal to $\begin{bmatrix}-1, \frac{4}{3} \end{bmatrix}$.

Solution

We have,

and

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

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

$\therefore$

$ \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,

$ \begin{aligned} 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 (domain of } f \cap \text { domain of } g \text { ) } \\ \text { i.e., }\lbrace2,5,8,10\rbrace \cap\lbrace2,7,8,10,11\rbrace \Rightarrow\lbrace2,8,10\rbrace \\ \text { (i) }(f-g)(2)=f(2)-g(2)=4-5=-1 \\ (f-g)(8)=f(8)-g(8)=-1-4=-5 \\ (f-g)(10)=f(10)-g(10)=-3-13=-16 \\ \therefore \quad f-g=\lbrace(2,-1),(8,-5),(10,-16)\rbrace \\ (f+g)(2)=f(2)+g(2)=4+5=9 \\ (f+g)(8)=f(8)+g(8)=-1+4=3 \\ (f+g)(10)=f(10)+g(10)=-3+13=10 \\ \therefore \quad f+g=\lbrace(2,9),(8,3),(10,10)\rbrace \end{aligned} $

(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 g & =\lbrace(2,20),(8,-4),(10,-39)\rbrace \end{aligned} $

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

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

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

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

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

Solution

False

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

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

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

Solution

False

We have, $\quad P=\lbrace1,2\rbrace$ 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.

Solution

True $\frac {\sin\theta}{cos \theta}$ We have, $\quad A=\lbrace1,2,3\rbrace, B=\lbrace3,4\rbrace$ 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, $\quad(x-2, y+5)=(-2, \frac{1}{3})$

$\begin{aligned} \Rightarrow & x-2 =-2, y+5=\frac{1}{3} \Rightarrow x=-2+2, y=\frac{1}{3}-5 \\ \therefore & x =0, y=\frac{-14}{3}\end{aligned}$

Solution

True

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

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

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