Solution

$A=\{-1,2,3\}$ and $B=\{1,3\}$

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

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

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

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

Solution

We have,

and

$\begin{array}{rl}P& =\{x:x<3,x\in N\}=\{1,2\}\\ Q& =\{x:x\le 2,x\in W\}=\{0,1,2\}\\ P\cup Q& =\{0,1,2\}\text{ and }P\cap Q=\{1,2\}\\ (P\cup Q)\times (P\cap Q)& =\{0,1,2\}\times \{1,2\}\\ & =\{(0,1),(0,2),(1,1),(1,2),(2,1),(2,2)\}\end{array}$

$\therefore P\cup Q=\{0,1,2\}\text{ and }P\cap Q=\{1,2\}$

Solution

We have,

and

$\begin{array}{rl}A& =\{x:x\in W,x<2\}=\{0,1\}\\ B& =\{x:x\in N,1<x<5\}\\ & =\{2,3,4\}\text{ and }C=\{3,5\}\end{array}$

(i) $:$

$B\cap C=\{3\}$

$\therefore A\times (B\cap C)=\{0,1\}\times \{3\}=\{(0,3),(1,3)\}$

(ii) $\because (B\cup C)=\{2,3,4,5\}$

$\begin{array}{rl}\therefore A\times (B\cup C)& =\{0,1\}\times \{2,3,4,5\}\\ & =\{(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)\}\end{array}$

Solution

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

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

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

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

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

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

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

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

(ii) We have, $\left(\begin{array}{c}\frac{a}{4},a-2b\end{array}\right)=(0,6+b)$

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

Solution

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

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

$\{(1,4),(2,3),(3,2),(4,1)\}$.

(ii) The set of ordered pairs satisfying $x+y<5$ is $\{(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)\}$.

(iii) The set of ordered pairs satisfying $x+y>8$ is $\{(4,5),(5,4),(5,5)\}$.

Solution

We have,

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

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

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

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

Solution

We have,

$\begin{array}{rl}{R}_1& =\{(x,y)\mid y=2x+7,\text{ where }x\in R\text{ and }-5\le x\le 5\}\\ \text{ Domain of }{R}_1& =\{-5\le x\le 5,x\in R\}\\ & =[-5,5]\\ y& =2x+7\\ y& =2(-5)+7=-3\\ y& =2(5)+7=17\\ \text{ Range of }{R}_1& =\{-3\le y\le 17,y\in R\}\\ & =[-3,17]\end{array}$

$\begin{array}{rl}& \text{ When }x=-5\text{, then }\\ & \text{ When }x=5\text{, then }\\ & \therefore \text{ Range of }{R}_1=\{-3\le y\le 17,y\in R\}\end{array}$

Solution

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

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

$\therefore R_2=(0,8),(0,-8),(8,0),(-8,0)\}$

Solution

We have

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

Clearly, domain of

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

$\therefore \text{ Range of }{R}_3=R^{+}\cup 0\}\text{ or }(0,\mathrm{\infty })$

Solution

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

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

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

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

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

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, $s=\{(x,x^2)\mid x.$ is a positive integer $\}$

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

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 $g(x)=3x+5$.

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

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

(iii) $f(-2)=(-2{)}^2+7=4+7=11$ and $g(-1)=3(-1)+5=-3+5=2$

$\therefore f(-2)+g(-1)=11+2=13$ (iv) $f(t)=t^2+7$ and $f(-2)=(-2{)}^2+7=4+7=11$

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

(v) $f(t)=t^2+7$ and $f(5)=5^2+7=25+7=32$

$\begin{array}{rl}\therefore \frac{f(t)-f(5)}{t-5},\text{ if }t\ne 5& \\ & =\frac{t^2+7-32}{t-5}\\ & =\frac{t^2-25}{t-5}=\frac{(t-5)(t+5)}{(t-5)}\\ & =t+5\end{array}$

Solution

We have,

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

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

Solution

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

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

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

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

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

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

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

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

Solution

We have,

$f:X\Rightarrow R,f(x)=x^3+1$

Where

$X=\{-1,0,3,9,7\}$

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

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

$\therefore$ Range of $f=\{0,1,28,730,344\}$

Solution

$f(x)=g(x)$

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

Solution

We have,

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

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

Now

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

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

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

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 $1-\cos x\ne 0$

$\cos x\ne 1$

$x\ne 2n\pi -\mathrm{\forall }n\in Z$

$\therefore$ Domain of $f=R-\{2n\pi :n\in Z\}$

(ii) We have,

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

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

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

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

(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

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

(v) We have,

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

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

$\Rightarrow x\ne 28$

$\therefore$ Domain of $f=R-\{28\}$

Solution

(i) We have,

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

Let

Then,

$\Rightarrow$

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

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

(ii) We know that, $\left|\begin{array}{c}x-2\end{array}\right|\ge 0\Rightarrow$ $-\left|\begin{array}{c}x-2\end{array}\right|$

$\Rightarrow$ 1- $\left|\begin{array}{c}x-2\end{array}\right|\le 1\Rightarrow$ $f(x)\le 1$

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

(iii) We know that,

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

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

(iv) We know that,

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

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

Solution

Since,

$\text{ and }$

$\begin{array}{r}Since,|x-2|=-(x-2),x<2\\ x-2,x\ge \ge 2\\ And|2+x|=-(2+x),x<-2\\ (2+x),x\ge -2\\ f(x)=|x-2|+|2+x|,-3\le x\le 3\\ =\{\begin{array}{l}-(x-2)-(2+x),-3\le x<-2\\ -(x-2)+2+x,-2\le x<2\\ x-2+2+x,2\le x\le 3\end{array}\\ =\{\begin{array}{l}-2x,-3\le x<-2\\ =-2\le x<2\\ 2,2\le x\le 3\end{array}\end{array}$

Solution

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

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

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

Solution

We have, $f(x)=\sqrt{x}$ and $g(x)=x$ be two function defined in the domain $R^{+}\cup \{0\}$.

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

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

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

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

Solution

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

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

$\therefore$ Domain of $f=(5,\mathrm{\infty })$

Let

$f(x)=y$

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

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

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

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

Solution

We have,

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

$\begin{array}{rl}\therefore f(y)& =\frac{ay-b}{cy-a}=\frac{a(\frac{ax-b}{cx-a})-b}{c(\frac{ax-b}{cx-a})-a}\\ & =\frac{a(ax-b)-b(cx-a)}{c(ax-b)-a(cx-a)}=\frac{a^{2}x-ab-bcx+ab}{acx-bc-acx+a^2}\\ & =\frac{a^{2}x-bcx}{a^2-bc}=\frac{x(a^2-bc)}{(a^2-bc)}=x\\ \therefore f(y)& =x\end{array}$

Hence proved.

Solution

(d) We have,

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

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

Solution

(c) We have,

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

Solution(b) We know that,

$-1\le -\cos x\le 1$

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

Solution

(c) We have,

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

$\begin{array}{cc}\because & \sqrt{1+x^2{y}^2}\le \sqrt{1+x^2+y^2+x^2{y}^2}\\ \Rightarrow & f(xy)\le f(x)\cdot f(y)\end{array}$

Solution

(b) Let

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

$f(x)$ is defined, if

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

$\Rightarrow x^2-a^2\le 0$

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

$\Rightarrow -a\le x\le a$

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

Solution

(b) We have,

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

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

$\begin{array}{rl}4-x& \ge 0\text{ or }{x}^2-1>0\\ x-4& \le 0\text{ or }(x+1)(x-1)>0\\ x& \le 4\text{ or }x<-1\text{ and }x>1\end{array}$

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

Solution

(c) We have,

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

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

$\therefore$ Domain of $f=R-\{4\}$

Let $f(x)=y$

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