If $R=\left\{(x, y): x, y \in W, x^2+y^2=25\right\}$, then find the domain and range of $R$.
We have,
$\begin{aligned} R & =\left\{(x, y): x, y \in W, x^2+y^2=25\right\} \\ & =\{(0,5),(3,4),(4,3),(5,0)\}\end{aligned}$
$\begin{aligned} \text{Domain of }\quad R & =\text { Set of first element of ordered pairs in } R \\ & =\{0,3,4,5\}\end{aligned}$
$\begin{aligned} \text{Range of }\quad R & =\text { Set of second element of ordered pairs in } R \\ & =\{5,4,3,0,\}\end{aligned}$
If $R_1=\{(x, y) \mid y=2 x+7$, where $x \in R$ and $-5 \leq x \leq 5\}$ is a relation. Then, find the domain and range of $R_1$.
We have, $\quad R_1=\{x, y) \mid y=2 x+7$, where $x \in R$ and $\left.-5 \leq x \leq 5\right\}$
Domain of $R_1=\{-5 \leq x \leq 5, x \in R\}$
$=[-5,5]$
$\begin{array}{ll}\because & y=2 x+7 \\ \text { When } x=-5 \text {, then } & y=2(-5)+7=-3 \\ \text { When } x=5 \text {, then } & y=2(5)+7=17\end{array}$
$$\begin{aligned} & \therefore \quad \text { Range of } R_1=\{-3 \leq y \leq 17, y \in R\} \\ & =[-3,17] \end{aligned}$$
If $R_2=\{x, y) \mid x$ and $y$ are integers and $\left.x^2+y^2=64\right\}$ is a relation, then find the value of $R_2$.
We have, $\mathrm{R}_2=\{(x, y)\} x$ and $y$ are integers and $\left.x^2+y^2=64\right\}$
Since, 64 is the sum of squares of 0 and $\pm 8$.
When $x=0$, then $y^2=64 \Rightarrow y= \pm 8$
$$\begin{aligned} x & =8 \text {, then } y^2=64-8^2 \Rightarrow 64-64=0 \\ x & =-8, \text { then } y^2=64-(-8)^2=64-64=0 \\ \therefore \quad R_2 & =\{(0,8),(0,-8),(8,0),(-8,0)\} \end{aligned}$$
If $R_3=\{(x,|x|) \mid x$ is a real number $\}$ is a relation, then find domain and range of $R_3$.
We have, $R_3=\{(x,|x|) \mid x$ is real number$\}$
clearly, domain of $R_3=R$
Since, image of any real number under $R_3$ is positive real number or zero.
$$\therefore \quad \text { Range of } R_3=R^{+} \cup\{0\} \text { or }(0, \infty)$$
Is the given relation a function? Give reason for your answer.
(i) $h=\{(4,6),(3,9),(-11,6),(3,11)\}$
(ii) $f=\{(x, x) \mid x$ is a real number $\}$
(iii) $g=\left\{\left(x, \frac{1}{x}\right) x\right.$ is a positive integer $\}$
(iv) $s=\left\{\left(x, x^2\right) \mid x\right.$ is a positive integer $\}$
(v) $t=\{(x, 3) \mid x$ is a real number $\}$
(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, $g=\left\{\left.\left(x, \frac{1}{x}\right) \right\rvert\, x\right.$ 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, $s=\left\{\left(x, x^2\right) \mid x\right.$ 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.