Question

$\mathrm{R}=\left\{(\mathrm{x},\mathrm{y}):x,\mathrm{y}\in \mathrm{I},{\mathrm{x}}^2+{\mathrm{y}}^2\le 4\right\}$ is a relation in $\mathrm{I}$ then domain of $\mathrm{R}$ is:

• A. $\left\{0,1,2\right\}$
• B. $\left\{-2,-1,0\right\}$
• C. $\left\{-2,-1,0,1,2\right\}$
• D. $\left\{-2,-1\right\}$

Answer 1

The correct option is C

$\left\{-2,-1,0,1,2\right\}$

Explanation for the correct option:

Finding the domain of $\mathrm{R}$:

Given relation $\mathrm{R}=\left\{(\mathrm{x},\mathrm{y}):x,\mathrm{y}\in \mathrm{I},{\mathrm{x}}^2+{\mathrm{y}}^2\le 4\right\}$

Now,

$$\begin{gathered} {\mathrm{x}}^2+{\mathrm{y}}^2\le 4 \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2\le 2^2...\left(i\right) \end{gathered}$$

It is nothing but equation of circle having Centre $(0,0)$ and radius $2$.

Therefore, the integers in the interval $[-2,2]$are the appropriate values of $(\mathrm{x},\mathrm{y})$

When we put $x=0$we get $y=2$ which lies inside $[-2,2]$

When we put $x=\pm 1$we get $y=\sqrt{3}$ which lies inside $[-2,2]$

When we put $x=\pm 2$we get $y=0$ which lies inside $[-2,2]$

Thus, the domain of $\mathrm{R}$ is $\{0,\pm 1,\pm 2\}$ which is $\{-2,-1,0,1,2\}$

Hence, the correct answer is option (C).