Question

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

• A. None of these
• B. $\{0,-1,-2\}$
• C. $\{-2,-1,0,1,2\}$
• D. $\{0,1,2\}$

Answer 1

The correct option is C $\{-2,-1,0,1,2\}$

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

When $x=0$, then

$x^2+y^2\le 4$

$\Rightarrow y^2\le 4$

$\Rightarrow y\in [-2,2]$

$\Rightarrow y=-2,-1,0,1,2$ $[\because y\in Z]$

When $x=\pm 1,$ then

$x^2+y^2\le 4$

$\Rightarrow y^2\le 3$

$\Rightarrow y\in [\sqrt{-3},\sqrt{3}]$

$\Rightarrow y=-1,0,1$ $[\because y\in Z]$

When $x=\pm 2,$ then

$x^2+y^2\le 4$

$\Rightarrow y^2\le 0$

$\Rightarrow y=0$ $[\because y\in Z]$

When $x=\pm 3,$ then

$x^2+y^2\le 4$

$\Rightarrow y^2\le -5$

Which is not possible.

So, the domain of $R$ is all possible values of $x$

Hence, domain $=\{-2,1,0,1,2\}$

Therefore, the correct option is $\left(c\right)$