Question

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

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

Answer 1

Answer: (C)

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

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

Let $x=0\therefore x^2+y^2\le 4\Rightarrow y^2\le 4\Rightarrow y=0,\pm 1,\pm 2$

Let $x=\pm 1\therefore x^2+y^2\le 4\Rightarrow y^2\le 3\Rightarrow y=0,\pm 1$

Let $x=\pm 2\therefore x^2+y^2\le 4\Rightarrow y^2\le 0\Rightarrow y=0$

$\therefore R=\left\{\right(0,0),(0,-1),(0,1),(0,-2),(0,2),(-1,0),(1,0),(1,1),(1,-1),(-1,1),(2,0),(-2,0\left)\right\}$

$\therefore$ Domain of $R=\{x:(x,y\left)ϵR\right\}=\{0,-1,1,-2,2\}$.

$\therefore$ The correct answer is $C$.