Question

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

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

Answer 1

The correct option is C

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

Find the domain of the function:

Step 1: Substitute $x=0$ in the given equation

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

$\begin{array}{rcl}{y}^2& \le & 4\\ & \Rightarrow & y=\pm 2,\pm 1,0\end{array}$

Step 2: Again, substitute $x=1$or $x=-1$ in the given equation

$\begin{array}{c}1+y^2\le 4\text{}\\ \Rightarrow y^2\le 3\text{}\\ \Rightarrow y=0,\pm 1\end{array}$

Step 3: $$Again, substitute $x=2$ or $x=-2$in the given equation

$\begin{array}{c}4+y^2\le 4\text{}\\ \Rightarrow y^2\le 0\text{}\\ \Rightarrow y=0\end{array}$

Therefore, the relation $R=\{(0,0),(0,-1),(0,1),(0,-2),(0,2),(-1,0),(1,0),(1,1),(-1,1),(1,-1)(-2,0),(2,0)\}$
Thus, the domain of $R$ is $\{-2,-1,0,1,2\}$.

Hence, the correct option is (C).