Question

If $R=\left\{(x,y)|x,y\in Z,x^2+3y^2\le 8\right\}$ is a relation on the set of integers $Z$, then the domain of $R^{-1}$ is

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

Answer 1

The correct option is A

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

Explanation for the correct option:

Step 1: Find the range of $R$, because the range of R is the domain of $R^{-1}$.

Substitute $x=0$ in the above equation, then

Given, $R=\left\{(x,y)|x,y\in Z,x^2+3y^2\le 8\right\}$

$\begin{array}{rcl}3y^2& \le & 8\\ & \Rightarrow & y^2\le \frac{8}{3}\\ & \Rightarrow & y=\pm 1,0\end{array}$

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

$\begin{array}{c}1+3y^2\le 8\text{}\\ \Rightarrow y^2\le \frac{7}{3}\text{}\\ \Rightarrow y=0,\pm 1\end{array}$

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

$\begin{array}{c}4+3y^2\le 8\text{}\\ \Rightarrow y^2\le \frac{4}{3}\text{}\\ \Rightarrow y=0\end{array}$

Therefore, the relation $R=\{(0,0),(0,-1),(0,1),(1,0),(1,-1),(1,1),(-1,0),(1,-1),(-1,1),(2,0),\left(-2,0\right)\}$

Thus, $R$ can be written as $\{-2,-1,0,1,2\}\to \{-1,0,1\}$

Where, the range of $R$ is $\{-1,0,1\}$.

Since the domain of $R^{-1}$ is the range of $R$, then $R^{-1}$ can be written as $\{-1,0,1\}\to \{-2,-1,0,1,2\}$.

Therefore the domain of $R^{-1}$ is $\{-1,0,1\}$.

Hence, the correct option is (A).