If $R = {(x, y) : x, y \in Z, x^2 + 3y^2 \le 8}$ is a relation on the set of integers $Z$, then the domain of $R^{-1}$ is :
Correct Answer:
C
Explanation
Given $R = {(x, y) : x, y \in Z, x^2 + 3y^2 \le 8}$.
So $R = {(0,1), (0,-1), (1,0), (-1,0), (1,1), (1,-1), (-1,1), (-1,-1), (2,0), (-2,0), (2,1), (2,-1), (-2,1), (-2,-1)}$.
$\Rightarrow R : {-2, -1, 0, 1, 2} \rightarrow {-1, 0, 1}$.
$\therefore R^{-1} : {-1, 0, 1} \rightarrow {-2, -1, 0, 1, 2}$.
$\therefore$ Domain of $R^{-1} = {-1, 0, 1}$.