If R={(x, y): x, y in Z, x^{2}+3 y^{2} leq 8} | vedclass

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(B) The domain of $R^{-1}$ is the range of the relation $R$. The range of $R$ consists of all possible integer values of $y$ such that there exists at

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

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(B) The domain of $R^{-1}$ is the range of the relation $R$. The range of $R$ consists of all possible integer values of $y$ such that there exists at least one integer $x$ satisfying $x^{2} + 3y^{2} \leq 8$.Given the inequality $3y^{2} \leq 8 - x^{2}$,since $x^{2} \geq 0$,we must have $3y^{2} \leq 8$,which implies $y^{2} \leq \frac{8}{3} \approx 2.66$.Since $y$ must be an integer,the possible values for $y$ are $y \in \{-1, 0, 1\}$.Let us verify these values:If $y = 0$,$x^{2} \leq 8 \Rightarrow x \in \{-2, -1, 0, 1, 2\}$.If $y = 1$,$x^{2} + 3(1)^{2} \leq 8 \Rightarrow x^{2} \leq 5 \Rightarrow x \in \{-2, -1, 0, 1, 2\}$.If $y = -1$,$x^{2} + 3(-1)^{2} \leq 8 \Rightarrow x^{2} \leq 5 \Rightarrow x \in \{-2, -1, 0, 1, 2\}$.Thus,the set of all possible values for $y$ is $\{-1, 0, 1\}$.Therefore,the domain of $R^{-1}$ is $\{-1, 0, 1\}$.

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

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