Kerala KEAM 5 June 2024 Solved Paper

We are given the relation: $$R = \{(x, y) : x, y \in \mathbb{Z}, x^2 + 3y^2 \leq 7 \}.$$This is a set of pairs of integers $(x, y)$ where $x^2 + 3y^2 \leq 7$.Step 1: We will find the possible values for $y$ for different integer values of $x$.- For $x = 0$:$$0^2 + 3y^2 \leq 7 \quad \Rightarrow \quad 3y^2 \leq 7 \quad \Rightarrow \quad y^2 \leq \frac{7}{3} \approx 2.33.$$Therefore, $y^2 \leq 2$, so $y \in \{-1, 0, 1\}$.- For $x = \pm 1$:$$1^2 + 3y^2 \leq 7 \quad \Rightarrow \quad 1 + 3y^2 \leq 7 \quad \Rightarrow \quad 3y^2 \leq 6 \quad \Rightarrow \quad y^2 \leq 2.$$Therefore, $y \in \{-1, 0, 1\}$.- For $x = \pm 2$:$$2^2 + 3y^2 \leq 7 \quad \Rightarrow \quad 4 + 3y^2 \leq 7 \quad \Rightarrow \quad 3y^2 \leq 3 \quad \Rightarrow \quad y^2 \leq 1.$$Therefore, $y \in \{-1, 0, 1\}$.- For $x = \pm 3$:$$3^2 + 3y^2 \leq 7 \quad \Rightarrow \quad 9 + 3y^2 \leq 7 \quad \Rightarrow \quad 3y^2 \leq -2.$$This results in no valid solutions for $y$.Step 2: From the above analysis, we see that the possible values for $y$ are $\{-1, 0, 1\}$.Therefore, the range of the relation $R$, which consists of the set of all possible values of $y$, is: $$\{0, -1, 1\}.$$Thus, the correct answer is option (E). Quick Tip: To find the range of a relation, determine all possible values of the second element in the ordered pairs that satisfy the given condition. In this case, solve for $y$ for different integer values of $x$.