Class JEE Mathematics Sets, Relations, and Functions Q #1003
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4 Marks 2025 JEE Main 2025 (Online) 3rd April Morning Shift MCQ SINGLE
Let $A = {-3, -2, -1, 0, 1, 2, 3}$. Let R be a relation on A defined by $xRy$ if and only if $0 \le x^2 + 2y \le 4$. Let $l$ be the number of elements in R and $m$ be the minimum number of elements required to be added in R to make it a reflexive relation. Then $l + m$ is equal to
(A) 18
(B) 20
(C) 17
(D) 19
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Correct Answer: A
Explanation
First, we determine the pairs $(x, y)$ that satisfy the condition $0 \le x^2 + 2y \le 4$ for the given set $A = \{-3, -2, -1, 0, 1, 2, 3\}$.

For $y = -3$, $x^2 + 2(-3) \le 4 \implies x^2 \le 10$, so $x$ can be $-3$ or $3$.
For $y = -2$, $x^2 + 2(-2) \le 4 \implies x^2 \le 8$, so $x$ can be $-2$ or $2$.
For $y = -1$, $x^2 + 2(-1) \le 4 \implies x^2 \le 6$, so $x$ can be $-2$ or $2$.
For $y = 0$, $x^2 + 2(0) \le 4 \implies x^2 \le 4$, so $x$ can be $-2, -1, 0, 1, 2$.
For $y = 1$, $x^2 + 2(1) \le 4 \implies x^2 \le 2$, so $x$ can be $-1, 0, 1$.
For $y = 2$, $x^2 + 2(2) \le 4 \implies x^2 \le 0$, so $x$ can only be $0$.

Thus, the relation R consists of the following pairs:
$R = \{(3, -3), (-3, -3), (-2, -2), (2, -2), (-2, -1), (2, -1), (-2, 0), (-1, 0), (0, 0), (1, 0), (2, 0), (-1, 1), (0, 1), (1, 1), (0, 2)\}$.
The number of elements in R is $l = 15$.

To make R reflexive, it must include all pairs $(x, x)$ for every $x \in A$. The set A has 7 elements, so the reflexive relation must contain $(-3, -3), (-2, -2), (-1, -1), (0, 0), (1, 1), (2, 2), (3, 3)$. Currently, R has $(-3,-3), (-2, -2), (0,0), (1,1)$. So the missing pairs are $(-1, -1), (2, 2), (3, 3)$, meaning $m = 3$.
Therefore, $l + m = 15 + 3 = 18$.

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