JEE Main 2025 Apr 3 Shift 1, Mathematics Q5: Relations

JEE Main2025Apr 3, Shift 1Mathematics

Let $A = {- 3, - 2, - 1, 0, 1, 2, 3}$ . Let $R$ be a relation on $A$ defined by $x R y$ if and only if $0 \leq x^{2} + 2 y \leq 4$ . 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
19
B
20
C
17
D
18

Solution

The constraint rewrites as $- 2 y \leq x^{2} \leq 4 - 2 y$ . For each $y \in A$ , find admissible $x \in A$ .

$y = - 3$ : $6 \leq x^{2} \leq 10 \Rightarrow x \in {- 3, 3}$ (2 values).

$y = - 2$ : $4 \leq x^{2} \leq 8 \Rightarrow x \in {- 2, 2}$ (2 values).

$y = - 1$ : $2 \leq x^{2} \leq 6 \Rightarrow x \in {- 2, 2}$ (2 values).

$y = 0$ : $0 \leq x^{2} \leq 4 \Rightarrow x \in {- 2, - 1, 0, 1, 2}$ (5 values).

$y = 1$ : $- 2 \leq x^{2} \leq 2 \Rightarrow x \in {- 1, 0, 1}$ (3 values).

$y = 2$ : $- 4 \leq x^{2} \leq 0 \Rightarrow x \in {0}$ (1 value).

$y = 3$ : $- 6 \leq x^{2} \leq - 2$ : impossible (0 values).

Total: $l = 2 + 2 + 2 + 5 + 3 + 1 + 0 = 15.$

For reflexivity we need every pair $(x, x)$ for $x \in A$ . Check which are already present: $(- 2, - 2), (0, 0), (1, 1), (- 3, - 3)$ are in $R$ . Missing: $(- 1, - 1), (2, 2), (3, 3)$ . So $m = 3$ .

$l + m = 15 + 3 = 18.$

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JEE Main 2025 Apr 3 Shift 1, Mathematics Q5: Relations

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