JEE Main 2026 Jan 21 Shift 2, Mathematics Q17: Relations

JEE Main2026Jan 21, Shift 2Mathematics

Let $A = {2, 3, 5, 7, 9}$ . Let R be the relation on A defined by $x R y$ if and only if $2 x \leq 3 y$ . 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 symmetric relation. Then $ℓ + m$ is equal to:

A
23
B
25
C
21
D
27

Solution

The condition $x R y \Leftrightarrow 2 x \leq 3 y \Leftrightarrow y \geq \frac{2 x}{3}$ . List valid pairs for each $x \in A$ :

$x = 2$ : $y \geq 4/3$ , so $y \in {2, 3, 5, 7, 9}$ — 5 pairs.

$x = 3$ : $y \geq 2$ , so $y \in {2, 3, 5, 7, 9}$ — 5 pairs.

$x = 5$ : $y \geq 10/3$ , so $y \in {5, 7, 9}$ — 3 pairs.

$x = 7$ : $y \geq 14/3$ , so $y \in {5, 7, 9}$ — 3 pairs.

$x = 9$ : $y \geq 6$ , so $y \in {7, 9}$ — 2 pairs.

Total $ℓ = 5 + 5 + 3 + 3 + 2 = 18$ .

For symmetry, every $(a, b) \in R$ needs $(b, a) \in R$ . The pairs whose reverses are missing: $(2, 5), (2, 7), (2, 9), (3, 5), (3, 7), (3, 9), (5, 9)$ — checking each reverse: $(5, 2), (7, 2), (9, 2), (5, 3), (7, 3), (9, 3), (9, 5)$ are not in R. That gives $m = 7$ pairs to add.

$ℓ + m = 18 + 7 = 25$ .

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JEE Main 2026 Jan 21 Shift 2, Mathematics Q17: Relations

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