JEE Main 2025 Apr 3 Shift 2, Mathematics Q6: Relations

JEE Main2025Apr 3, Shift 2Mathematics

Let $A = {- 2, - 1, 0, 1, 2, 3}$ . Let $R$ be a relation on $A$ defined by $x R y$ if and only if $y = max {x, 1}$ . Let $l$ be the number of elements in $R$ . Let $m$ and $n$ be the minimum number of elements required to be added in $R$ to make it reflexive and symmetric relations, respectively. Then $l + m + n$ is equal to

A
$12$
B
$11$
C
$13$
D
$14$

Solution

For each $x \in A$ , compute $max {x, 1}$ :

$x = - 2 \to 1$ , $x = - 1 \to 1$ , $x = 0 \to 1$ , $x = 1 \to 1$ , $x = 2 \to 2$ , $x = 3 \to 3$ .

So $R = {(- 2, 1), (- 1, 1), (0, 1), (1, 1), (2, 2), (3, 3)}$ , giving $l = 6$ .

For reflexivity, $R$ must contain $(x, x)$ for every $x \in A$ . The pairs $(1, 1)$ , $(2, 2)$ , $(3, 3)$ are already present; the missing reflexive pairs are $(- 2, - 2), (- 1, - 1), (0, 0)$ . So $m = 3$ .

For symmetry, for each $(a, b) \in R$ with $a \neq = b$ we need $(b, a)$ . The asymmetric pairs are $(- 2, 1), (- 1, 1), (0, 1)$ ; their reverses $(1, - 2), (1, - 1), (1, 0)$ are absent. So $n = 3$ .

$∴ l + m + n = 6 + 3 + 3 = 12$ .

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JEE Main 2025 Apr 3 Shift 2, Mathematics Q6: Relations

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