JEE Main 2026 — Jan 23, Shift 1 — Mathematics solution

JEE Main2026Jan 23, Shift 1Mathematics

Let $A = {- 2, - 1, 0, 1, 2, 3, 4}$ . Let R be a relation on A defined by $x R y$ if and only if $2 x + y \leq 2$ . 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
$32$
B
$34$
C
$33$
D
$35$

Solution

Count elements in R. For each $x \in A$ , count $y \in A$ with $y \leq 2 - 2 x$ :

$x = - 2 \Rightarrow y \leq 6$ : 7 choices. $x = - 1 \Rightarrow y \leq 4$ : 7. $x = 0 \Rightarrow y \leq 2$ : 5. $x = 1 \Rightarrow y \leq 0$ : 3. $x = 2 \Rightarrow y \leq - 2$ : 1. $x = 3, 4 \Rightarrow$ no $y$ works.

$ℓ = 7 + 7 + 5 + 3 + 1 = 23$ .

Reflexive (m). Need $(a, a)$ for each $a \in A$ . Check which already lie in R: $2 a + a \leq 2 \Rightarrow a \leq 2/3 \Rightarrow a \in {- 2, - 1, 0}$ . So missing: $(1, 1), (2, 2), (3, 3), (4, 4)$ , giving $m = 4$ .

Symmetric (n). For every $(x, y) \in R$ with $x \neq = y$ , check whether $(y, x) \in R$ . Identifying pairs where the reverse is missing yields 6 needed additions (e.g. $(3, - 2), (4, - 2), (2, - 1), (2, 0), (3, - 1), (4, - 1)$ ), so $n = 6$ .

$ℓ + m + n = 23 + 4 + 6 = 33$ .

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