JEE Main 2026 — Jan 24, Shift 1 — Mathematics solution

JEE Main2026Jan 24, Shift 1Mathematics

Let $R$ be a relation defined on the set ${1, 2, 3, 4} \times {1, 2, 3, 4}$ by

$R = {((a, b), (c, d)) : 2 a + 3 b = 3 c + 4 d}$ .

Then the number of elements in $R$ is

A
$6$
B
$18$
C
$12$
D
$15$

Solution

For each $(a, b)$ with $a, b \in {1, 2, 3, 4}$ , compute $N = 2 a + 3 b$ and count $(c, d) \in {1, 2, 3, 4}^{2}$ with $3 c + 4 d = N$ .

The achievable values of $3 c + 4 d$ over ${1, 2, 3, 4}^{2}$ are: $7, 10, 11, 13, 14, 15, 16, 17, 18, 19, 20, 22$ (mostly unique).

Pairing each $(a, b)$ with its matching $(c, d)$ where one exists:

$(1, 3) \to (1, 2)$ [ $N = 11$ ]; $(1, 4) \to (2, 2)$ [ $N = 14$ ]; $(2, 1) \to (1, 1)$ [ $N = 7$ ]; $(2, 2) \to (2, 1)$ [ $N = 10$ ]; $(2, 3) \to (3, 1)$ [ $N = 13$ ]; $(2, 4) \to (4, 1)$ [ $N = 16$ ]; $(3, 3) \to (1, 3)$ [ $N = 15$ ]; $(3, 4) \to (2, 3)$ [ $N = 18$ ]; $(4, 1) \to (1, 2)$ [ $N = 11$ ]; $(4, 2) \to (2, 2)$ [ $N = 14$ ]; $(4, 3) \to (3, 2)$ [ $N = 17$ ]; $(4, 4) \to (4, 2)$ [ $N = 20$ ].

The remaining $(a, b)$ values $(1, 1), (1, 2), (3, 1), (3, 2)$ produce $N \in {5, 8, 9, 12}$ , none of which are representable as $3 c + 4 d$ with $c, d \in {1, 2, 3, 4}$ .

Total pairs in $R$ : $12$ .

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