JEE Main 2026 — Jan 22, Shift 1 — Mathematics solution

JEE Main2026Jan 22, Shift 1Mathematics

Let the relation R on the set $M = {1, 2, 3, \dots, 16}$ be given by $R = {(x, y) : 4 y = 5 x - 3, x, y \in M}$ . Then the minimum number of elements required to be added in R, in order to make the relation symmetric, is equal to

Solution

Find pairs $(x, y) \in M \times M$ with $y = \frac{5 x - 3}{4}$ — need $5 x - 3$ divisible by 4, so $x \equiv 3 (mod 4)$ .

Try $x = 3 \Rightarrow y = 3$ ; $x = 7 \Rightarrow y = 8$ ; $x = 11 \Rightarrow y = 13$ ; $x = 15 \Rightarrow y = 18$ (out of M).

So $R = {(3, 3), (7, 8), (11, 13)}$ .

For symmetry, the reverses of $(7, 8)$ and $(11, 13)$ must be in $R$ . So we need to add $(8, 7)$ and $(13, 11)$ — 2 elements.

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