JEE Main 2025 Apr 4 Shift 2, Mathematics Q4: Relations

JEE Main2025Apr 4, Shift 2Mathematics

Let $A = {- 3, - 2, - 1, 0, 1, 2, 3}$ and $R$ be a relation on $A$ defined by $x R y$ if and only if $2 x - y \in {0, 1}$ . 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
18
B
17
C
15
D
16

Solution

List the pairs $(x, y) \in A \times A$ satisfying $2 x - y = 0$ or $2 x - y = 1$ .

$2 x - y = 0$ (i.e. $y = 2 x$ ): $(0, 0), (- 1, - 2), (1, 2)$ .

$2 x - y = 1$ (i.e. $y = 2 x - 1$ ): $(0, - 1), (1, 1), (2, 3), (- 1, - 3)$ .

So $R$ has $l = 7$ ordered pairs.

Reflexive completion ( $m$ ): $R$ already contains $(0, 0)$ and $(1, 1)$ . The missing diagonal elements are $(- 3, - 3), (- 2, - 2), (- 1, - 1), (2, 2), (3, 3)$ , so $m = 5$ .

Symmetric completion ( $n$ ): for each non-symmetric pair we need to add its reverse. The non-diagonal pairs in $R$ are $(- 1, - 2), (1, 2), (0, - 1), (2, 3), (- 1, - 3)$ , none of whose reverses already lies in $R$ . So $n = 5$ .

$∴ l + m + n = 7 + 5 + 5 = 17$ .

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JEE Main 2025 Apr 4 Shift 2, Mathematics Q4: Relations

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