JEE Main 2026 Apr 8 Shift 2, Mathematics Q1: Relations

JEE Main2026Apr 8, Shift 2Mathematics

Consider the relation $R$ on the set ${- 2, - 1, 0, 1, 2}$ defined by $(a, b) \in R$ if and only if $1 + ab > 0$ . Then among the statements :

I. The number of elements in $R$ is $17$

II. $R$ is an equivalence relation

A
Only I is true
B
Only II is true
C
Both I and II are true
D
Neither I nor II is true

Solution

Counting the pairs: Out of $5 \times 5 = 25$ ordered pairs, we must remove those for which $1 + ab \leq 0$ , i.e. $ab \leq - 1$ . These are the pairs $(- 2, - 1), (- 1, - 2), (- 2, 1), (1, - 2), (- 2, 2), (2, - 2), (- 1, 2), (2, - 1)$ — a total of $8$ pairs.

$∴$ $∣ R ∣ = 25 - 8 = 17$ . Statement I is true.

Equivalence check:

Reflexive: $1 + a^{2} > 0$ for every $a$ , so $(a, a) \in R$ . $✓$

Symmetric: $1 + ab = 1 + ba$ , so if $(a, b) \in R$ then $(b, a) \in R$ . $✓$

Transitive: Take $a = - 2, b = 0, c = 2$ . Then $(- 2, 0) \in R$ and $(0, 2) \in R$ , but $1 + (- 2) (2) = - 3 < 0$ , so $(- 2, 2) \in / R$ . Transitivity fails. $\times$

Hence $R$ is not an equivalence relation. Statement II is false.

$∴$ Only I is true.

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JEE Main 2026 Apr 8 Shift 2, Mathematics Q1: Relations

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