JEE Main 2026 — Jan 22, Shift 2 — Mathematics solution

JEE Main2026Jan 22, Shift 2Mathematics

Let $α, β$ be the roots of the quadratic equation $12 x^{2} - 20 x + 3 λ = 0$ , $λ \in Z$ . If $\frac{1}{2} \leq ∣ β - α ∣ \leq \frac{3}{2}$ , then the sum of all possible values of $λ$ is :

Solution

Square the inequality: $\frac{1}{4} \leq (β - α)^{2} \leq \frac{9}{4}$ .

Using $(β - α)^{2} = (α + β)^{2} - 4 α β$ with $α + β = \frac{20}{12} = \frac{5}{3}$ and $α β = \frac{λ}{4}$ :

$\frac{1}{4} \leq \frac{25}{9} - λ \leq \frac{9}{4}$ .

Solving: $\frac{19}{36} \leq λ \leq \frac{91}{36}$ , so $λ \in {1, 2}$ (integer values).

Sum $= 1 + 2 = 3$ .

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