JEE Main 2026 — Jan 21, Shift 2 — Mathematics solution

JEE Main2026Jan 21, Shift 2Mathematics

Let $α$ and $β$ be the roots of equation $x^{2} + 2 a x + (3 a + 10) = 0$ such that $α < 1 < β$ . Then the set of all possible values of a is :

A
$(- \infty, \frac{- 11}{5}) \cup (5, \infty)$
B
$(- \infty, - 2) \cup (5, \infty)$
C
$(- \infty, - 3)$
D
$(- \infty, \frac{- 11}{5})$

Solution

Let $f (x) = x^{2} + 2 a x + (3 a + 10)$ . With $α < 1 < β$ (both roots real and 1 lies strictly between), the parabola (opening upward) must satisfy $f (1) < 0$ .

$f (1) = 1 + 2 a + 3 a + 10 = 5 a + 11 < 0 \Rightarrow a < - \frac{11}{5}$ .

So $a \in (- \infty, - \frac{11}{5})$ .

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