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 :

Mathematics · Quadratic Equations

JEE Main 2026 — 21 January, Evening Shift — Question 9

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

Option A:
$\left(-\infty, \frac{-11}{5}\right) \cup(5, \infty)$
Option B:
$(-\infty,-2) \cup(5, \infty)$
Option C:
$(-\infty,-3)$
Option D:
$\left(-\infty, \frac{-11}{5}\right)$
Correct

Answer: D

Step-by-step solution

Given $\alpha < 1 < \beta$ . Let $f(x) = x^2 + 2ax + (3a+10)$ . Since the coefficient of $x^2$ is positive, the parabola opens upward. For one root less than 1 and the other greater than 1, we require $f(1) < 0$ . Compute $f(1) = 1 + 2a + 3a + 10 = 5a + 11$ . Set $5a + 11 < 0$ ⇒ $a < -\frac{11}{5}$ . Thus the set of all possible values of $a$ is $(-\infty, -\frac{11}{5})$ .

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Exam: JEE Main 2026
Paper: 21 January, Evening Shift
Subject: Mathematics
Chapter: Quadratic Equations
Topic: Location of roots of a Quadratic Equation

The positive integer n , for which the solutions of the equation x( x+2)+( x+2)( x+4)+ldots .+( x+2 n-2)( x+2 n) =frac8 n3 are two…

Answer: D

Step-by-step solution

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