JEE Main 2026 · 6 Apr · Shift 1 · Mathematics Q3

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MathematicsSingle correct+4 / -1Question 3 of 75

Let $e_1$ and $e_2$ be two distinct roots of the equation $x^2-ax+2=0$ . Let the sets
$\{a\in\mathbb R:e_1\text{ and }e_2\text{ are the eccentricities of hyperbolas}\}=(\alpha,\beta)$ , and
$\{a\in\mathbb R:e_1\text{ and }e_2\text{ are the eccentricities of an ellipse and a hyperbola, respectively}\}=(\gamma,\infty)$ .
Then $\alpha^2+\beta^2+\gamma^2$ is equal to:

Answer and worked solution

Official answer: C

Roots of $x^2-ax+2$ : product 2, sum a. Write $p(x)=x^2-ax+2$ .

Both > 1 (two hyperbolas). Need:

$a^2>8$ ;
$p(1)=3-a>0$ ;
$\tfrac a2>1$ .

So $a\in(2\sqrt2,3)$ , with $\alpha^2+\beta^2=8+9=17$ .

One in (0,1), one > 1 (an ellipse and a hyperbola). $p(1)=3-a<0$ , and $p(0)=2>0$ keeps the smaller root positive. So $a>3$ and $\gamma=3$ .

Total: $17+9=26$ .

(C)

Topics:Location of roots Hyperbola Ellipse

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JEE Main 2026 · 6 Apr · Shift 1, Mathematics question 3