JEE Mathematics 2026 Question with Solution

Given: The quadratic is $f(x) = x^2 + 2ax + (3a + 10)$ and its roots $\alpha, \beta$ satisfy $\alpha < 1 < \beta$.

Find: The set of all possible values of $a$.

Since the coefficient of $x^2$ is positive, the parabola opens upward. If $1$ lies between the two roots, then the value of the quadratic at $x = 1$ must be negative.

Substituting $x = 1$:

Thus,

The condition $f(1) < 0$ for an upward-opening parabola ensures that the roots are real and distinct, so the required condition is satisfied.

Therefore, the correct option is D.

Given: $f(x) = x^2 + 2ax + (3a + 10)$ and $\alpha < 1 < \beta$.

Find: The permissible values of $a$.

Use the standard result: if a number $k$ lies between the roots of a quadratic $Ax^2 + Bx + C$, then

Here $A = 1 > 0$ and $k = 1$, so we only need

Now,

Hence,

So the set of all possible values is $(-\infty, -\frac{11}{5})$, and the correct option is D.