JEE Mathematics 2025 Question with Solution

Given:

Find: The numbers of local maximum and local minimum points of $f(x)$.

Using differentiation under the integral sign,

because the integrand evaluated at $t=x^2$ is

Hence,

Since $e^{x^2}>0$ for all real $x$, the sign of $f'(x)$ depends on

Critical points are obtained from

which gives

Now check sign changes of $f'(x)$ on the intervals determined by these points:

• For $x< -\sqrt{5}$, $f'(x)<0$
• For $-\sqrt{5}<x< -\sqrt{3}$, $f'(x)>0$
• For $-\sqrt{3}<x<0$, $f'(x)<0$
• For $0<x<\sqrt{3}$, $f'(x)>0$
• For $\sqrt{3}<x<\sqrt{5}$, $f'(x)<0$
• For $x>\sqrt{5}$, $f'(x)>0$

From the sign chart:

• At $x=-\sqrt{5}$, $f'(x)$ changes from negative to positive, so this is a local minimum.
• At $x=-\sqrt{3}$, $f'(x)$ changes from positive to negative, so this is a local maximum.
• At $x=0$, $f'(x)$ changes from negative to positive, so this is a local minimum.
• At $x=\sqrt{3}$, $f'(x)$ changes from positive to negative, so this is a local maximum.
• At $x=\sqrt{5}$, $f'(x)$ changes from negative to positive, so this is a local minimum.

Therefore, the function has $2$ local maxima and $3$ local minima. However, the solution explicitly marks Option D as correct. Among the given options, the recorded correct option is D.