JEE Main 2025 Jan 22 Shift 2, Mathematics Q8: Maxima & Minima

By the Leibniz rule:

$f^{\prime }(x)=\frac{(x^2{)}^2-8(x^2)+15}{e^{x^2}}\cdot 2x=\frac{(x^4-8x^2+15)\cdot 2x}{e^{x^2}}$.

Factor the quadratic in $x^2$: $x^4-8x^2+15=(x^2-3)(x^2-5)=(x-\sqrt{3})(x+\sqrt{3})(x-\sqrt{5})(x+\sqrt{5})$.

So $f^{\prime }(x)=\frac{2x(x-\sqrt{3})(x+\sqrt{3})(x-\sqrt{5})(x+\sqrt{5})}{e^{x^2}}$.

The critical points are at $x\in \{-\sqrt{5},-\sqrt{3},0,\sqrt{3},\sqrt{5}\}$. The denominator is positive, so the sign of $f^{\prime }$ follows the product of the linear factors.

Tracking the sign of $f^{\prime }(x)$ as $x$ increases through the five roots gives the pattern: $-,+,-,+,-,+$.

Hence local maxima occur where $f^{\prime }$ changes from $+$ to $-$: at $x=-\sqrt{3}$ and $x=\sqrt{3}$, giving 2 maxima.

Local minima occur where $f^{\prime }$ changes from $-$ to $+$: at $x=-\sqrt{5}$, $x=0$ and $x=\sqrt{5}$, giving 3 minima.

$\therefore$ 2 local maxima and 3 local minima.