Mathematics · Application of Derivatives

JEE Main 2025 — 22 January, Evening Shift — Question 8

Let $f(x)=\int_{0}^{x^{2}} \frac{t^{2}-8 t+15}{e^{t}} d t, x \in \mathbf{R}$ . Then the numbers of local maximum and local minimum points of

f, respectively, are :

Answer: A

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

$=\frac{\left(\mathrm{x}^{2}-3\right)\left(\mathrm{x}^{2}-5\right)(2 \mathrm{x})}{\mathrm{e}^{\mathrm{x}^{2}}}$

$=\frac{(\mathrm{x}-\sqrt{3})(\mathrm{x}+\sqrt{3})(\mathrm{x}-\sqrt{5})(\mathrm{x}+\sqrt{5}) 2 \mathrm{x}}{\mathrm{e}^{\mathrm{x}^{2}}}$

Maxima at $\mathrm{x} \in\{-\sqrt{3}, \sqrt{3}\}$

Minima at $\mathrm{x} \in\{-\sqrt{5}, 0, \sqrt{5}\}$

2 points of maxima and 3 points of minima.

Answer key and solution verified before publishing.

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