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 :

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Application of Derivatives

Let

f (x) = \int_{0}^{x^{2}} \frac{t ^{2} - 8 t + 15}{e ^{t}} d t, x \in R

. Then the numbers of local maximum and local minimum points of f, respectively, are :

Answer: (A)

By Leibniz's rule,

f^{'} (x) = \frac{( x ^{2} ) ^{2} - 8 x ^{2} + 15}{e ^{x^{2}}} \cdot 2 x = \frac{2 x ( x ^{4} - 8 x ^{2} + 15 )}{e ^{x^{2}}} = \frac{2 x ( x ^{2} - 3 ) ( x ^{2} - 5 )}{e ^{x^{2}}} .

Since

e^{x^{2}} > 0

always, the sign of

f^{'}

is the sign of

g (x) = x (x^{2} - 3) (x^{2} - 5)

, which has zeros at

x = 0, \pm 3, \pm 5

.

Checking the sign of g in each interval (testing

x = - 3, - 2, - 1, 1, 2, 3

) gives the pattern

-, +, -, +, -, +

across

(- \infty, - 5), (- 5, - 3), (- 3, 0), (0, 3), (3, 5), (5, \infty)

.

A sign change from

-

+

gives a local minimum, and from

+

-

a local maximum:

The numbers of local maximum and local minimum points, respectively, are 2 and 3.

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Single correct option