Let f(x) = int_0^{x^2} frac{t^2-8t+15}{e^t} d | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Using the Leibniz rule for differentiation under the integral sign,we have:$f'(x) = \frac{(x^2)^2 - 8(x^2) + 15}{e^{x^2}} \cdot \frac{d}{dx}(x^2)

Vedclass · Accepted Answer

$2$ and $3$

Vedclass · Answer

(A) Using the Leibniz rule for differentiation under the integral sign,we have:$f'(x) = \frac{(x^2)^2 - 8(x^2) + 15}{e^{x^2}} \cdot \frac{d}{dx}(x^2) - \frac{0^2 - 8(0) + 15}{e^0} \cdot \frac{d}{dx}(0)$$f'(x) = \frac{x^4 - 8x^2 + 15}{e^{x^2}} \cdot (2x)$$f'(x) = \frac{(x^2 - 3)(x^2 - 5)(2x)}{e^{x^2}}$$f'(x) = \frac{2x(x - \sqrt{3})(x + \sqrt{3})(x - \sqrt{5})(x + \sqrt{5})}{e^{x^2}}$The critical points are $x = -\sqrt{5}, -\sqrt{3}, 0, \sqrt{3}, \sqrt{5}$.Analyzing the sign of $f'(x)$ across these intervals:For $x For $-\sqrt{5}  0$. (Local Min at $-\sqrt{5}$)For $-\sqrt{3} For $0  0$. (Local Min at $0$)For $\sqrt{3} For $x > \sqrt{5}$,$f'(x) > 0$. (Local Min at $\sqrt{5}$)Thus,there are $2$ local maximum points and $3$ local minimum points.

Let $f(x) = \int_0^{x^2} \frac{t^2-8t+15}{e^t} dt$,$x \in R$. Then the numbers of local maximum and local minimum points of $f$,respectively,are:

Similar Questions

An enemy fighter jet is flying along the curve given by $y = x^2 + 2$. $A$ soldier is placed at $(3, 2)$ and wants to shoot down the jet when it is nearest to him. The nearest distance is:

If $y=a \log x+b x^2+x$ has its extreme value at $x=-1$ and $x=2$,then the value of $a+b$ is

If $m$ and $n$ respectively are the number of local maximum and local minimum points of the function $f(x) = \int_{0}^{x^{2}} \frac{t^{2}-5t+4}{2+e^{t}} dt$,then the ordered pair $(m, n)$ is equal to

$\max _{0 \leq x \leq \pi}\left\{x-2 \sin x \cos x+\frac{1}{3} \sin 3 x\right\}=$

The minimum value of the function $f(x)=2|x-1|+|x-2|$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module