Let $\mathrm{f}(\mathrm{x})=\int_0^{x^2} \frac{t^2-8 t+15}{e^t} d t, \mathrm{x} \in \mathrm{R}$, the number of local maximum and minimum point of $f(x)$ respectively are (22 Jan, Shift II, Memory Based)

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Correct answer: b) 5. $ \begin{aligned} & \because f(x)=\int_0^{x^2} \frac{t^2-8 t+15}{e^t} \ & f^{\prime}(x)=\frac{2 x\left(x^4-8 x^2+15\right)}{e^{x^2}} \ & =\frac{2 x\left(x^2-5\right)\left(x^2-3\right)}{e^{x^2}} \end{aligned} $ The extremum value of $f(x)$ are $x=0, \pm \sqrt{5}, \pm \sqrt{3}$ $\therefore \quad$ Number of extremum points are 5.

Let \(\mathrm{f}(\mathrm{x})=\int_0^{x^2} \frac{t^2-8 t+15}{e^t} d t, \mathrm{x} \in \mathrm{R}\), the number of local m…

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