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

We need to find the number of local maximum and local minimum points of $$f(x) = \int_0^{x^2} \f\frac{t^2 - 8t + 15}{e^t}\,dt\,. $$

By the Leibniz rule (chain rule with integral) we have $$f'(x) = \f\frac{(x^2)^2 - 8(x^2) + 15}{e^{x^2}}\cdot 2x = \f\frac{2x\bigl(x^4 - 8x^2 + 15\bigr)}{e^{x^2}}\,. $$ Factoring the quartic by setting $$u = x^2$$ gives $$x^4 - 8x^2 + 15 = u^2 - 8u + 15 = (u - 3)(u - 5) = (x^2 - 3)(x^2 - 5)\,. $$ This yields $$f'(x) = \f\frac{2x\,(x^2 - 3)\,(x^2 - 5)}{e^{x^2}}\,. $$

Since $$e^{x^2}\gt 0$$ always, the critical points occur when the numerator is zero, namely when $$2x=0$$ or $$x^2-3=0$$ or $$x^2-5=0\,. $$ Hence the critical points are $$x=0,\quad x=\pm\sqrt{3},\quad x=\pm\sqrt{5}\,, $$ giving five critical values in all.

To determine the nature of each critical point, we examine the sign of $$f'(x)=\f\frac{2x\,(x^2-3)\,(x^2-5)}{e^{x^2}}\,. $$ For $$x\lt -\sqrt{5}$$ all three factors $$2x$$, $$(x^2-3)$$ and $$(x^2-5)$$ have signs $$-$$, $$+$$, $$+$$ respectively, so $$f'(x)\lt 0\,. $$ In the interval $$-\sqrt{5}0\,. $$ For $$-\sqrt{3}0\,. $$ In $$\sqrt{3}\sqrt{5}$$ all three factors are positive and $$f'(x)\gt 0\,. $$

Since $$f'(x)$$ changes from negative to positive at $$x=-\sqrt{5},\;0,\;\sqrt{5}$$ these are local minima, and since it changes from positive to negative at $$x=-\sqrt{3},\;\sqrt{3}$$ these are local maxima. Thus there are 2 local maxima and 3 local minima. The answer is Option A: 2 and 3.