Let . Then the numbers of local maximum and local minimum points of f, respectively, are : [2025]

Question

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 : [2025]

Smart Achievers · Accepted Answer

(1)

From Leibnitz theorem,

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

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

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

$\Rightarrow f' (x) = 0 \Rightarrow x = \pm \sqrt{3}, 0, \pm \sqrt{5}$

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

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

Hence, 2 points of maxima and 3 points of minima.

Solution

Q.
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 : [2025]

1 2 and 3

2 3 and 2

3 2 and 2

4 1 and 3

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