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

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a

(2,3)

b

(3,2)

c

(2,2)

d

(3,4)

answer is B.

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Detailed Solution

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$\begin{array}{l} f (x) = \int_{0}^{x^{2}} \frac{t^{2} - 5 t + 4}{2 + e^{t}} d t \\ f^{'} (x) = \frac{(x^{4} - 5 x^{2} + 4) 2 x}{2 + e^{x^{2}}} = \frac{2 x (x^{2} - 1) (x^{2} - 4)}{2 + e^{x^{2}}} \\ = \frac{2 x (x - 1) (x + 1) (x - 2) (x + 2)}{2 + e^{x^{2}}} \end{array}$ Question Image
$so, m = local maximum point = 2 and n = localminimum point = 3$

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If m and n respectively are the number of local maximum and local minimum points of the function f(x)=∫0x2t2−5t+42+etdt, then the ordered pair (m,n) is equal to