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If $\mathrm{m\: and \: n}$ respectively are the number of local maximum and local minimum points of the function $\mathrm{f(x)=\int_{0}^{x^{2}} \frac{t^{2}-5 t+4}{2+e^{t}} d t}$ , then the ordered pair $\mathrm{\left ( m,n \right )}$ is
Option: 1
$\mathrm{\left ( 3,2 \right )}$

Option: 2
$\mathrm{\left ( 2,3 \right )}$

Option: 3
$\mathrm{\left ( 2,2 \right )}$

Option: 4
$\mathrm{\left ( 3,4 \right )}$

Answers (1)

best_answer

$\mathrm{f^{\prime}(a) =\frac{2 x-\left(\left(x^{2}\right)^{2}-5 x^{2}+4\right)}{2+e^{x^{2}}}} \\$

$\mathrm{=\frac{2 x\left[\left(x^{2}-1\right)\left(x^{2}-4\right)\right]}{2+e^{x^{2}}}}$

$\mathrm{=\frac{2(x+2)(x+1) x(x-1)(x-2)}{2+e^{x^{2}}}}$

So $\mathrm{\{-2,0,2\}}$ are poits of Minima $\mathrm{\Rightarrow n=3 }$

and $\mathrm{\{-1,1\}}$ are poits is Maxixma $\mathrm{\Rightarrow m=2 }$

$\mathrm{(m,n)= (2,3) }$

Hence the correct answer is option 2

Posted by

Deependra Verma

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