Question

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

• A. $(2,2)$
• B. $(3,2)$
• C. $(3,4)$
• D. $(2,3)$

Answer 1

The correct option is D $(2,3)$

Given, $f\left(x\right)=\underset{0}{\overset{x^2}{\int }}\frac{t^2-5t+4}{2+e^t}dt$
$\Rightarrow f^{\prime }\left(x\right)=2x\left(\frac{x^4-5x^2+4}{2+e^{x^2}}\right)$
$\Rightarrow f^{\prime }\left(x\right)=\frac{2x(x+1)(x-1)(x+2)(x-2)}{(e^{x^2}+2)}$
For max/min, $f^{\prime }\left(x\right)=0$
$\Rightarrow x(x^2-4)(x^2-1)=0$
$\Rightarrow x=0,\pm 1,\pm 2$

sign convention of $f^{\prime }\left(x\right)$:


$f^{\prime }\left(x\right)$ changes sign from positive to negative at
$x=-1,1.$ So, the number of local maximum points $=2$

and $f^{\prime }\left(x\right)$ changes sign from negative to positive at $x=-2,0,2.$ So, number of local minimum points $=3$
$\therefore m=2,n=3$