Question

The function $f\left(x\right)=\frac{x^2}{e^x}$ is monotonically increasing if

• A. $x<0$ only
• B. $x>2$ only
• C. $0<x<2$
• D. $x\in (-\infty ,0)\cup (2,\infty )$

Answer 1

Answer: (C)

$0<x<2$

Solution:

We have, $f\left(x\right)=x^2{e}^{-x}$

$⟹f^{\prime }\left(x\right)=x^2{e}^{-x}(-1)+e^{-x}.2x$

$=x.e^{-x}(2-x)$

Since, $f\left(x\right)$ is monotonically increasing.

$\therefore f^{\prime }\left(x\right)>0$

$⟹x.e^{-x}(2-x)>0⟹x(x-2)<0$

Hence, $0<x<2$

Hence, C is the correct option.