Question

The interval in which $y=x^2{e}^{-x}$ is increasing, is

• A. $(-\infty ,\infty )$
• B. $(-2,0)$
• C. $(2,\infty )$
• D. $(0,2)$

Answer 1

Answer: (D)

Given,
$y=x^2{e}^{-x}\Rightarrow \frac{dy}{dx}=x^2{e}^{-x}(-1)+e^{-x}(2x)$
$=x{e}^{-x}(-x+2)=x(2-x)e^{-x}$
For increasing function, $\frac{dy}{dx}>0\Rightarrow x{e}^{-x}(2-x)>0$
Case I
$\Rightarrow x>0\text{ and }2-x>0$
$\Rightarrow x>0\text{ and }x<2$
$\Rightarrow 0<x<2$
Case II
$\Rightarrow x<0\text{ and }2-x<0$
$\Rightarrow x<0$ and $x>2$
Hence, there is no value of $x$ exist.
Clearly, it is increasing in $(0,2)$. So, correct answer is $(d)$.