Question

Let $y=x^2{e}^{-x},$ then the interval in which $y$ increases with respect to $x$ is

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

Answer 1

The correct option is D

$(0,2)$

Explanation for the correct option:

Finding the value at which $y$ increases:

Given

$$\begin{gathered} y=x^2{e}^{-x} \\ \text{Considering}u=x^2,v=e^{-x} \\ y\text{'}=-x^2{e}^{-x}+2x{e}^{-x} \end{gathered}$$

Now, if $y$increases then $y\text{'}>0$

$$\begin{gathered} -x^2{e}^{-x}+2x{e}^{-x}>0 \\ {e}^{-x}(2x-x^2)>0 \\ 2x-x^2>0 \\ {x}^2-2x<0 \\ x(x-2)<0 \\ \Rightarrow x\in (0,2) \end{gathered}$$

Therefore,$\Rightarrow x\in (0,2)$

In the interval $(0,2)$ the function is increasing.

Hence the correct answer is option (D).