Question

The function $f\left(x\right)=x^2{e}^{-x}$ increases in the interval.

• A. $(0,2)$
• B. $(2,3)$
• C. $(3,4)$
• D. $(4,5)$

Answer 1

The correct option is A

$(0,2)$

Explanation of correct option.

Applying condition for increasing function:

For increasing function $f\text{'}\left(x\right)>0$

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

$$\begin{gathered} \Rightarrow f\text{'}\left(x\right)=2x\left(e^{-x}\right)+x^2(-e^{-x}) \\ \Rightarrow f\text{'}\left(x\right)=x{e}^{-x}(2-x) \end{gathered}$$

Put, $f\text{'}\left(x\right)>0$

$$\begin{gathered} \Rightarrow f\text{'}\left(x\right)=x{e}^{-x}(2-x) \\ \Rightarrow x{e}^{-x}(2-x)>0 \\ \Rightarrow x(2-x)>0 \end{gathered}$$

So, $x>0$ and $x<2$.

Here, $x\in (0,2)$ for $f\left(x\right)$to be increasing function.

Therefore, $f\left(x\right)$ is increasing in the region $(0,2)$.

Hence, the correct option is option(A)