Question

Select the correct statement for the function $f\left(x\right)=x{e}^{1-x}$.

• A. Strictly increases in the interval $\left(\frac{1}{2},2\right)$
• B. Increases in the interval $\left(0,\infty \right)$
• C. Decreases in the interval $\left(0,2\right)$
• D. Strictly decreases in the interval $\left(1,\infty \right)$

Answer 1

The correct option is D

Strictly decreases in the interval $\left(1,\infty \right)$

Analyzing the behaviour of $f\left(x\right)=x{e}^{1-x}$

Given , $f\left(x\right)=x{e}^{1-x}$

To see how $f\left(x\right)$ is varying and to get its maximum or minimum points. We have to differentiate $f\left(x\right)$ once. and equate it to $0$.

So, $f\text{'}\left(x\right)=e^{1-x}-x{e}^{1-x}$

$=e^{1-x}(1-x)$

Equating, $f\text{'}\left(x\right)=0$

$\Rightarrow e^{1-x}\left(1-x\right)=0$

$\Rightarrow$ $(1-x)=0$

$\therefore$ $x=1$.

To know, whether $x=1$is a maximum point or minimum point, we have to differentiate $f\text{'}\left(x\right)$ once.

So, $f\text{'}\text{'}\left(x\right)=-e^{1-x}(1-x)-e^{1-x}$

$=-e^{1-x}(2-x)$

Putting $x=1$ in $f\text{'}\text{'}\left(x\right)$ , we get $f\text{'}\text{'}\left(x\right)=-1<0$.

So, $x=1$is the maximum point i.e. $f\text{'}\left(x\right)<0\forall \left(1,\infty \right)$.

Thus, $f\left(x\right)$is strictly decreasing in the interval $\left(1,\infty \right)$.

Hence, option D is correct.