Question

The absolute difference between the greatest and the least values of the function $f\left(x\right)=x(\ln x-2)$ on $[1,e^2]$ is

• A. $2$
• B. $e$
• C. $e^2$
• D. $1$

Answer 1

The correct option is B $e$

$f\left(x\right)=x(\ln x-2)$
$\Rightarrow f^{\prime }\left(x\right)=x\left(\frac{1}{x}\right)+(\ln x-2)=\ln x-1$
$\Rightarrow f^{\prime }\left(x\right)=\ln x-1$
$f^{\prime }\left(x\right)=0\Rightarrow x=e$
For $x\in [1,e),f^{\prime }\left(x\right)<0$ and for $x\in (e,e^2],f^{\prime }\left(x\right)>0$
So, the minimum value of $f\left(x\right)$ occurs at $x=e$

Now, $f\left(1\right)=-2$
$\Rightarrow f\left(e\right)=-e$ (least)
$\Rightarrow f\left(e^2\right)=0$ (greatest)

Hence, $|f\left(e^2\right)-f\left(e\right)|=|0-(-e)|=e$