Question

The maximum of $f\left(x\right)=\frac{\log x}{x^2}(x>0)$ occurs, when $x$ is equal to

• A. $\frac{1}{\sqrt{e}}$
• B. $e$
• C. $\sqrt{e}$
• D. $\frac{1}{e}$

Answer 1

The correct option is C $\sqrt{e}$

$f\left(x\right)=\frac{\log x}{x^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{x^2\cdot \frac{1}{x}-\left(\log x\right)2x}{(x^2{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{-\left(\log \right(x^2)-1)}{x^3}$
From $f^{\prime }\left(x\right)=0$
$\Rightarrow \log \left(x^2\right)-1=0$
$\Rightarrow x^2=e\Rightarrow x=\sqrt{e}$
There is only one critical point $x=\sqrt{e},(\because x>0)$

$f\left(x\right)$ changes sign from positive to negative as $x$ crosses $\sqrt{e}$ from left to right.
Hence, $f\left(x\right)$ is maximum at $x=\sqrt{e}$