Question

Local maximum value of the function $\frac{{\log }_{e}x}{x}$ is

• A. $1$
• B. $e$
• C. $\frac{1}{e}$
• D. None of these

Answer 1

The correct option is C $\frac{1}{e}$

Let $f\left(x\right)=\frac{{\log }_{e}x}{x}$
On differentiating, w.r.t. $x$, we get
$f^{\prime }\left(x\right)=\frac{1}{x^2}-\frac{{\log }_{e}x}{x^2}$
For maximum or minimum value of $f\left(x\right)$,
Put $f^{\prime }\left(x\right)=0$
$\Rightarrow \frac{1-{\log }_{e}x}{x^2}=0$
$\Rightarrow {\log }_{e}x=1$
$\Rightarrow x=e$, which lies in $(0,\infty )$
For $x=e,f^{\prime \prime }\left(x\right)=-ve$
Hence, $y$ is maximum at $x=e$ and its maximum value $=\frac{{\log }_{e}e}{e}=\frac{1}{e}$.