Answer:
Option A
Explanation:
Let f(x)= $\frac{x}{\log x}$
$\Rightarrow$ $f'(x)=\frac{\log x-1}{(\log x)^{2}}$
For maxima and minima , put f'(x)=0
$\log x-1=0$
$\Rightarrow$ $x=e$
Now, f"(x)
= $\frac{(\log x)^{2}.\frac{1}{x}-(\log x-1).\frac{2 \log x}{x}}{(\log x)^{4}}$
$\Rightarrow$ $f"(e)=\frac{\frac{1}{e}-0}{1}=\frac{1}{e} >0$
$\therefore$ f(x) is minimum at x=e
Hence, minimum value of f(x) at x=e is
$f(e) =\frac{e}{\log e}=e$
