Question

The maximum value of $ f(x) = \frac{\log x}{x} , 0 < x < \infty$ is

• A. $\frac{2}{e}$
• B. $\frac{1}{e}$
• C. $\sqrt{e} $
• D. $e$

Answer 1

The Correct Option is B

We have, $ f(x) = \frac{\log x}{x} , 0 < x < \infty$
Maximum or minimum point is given by $f'(x) =0$
$f'\left(x\right) =\frac{x \frac{1}{x} -\log x.1}{x^{2}} = 0 \Rightarrow \frac{1-\log x}{x^{2}} =0$
$\Rightarrow 1=\log x \Rightarrow x-e$
Now $ f" \left(x\right) =\frac{x^{2} \left(\frac{-1}{x}\right) -\left(1-\log x\right)2x}{x^{4}} $
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = \frac{-1-2+2\log x}{x^{3}} =\frac{-3+2\log x}{x^{3}} f"\left(x\right)_{x=e} =\frac{-1}{e^{3} } <0$
$\Rightarrow x = e$ is a maximum.point and maximum value of $f(x)$ is given by , $f(e) = \frac{\log \: e}{e} = \frac{1}{e}$