The maximum value of $f(x)=\frac{\log x}{x}$ for x in $[2, \infty]$ is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) - $\frac{1}{e}$
$f(x)=\frac{\log x}{x}$
$f'(x)=\frac{1}{x^2}-\frac{\log x}{x^2}=0$
$⇒x=e$
so e → point of maxima
max value = $f(x)=\frac{1}{e}$