Question

The maximum value of $\frac{logx}{x}$ in $(2,\infty )$ is

• A. $5$
• B. $\frac{5}{e}$
• C. $e^2$
• D. $\frac{1}{e}$

Answer 1

Answer: (D)

$\frac{1}{e}$

Let $f=\frac{logx}{x}$

Derivative is

$\frac{df}{dx}=\frac{x\times \frac{1}{x}-logx\times 1}{x^2}=\frac{1-logx}{x^2}$

maximum value occurs when derivative is zero.

So $\Rightarrow \frac{1-logx}{x^2}=0$

$\Rightarrow logx=1$

$\Rightarrow x=e$