Question

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

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

Answer 1

Answer: (D)

Let $y=\frac{logx}{x}$ and
$\therefore \frac{dy}{dx}=\frac{1-logx}{x^2}$ and
$\frac{d^{2}y}{d{x}^2}=\frac{x^2\left(-\frac{1}{x}\right)-\left(1-logx\right)2x}{x^4}$
$=\frac{-1-\left(1-logx\right)2}{x^3}$
$=\frac{2logx-3}{x^3}$
$y$ is Max. or Min. when $\frac{dy}{dx}=0$
$\Rightarrow logx=1$
$\Rightarrow x=e$ For $x=e$, $\frac{d^{2}y}{d{x}^2}<0$
$\therefore y$ is Max. of $x=e$
$\therefore$ Max. of $y=\frac{loge}{e}=\frac{1}{e}$