Question

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

• A. 1
• B. $e$
• C. $2/e$
• D. $1/e$

Answer 1

Answer: (D)

Let $y=\frac{\ln x}{x}$
$\frac{dy}{dx}=\frac{x\cdot \frac{1}{x}-\ln x\cdot 1}{x^2}=\frac{1-\log x}{x^2}$
For maxima, put $\frac{dy}{dx}=0$
$\Rightarrow \frac{1-\ln x}{x^2}=0\Rightarrow x=e$
Now, $\frac{d^{2}y}{d{x}^2}=\frac{x^2\left(-\frac{1}{x}\right)-(1-\ln x)2x}{{\left(x^2\right)}^2}$
At $x=e,\frac{d^{2}y}{d{x}^2}<0$
$\therefore$ The maximum value at $x=e$ is $y=\frac{1}{e}$