Question

The value of x for which the function $\frac{\log x}{x}\in 0<x<\infty$ attains its maximum value is

• A. $1$
• B. $e$
• C. $1/e$
• D. $\infty$

Answer 1

The correct option is B $e$

$y=\frac{\log x}{x}$
$\frac{dy}{dx}=\frac{1}{x^2}(1-\log x)$
For maximum or minimum ,
$\frac{dy}{dx}=0$
$\therefore \log x=1$
$\Rightarrow x=e$
Now, $\frac{d^{2}y}{d{x}^2}=\frac{x^2(-\frac{1}{x})-(1-\log x)2x}{x^4}=\frac{-3x+2x\log x}{x^4}$
$\frac{d^{2}y}{d{x}^2}=-\frac{1}{e^3}<0$
Hence, $y$ has a maximum at $x=e$
Maximum value $=\frac{1}{e}$