Question

The maximum value of $\frac{\log x}{x}$ is

• A. $0$
• B. $2$
• C. $\frac{1}{e}$
• D. $-1$

Answer 1

The correct option is C $\frac{1}{e}$

Let $y=\frac{\log x}{x}$
On differentiating w.r.t. x, we get
$\frac{dy}{dx}=\frac{x\cdot \frac{1}{x}-\log x}{x^2}$
$\frac{dy}{dx}=\frac{(1-\log x)}{x^2}$
For maximum or minimum, put $\frac{dy}{dx}=0$
$\Rightarrow 1-\log x=0$
$\Rightarrow x=e$
Again, differentiating w.r.t. x, we get
$\frac{d^{2}y}{d{x}^2}=\frac{x^2(-\frac{1}{x})-(1-\log x)2x}{(x^2{)}^2}$
$=-\frac{-x[1+2-2\log x]}{x^4}$
$=-\frac{(3-2\log x}{x^3}$
At $x=e,\frac{d^{2}y}{d{x}^2}=-ve<0$
It is maximum at $x=e$.
$\therefore$ The maximum value at $x=e$ is
$y=\frac{\log e}{e}=\frac{1}{e}$