Question

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

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

Answer 1

Answer: (D)

$\frac{1}{e}$

Answer 2

Let $y = \frac{log\,x}{x}$
On differentiating both sides w.r.t, $x$, we get
$\frac{dy}{dx} = \frac{x\frac{d}{dx} log\,x - log\,x \frac{d}{dx} \,x}{x^2}$
$= \frac{x\cdot \frac{1}{x} - log\,x\cdot 1}{x^2}$
$\therefore \frac{dy}{dx} = \frac{1 - log\,x }{x^2} \,\,...(i)$
For maximum or minimum values,
$\frac{dy}{dx} = 0$
$\Rightarrow \frac{1- log \,x}{x^2} = 0$
$\Rightarrow 1 - log \,x = 0$
$\Rightarrow log\,x = 0$
$x = e$
Now, differentiate E $(i)$ w.r.t. $'x'$ we get
$\frac{d^2y}{dx^2} = \frac{x^2(-\frac{1}{x}) - (1 - log\,x) \cdot 2x}{x^4}$
$= \frac{-x - 2x + 2x \,log\,x}{x^4}$
$= \frac{-3x + 2x\,log\,x}{x^4}$
$\left(\frac{d^{2}y}{dx^{2}}\right)_{x =e} = \frac{-3e + 2e\, log \,x}{e^{4}}$
$= \frac{-3e+2e}{e^{4}}$
$= -\frac{e}{e^{4}} = -\frac{1}{e^{3}} < 0$
$\therefore y = \frac{log x}{x}$ is maximum at $x = e$
$\Rightarrow y_{max} = \frac{log\, e}{e} = \frac{1}{e}$