Question

The maximum value of $\frac{logx}{x}$ is:

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

Answer 1

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

Let $y=\frac{logx}{x}$

1) Differentiate y w.r.t. x and equate to zero.

$\therefore \frac{dy}{dx}=\frac{d}{dx}\left[\frac{logx}{x}\right]$

By using the rule of $\frac{d}{dx}\left(\frac{u}{v}\right)=\left[\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}\right]$, we get,

$\frac{dy}{dx}=\frac{x\frac{d}{dx}\left(logx\right)-logx\frac{d}{dx}\left(x\right)}{x^2}$

$\therefore \frac{dy}{dx}=\frac{x\left(\frac{1}{x}\right)-logx\left(1\right)}{x^2}$

$\therefore \frac{dy}{dx}=\frac{1-logx}{x^2}$

For the function to be maximum or minimum, $\frac{dy}{dx}=0$

$\therefore \frac{1-logx}{x^2}=0$

$\therefore 1-logx=0$

$\therefore logx=1$

$\therefore x=e^1$

$\therefore x=e$

Thus, at $x=e$, function will be maximum.

$\therefore y_{max}=\frac{log\left(e\right)}{e}$

$\therefore y_{max}=\frac{1}{e}$ $(\because loge=1)$