Question

Local maximum value of the function $\frac{logx}{x}$ is

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

Answer 1

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

Let $f\left(x\right)=\frac{logx}{x}\Rightarrow f^{\prime }\left(x\right)=\frac{1}{x^2}-\frac{logx}{x^2}$
For maximum or minimum value of f(x), f'(x) =0
$\Rightarrow f^{\prime }\left(x\right)=\frac{1-lo{g}_{e}x}{x^2}=0$
$\therefore lo{g}_{e}x=1$ or x=e, which lie in $(0,\infty ).$
For $x=e,\frac{d^{2}y}{d{x}^2}=-\frac{1}{e^3}$, which is -ve.
Hence y is maximum at x=e and its maximum value =$\frac{loge}{e}=\frac{1}{e}$