Question

The maximum value of $f(x)=\frac{\log x}{x},0<x<\infty$ is

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

Answer 1

Answer: (B)

We have, $f(x)=\frac{\log x}{x},0<x<\infty$
Maximum or minimum point is given by $f^{\prime }(x)=0$
$f^{\prime }\left(x\right)=\frac{x\frac{1}{x}-\log x.1}{x^2}=0\Rightarrow \frac{1-\log x}{x^2}=0$
$\Rightarrow 1=\log x\Rightarrow x-e$
Now $f"\left(x\right)=\frac{x^2\left(\frac{-1}{x}\right)-\left(1-\log x\right)2x}{x^4}$
$=\frac{-1-2+2\log x}{x^3}=\frac{-3+2\log x}{x^3}f"{\left(x\right)}_{x=e}=\frac{-1}{e^3}<0$
$\Rightarrow x=e$ is a maximum.point and maximum value of $f(x)$ is given by , $f(e)=\frac{\log e}{e}=\frac{1}{e}$