Question

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

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

Answer 1

The correct option is C

$\frac{1}{e}$

Explanation for correct answer:

Finding the local maximum:

Let $f\left(x\right)=\frac{\log x}{x}$

Finding the local maximum,

$$\begin{gathered} f\left(x\right)=\frac{\log x}{x} \\ =\frac{x.{\displaystyle \frac{1}{x}}-\log x.1}{x^2}\left[\because \frac{d}{dx}\frac{u}{v}=\frac{v.du-u.dv}{v^2}\right] \\ =\frac{1-\log x}{x^2} \end{gathered}$$

equating $f\text{'}\left(x\right)=0$,

$$\begin{gathered} \frac{1-\log x}{x^2}=0 \\ 1-\log x=0 \\ \log x=1 \\ {e}^{\log x}=e^1\left[\text{raising}e\text{power on both side}\right] \\ x=e^1\left[\because e^{\log }=1\right] \end{gathered}$$

$x=e$ is the critical points which lie in $\left(0,\infty \right)$

Substitute these points in $f\left(x\right)$.

$f\left(e\right)=\frac{\log e}{e}=\frac{1}{e}$ is the maximum value of $\frac{\log x}{x}$

Hence, the correct answer is option (C).