Question

The maximum value of $\frac{\log x}{x}$ in $\left(2,\infty \right)$ is

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

Answer 1

The correct option is D

$\frac{1}{e}$

Explanation for the correct option

Solve for the maximum value of $\frac{\log x}{x}$ in $\left(2,\infty \right)$

$$\begin{gathered} y=\frac{\log x}{x} \\ \frac{dy}{dx}=\frac{\left[1-\log x\right]}{x^2}\mathbf{}\left[\because \frac{d}{\mathrm{dx}}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{g\left(x\right)f\text{'}\left(x\right)-f\left(x\right)g\text{'}\left(x\right)}{{\left[g\right(x\left)\right]}^2}\right] \end{gathered}$$

Put $\frac{dy}{dx}=0$

$$\begin{gathered} \frac{\left[1-\log x\right]}{x^2}=0 \\ 1-\log x=0 \\ \Rightarrow \log x=1 \\ \Rightarrow x=e \\ Now, \\ \frac{d^{2}y}{d{x}^2}=\frac{\left[x^2\left({\displaystyle \frac{-1}{x}}\right)-(1-\log x)2x\right]}{{\left[x^2\right]}^2}\left[\because \frac{d}{\mathrm{dx}}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{g\left(x\right)f\text{'}\left(x\right)-f\left(x\right)g\text{'}\left(x\right)}{{\left[g\right(x\left)\right]}^2}\right] \end{gathered}$$

At $x=e,\frac{d^{2}y}{d{x}^2}\le 0,$ maxima

The maximum value at x $=e$ is $y=\frac{\log e}{e}=\frac{1}{e}$

Hence the maximum value of $\frac{\log x}{x}$ in $\left(2,\infty \right)$ is $\frac{1}{e}$.

Hence, option (D) is correct answer.