Question

The function $f\left(x\right)=\frac{\log \left(x\right)}{x}$ is increasing in the interval

• A. $\left(1,2e\right)$
• B. $\left(0,e\right)$
• C. $\left(2,2e\right)$
• D. $\left(\frac{1}{e},2e\right)$

Answer 1

The correct option is B

$\left(0,e\right)$

Find the interval in which the given function is increasing

We know that a function $f\left(x\right)$ is increasing then $f\text{'}\left(x\right)>0$

$$\begin{gathered} f\left(x\right)=\frac{\log \left(x\right)}{x} \\ \Rightarrow f\text{'}\left(x\right)=\frac{x{\displaystyle \frac{d\log \left(x\right)}{dx}}-\log \left(x\right){\displaystyle \frac{dx}{dx}}}{x^2} \\ \Rightarrow f\text{'}\left(x\right)=\frac{x·{\displaystyle \frac{1}{x}}-\log \left(x\right)}{x^2} \\ \Rightarrow f\text{'}\left(x\right)=\frac{1-\log \left(x\right)}{x^2} \end{gathered}$$

For increasing function $f\text{'}\left(x\right)>0$

$$\begin{gathered} \Rightarrow \frac{1-\log \left(x\right)}{x^2}>0 \\ \Rightarrow 1-\log \left(x\right)>0 \\ \Rightarrow \log \left(x\right)<1 \\ \Rightarrow e>x \end{gathered}$$

$f\left(x\right)$ is increasing in the interval $\left(0,e\right)$

Hence, option $\left(\mathrm{B}\right)$ is correct.