If $\mathrm{f}(x)=\frac{\log x}{x}(x\gt0)$, then it is increasing in

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{f}(x)=\frac{\log x}{x} \\ & \therefore \quad \mathrm{f}^{\prime}(x)=\frac{1}{x^2}-\frac{\log x}{x^2}=\frac{1-\log x}{x^2} \end{aligned}$
For $\mathrm{f}(x)$ to be increasing, $\mathrm{f}^{\prime}(x) \gt 0$ $\Rightarrow 1-\log x \gt 0 \Rightarrow 1 \gt \log x \Rightarrow \mathrm{e} \gt x$ $\therefore \quad \mathrm{f}(x)$ is increasing in the interval $(0, \mathrm{e})$.

If $\mathrm{f}(x)=\frac{\log x}{x}(x\gt0)$, then it is increasing in

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