The maximum value of the function $\frac{\log x}{x}, x \neq 0$ is

The maximum value of the function $\frac{\log x}{x}, x \neq 0$ is
  1. $e^{2}$
  2. $\frac{1}{e}$
  3. $\frac{1}{e^{2}}$
  4. $e$

Solution

Let $y=\frac{\log x}{x}$ $\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x} \cdot \frac{1}{\mathrm{x}}-\log \mathrm{x}}{\mathrm{x}^{2}}=\frac{1-\log \mathrm{x}}{\mathrm{x}^{2}}$ Put $\frac{d y}{d x}=0$, we get $\begin{aligned} 1-\log x=0 & \Rightarrow \log x=1 \Rightarrow \log x=\log e \Rightarrow x=e \\ \text { Now } \frac{d^{2} y}{d x^{2}} &=\frac{x^{2}\left(-\frac{1}{x}\right)-(1-\log x)(2 x)}{x^{4}}=\frac{-x-2 x+2 x \log x}{x^{4}} \\ &=\frac{2 \log x-3}{x^{3}} \end{aligned}$ $\text { At } x=e, \frac{d^{2} y}{d x^{2}} < 0$ from (1), maximum value is $y=\frac{1}{\mathrm{e}}$

Asked in: MHT CET 2020 (13 Oct Shift 2)

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