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Previous Year Question (PYQs)

The maximum value of $\frac{\log x}{x}$ is




Solution

Let $f(x)=\frac{\log x}{x}$ $f'(x)=\frac{1-\log x}{x^2}$ Set $f'(x)=0 \Rightarrow \log x=1 \Rightarrow x=e$ Maximum value $= \frac{\log e}{e}=\frac{1}{e}$

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