The maximum value of f(x) = frac{log x}{x} (x | vedclass

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(B) Given function is $f(x) = \frac{\log x}{x}$.To find the maximum value,we find the derivative $f'(x)$ using the quotient rule:$f'(x) = \frac{x \cdo

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$\frac{1}{e}$

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(B) Given function is $f(x) = \frac{\log x}{x}$.To find the maximum value,we find the derivative $f'(x)$ using the quotient rule:$f'(x) = \frac{x \cdot \frac{d}{dx}(\log x) - \log x \cdot \frac{d}{dx}(x)}{x^2} = \frac{x \cdot \frac{1}{x} - \log x \cdot 1}{x^2} = \frac{1 - \log x}{x^2}$.For critical points,set $f'(x) = 0$:$\frac{1 - \log x}{x^2} = 0 \Rightarrow 1 - \log x = 0 \Rightarrow \log x = 1 \Rightarrow x = e$.To confirm this is a maximum,we check the second derivative or the sign change of $f'(x)$. Since $f'(x) > 0$ for $x  e$,$x = e$ is a point of local maximum.The maximum value is $f(e) = \frac{\log e}{e} = \frac{1}{e}$.

The maximum value of $f(x) = \frac{\log x}{x}$ $(x > 0, x \neq 1)$ is

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