A) e
B) \[\sqrt{e}\]
C) \[\frac{1}{e}\]
D) \[\frac{2}{e}\]
Correct Answer: C
Solution :
\[f(x)=\frac{\log x}{x}\] \[f'(x)=\frac{x.\frac{1}{x}-\log x.1}{{{x}^{2}}}\] \[\Rightarrow \] \[f'(x)=\frac{1-\log x}{{{x}^{2}}}\] For maxima and minima \[f'(x)=0\] \[\Rightarrow \] \[\frac{1-\log x}{{{x}^{2}}}=0\] \[\Rightarrow \] \[\log x=1\] \[\Rightarrow \] \[x=e\] \[f''(x)=\frac{{{x}^{2}}\left( -\frac{1}{x} \right)-1(1-\log x)2x}{{{x}^{4}}}\] \[=\frac{-x-2x(1-\log x)}{{{x}^{4}}}\] At \[x=e,\]\[f''(e)=\frac{-e-2e(1-\log e)}{{{e}^{4}}}=-\frac{1}{{{e}^{3}}}<0\] \[\therefore \] \[f(x)\]is maximum at\[x=e\] Hence, maximum value \[=\frac{\log e}{e}=\frac{1}{e}\]
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