A) \[\frac{2}{e}\]
B) \[\frac{1}{e}\]
C) \[e\]
D) \[1\]
E) \[\frac{e}{2}\]
Correct Answer: B
Solution :
Let \[y=\frac{\log x}{x}\] On differentiating w.r.t.\[x,\]we get \[\frac{dy}{dx}=\frac{x.\frac{1}{x}-\log x}{{{x}^{2}}}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{1-\log x}{{{x}^{2}}}\] Put \[\frac{dy}{dx}=0\] for maxima or minima \[\therefore \] \[x=e\] Now, \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{{{x}^{2}}\left( -\frac{1}{x} \right)-(1-\log x)2x}{{{({{x}^{2}})}^{2}}}\] \[=\frac{-3+2\log x}{{{x}^{3}}}\] and \[\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right){{ & }_{x=e}}=-\frac{1}{{{e}^{3}}}<0\] \[\therefore \]Function is maximum at\[x=e\]. \[\therefore \] \[{{y}_{\max }}=\frac{\log e}{e}=\frac{1}{e}\]
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