A) \[\infty \]
B) \[e\]
C) \[1\]
D) \[{{e}^{-1}}\]
Correct Answer: D
Solution :
Let \[y=\frac{\log x}{x}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{x\cdot \frac{1}{x}\log x}{{{x}^{2}}}=\frac{1-\log x}{{{x}^{2}}}\] Put \[\frac{dy}{dx}=0\] \[\Rightarrow \] \[\log x=1\] \[\Rightarrow \] \[x=e\] Now, \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{{{x}^{2}}\left( -\frac{1}{x} \right)-(1-\log x)2x}{{{({{x}^{2}})}^{2}}}\] \[=-\frac{(3-2\log x)}{{{x}^{3}}}\] \[\Rightarrow \]\[{{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}_{(x=e)}}=\frac{-(3-2)}{{{e}^{3}}}=-\frac{1}{{{e}^{3}}}<0\], maxima Hence, maximum value at\[x=e\]is\[\frac{1}{e}\].
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