A) \[e\]
B) \[{{\log }_{x}}e\]
C) \[1/e\]
D) \[e/2\]
Correct Answer: C
Solution :
\[y=\frac{{{\log }_{e}}x}{x}\Rightarrow \frac{dy}{dx}=\frac{x.\frac{1}{x}-{{\log }_{e}}x.1}{{{x}^{2}}}\] \[\Rightarrow \]\[\frac{dy}{dx}=\frac{1-{{\log }_{e}}x}{{{x}^{2}}}\] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{{{x}^{2}}\left( -\frac{1}{x} \right)-(1-{{\log }_{e}}x)2x}{{{x}^{4}}}\] \[\Rightarrow \]\[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{-x-2x+2x{{\log }_{e}}x}{{{x}^{4}}}\] \[\Rightarrow \]\[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{-3x+2x{{\log }_{e}}x}{{{x}^{4}}}\] \[\Rightarrow \]\[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{-3+2{{\log }_{e}}x}{{{x}^{3}}}\]put \[\frac{dy}{dx}=0\] \[\frac{1-{{\log }_{e}}x}{{{x}^{2}}}\Rightarrow 1-{{\log }_{e}}x=0\] \[\Rightarrow \]\[1={{\log }_{e}}x\] put the value of \[{{\log }_{e}}x\]in \[\frac{{{d}^{2}}y}{d{{x}^{2}}}\] \[=\frac{-3+2\times 1}{{{e}^{2}}}=-\]and \[y\]is maximum at \[x=e\]there \[y=\frac{{{\log }_{e}}e}{e}=1/e\]
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