A) e
B) 1
C) \[\frac{1}{e}\]
D) 2e
Correct Answer: C
Solution :
Let \[f(x)=\frac{\log x}{x}\Rightarrow f'(x)=\frac{1}{{{x}^{2}}}-\frac{\log x}{{{x}^{2}}}\] For maximum or minimum value of \[f(x),\,\,f'(x)=0\] Þ \[f'(x)=\frac{1-{{\log }_{e}}x}{{{x}^{2}}}=0\] or \[\frac{1-{{\log }_{e}}x}{{{x}^{2}}}=0\] \[\therefore {{\log }_{e}}x=1\]or \[x=e\], which lie in \[(0,\infty )\]. For \[x=e,\,\,\frac{{{d}^{2}}y}{d{{x}^{2}}}=-\frac{1}{{{e}^{3}}}\], which is \[-ve\]. Hence y is maximum at \[x=e\] and its maximum value \[=\frac{\log e}{e}=\frac{1}{e}\].
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