A) \[\frac{\log 2}{2}\]
B) Zero
C) \[\frac{1}{e}\]
D) Does not exist
Correct Answer: D
Solution :
Let \[y=\frac{\log x}{x}\] Þ \[\frac{dy}{dx}=\frac{x.\frac{1}{x}-\log x}{{{x}^{2}}}\]\[=\frac{1-\log x}{{{x}^{2}}}\] Put \[\frac{dy}{dx}=0\Rightarrow \frac{1-\log x}{{{x}^{2}}}=0\] Þ \[1-\log x=0\] Þ \[x=e\] and \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{-3x+2x\log x}{{{x}^{4}}}\] At \[x=e\], \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{1}{-{{e}^{3}}}<0\] \ In [2, ¥) the function \[{{p}^{2}}=q\] will be maximum and minimum value does not exist.
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