A) \[1\]
B) \[\frac{2}{e}\]
C) \[e\]
D) \[\frac{1}{e}\]
Correct Answer: D
Solution :
Let \[y=\frac{1ogx}{x}\] On differentiating w.r.t. x, we get \[\frac{dy}{dx}=\frac{x.\frac{1}{x}-\log x.1}{{{x}^{2}}}=\frac{1-\log x}{{{x}^{2}}}\] For maxima, put \[\frac{dy}{dx}=0\] \[\Rightarrow \] \[\frac{1-\log x}{{{x}^{2}}}=0\] \[\Rightarrow \] \[\log \,x=1\] \[\Rightarrow \] \[x=e\] Now, \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{{{x}^{2}}\left( -\frac{1}{x} \right)-(1-\log x)2x}{{{({{x}^{2}})}^{2}}}\] At \[x=e,\] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}\le 0,\] maxima \[\therefore \] The maximum value at \[x=e\]is \[y=\frac{{{\log }_{e}}}{e}=\frac{1}{e}\]
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