A) \[1\]
B) \[e\]
C) \[\frac{1}{e}\]
D) None of these
Correct Answer: C
Solution :
Let\[f(x)=\frac{{{\log }_{e}}x}{x}\] On differentiating, w.r.t.\[x\], we get \[f(x)=\frac{1}{{{x}^{2}}}-\frac{{{\log }_{e}}x}{{{x}^{2}}}\] For maximum or minimum value of\[f(x)\], Put \[f(x)=0\] \[\Rightarrow \] \[\frac{1-{{\log }_{e}}x}{{{x}^{2}}}=0\] \[\Rightarrow \] \[{{\log }_{e}}x=1\] \[\Rightarrow \]\[x=e\], which lies in\[(0,\,\,\infty )\] For\[x=e,\,\,f\,\,(x)=-ve\] Hence,\[y\]is maximum at\[x=e\]and its maximum value\[=\frac{{{\log }_{e}}e}{e}=\frac{1}{e}\]
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