A) 1
B) 2
C) 0
D) \[e\]
E) \[1/e\]
Correct Answer: E
Solution :
\[\because \] \[f(x)={{\log }_{x}}({{\log }_{e}}x)=\frac{{{\log }_{e}}{{\log }_{e}}x}{{{\log }_{e}}x}\] \[\therefore \] \[f(x)=\frac{\left[ \begin{align} & {{\log }_{e}}x\frac{1}{{{\log }_{e}}x}.\frac{1}{x}- \\ & \,\,\,\,\,\,\,\,\,\,\,\,\log {{\log }_{e}}x.\frac{1}{x} \\ \end{align} \right]}{{{({{\log }_{e}}x)}^{2}}}\] \[\Rightarrow \] \[f(x)=\frac{1-{{\log }_{e}}{{\log }_{e}}x}{x{{({{\log }_{e}}x)}^{2}}}\] \[\Rightarrow \] \[f(e)=\frac{1-{{\log }_{e}}{{\log }_{e}}e}{e({{\log }_{e}}e)}=\frac{1-{{\log }_{e}}1}{e}\] \[=\frac{1}{e}\]
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