A) \[\{x:x\ge e\}\]
B) empty
C) \[\{x:x<e\}\]
D) \[\{1\}\]
Correct Answer: A
Solution :
\[f(x)=\frac{x}{\log x}\] \[f(x)=\frac{\log x.1-x.\frac{1}{x}}{{{(\log x)}^{2}}}=\frac{(\log x-1)}{{{(\log x)}^{2}}}\] We know that, \[f(x)\]is increasing (strictly) When \[f'(x)>0\] \[\Rightarrow \] \[\frac{(\log x-1)}{{{(\log x)}^{2}}}>0\] \[\Rightarrow \] \[(\log x-1)>0\] \[\Rightarrow \] \[\log x>1\] \[\Rightarrow \] \[{{\log }_{e}}x>{{\log }_{e}}e\] \[\Rightarrow \] \[x>e\] Hence, \[x:x\ge e\]
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