A) \[x>\frac{1}{e}\]
B) \[x<\frac{1}{e}\]
C) \[x<0\]
D) For all real x
Correct Answer: A
Solution :
Let \[y={{x}^{x}}\] Þ \[\frac{dy}{dx}={{x}^{x}}(1+\log x)\] For \[\frac{dy}{dx}>0\]; \[{{x}^{x}}(1+\log x)>0\] Þ \[1+\log x>0\Rightarrow {{\log }_{e}}x>{{\log }_{e}}\frac{1}{e}\] For this to be positive, x should be greater than \[\frac{1}{e}\].
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