A) \[{{e}^{1/e}}\]
B) \[{{e}^{-e}}\]
C) \[{{e}^{-1/e}}\]
D) None of these
Correct Answer: C
Solution :
Idea Here \[y={{x}^{x}}\Rightarrow \log y=x\log x=S\]Use above condition and then differential with respect to x. \[\because \] We know that for minimum \[\frac{{{d}^{2}}S}{d{{x}^{2}}}>0\] Let\[f(x)={{(\sin x)}^{\sin x}}\] \[S=\log f(x)=\sin x\log (\sin x)\] \[\frac{dS}{dx}=\sin x\frac{\cos x}{\sin x}+\log (\sin x)\cos x\] \[=\cos x+\cos x\log \sin x\] \[=\cos x+(1+\log \sin x)\] \[\frac{{{d}^{2}}S}{d{{x}^{2}}}=\cos x\frac{\cos x}{\sin x}+(1+\log \sin x)(-\sin x)\] \[\frac{{{d}^{2}}S}{d{{x}^{2}}}=\frac{{{\cos }^{2}}x}{\sin x}-\sin x(1+\log \sin x)\] \[\frac{dS}{dx}=0\]\[\cos x(1+\log \sin x)=0\] \[\Rightarrow \]\[\cos x=0\]or\[1+\log \sin x=0\] \[x=\pi /2\]or \[\log \sin x=-1\] \[x=\frac{\pi }{2}\] or \[\sin x={{e}^{-1}}\] \[x=\frac{\pi }{2}\] or \[\sin x=\frac{1}{e}\] \[{{\left( \frac{{{d}^{2}}S}{d{{x}^{2}}} \right)}_{\sin x=\frac{1}{e}}}=\frac{\left( 1-\frac{1}{{{e}^{2}}} \right)}{\left( \frac{1}{e} \right)}-\frac{1}{e}(1+\log {{e}^{-1}})\] \[=e\left( 1-\frac{1}{{{e}^{2}}} \right)-\frac{1}{e}(1-1)=e\left( 1-\frac{1}{{{e}^{2}}} \right)>0\] \[\therefore \]f (x) has minimum value at \[x=\frac{1}{e}.\] \[\therefore \]Minimum value \[{{\left( \frac{1}{e} \right)}^{1/e}}={{e}^{-1/e}}\] TEST Edge The function is maximum or minimum and maximum or minimum value related Question are asked. Students are suggested that learn of maximum or minimum and also acquainted yourself the domain of some functions.
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