A) \[e\]
B) \[{{(e)}^{1/e}}\]
C) \[{{\left( \frac{1}{e} \right)}^{e}}\]
D) none of these
Correct Answer: B
Solution :
We have,\[f(x)={{\left( \frac{1}{x} \right)}^{x}}\] On taking log on both sides, we get \[\log f(x)=x\log \left( \frac{1}{x} \right)\] \[\Rightarrow \] \[\log f(x)=-x\log x\] On differentiating both sides, we get \[\frac{1}{f(x)}f'(x)=-(1+\log x)\] \[\Rightarrow \] \[f'(x)=-f(x)(1+\log x)\] ? (i) For maximum or minimum, put\[f'(x)=0\] \[\Rightarrow \] \[-(1+\log x)f(x)=0\] \[\Rightarrow \] \[1+\log x=0\] \[\Rightarrow \] \[\log x=\log \left( \frac{1}{e} \right)\] \[\Rightarrow \] \[x=\frac{1}{e}\] \[f'\,\,'(x)=-\left[ f'(x)(1+\log x)+\frac{f(x)}{x} \right]\] At\[x=\frac{1}{e},\,\,f'\,\,'(x)<0\], maximum \[\therefore \]Maximum value of\[f(x)={{\left( \frac{1}{1/e} \right)}^{1/e}}={{e}^{1/e}}\] Note: If\[f'\,'(x)>0\], then the function is minimum.
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