A) 1
B) 1/e
C) 2/e
D) \[\frac{2}{{{e}^{2}}}\]
Correct Answer: B
Solution :
\[f(x)={{\cos }^{-1}}\left[ \frac{1-{{(\log x)}^{2}}}{1+{{(\log x)}^{2}}} \right]\]\[=2{{\tan }^{-1}}(\log x)\] Þ \[{f}'(x)=2.\frac{1}{1+{{(\log x)}^{2}}}.\frac{1}{x}.\text{Therefore }{f}'(e)=\frac{1}{e}\].You need to login to perform this action.
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