A) \[{{\log }_{e}}3\]
B) \[{{\log }_{e}}2\]
C) \[{{\log }_{e}}e\]
D) \[{{\log }_{e}}1\]
Correct Answer: D
Solution :
\[g(f(x))=\ell n(f(x))=\ell n\left( \frac{2-x.\cos x}{2+x.\cos x} \right)\] \[\therefore \]\[I=\int\limits_{-\pi /4}^{\pi /4}{\ell n}\left( \frac{2-x.\cos x}{2+x.\cos x} \right)dx\] \[=\int\limits_{0}^{\pi /4}{\left( \ell n\left( \frac{2-x.\cos x}{2+x.\cos x} \right)+\ell n\left( \frac{2+x.\cos x}{2-x.\cos x} \right) \right)dx}\] \[=\int\limits_{0}^{\pi /4}{(0)dx=0={{\log }_{e}}}(1)\]
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