A) \[\frac{\pi }{12}\]
B) \[\frac{\pi }{6}\]
C) \[-\frac{\pi }{12}\]
D) \[\frac{\pi }{4}\]
Correct Answer: A
Solution :
| \[2f\left( \tan x \right)+\left( \cot x \right)\]\[=x;x\in R-\left\{ \frac{n\pi }{2},n\in I \right\}\] |
| Replacing, \[x\]by \[\left( \frac{\pi }{2}-x \right)\] |
| \[2f\left( \cot \,x \right)+f\left( \tan \,x \right)=\frac{\pi }{2}-x\] |
| Solving, we get \[f\left( \tan \,x \right)=x-\frac{\pi }{6}\] |
| \[\Rightarrow f\left( x \right)={{\tan }^{-1}}x-\frac{\pi }{6}\] |
| \[\int\limits_{0}^{1}{f\left( x \right)dx=\int\limits_{0}^{1}{\left( {{\tan }^{-1}}x-\frac{\pi }{6} \right)}dx}\]\[=\left( x{{\tan }^{-1}}x \right)_{0}^{1}\left. -\frac{1}{2}\ell n\left( 1+{{x}^{2}} \right) \right|_{0}^{1}-\frac{\pi }{6}\] |
| \[=\frac{\pi }{4}-\frac{1}{2}\ell n2-\frac{\pi }{4}=\frac{\pi }{12}-\frac{1}{2}\ell n2\] |
| \[\therefore \int\limits_{0}^{1}{f\left( x \right)dx+\frac{1}{2}\ell n2=\frac{\pi }{12}.}\] |
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