A) \[{{\tan }^{-1}}(\log x)\]
B) \[0\]
C) \[\frac{1}{2}\]
D) None of these
Correct Answer: B
Solution :
We have\[f(x)={{\tan }^{-1}}\left( \frac{\log \frac{e}{{{x}^{2}}}}{\log e{{x}^{2}}} \right)\] \[+{{\tan }^{-1}}\left( \frac{1+2\log x}{1-2\log x} \right)\] \[\Rightarrow \] \[f(x)={{\tan }^{-1}}\left\{ \frac{\log e-\log {{x}^{2}}}{\log e+\log {{x}^{2}}} \right\}\] \[+{{\tan }^{-1}}\left( \frac{1+\log {{x}^{2}}}{1-\log {{x}^{2}}} \right)\] \[\Rightarrow \] \[f(x)={{\tan }^{-1}}\left( \frac{1-\log {{x}^{2}}}{1+\log {{x}^{2}}} \right)\] \[+{{\tan }^{-1}}\left( \frac{1+\log {{x}^{2}}}{1-\log {{x}^{2}}} \right)\] \[\Rightarrow \] \[f(x)={{\tan }^{-1}}\left( \frac{1-\log {{x}^{2}}}{1+\log {{x}^{2}}} \right)\] \[+{{\cot }^{-1}}\left( \frac{1-\log {{x}^{2}}}{1+\log {{x}^{2}}} \right)\] \[\Rightarrow \] \[f(x)=\frac{\pi }{2}\] \[\left[ \because \,\,{{\tan }^{-1}}x={{\cot }^{-1}}\left( \frac{1}{x} \right) \right]\] \[\therefore \] \[f(x)=0\]
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