A) \[|x|\,\,<1\]
B) \[|x|\,\,>1\]
C) \[|x|\,\,=1\]
D) \[\phi \]
Correct Answer: B
Solution :
We know that; if\[|x|\,\,\le 1\], then\[2{{\tan }^{-1}}x={{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)\] and if\[|x|\,\,>1\], then\[\pi -2{{\tan }^{-1}}x={{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)\] thus, if\[|x|\,\,>1\] \[2{{\tan }^{-1}}x+{{\sin }^{-1}}\frac{2x}{1+{{x}^{2}}}=2{{\tan }^{-1}}x+\pi \] \[-2{{\tan }^{-1}}x=\pi \] which is independent of\[x\]
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