| Direction: Question Based on the following paragraph. |
| Let \[f:A\to B\]be a function defined by y=f(x) such that f is both one-one (Injective) and onto Surjective), then there exists a unique function \[g:B\to A\]such that \[f(x)=y\leftrightarrow g\] \[(y)=x,\forall x\in A\]and \[y\in B,\]then g is said to be inverse off Thus,\[g={{f}^{-1}}:B\to A=[\{f(x),x\}:\{x,f(x)\}\in {{f}^{-1}}]\] If no branch of an inverse trigonometric function is mentioned, then it means the principal value branch of that function. |
A) \[\frac{\pi }{2}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{3\pi }{2}\]
D) \[\pi \]
Correct Answer: D
Solution :
Since, \[2{{\tan }^{-1}}x=\pi -{{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right),x>1\] \[\therefore \] \[2{{\tan }^{-1}}x+{{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)\] \[=\pi -{{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)+{{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)\]\[=\pi \]
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