A) \[\left( -\frac{\pi }{2},\frac{\pi }{2} \right)\]
B) \[\left[ -\frac{\pi }{2},\frac{\pi }{2} \right]\]
C) \[\left[ 0,\frac{\pi }{2} \right)\]
D) \[\left( 0,\frac{\pi }{2} \right)\]
Correct Answer: A
Solution :
Since, \[x\in (-1,1)\] \[\Rightarrow \]\[{{\tan }^{-1}}x\in \left( -\frac{\pi }{4},\frac{\pi }{4} \right)\]\[\Rightarrow \]\[2{{\tan }^{-1}}x\in \left( -\frac{\pi }{2},\frac{\pi }{2} \right)\] and \[f(x)={{\tan }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right)=2{{\tan }^{-1}}x,({{x}^{2}}<1)\] So, \[f(x)\in \left( -\frac{\pi }{2},\frac{\pi }{2} \right)\] \[\therefore \]Function is one-to-one onto.You need to login to perform this action.
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