question_answer

Let\[f:(-1,1)\to B\] be a function defined by\[f(x)={{\tan }^{-1}}\frac{2x}{1-{{x}^{2}}},\] then f is both one-one and onto when B is the interval     AIEEE  Solved  Paper-2005

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, and    So,     Function is one-one onto.