Let `f:(-1,1)vecB` be a function defined by `f(x)=tan^(-1)(2x)/(1-x^2)` . Then `f` is both one-one and onto when `B` is the interval. `[0,pi/2)` (b) `(0,pi/2)` `(-pi/2,pi/2)` (d) `[-pi/2,pi/2]`

Options

A. `(-pi//2,pi//2)`
B. `[-pi//2,pi//2]`
C. `[0,pi//2]`
D. `(0,pi//2)`

Answer

A

Explanation

We have
`f(x)=tan^(-1)((2x)/(1-x^(2)))=2tan^(-1) x,"if"x in (-1,1)`

Since, f(x) is both one-one both and onto. Therefore,
B=Range `(f)=(f(-1)),f(1))=(-pi//2,pi//2)`