If `f: Rvec(-1,1)` is defined by `f(x)=-(x|x|)/(1+x^2),t h e nf^(-1)(x)` equals `sqrt((|x|)/(1-|x|))` (b) `-sgn(x)sqrt((|x|)/(1-|x|))` `-sqrt(x/(1-x))` (d) none of these

Explanation

Clearly, `f: R ->(-1,1)`, given by `f(x)=frac{-x|x|}{1+x^{2}}` is a bijection


Now `f circ f^{-1}(x)=x`



`Rightarrow {f}({f}^{-1}({x}))={x} `


`Rightarrow frac{-{f}^{-1}({x})|{f}^{-1}({x})|}{1+({f}^{-1}({x}))^{2}}={x} `


`Rightarrow-frac{({f}^{-1}({x}))^{2}}{1+({f}^{-1}({x}))^{2}}={x}, quad text { if } quad {f}^{-1}({x}) leq 0`



and `frac{(f^{-1}(x))^{2}}{1+(f^{-1}(x))^{2}}=x` if `f^{-1}(x) geq 0`


...