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 theseA. `sqrt((x)/(1-|x|))`B. `-"sign"(x)sqrt((|x|)/(1-|x|))`C. `sqrt((x)/(1-x))`D. None of these

If `f: Rvec(-1,1)`is defined by `f(x)=-(x|x|)/(1+x^2),t h e nf^(-1)(x)

1.	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 theseA. `sqrt((x)/(1-\|x\|))`B. `-"sign"(x)sqrt((\|x\|)/(1-\|x\|))`C. `sqrt((x)/(1-x))`D. None of these
Answer» Correct Answer - B Clearly, `f:R to (-1,1)` given by f(x)=`(-x\|x\|)/(1+x^(2))` is a bijection. Now, `fof^(-1)(x)=x` `Rightarrow f(f^(-1)(x))=x` `Rightarrow (f^(-1)(x)\|f^(-1)(x)\|)/(1+(f^(-1)(x))^(2))=x` `Rightarrow (-(f^(-1)(x)^(2))\|)/(1+(f^(-1)(x))^(2))=x, if f^(-1)(x) ge 0` and `(f^(-1)(x))^(2)/(1+(f^(-1)(x))^(2))=x,"if",f^(-1)(x) lt 0` `Rightarrow f^(-1)(x)={{:(,sqrt((x)/(1+x)),"if"xle 0),(,-sqrt((x)/(1-x)),"if"x gt 0):}` `Rightarrow f^(-1)(x)={{:(,sqrt((\|x\|)/(1-\|x\|)),"if"x le 0),(,-sqrt((\|x\|)/(1-\|x\|)),"if"x gt 0):}` `Rightarrow f^(-1)(x)=-sgn(x) sqrt((\|x\|)/(1-\|x\|))`