If `f: R->(-1,\ 1)` is defined by `f(x)=(-x|x|)/(1+x^2)` , then `f^(-1)(x)` equals `sqrt((|x|)/(1-|x|))` (b) ` Sgn(x)sqrt((|x|)/(1-|x|))` (c) `-sqrt(x/(1-x))` (d) none of these

Options

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

Answer

B

Explanation

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|))`