Question

If $f:R\to (-1,1)$ is defined by $f\left(x\right)=\frac{-x\left|x\right|}{1+x^2}$ then $f^{-1}\left(x\right)$ equals

• A. $\sqrt{\frac{\left|x\right|}{1-\left|x\right|}}$
• B. $-sgn\left(x\right)\sqrt{\frac{\left|x\right|}{1-\left|x\right|}}$
• C. $\sqrt{\frac{x}{1-x}}$
• D. None of these

Answer 1

Answer: (B)

$-sgn\left(x\right)\sqrt{\frac{\left|x\right|}{1-\left|x\right|}}$

Clearly, $f:R\to (-1,1)$, given by $f\left(x\right)=\frac{-x\left|x\right|}{1+x^2}$ is a bijection
Now $f\circ f^{-1}\left(x\right)=x$
$\Rightarrow f\left(f^{-1}\right(x\left)\right)=x$
$\Rightarrow \frac{-f^{-1}\left(x\right)\left|f^{-1}\right(x\left)\right|}{1+{\left(f^{-1}\right(x\left)\right)}^2}=x$
$\Rightarrow -\frac{{\left(f^{-1}\right(x\left)\right)}^2}{1+{\left(f^{-1}\right(x\left)\right)}^2}=x,if{f}^{-1}\left(x\right)\le 0$
and $\frac{{\left(f^{-1}\right(x\left)\right)}^2}{1+{\left(f^{-1}\right(x\left)\right)}^2}=xif{f}^{-1}\left(x\right)\ge 0$
$\therefore f^{-1}\left(x\right)=\{\begin{array}{c}\sqrt{\frac{-\left|x\right|}{1-\left|x\right|}},ifx\le 0\\ -\sqrt{\frac{-\left|x\right|}{1-\left|x\right|}},ifx>0\end{array}$
$\therefore f^{-1}\left(x\right)=-sgn\left(x\right)\sqrt{\frac{\left|x\right|}{1-\left|x\right|}}$