Question

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

(b) $-\mathrm{Sgn}\left(x\right)\sqrt{\frac{\left|x\right|}{1-\left|x\right|}}$

$$\begin{gathered} \mathrm{We}\mathrm{have},f\left(x\right)=\frac{-x\left|x\right|}{1+x^2}x\in \left(-1,1\right) \\ \mathrm{Case}-\left(\mathrm{I}\right) \\ \mathrm{When},x<0, \\ \mathrm{Then},\left|x\right|=-x \\ \mathrm{And}f\left(x\right)>0 \\ \mathrm{Now}, \\ f\left(x\right)=\frac{-x\left(-x\right)}{1+x^2} \\ \Rightarrow y=\frac{x^2}{1+x^2} \\ \Rightarrow \frac{y}{1}=\frac{x^2}{1+x^2} \\ \Rightarrow \frac{y+1}{y-1}=\frac{x^2+1+x^2}{x^2-1-x^2}\left[\mathrm{Using}\mathrm{Componendo}\mathrm{and}\mathrm{dividendo}\right] \\ \Rightarrow \frac{y+1}{y-1}=\frac{2x^2+1}{-1} \\ \Rightarrow -\frac{y+1}{y-1}=2x^2+1 \\ \Rightarrow \frac{2y}{1-y}=2x^2 \\ \Rightarrow \frac{y}{1-y}=x^2 \\ \Rightarrow x=-\sqrt{\frac{y}{1-y}}\left(\mathrm{As}x<0\right) \\ \Rightarrow x=-\sqrt{\frac{\left|y\right|}{1-\left|y\right|}} \\ \left[\mathrm{As}y>0\right] \\ \mathrm{To}\mathrm{find}\mathrm{the}\mathrm{inverse}\mathrm{interchanging}x\mathrm{and}y\mathrm{we}\mathrm{get}, \\ {f}^{-1}\left(x\right)=-\sqrt{\frac{\left|x\right|}{1-\left|x\right|}}...\left(\mathrm{i}\right) \\ \mathrm{Case}-\left(\mathrm{II}\right) \\ \mathrm{When},x>0, \\ \mathrm{Then},\left|x\right|=x \\ \mathrm{And}f\left(x\right)<0 \\ \mathrm{Now}, \\ f\left(x\right)=\frac{-x\left(x\right)}{1+x^2} \\ \Rightarrow y=\frac{-x^2}{1+x^2} \\ \Rightarrow \frac{y}{1}=\frac{-x^2}{1+x^2} \\ \Rightarrow \frac{y+1}{y-1}=\frac{-x^2+1+x^2}{-x^2-1-x^2}\left[\mathrm{Using}\mathrm{Componendo}\mathrm{and}\mathrm{dividendo}\right] \\ \Rightarrow \frac{y+1}{y-1}=\frac{1}{-2x^2-1} \\ \Rightarrow \frac{1+y}{1-y}=\frac{1}{2x^2+1} \\ \Rightarrow \frac{1-y}{1+y}=2x^2+1 \\ \Rightarrow \frac{-2y}{1+y}=2x^2 \\ \Rightarrow x^2=\frac{-y}{1+y} \\ \Rightarrow x=\sqrt{\frac{-y}{1+y}}\left(\mathrm{As}x>0\right) \\ \Rightarrow x=\sqrt{\frac{\left|y\right|}{1-\left|y\right|}} \\ \left[\mathrm{As}y<0\right] \\ \mathrm{To}\mathrm{find}\mathrm{the}\mathrm{inverse}\mathrm{interchanging}x\mathrm{and}y\mathrm{we}\mathrm{get}, \\ {f}^{-1}\left(x\right)=\sqrt{\frac{\left|x\right|}{1-\left|x\right|}}...\left(\mathrm{ii}\right) \\ \mathrm{Case}-\left(\mathrm{III}\right) \\ \mathrm{When},x=0, \\ \mathrm{Then},f\left(x\right)=0 \\ \mathrm{Hence},f^{-1}\left(x\right)=0...\left(\mathrm{iii}\right) \\ \mathrm{Combinig}\mathrm{equation}\left(\mathrm{i}\right),\left(\mathrm{ii}\right)\mathrm{and}\left(\mathrm{iii}\right)\mathrm{we}\mathrm{get}, \\ {f}^{-1}\left(x\right)=-\mathrm{Sgn}\left(x\right)\sqrt{\frac{\left|x\right|}{1-\left|x\right|}} \end{gathered}$$