# If f : R→ (−1, 1) is defined by

Question:

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

(a) $\sqrt{\frac{|x|}{1-|x|}}$

(b) $\operatorname{Sgn}(x) \sqrt{\frac{|x|}{1-|x|}}$

(c) $-\sqrt{\frac{x}{1-x}}$

(d) None of these

Solution:

(b) $-\operatorname{Sgn}$ $(x) \sqrt{\frac{|x|}{1-|x|}}$

We have, $f(x)=\frac{-x|x|}{1+x^{2}} \quad x \in(-1,1)$

Case - (I)

When, $x<0$,

Then, $|x|=-x$

And $f(x)>0$

Now,

$f(x)=\frac{-x(-x)}{1+x^{2}}$

$\Rightarrow y=\frac{x^{2}}{1+x^{2}}$

$\Rightarrow \frac{y}{1}=\frac{x^{2}}{1+x^{2}}$

$\Rightarrow \frac{y+1}{y-1}=\frac{x^{2}+1+x^{2}}{x^{2}-1-x^{2}} \quad[$ Using Componendo and dividendo]

$\Rightarrow \frac{y+1}{y-1}=\frac{2 x^{2}+1}{-1}$

$\Rightarrow-\frac{y+1}{y-1}=2 x^{2}+1$

$\Rightarrow \frac{2 y}{1-y}=2 x^{2}$

$\Rightarrow \frac{y}{1-y}=x^{2}$

$\Rightarrow x=-\sqrt{\frac{y}{1-y}}$                           (As $x<0$ )

$\Rightarrow x=-\sqrt{\frac{|y|}{1-|y|}}$

$[$ As $y>0]$

To find the inverse interchanging $x$ and $y$ we get,

$f^{-1}(x)=-\sqrt{\frac{|x|}{1-|x|}}$

Case - (II)

When, $x>0$,

Then, $|x|=x$

And $f(x)<0$

Now,

$f(x)=\frac{-x(x)}{1+x^{2}}$

$\Rightarrow y=\frac{-x^{2}}{1+x^{2}}$

$\Rightarrow \frac{y}{1}=\frac{-x^{2}}{1+x^{2}}$

$\Rightarrow \frac{y+1}{y-1}=\frac{-x^{2}+1+x^{2}}{-x^{2}-1-x^{2}} \quad$ [Using Componendo and dividendo]

$\Rightarrow \frac{y+1}{y-1}=\frac{1}{-2 x^{2}-1}$

$\Rightarrow \frac{1+y}{1-y}=\frac{1}{2 x^{2}+1}$

$\Rightarrow \frac{1-y}{1+y}=2 x^{2}+1$

$\Rightarrow \frac{-2 y}{1+y}=2 x^{2}$

$\Rightarrow x^{2}=\frac{-y}{1+y}$

$\Rightarrow x=\sqrt{\frac{-y}{1+y}}$                     $(\operatorname{As} x>0)$

$\Rightarrow x=\sqrt{\frac{|y|}{1-|y|}}$

$[$ As $y<0]$

To find the inverse interchanging $x$ and $y$ we get,

$f^{-1}(x)=\sqrt{\frac{|x|}{1-|x|}}$ ... (ii)

Case - (III)

When, $x=0$,

Then, $f(x)=0$

Hence, $f^{-1}(x)=0 \quad \ldots$ (iii)

Combinig equation (i), (ii) and (iii) we get,

$f^{-1}(x)=-\operatorname{Sgn}(x) \sqrt{\frac{|x|}{1-|x|}}$