Question

If $f:R\to \left( -1,1 \right)$ is defined by $f\left( x \right)=\dfrac{-x\left| x \right|}{1+{{x}^{2}}}$, then ${{f}^{-1}}\left( x \right)$ equals
(a) $\sqrt{\dfrac{\left| x \right|}{1-\left| x \right|}}$
(b) $-\left( x \right)\sqrt{\dfrac{\left| x \right|}{1-\left| x \right|}}$
(c) $\sqrt{\dfrac{x}{1-x}}$
(d) None of these

Answer 1

- Hint:In this question first of all, expand $\left| x \right|$ by writing it x for $x\ge 0$ and – x for x < 0. Then express x in terms of f(x) and replace x by ${{f}^{-1}}\left( x \right)$ and f(x) by x. Also the domain and range of f(x) and ${{f}^{-1}}\left( x \right)$ would be interchanged

Complete step-by-step solution -

We are given function $f:R\to \left( -1,1 \right)$ such that $f\left( x \right)=\dfrac{-x\left| x \right|}{1+{{x}^{2}}}$.
Here, we have to find the value of ${{f}^{-1}}\left( x \right)$. We know that if $f:A\to B$, then ${{f}^{-1}}:B\to A$.
Here, as $f:R\to \left( -1,1 \right)$, we get ${{f}^{-1}}\left( -1,1 \right)\to R$.
Now, to find ${{f}^{-1}}\left( x \right)$, we must express x in terms of f(x) and then replace x by ${{f}^{-1}}\left( x \right)$ and f(x) by x as follows:
First of all, we take,
$f\left( x \right)=\dfrac{-x\left| x \right|}{1+{{x}^{2}}}$
We know that for $x\ge 0,\left| x \right|=x$
And for $x<0,\left| x \right|=-x$
Hence, for $x\ge 0$ and $-1< f\left( x \right)<0$
We get, $f\left( x \right)=\dfrac{\left( -x \right).x}{1+{{x}^{2}}}$
$f\left( x \right)=\dfrac{-{{x}^{2}}}{1+{{x}^{2}}}$
Now, by cross multiplying above equation we get,
$f\left( x \right)\left( 1+{{x}^{2}} \right)=-{{x}^{2}}$
$f\left( x \right)+f\left( x \right).{{x}^{2}}=-{{x}^{2}}$
$f\left( x \right).{{x}^{2}}+{{x}^{2}}=-f\left( x \right)$
${{x}^{2}}\left[ 1+f\left( x \right) \right]=-f\left( x \right)$
${{x}^{2}}=\dfrac{-f\left( x \right)}{\left[ 1+f\left( x \right) \right]}$
$x=\sqrt{\dfrac{-f\left( x \right)}{1+f\left( x \right)}}$
Now, to get ${{f}^{-1}}\left( x \right)$, we will replace x by ${{f}^{-1}}\left( x \right)$ and f(x) by x.
Hence, we get
${{f}^{-1}}\left( x \right)=\sqrt{\dfrac{-x}{1+x}}$ defined for $-1< x\le 0$
Now, for x < 0, we get,
$f\left( x \right)=\dfrac{\left( -x \right).\left( -x \right)}{1+{{x}^{2}}}$
$f\left( x \right)=\dfrac{{{x}^{2}}}{1+{{x}^{2}}}$
Now, by cross multiplying above equation, we get,
$f\left( x \right)\left[ 1+{{x}^{2}} \right]={{x}^{2}}$
$f\left( x \right)+{{x}^{2}}f\left( x \right)={{x}^{2}}$
${{x}^{2}}f\left( x \right)-{{x}^{2}}=-f\left( x \right)$
${{x}^{2}}\left[ f\left( x \right)-1 \right]=-f\left( x \right)$
${{x}^{2}}=\dfrac{-f\left( x \right)}{\left( f\left( x \right)-1 \right)}$
Multiplying by (– 1) in numerator and denominator, we get,
${{x}^{2}}=\dfrac{+f\left( x \right)}{1-f\left( x \right)}$
$x=\sqrt{\dfrac{f\left( x \right)}{1-f\left( x \right)}}$
Now, to get ${{f}^{-1}}\left( x \right)$, we will replace x by ${{f}^{-1}}\left( x \right)$ and f(x) by x.
Hence, we get
${{f}^{-1}}\left( x \right)=\sqrt{\dfrac{x}{1-x}}$ defined for $0\le x<1$
Therefore, for $x\ge 0$,
${{f}^{-1}}\left( x \right)=\sqrt{\dfrac{-x}{1+x}}$; – 1 < x < 0
And x < 0, ${{f}^{-1}}\left( x \right)=\sqrt{\dfrac{x}{1-x}};0\le x<1$
We know that $\left| x \right|=x$ when $x\ge 0$ and $\left| x \right|=-x$ when x < 0.
Therefore, we get ${{f}^{-1}}\left( x \right)=\sqrt{\dfrac{\left| x \right|}{1-\left| x \right|}}$ for – 1 < x < 1 as it satisfies both ${{f}^{-1}}\left( x \right)$.
Therefore, option (a) is the correct answer.

Note: Here students must be careful about modulus function and should remember that in f(x) and ${{f}^{-1}}\left( x \right)$, the range and domain interchange. Students must take care that when x was greater than 0 in f(x), at that time, when we found ${{f}^{-1}}\left( x \right)$, x was less than 0. Similarly when x was less than 0 in f(x), at that time when we found ${{f}^{-1}}\left( x \right)$, x was greater than 0.