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.