Question

If  \[f : R \to \left( - 1, 1 \right)\] is defined by

\[f\left( x \right) = \frac{- x|x|}{1 + x^2}, \text{ then } f^{- 1} \left( x \right)\] equals

 

Options

• \[\sqrt{\frac{\left| x \right|}{1 - \left| x \right|}}\]

• \[\text{ Sgn } \left( x \right) \sqrt{\frac{\left| x \right|}{1 - \left| x \right|}}\]

•  \[- \sqrt{\frac{x}{1 - x}}\]

• None of these

Answer 1

(b)  \[- Sgn \left( x \right) \sqrt{\frac{\left| x \right|}{1 - \left| x \right|}}\]

\[\text{We have}, f\left( x \right) = \frac{- x|x|}{1 + x^2} x \in \left( - 1, 1 \right)\] 
\[\text{Case} - \left( I \right)\] 
\[\text{When}, x < 0, \] 
\[\text{Then}, \left| x \right| = - x\] 
\[\text{And} f\left( x \right) > 0\] 
\[\text{Now}, \] 
\[f\left( x \right) = \frac{- x\left( - x \right)}{1 + x^2}\] 
\[ \Rightarrow y = \frac{x^2}{1 + x^2}\] 
\[ \Rightarrow \frac{y}{1} = \frac{x^2}{1 + x^2}\] 
\[ \Rightarrow \frac{y + 1}{y - 1} = \frac{x^2 + 1 + x^2}{x^2 - 1 - x^2} \left[ \text { Using Componendo and dividendo } \right]\] 
\[ \Rightarrow \frac{y + 1}{y - 1} = \frac{2 x^2 + 1}{- 1}\] 
\[ \Rightarrow - \frac{y + 1}{y - 1} = 2 x^2 + 1\] 
\[ \Rightarrow \frac{2y}{1 - y} = 2 x^2 \] 
\[ \Rightarrow \frac{y}{1 - y} = x^2 \] 

\[ \Rightarrow x = - \sqrt{\frac{y}{1 - y}} \left( \text{ As } x < 0 \right)\] 
\[ \Rightarrow x = - \sqrt{\frac{\left| y \right|}{1 - \left| y \right|}} \] 
\[ \left[ \text{ As } y > 0 \right]\] 
\[\text{To find the inverse interchanging x and y we get}, \] 
\[ f^{- 1} \left( x \right) = - \sqrt{\frac{\left| x \right|}{1 - \left| x \right|}} . . . \left( i \right)\] 
\[\text{Case} - \left( II \right)\] 
\[\text{When}, x > 0, \] 
\[\text{Then}, \left| x \right| = x\] 
\[\text{And} f\left( x \right) < 0\] 
\[\text{Now}, \] 
\[f\left( x \right) = \frac{- x\left( x \right)}{1 + x^2}\] 
\[ \Rightarrow y = \frac{- x^2}{1 + x^2}\] 
\[ \Rightarrow \frac{y}{1} = \frac{- x^2}{1 + x^2}\] 
\[ \Rightarrow \frac{y + 1}{y - 1} = \frac{- x^2 + 1 + x^2}{- x^2 - 1 - x^2} \left[ \text{Using Componendo and dividendo} \right]\] 
\[ \Rightarrow \frac{y + 1}{y - 1} = \frac{1}{- 2 x^2 - 1}\] 
\[ \Rightarrow \frac{1 + y}{1 - y} = \frac{1}{2 x^2 + 1}\] 
\[ \Rightarrow \frac{1 - y}{1 + y} = 2 x^2 + 1\] 
\[ \Rightarrow \frac{- 2y}{1 + y} = 2 x^2 \] 

\[ \Rightarrow x^2 = \frac{- y}{1 + y}\] 
\[ \Rightarrow x = \sqrt{\frac{- y}{1 + y}} \left( \text{As} x > 0 \right)\] 
\[ \Rightarrow x = \sqrt{\frac{\left| y \right|}{1 - \left| y \right|}} \] 
\[ \left[ \text{ As } y < 0 \right]\] 
\[\text{To find the inverse interchanging x and y we get}, \] 
\[ f^{- 1} \left( x \right) = \sqrt{\frac{\left| x \right|}{1 - \left| x \right|}} . . . \left( ii \right)\] 
\[\text{Case} - \left( III \right)\] 
\[\text{When}, x = 0, \] 
\[\text{Then}, f\left( x \right) = 0\] 
\[\text{Hence}, f^{- 1} \left( x \right) = 0 . . . \left( iii \right)\] 
\[\text{Combinig equation} \left( i \right) , \left( ii \right) \text{and} \left( iii \right) \text{we get}, \] 
\[ f^{- 1} \left( x \right) = - Sgn\left( x \right)\sqrt{\frac{\left| x \right|}{1 - \left| x \right|}}\]