Question-26
Let \(\displaystyle f:\mathbb{R}\rightarrow \mathbb{R}\)
\[ \begin{equation*} f( x) =\frac{e^{x} -e^{-x}}{2} \end{equation*} \]
Consider the following statements:
- \(\displaystyle f\) is onto
- \(\displaystyle f\) is one-one
- \(\displaystyle f^{-1}\) exists
- \(\displaystyle f^{-1}( x) = \ln \left(x+\sqrt{x^{2} +1}\right)\)
- \(\displaystyle f^{-1}\) doesn’t exist
- \(\displaystyle f( x) \geqslant 0\) for all \(\displaystyle x\in \mathbb{R}\)
Find the number of true statements.
\(4\)
One-one
\[ \begin{equation*} \begin{aligned} f( x_{1}) & =f( x_{2})\\ e^{x_{1}} -e^{x_{2}} & =e^{-x_{1}} -e^{-x_{2}}\\ \left[ e^{x_{1}} -e^{x_{2}}\right] e^{x_{1} +x_{2}} & =e^{x_{2}} -e^{x_{1}}\\ \left[ e^{x_{1}} -e^{x_{2}}\right]\left( e^{x_{1} +x_{2}} +1\right) & =0\\ e^{x_{1}} & =e^{x_{2}}\\ \Longrightarrow x_{1} & =x_{2} \end{aligned} \end{equation*} \]
We see that \(\displaystyle f\) is indeed one-one.
Onto
Setting \(\displaystyle z:=e^{x}\) and \(\displaystyle y:=f( x)\), we have:
\[ \begin{equation*} \begin{aligned} 2y & =z-\frac{1}{z}\\ z^{2} -( 2y) z-1 & =0 \end{aligned} \end{equation*} \]
This gives us:
\[ \begin{equation*} z=y\pm \sqrt{y^{2} +1} \end{equation*} \]
Since \(\displaystyle z=e^{x} >0\), we have:
\[ \begin{equation*} z=y+\sqrt{y^{2} +1} \end{equation*} \]
Substituting \(\displaystyle z:=e^{x}\) back:
\[ \begin{equation*} e^{x} =y+\sqrt{y^{2} +1} \Longrightarrow x=\ln\left( y+\sqrt{y^{2} +1}\right) \end{equation*} \]
This is well defined as \(\displaystyle y+\sqrt{y^{2} +1}\) is always positive. Since \(\displaystyle y\) can assume any real value and there is a corresponding \(\displaystyle x\) for every such \(\displaystyle y\), \(\displaystyle f\) is onto.
Inverse
Since \(\displaystyle f\) is both one-one and onto, \(\displaystyle f\) is bijective. The inverse exists, is unique and is given by:
\[ \begin{equation*} f^{-1}( x) =\ln\left( x+\sqrt{1+x^{2}}\right) \end{equation*} \]