2021 EJC Promo Q8

The function \(\mathrm{f}\) is defined by \(\mathrm{f}:x\mapsto {{\mathrm{e}}^{\left| 2-x \right|}}\), \(x\in \mathbb{R}\), \(x>a\), where \(a\) is a constant.

It is given that \(a\) is the least possible value such that \({{\mathrm{f}}^{-1}}\) exists.

1. State the value of \(a\).[1]
2. Hence find \({{\mathrm{f}}^{-1}}\left( x \right)\) and state the domain of \({{\mathrm{f}}^{-1}}\).[4]
3. Sketch on the same diagram the graphs of \(y=\mathrm{f}\left( x \right)\), \(y={{\mathrm{f}}^{-1}}\left( x \right)\) and \(y=\mathrm{f}{{\mathrm{f}}^{-1}}\left( x \right)\), giving the coordinates of all end points.[3]

It is now given that \(a\) is replaced by a constant such that the functions \(\mathrm{f}{{\mathrm{f}}^{-1}}\) and \({{\mathrm{f}}^{-1}}\mathrm{f}\) are equal.

4. Explain why the new value of \(a\) satisfies the equation \(a=\mathrm{f}\left( a \right)\), and find \(a\).[2]
5. The function \(\mathrm{g}\) is defined by \(\mathrm{g}:x\mapsto {{\mathrm{e}}^{\left| 2-x \right|}}\), \(x\in \mathbb{Z}\). Explain why the composite function \({{\mathrm{g}}^{2}}\) does not exist.[1]