2018 AJC P1 Q9

The functions \(\mathrm{f}\) and \(\mathrm{g}\) are defined by

\(\mathrm{f}:x\mapsto \left( x+3 \right)\left| x-2 \right|\),

\(x\in \mathbb{R}\), \(\mathrm{g}:x\mapsto \frac{24}{{{x}^{2}}+3}-1\), \(x\in \mathbb{R}\), \(x>0\).

1. Explain why the function \({{\mathrm{f}}^{-1}}\) does not exist.[1]

In the rest of the question, the domain of \(\mathrm{f}\) is restricted to \(\left[ k,2 \right]\) where \(k\) is the least value such that \({{\mathrm{f}}^{-1}}\) exists.

2. Write down the value of \(k\). Find \({{\mathrm{f}}^{-1}}\left( x \right)\) and state the domain of \({{\mathrm{f}}^{-1}}\).[4]
3. Sketch the graphs of \(y=\mathrm{f}\left( x \right)\) and \(y={{\mathrm{f}}^{-1}}\left( x \right)\)on the same diagram, showing clearly the relationship between the two graphs. Hence, find the exact solution of the equation \(\mathrm{f}\left( x \right)={{\mathrm{f}}^{-1}}\left( x \right)\).[4]
4. Determine whether \({{\mathrm{g}}^{-1}}\mathrm{f}\) exists.[2]
5. Show that \(\mathrm{g}\) is a strictly decreasing function. Hence, without finding \({{\mathrm{g}}^{-1}}\), find the range of values of \(x\) for which \({{\mathrm{g}}^{-1}}\left( \mathrm{f}\left( x \right) \right)<1\), giving your answer in exact form.[3]