2022 ACJC Promo Q9

It is given that

\(\mathrm{f}\left( x \right)={{x}^{2}}-6x+5\), for \(1 < x\le 5\),

\(\mathrm{g}\left( x \right)={{x}^{2}}-6x+5\), for \(1 < x < 3\),

\(\mathrm{h}\left( x \right)=\left\{ \begin{matrix} \ln x\,\mathrm{for}0<x\le 1, \\ {{x}^{2}}-6x+5\mathrm{for}1<x\le 5, \\ \end{matrix} \right.\)

and that \(\mathrm{h}\left( x \right)=\mathrm{h}\left( x-5 \right)\) for all real values of \(x\).

1. Find the range of \(\mathrm{f}\).[1]
2. Show that \({{\mathrm{g}}^{-1}}\) exists.[1]
3. Find \({{\mathrm{g}}^{-1}}\left( x \right)\) and state the domain of \({{\mathrm{g}}^{-1}}\).[3]
4. Find the exact value of \(\mathrm{h}\left( \frac{11}{2} \right)+\mathrm{h}\left( -2 \right)\).[2]
5. Sketch the graph of \(y=\mathrm{h}\left( x \right)\) for \(-4\le x\le 6\), stating the equations of any asymptotes and the coordinates of any points of intersection with the axes.[3]