2019 ASRJC P2 Q2

2019 ASRJC P2 Q2

5 marks
Prelims

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

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

\(x\in \mathbb{R}\) \(\mathrm{g}:x\mapsto a{{\mathrm{e}}^{-x}}+1\), \(x\ge 0\)

where \(a\) is a positive constant and \(a\ge 5\).

  1. Sketch the graph of \(y=\mathrm{g}(x)\) and \(y=\mathrm{f}(x)\) on separate diagrams.
    Hence find the range of the composite function \(\mathrm{fg}\), leaving your answer in exact form in terms of \(a\).[5]
  2. Given that the domain of \(\mathrm{f}\) is restricted to the subset of \(\mathbb{R}\) for which \(x\ge k\), find the smallest value of \(k\), in terms of \(a\), for which \({{\mathrm{f}}^{-1}}\) exists. Hence, without finding the expression for \({{\mathrm{f}}^{\mathrm{-1}}}(x)\), sketch the graphs of \(y=\mathrm{f}(x)\) and \(y={{\mathrm{f}}^{-1}}(x)\) on a single diagram, stating the exact coordinates of any points of intersections with the axes.[3]
  3. With the restricted domain of \(\mathrm{f}\) in part (ii), find with clear working, the set of values of \(x\) such that \(\mathrm{f}{{\mathrm{f}}^{-1}}(x)={{\mathrm{f}}^{\mathrm{-1}}}\mathrm{f}(x)\).[2]

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Answer:(i) \(R_{fg}=\left[0,\frac{a^2}{4}\right]\); (ii) smallest value \(a\); (iii) \([a,\infty)\)

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