2024 ASRJC P1 Q8

2024 ASRJC P1 Q8

12 marks
Prelims

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

\(\mathrm{f}:x\mapsto \left| 4+2x-{{x}^{2}} \right|\), \(x\in \mathbb{R}\), \(x\ge 3.5\),

\(\mathrm{g}:x\mapsto 4+\mathrm{e}{{\,}^{ax}}\), \(x\in \mathbb{R}\), \(x\ge -1\),

where \(a > 0\).

  1. Find \({{\mathrm{f}}^{-1}}\left( x \right)\) and state its domain.[3]
  2. Find the value of \(x\) for which \({{\mathrm{f}}^{-1}}\left( x \right)=\mathrm{f}\left( x \right)\).[2]
  3. Show that the composite function \(\mathrm{fg}\) exists and express the exact range of \(\mathrm{fg}\) in the form of \(A+B{{\mathrm{e}}^{-a}}+C{{\mathrm{e}}^{-2a}}\), where \(A\), \(B\) and \(C\) are real constants.[4]
  4. Without the use of a graphing calculator, solve the inequality \(\frac{\mathrm{g}\left( x \right)}{{{x}^{2}}-2x-2}\ge 0\).
    Leave your answer in exact form.[3]

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Answer:(a) \(f^{-1}(x)=1+\sqrt{x+5}\), \(\mathrm D_{f^{-1}}=[1.25,\infty)\) (b) \(x=4\) (c) \(\mathrm R_{fg}=[4+6\mathrm e^{-a}+\mathrm e^{-2a},\infty)\) (d) \(-1\le x<1-\sqrt3\) or \(x>1+\sqrt3\)

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