2024 DHS Promo Q10

2024 DHS Promo Q10

11 marks

The function \(\mathrm{f}\) is defined by

\[\mathrm{f} : x \mapsto 3\sin \left( 2x - \frac{1}{6}\pi \right), \, 0 \le x \le k\]

  1. Show that the largest exact value of \(k\) such that \(\mathrm{f}^{-1}\) exists is \(\frac{1}{3}\pi\). Find \(\mathrm{f}^{-1}(x)\).[3]
  2. It is given that \(k = \frac{1}{3}\pi\). In a single diagram, sketch the graphs of \(y = \mathrm{f}(x)\) and \(y = \mathrm{f}^{-1}(x)\), labelling your graphs clearly.[3]

The function \(\mathrm{h}\) is defined by \(\mathrm{h} : x \mapsto 3\sin \left( 2x - \frac{1}{6}\pi \right), \, 0 \le x \le \frac{1}{3}\pi, \, x \neq \frac{1}{12}\pi\).

Another function \(\mathrm{g}\) is defined by \(\mathrm{g} : x \mapsto \left| 3 - \frac{1}{x} \right|, \, -3 \le x \le 3, \, x \neq 0\).

  1. Explain clearly why \(\mathrm{gh}\) exists. Find \(\mathrm{gh}(x)\) and its range.[3]
  2. Supposing \((\mathrm{gh})^{-1}\) exists for a restricted domain. Find the exact value of \(x\) for which \((\mathrm{gh})^{-1}(x) = 0\). Show your working clearly.[2]

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Answer:(a) \(k=\frac\pi3\), \(\mathrm f^{-1}(x)=\frac12\left[\sin^{-1}(x/3)+\frac\pi6\right]\) (b) See graph. (c) \(\mathrm{gh}(x)=\left|3-\dfrac1{3\sin(2x-\pi/6)}\right|\), range \([0,\infty)\) (d) \(x=\frac{11}{3}\)

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