2026 ACJC P1 Q7

2026 ACJC P1 Q7

Junior College 2
8 marks

It is given that \(\mathrm{f}:x\mapsto(x-4)^2-|x-4|\), \(0\leq x\leq\lambda\), where \(\lambda\) is a constant.

  1. Find the greatest value of \(\lambda\) for which the function \(\mathrm{f}^{-1}\) exists.[1]
  2. The function \(\mathrm{g}(x)\) is defined by \[\mathrm{g}(x)=\begin{cases}x+2&\text{for}\hspace{0.5em}-2\leq x\leq0,\\-x-2&\text{for}\hspace{0.5em}0<x\leq13.\end{cases}\]
    Sketch the graph of \(y=\mathrm{g}(x)\), stating the coordinates of the endpoints.[2]
  3. Considering the result in part (a), show that the composite function \(\mathrm{gf}\) exists.[1]
  4. Find the range of function \(\mathrm{gf}\).[1]
  5. Without finding \(\mathrm{gf}\) and \(\mathrm{g}^{-1}\), find the exact value of \(x\) for which \(\mathrm{gf}(x)=-3\).[3]

Solution:

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Answer:(a) \(\lambda=\frac72\) (b) endpoints \((-2,0)\), \((0,2)\), open \((0,-2)\), \((13,-15)\) (c) \(\mathrm{gf}\) exists (d) \([-14,-2)\cup[\frac74,2]\) (e) \(x=\frac{7-\sqrt5}{2}\)

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