2024 EJC P1 Q9

2024 EJC P1 Q9

11 marks
Prelims

The function \(\mathrm{f}\) is given by \(\mathrm{f}\left( x \right)=\frac{1}{{{x}^{2}}-4}\), \(x\in \mathbb{R}\), \(x\ne -2\), \(x\ne 2\).

  1. Sketch the graph of \(y=\mathrm{f}\left( x \right)\). Give the equations of any asymptotes and the coordinates of any turning points.[2]
  2. If the domain of \(\mathrm{f}\) is further restricted to \(x\ge k\), state the least value of \(k\) for which the function \({{\mathrm{f}}^{-1}}\) exists. For this value of \(k\), find \({{\mathrm{f}}^{-1}}\left( x \right)\) and state the domain of \({{\mathrm{f}}^{-1}}\).[4]

For the rest of the question, the domain of \(\mathrm{f}\) is \(x\in \mathbb{R}\), \(x\ne -2\), \(x\ne 2\) as originally defined.

The function \(\mathrm{g}\) is given by \(\mathrm{g}\left( x \right)=\frac{1}{x-1}\), \(x\in \mathbb{R}\), \(x\ne \frac{1}{2}\), \(x\ne 1\), \(x\ne \frac{3}{2}\).

  1. Find \(\mathrm{fg}\left( x \right)\).[2]
  2. Find the range of \(\mathrm{fg}\).[3]

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Answer:(a) Asymptotes \(x=-2\), \(x=2\), \(y=0\); turning point \(\left(0,-\frac14\right)\) (b) \(k=0\), \(f^{-1}(x)=\sqrt{\frac1x+4}\), \(\mathrm D_{f^{-1}}=\left(-\infty,-\frac14\right]\cup(0,\infty)\) (c) \(fg(x)=\frac{x^2-2x+1}{-4x^2+8x-3}\) (d) \(\mathrm R_{fg}=\left(-\infty,-\frac14\right)\cup(0,\infty)\)

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