2018 MI Promo Q10

2018 MI Promo Q10

14 marks
Promo

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

\(\mathrm{f}:x\mapsto \frac{1}{x-2},\) \(x\in \mathbb{R},\) \(x\ne 2,\)

\(\mathrm{g}:x\mapsto {{x}^{2}}-4x\), \(x\in \mathbb{R}.\)

  1. Determine whether the composite functions \({{\mathrm{f}}^{2}}\), \(\mathrm{fg}\) and \(\mathrm{gf}\) exist.[4]
  2. Give a definition (including the domain) of the composite function that exists, and find the range of this composite function.[2]

In the rest of the equation, the domain of \(\mathrm{f}\) is \(x\in \mathbb{R},\) \(x < 2\).

  1. Define \({{\mathrm{f}}^{-1}}\) in similar form.[3]
  2. Sketch on a single clearly labelled diagram, the graphs of \(y=\mathrm{f}\left( x \right)\), \(y={{\mathrm{f}}^{-1}}\left( x \right)\) and \(y={{\mathrm{f}}^{-1}}\mathrm{f}\left( x \right)\). Your sketch should indicate the position of the graphs in relation to the origin.[3]
  3. Without using a graphing calculator, solve the equation \(\mathrm{f}\left( x \right)={{\mathrm{f}}^{-1}}\left( x \right)\).[2]

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Answer:\(gf\) exists; \(gf(x) = \frac{9 - 4x}{(x - 2)^2}\); \(f^{-1}(x) = 2 + \frac{1}{x}\); \(x = 1 - \sqrt{2}\)

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