N2022 P1 Q6

N2022 P1 Q6

Junior College 2
9 marks
Free

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

\[\mathrm{f}:x\mapsto\frac{ax+k}{x-a},\quad x\in\mathbb{R},\ x\ne a\]

where \(a\) and \(k\) are constants.

  1. Describe fully a sequence of transformations which transforms the curve \(y=\frac1x\) onto the curve \(y=\mathrm{f}(x)\).[4]
  2. Find \(\mathrm{f}^{-1}(x)\).[2]
  3. Hence, or otherwise, find \(\mathrm{f}^{2}(x)\).[1]
  4. Find \(\mathrm{f}^{2023}(1)\) in terms of \(a\) and \(k\).[2]

Default solution

  1. \(\mathrm{f}(x)=a+\frac{a^2+k}{x-a}\)
    Translate \(y=1/x\) by \(a\) units parallel to the \(x\)-axis; scale by factor \(a^2+k\) parallel to the \(y\)-axis; translate by \(a\) units parallel to the \(y\)-axis.
  2. \(y=\frac{ax+k}{x-a}\Rightarrow yx-ay=ax+k\Rightarrow x(y-a)=ay+k\)
    \(\mathrm{f}^{-1}(x)=\frac{ax+k}{x-a}\), provided \(a^2+k\ne0\).
  3. \(\mathrm{f}=\mathrm{f}^{-1}\Rightarrow \mathrm{f}^{2}(x)=x\), provided \(a^2+k\ne0\).
  4. \(2023\) is odd, so \(\mathrm{f}^{2023}(1)=\mathrm{f}(1)=\frac{a+k}{1-a}\).
    This requires \(a^2+k\ne0\) and \(a\ne1\); the printed question omits these conditions.
Answer:(a) Right \(a\), vertical scale \(a^2+k\), up \(a\) (b) \(\mathrm{f}^{-1}(x)=\frac{ax+k}{x-a}\) (c) \(\mathrm{f}^{2}(x)=x\) (d) \(\frac{a+k}{1-a}\) (where defined)

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