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2007 Paper 2 Q5
D: 1600.0 B: 1488.1
trigonometry function iteration tan addition formula composite functions periodicity inverse trigonometric functions substitution

In this question, \(\f^2(x)\) denotes \(\f(\f(x))\), \(\f^3(x)\) denotes \(\f( \f (\f(x)))\,\), and so on.

  1. The function \(\f\) is defined, for \(x\ne \pm 1/ \sqrt3\,\), by $$ \f(x) = \ds \frac{x+\sqrt3} {1-\sqrt3\, x }\,. $$ Find by direct calculation \(\f^2(x) \) and \(\f^3(x)\), and determine \(\f^{2007}(x)\,\).
  2. Show that \(\f^n(x) = \tan(\theta + \frac 13 n\pi)\), where \(x=\tan\theta\) and \(n\) is any positive integer.
  3. The function \(\g(t)\) is defined, for \(\vert t\vert\le1\) by \(\g(t) = \frac {\sqrt3}2 t + \frac 12 \sqrt {1-t^2}\,\). Find an expression for \(\g^n(t)\) for any positive integer \(n\).

Solution:

  1. \(\,\) \begin{align*} && f(x) &= \frac{x+\sqrt3}{1-\sqrt3x} \\ \Rightarrow && f(f(x)) &= \frac{f(x)+\sqrt3}{1-\sqrt3f(x)} \\ &&&= \frac{\frac{x+\sqrt3}{1-\sqrt3x}+\sqrt3}{1-\sqrt3 \frac{x+\sqrt3}{1-\sqrt3x}} \\ &&&= \frac{x+\sqrt{3}+\sqrt3(1-\sqrt3x)}{1-\sqrt3x-\sqrt3(x+\sqrt3)} \\ &&&= \frac{-2x+2\sqrt3}{-2-2\sqrt3x} \\ &&&= \frac{x-\sqrt3}{1+\sqrt3 x} \\ \\ && f^3(x) &= f^2(f(x)) \\ &&&= \frac{f(x)-\sqrt3}{1+\sqrt3 f(x)} \\ &&&=\frac{\frac{x+\sqrt3}{1-\sqrt3x}-\sqrt3}{1+\sqrt3 \frac{x+\sqrt3}{1-\sqrt3x}} \\ &&&= \frac{(x+\sqrt3)-\sqrt3(1-\sqrt3 x)}{(1-\sqrt3x)+\sqrt3 (x+\sqrt3)} \\ &&&= \frac{-2x}{-2} = x \\ \\ && f^{2007}(x) &= x \end{align*}
  2. If \(x = \tan \theta\) then \(f(x) = \frac{\tan \theta + \tan \frac{\pi}{3}}{1 - \tan \frac{\pi}{3} \tan \theta} = \tan (\theta + \frac{\pi}{3})\) and hence \(f^n(x) = \tan (\theta + \frac{n \pi}{3})\)
  3. Note that if \(t = \sin \theta\) then \(g(t) = \cos \frac{\pi}{6} t\sin \theta + \frac12 \cos \theta = \sin(\theta + \frac{\pi}6)\) therefore \(g^n(t) = \sin(\sin^{-1}(t) + \frac{n\pi}{6})\)

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