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STEP II 2007 question 5 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 2007 question 5 solution

(i)

\displaystyle f(x) = \frac{x+\sqrt{3}}{1-\sqrt{3} x}

\displaystyle f^2(x) = \left( \frac{x+\sqrt{3}}{1-\sqrt{3} x} + \sqrt{3} \right) / \left( 1 - \sqrt{3} \frac{x+\sqrt{3}}{1-\sqrt{3} x} \right)

\displaystyle = \left( \frac{x+\sqrt{3}+\sqrt{3} -3x}{1-\sqrt{3} x}\right) / \left(\frac{1-\sqrt{3} x - \sqrt{3} x-3}{1-\sqrt{3} x} \right)

\displaystyle = \frac{x+\sqrt{3}+\sqrt{3} -3x}{1-\sqrt{3} x - \sqrt{3} x-3}

\displaystyle = \frac{2\sqrt{3} -2x}{-2-2\sqrt{3} x} = \frac{x - \sqrt{3}}{1 + \sqrt{3} x} .

\displaystyle f^3(x) = \left( \frac{x-\sqrt{3}}{1+\sqrt{3} x} + \sqrt{3} \right) / \left( 1 - \sqrt{3} \frac{x-\sqrt{3}}{1+\sqrt{3} x} \right)

\displaystyle = \left( \frac{x-\sqrt{3}+\sqrt{3} + 3x}{1+\sqrt{3} x} + \sqrt{3} \right) / \left(\frac{1+\sqrt{3} x-\sqrt{3} x+3}{1+\sqrt{3} x} \right)

\displaystyle = \frac{x-\sqrt{3}+\sqrt{3} + 3x}{1+\sqrt{3} x-\sqrt{3} x+3}

\displaystyle = \frac{4x}{4} = x.

\therefore x = f^3(x) = f^3(f^3(x)) = f^3(f^3(f^3(x))) = \dots

\therefore x = f^3(x) = f^6(x) = f^9(x) = \dots = f^{3n} (x) = \dots (or induction if you're bored)

\therefore x = f^{2007}(x) as 3 divides 2007.

(ii)

tan is periodic: since \tan (a+n\pi) = \tan a, it suffices to prove the result for n = 0, 1, 2 (and then, again, be as fussy as you like about the induction, which I won't).

\tan (\theta + \frac{1}{3}\cdot 0\pi ) = \tan \theta = x = f^0(x) as required.

\displaystyle \tan (\theta + \frac{1}{3} \pi ) = \frac{\tan \theta + \tan \tfrac{1}{3} \pi}{1 - \tan \theta \tan \frac{1}{3} \pi} = \frac{x + \sqrt{3}}{1 - \sqrt{3} x} = f(x) as required.

\displaystyle \tan (\theta + \frac{2}{3} \pi ) = \frac{\tan \theta + \tan \tfrac{2}{3} \pi}{1 - \tan \theta \tan \frac{2}{3} \pi} = \frac{x - \sqrt{3}}{1+\sqrt{3} x} = f^2(x) as required.

(Cue tedious inductive step, or rather obvious conclusion as f^n(x) = f^{n+3}(x) and \tan (\theta + \frac{1}{3} n\pi ) = \tan (\theta + \frac{1}{3} (n+3)\pi ).)

(iii)

Bit of guesswork involved, but several hints give it away; |t| [U]<[/U] 1, the use of tan in part (ii), and the form of the expression suggest an expression in sin or cos. I chose sin:

Let t = \sin \phi , so that \sqrt{1 - t^2} = \cos\phi .

Note also that \frac{\sqrt{3}}{2} = \cos \frac{\pi}{6} , \quad \frac{1}{2} = \sin \frac{\pi}{6} .

Then the expression for g can be rewritten as: g(t) = \sin\phi\cos\frac{\pi}{6} + \cos\phi\sin\frac{\pi}{6}

g(t) = \sin (\phi +\frac{\pi}{6} )

And then another small inductive step... let s = \sin (\theta ), \quad \theta = \phi + \frac{\pi}{6}, so that s = g(t) . Then:

g(s) = \sin (\theta + \frac{\pi}{6} )

g(g(t)) = \sin ((\phi + \frac{\pi}{6} ) + \frac{\pi}{6} )

g^2(t) = \sin (\phi + \frac{\pi}{3} )

etc.

Hence g^n(t) = \sin (\phi + \frac{n\pi}{6} ).

Solution by generalebriety.

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