STEP II 2007 question 5 solution

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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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