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2003 Paper 3 Q2
D: 1700.0 B: 1484.0
generalised binomial theorem binomial series combinatorial identity summation central binomial coefficient series convergence proof

Show that $\ds ^{2r} \! {\rm C}_r =\frac{1\times3\times\dots\times (2r-1)}{r!} \, \times 2^r \;, $ for \(r\ge1\,\).

  1. Give the first four terms of the binomial series for \(\l 1 - p \r^{-\frac12}\). By choosing a suitable value for \(p\) in this series, or otherwise, show that $$ \displaystyle \sum_{r=0}^\infty \frac{ {\vphantom {\A}}^{2r} \! {\rm C}_r }{ 8^r} = \sqrt 2 \; .$$
  2. Show that $$ \displaystyle \sum_{r=0}^\infty \frac{\l 2r + 1 \r \; {\vphantom{A}}^{2r} \! {\rm C} _r }{ 5^r} =\big( \sqrt 5\big)^3 \;. $$
[{\bf Note: } $ {\vphantom{A}}^n {\rm C}_r $ is an alternative notation for $\ds \ \binom n r \, \( for \)r\ge1\,\(, and \) {\vphantom{A}}^0 {\rm C}_0 =1 $ .]

Solution: \begin{align*} \binom{2r}{r} &= \frac{(2r)!}{r!r!} \\ &= \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdots (2r-1)(2r)}{r! r!} \\ &= \frac{1 \cdot 3 \cdot 5 \cdots (2r-1) \cdot (2 \cdot 1) \cdot (2 \cdot 2) \cdots (2 \cdot r)}{r!}{r!} \\ &= \frac{1\cdot 3 \cdots (2r-1) \cdot 2^r \cdot 1 \cdot 2 \cdots r}{r!r!} \\ &= \frac{1\cdot 3 \cdots (2r-1) \cdot 2^r \cdot r!}{r!r!} \\ &= \frac{1\cdot 3 \cdots (2r-1)}{r!} \cdot 2^r \end{align*} which is what we wanted to show

  1. \begin{align*} (1 - p)^{-\frac12} &= 1 + \left ( -\frac12 \right )(-p) + \frac{1}{2!} \left (-\frac12 \right )\left (-\frac32 \right )(-p)^2 + \ldots \\ & \quad \quad \quad \cdots +\frac{1}{3!} \left (-\frac12 \right )\left (-\frac32 \right )\left (-\frac52 \right )(-p)^3 + O(p^4) \\ &= \boxed{1 + \frac{1}{2}p + \frac{3}{8}p^2 + \frac{5}{16}p^3} + O(p^4) \end{align*} More generally: \begin{align*} \binom{-\frac{1}{2}}{k} &=\frac{(-\frac{1}{2})\cdot(-\frac{1}{2} -1)\cdots(-\frac12 -k+1)}{k!} \\ &= \frac{(-1)(-3)(-5)\cdots(-(2k-1))}{k!2^k} \\ &= \frac{(-1)^k(1)(3)(5)\cdots((2k-1))}{k!2^k} \\ &= (-1)^k \frac{1}{4^k}\binom{2k}{k} \\ \end{align*} Therefore, \begin{align*} \sqrt{2} = \left (1-\frac12 \right)^{-\frac12} &= \sum_{r=0}^{\infty} \binom{-\frac12}{r} \left (-\frac12 \right )^r \tag{\(\frac12 < 1\) so series is valid} \\ &= \sum_{r=0}^{\infty} (-1)^r \frac{1}{4^r}\binom{2r}{r} \left (-\frac12 \right )^r \\ &= \sum_{r=0}^{\infty} \frac{1}{8^r}\binom{2r}{r} \end{align*}, which is what we wanted to show.
  2. \begin{align*} p(1-p^2)^{-\frac12} &= \sum_{r=0}^{\infty} \binom{-\frac12}{r} \left (-p^2 \right )^rp \\ &= \sum_{r=0}^{\infty} \frac{1}{4^r}\binom{2r}{r} p^{2r+1} \end{align*} Differentiating with respect to \(p\), \begin{align*} (1-p^2)^{-\frac12} +p^2(1-p^2)^{-\frac32} &= \sum_{r=0}^{\infty} \frac{1}{4^r}(2r+1)\binom{2r}{r} p^{2r} \\ (1-p^2)^{-\frac32} &= \sum_{r=0}^{\infty} \frac{1}{4^r}(2r+1)\binom{2r}{r} p^{2r} \end{align*} Letting \(p = \frac{2}{\sqrt{5}}\), and \(|\frac2{\sqrt{5}}| < 1\) we have \begin{align*} \left (1-\frac45 \right )^{-\frac32} &= \sum_{r=0}^{\infty} \frac{1}{5^r}(2r+1)\binom{2r}{r} \\ (\sqrt{5})^3 &= \sum_{r=0}^{\infty} \frac{1}{5^r}(2r+1)\binom{2r}{r} \end{align*} (Alternative) \begin{align*} (\sqrt5)^3 &= \left ( \frac{1}{5} \right )^{-\frac32} \\ &= \left ( 1- \frac{4}{5} \right )^{-\frac32} \\ &= \sum_{r=0}^{\infty} \binom{-\frac32}{r} \left (-\frac45 \right)^r \\ &= \sum_{r=0}^{\infty} \binom{-\frac12}{r} \frac{-\frac32-(r-1)}{-\frac12} \left (-\frac45 \right)^r \\ &= \sum_{r=0}^{\infty} \binom{-\frac12}{r} (2r+1) \left (-\frac45 \right)^r \\ &= \sum_{r=0}^{\infty} (-1)^r \frac{1}{4^r}\binom{2r}{r} (2r+1) \left (-\frac45 \right)^r \\ &= \sum_{r=0}^{\infty}(2r+1)\binom{2r}{r} \left (\frac15 \right)^r \\ (\sqrt{5})^3 &= \sum_{r=0}^{\infty}\frac{1}{5^r}(2r+1)\binom{2r}{r} \\ \end{align*}

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2002 Paper 3 Q14
D: 1700.0 B: 1500.0
probability variance covariance binomial distribution discrete random variables bivariate data proof correlation coefficient chess

Prove that, for any two discrete random variables \(X\) and \(Y\), \[ \mathrm{Var} \left(X + Y \right) = \mathrm{Var}(X) + \mathrm{Var}(Y) + 2 \, \mathrm{Cov}(X,Y), \] where \(\mathrm{Var}(X)\) is the variance of \(X\) and \(\mathrm{Cov}(X,Y)\) is the covariance of \(X\) and \(Y\). When a Grandmaster plays a sequence of \(m\) games of chess, she is, independently, equally likely to win, lose or draw each game. If the values of the random variables \(W\), \(L\) and \(D\) are the numbers of her wins, losses and draws respectively, justify briefly the following claims:

  1. \(W + L + D\) has variance \(0\,\);
  2. \(W + L\) has a binomial distribution.
Find the value of \(\displaystyle {\mathrm{Cov}(W,L) \over \sqrt{\mathrm{Var}(W) \mathrm{Var}(L)}}\;\).

Solution: \begin{align*} && \var[X+Y] &= \E\left [(X+Y-\E[X+Y])^2 \right] \\ &&&= \E \left [ (X - \E[X] + Y - \E[Y])^2 \right] \\ &&&= \E \left [(X - \E[X])^2 + (Y-\E[Y])^2 + 2(X-\E[X])(Y-\E[Y]) \right] \\ &&&= \E \left [(X - \E[X])^2 \right]+\E \left [(Y-\E[Y])^2 \right]+\E \left [2(X-\E[X])(Y-\E[Y]) \right] \\ &&&= \var[X] + \var[Y] + 2 \mathrm{Cov}(X,Y) \end{align*}

  1. \(W+L+D = m\) where \(m\) is the number of games, which has variance \(0\). Therefore \(W+L+D\) has variance \(0\).
  2. The probability of a decisive game is \(\frac23\) and \(W+L\) is the number of decisive games. Each game is independent so this meets the criteria for a binomial distribution.
Notice \(W+L \sim B(m, \tfrac23)\) and \(W, L, D \sim B(m, \tfrac13)\), in particular \(\var[W+L] = m \tfrac23 \tfrac13 = \tfrac29m\) and \(\var[W] = \var[D] = \var[D] = m \tfrac13 \tfrac13 = \tfrac29m\) \begin{align*} && \var[W+L] &= \var[W] + \var[L] + 2\mathrm{Cov}(W,L) \\ \Rightarrow && \mathrm{Cov}(W,L) &= -\tfrac19m \\ \Rightarrow && \frac{\mathrm{Cov}(W,L) }{\sqrt{\var[W]\var[L]}} &= -\frac12 \end{align*}

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2001 Paper 2 Q2
D: 1600.0 B: 1500.0
curve sketching floor function summation geometric series limits integer part proof

Sketch the graph of the function \([x/N]\), for \(0 < x < 2N\), where the notation \([y]\) means the integer part of \(y\). (Thus \([2.9] = 2\), \ \([4]=4\).)

  1. Prove that \[ \sum_{k=1}^{2N} (-1)^{[k/N]} k = 2N-N^2. \]
  2. Let \[ S_N = \sum_{k=1}^{2N} (-1)^{[k/N]} 2^{-k}. \] Find \(S_N\) in terms of \(N\) and determine the limit of \(S_N\) as \(N\to\infty\).

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2001 Paper 2 Q4
D: 1600.0 B: 1484.0
trigonometry sin nx quadratic in cos roots of equations discriminant factorization proof trigonometric identities

Let $$ \f(x) = P \, {\sin x} + Q\, {\sin 2x} + R\, {\sin 3x} \;. $$ Show that if \(Q^2 < 4R(P-R)\), then the only values of \(x\) for which \(\f(x) = 0\) are given by \(x=m\pi\), where \(m\) is an integer. \newline [You may assume that \(\sin 3x = \sin x(4\cos^2 x -1)\).] Now let $$ \g(x) = {\sin 2nx} + {\sin 4nx} - {\sin 6nx}, $$ where \(n\) is a positive integer and \(0 < x < \frac{1}{2}\pi \). Find an expression for the largest root of the equation \(\g(x)=0\), distinguishing between the cases where \(n\) is even and \(n\) is odd.

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2000 Paper 2 Q1
D: 1600.0 B: 1516.0
proof unit fractions number theory prime numbers factorisation uniqueness algebraic manipulation Erdos-Straus conjecture

A number of the form \(1/N\), where \(N\) is an integer greater than 1, is called a unit fraction. Noting that \[ \frac1 2 =\frac13 + \frac16\\\ \mbox{ and } \frac13 = \frac14 + \frac1{12}, \] guess a general result of the form $$ \frac1N =\frac1a +\frac1b \tag{*} $$ and hence prove that any unit fraction can be expressed as the sum of two distinct unit fractions. By writing \((*)\) in the form \[ (a-N)(b-N)=N^2 \] and by considering the factors of \(N^2\), show that if \(N\) is prime, then there is only one way of expressing \(1/N\) as the sum of two distinct unit fractions. Prove similarly that any fraction of the form \(2/N\), where \(N\) is prime number greater than 2, can be expressed uniquely as the sum of two distinct unit fractions.

Solution: Notice that \(\frac{1}{N} = \frac{1}{N+1} + \frac{1}{N(N+1)}\), so any unit fraction can be expressed as the sum of two distinct unit fractions. \begin{align*} && \frac{1}N &= \frac1a + \frac1b \\ \Leftrightarrow && ab&= Nb+Na \\ \Leftrightarrow && 0 &= (a-N)(b-N)-N^2 \\ \Leftrightarrow && N^2 &= (a-N)(b-N) \end{align*} If \(N\) is prime then the only factors of \(N^2\) are \(1,N\) and \(N^2\). if \(a-N = b-N = N\) then \(a=b\) and we don't have distinct fractions. Therefore \(a-N = 1\) and \(b-N = N^2\) and we obtain the decomposition earlier (and it must be the only solution). \begin{align*} && \frac2N &= \frac1a+\frac1b \\ \Leftrightarrow && 2ab &= Nb+Na \\ \Leftrightarrow && 4ab &= 2Na+2Nb \\ \Leftrightarrow && N^2 &= (2a-N)(2b-N) \end{align*} Therefore for \(a,b\) to be distinct we must have \(2a = N+1\) and \(2b = N+N^2\) as the only possible factorisation. Both of the right hand sides are even so we can write \[ \frac{1}{N} = \frac{1}{\frac{N+1}{2}} + \frac{1}{\frac{N(N+1)}{2}} \] and this is unique

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2000 Paper 2 Q2
D: 1600.0 B: 1503.8
polynomials repeated roots differentiation factor theorem proof quartic factor

Prove that if \({(x-a)^{2}}\) is a factor of the polynomial \(\p(x)\), then \(\p'(a)=0\). Prove a corresponding result if \((x-a)^4\) is a factor of \(\p(x).\) Given that the polynomial $$ x^6+4x^5-5x^4-40x^3-40x^2+32x+k $$ has a factor of the form \({(x-a)}^4\), find \(k\).

Solution: First notice that \(p(x) = (x-a)^2q(x)\) so \(p'(x) = 2(x-a)q(x) + (x-a)^2q'(x) = (x-a)(2q(x)+(x-a)q'(x))\), in particular \(p'(a) = 0\) so \(x-a\) is a root of \(p'(x)\). If \((x-a)^4\) is a root of \(p(x)\) then \(p^{(3)}(a)= 0\). The proof is similar. Differentiating \(3\) times we obtain: \(6 \cdot 5 \cdot 4 x^3 + 4 \cdot 5 \cdot 4 \cdot 3 x^2 - 5\cdot4 \cdot 3 \cdot 2 x-40 \cdot 3 \cdot 2 \cdot 1 = 5!(x^3+2x^2-x-2) = 5!(x+2)(x^2-1)\). So our possible (repeated) roots are \(x=-2,-1,1\). We can check \(p'(x) = 6x^5+20x^4-20x^3-120x^2-80x+32\), and see \(p'(1) = 36 - 200 \neq 0\), \(p'(-1) = -6+20+20-120+80+32 \neq 0\), therefore \(a = -2\)

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1999 Paper 1 Q10
D: 1500.0 B: 1500.0
SHM elastic spring Hooke's law vertical oscillation period modulus of elasticity proof

A particle is attached to a point \(P\) of an unstretched light uniform spring \(AB\) of modulus of elasticity \(\lambda\) in such a way that \(AP\) has length \(a\) and \(PB\) has length \(b\). The ends \(A\) and \(B\) of the spring are now fixed to points in a vertical line a distance \(l\) apart, The particle oscillates along this line. Show that the motion is simple harmonic. Show also that the period is the same whatever the value of \(l\) and whichever end of the string is uppermost.

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1999 Paper 2 Q2
D: 1600.0 B: 1516.0
quadratics discriminant inequality no real roots proof parameter analysis

Consider the quadratic equation $$ nx^2+2x \sqrt{pn^2+q} + rn + s = 0, \tag{*} $$ where \(p>0\), \(p\neq r\) and \(n=1\), \(2\), \(3\), \(\ldots\) .

  1. For the case where \(p=3\), \(q=50\), \(r=2\), \(s=15\), find the set of values of \(n\) for which equation \((*)\) has no real roots.
  2. Prove that if \(p < r\) and \(4q(p-r) > s^2\), then \((*)\) has no real roots for any value of \(n\).
  3. If \(n=1\), \(p-r=1\) and \(q={s^2}/8\), show that \((*)\) has real roots if, and only if, \(s \le 4-2\sqrt{2}\ \) or \(\ s \ge 4+2\sqrt{2}\).

Solution:

  1. \(\,\) \begin{align*} && 0 &= nx^2 + 2\sqrt{3n^2+50}x + 2n + 15 \\ && 0 &> \Delta = 4(3n^2+5) - 4\cdot n \cdot (2n + 15) \\ \Leftrightarrow && 0 &> n^2-15n+5\\ \text{cv}: && n &= \frac{15 \pm \sqrt{225 - 20}}{2} \\ &&&\approx \frac{15\pm14.x}{2}\\ \Leftrightarrow &&n &\in \{1, 2, \cdots, 14\} \end{align*}
  2. \(\,\) \begin{align*} && 0 &> \Delta = 4(pn^2+q) - 4\cdot n \cdot (rn+s) \\ \Leftrightarrow && 0&>(p-r)n^2-sn+q \end{align*} Which is always true if \(r > p\) and \(s^2 < 4q(p-r)\)
  3. \(\,\) \begin{align*} && 0 &= x^2 + 2\sqrt{p+q}x+ r+s \\ && 0 &\leq \Delta = 4(p+q) - 4(r+s) \\ && 0 &\leq 1 + s^2/8 - s \\ \text{c.v}: && s &= \frac{1 \pm \sqrt{1-4 \cdot \frac{1}{8}}}{2\cdot\frac18} \\ &&&= 4 \pm 4\sqrt{\frac12} \\ &&&= 4 \pm 2\sqrt{2} \\ \Rightarrow && s \leq 4 - 2\sqrt{2} &\text{ or } s \geq 4 + 2\sqrt{2} \end{align*}

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1999 Paper 2 Q5
D: 1600.0 B: 1500.0
trigonometry R-formula t-substitution Weierstrass substitution quadratic equation inequality cosine proof

Show that if \(\alpha\) is a solution of the equation $$ 5{\cos x} + 12{\sin x} = 7, $$ then either $$ {\cos }{\alpha} = \frac{35 -12\sqrt{120}}{169} $$ or \(\cos \alpha\) has one other value which you should find. Prove carefully that if \(\frac{1}{2}\pi< \alpha < \pi\), then \(\alpha < \frac{3}{4}\pi\).

Solution: \begin{align*} && 5 \cos x + 12\sin x &= 7 \\ \Rightarrow && 5 \cos x - 7 &= -12 \sin x \\ \Rightarrow && 25 \cos^2 x - 70\cos x + 49 &= 144 \sin^2 x \\ \Rightarrow && 25 \cos^2 x - 70\cos x + 49 &= 144 (1-\cos^2 x) \\ \Rightarrow && 169 \cos^2 x - 70 \cos x -95 &= 0 \\ \Rightarrow && \cos \alpha &= \frac{70 \pm \sqrt{70^2 - 4 \cdot 169 \cdot (-95)}}{2 \cdot 169} \\ &&&= \frac{35 \pm \sqrt{35^2 + 169 \cdot 95} }{169} \\ &&&= \frac{35 \pm 12\sqrt{120}}{169} \end{align*} If \(\frac12 \pi < \alpha < \pi\) then \(\cos \alpha\) is negative, in particular \(\cos \alpha = \frac{35 -12\sqrt{120}}{169}\). Since \(\cos\) is decreasing over this range, if \(\cos \alpha > \cos \frac34 \pi = -\frac{\sqrt{2}}2\), then we will have shown \(\alpha < \frac34 \pi\) \begin{align*} && \cos \alpha &= \frac{35 - 12 \sqrt{120}}{169} \\ &&&> \frac{35 - 12 \cdot \sqrt{121}}{169} \\ &&&= \frac{35 - 12 \cdot 11}{169} \\ &&&= \frac{35 - 132}{169} \\ &&&= -\frac{97}{169} \\ &&&> -\frac{8}{13} \end{align*} but \(\left ( \frac{8}{13} \right)^2 = \frac{64}{169} < \frac12\), so we are done.

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1999 Paper 2 Q8
D: 1600.0 B: 1500.0
trigonometric series summation differentiation small angle approximation proof sin sum formula asymptotic approximation product-to-sum

Prove that $$ \sum_{k=0}^n \sin k\theta = \frac { \cos \tfrac12\theta - \cos (n+ \tfrac12) \theta} {2\sin \tfrac12\theta}\;. \tag{*}$$

  1. Deduce that, when \(n\) is large, \[ \sum_{k=0}^n \sin \left(\frac{k\pi}{n}\right) \approx \frac{2n}\pi\;. \]
  2. By differentiating \((*)\) with respect to \(\theta\), or otherwise, show that, when \(n\) is large, \[ \sum_{k=0}^n k \sin^2 \left(\frac{k\pi}{2n}\right) \approx \left(\frac{1}4 +\frac{1}{\pi^2} \right)n^2\;. \]

Solution: \begin{align*} && \sum_{k=0}^n \sin k\theta &= \textrm{Im} \left ( \sum_{k=0}^n e^{i k \theta}\right)\\ &&&= \textrm{Im} \left ( \frac{e^{i(n+1)\theta}-1}{e^{i \theta}-1} \right)\\ &&&= \textrm{Im} \left ( \frac{e^{i(n+\tfrac12)\theta}-e^{-i\theta/2}}{e^{i \theta/2}-e^{-i \theta/2}} \right)\\ &&&= \textrm{Im} \left ( \frac{(e^{i(n+\tfrac12)\theta}-e^{-i\theta/2})/2i}{(e^{i \theta/2}-e^{-i \theta/2})/2i} \right)\\ &&&= \textrm{Im} \left ( \frac{(e^{i(n+\tfrac12)\theta}-e^{-i\theta/2})/i}{2\sin \tfrac12 \theta} \right)\\ &&&= \frac{\cos \tfrac12 \theta - \cos(n+ \tfrac12)\theta}{2\sin \tfrac12 \theta} \end{align*}

  1. When \(n\) is large we have \begin{align*} &&\sum_{k=0}^n \sin \left(\frac{k\pi}{n}\right) &= \frac{\cos \frac{\pi}{2n} - \cos \frac{(2n + 1)\pi}{2n}}{2 \sin \frac{\pi}{2n}} \\ &&&= \frac{\cos \frac{\pi}{2n} +\cos \frac{\pi}{2n}}{2 \sin \frac{\pi}{2n}} \\ &&&\approx \frac{1- \frac{\pi^2}{4n^2}}{\frac{\pi}{2n}} \\ &&&= \frac{2n}{\pi} - \frac{\pi}{2n} \\ &&&\approx \frac{2n}{\pi} \end{align*}
  2. \(\,\) \begin{align*} && \sum_{k=0}^n \sin k\theta &= \frac { \cos \tfrac12\theta - \cos (n+ \tfrac12) \theta} {2\sin \tfrac12\theta} \\ \frac{\d }{\d \theta}: && \sum_{k=0}^n k\cos k\theta &= \frac { (-\tfrac12\sin\tfrac12\theta +(n+\tfrac12) \sin (n+ \tfrac12) \theta)2 \sin \tfrac12 \theta - (\cos \tfrac12\theta - \cos (n+ \tfrac12) \theta) \cos \tfrac12 \theta} {4\sin^2 \tfrac12\theta} \\ &&&= \frac{-\sin^2 \tfrac12 \theta+(2n+1)\sin(n+\tfrac12)\theta \sin \tfrac12 \theta-\cos^2\tfrac12 \theta+\cos(n+\frac12)\theta \cos\tfrac12 \theta}{4 \sin^2 \tfrac12 \theta} \\ &&&= \frac{(2n+1)\sin(n+\tfrac12)\theta \sin \tfrac12 \theta+\cos(n+\frac12)\theta \cos\tfrac12 \theta-1}{4 \sin^2 \tfrac12 \theta} \\ &&&= \frac{2n\sin(n+\tfrac12)\theta \sin \tfrac12 \theta+\cos n \theta-1}{4 \sin^2 \tfrac12 \theta} \\ \\ \Rightarrow && \sum_{k=0}^n k\left ( 1-2\sin^2 \left ( \frac{k\theta}{2} \right) \right) &= \frac{2n\sin(n+\tfrac12)\theta \sin \tfrac12 \theta+\cos n \theta-1}{4 \sin^2 \tfrac12 \theta} \\ \Rightarrow && \sum_{k=0}^n k\sin^2 \left ( \frac{k\theta}{2} \right) &= \frac{n(n+1)}{4} - \frac{2n\sin(n+\tfrac12)\theta \sin \tfrac12 \theta+\cos n \theta-1}{8 \sin^2 \tfrac12 \theta} \\ \theta = \frac{\pi}{n}: && \sum_{k=0}^n k\sin^2 \left ( \frac{k\pi}{2n} \right) &= \frac{n(n+1)}{4} - \frac{2n\sin(n+\tfrac12)\frac{\pi}{n}\sin \tfrac12 \frac{\pi}{n}+\cos n \frac{\pi}{n}-1}{8 \sin^2 \tfrac12 \frac{\pi}{n}} \\ &&&= \frac{n(n+1)}{4} - \frac{-2n \sin^2 \frac{\pi}{2n}-2}{8 \sin^2 \tfrac12 \frac{\pi}{n}} \\ &&&= \frac{n^2+n}{4} + \frac{n}{4} + \frac{1}{4\sin^2\frac{\pi}{2n}} \\ &&&\approx \frac{n^2}4 + \frac{n}{2}+ \frac{n^2}{\pi^2} \\ &&&= \left ( \frac14 + \frac{1}{\pi^2} \right)n^2 + \frac{n}{4} \\ &&&\approx \left ( \frac14 + \frac{1}{\pi^2} \right)n^2 \end{align*}

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1999 Paper 3 Q4
D: 1700.0 B: 1516.0
proof Euler's formula polyhedron combinatorial argument enumeration regular polyhedra Platonic solids inequality

A polyhedron is a solid bounded by \(F\) plane faces, which meet in \(E\) edges and \(V\) vertices. You may assume \textit{Euler's formula}, that \(V-E+F=2\). In a regular polyhedron the faces are equal regular \(m\)-sided polygons, \(n\) of which meet at each vertex. Show that $$ F={4n\over h} \,, $$ where \(h=4-(n-2)(m-2)\). By considering the possible values of \(h\), or otherwise, prove that there are only five regular polyhedra, and find \(V\), \(E\) and \(F\) for each.

Solution: Note that each of the \(F\) faces have \(m\) edges. If we count edges from each face we will get \(mF\) edges, but this counts each edge twice, so \(E = \frac{m}{2}F\). Similarly each edge goes into \(2\) vertices, but this counts each vertex \(n\) times, therefore \(V = \frac{2E}{n} = \frac{m}{n}F\) Therefore \begin{align*} && 2 &= V - E + F \\ &&&= \frac{m}{n}F - \frac{m}{2}F + F \\ &&&= F \left ( \frac{2m-nm+2n}{2n} \right) \\ &&&= F \left ( \frac{4-(n-2)(m-2)}{2n} \right) \\ \Rightarrow && F &= \frac{4n}{4-(n-2)(m-2)} = \frac{4n}{h} \end{align*} Notice that \(1 \leq h \leq 4\), so If \(h = 4\), then \(n = 2\) or \(m = 2\) but we can't have two-sided polygons or polyhedra with two faces making a vertex so this is not possible. If \(h = 3\) then \((n-2)(m-2) = 1\) and \(3 \mid n\), so we have \(n = 3, m = 3\). ie we have triangular faces meeting at \(3\) per point. We also have \(F = \frac{4 \cdot 3 }3\) which is \(4\) faces so this is a tetrahedron and \((V,E, F) = (4, 6, 4)\) If \(h = 2\) then \((n-2)(m-2) = 2\): Case 1: \(n = 4, m = 3\) which is four triangles meeting at each vertex with \((V,E,F) = (6, 12, 8)\), ie an octohedron. Case 2: \(n = 3, m = 4\) so we have three squares meeting at each vertex, with \((V,E,F) = (8,12,6)\) which is a cube. If \(h = 1\) then \((n-2)(m-2) = 3\) Case 1: \(n = 5, m = 3\) which is five triangles meeting at each vertex, with \((V,E,F) = ( 12,30, 20)\) ie an icosahedron. Case 2: \(n = 3, m = 5\) which is three pentagons meeting at each vertex, with \((V,E,F) = (20, 30,12)\) which is a dodecahedron. These are all the possible cases and hence we have found the five platonic solids.

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1998 Paper 1 Q3
D: 1500.0 B: 1500.0
true-false logarithms exponentials trigonometric identities polynomial approximation inequality AM-GM inequality proof

Which of the following statements are true and which are false? Justify your answers.

  1. \(a^{\ln b}=b^{\ln a}\) for all \(a,b>0\).
  2. \(\cos(\sin\theta)=\sin(\cos\theta)\) for all real \(\theta\).
  3. There exists a polynomial \(\mathrm{P}\) such that \(|\mathrm{P}(\theta)-\cos\theta|\leqslant 10^{-6}\) for all real \(\theta\).
  4. \(x^{4}+3+x^{-4}\geqslant 5\) for all \(x>0\).

Solution:

  1. True. \begin{align*} && \ln a \cdot \ln b &= \ln b \cdot \ln a \\ \Leftrightarrow && \exp ( \ln a \cdot \ln b) &= \exp ( \ln b \cdot \ln a) \\ \Leftrightarrow && \exp ( \ln a )^{\ln b} &= \exp ( \ln b )^{\ln a} \\ \Leftrightarrow && a^{\ln b} &= b^{\ln a} \\ \end{align*}
  2. False. Consider \(\theta = 0\). We'd need \(\cos 0 = 1 = \sin 1\), but \(0 < 1 < \frac{\pi}{2}\) so \(\sin 1 \neq 1\)
  3. False. If the polynomial has positive degree, then as \(n \to \infty\), \(\P(x) \to \pm \infty\), in particular it must be well outside the interval \([-1,1]\). Therefore it can't be within \(10^{-6}\) of \(\cos \theta\) which is confined to that interval. The only polynomial which is restricted to that range are constants, but then \(|\cos 0 - c| \leq 10^{-6}\) and \(|\cos \pi - c| \leq 10^{-6}\) \(2 = |1-(-1)| \leq |1-c| + |-1-c| \leq 2\cdot 10^{-6}\) contradiction.
  4. True. \begin{align*} && (x^2-x^{-2})^2 &\geq 0 \\ \Leftrightarrow && x^4-2+x^{-4} &\geq0 \\ \Leftrightarrow && x^4+3+x^{-4} &\geq 5 \\ \end{align*}

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1998 Paper 2 Q1
D: 1600.0 B: 1500.0
proof proof by contradiction number theory divisibility parity cubic equations integer solutions Diophantine equation

Show that, if \(n\) is an integer such that $$(n-3)^3+n^3=(n+3)^3,\quad \quad {(*)}$$ then \(n\) is even and \(n^2\) is a factor of \(54\). Deduce that there is no integer \(n\) which satisfies the equation \((*)\). Show that, if \(n\) is an integer such that $$(n-6)^3+n^3=(n+6)^3, \quad \quad{(**)}$$ then \(n\) is even. Deduce that there is no integer \(n\) which satisfies the equation \((**)\).

Solution: \begin{align*} && n^3 &= (n+3)^3 - (n-3)^3 \\ &&&= n^3 + 9n^2+27n + 27 - (n^3 - 9n^2+27n-27) \\ &&&= 18n^2+54 \end{align*} Therefore since \(2 \mid 2(9n^2 + 27)\), \(2 \mid n^3 \Rightarrow 2 \mid n\), so \(n\) is even. Since \(n^2 \mid n^3\), \(n^2 \mid 54 = 2 \cdot 3^3\), therefore \(n = 1\) or \(n = 3\). \((1-3)^3 + 1^3 < 0 < (1+3)^3\). So \(n = 1\) doesn't work. \((3 - 3)^3 + 3^3 < (3+3)^3\) so \(n = 3\) doesn't work. Therefore there are no solutions. \begin{align*} && n^3 &= (n+6)^3 - (n-6)^3 \\ &&&= n^3 + 18n^2 + 180n + 6^3 - (n^3 - 18n^2 + 180n - 6^3 ) \\ &&&= 36n^2+2 \cdot 6^3 \end{align*} Therefore \(n^2 \mid 2 \cdot 6^3 = 2^4 \cdot 3^3\), therefore \(n = 1, 2, 3, 4, 6, 12\). \(n = 1\), \(1^3 <36+2\cdot 6^3\) \(n = 2\), \(2^3 <36 \cdot 4 + 2 \cdot 6^3\) \(n = 3\), \(3^3 <36 \cdot 9 + 2 \cdot 6^3\) \(n = 4\), \(4^3 < 36 \cdot 16 + 2 \cdot 6^3\) \(n = 6\), \(6^3 < 36\cdot 6^2+ 2 \cdot 6^3\) \(n = 12\), \(12^3 < 36 \cdot 12^2 + 2 \cdot 6^3\) Therefore there are no solutions \(n\) to the equation. These are both special cases of Fermat's Last Theorem, when \(n = 3\)

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1998 Paper 2 Q7
D: 1600.0 B: 1458.4
differentiation trigonometric inequality proof monotonic functions logarithmic differentiation stationary points

\begin{eqnarray*} {\rm f}(x)&=& \tan x-x,\\ {\rm g}(x)&=& 2-2\cos x-x\sin x,\\ {\rm h}(x)&=& 2x+x\cos 2x-\tfrac{3}{2}\sin 2x,\\ {\rm F}(x)&=& {x(\cos x)^{1/3}\over\sin x}. \end{eqnarray*} \vspace{1mm}

  1. By considering \(\f(0)\) and \(\f'(x)\), show that \(\f(x)>0\) for \(0
  2. Show similarly that \(\g(x)>0\) for \(0
  3. Show that \(\h(x)>0\) for \(00\] for \(0
  4. By considering \(\displaystyle {{\rm F}'(x)\over {\rm F}(x)}\), show that \({\rm F}'(x)<0\) for \(0

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1997 Paper 2 Q8
D: 1600.0 B: 1500.0
integration inequality proof integration as area power functions comparison of integrals monotonicity

If \({\rm f}(t)\ge {\rm g}(t)\) for \(a\le t\le b\), explain very briefly why \(\displaystyle \int_a^b {\rm f}(t) \d t \ge \int_a^b {\rm g}(t) \d t\). Prove that if \(p>q>0\) and \(x\ge1\) then $$\frac{x^p-1}{ p}\ge\frac{x^q-1}{ q}.$$ Show that this inequality also holds when \(p>q>0\) and \(0\le x\le1\). Prove that, if \(p>q>0\) and \(x\ge0\), then $$\frac{1}{ p}\left(\frac{x^p}{ p+1}-1\right)\ge \frac{1}{q}\left(\frac{x^q}{ q+1}-1\right).$$

Solution: This is just the result that all of the area beneath \(g(t)\) is also below \(f(t)\) If \(p > q > 0, x \geq 1 \Rightarrow x^p \geq x^q\), therefore applying the result we have \begin{align*} && \int_1^x x^p\, \d t & \geq \int_1^x x^q\, \d t \\ \Rightarrow && \frac{x^p-1}{p} & \geq \frac{x^q-1}{q} \end{align*} When \(p > q > 0, 0 \leq x \leq 1\) we have \(x^p \leq x^q\), ie \begin{align*} && \int_x^1 x^q\, \d t & \geq \int_{x}^1 x^p\, \d t \\ \Rightarrow && \frac{1-x^q}{q} & \geq \frac{1-x^p}{p} \\ \Rightarrow && \frac{x^p-1}{p} &\geq \frac{x^q-1}{q} \end{align*} Now looking at the functions \(f(x) = \frac{x^p-1}{p}, g(x) = \frac{x^q-1}{q}\) and \(x \geq 1\) we have \begin{align*} && \int_0^x \frac{t^p-1}{p} \d t & \geq \int_0^x \frac{t^q-1}{q} \d t\\ \Rightarrow &&\frac1p \left[\frac{t^{p+1}}{p+1} - t \right]_0^x &\geq\frac1q \left[\frac{t^{q+1}}{q+1} - t \right]_0^x \\ \Rightarrow &&\frac1p \left(\frac{x^{p+1}}{p+1} -x\right) &\geq\frac1q \left(\frac{x^{q+1}}{q+1} - x \right)\\ \Rightarrow &&\frac1p \left(\frac{x^{p}}{p+1} -1\right) &\geq\frac1q \left(\frac{x^{q}}{q+1} - 1 \right)\\ \end{align*}

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