Problems

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Solution: \begin{align*} && \frac{1+x}{(1-x)^2(1+x^2)} &= \frac{A}{1-x} + \frac{B}{(1-x)^2} + \frac{Cx+D}{1+x^2} \\ \Rightarrow && 1 + x &= A(1-x)(1+x^2) + B(1+x^2) + Cx(1-x)^2 + D(1-x)^2 \\ \Rightarrow && 2 &= 2B \tag{\(x = 1\)} \\ \Rightarrow && 1 &= B \\ \Rightarrow && 1 &= A+B+D \tag{\(x = 0\)}\\ \Rightarrow && A &= -D \\ \Rightarrow && 0 &= 4A+2B-4C+4D \tag{\(x = -1\)}\\ \Rightarrow && C &= \frac12\\ \Rightarrow && 3 &= -5A+5B+2C+D \tag{\(x=2\)} \\ \Rightarrow && 3 &= -6A+6 \\ \Rightarrow && A,D &=-\frac12,\frac12 \\ \Rightarrow && \frac{1+x}{(1-x)^2(1+x^2)} &= \frac{1}{(1-x)^2} +\frac{1}{2(1-x)}+ \frac{x-1}{2(1+x^2)} \\ &&&=\sum_{k=0}^{\infty}(k+1)x^k + \sum_{k=0}^{\infty}\frac12 x^k + \sum_{k=0}^{\infty}\frac12 (-1)^kx^{2k+1} - \sum_{k=0}^{\infty}\frac12 (-1)^kx^{2k} \end{align*} Therefore the coefficient of \(x^n\) is \(n+1\) or \(n+2\) depending on whether the coefficients from the final series add constructively \(n \equiv 1, 2 \pmod{4}\) or destructively. \begin{align*} \frac{11\, 000}{8181} &= \frac{(10+1) \cdot 1000}{(10-1)^2 \cdot (100+1)} \\ &= \frac{(1+\frac{1}{10})}{(1-\frac{1}{10})^2 \cdot (1+\frac1{10})} \\ &= 1 + \frac3{10} + \frac4{10^2} + \frac{4}{10^3}+\frac{5}{10^4} + \frac{7}{10^5} + \frac{8}{10^6} + \cdots \\ & \quad \quad \cdots + \frac{8}{10^7} + \frac{9}{10^8} + \frac{11}{10^9} + \frac{12}{10^{10}} + \cdots \\ &= 1.34457890 + \frac{12}{10^{10}} + \cdots \end{align*} \begin{align*} && \sum_{k=m}^{\infty} (k+2)x^k &= x^m \sum_{k=0}^{\infty} (k+m+2)x^{k} \\ && &= \frac{x^k}{(1-x)^2} + \frac{(m+2)x^k}{1-x} \\ \Rightarrow && |\sum_{k=m}^{\infty} a_k \left ( \frac1{10} \right )^k | &\leq \frac{1}{10^m}\left ( \frac{1}{(1-\frac1{10})^2} + \frac{m+2}{1-\frac1{10}} \right) \\ &&&= \frac{1}{10^{m-1}} \left ( \frac{9m+28}{81}\right ) \end{align*} Therefore for this will be less than \(10^{-9}\), when \(m = 11\), so our approximation is valid to 9sf

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Solution:

1. \(\,\) \begin{align*} && y & = 27x^3 - 27x^2 + 4 \\ \Rightarrow && \frac{\d y}{\d x} &= 81x^2 - 54x \\ \Rightarrow && x &= 0, \frac23 \\ \Rightarrow && (x,y) &= (0, 4), \left (\frac23, 0 \right) \end{align*}
   

   + - ↺
   Since \(f(x) \geq 0\) for \(x \geq 0\) we must have \(27x^2(1-x) \leq 4 \Rightarrow x^2(1-x) \leq \frac{4}{27}\) Suppose for contradiction that \(bc(1-a) > \frac{4}{27}, ca(1-b) > \frac{4}{27}, ab(1-c) > \frac{4}{27}\) then taking the product we see \begin{align*} && \left ( \frac{4}{27} \right)^3 &< bc(1-a) \cdot ca(1-b) \cdot ab(1-c) \\ &&&= a^2(1-c) \cdot b^2(1-b) \cdot c^2(1-c) \leq \left ( \frac{4}{27}\right)^3 \end{align*} which is a contradiction.
2. Notice that \(f(x) = x(1-x)\) has a turning point at \((\frac12, \frac14)\), and so \(f(x) \leq \frac14\). Suppose for contradiction that both \(p(1-q)\) and \(q(1-p)\) are larger than \(1/4\) \begin{align*} && \left ( \frac14 \right)^2 &< p(1-q) \cdot q(1-p) \\ &&&= p(1-p) \cdot q(1-q) \\ &&&\leq \left ( \frac14 \right)^2 \end{align*} which is a contradiction.

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Solution: \begin{align*} && 1 &= x^2 + y^2 + 2axy \\ \frac{\d}{\d x}: && 0 &= 2x + 2y \frac{\d y}{\d x} + 2ay + 2ax \frac{\d y}{\d x} \\ &&&= (2x+2ay) + \frac{\d y}{\d x} \left (2ax + 2y \right) \\ \Rightarrow && \frac{\d y}{\d x} &= -\frac{x+ay}{ax+y} \end{align*}

The gradient of \(OP\) is \(\frac{y}{x}\). The gradient of the normal is \(\frac{ax+y}{x+ay}\) Therefore (noting the absolute values in case they are on opposite sides to this diagram: \begin{align*} && \tan \theta &= \Big |\tan \left ( \tan^{-1} \frac{ax+y}{x+ay} - \tan^{-1} \frac{y}{x} \right) \Big | \\ &&&= \Big | \frac{\frac{ax+y}{x+ay} - \frac{y}{x}}{1+\frac{ax+y}{x+ay}\frac{y}{x} } \Big | \\ &&&= \Big | \frac{(ax+y)x - y(x+ay)}{x(x+ay)+y(ax+y)} \Big | \\ &&&= \Big | \frac{ax^2 - ay^2}{x^2+y^2+2ayx} \Big | \\ &&&= a \frac{|y^2-x^2|}{1} \\ &&&= a|y^2-x^2| \end{align*}

1. \(\,\) \begin{align*} && \sec^2 \theta \frac{\d \theta}{\d x} &= \pm a \left (2y \frac{\d y}{\d x} - 2 x\right) \\ \Rightarrow && 0 &= a \left (y \cdot \frac{x+ay}{ax+y} + x \right) \\ &&&=a \left ( \frac{xy+ay^2+ax^2+xy}{ax+y} \right) \\ \Rightarrow && 0 &= a(x^2+y^2)+2xy \end{align*}
2. \(\,\) \begin{align*} && 0 &=a(x^2+y^2)+2xy \\ && 1 &= x^2+y^2 + 2axy \\ \Rightarrow && 1 &= (a+1)(x^2+y^2) + (a+1)(2xy) \\ &&&= (a+1)(x^2+y^2+2xy) \end{align*}
3. \(\,\) \begin{align*} && 1 &= (a+1)(x+y)^2 \\ \Rightarrow && x +y &= \pm \frac{1}{\sqrt{1+a}} \\ && 0 &=a(x^2+y^2)+2xy \\ && 1 &= x^2+y^2 + 2axy \\ \Rightarrow && 1 &= (1-a)(x^2+y^2) + (a-1)(2xy) \\ &&&= (1-a)(x^2+y^2-2xy)\\ \Rightarrow && x-y &= \pm \frac{1}{\sqrt{1-a}} \\ \Rightarrow && \frac{\d \theta}{\d x} &= a|y^2-x^2| \\ &&&= a|(y-x)(x+y)| \\ &&&= \frac{a}{\sqrt{1-a^2}} \end{align*}

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Solution: \begin{align*} && I &= \int_0^{\frac{1}{2}\pi} \frac{\sin 2x}{1+\sin^2x} \d x \\ &&&= \int_0^{\frac{1}{2}\pi} \frac{2 \sin x \cos x}{1+\sin^2x} \d x \\ &&&= \left [\ln (1 + \sin^2 x) \right]_0^{\pi/2} \\ &&&= \ln 2 \\ \\ && J &= \int_0^{\frac{1}{2}\pi} \frac{\sin x}{1+\sin^2x} \d x \\ &&&= \int_0^{\frac{1}{2}\pi} \frac{\sin x}{2-\cos^2x} \d x \\ &&&= \frac{1}{2\sqrt{2}}\int_0^{\frac{1}{2}\pi} \left ( \frac{\sin x}{\sqrt{2}-\cos x}+ \frac{\sin x}{\sqrt{2}+\cos x} \right) \d x \\ &&&= \frac{1}{2\sqrt{2}} \left [\ln (\sqrt{2}-\cos x) - \ln (\sqrt{2}+\cos x) \right]_0^{\pi/2} \\ &&&= \frac{1}{2\sqrt{2}} \left (-\ln(\sqrt{2}-1)+\ln(\sqrt{2}+1) \right) \\ &&&= \frac1{2\sqrt{2}} \ln \left (\frac{\sqrt{2}+1}{\sqrt{2}-1} \right)\\ &&&= \frac1{\sqrt{2}} \ln (\sqrt{2}+1) \end{align*} \begin{align*} && (1+\sqrt{2})^5 + (1-\sqrt{2})^5 &= 2(1+10\cdot2+5\cdot2^2) \\ &&&= 82 \\ && |(1-\sqrt{2})^5| & < 1 \\ && (1+\sqrt{2})^5 &< 83 < 99 \\ \\ && 1.4^2 &= 1.96 \\ &&&< 2 \\ \Rightarrow && 1.4 &<\sqrt{2} \\ \\ \Rightarrow && 2^{\sqrt{2}} &> 2^{1.4} \\ &&&=2^{7/5} \\ &&&= {128}^{1/5} \\ &&&>99^{1/5} \\ &&&>1+\sqrt{2} \end{align*} \begin{align*} && \ln 2 & > \frac{1}{\sqrt{2}} \ln(\sqrt{2}+1) \\ \Leftrightarrow && \sqrt{2} \ln 2 &> \ln(\sqrt{2}+1) \\ \Leftrightarrow && 2^{\sqrt{2}} &> 1+\sqrt{2} \end{align*} which we have already shown, so the first integral is larger.

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Solution: \begin{align*} && f(x) &= \cos \left( 2x+ \frac \pi 3\right) + \sin \left ( \frac{3x}2 - \frac\pi 4\right) \\ &&&= \cos \left( 2x+ \frac \pi 3\right) + \cos\left (\frac{\pi}{2} - \left ( \frac{3x}2 - \frac\pi 4\right) \right)\\ &&&= \cos \left( 2x+ \frac \pi 3\right) + \cos\left (\frac{3\pi}{4} - \frac{3x}2 \right)\\ &&&= 2 \cos \left (\frac{2x+ \frac \pi 3+\frac{3\pi}{4} - \frac{3x}2}{2} \right) \cos \left ( \frac{\left (2x+ \frac \pi 3 \right) - \left (\frac{3\pi}{4} - \frac{3x}2 \right)}{2} \right)\\ &&&= 2 \cos \left (\frac{\frac{x}{2}+ \frac {13\pi}{12}}{2} \right) \cos \left ( \frac{\frac{7x}{2}- \frac {5\pi}{12}}{2} \right)\\ &&&= 2 \cos \left (\frac{x}{4}+ \frac {13\pi}{24} \right) \cos \left ( \frac{7x}{4}- \frac {5\pi}{24} \right)\\ \end{align*}

1. The period of \(f\) will be the LCM of \(\frac{2\pi}{\pi}\) and \(\frac{2\pi}{\frac32} = \frac{4\pi}{3}\) which is \(4\pi\). (This is also clear from the factorised form).
2. \(f(x) = 0\) means one of those two factors is zero, ie \begin{align*} \text{first factor}: && 0 &= \cos \left (\frac{x}{4}+ \frac {13\pi}{24} \right) \\ &&n\pi + \frac{\pi}{2}&= \frac{x}{4}+ \frac {13\pi}{24} \\ \Rightarrow && x &= 4n\pi - \frac{\pi}{6} \\ \Rightarrow && x &= -\frac{\pi}{6} \\ \\ \text{second factor}: && 0 &= \cos \left ( \frac{7x}{4}- \frac {5\pi}{24} \right) \\ && n\pi + \frac{\pi}{2} &= \frac{7x}{4}- \frac {5\pi}{24} \\ \Rightarrow && 7x &= 4n\pi + \frac{17}{6}\pi \\ \Rightarrow && x &= \frac{4n}7\pi + \frac{17}{42}\pi \\ \Rightarrow && x &= -\frac{31}{42} \pi, -\frac16\pi, \frac{17}{42}\pi, \frac{41}{42}\pi \end{align*} Therefore all solutions are \(-\frac{31}{42} \pi, -\frac16\pi, \frac{17}{42}\pi, \frac{41}{42}\pi\) We can see that \(-\frac{\pi}{6}\) is a repeated root, therefore it touches the axis and does not cross.
3. \(f(x) = 2\) requires both factors to be \(1\) or \(-1\). \begin{align*} \text{first factor}: && \pm1 &= \cos \left (\frac{x}{4}+ \frac {13\pi}{24} \right) \\ &&n\pi &= \frac{x}{4}+ \frac {13\pi}{24} \\ \Rightarrow && x &= 4n\pi - \frac{13\pi}{6} \\ \Rightarrow && x &= \frac{11}{6}\pi \\ \end{align*} We only need to test this value, where it's \(-1\), so we look at \( \cos \left ( \frac{77\pi}{24}- \frac {5\pi}{24} \right) = \cos (3\pi) = -1\), so the only value is \(\frac{11}{6}\pi\)

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Solution:

\(\mathbf{p} = \lambda \mathbf{a} +(1- \lambda) \mathbf{b}\) Applying the Sine rule, we can see that: \begin{align*} && \frac{OA}{\sin \angle APO} = \frac{AP}{\sin \angle AOP} \\ && \frac{OB}{\sin \angle BPO} = \frac{BP}{\sin \angle BOP} \\ \end{align*} But \(\angle AOP = \angle BOP = \frac12 \angle AOP\) (since \(OP\) bisects the angle) and \(\angle APO = 180^{\circ} -\angle BPO\), so their sines are also equal. Therefore \begin{align*} && \frac{a}{AP} &= \frac{b}{BP} \\ \Rightarrow && \frac{a}{b} &= \frac{1-\lambda}{\lambda} \\ \Rightarrow && \lambda &= \frac{b}{a+b} \end{align*} Note that \(\mathbf{p} = \frac{b\mathbf{a} + a \mathbf{b}}{a+b} = \mathbf{a} + \frac{a}{a+b}(\mathbf{b}-\mathbf{a})\) and \(\mathbf{q} = \mathbf{b} +\frac{a}{a+b}(\mathbf{a}-\mathbf{b}) = \frac{a\mathbf{a} +b \mathbf{b}}{a+b}\) Therefore \begin{align*} && OQ^2 &= \frac{1}{(a+b)^2} \left (a\mathbf{a} + b \mathbf{b} \right)\cdot \left (a\mathbf{a} + b \mathbf{b} \right) \\ &&&= \frac{a^4+b^4+2ab\mathbf{a}\cdot\mathbf{b}}{(a+b)^2} \\ &&OP^2 &= \frac{1}{(a+b)^2} \left (b\mathbf{a} + a \mathbf{b} \right)\cdot \left (b\mathbf{a} + a \mathbf{b} \right) \\ &&&= \frac{2a^2b^2+2ab\mathbf{a}\cdot\mathbf{b}}{(a+b)^2} \\ \\ && OQ^2 - OP^2 &= \frac{a^4+b^4-2a^2b^2}{(a+b)^2} \\ &&&= \frac{(a^2-b^2)^2}{(a+b)^2} \\ &&&= (a-b)^2 \end{align*}

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Solution:

1. The initial velocity is \(u = \binom{u \cos\alpha}{u \sin \alpha}\), therefore \(v = \binom{v_x}{u \sin \alpha}\). Newton's experimental law tells us \(v_x = -e u_x = -eu \cos\alpha\), therefore \(v = \binom{-eu \cos \alpha}{u \sin \alpha} = \binom{-v \sin \beta}{v\cos \beta} \Rightarrow -\tan \beta = -e \cot \alpha \Rightarrow \tan \alpha \tan \beta = e\). There is nothing special about the result here, and so it must also be the case that \(\tan \beta \tan \gamma = e \Rightarrow \tan \gamma = \tan \alpha\)
2. \(\tan \alpha = \frac{XB}{BA}\) so the point \(X\) is at \((a, \tan \alpha a)\). The point \(Y\) satisfies \(\tan \beta = \frac{CY}{CX} = \frac{CY}{2b - \tan \alpha a}\) so the point \(Y\) is \((a-(2b - a \tan \alpha)\tan \beta,2b) = (a - 2b\tan \beta + ea, 2b) = ((1+e)a-2b\tan \beta, 2b)\). Finally, the point \(M\) is the midpoint, so \(\tan \gamma = \frac{DM}{DY}\) so \(M\) is the point \((0, 2b - ((1+e)a-2b\tan \beta)\tan \gamma) = (0, 2b - (1+e)a \tan \gamma - 2b e) = (0, (2b(1-e) - (1+e)a \tan \gamma)\), but \(M\) is the point \((0, b)\), ie \begin{align*} && b &= 2b(1-e) - (1+e)a \tan \gamma \\ \Rightarrow && b+2eb &= (1+e)a \tan \gamma \\ \Rightarrow && \tan \gamma &= \frac{(1+2e)b}{(1+e)a} \\ \Rightarrow && \tan \alpha &= \frac{(1+2e)b}{(1+e)a} \end{align*} Since \( \frac{(1+2e)b}{(1+e)a} = \frac{b}{a} + \frac{e}{1+e}b\) which is clearly an increasing function of \(e\) on \([0,1]\), so \(\tan \alpha \in \left [\frac{b}{a}, \frac{3b}{2a} \right]\) which are all all angles which place \(X\) in sensible places, therefore we can always hit the middle pocket. (Except \(e = 0\), where we would put the ball in \(N\), but we are given \(e > 0\)).
3. After the first collision the velocity is \(\binom{-eu \cos \alpha}{u \sin \alpha}\) after the second collision the velocity is \(\binom{-eu \cos \alpha}{-eu \sin \alpha}\). Initial kinetic energy is therefore \(\frac12 m u^2\) and final kinetic energy is \(\frac12 m e^2u^2\) therefore the fraction lost is \(\frac{\frac12 m u^2 - \frac12 m e^2u^2}{\frac12 m u^2} = 1-e^2\)

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Solution: \begin{align*} && \mathbb{E}(X) &= p + \frac{p}{3} + p^2 = \frac43p+p^2 \\ \Rightarrow && \frac43 &= \frac43p+p^2 \\ \Rightarrow && 0 &= 3p^2+4p-4 \\ &&&= (3p-2)(p+2) \\ \Rightarrow && p &= \frac23, -2 \end{align*} Since \(p \in [0,1]\) we must have \(p = \frac23\). \begin{align*} && \mathbb{P}(\text{correct verdict}) &= t p+ (1-t) (1-p) \\ &&&= t(2p-1)+(1-p)\\ \Rightarrow && \frac12 &= t(2p-1)+(1-p) \\ \Rightarrow && t &= \frac{\frac12-(1-p)}{2p-1} \\ &&&= \frac{2p-1}{2(2p-1)} = \frac12 \end{align*} (so it doesn't depend at all on what the judges are doing, the only way to be fair is if the trials happen at random!)

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Solution:

1. \(\,\) \begin{align*} \mathbb{P}(\text{black counter in }P) &= \mathbb{P}(\text{black counter moves twice})+\mathbb{P}(\text{black counter doesn't move}) \\ &= \mathbb{P}(\text{black counter moves out})\mathbb{P}(\text{black counter moves back}) + (1-\mathbb{P}(\text{black counter moves out})) \\ &= \frac{k}n\cdot \frac{k}{n+k}+\frac{n-k}{n} \\ &= \frac{k^2+n^2-k^2}{n(n+k)} \\ &= \frac{n^2}{n(n+k)} = \frac{n}{n+k} \end{align*} This is maximised if \(k\) is as small as possibe, ie \(k = 0\) (ie it doesn't leave it's bag)
2. \(\,\) \begin{align*} && \mathbb{P}(\text{both counters in same bag}) &= \mathbb{P}(\text{both in }P)+ \mathbb{P}(\text{both in }Q) \\ &&&= \mathbb{P}(B_P \to Q \to P, B_Q \to P)+\mathbb{P}(B_P \text{ stays}, B_Q \to P)+\mathbb{P}(B_P \to Q, \text{both stay}) \\ &&&= \frac{k}{n} \cdot \frac{k(k-1)}{(n+k)(n+k-1)} + \frac{n-k}{n} \frac{k}{n+k} + \frac{k}{n} \frac{n(n-1)}{(n+k)(n+k-1)} \\ &&&= \frac{(k^3-k^2)+(n-k)k(n+k-1)+kn(n-1)}{n(n+k)(n+k-1)}\\ &&&= \frac{2kn(n-1)}{n(n+k)(n+k-1)}\\ &&&= \frac{2k(n-1)}{(n+k)(n+k-1)} \end{align*} \begin{align*} && \frac{P_{k+1}}{P_k} &= \frac{2(k+1)(n-1)}{(n+k+1)(n+k)} \frac{(n+k)(n+k-1)}{2k(n-1)} \\ &&&= \frac{(k+1)(n+k-1)}{k(n+k+1)} \\ &&& \geq 1 \\ \Leftrightarrow && (k+1)(n+k-1) &\geq k(n+k+1) \\ \Leftrightarrow && n-1 &\geq k \\ \end{align*} Therefore this probability is increasing while \(k \leq n-1\), ie it's maximised \(k = n-1\) or \(k=n\)