Problems

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Solution: \begin{align*} n = -1: && (-1)^2 &= s - r+q -p \\ n = 0: && 1 + 0 &= s \\ n = 1: && 1 + 1 &= s + r + q + p \\ n = 2: && 2 + 2^2 &= s + 2r + 4q + 8p \\ \Rightarrow &&& \begin{cases} 1 &= s \\ 1 &= s - r + q - p \\ 2 &= s + r + q + p \\ 6 &= s + 2r + 4q + 8p \end{cases} \\ \Rightarrow && s &= 1 \\ && q &= \frac12 \\ &&& \begin{cases} \frac12 &= r + p \\ 3 &= 2r + 8p \end{cases} \\ \Rightarrow && r &= \frac16 \\ && p &= \frac13 \\ \Rightarrow && \sum_{r=0}^n r^2 &= 1 + \frac16 n + \frac12 n^2 + \frac13 n^3 - (-1)^2 \\ &&&= \frac{n}{6} \l 1 + 3n + 2n^2 \r \\ &&&= \frac{n(n+1)(2n+1)}{6} \end{align*} Similarly, \begin{align*} n = -2: && (-2)^3 &= e - 2d + 4c - 8b + 16a \\ n = -1: && -8 + (-1)^3 &= e -d+c-b+a \\ n = 0: && -9 + 0^3 &= e \\ n = 1: && -9 + 1^3 &= e+d+c+b+a \\ n = 2: && -8 + 2^3 &= e+2d+4c+8b+16a \\ \Rightarrow &&& \begin{cases} -9 &= e \\ -9 &= e - d+c -b + a \\ -8 &= e +d+c+b+a \\ -8 &= e-2d+4c-8b+16a \\ 0 &= e+2d+4c+8b+16a \\ \end{cases} \\ \Rightarrow && e &= -9 \\ \Rightarrow &&& \begin{cases} 1 &= 2c+2a \\ 10 &= 8c+32a \\ 1 &= 2d+2b \\ 8 &= 4d+16b \\ \end{cases} \\ \Rightarrow && a &= \frac14 \\ && c &= \frac14 \\ && b &= \frac12 \\ && d &= 0 \\ \\ \Rightarrow && \sum_{r=0}^n r^3 &= -9 + \frac14n^2 + \frac12 n^3+\frac14 n^4 -((-1)^3+(-2)^3) \\ &&&= \frac14n^2 \l1 + 2n+n^2\r \\ &&&= \frac{n^2(n+1)^2}{4} \end{align*} as required

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Solution: Suppose \begin{align*} && \frac{1}{a+b} &= \frac{1}{a} + \frac{1}{b} \\ \Rightarrow && ab &= b(a+b)+a(a+b) \\ &&&= (a+b)^2 \\ \Rightarrow && 0 &= a^2+ab + b^2 \\ &&&= \tfrac12 (a^2+(a+b)^2+b^2) \end{align*} Which clearly has no solution for non-zero \(a,b\). Suppose \begin{align*} && \frac{1}{a+b+c} &= \frac1a + \frac1b+\frac1c \\ \Leftrightarrow && abc &= (a+b+c)(bc+ca+ab) \\ \Leftrightarrow && 0 &= (a+b+c)(bc+ca+ab) - abc \\ &&&= (a+b)(b+c)(c+a) \end{align*} Therefore it is true iff \(a+b = 0\) or \(b+c=0\) or \(c+a =0\) as required.

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Solution:

1. \(\,\) \begin{align*} && 2 \sin (\tfrac12 \theta) &= \sin \theta \\ \Leftrightarrow && 2 \sin (\tfrac12 \theta) &= 2\sin (\tfrac12 \theta) \cos (\tfrac12 \theta) \\ \Leftrightarrow && 0 &= 2\sin(\tfrac12\theta)(1-\cos(\tfrac12 \theta)) \\ \Leftrightarrow && 0 = \sin(\tfrac12 \theta) &\text{ or } 1 = \cos(\tfrac12 \theta) \\ \Leftrightarrow && 0 &= \sin(\tfrac12 \theta) \end{align*}
2. Let \(= \tan(\tfrac12 \theta)\), then \begin{align*} && 2t &= \frac{2t}{1-t^2} \\ \Leftrightarrow && 0 &= \frac{2t(1-(1-t^2)}{1-t^2} \\ &&&= \frac{2t^3}{1-t^2} \\ \Leftrightarrow && t&= 0 \\ \Leftrightarrow && \frac12\theta &= n \pi \\ \Leftrightarrow && \theta &= 2n\pi \end{align*}
3. Let \(c = \cos(\tfrac12 \theta)\), then \begin{align*} && 2c &= 2c^2 - 1 \\ && 0 &= 2c^2-2c-1 \\ \Leftrightarrow && c &= \frac{2 \pm \sqrt{4+8}}{4} \\ &&&= \frac{1 \pm \sqrt{3}}{2} \\ \Leftrightarrow && c &= \frac{1 - \sqrt{3}}{2} \\ \Leftrightarrow && \frac12 \theta &= \pm \cos^{-1} \frac{1 - \sqrt{3}}{2} + 2n \pi \\ &&&= \mp (\phi+\pi) + 2n \pi \\ \Leftrightarrow && \theta &= (4n+2)\pi \pm 2\phi \end{align*}

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Solution:

\(\theta \in \{0\} \cup (\frac{\pi}{2}, 2\pi]\). In \((0, \frac{\pi}{2})\) \(\cos \theta < 1 + \sin \theta\), and then it's either negative or greater than \(1+ \sin \theta\)

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Solution:

1. The terms will all be of the form \(x^{6-3k}\), so we are looking for \(\binom{6}{2}2^{4} \cdot 1^2 = 6 \cdot 5 \cdot 8 = 240\) The terms will be \(x^{(5n-k)3 -2k} = x^{15n-5k}\), so we want \(k = 3n\), \(\binom{5n}{3n}a^{5n-3n}b^{5n-2n} = \binom{5n}{3n}a^{2n}b^{3n}\)
2. Let \(f(x) = (x^6+3x^5)^{1/2}\) If \(|x| > 3\), then consider \begin{align*} && f(x) &= x^3(1+3/x)^{1/2} \\ &&&= x^3 \left (1 + \frac12 \frac{3}{x} +\frac{1}{2!} \frac12\cdot \frac{-1}{2} \left ( \frac{3}{x} \right)^2 + \frac{1}{3!} \frac{1}{2} \cdot \frac{-1}{2} \cdot \frac{-3}{2} \left (\frac{3}{x} \right)^3 + \cdots \right) \\ &&&= \cdots \frac{1}{6} \frac{3}{8} 27x^0 + \cdots \\ &&&= \cdots + \frac{27}{16} + \cdots \end{align*} If \(|x| < 3\) we can consider \(f(0) = 0\) and notice that there must be no term independent of \(x\)

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Solution:

1. \(\,\) \begin{align*} && I &= \int_0^1 \frac 1 {x^2 + (a+2)x +2a} \; \d x \\ &&&= \int_0^1 \frac{1}{(x+a)(x+2)} \d x\\ \end{align*} Case 1: \(a = 2\) \begin{align*} && I &= \int_0^1 \frac{1}{(x+2)^2} \d x \\ &&&= \left [ -(x+2)^{-1}\right]_0^1 = \frac12 - \frac13 = \frac16 \end{align*} Case 2: \(a \neq 2, a \not \in [0,1]\) \begin{align*} && I &=\frac{1}{a-2} \int_0^1 \left ( \frac{1}{x+2} - \frac{1}{x+a} \right) \d x \\ &&&= \frac{1}{a-2} \left [ \ln |x+2| - \ln |x + a|\right]_0^1 \\ &&&= \frac{1}{a-2} \left ( \ln \frac{3}{|1+a|} - \ln \frac{2}{|a|} \right) \\ &&&= \frac{1}{a-2} \ln \frac{3|a|}{2|a+1|} \end{align*} Case 3: \(a \in [0, 1]\), \(I\) does not converge
2. \(\,\) \begin{align*} && J &= \int _{1}^2\frac 1 {u^2 +au +a-1} \; \d u \\ &&&= \int_1^2 \frac{1}{(u+a-1)(u+1)} \d u \\ x = u-1:&&&= \int_0^1 \frac{1}{(x+a)(x+2)} \d x \end{align*} So it's the same as the previous integral

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Solution: First note that \(10k - k^2 = 25 - (5-k)^2\) which is clearly bounded above by \(25\). The smallest it can be is when \(|5-k|\) is as large as possible, ie when \(k =0\) and we get a lower bound of \(0\). For \((k-1)k(k+1)\) notice this is the product of \(3\) consecutive integers, and therefore must be divisible by \(3\). (In fact, it's divisible by six, since \(\binom{k+1}{3}\) is the number of ways to choose \(3\) objects from \(k+1\). Let \(N = 100a + 10b + c\) where \(0 \leq a,b,c < 10\) and \(1 \leq a\). \(S = a + b^2 + c^3\) we want to find \begin{align*} && 100a +10b + c &= a + b^2 + c^3 \\ \Rightarrow && 0 &= \underbrace{99a}_{3 \mid 99 } + 10b - b^2 -\underbrace{c(c+1)(c-1)}_{3 \mid c(c+1)(c-1)} \\ \end{align*} Therefore \(3 \mid 10b - b^2 = b(10-b)\). Therefore \(3 \mid b\) or \(3 \mid 10-b\) so \(b = 0, 3, 6, 1, 4, 7\) We also have \(99a \geq 99\) and \(10b-b^2 \in [0, 25]\) so we need \(c^3-c \geq 99\), so \(c \geq 5\) Case \(c = 5\), Then \(c^3-c = 120\) so \(a = 1\) and \(10b-b^2 = 21\) so \(b= 3, 7\) \(N = 135, 175\) Case \(c = 6\), so \(c^3 - c = 210\) so \(a = 2\) and \(25-(5-k)^2 = 12\) so no solutions. Case \(c = 7\), so \(7^3 - 7 = 336\) so \(a = 3\) and \(25-(5-k)^2 = 39\) so no solutions. Case \(c = 8\) so \(8^3-8 = 504\) so \(a = 5\) and \(25-(5-k)^2 = 9\), so \(b = 1, 9\) and \(N = 518, 598\) Case \(c = 9\) so \(9^3 - 9 = 720\), so \(a = 7\) and \(25-(5-k)^2 = 27\) so no solutions. Therefore all the solutions are \(N = 135, 175, 518, 598\)

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Solution: We have \(\dot{A} = -kVxy\) or \(\dot{x}V = -kVxy\), ie \(\dot{x} = -kxy\) and similarly \(\dot{y} = kxy = k(1-y)y\). \begin{align*} && \frac{\d y}{\d t} &= ky(1-y) \\ \Rightarrow && \int k \d t &= \int \frac{1}{y(1-y)} \d y \\ \Rightarrow && kt &= \int \left ( \frac{1}{y} + \frac{1}{1-y} \right) \d y \\ &&&= \ln y - \ln (1-y) + C\\ \Rightarrow && kt &= \ln \frac{y}{D(1-y)} \\ \Rightarrow && De^{kt} &= \frac{y}{1-y} \\ \Rightarrow && y(1+De^{kt}) &= De^{kt} \\ \Rightarrow && y &= \frac{De^{kt}}{1+De^{kt}} \end{align*} As \(t \to \infty\) \(y \to \frac{D}{D} = 1\) so depending on how fine grained we want to go we might say that 'yes it completely converts' when there is an immeasurably small amount of \(A\) left, or we might say it doesn't since it only tends to \(1\) and never actually reaches it.

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Solution:

\begin{array}{c|c|c|c} & ABX & ABCD & AXCD \\ \hline \text{area} & \frac12 q r & pq & q(p - \frac12 r) \\ \text{com} & \binom{\frac{r}{3}}{\frac{2q}{3}} & \binom{p/2}{q/2} & \binom{a}{b} \end{array} \begin{align*} && q(p-\frac12 r) \binom{a}{b} &= pq\binom{p/2}{q/2} - \frac12 q r \binom{\frac{r}{3}}{\frac{2q}{3}} \\ \Rightarrow && \binom{a}{b} &= \frac{2}{2p-r}\binom{p^2/2-\frac16r^2}{pq/2-\frac13qr} \\ &&&= \binom{\frac{p^2-\frac13 r^2}{2p-r}}{\frac{pq-\frac23qr}{2p-r}} \end{align*} We must have: \begin{align*} && 1 &= \frac{p^2-\frac13r^2}{pq-\frac23qr} \\ \Rightarrow && pq-\frac23qr &= p^2 - \frac13 r^2 \\ \Rightarrow && 0 &=r^2-2q r + 3p(q-p) \\ \Rightarrow && 0 &= (r-q)^2 -q^2+3pq-3p^2 \\ \Rightarrow && r&= q \pm \sqrt{q^2-3pq+3p^2} \end{align*} Suppose \(r > q\), then \(p > q > r\) and we have a shape which looks like this which definitely wouldn't have \(G\) hanging below \(A\).

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Solution: We can `ignore' the fact that they are both accelerating, because the acceleration is the same for both object so it will "cancel" out. Therefore the time taken is the same as if the object has to travel distance \(d\) at speed \(u\), ie \(d/u\). \begin{align*} && u_A &= u - g \frac{d}{u} \\ && u_B &= -g\frac{d}{u} \end{align*}

The speed of approach is \(u\), therefore the speed of separation is \(eu\), in particular \(v_B = v_A + eu\) \begin{align*} \text{COM}: && m\left (u-g\frac{d}{u} \right)+m\left (-g\frac{d}{u} \right) &= mv_A + m(v_A + eu) \\ \Rightarrow && 2v_A &= u - 2g\frac{d}{u}-eu \\ \Rightarrow && v_A &= \frac12 (1-e)u - \frac{gd}{u} \\ \Rightarrow && v_B &= \frac12 (1+e)u - \frac{gd}{u} \\ \\ && \text{initial k.e.} &= \frac12 m \left (u-g\frac{d}{u} \right)^2 + \frac12 m \left (-g\frac{d}{u} \right)^2 \\ &&&= \frac12m \left (u^2 -2gd + \frac{2g^2d^2}{u^2} \right) \\ && \text{final k.e.} &= \frac12 m \left ( \frac12 (1-e)u - \frac{gd}{u}\right)^2 + \frac12 m \left ( \frac12 (1+e)u - \frac{gd}{u}\right)^2 \\ &&&= \frac12 m \left (\frac14 \left ( (1-e)^2+(1+e)^2\right)u^2 - gd \left ((1-e)+(1+e) \right) +\frac{2g^2d^2}{u^2}\right) \\ &&&= \frac12 m \left (\frac12(1+e^2)u^2-2gd+ \frac{2g^2d^2}{u^2}\right) \\ \Rightarrow && \text{loss k.e.} &= \frac12m \left ( u^2 - \frac12(1+e^2)u^2\right) \\ &&&= \frac14mu^2(1-e^2) \end{align*}

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Solution:

1. Since the second draw is independently of the first draw \(\mathbb{P}(F=r, S=s) = \mathbb{P}(F=r)\mathbb{P}(S=s) = \frac{1}{n^2}\) and so \begin{align*} && \E[FS] &= \sum_{r=1}^n \sum_{s=1}^n rs \mathbb{P}(F=r, S = s) \\ &&&= \frac{1}{n^2} \sum_{r=1}^n \sum_{s=1}^n rs \\ &&&= \frac{1}{n^2} \left ( \sum_{r=1}^n r \right)\left ( \sum_{s=1}^n s \right) \\ &&&= \frac{1}{n^2} \left ( \frac{n(n+1)}{2} \right)^2 \\ &&&= \frac{(n+1)^2}{4} \end{align*}
2. Under this methodology, \(\mathbb{P}(F=r, S=s) = \frac{1}{n(n-1)}\) as long as \(r \neq s\), therefore \begin{align*} && \E[FS] &= \sum_{r=1}^n \sum_{s\neq r} rs \mathbb{P}(F = r, S = s) \\ &&&= \frac{1}{n(n-1)} \sum_{r=1}^n \left ( \sum_{s=1}^n rs - r^2 \right) \\ &&&= \frac{1}{n(n-1)} \left ( \frac{n^2(n+1)^2}{4} - \frac{n(n+1)(2n+1)}{6} \right) \\ &&&= \frac{n+1}{12(n-1)} \left ( 3n(n+1)-2(2n+1) \right) \\ &&&= \frac{n+1}{12(n-1)} \left ( 3n^2 -n -2 \right) \\ &&&= \frac{n+1}{12(n-1)} (3n+2)(n-1)\\ &&&= \frac{(n+1)(3n+2)}{12} \end{align*}

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Solution: Note that in the first \(10\)-short period the number of goals scored by each team is \(B(5, \p_i)\). For them to be equal they must both have scored the same number of goals, ie \begin{align*} && \alpha &= \sum_{i=0}^5 \mathbb{P}(\text{both teams score }5-i) \\ &&&= \sum_{i=0}^5 \binom{5}{i} (1-p_A)^ip_A^{5-i} \binom{5}{i} (1-p_B)^i p_B^{5-i} \\ &&&= \sum_{i=0}^5 \binom{5}{i} ^2(1-p_A)^i (1-p_B)^i p_A^{5-i} p_B^{5-i} \\ \end{align*} Suppose we make it to the end of the shoot out with scores tied. The probability that we finish each round is \(p_A(1-p_B) + p_B(1-p_A)\) (the probability \(A\) wins or \(B\) wins). This is \(p_A + p_B - 2p_Ap_B = \beta\)). Therefore the number of additional rounds is geometric with parameter \(\beta\) and the expected number of rounds is \(\frac{1}{\beta}\). Each round has two shots, and there is a probability \(\alpha\) of this occuring, ie \(\frac{2\alpha}{\beta}\). Added to the \(10\) guaranteed shots we get the desired result