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2004 Paper 3 Q10
A particle \(P\) of mass \(m\) is attached to points \(A\) and \(B\), where \(A\) is a distance \(9a\) vertically above \(B\), by elastic strings, each of which has modulus of elasticity \(6mg\). The string \(AP\) has natural length \(6a\) and the string \(BP\) has natural length \(2a\). Let \(x\) be the distance \(AP\). The system is released from rest with \(P\) on the vertical line \(AB\) and \(x = 6a\). Show that the acceleration \(\ddot{x}\) of \(P\) is \(\ds{4g \over a}(7a - x)\) for \(6a < x < 7a\) and \(\ds{g \over a}(7a - x)\) for \(7a < x < 9a\,\). Find the time taken for the particle to reach \(B\).
2004 Paper 3 Q11
Particles \(P\), of mass \(2\), and \(Q\), of mass \(1\), move along a line. Their distances from a fixed point are \(x_1\) and \(x_2\), respectively where \(x_2>x_1\,\). Each particle is subject to a repulsive force from the other of magnitude \(\displaystyle {2 \over z^3}\), where \(z = x_2-x_1 \,\). Initially, \(x_1=0\), \(x_2 = 1\), \(Q\) is at rest and \(P\) moves towards \(Q\) with speed 1. Show that \(z\) obeys the equation \(\displaystyle {\mathrm{d}^2 z \over \mathrm{d}t^2} = {3 \over z^3}\). By first writing \(\displaystyle {\mathrm{d}^2 z \over \mathrm{d}t^2} = v {\mathrm{d}v \over \mathrm{d}z} \,\), where \(\displaystyle v={\mathrm{d}z \over \mathrm{d}t}\,\), show that \(z=\sqrt{4t^2-2t+1}\,\). By considering the equation satisfied by \(2x_1+x_2\,\), find \(x_1\) and \(x_2\) in terms of \(t \,\).
Solution: \begin{align*} \text{N2}: && 2\ddot{x}_1 &= -\frac{2}{(x_2-x_1)^3}\\ \text{N2}: && \ddot{x}_2 &= \frac{2}{(x_2-x_1)^3}\\ \Rightarrow && \ddot{x}_2 - \ddot{x}_1 &= \frac{3}{(x_1-x_2)^3} \\ \Rightarrow && \frac{\d^2 z}{\d t^2} &= \frac{3}{z^3} \\ \Rightarrow && v \frac{\d v}{\d z} &= \frac{3}{z^3} \\ \Rightarrow && \int v \d v &= \int \frac{3}{z^3} \d z \\ \Rightarrow && \frac{v^2}{2} &= -\frac{3}{2}z^{-2} + C \\ \Rightarrow && v^2 &= -3 z^{-2} + C' \\ t=0,z=1,v=-1: && 1 &= -3+C \Rightarrow C = 4 \\ \Rightarrow && \frac{\d z}{\d t} &= -\sqrt{4-3z^{-2}} \\ \Rightarrow && \int \d t &= -\int \frac{1}{\sqrt{4-3z^{-2}}} \d z \\ \Rightarrow && t &= \int \frac{z}{\sqrt{4z^2-3}} \d z \\ \Rightarrow && t &= -\frac14\sqrt{4z^2-3} + C \\ t=0, z = 1: && 0 &= -\frac14+C \\ \Rightarrow && C &= \frac14\\ \Rightarrow && 4t -1 &= -\sqrt{4z^2-3} \\ \Rightarrow && 16t^2+1-8t &= 4z^2-3 \\ \Rightarrow && z &= \sqrt{4t^2-2t+1} \end{align*} \begin{align*} && 2\ddot{x}_1 + \ddot{x}_2 &= 0 \\ \Rightarrow && 2x_1+x_2 &= At + B \\ t = 0, v = -1: && 2x_1+x_2 &= -t+1 \\ \\ \Rightarrow && x_2-x_1 &= \sqrt{4t^2-2t+1}\\ && 2x_1+x_2 &= 1-t \\ \Rightarrow && x_1 &= \frac13 \left (1-t-\sqrt{4t^2-2t+1} \right) \\ && x_2 &= \frac13(1-t + \sqrt{4t^2-2t+1}) \end{align*} This method of considering the relative position and considering the motion of the centre of mass is extremely common for solving systems of particles problems.
2004 Paper 3 Q12
A team of \(m\) players, numbered from \(1\) to \(m\), puts on a set of a \(m\) shirts, similarly numbered from \(1\) to \(m\). The players change in a hurry, so that the shirts are assigned to them randomly, one to each player. Let \(C_i\) be the random variable that takes the value \(1\) if player \(i\) is wearing shirt \(i\), and 0 otherwise. Show that \(\mathrm{E}\left(C_1\right)={1 \over m}\) and find \(\var \left(C_1\right)\) and \(\mathrm{Cov}\left(C_1 \, , \; C_2 \right) \,\). Let \(\, N = C_1 + C_2 + \cdots + C_m \,\) be the random variable whose value is the number of players who are wearing the correct shirt. Show that \(\mathrm{E}\left(N\right)= \var \left(N\right) = 1 \,\). Explain why a Normal approximation to \(N\) is not likely to be appropriate for any \(m\), but that a Poisson approximation might be reasonable. In the case \(m = 4\), find, by listing equally likely possibilities or otherwise, the probability that no player is wearing the correct shirt and verify that an appropriate Poisson approximation to \(N\) gives this probability with a relative error of about \(2\%\). [Use \(\e \approx 2\frac{72}{100} \,\).]
Solution: There are \(m!\) different ways of assigning the shirts, and in \((m-1)!\) of them player \(1\) gets their own shirt, ie \(\mathbb{E}(C_1) = \mathbb{P}(\text{player }1\text{ gets own shirt}) = \frac{(m-1)!}{m!} = \frac{1}{m}\). \(\var(C_1) = \mathbb{E}(C_1^2) - [\mathbb{E}(C_1)]^2 = \frac{1}{m} - \frac{1}{m^2} = \frac{m-1}{m^2}\). If we have two players, there are \((m-2)!\) ways they both get their own shirts, therefore \(\textrm{Cov}(C_1,C_2) = \mathbb{E}(C_1C_2) - \mathbb{E}(C_1)\mathbb{E}(C_2) = \frac{(m-2)!}{m!} - \frac{1}{m^2} = \frac{1}{m(m-1)} - \frac{1}{m^2} = \frac{m-m+1}{m^2(m-1)} = \frac{1}{m^2(m-1)}\). \begin{align*} \mathbb{E}(N) &= \mathbb{E}(C_1 + C_2 + \cdots + C_m) \\ &= \mathbb{E}(C_1) + \mathbb{E}(C_2) + \cdots + \mathbb{E}(C_m) \\ &= \frac{1}{m} + \frac{1}{m} +\cdots+ \frac1m \\ &= 1 \\ \\ \var(N) &= \sum_{r=1}^m \var(C_r) + 2\sum_{r=1}^{m-1} \sum_{s=2}^{m} \textrm{Cov}(C_r,C_s) \\ &= m \frac{m-1}{m^2} + 2 \frac{m(m-1)}{2}\frac{1}{m^2(m-1)} \\ &=\frac{m-1}{m} + \frac{1}{m} \\ &= 1 \end{align*} If we were to take a normal approximation, we would want to take \(N(1,1)\), but this would say things like \(-1\) is as likely as \(3\) shirts being correct, which is clearly a bad model. A Poisson is much more likely to be a sensible model as they have the same mean and variance as the parameter, and if \(m\) is large, the covariance between shirts is going to be very small, so it will appear similar to random events occurring. We can have \begin{align*} BADC \\ BCDA \\ BDAC \\ CADB \\ CDAB\\ CDBA \\ DABC\\ DCAB \\ DCBA \end{align*} Ie \(\frac{9}{24}\) ways to have no player wearing their own shirt with \(4\) players. \(Po(1)\) would say this probability is \(e^{-1}\), giving a relative error of: \begin{align*} \frac{e^{-1}-\frac{9}{24}}{\frac9{24}} &\approx \frac{\frac{100}{272} - \frac{9}{24}}{\frac9{24}} \\ &= -\frac{1}{51} \\ &\approx -2\% \end{align*}
2004 Paper 3 Q13
A men's endurance competition has an unlimited number of rounds. In each round, a competitor has, independently, a probability \(p\) of making it through the round; otherwise, he fails the round. Once a competitor fails a round, he drops out of the competition; before he drops out, he takes part in every round. The grand prize is awarded to any competitor who makes it through a round which all the other remaining competitors fail; if all the remaining competitors fail at the same round the grand prize is not awarded. If the competition begins with three competitors, find the probability that:
- all three drop out in the same round;
- two of them drop out in round \(r\) (with \(r \ge 2\)) and the third in an earlier round;
- the grand prize is awarded.
Solution:
- This is the same as the sum of the probability that they all drop out in the \(k\)th round for all values of \(k\), ie \begin{align*} \mathbb{P}(\text{all drop in same round}) &= \sum_{k=0}^\infty \mathbb{P}(\text{all drop out in the }k+1\text{th round}) \\ &= \sum_{k=0}^{\infty}(p^k(1-p))^3 \\ &= (1-p)^3 \sum_{k=0}^{\infty}p^{3k} \\ &= \frac{(1-p)^3}{1-p^3} \\ &= \frac{1+3p(1-p)(p-(1-p))-p^3}{1-p^3} \\ &= \frac{1-p^3-3p(1-p)(1-2p)}{1-p^3} \end{align*}
- There are \(3\) ways to choose the person who drops out earlier, and then they can drop out in round \(0, 1, \cdots r-1\) \begin{align*} \mathbb{P}(\text{exactly two drop out in round }r\text{ and one before}) &= 3\sum_{k=0}^{r-2} (p^{r-1}(1-p))^2p^k(1-p) \\ &= 3p^{2r-2}(1-p)^3 \sum_{k=0}^{r-2}p^k \\ &= 3p^{2r-2}(1-p)^3 \frac{1-p^{r-1}}{1-p} \\ &= 3p^{2r-2}(1-p)^2(1-p^{r-1}) \end{align*}
- The probability exactly \(2\) finish after the third \begin{align*} \mathbb{P}(\text{exactly two drop out after third}) &= \sum_{r=2}^{\infty}\mathbb{P}(\text{exactly two drop out in round }r\text{ and one before}) \\ &= \sum_{r=2}^{\infty}3p^{2r-2}(1-p)^2(1-p^{r-1}) \\ &= 3(1-p)^2p^{-2}\sum_{r=2}^{\infty}(p^{2r}-p^{3r-1}) \\ &= 3(1-p)^2p^{-2} \left( \frac{p^4}{1-p^2} - \frac{p^5}{1-p^3} \right) \\ &= \frac{3(1-p)^2(p^2(1-p^3)-p^3(1-p^2))}{(1-p^2)(1-p^3)}\\ &= \frac{3(1-p)^3p^2}{(1-p^2)(1-p^3)}\\ \end{align*} Therefore the probability the grand prize is not awarded is \begin{align*} P &= 1 - \frac{(1-p)^3}{1-p^3} - \frac{3(1-p)^3p^2}{(1-p^2)(1-p^3)} \\ &= \frac{(1-p^3)(1-p^2) - (1-p)^3(1-p^2)-3(1-p)^3p^2}{(1-p^2)(1-p^3)} \\ &= \frac{(1-p^3)(1-p^2) - (1-p)^3(1+2p^2)}{(1-p^2)(1-p^3)} \\ \end{align*}
2004 Paper 3 Q14
\textit{In this question, \(\Phi(z)\) is the cumulative distribution function of a standard normal random variable.} A random variable is known to have a Normal distribution with mean \(\mu\) and standard deviation either \(\sigma_0\) or \(\sigma_1\), where \(\sigma_0 < \sigma_1\,\). The mean, \(\overline{X}\), of a random sample of \(n\) values of \(X\) is to be used to test the hypothesis \(\mathrm{H}_0: \sigma = \sigma_0\) against the alternative \(\mathrm{H}_1: \sigma = \sigma_1\,\). Explain carefully why it is appropriate to use a two sided test of the form: accept \(\mathrm{H}_0\) if \phantom{} \(\mu - c < \overline{X} < \mu+c\,\), otherwise accept \(\mathrm{H}_1\). Given that the probability of accepting \(\mathrm{H}_1\) when \(\mathrm{H}_0\) is true is \(\alpha\), determine \(c\) in terms of \(n\), \(\sigma_0\) and \(z_{\alpha}\), where \(z_\alpha \) is defined by \(\ds\Phi(z_{\alpha}) = 1 - \tfrac{1}{2}\alpha\). The probability of accepting \(\mathrm{H}_0\) when \(\mathrm{H}_1\) is true is denoted by \(\beta\). Show that \(\beta\) is independent of \(n\). Given that \(\Phi(1.960)\approx 0.975\) and that \(\Phi(0.063) \approx 0.525\,\), determine, approximately, the minimum value of \(\ds \frac{\sigma_1}{\sigma_0}\) if \(\alpha\) and \(\beta\) are both to be less than \(0.05\,\).
2003 Paper 1 Q1
It is given that \(\sum\limits_{r=-1}^ {n} r^2\) can be written in the form \(pn^3 +qn^2+rn+s\,\), where \(p\,\), \(q\,\), \(r\,\) and \(s\) are numbers. By setting \(n=-1\), \(0\), \(1\) and \(2\), obtain four equations that must be satisfied by \(p\,\), \(q\,\), \(r\,\) and \(s\) and hence show that \[ { \sum\limits_{r=0} ^n} r^2= {\textstyle \frac16} n(n+1)(2n+1)\;. \] Given that \(\sum\limits_{r=-2}^ nr^3\) can be written in the form \(an^4 +bn^3+cn^2+dn +e\,\), show similarly that \[ { \sum\limits_{r=0} ^n} r^3= {\textstyle \frac14} n^2(n+1)^2\;. \]
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
2003 Paper 1 Q2
The first question on an examination paper is: \hspace*{3cm} Solve for \(x\) the equation \ \ \ $\ds \frac 1x = \frac 1 a + \frac 1b \;. $ \noindent where (in the question) \(a\) and \(b\) are given non-zero real numbers. One candidate writes \(x=a+b\) as the solution. Show that there are no values of \(a\) and \(b\) for which this will give the correct answer. The next question on the examination paper is: \hspace*{3cm} Solve for \(x\) the equation \ \ \ $\ds \frac 1x = \frac 1 a + \frac 1b +\frac 1c \;. $ \noindent where (in the question) \(a\,\), \(b\) and \(c\) are given non-zero numbers. The candidate uses the same technique, giving the answer as $\ds x = a + b +c \;. $ Show that the candidate's answer will be correct if and only if \(a\,\), \(b\) and \(c\) satisfy at least one of the equations \(a+b=0\,\), \(b+c=0\) or \(c+a=0\,\).
2003 Paper 1 Q3
- Show that \( 2\sin(\frac12\theta)=\sin \theta\) if and only if \(\sin(\frac12\theta)=0\,\).
- Solve the equation \(2\tan (\frac12\theta) = \tan\theta\,\).
- Show that \(2\cos(\frac12\theta)=\cos \theta\) if and only if \(\theta=(4n+2)\pi\pm 2\phi\) where \(\phi\) is defined by \(\cos \phi=\frac12(\sqrt 3-1)\;\), \(0\le \phi\le \frac{1}{2}\pi\), and \(n\) is any integer.
Solution:
- \(\,\) \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*}
- 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*}
- 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*}
2003 Paper 1 Q4
Solve the inequality $$\frac{\sin\theta+1}{\cos\theta}\le1\;$$ where \(0\le\theta<2\pi\,\) and \(\cos\theta\ne0\,\).
Solution:
2003 Paper 1 Q5
- In the binomial expansion of \((2x+1/x^{2})^{6}\;\) for \(x\ne0\), show that the term which is independent of \(x\) is \(240\). Find the term which is independent of \(x\) in the binomial expansion of \((ax^3+b/x^{2})^{5n}\,\).
- Let \(\f(x) =(x^6+3x^5)^{1/2}\,\). By considering the expansion of \((1+3/x)^{1/2}\,\) show that the term which is independent of \(x\) in the expansion of \(\f(x)\) in powers of \(1/x\,\), for \( \vert x\vert>3\,\), is \(27/16\,\). Show that there is no term independent of \(x\,\) in the expansion of \(\f(x)\) in powers of \(x\,\), for \( \vert x\vert<3\,\).
Solution:
- 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 \(k = 3n\), \(\binom{5n}{3n}a^{5n-3n}b^{5n-2n} = \binom{5n}{3n}a^{2n}b^{3n}\)
- 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 \(x\)