Problems
Filters
1531 problems found
2004 Paper 3 Q5
Show that if \(\, \cos(x - \alpha) = \cos \beta \,\) then either \(\, \tan x = \tan ( \alpha + \beta)\,\) or \(\; \tan x = \tan ( \alpha - \beta)\,\). By choosing suitable values of \(x\), \(\alpha\) and \(\beta\,\), give an example to show that if \(\,\tan x = \tan ( \alpha + \beta)\,\), then \(\,\cos(x - \alpha) \, \) need not equal \( \cos \beta \,\). Let \(\omega\) be the acute angle such that \(\tan \omega = \frac 43\,\).
- For \(0 \le x \le 2 \pi\), solve the equation \[ \cos x -7 \sin x = 5 \] giving both solutions in terms of \(\omega\,\).
- For \(0 \le x \le 2 \pi\), solve the equation \[ 2\cos x + 11 \sin x = 10 \] showing that one solution is twice the other and giving both in terms of \(\omega\,\).
2004 Paper 3 Q6
Given a sequence \(w_0\), \(w_1\), \(w_2\), \(\ldots\,\), the sequence \(F_1\), \(F_2\), \(\ldots\) is defined by $$F_n = w_n^2 + w_{n-1}^2 - 4w_nw_{n-1} \,.$$ Show that $\; F_{n}-F_{n-1} = \l w_n-w_{n-2} \r \l w_n+w_{n-2}-4w_{n-1} \r \; \( for \)n \ge 2\,$.
- The sequence \(u_0\), \(u_1\), \(u_2\), \(\ldots\) has \(u_0 = 1\), and \(u_1 = 2\) and satisfies \[ u_n = 4u_{n-1} -u_{n-2} \quad (n \ge 2)\;. \] Prove that \ $ u_n^2 + u_{n-1}^2 = 4u_nu_{n-1}-3 \; $ for \(n \ge 1\,\).
- A sequence \(v_0\), \(v_1\), \(v_2\), \(\ldots\,\) has \(v_0=1\) and satisfies \begin{equation*} v_n^2 + v_{n-1}^2 = 4v_nv_{n-1}-3 \quad (n \ge 1). \tag{\(\ast\)} \end{equation*} \makebox[7mm]{(a) \hfill}Find \(v_1\) and prove that, for each \(n\ge2\,\), either \(v_n= 4v_{n-1} -v_{n-2}\) or \(v_n=v_{n-2}\,\). \makebox[7mm]{(b) \hfill}Show that the sequence, with period 2, defined by \begin{equation*} v_n = \begin{cases} 1 & \mbox{for \(n\) even} \\ 2 & \mbox{for \(n\) odd} \end{cases} \end{equation*} \makebox[7mm]{\hfill}satisfies \((\ast)\). \makebox[7mm]{(c) \hfill}Find a sequence \(v_n\) with period 4 which has \(v_0=1\,\), and satisfies~\((\ast)\).
2004 Paper 3 Q7
For \(n=1\), \(2\), \(3\), \(\ldots\,\), let \[ I_n = \int_0^1 {t^{n-1} \over \l t+1 \r^n} \, \mathrm{d} t \, . \] By considering the greatest value taken by \(\displaystyle {t \over t+1}\) for \(0 \le t \le 1\) show that \(I_{n+1} < {1 \over 2} I_{n}\,\). Show also that \(\; \displaystyle I_{n+1}= - \frac 1{\; n\, 2^n} + I_{n}\,\). Deduce that \(\; \displaystyle I_n < \frac1 {\; n \, 2^{n-1}}\,\). Prove that \[ \ln 2 = \sum_{r=1}^n {1 \over \; r\, 2^r} + I_{n+1} \] and hence show that \({2 \over 3} < \ln 2 < {17 \over 24}\,\).
Solution: \begin{align*} && \frac{t}{t+1} &= 1 - \frac{1}{t+1} \geq \frac12 \\ \Rightarrow && I_{n+1} &= \int_0^1 \frac{t^{n}}{(t+1)^{n+1}} \d t \\ &&&= \int_0^1\frac{t}{t+1} \frac{t^{n-1}}{(t+1)^{n}} \d t \\ &&&< \int_0^1\frac12\frac{t^{n-1}}{(t+1)^{n}} \d t \\ &&&= \frac12 I_n \\ \\ && I_{n+1} &= \int_0^1 \frac{t^{n}}{(t+1)^{n+1}} \d t \\ &&&= \left [ t^n \frac{(1+t)^{-n}}{-n} \right]_0^1 +\frac1n \int_0^1 n t^{n-1}(1+t)^{-n} \d t \\ &&&= -\frac{1}{n2^n} + I_n \\ \Rightarrow && \frac12 I_n &> -\frac1{n2^n} + I_n \\ \Rightarrow && \frac{1}{n2^{n-1}} &> I_n \end{align*} \begin{align*} && \ln 2 &= \int_0^1 \frac{1}{1+t} \d t \\ &&&= I_1 \\ &&&= \frac1{2} + I_2 \\ &&&= \frac1{2} + \frac{1}{2 \cdot 2^2} + I_3 \\ &&&= \sum_{r=1}^n \frac{1}{r2^r} + I_{n+1} \\ \\ && \ln 2 &= \frac12 + \frac18 + \frac1{24} + I_4 \\ \Rightarrow && \ln 2 &> \frac12 + \frac18 + \frac1{24} = \frac{12+3+1}{24} = \frac{16}{24} = \frac23 \\ \Rightarrow && \ln 2 &= \frac12 + \frac18 + I_3 \\ &&&< \frac12 + \frac18 +\frac{1}{3 \cdot 4} \\ &&&< \frac{12}{24} + \frac{3}{24} + \frac{2}{24} = \frac{17}{24} \end{align*}
2004 Paper 3 Q8
Show that if \[ {\mathrm{d}y \over \mathrm{d} x}=\f(x)y + {\g(x) \over y} \] then the substitution \(u = y^2\) gives a linear differential equation for \(u(x)\,\). Hence or otherwise solve the differential equation \[ {\mathrm{d}y \over \mathrm{d} x}={y \over x} - {1 \over y}\;. \] Determine the solution curves of this equation which pass through \((1 \,, 1)\,\), \((2\, , 2)\) and \((4 \, , 4)\) and sketch graphs of all three curves on the same axes.
Solution: \begin{align*} && \frac{\d y}{\d x} &= f(x) y + \frac{g(x)}{y} \\ && y \frac{\d y}{\d x} &= f(x) y^2 + g(x) \\ u = y^2: && \frac12 \frac{\d u}{\d x} &= f(x) u + g(x) \end{align*} Which is a linear differential equation for \(u\). \begin{align*} && \frac12 u' &= \frac1x u -1 \\ \Rightarrow && u' - \frac2xu &= -1 \\ \Rightarrow && \frac{1}{x^2} u' - \frac{2}{x^3} u &= -\frac{1}{x^2} \\ \Rightarrow && (\frac{u}{x^2})' &= - \frac{1}{x^2} \\ \Rightarrow && \frac{u}{x^2} &= \frac1x + C \\ \Rightarrow && u &= Cx^2 + x \\ \Rightarrow && y^2 &= Cx^2 + x \end{align*} If \((1,1)\) is on the curve then \(1 = C + 1 \Rightarrow C = 0 \Rightarrow y^2 = x\). If \((2,2)\) is on the curve then \(4 = 4C + 2 \Rightarrow C = \frac12 \Rightarrow y^2 = x + \frac12 x^2\). If \((3,3)\) is on the curve then \(9 = 9C + 3 \Rightarrow C = \frac23 \Rightarrow y^2 = x + \frac23 x^2\)
2004 Paper 3 Q9
A circular hoop of radius \(a\) is free to rotate about a fixed horizontal axis passing through a point \(P\) on its circumference. The plane of the hoop is perpendicular to this axis. The hoop hangs in equilibrium with its centre, \(O\), vertically below \(P\). The point \(A\) on the hoop is vertically below \(O\), so that \(POA\) is a diameter of the hoop. A mouse \(M\) runs at constant speed \(u\) round the rough inner surface of the lower part of the hoop. Show that the mouse can choose its speed so that the hoop remains in equilibrium with diameter \(POA\) vertical. Describe what happens to the hoop when the mouse passes the point at which angle \(AOM = 2 \arctan \mu\,\), where \(\mu\) is the coefficient of friction between mouse and hoop.
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
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 \(\mu - c < \overline{X} < \mu+c\,\), 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 \(\displaystyle\Phi(z_{\alpha}) = 1 - \tfrac{1}{2}\alpha\). 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 \(\displaystyle \frac{\sigma_1}{\sigma_0}\) if \(\alpha\) and \(\beta\) are both to be less than \(0.05\,\).
Solution: If \(\sigma\) is smaller we should expect our sample to have a mean closer to the true mean. Therefore we should use a two sided test which accepts \(\mathrm{H}_0\) if the mean is very close to the true mean. Suppose \(\textrm{H}_0\) is true, ie \(\sigma = \sigma_0\), then note that \(X \sim N(\mu, \frac{\sigma_0^2}{n})\) \begin{align*} && 1-\alpha &= \mathbb{P}(\mu - c < X < \mu + c) \\ &&&= \mathbb{P}(\mu - c < \frac{\sigma_0}{\sqrt{n}} Z + \mu < \mu + c) \\ &&&= \mathbb{P}(- \frac{c\sqrt{n}}{\sigma_0} < Z<\frac{\sqrt{n}c}{\sigma_0}) \\ &&&= \mathbb{P}(Z<\frac{\sqrt{n}c}{\sigma_0}) -\mathbb{P}( Z<-\frac{\sqrt{n}c}{\sigma_0}) \\ &&&= \mathbb{P}(Z<\frac{\sqrt{n}c}{\sigma_0}) -(1-\mathbb{P}( Z<\frac{\sqrt{n}c}{\sigma_0})) \\ &&&= 2\mathbb{P}(Z<\frac{\sqrt{n}c}{\sigma_0})-1 \\ \Rightarrow && \Phi(\frac{\sqrt{n}c}{\sigma_0})&=1 - \tfrac12 \alpha \\ \Rightarrow && \frac{\sqrt{n}c}{\sigma_0} &= z_{\alpha} \\ && c &= \frac{\sigma_0 z_{\alpha}}{\sqrt{n}} \end{align*} Under \(\mathrm{H}_1\), \(\sigma = \sigma_1\) so \begin{align*} && \beta &= \mathbb{P}(\mu - c < X < \mu + c) \\ &&&= \mathbb{P}(-\frac{c\sqrt{n}}{\sigma_1} < Z < \frac{\sqrt{n}c}{\sigma_1}) \\ &&&= \mathbb{P}(-\frac{\sigma_0}{\sigma_1} z_{\alpha}< Z < \frac{\sigma_0}{\sigma_1} z_{\alpha}) \\ &&&= 2\Phi(\frac{\sigma_0}{\sigma_1} z_{\alpha})-1 \end{align*} which does not depend on \(n\). Suppose both \(\alpha<0.05\) and \(\beta<0.05\), then \(z_{\alpha} > 1.96\) and \(\Phi(\frac{\sigma_0}{\sigma_1}1.96)<0.525 \Rightarrow \frac{\sigma_0}{\sigma_1}1.96 < 0.063 \Rightarrow \frac{\sigma_1}{\sigma_0} > \frac{1.96}{0.063} = 31.1 \) so the ratio of variances needs to be larger than \(31.1\).
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: Solve for \(x\) the equation \(\displaystyle \frac 1x = \frac 1 a + \frac 1b \;.\) 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: Solve for \(x\) the equation \(\displaystyle \frac 1x = \frac 1 a + \frac 1b +\frac 1c \;.\) where (in the question) \(a\,\), \(b\) and \(c\) are given non-zero numbers. The candidate uses the same technique, giving the answer as \(\displaystyle x = a + b +c \;.\) 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 \(c+a=0\,\).
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.
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 \(x\)