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2000 Paper 2 Q7
D: 1600.0 B: 1486.1

The line \(l\) has vector equation \({\bf r} = \lambda {\bf s}\), where \[ {\bf s} = (\cos\theta+\sqrt3\,) \; {\bf i} +(\surd2\;\sin\theta)\;{\bf j} +(\cos\theta-\sqrt3\,)\;{\bf k} \] and \(\lambda\) is a scalar parameter. Find an expression for the angle between \(l\) and the line \mbox{\({\bf r} = \mu(a\, {\bf i} + b\,{\bf j} +c\, {\bf k})\)}. Show that there is a line \(m\) through the origin such that, whatever the value of \(\theta\), the acute angle between \(l\) and \(m\) is \(\pi/6\). A plane has equation \(x-z=4\sqrt3\). The line \(l\) meets this plane at \(P\). Show that, as \(\theta\) varies, \(P\) describes a circle, with its centre on \(m\). Find the radius of this circle.

2000 Paper 2 Q8
D: 1600.0 B: 1500.1

  1. Let \(y\) be the solution of the differential equation \[ \frac{\d y}{\d x} + 4x\e^{-x^2} {(y+3)}^{\frac12} = 0 \qquad (x \ge 0), \] that satisfies the condition \(y=6\) when \(x=0\). Find \(y\) in terms of \(x\) and show that \(y\to1\) as \(x \to \infty\).
  2. Let \(y\) be any solution of the differential equation \[ \frac{\d y}{\d x} -x\e^{6 x^2} (y+3)^{1-k} = 0 \qquad (x \ge 0). \] %that satisfies the condition \(y=6\) %when \(x=0\). Find a value of \(k\) such that, as \(x \to \infty\), \(\e^{-3x^2}y\) tends to a finite non-zero limit, which you should determine.
\noindent [The approximations, valid for small \(\theta\), \(\sin\theta \approx \theta\) and \(\cos\theta \approx 1-{\textstyle\frac12}\,\theta^2\) may be assumed.]

2000 Paper 2 Q9
D: 1600.0 B: 1500.0

In an aerobatics display, Jane and Karen jump from a great height and go through a period of free fall before opening their parachutes. While in free fall at speed \(v\), Jane experiences air resistance \(kv\) per unit mass but Karen, who spread-eagles, experiences air resistance \mbox{\(kv + (2k^2/g)v^2\)} per unit mass. Show that Jane's speed can never reach \(g/k\). Obtain the corresponding result for Karen. Jane opens her parachute when her speed is \(g/(3{k})\). Show that she has then been in free fall for time \(k^{-1}\ln (3/2)\). Karen also opens her parachute when her speed is \(g/(3{k})\). Find the time she has then been in free fall.


Solution: Looking at the forces on Jane, \(kv < g \Rightarrow v < \frac{g}{k}\). For Karen we have \begin{align*} kv + (2k^2/g)v^2 &< g\\ -g^2 + gkv + (2k^2)v^2 &< 0 \\ (2kv-g)(kv+g) &< 0\\ \Rightarrow v &< \frac{g}{2k} \end{align*} \begin{align*} && \dot{v} &= g - kv \\ \Rightarrow && \frac{\dot{v}}{g - kv} &= 1 \\ \Rightarrow && T &= \int_0^{g/(3k)} \frac{1}{g - kv} dv \\ && &= \int_0^{g/(3k)} \frac{1}{g - kv} dv\\ && &= \int_0^{g/(3k)} \frac{1}{g - kv} dv \\ && &= \left [-\frac{1}{k} \ln \l g - kv \r \right ]_0^{g/(3k)} \\ && &= \frac{1}{k} \ln \l g \r - \frac{1}{k} \ln \l \frac{2}{3}g \r\\ &&&= \frac{1}{k} \ln \l \frac{3}{2} \r \end{align*} \begin{align*} && \dot{v} &= g - kv - (2k^2/g)v^2 \\ \Rightarrow && \frac{\dot{v}}{g - kv - (2k^2/g)v^2} &= 1 \\ \Rightarrow && T &= \int_0^{g/(3k)} \frac{1}{g - kv - (2k^2/g)v^2} dv \\ && &= \int_0^{g/(3k)} \frac{g}{(g-2kv)(kv+g)} dv\\ && &= \int_0^{g/(3k)} \l \frac{2}{3(g-2kv)} + \frac{1}{3(kv+g)} \r dv\\ && &= \left [ \l -\frac{1}{3k} \ln (g-2kv) + \frac{1}{3k}\ln(kv+g) \r \right ]_0^{g/(3k)} \\ && &= \left [ \l -\frac{1}{3k}\ln \l \frac{g}{3} \r + \frac{1}{3k}\ln \l \frac{4g}{3} \r \r \right ] - \left [- \frac1{3k} \ln(g) + \frac{1}{3k} \ln (g) \right ] \\ && &= \frac{1}{3k} \ln \l 4 \r \end{align*} NB: \(\sqrt[3]{4} \approx 1.58 > \frac{3}{2}\) so Karen has been in free-fall for longer, but not \emph{much} longer than Jane.

2000 Paper 2 Q10
D: 1600.0 B: 1500.0

A long light inextensible string passes over a fixed smooth light pulley. A particle of mass 4~kg is attached to one end \(A\) of this string and the other end is attached to a second smooth light pulley. A long light inextensible string \(BC\) passes over the second pulley and has a particle of mass 2 kg attached at \(B\) and a particle of mass of 1 kg attached at \(C\). The system is held in equilibrium in a vertical plane. The string \(BC\) is then released from rest. Find the accelerations of the two moving particles. After \(T\) seconds, the end \(A\) is released so that all three particles are now moving in a vertical plane. Find the accelerations of \(A\), \(B\) and \(C\) in this second phase of the motion. Find also, in terms of \(g\) and \(T\), the speed of \(A\) when \(B\) has moved through a total distance of \(0.6gT^{2}\)~metres.

2000 Paper 2 Q11
D: 1600.0 B: 1484.0

The string \(AP\) has a natural length of \(1\!\cdot5\!\) metres and modulus of elasticity equal to \(5g\) newtons. The end \(A\) is attached to the ceiling of a room of height \(2\!\cdot\!5\) metres and a particle of mass \mbox{\(0\!\cdot\!5\) kg} is attached to the end \(P\). The end \(P\) is released from rest at a point \(0\!\cdot\!5\) metres above the floor and vertically below \(A\). Show that the string becomes slack, but that \(P\) does not reach the ceiling. Show also that while the string is in tension, \(P\) executes simple harmonic motion, and that the time in seconds that elapses from the instant when \(P\) is released to the instant when \(P\) first returns to its original position is $$ \left(\frac8{3g}\right)^{\!\frac12}+ \left(\frac3 {5g}\right)^{\!\frac12} {\Big(\pi - \arccos (3/7)\Big)}. $$ \noindent [Note that \(\arccos x\) is another notation for \(\cos^{-1} x\).]

2000 Paper 2 Q12
D: 1600.0 B: 1487.4

Tabulated values of \({\Phi}(\cdot)\), the cumulative distribution function of a standard normal variable, should not be used in this question. Henry the commuter lives in Cambridge and his working day starts at his office in London at 0900. He catches the 0715 train to King's Cross with probability \(p\), or the 0720 to Liverpool Street with probability \(1-p\). Measured in minutes, journey times for the first train are \(N(55,25)\) and for the second are \(N(65,16)\). Journey times from King's Cross and Liverpool Street to his office are \(N(30,144)\) and \(N(25,9)\), respectively. Show that Henry is more likely to be late for work if he catches the first train. Henry makes \(M\) journeys, where \(M\) is large. Writing \(A\) for \(1-{\Phi}(20/13)\) and \(B\) for \(1-{\Phi}(2)\), find, in terms of \(A\), \(B\), \(M\) and \(p\), the expected number, \(L\), of times that Henry will be late and show that for all possible values of \(p\), $$BM \le L \le AM.$$ Henry noted that in 3/5 of the occasions when he was late, he had caught the King's Cross train. Obtain an estimate of \(p\) in terms of \(A\) and \(B\). [A random variable is said to be \(N\left({{\mu}, {\sigma}^2}\right)\) if it has a normal distribution with mean \({\mu}\) and variance \({\sigma}^2\).]


Solution: If Henry catches the first train, his journey time is \(N(55+30,25+144) = N(85,13^2)\). He is on time if the journey takes less than \(105\) minutes, \(\frac{20}{13}\) std above the mean. If he catches the second train, his journey times is \(N(65+25, 16+9) = N(90, 5^2)\). He is on time if his journey takes less than \(80\) minutes, ie \(\frac{10}{5} = 2\) standard deviations above the mean. This is more likely than from the first train. \(A = 1 - \Phi(20/13)\) is the probability he is late from the first train. \(B = 1 - \Phi(2)\) is the probability he is late from the second train. The expected number of lates is \(L = M \cdot p \cdot A + M \cdot (1-p) \cdot B\), since \(B \leq A\) we must have \(BM \leq L \leq AM\) \begin{align*} && \frac35 &= \frac{pA}{pA + (1-p)B} \\ \Rightarrow && 3(1-p)B &= 2pA \\ \Rightarrow && p(2A+3B) &= 3B \\ \Rightarrow && p &= \frac{3B}{2A+3B} \end{align*}

2000 Paper 2 Q13
D: 1600.0 B: 1594.9

A group of biologists attempts to estimate the magnitude, \(N\), of an island population of voles ({\it Microtus agrestis}). Accordingly, the biologists capture a random sample of 200 voles, mark them and release them. A second random sample of 200 voles is then taken of which 11 are found to be marked. Show that the probability, \(p_N\), of this occurrence is given by $$ p_N = k{{{\big((N-200)!\big)}^2} \over {N!(N-389)!}}, $$ where \(k\) is independent of \(N\). The biologists then estimate \(N\) by calculating the value of \(N\) for which \(p_N\) is a maximum. Find this estimate. All unmarked voles in the second sample are marked and then the entire sample is released. Subsequently a third random sample of 200 voles is taken. Write down the probability that this sample contains exactly \(j\) marked voles, leaving your answer in terms of binomial coefficients. Deduce that $$ \sum_{j=0}^{200}{389 \choose j}{3247 \choose {200-j}} = {3636 \choose 200}. $$


Solution: There will be \(200\) marked vols out of \(N\), and we are finding \(11\) of them. There are \(\binom{200}{11}\) ways to chose the \(11\) marked voles and \(\binom{N - 200}{200-11}\) ways to choose the unmarked voles. The total number of ways to choose \(200\) voles is \(\binom{N}{200}\). Therefore the probability is \begin{align*} p_N &= \frac{\binom{200}{11} \cdot \binom{N - 200}{200-11}}{\binom{N}{200}} \\ &= \binom{200}{11} \cdot \frac{ \frac{(N-200)!}{(189)!(N - 389)!} }{\frac{N!}{(N-200)!(200)!}} \\ &= \binom{200}{11} \frac{200!}{189!} \frac{\big((N-200)!\big)^2}{N!(N-389)!} \end{align*} As required and \(k = \binom{200}{11} \frac{200!}{189!}\). We want to maximise \(\frac{(N-200)!^2}{N!(N-389)!}\), we will do this by comparing consecutive \(p_N\). \begin{align*} \frac{p_{N+1}}{p_N} &= \frac{\frac{(N+1-200)!^2}{(N+1)!(N+1-389)!}}{\frac{(N-200)!^2}{N!(N-389)!}} \\ &= \frac{(N-199)!^2 \cdot N! \cdot (N-389)!}{(N+1)!(N-388)!(N-200)!^2} \\ &= \frac{(N-199)^2 \cdot 1 \cdot 1}{(N+1) \cdot (N-388)\cdot 1} \\ \end{align*} \begin{align*} && \frac{p_{N+1}}{p_N} &> 1 \\ \Leftrightarrow && \frac{(N-199)^2 \cdot 1 \cdot 1}{(N+1) \cdot (N-388)\cdot 1} & > 1 \\ \Leftrightarrow && (N-199)^2 & > (N+1) \cdot (N-388) \\ \Leftrightarrow && N^2-2\cdot199N+199^2 & > N^2 - 387N -388 \\ \Leftrightarrow && -398N+199^2 & > - 387N -388 \\ \Leftrightarrow && 199^2+388 & > 11N\\ \Leftrightarrow && \frac{199^2+388}{11} & > N\\ \Leftrightarrow && 3635\frac{4}{11} & > N\\ \end{align*} Therefore \(p_N\) is increasing if \(N \leq 3635\), so we should take \(N = 3636\). \[ \P(\text{exactly } j \text{ marked voles}) = \frac{\binom{389}{j} \cdot \binom{3636 - 389}{200-j}}{\binom{3636}{200}}\] Since \begin{align*} && 1 &= \sum_{j=0}^{200} \P(\text{exactly } j \text{ marked voles}) \\ && &= \sum_{j=0}^{200} \frac{\binom{389}{j} \cdot \binom{3247}{200-j}}{\binom{3636}{200}} \\ \Leftrightarrow&& \binom{3636}{200} &= \sum_{j=0}^{200} \binom{389}{j} \cdot \binom{3247}{200-j} \end{align*}

2000 Paper 2 Q14
D: 1600.0 B: 1484.0

The random variables \(X_1\), \(X_2\), \(\ldots\) , \(X_{2n+1}\) are independently and uniformly distributed on the interval \(0 \le x \le 1\). The random variable \(Y\) is defined to be the median of \(X_1\), \(X_2\), \(\ldots\) , \(X_{2n+1}\). Given that the probability density function of \(Y\) is \(\g(y)\), where \[ \mathrm{g}(y)=\begin{cases} ky^{n}(1-y)^{n} & \mbox{ if }0\leqslant y\leqslant1\\ 0 & \mbox{ otherwise} \end{cases} \] use the result $$ \int_0^1 {y^{r}}{{(1-y)}^{s}}\,\d y = \frac{r!s!}{(r+s+1)!} $$ to show that \(k={(2n+1)!}/{{(n!)}^2}\), and evaluate \(\E(Y)\) and \({\rm Var}\,(Y)\). Hence show that, for any given positive number \(d\), the inequality $$ {\P\left({\vert {Y - 1/2} \vert} < {d/{\sqrt {n}}} \right)} < {\P\left({\vert {{\bar X} - 1/2} \vert} < {d/{\sqrt {n}}} \right)} $$ holds provided \(n\) is large enough, where \({\bar X}\) is the mean of \(X_1\), \(X_2\), \(\ldots\) , \(X_{2n+1}\). [You may assume that \(Y\) and \(\bar X\) are normally distributed for large \(n\).]

2000 Paper 3 Q1
D: 1700.0 B: 1485.5

Sketch on the same axes the two curves \(C_1\) and \(C_2\), given by

\begin{align*} C_1: && x y & = 1 \\ C_2: && x^2-y^2 & = 2 \end{align*}
The curves intersect at \(P\) and \(Q\). Given that the coordinates of \(P\) are \((a,b)\) (which you need not evaluate), write down the coordinates of \(Q\) in terms of \(a\) and \(b\). The tangent to \(C_1\) through \(P\) meets the tangent to \(C_2\) through \(Q\) at the point \(M\), and the tangent to \(C_2\) through \(P\) meets the tangent to \(C_1\) through \(Q\) at \(N\). Show that the coordinates of \(M\) are \((-b,a)\) and write down the coordinates of \(N\). Show that \(PMQN\) is a square.


Solution:

TikZ diagram
\(Q = (-a,-b)\) \begin{align*} && \frac{\d y}{\d x} &= -\frac{1}{x^2} \\ \Rightarrow && \frac{y-b}{x-a} &= -\frac{1}{a^2} \\ \Rightarrow && 0 &= a^2y+x-a^2b-a \\ &&&= a^2y+x - 2a\\ \\ && 2x - 2y \frac{\d y}{\d x} &= 0 \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{x}{y} \\ \Rightarrow && \frac{y+b}{x+a} &= \frac{a}{b} \\ \Rightarrow && 0 &= by-ax+b^2 - a^2 \\ &&&= by - ax -2 \end{align*} Notice that \((-b,a)\) is on both lines, therefore it is their point of intersection. The coordinates of \(N\) will be \((a,-b)\). We can see this is a square by noting each point is a rotation (centre the origin) of \(90^\circ\) of each other.

2000 Paper 3 Q2
D: 1700.0 B: 1484.2

Use the substitution \(x = 2-\cos \theta \) to evaluate the integral $$ \int_{3/2}^2 \left(x - 1 \over 3 - x\right)^{\!\frac12}\! \d x. $$ Show that, for \(a < b\), $$ \int_p^q \left( x - a \over b - x\right)^{\!\frac12} \!\d x = \frac{(b-a)(\pi +3{\surd3} -6)}{12}, $$ where \(p= {(3a+b)/4}\) and \(q={(a+b)/2}\).