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2001 Paper 2 Q7
In an Argand diagram, \(O\) is the origin and \(P\) is the point \(2+0\mathrm{i}\). The points \(Q\), \(R\) and \(S\) are such that the lengths \(OP\), \(PQ\), \(QR\) and \(RS\) are all equal, and the angles \(OPQ\), \(PQR\) and \(QRS\) are all equal to \({5{\pi}}/6\), so that the points \(O\), \(P\), \(Q\), \(R\) and \(S\) are five vertices of a regular 12-sided polygon lying in the upper half of the Argand diagram. Show that \(Q\) is the point \(2 + \sqrt 3 + \mathrm{i}\) and find \(S\). The point \(C\) is the centre of the circle that passes through the points \(O\), \(P\) and \(Q\). Show that, if the polygon is rotated anticlockwise about \(O\) until \(C\) first lies on the real axis, the new position of \(S\) is $$ - \tfrac{1}{2} (3\sqrt 2+ \sqrt6)(\sqrt3-\mathrm{i})\;. $$
2001 Paper 2 Q8
The function \(\f\) satisfies \(\f(x+1)= \f(x)\) and \(\f(x)>0\) for all \(x\).
- Give an example of such a function.
- The function \(\F\) satisfies \[ \frac{\d \F}{\d x} =\f(x) \] and \(\F(0)=0\). Show that \(\F(n) = n\F(1)\), for any positive integer \(n\).
- Let \(y\) be the solution of the differential equation \[ \frac{\d y}{\d x} +\f(x) y=0 \] that satisfies \(y=1\) when \(x=0\). Show that \(y(n) \to 0\) as \(n\to\infty\), where \(n= 1,\,2,\, 3,\, \ldots\)
Solution:
- \(f(x) = \lfloor x \rfloor+1\)
- Clearly \(\displaystyle F(x) = \int_0^x f(t) \d t\), in particular: \begin{align*} && F(n) &= \int_0^n f(t) \d t \\ &&&= \sum_{i=1}^n \int_{i-1}^i f(t) \d t \\ &&&= \sum_{i=1}^n \int_{0}^1 f(t-i+1) \d t \\ &&&= \sum_{i=1}^n \int_{0}^1 f(t) \d t \\ &&&= n \int_{0}^1 f(t) \d t\\ &&&= n F(1) \end{align*}
- \(\,\) \begin{align*} && 0 &= \frac{\d y}{\d x} +f(x) y \\ \Rightarrow && \int -f(x) \d x &= \int \frac1y \d y\\ \Rightarrow && -F(x) & = \ln y + C \\ x=0,y=1: && C &= -F(0) \\ \Rightarrow && y &= \exp(F(0)-F(x)) \end{align*} Well this \(F(0)-F(x)\) is equivalent to \(-F(x)\) where \(F(0) = 0\), in particular \(F(n) = nF(1)\), so \(y(n) = e^{-nF(1)}\) which tends to zero as long as \(F(1) > 0\), but since \(f(x) > 0\) for all \(x\) this must be true.
2001 Paper 2 Q9
A particle of unit mass is projected vertically upwards with speed \(u\). At height \(x\), while the particle is moving upwards, it is found to experience a total force \(F\), due to gravity and air resistance, given by \(F=\alpha \e^{-\beta x}\), where \(\alpha\) and \(\beta\) are positive constants. Calculate the energy expended in reaching this height. Show that \[ F= {\textstyle \frac12} \beta v^2+ \alpha - {\textstyle \frac12} \beta u^2 \;, \] where \(v\) is the speed of the particle, and explain why \( \alpha = \frac12 \beta u^2 +g\), where \(g\) is the acceleration due to gravity. Determine an expression, in terms of \(y\), \(g\) and \(\beta\), for the air resistance experienced by the particle on its downward journey when it is at a distance \(y\) below its highest point.
Solution: Considering the energy of the particle, we have initial kinetic energy of \(\frac12 u^2\) and final energy is \(\frac12 v^2\), the change in energy is the work done by the force, \begin{align*} &&\text{Work done against resistance} &= \text{loss in kinetic energy} \\ &&\int F \, \d x &= \int \alpha e^{-\beta x} \, \d x \\ &&&= \frac{\alpha}{\beta} \l 1 - e^{-\beta x} \r \\ &&&= \frac{1}{\beta} \l \alpha - F\r \\ &&&= \frac12 u^2 - \frac12 v^2 \\ \Rightarrow && F &= \frac12 \beta v^2 + \alpha - \frac12 \beta u^2 \end{align*} When \(v = 0\) there is no air resistance, ie \(F = g\), but \(g = 0 + \alpha - \frac12 \beta u^2 \Rightarrow \alpha = g + \frac12 \beta u^2\) \(F = \frac12 \beta v^2 + g\), ie air resistance is \(\frac12 \beta v^2\) Looking at forces acting on the particle when it's descending, \begin{align*} && v \frac{dv}{dx} &= g - \frac12 \beta v^2 \\ \Rightarrow && \frac{v}{g - \frac12 \beta v^2} \frac{dv}{dx} &= 1 \\ \Rightarrow && \int \frac{v}{g - \frac12 \beta v^2} \, dv &= \int dx \\ \Rightarrow && \frac1{\beta}\l\ln(g - \frac12\beta v^2) - \ln(g)\r &= y\\ \Rightarrow && \ln \l 1 - \frac12 \frac{\beta}{g}v^2 \r &= \beta y \\ \Rightarrow && \frac{g}{\beta} \l 1-e^{-\beta y} \r = \frac12 v^2 \end{align*} Since force is the rate of change of work, we can say that the force is \(ge^{-\beta y}\) and the air resistance is \(g \l 1-e^{-\beta y} \r\)
2001 Paper 2 Q10
Two particles \(A\) and \(B\) of masses \(m\) and \(km\), respectively, are at rest on a smooth horizontal surface. The direction of the line passing through \(A\) and \(B\) is perpendicular to a vertical wall which is on the other side of \(B\) from \(A\). The particle \(A\) is now set in motion towards \(B\) with speed \(u\). The coefficient of restitution between \(A\) and \(B\) is \(e_1\) and between \(B\) and the wall is \(e_2\). Show that there will be a second collision between \(A\) and \(B\) provided $$ k< \frac {1+e_2(1+e_1)} {e_1}\;. $$ Show that, if \(e_1=\frac13\), \(e_2=\frac12\) and \(k<5\), then the kinetic energy of \(A\) and \(B\) immediately after \(B\) rebounds from the wall is greater than \(mu^2/27\).
Solution: First collision:
2001 Paper 2 Q11
A two-stage missile is projected from a point \(A\) on the ground with horizontal and vertical velocity components \(u\) and \(v\), respectively. When it reaches the highest point of its trajectory an internal explosion causes it to break up into two fragments. Immediately after this explosion one of these fragments, \(P\), begins to move vertically upwards with speed \(v_e\), but retains the previous horizontal velocity. Show that \(P\) will hit the ground at a distance \(R\) from \(A\) given by $$ \frac{gR}u = v+v_e + \sqrt{v_e^2 +v^2}\, . $$ It is required that the range \(R\) should be greater than a certain distance \(D\) (where \(D> 2uv/g\)). Show that this requirement is satisfied if \[ v_e> \frac{gD}{2u}\left( \frac{gD-2uv}{gD-uv}\right). \] \noindent[{\sl The effect of air resistance is to be neglected.}]
2001 Paper 2 Q12
The national lottery of Ruritania is based on the positive integers from \(1\) to \(N\), where \(N\) is very large and fixed. Tickets cost \(\pounds1\) each. For each ticket purchased, the punter (i.e. the purchaser) chooses a number from \(1\) to \(N\). The winning number is chosen at random, and the jackpot is shared equally amongst those punters who chose the winning number. A syndicate decides to buy \(N\) tickets, choosing every number once to be sure of winning a share of the jackpot. The total number of tickets purchased in this draw is \(3.8N\) and the jackpot is \(\pounds W\). Assuming that the non-syndicate punters choose their numbers independently and at random, find the most probable number of winning tickets and show that the expected net loss of the syndicate is approximately \[ N\; - \; %\textstyle{ \frac{5 \big(1- e^{-2.8}\big)}{14} \;W\;. \]
2001 Paper 2 Q13
The life times of a large batch of electric light bulbs are independently and identically distributed. The probability that the life time, \(T\) hours, of a given light bulb is greater than \(t\) hours is given by \[ \P(T>t) \; = \; \frac{1}{(1+kt)^\alpha}\;, \] where \(\alpha\) and \(k\) are constants, and \(\alpha >1\). Find the median \(M\) and the mean \(m\) of \(T\) in terms of \(\alpha\) and \(k\). Nine randomly selected bulbs are switched on simultaneously and are left until all have failed. The fifth failure occurs at 1000 hours and the mean life time of all the bulbs is found to be 2400 hours. Show that \(\alpha\approx2\) and find the approximate value of \(k\). Hence estimate the probability that, if a randomly selected bulb is found to last \(M\) hours, it will last a further \(m-M\) hours.
2001 Paper 2 Q14
Two coins \(A\) and \(B\) are tossed together. \(A\) has probability \(p\) of showing a head, and \(B\) has probability \(2p\), independent of \(A\), of showing a head, where \(0 < p < \frac12\). The random variable \(X\) takes the value 1 if \(A\) shows a head and it takes the value \(0\) if \(A\) shows a tail. The random variable \(Y\) takes the value 1 if \(B\) shows a head and it takes the value \(0\) if \(B\) shows a tail. The random variable \(T\) is defined by \[ T= \lambda X + {\textstyle\frac12} (1-\lambda)Y. \] Show that \(\E(T)=p\) and find an expression for \(\var(T)\) in terms of \(p\) and \(\lambda\). Show that as \(\lambda\) varies, the minimum of \(\var(T)\) occurs when \[ \lambda =\frac{1-2p}{3-4p}\;. \] The two coins are tossed \(n\) times, where \(n>30\), and \(\overline{T}\) is the mean value of \(T\). Let \(b\) be a fixed positive number. Show that the maximum value of \(\P\big(\vert \overline{T}-p\vert < b\big)\) as \(\lambda\) varies is approximately \(2\Phi(b/s)-1\), where \(\Phi\) is the cumulative distribution function of a standard normal variate and \[ s^2= \frac{p(1-p)(1-2p)}{(3-4p)n}\;. \]
Solution: \begin{align*} && \E[T] &= \E[\lambda X + \tfrac12(1-\lambda)Y] \\ &&&= \lambda \E[X] + \tfrac12(1-\lambda) \E[Y] \\ &&&= \lambda p + \tfrac12 (1-\lambda) 2p \\ &&&= p \\ \\ && \var[T] &= \var[\lambda X + \tfrac12(1-\lambda)Y] \\ &&&= \lambda^2 \var[X] + \tfrac14(1-\lambda)^2 \var[Y] \\ &&&= \lambda^2 p(1-p) + \tfrac14(1-\lambda)^22p(1-2p) \\ &&&= p(\lambda^2 + \tfrac12(1-\lambda)^2) -p^2(\lambda^2+(1-\lambda)^2)\\ &&&= p(\tfrac32\lambda^2 - \lambda + \tfrac12) -p^2(2\lambda^2 -2\lambda + 2) \end{align*} Differentiating \(\var[T]\) with respect to \(\lambda\), and noting it is a quadratic with positive leading coefficient, we get \begin{align*} && \frac{\d \var[T]}{\d \lambda} &= p(2\lambda -(1-\lambda)) - p^2(2 \lambda -2(1-\lambda)) \\ &&&= p(3\lambda - 1)-p^2(4\lambda - 2) \\ \Rightarrow && \lambda(4p-3) &= 2p-1 \\ \Rightarrow && \lambda &= \frac{1-2p}{3-4p} \end{align*} By the central limit theorem \(\overline{T} \sim N(p, \frac{\sigma^2}{n})\) in particular, \(\mathbb{P}(|\overline{T} - p| < b) = \mathbb{P}(\left \lvert |\frac{\overline{T}-p}{\frac{\sigma}{\sqrt{n}}} \right \lvert < \frac{b}{\frac{\sigma}{\sqrt{n}}}) = \mathbb{P}(|Z| < \frac{b\sqrt{n}}{\sigma}) = 2\Phi(b/s) - 1\) where \(s = \frac{\sigma}{\sqrt{n}}\) so \begin{align*} && s^2 &= \frac1n \sigma^2 \\ &&&= \frac1n \left ( \left (\left ( \frac{1-2p}{3-4p} \right)^2 + \tfrac12 \left (1-\frac{1-2p}{3-4p} \right)^2 \right)p - \left ( \left ( \frac{1-2p}{3-4p} \right)^2 + \left (1-\frac{1-2p}{3-4p} \right)^2\right)p^2 \right) \\ &&&= \frac1n \left ( \left (\left ( \frac{1-2p}{3-4p} \right)^2 + \tfrac12 \left (\frac{2-2p}{3-4p} \right)^2 \right)p - \left ( \left ( \frac{1-2p}{3-4p} \right)^2 + \left (\frac{2-2p}{3-4p} \right)^2\right)p^2 \right) \\ &&&= \frac{p}{n(3-4p)^2} \left ( (1 -4p + 4p^2 + 2-4p+2p^2) - (1-4p+4p^2+4-8p+4p^2)p \right) \\ &&&= \frac{p}{n(3-4p)^2} \left (3-13p+18p^2-8p^3 \right) \\ &&&= \frac{p}{n(3-4p)^2} (3-4p)(1-2p)(1-p) \\ &&&= \frac{p(1-p)(1-2p)}{(3-4p)n} \end{align*}
2001 Paper 3 Q1
Given that \(y = \ln ( x + \sqrt{x^2 + 1})\), show that \( \displaystyle \frac{\d y}{\d x} = \frac1 {\sqrt{x^2 + 1} }\;\). Prove by induction that, for \(n \ge 0\,\), \[ \l x^2 + 1 \r y^{\l n + 2 \r} + \l 2n + 1 \r x y^{\l n + 1 \r} + n^2 y^{\l n \r} = 0\;, \] where \(\displaystyle y^{(n)} = \frac{\d^n y}{\d x^n}\) and \(y^{(0)} =y\,\). Using this result in the case \(x = 0\,\), or otherwise, show that the Maclaurin series for \(y\) begins \[ x - {x^3 \over 6} +{3 x^5 \over 40} \] and find the next non-zero term.
Solution: \begin{align*} && y & = \ln ( x + \sqrt{x^2+1}) \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{1}{x+\sqrt{x^2+1}} \cdot \frac{\d }{\d x} \left ( x + \sqrt{x^2+1} \right) \\ &&&= \frac{1}{x+\sqrt{x^2+1}} \left (1 + \frac12 \frac{2x}{\sqrt{x^2+1}} \right) \\ &&&= \frac{1}{x+\sqrt{x^2+1}} \left ( \frac{\sqrt{x^2+1} + x}{\sqrt{x^2+1}}\right) \\ &&&= \frac{1}{\sqrt{x^2+1}} \end{align*} Note that \(\displaystyle y^{(2)} = - \frac12 \frac{2x}{(x^2+1)^{3/2}} = - \frac{x}{(x^2+1)^{3/2}}\), and in particular \((x^2+1)y^{(2)} + xy^{(1)} = 0\). Now applying Leibnitz formula: \begin{align*} 0 &= \left ( (x^2+1)y^{(2)} + xy^{(1)} \right )^{(n)} \\ &= \left ( (x^2+1)y^{(2)}\right )^{(n)} + \left (xy^{(1)} \right )^{(n)} \\ &= (x^2+1)y^{(n+2)} +n2xy^{(n+1)} + \binom{n}{2}2y^{(n)} + xy^{(n+1)} + n y^{(n)} \\ &= (x^2+1)y^{(n+2)} + (2n+1)xy^{(n+1)} + (n^2-n+n)y^{(n)} \\ &= (x^2+1)y^{(n+2)} + (2n+1)xy^{(n+1)} + n^2y^{(n)} \end{align*} as required. When \(x = 0\): \begin{align*} && y(0) &= \ln(0 + \sqrt{0^2+1}) \\ &&&= \ln 1 = 0 \\ && y'(0) &= \frac{1}{\sqrt{0^2+1}} = 1 \\ && y^{(n+2)} &= -n^2 y^{(n)} \\ && y^{(2k)} &= 0 \\ && y^{(3)} &= -1 \\ && y^{(5)} &= 3^2 \\ && y^{(7)} &= - 5^2 \cdot 3^2 \\ \end{align*} Therefore the Maclaurin series about \(x = 0\) is \begin{align*} y &= x - \frac{1}{3!} x^3 + \frac{3^2}{5!} x^5 - \frac{3^2 \cdot 5^2}{7!} x^7 + \cdots \\ &= x - \frac{1}{6} x^3 + \frac{3}{1 \cdot 2 \cdot 4 \cdot 5} x^5 - \frac{5}{1 \cdot 2 \cdot 4 \cdot 2 \cdot 7} x^7 + \cdots \\ &= x - \frac{1}{6}x^3 + \frac{3}{40} x^5 - \frac{5}{56} x^7 + \cdots \end{align*}
2001 Paper 3 Q2
Show that \( \cosh^{-1} x = \ln ( x + \sqrt{x^2-1})\). Show that the area of the region defined by the inequalities \(\displaystyle y^2 \ge x^2-8\) and \(\displaystyle x^2\ge 25y^2 -16 \) is \((72/5) \ln 2\).
Solution: \begin{align*} && x &= \cosh y \\ \Rightarrow && x &= \tfrac12 (e^y + e^{-y} ) \\ \Rightarrow && 0 &= e^{2y} - 2xe^y + 1 \\ \Rightarrow && e^y &= \frac{2x \pm \sqrt{4x^2-4}}{2} \\ &&&= x \pm \sqrt{x^2-1} \\ \Rightarrow &&e^y &= x + \sqrt{x^2-1} \tag{by convention \(\cosh^{-1} > 0\)} \\ \Rightarrow && y &= \ln (x + \sqrt{x^2-1}) \end{align*}
