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1995 Paper 2 Q1
D: 1600.0 B: 1484.0

  1. By considering \((1+x+x^{2}+\cdots+x^{n})(1-x)\) show that, if \(x\neq1\), \[ 1+x+x^{2}+\cdots+x^{n}=\frac{1-x^{n+1}}{1-x}. \]
  2. By differentiating both sides and setting \(x=-1\) show that \[ 1-2+3-4+\cdots+(-1)^{n-1}n \] takes the value \(-n/2\) is \(n\) is even and the value \((n+1)/2\) if \(n\) is odd.
  3. Show that \[ 1^{2}-2^{2}+3^{2}-4^{2}+\cdots+(-1)^{n-1}n^{2}=(-1)^{n-1}(An^{2}+Bn) \] where the constants \(A\) and \(B\) are to be determined.

Solution:

  1. \begin{align*} && (1+x+x^{2}+\cdots+x^{n})(1-x) &= 1-x+x-x^2+\cdots -x^n+x^n-x^{n+1} \\ &&&= 1-x^{n+1} \\ \Rightarrow && 1+x+x^2+\cdots+x^n &= \frac{1-x^{n+1}}{1-x} \tag{dividing by \(1-x\)} \end{align*}
  2. \begin{align*} \frac{\d}{\d x}: && 0+1+2x+\cdots+nx^{n-1} &= \frac{(n+1)(1-x)x^n+(1-x^{n+1})}{(1-x)^2} \\ \Rightarrow && 1-2x+\cdots+(-1)^n n &= \frac{-(n+1)2(-1)^n+(1-(-1)^{n+1})}{4} \\ &&&= \begin{cases} \frac{-(n+1)\cdot2\cdot1+(1-(-1)}{4} & \text{if }n\text{ is even} \\ \frac{-(n+1)\cdot 2 \cdot(-1)+(1-1)}{4} & \text{if }n\text{ is odd}\end{cases} \\ &&&= \begin{cases} \frac{-n}{2} & \text{if }n\text{ is even}\\ \frac{n+1}{2} & \text{if }n\text{ is odd}\end{cases} \\ \end{align*}
  3. \begin{align*} x: && x+2x^2+\cdots+nx^{n} &= \frac{(n+1)(1-x)x^{n+1}+x(1-x^{n+1})}{(1-x)^2} \\ &&&= \frac{x+(n+1)x^{n+1}-nx^{n+2}}{(1-x)^2}\\ \frac{\d}{\d x}: && 1^2+2^2x + \cdots + n^2x^{n-1} &= \frac{(1-x)^2(1+(n+1)^2x^{n}-n(n+2)x^{n+1}) +2(1-x)(x+(n+1)x^{n+1}-nx^{n+2})}{(1-x)^4} \\ &&&= \frac{1 + x - (1 + n)^2 x^n + (2 n^2+2n-1) x^{n+1} - n^2 x^{n+2}}{(1-x)^3} \\ \Rightarrow && 1^2-2^2 + \cdots + (-1)^{n-1}n^2 &= \frac{(-1)^n \l - (1 + n)^2- (2 n^2+2n-1) - n^2 \r}{8} \\ &&&= \frac{(-1)^n(-4n^2-4n}{8} \\ &&&= \frac{(-1)^{n-1}(n^2+n)}{2} \end{align*}

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1995 Paper 2 Q2
D: 1600.0 B: 1516.0

I have \(n\) fence posts placed in a line and, as part of my spouse's birthday celebrations, I wish to paint them using three different colours red, white and blue in such a way that no adjacent fence posts have the same colours. (This allows the possibility of using fewer than three colours as well as exactly three.) Let \(r_{n}\) be the number of ways (possibly zero) that I can paint them if I paint the first and the last post red and let \(s_{n}\) be the number of ways that I can paint them if I paint the first post red but the last post either of the other two colours. Explain why \(r_{n+1}=s_{n}\) and find \(r_{n}+s_{n}.\) Hence find the value of \(r_{n+1}+r_{n}\) for all \(n\geqslant1.\) Prove, by induction, that \[ r_{n}=\frac{2^{n-1}+2(-1)^{n-1}}{3}. \] Find the number of ways of painting \(n\) fence posts (where \(n\geqslant3\)) placed in a circle using three different colours in such a way that no adjacent fence posts have the same colours.

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1995 Paper 2 Q3
D: 1600.0 B: 1500.0

The Tour de Clochemerle is not yet as big as the rival Tour de France. This year there were five riders, Arouet, Barthes, Camus, Diderot and Eluard, who took part in five stages. The winner of each stage got 5 points, the runner up 4 points and so on down to the last rider who got 1 point. The total number of points acquired over the five states was the rider's score. Each rider obtained a different score overall and the riders finished the whole tour in alphabetical order with Arouet gaining a magnificent 24 points. Camus showed consistency by gaining the same position in four of the five stages and Eluard's rather dismal performance was relieved by a third place in the fourth stage and first place in the final stage. Explain why Eluard must have received 11 points in all and find the scores obtained by Barthes, Camus and Diderot. Where did Barthes come in the final stage?

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1995 Paper 2 Q4
D: 1600.0 B: 1504.3

Let \[ u_{n}=\int_{0}^{\frac{1}{2}\pi}\sin^{n}t\,\mathrm{d}t \] for each integer \(n\geqslant0\). By integrating \[ \int_{0}^{\frac{1}{2}\pi}\sin t\sin^{n-1}t\,\mathrm{d}t \] by parts, or otherwise, obtain a formula connecting \(u_{n}\) and \(u_{n-2}\) when \(n\geqslant2\) and deduce that \[ nu_{n}u_{n-1}=\left(n-1\right)u_{n-1}u_{n-2} \] for all \(n\geqslant2\). Deduce that \[ nu_{n}u_{n-1}=\tfrac{1}{2}\pi. \] Sketch graphs of \(\sin^{n}t\) and \(\sin^{n-1}t\), for \(0\leqslant t\leqslant\frac{1}{2}\pi,\) on the same diagram and explain why \(0 < u_{n} < u_{n-1}.\) By using the result of the previous paragraph show that \[ nu_{n}^{2} < \tfrac{1}{2}\pi < nu_{n-1}^{2} \] for all \(n\geqslant1\). Hence show that \[ \left(\frac{n}{n+1}\right)\tfrac{1}{2}\pi < nu_{n}^{2} < \tfrac{1}{2}\pi \] and deduce that \(nu_{n}^{2}\rightarrow\tfrac{1}{2}\pi\) as \(n\rightarrow\infty\).

Solution: \begin{align*} && u_n &= \int_0^{\tfrac12 \pi} \sin^{n} t \, \d t \\ && &= \int_0^{\tfrac12 \pi} \sin t \sin^{n-1} t \, \d t \\ && &= \left [ -\cos t \sin^{n-1} t \right]_0^{\tfrac12 \pi} + \int_0^{\tfrac12 \pi} \cos t (n-1) \sin^{n-2} t \cos t \d t \\ && &= 0 + (n-1)\int_0^{\tfrac12 \pi} \cos^2 t \sin^{n-2} t \d t \\ && &= (n-1) \int_0^{\tfrac12 \pi}(1-\sin^2 t) \sin^{n-2} t \d t \\ && &= (n-1)u_{n-2} - (n-1)u_n \\ \Rightarrow && n u_n &= (n-1)u_{n-2} \\ \end{align*} Mutplying both sides by \(u_{n-1}\) we obtain \(nu_{n}u_{n-1}=\left(n-1\right)u_{n-1}u_{n-2}\). Therefore \(nu_nu_{n-1}\) is constant, ie is equal to \(\displaystyle u_1u_0 = \int_0^{\tfrac12 \pi} \sin^{1} t \, \d t \int_0^{\tfrac12 \pi} \sin^{0} t \, \d t = 1 \cdot \frac{\pi}{2} = \frac{\pi}{2}\)

TikZ diagram
Since \(0 < \sin t < 1\) for \(t \in (0, \tfrac{\pi}{2})\) we must have \(0 < \sin^n t < \sin^{n-1} t\), in particular \(0 < u_n < u_{n-1}\) Therefore \begin{align*} && nu_{n}u_{n-1} &= \tfrac{1}{2}\pi \\ \Rightarrow && nu_n u_n &< \tfrac{1}{2}\pi \tag{\(u_n < u_{n-1}\)} \\ \Rightarrow && nu_{n-1} u_{n-1} &> \tfrac{1}{2}\pi \tag{\(u_n < u_{n-1}\)} \\ \Rightarrow && nu_n^2 &< \tfrac12 \pi < n u_{n-1}^2 \end{align*} However we also have \(\tfrac12 \pi < (n+1)u_n^2\) (by considering the next inequality), so \(\left ( \frac{n}{n+1}\right) \tfrac12 \pi < n u_n^2 < \tfrac12 \pi\) but since as \(n \to \infty\) the right hand bound is constant and the left hand bound tends to \(\tfrac12 \pi\) therefore \(n u_n^2 \to \tfrac12 \pi\)

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1995 Paper 2 Q5
D: 1600.0 B: 1545.6

The famous film star Birkhoff Maclane is sunning herself by the side of her enormous circular swimming pool (with centre \(O\)) at a point \(A\) on its circumference. She wants a drink from a small jug of iced tea placed at the diametrically opposite point \(B\). She has three choices:

  1. to swim directly to \(B\).
  2. to choose \(\theta\) with \(0<\theta<\pi,\) to run round the pool to a point \(X\) with \(\angle AOX=\theta\) and then to swim directly from \(X\) to \(B\).
  3. to run round the pool from \(A\) to \(B\).
She can run \(k\) times as fast as she can swim and she wishes to reach her tea as fast as possible. Explain, with reasons, which of (i), (ii) and (iii) she should choose for each value of \(k\). Is there one choice from (i), (ii) and (iii) she will never take whatever the value of \(k\)?

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1995 Paper 2 Q6
D: 1600.0 B: 1516.0

If \(u\) and \(v\) are the two roots of \(z^{2}+az+b=0,\) show that \(a=-u-v\) and \(b=uv.\) Let \(\alpha=\cos(2\pi/7)+\mathrm{i}\sin(2\pi/7).\) Show that \(\alpha\) is a root of \(z^{6}-1=0\) and express the roots in terms of \(\alpha.\) The number \(\alpha+\alpha^{2}+\alpha^{4}\) is a root of a quadratic equation \[ z^{2}+Az+B=0 \] where \(A\) and \(B\) are real. By guessing the other root, or otherwise, find the numerical values of \(A\) and \(B\). Show that \[ \cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{8\pi}{7}=-\frac{1}{2}, \] and evaluate \[ \sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}, \] making it clear how you determine the sign of your answer.

Solution: \begin{align*} 0 &= z^2+az+b \\ &= (z-u)(z-v) \\ &= z^2-(u+v)z+uv \end{align*} Therefore by comparing coefficients, \(a = -u-v\) and \(b = uv\). Suppose \(\alpha = \cos(2\pi/7) + i \sin (2\pi/7)\), then by De Moivre, \(\alpha^7 = \cos(2\pi) + i \sin (2\pi) = 1\), ie \(\alpha^7-1 = 0\). Notice that \((\alpha+\alpha^2 + \alpha^4) + (\alpha^3+\alpha^5+\alpha^6) = -1\) and \begin{align*} P &= (\alpha+\alpha^2 + \alpha^4)(\alpha^3+\alpha^5+\alpha^6) \\ &= \alpha^4 + \alpha^6 + \alpha^7 + \alpha^5 + \alpha^7 + \alpha^8 + \alpha^{7}+\alpha^{9}+\alpha^{10} \\ &= 3 + \alpha+ \alpha^2 + \alpha^3 + \alpha^4 + \alpha^5 + \alpha^6 \\ &= 2 \end{align*} Therefore it is a root of \(x^2+x+2 = 0 \Rightarrow x = \frac{-1 \pm i\sqrt{7}}{2}\) Therefore $\cos\frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{8\pi}{7} = \textrm{Re}(\alpha + \alpha^2 + \alpha^4) = -\frac12$ And \(\sin\frac{2\pi}{7} + \sin\frac{4\pi}{7} + \sin\frac{8\pi}{7} = \textrm{Im}(\alpha + \alpha^2 + \alpha^4) = \pm\frac{\sqrt{7}}2\) since it is positive it is \(\frac{\sqrt{7}}{2}\)

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1995 Paper 2 Q7
D: 1600.0 B: 1516.7

The diagram shows a circle, of radius \(r\) and centre \(I\), touching the three sides of a triangle \(ABC\). We write \(a\) for the length of \(BC\) and \(\alpha\) for the angle \(\angle BAC\) and so on. Let \(s=\frac{1}{2}\left(a+b+c\right)\) and let \(\triangle\) be the area of the triangle.

TikZ diagram
  1. By considering the area of the triangles \(AIB,\) \(BIC\) and \(CIA\), or otherwise, show that \(\Delta=rs\).
  2. By using the formula \(\Delta=\frac{1}{2}bc\sin\alpha\), show that \[ \Delta^{2}=\tfrac{1}{16}[4b^{2}c^{2}-\left(2bc\cos\alpha\right)^{2}]. \] Now use the formula \(a^{2}=b^{2}+c^{2}-2bc\cos\alpha\) to show that \[ \Delta^{2}=\tfrac{1}{16}[(a^{2}-\left(b-c\right)^{2})(\left(b+c\right)^{2}-a^{2})] \] and deduce that \[ \Delta=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}. \]
  3. A hole in the shape of the triangle \(ABC\) is cut in the top of a level table. A sphere of radius \(R\) rests in the hole. Find the height of the centre of the sphere above the level of the table top, expressing your answer in terms of \(a,b,c,s\) and \(R\).

Solution:

  1. \([AIB] = \frac12br\), \([BIC] = \frac12ar\), \([CIA] = \frac12 rc\), therefore \(\Delta = [AIB] +[BIC] + [CIA] = \frac12r(a+b+c) = sr\)
  2. \(\,\) \begin{align*} && \Delta &= \frac12 bc \sin \alpha \\ \Rightarrow && \Delta^2 &= \frac14 b^2c^2 \sin^2 \alpha \\ &&&= \frac14 \left (b^2c^2 - b^2c^2\cos^2 \alpha \right) \\ &&&= \frac1{16} \left (4b^2c^2 - (2bc\cos \alpha )^2\right) \\ \\ \Rightarrow && \Delta^2 &= \frac1{16} \left (4b^2c^2 - (b^2+c^2-a^2 )^2\right) \\ &&&= \frac1{16} (2bc-b^2-c^2+a^2)(2bc+b^2+c^2-a^2) \\ &&&= \frac{1}{16}(a^2-(b-c)^2)((b+c)^2-a^2) \\ &&&= \frac1{16}(a-b+c)(a+b-c)(b+c-a)(b+c+a) \\ &&&= (s - b)(s-c)(s-a)s \\ \Rightarrow && \Delta &= \sqrt{s(s-a)(s-b)(s-c)} \end{align*}
  3. We have the setting like this,
    TikZ diagram
    so \begin{align*} && h & = \sqrt{R^2-r^2} \\ &&&= \sqrt{R^2-\frac{\Delta^2}{s^2}} \\ &&&= \sqrt{R^2 - \frac{(s-a)(s-b)(s-c)}{s}} \end{align*}

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1995 Paper 2 Q8
D: 1600.0 B: 1500.8

If there are \(x\) micrograms of bacteria in a nutrient medium, the population of bacteria will grow at the rate \((2K-x)x\) micrograms per hour. Show that, if \(x=K\) when \(t=0\), the population at time \(t\) is given by \[ x(t)=K+K\frac{1-\mathrm{e}^{-2Kt}}{1+\mathrm{e}^{-2Kt}}. \] Sketch, for \(t\geqslant0\), the graph of \(x\) against \(t\). What happens to \(x(t)\) as \(t\rightarrow\infty\)? Now suppose that the situation is as described in the first paragraph, except that we remove the bacteria from the nutrient medium at a rate \(L\) micrograms per hour where \(K^{2}>L\). We set \(\alpha=\sqrt{K^{2}-L}.\) Write down the new differential equation for \(x\). By considering a new variable \(y=x-K+\alpha,\) or otherwise, show that, if \(x(0)=K\) then \(x(t)\rightarrow K+\alpha\) as \(t\rightarrow\infty\).

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1995 Paper 2 Q9
D: 1600.0 B: 1484.0

\noindent

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Two thin horizontal bars are parallel and fixed at a distance \(d\) apart, and the plane containing them is at an angle \(\alpha\) to the horizontal. A thin uniform rod rests in equilibrium in contact with the bars under one and above the other and perpendicular to both. The diagram shows the bards (in cross section and exaggerated in size) with the rod over one bar at \(Y\) and under the other at \(Z\). (Thus \(YZ\) has length \(d\).) The centre of the rod is at \(X\) and \(XZ\) has length \(l.\) The coefficient of friction between the rod and each bar is \(\mu.\) Explain why we must have \(l\leqslant d.\) Find, in terms of \(d,l\) and \(\alpha,\) the least possible value of \(\mu.\) Verify that, when \(l=2d,\) your result shows that \[ \mu\geqslant\tfrac{1}{3}\tan\alpha. \]

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1995 Paper 2 Q10
D: 1600.0 B: 1484.0

Three small spheres of masses \(m_{1},m_{2}\) and \(m_{3},\) move in a straight line on a smooth horizontal table. (Their order on the straight line is the order given.) The coefficient of restitution between any two spheres is \(e\). The first moves with velocity \(u\) towards the second whilst the second and third are at rest. After the first collision the second sphere hits the third after which the velocity of the second sphere is \(u.\) Find \(m_{1}\) in terms of \(m_{2},m_{3}\) and \(e\). deduce that \[ m_{2}e>m_{3}(1+e+e^{2}). \] Suppose that the relation between \(m_{1},m_{2}\) and \(m_{3}\) is that in the formula you found above, but that now the first sphere initially moves with velocity \(u\) and the other two spheres with velocity \(v\), all in the same direction along the line. If \(u>v>0\) use the first part to find the velocity of the second sphere after two collisions have taken place. (You should not need to make any substantial computations but you should state your argument clearly.)

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