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1996 Paper 1 Q2
D: 1484.0 B: 1500.0

  1. Show that \[ \int_{0}^{1}\left(1+(\alpha-1)x\right)^{n}\,\mathrm{d}x=\frac{\alpha^{n+1}-1}{(n+1)(\alpha-1)} \] when \(\alpha\neq1\) and \(n\) is a positive integer.
  2. Show that if \(0\leqslant k\leqslant n\) then the coefficient of \(\alpha^{k}\) in the polynomial \[ \int_{0}^{1}\left(\alpha x+(1-x)\right)^{n}\,\mathrm{d}x \] is \[ \binom{n}{k}\int_{0}^{1}x^{k}(1-x)^{n-k}\,\mathrm{d}x\,. \]
  3. Hence, or otherwise, show that \[ \int_{0}^{1}x^{k}(1-x)^{n-k}\,\mathrm{d}x=\frac{k!(n-k)!}{(n+1)!}\,. \]

Solution:

  1. \begin{align*} u = 1+(\alpha-1)x: && \int_0^1 (1 + (\alpha - 1)x)^n \d x &= \int_{u=1}^{u=\alpha} u^n \frac{1}{\alpha - 1} \d u \\ &&&= \left [\frac{u^{n+1}}{(n+1)(\alpha-1)} \right]_1^\alpha \\ &&&= \frac{\alpha^{n+1}-1}{(n+1)(\alpha-1)} \end{align*}
  2. \begin{align*} && \int_0^1 (\alpha x + (1-x))^n \d x &= \int_0^1 \sum_{k=0}^n \binom{n}{k} \alpha^k x^k (1-x)^{n-k} \d x \\ &&&= \sum_{k=0}^n \alpha^k \int_0^1 \binom{n}{k} x^k (1-x)^{n-k} \d x \end{align*} Therefore the coefficient of \(\alpha^k\) is \(\displaystyle \int_0^1 \binom{n}{k} x^k (1-x)^{n-k} \d x\)
  3. The coefficient of \(\alpha^{k}\) in \(\displaystyle \frac{\alpha^{n+1}-1}{(n+1)(\alpha-1)}\) is \(\displaystyle \frac1{n+1}\). Therefore \begin{align*} && \frac1{n+1} &= \binom{n}{k} \int_0^1 x^k(1-x)^{n-k} \d x \\ \Rightarrow && \int_0^1 x^k (1-x)^{n-k} \d x &= \frac{k!(n-k)!}{(n+1)n!} \\ &&&= \frac{k!(n-k)!}{(n+1)!} \end{align*}

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1996 Paper 1 Q3
D: 1500.0 B: 1486.0

Let \(n\) be a positive integer.

  1. Factorise \(n^{5}-n^{3},\) and show that it is divisible by 24.
  2. Prove that \(2^{2n}-1\) is divisible by 3.
  3. If \(n-1\) is divisible by 3, show that \(n^{3}-1\) is divisible by 9.

Solution:

  1. \(n^5 -n^3 = n^3(n-1)(n+1)\). If \(n\) is even then \(8 \mid n^3\). if \(n\) is odd then both of \(n-1\) and \(n+1\) are divisible by \(2\) and one is divisible by \(4\), so regardless \(8\) divides our expression. We can write \(n = 3k, 3k+1, 3k+2\) and in all cases our expression is divisible by \(3\). \(n = 3k \Rightarrow 3 \mid n\), \(n = 3k+1 \Rightarrow 3 \mid n-1\), \(n = 3k+2 \Rightarrow 3 \mid n+1\). Therefore \(3\) and \(8\) both divide our expression, and they are coprime so their product (24) divides our expression.
  2. \(2^{2n}-1 = (2^2-1) \cdot (1+2^2+\cdots + 2^{2n-2}) = 3 \cdot N\) therefore \(3\) divides our number.
  3. Suppose \(n-1 = 3k\) then \(n^3-1 = (3k+1)^3-1 = 27k^3 + 27k^2 + 9k\) which is clearly divisible by 9

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1996 Paper 1 Q4
D: 1484.0 B: 1500.0

Show that \[ \int_{0}^{1}\frac{1}{x^{2}+2ax+1}\,\mathrm{d}x=\begin{cases} \dfrac{1}{\sqrt{1-a^{2}}}\tan^{-1}\sqrt{\dfrac{1-a}{1+a}} & \text{ if }\left|a\right|<1,\\ \dfrac{1}{2\sqrt{a^{2}-1}}\ln |a+\sqrt{a^{2}-1}| & \text{ if }\left|a\right|>1. \end{cases} \]

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1996 Paper 1 Q5
D: 1484.0 B: 1500.0

  1. Find all rational numbers \(r\) and \(s\) which satisfy \[ (r+s\sqrt{3})^{2}=4-2\sqrt{3}. \]
  2. Find all real numbers \(p\) and \(q\) which satisfy \[ (p+q\mathrm{i})^{2}=(3-2\sqrt{3})+2(1-\sqrt{3})\mathrm{i}. \]
  3. Solve the equation \[ (1+\mathrm{i})z^{2}-2z+2\sqrt{3}-2=0, \] writing your solutions in as simple a form as possible.
{[}No credit will be given to answers involving use of calculators.{]}

Solution:

  1. Suppose \begin{align*} && 4 - 2\sqrt{3} &= (r+s\sqrt{3})^2 \\ &&&= r^2+3s^2+2sr \sqrt{3} \\ \Rightarrow && rs &= -1 \\ && r^2+3s^2 &= 4 \\ \Rightarrow && (r,s) &= (1,-1), (-1,1) \end{align*}
  2. \begin{align*} && (3-2\sqrt{3})+2(1-\sqrt{3})i &= (p+qi)^2 \\ &&&= p^2-q^2 + 2pq i \\ \Rightarrow && pq &= (1-\sqrt{3}) \\ && p^2 - q^2 &= 3-2\sqrt{3} \\ \Rightarrow &&3-2\sqrt{3} &= p^2 - \frac{(1-\sqrt{3})^2}{p^2} \\ \Rightarrow && 0 &= p^4-(3-2\sqrt{3})p^2-(4-2\sqrt{3}) \\ &&&= (p^2-(4-2\sqrt{3}))(p^2+1) \\ \Rightarrow && p &= \pm (1-\sqrt{3}) \\ && q &=\mp \frac12(1+\sqrt{3}) \end{align*}
  3. \begin{align*} && 0 &= (1+i)z^2 - 2z + 2(\sqrt{3}-1) \\ \Rightarrow && z &= \frac{2 \pm \sqrt{4-4(1+i)2(\sqrt{3}-1)}}{2(1+i)} \\ &&&= \frac{1 \pm \sqrt{1-(1+i)2(\sqrt{3}-1)}}{1+i} \\ &&&= \frac{1 \pm \sqrt{(3-2\sqrt{3})+(2-2\sqrt{3})i}}{1+i} \\ &&&= \frac{1 \pm (1 - \sqrt{3}) \mp \frac12 (1+\sqrt{3})i}{1+i} \\ &&&= \frac{5-\sqrt{3}}{4} + \frac{3-3\sqrt{3}}{4}i, \\ &&& \frac{\sqrt{3}-1}{4} + \frac{1+3\sqrt{3}}{4}i \end{align*}

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1996 Paper 1 Q6
D: 1500.0 B: 1500.0

Let \(\mathrm{f}(x)=\dfrac{\sin(n+\frac{1}{2})x}{\sin\frac{1}{2}x}\) for \(0 < x\leqslant\pi.\)

  1. Using the formula \[ 2\sin\tfrac{1}{2}x\cos kx=\sin(k+\tfrac{1}{2})x-\sin(k-\tfrac{1}{2})x \] (which you may assume), or otherwise, show that \[ \mathrm{f}(x)=1+2\sum_{k=1}^{n}\cos kx\,. \]
  2. Find \({\displaystyle \int_{0}^{\pi}\mathrm{f}(x)\,\mathrm{d}x}\) and \({\displaystyle \int_{0}^{\pi}\mathrm{f}(x)\cos x\,\mathrm{d}x}.\)

Solution:

  1. \(\,\) \begin{align*} && 2\sin \tfrac12 x \sum_{k=1}^n \cos kx &= \sum_{k=1}^n 2\sin \tfrac12 x \cos kx \\ &&&= \sum_{k=1}^n \left ( \sin(k+\tfrac12)x - \sin(k - \tfrac12)x \right) \\ &&&= \left ( \sin\tfrac32x - \sin\tfrac12x \right) + \\ &&&\quad \quad \left ( \sin\tfrac52x - \sin \tfrac32 x \right) + \\ &&&\quad \quad \quad +\cdots + \\ &&&\quad \quad \quad \quad +\left ( \sin(n+\tfrac12)x - \sin(n - \tfrac12)x \right) \\ &&&= \sin(n+\tfrac12)x - \sin \tfrac12 x \\ \Rightarrow && \sin(n+\tfrac12)x &= \sin \tfrac12 x + 2\sin \tfrac12 x \sum_{k=1}^n \cos kx \\ \Rightarrow && f(x) &= 1 + 2 \sum_{k=1}^n \cos kx \end{align*}
  2. \(\,\) \begin{align*} && \int_0^{\pi} f(x) \d x &= \int_0^{\pi} \left (1 + 2 \sum_{k=1}^n \cos kx \right) \d x \\ &&&= \pi + 2 \left [ \sum_{k=1}^n \frac{1}{k} \sin k x\right]_0^\pi \\ &&&= \pi \\ \\ && \int_0^{\pi} f(x) \cos x \d x &= \int_0^{\pi} \left (\cos x + 2 \sum_{k=1}^n \cos kx \cos x \right) \d x \\ &&&= 0 + \sum_{k=1}^n \left ( \int_0^{\pi} 2 \cos k x \cos x \d x \right) \\ &&&= \sum_{k=1}^n \left ( \int_0^{\pi} (\cos (k+1)x - \cos (k-1) x)\d x\right) \\ &&&= -\pi \end{align*}

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1996 Paper 1 Q7
D: 1484.0 B: 1469.7

  1. At time \(t=0\) a tank contains one unit of water. Water flows out of the tank at a rate proportional to the amount of water in the tank. The amount of water in the tank at time \(t\) is \(y\). Show that there is a constant \(b<1\) such that \(y=b^{t}.\)
  2. Suppose instead that the tank contains one unit of water at time \(t=0,\) but that in addition to water flowing out as described, water is added at a steady rate \(a>0.\) Show that \[ \frac{\mathrm{d}y}{\mathrm{d}t}-y\ln b=a, \] and hence find \(y\) in terms of \(a,b\) and \(t\).

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1996 Paper 1 Q8
D: 1500.0 B: 1500.0

  1. By using the formula for the sum of a geometric series, or otherwise, express the number \(0.38383838\ldots\) as a fraction in its lowest terms.
  2. Let \(x\) be a real number which has a recurring decimal expansion \[ x=0\cdot a_{1}a_{2}a_{2}\cdots, \] so that there exists positive integers \(N\) and \(k\) such that \(a_{n+k}=a_{n}\) for all \(n>N.\) Show that \[ x=\frac{b}{10^{N}}+\frac{c}{10^{N}(10^{k}-1)}\,, \] where \(b\) and \(c\) are integers to be found. Deduce that \(x\) is rational.

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1996 Paper 1 Q9
D: 1500.0 B: 1485.6

A bungee-jumper of mass \(m\) is attached by means of a light rope of natural length \(l\) and modulus of elasticity \(mg/k,\) where \(k\) is a constant, to a bridge over a ravine. She jumps from the bridge and falls vertically towards the ground. If she only just avoids hitting the ground, show that the height \(h\) of the bridge above the floor of the ravine satisfies \[ h^{2}-2hl(k+1)+l^{2}=0, \] and hence find \(h.\) Show that the maximum speed \(v\) which she attains during her fall satisfies \[ v^{2}=(k+2)gl. \]

Solution: \begin{align*} && \text{Energy at the top} &= mgh \\ && \text{Energy at the bottom} &= \frac12\frac{\lambda (h-l)^2}{l} \\ \Rightarrow && mgh & = \frac{\frac{mg}{k}(h-l)^2}{2l} \\ \Rightarrow && 2hkl &= (h-l)^2 \\ \Rightarrow && 0 &= h^2-2lh-2hlk+l^2 \\ &&0&= h^2-2hl(k+1)+l^2 \\ \Rightarrow && \frac{h}{l} &= \frac{2(k+1)\pm \sqrt{4(k+1)^2-4}}{2} \\ &&&= (k+1) \pm \sqrt{k^2+2k} \\ \Rightarrow && h &= l \left ( (k+1) \pm \sqrt{k^2+2k} \right) \end{align*} Since the negative root is less than \(1\), she would have not fully extended the cord. Therefore \(h = l \left ( (k+1) + \sqrt{k^2+2k} \right)\) Her maximum speed will be when her acceleration is \(0\), ie \(g = \text{force from cord}\) ie \(mg = \frac{\lambda x}{l}\) or \(x = \frac{mgl}{\lambda} = \frac{mglk}{mg} = kl\). At this point by conservation of energy we will have \begin{align*} && mgh &= mg(h-l-x) + \frac12 m v^2+\frac{1}{2} \frac{mgx^2}{kl} \\ \Rightarrow && mg\left ( l + kl \right) &= \frac12 m v^2 + \frac12 \frac{mgl^2k^2}{kl} \\ \Rightarrow && 2g\left ( l + kl \right) &= v^2 + glk \\ \Rightarrow && v^2 &= gl(2+k) \end{align*}

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1996 Paper 1 Q10
D: 1500.0 B: 1516.0

A spaceship of mass \(M\) is at rest. It separates into two parts in an explosion in which the total kinetic energy released is \(E\). Immediately after the explosion the two parts have masses \(m_{1}\) and \(m_{2}\) and speeds \(v_{1}\) and \(v_{2}\) respectively. Show that the minimum possible relative speed \(v_{1}+v_{2}\) of the two parts of the spaceship after the explosion is \((8E/M)^{1/2}.\)

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1996 Paper 1 Q11
D: 1484.0 B: 1484.0

A particle is projected under the influence of gravity from a point \(O\) on a level plane in such a way that, when its horizontal distance from \(O\) is \(c\), its height is \(h\). It then lands on the plane at a distance \(c+d\) from \(O\). Show that the angle of projection \(\alpha\) satisfies \[ \tan\alpha=\frac{h(c+d)}{cd} \] and that the speed of projection \(v\) satisfies \[ v^{2}=\frac{g}{2}\left(\frac{cd}{h}+\frac{(c+d)^{2}h}{cd}\right)\,. \]

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