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2004 Paper 2 Q7
D: 1600.0 B: 1516.0

The function f is defined by $$\f(x) = 2\sin x - x\,.$$ Show graphically that the equation \(\f(x)=0\) has exactly one root in the interval \([\frac12\pi ,\,{\pi}]\,\). This interval is denoted \(I_0\). In order to determine the root, a sequence of intervals \(I_1\), \(I_2, \,\ldots\) is generated in the following way. If the interval \(I_n=[a_n,b_n]\,\), and \(c_n=(a_n+b_n)/2\,\), then \begin{equation*} I_{n+1}= \begin{cases} [a_n,c_n] & \text{if \(\; \f(a_n)\f(c_n)<0 \,\)}; \\[5pt] [c_n,b_n] & \text{if \(\; \f(c_n)\f(b_n)<0 \,\)}. \end{cases} \end{equation*} By using the approximations \(\ds \frac 1{\sqrt{2}} \approx 0.7\) and \({\pi} \approx \sqrt{10} \,\), show that \(I_2=[\frac12{\pi},\,\frac58{\pi}]\) and find \(I_3\,\).

Solution: \begin{array}{c|c|c|c|c|c} n & a_n & b_n & c_n & f(a_n) & f(c_n) & f(b_n) \\ \hline 0 & \tfrac12 \pi & \pi & \tfrac34\pi & 2\sin(\tfrac12\pi)-\tfrac12\pi = 2-\tfrac12\pi & 2\sin(\tfrac34\pi)-\tfrac34\pi = \frac{2}{\sqrt{2}}-\tfrac34\pi & 2\sin(\pi)-\pi =-\pi \\ 0 & \tfrac12 \pi & \pi & \tfrac34\pi & >0 & 2-\frac{9}{16}10 < 0& <0 \\ \hline 1 & \frac12 \pi & \frac34\pi & \frac58\pi & >0 &2\sin \tfrac58\pi - \tfrac58\pi & < 0\\ 1 & \frac12 \pi & \frac34\pi & \frac58\pi & >0 & \approx 1.4 \cdot \sqrt{1.7} -\frac58\sqrt{10} < 0 & <0 \\ \hline 2 & \frac12 \pi & \frac58\pi & \frac9{16}\pi & >0 & 2\sin \frac{9}{16}\pi-\frac{9}{16}\pi & <0 \\ 2 & \frac12 \pi & \frac58\pi & \frac9{16}\pi & >0 & > 0 & <0 \\ \end{array} Threfore \(I_3 = [\frac9{16}\pi,\frac58\pi]\) \(\sin \frac{5\pi}{8} = \cos \frac{\pi}{8} = \sqrt{\frac12(\cos \frac{\pi}{4}+1)} = \frac{1}{\sqrt{2}}\sqrt{1 + \frac{1}{\sqrt{2}}} \approx 0.7 \cdot \sqrt{1.7}\) \(\sin \frac{9\pi}{16} = \cos \frac{\pi}{16} = \sqrt{\frac12\left ( \cos \frac{\pi}{8}+1 \right)} \) So we are comparing \(2\cos \frac{\pi}{16}\) with \(\frac{9}{16}\pi\) or \(4 \cos^2 \frac{\pi}{16} = 2\cos \frac{\pi}{8}+2\) with \(\frac{90}{16}\)

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2004 Paper 2 Q8
D: 1600.0 B: 1483.3

Let \(x\) satisfy the differential equation $$ \frac {\d x}{\d t} = {\big( 1-x^n\big)\vphantom{\Big)}}^{\!1/n} $$ and the condition \(x=0\) when \(t=0 \,\).

  1. Solve the equation in the case \(n=1\) and sketch the graph of the solution for \(t > 0 \,\).
  2. Prove that \(1-x < (1-x^2)^{1/2} \) for \(0 < x < 1 \,\). Use this result to sketch the graph of the solution in the case \(n=2\) for \(0 < t < \frac12 \pi \,\), using the same axes as your previous sketch. By setting \(x=\sin y\,\), solve the equation in this case.
  3. Use the result (which you need not prove) \[ (1-x^2)^{1/2} < (1-x^3)^{1/3} \text{ \ \ for \ \ } 0 < x < 1 \;, \] to sketch, without solving the equation, the graph of the solution of the equation in the case \(n=3\) using the same axes as your previous sketches. Use your sketch to show that \(x=1\) at a value of \(t\) less than \(\frac12 \pi \,\).

Solution:

  1. \(\,\) \begin{align*} && \dot{x} &= (1-x) \\ \Rightarrow &&\int \frac{1}{1-x} \d x &= \int \d t \\ \Rightarrow && -\ln |1-x| &= t + C \\ t=0, x = 0: && -\ln 1 &= C \Rightarrow C = 0\\ \Rightarrow && -\ln|1-x| &= t \\ \Rightarrow && 1-x&= e^{-t} \\ \Rightarrow && x &= 1-e^{-t} \end{align*}
    TikZ diagram
  2. Notice that \((1-x^2)^{1/2} = (1-x)^{1/2}(1+x)^{1/2} > (1-x)^{1/2} > 1-x\)
    TikZ diagram
    \begin{align*} && \dot{x} &= \sqrt{1-x^2} \\ \Rightarrow && \int \frac{1}{\sqrt{1-x^2}} \d x &= t + C \\ x = \sin y, \d x = \cos y && \int \frac{\cos y}{\cos y} \d y &= t + C \\ \Rightarrow && y &= t + C \\ \Rightarrow && \sin^{-1} x &= t + C \\ t = 0, x = 0: && x &= \sin t \end{align*}
  3. \(\,\)
    TikZ diagram
    We know the gradient is steeper, so the solution must always be above \(\sin t\), which means it reaches \(1\) before \(\frac{\pi}{2}\)

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2004 Paper 2 Q9
D: 1600.0 B: 1484.0

The base of a non-uniform solid hemisphere, of mass \(M,\) has radius \(r.\) The distance of the centre of gravity, \(G\), of the hemisphere from the base is \(p\) and from the centre of the base is \(\sqrt{p^2 + q^2} \,\). The hemisphere rests in equilibrium with its curved surface on a horizontal plane. A particle of mass \(m,\,\) where \(m\) is small, is attached to \(A\,\), the lowest point of the circumference of the base. In the new position of equilibrium, find the angle, \(\alpha\), that the base makes with the horizontal. The particle is removed and attached to the point \(B\) of the base which is at the other end of the diameter through \(A\,\). In the new position of equilibrium the base makes an angle \({\beta}\) with the horizontal. Show that $$\tan(\alpha-\beta)= \frac{2mMrp} {M^2\left(p^2+q^2\right)-m^2r^2}\;.$$

Solution:

TikZ diagram
In the coordinate system where \((0,0)\) is the centre base of the hemisphere, \(G\) is at \((p, q)\). Once the mass is attached at \(A\), the new centre of mass will satisfy: \(M \begin{pmatrix} p \\ q \end{pmatrix} + m \begin{pmatrix} r \\ 0 \end{pmatrix} = (M+m)\bar{x} \Rightarrow \bar{x} = \frac{1}{M+m} \begin{pmatrix} Mp+mr \\ Mq \end{pmatrix}\) The angle between the horizontal and \(AB\), \(\alpha\) will satisfy: $$\tan \alpha = \frac{Mp + mr}{Mq}$$ Similarly, when the mass is attached at \(B\), the new centre of mass will satisfy: \(M \begin{pmatrix} p \\ q \end{pmatrix} + m \begin{pmatrix} -r \\ 0 \end{pmatrix} = (M+m)\bar{x} \Rightarrow \bar{x} = \frac{1}{M+m} \begin{pmatrix} Mp-mr \\ Mq \end{pmatrix}\) The angle between the horizontal and \(AB\), \(\beta\) will satisfy: $$\tan \beta = \frac{Mp - mr}{Mq}$$ We are trying to find: \begin{align*} \tan \l \alpha - \beta \r &= \frac{\tan \alpha - \tan \beta}{1+ \tan \alpha \tan \beta} \\ &= \frac{\frac{Mp + mr}{Mq} - \frac{Mp - mr}{Mq}}{1 + \frac{Mp + mr}{Mq} \frac{Mp - mr}{Mq}} \\ &= \frac{(Mp + mr)Mq - (Mp - mr)Mq}{M^2q^2 + (Mp + mr)(Mp - mr)} \\ &= \frac{2Mmrp}{M^2(q^2+p^2) -m^2r^2} \\ \end{align*}

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2004 Paper 2 Q10
D: 1600.0 B: 1500.0

In this question take \(g = 10 ms^{-2}.\) The point \(A\) lies on a fixed rough plane inclined at \(30^{\circ}\) to the horizontal and \(\ell\) is the line of greatest slope through \(A\). A particle \(P\) is projected up \(\ell\) from \(A\) with initial speed \(6\)ms\(^{-1}\). A time \(T\) seconds later, a particle \(Q\) is projected from \(A\) up \(\ell\), also with speed \(6\)ms\(^{-1}\). The coefficient of friction between each particle and the plane is \(1/(5\sqrt{3})\,\) and the mass of each particle is \(4\)kg.

  1. Given that \(T<1+\sqrt{3/2}\), show that the particles collide at a time \((3-\sqrt6)T+1\) seconds after \(P\) is projected.
  2. In the case \(T=1+\sqrt{2/3}\,\), determine the energy lost due to friction from the instant at which \(P\) is projected to the time of the collision.

Solution: Since the particles are identical and are projected with the same speed, the only way they can reach the same point \(x\) at the same time, is if \(A\) has reached it's apex and started descending. Considering \(P\), we must have (setting the level of \(A\) to be the \(0\) G.P.E. level), suppose it travels a distance \(x\) before becoming stationary: \begin{align*} \text{N2}(\nwarrow): && R - 4g \cos(30^\circ) &= 0 \\ \Rightarrow && R &= 20\sqrt{3} \\ \Rightarrow && \mu R &= \frac1{5 \sqrt{3}} (20 \sqrt{3}) \\ &&&= 4 \\ \end{align*} Therefore in the two phases of the journey the particle is being accelerated down the slope by either \(6\) or \(4\). \(v^2 = u^2 + 2as \Rightarrow 0 = 36 - 12s \Rightarrow s = 3\). \(v = u + at \Rightarrow t = 1\). Therefore after \(1\) second \(P\) reaches its highest point having travelled \(3\) metres. It will pass back to the start in \(s = ut + \frac12 a t^2 \Rightarrow 3 = \frac12 4 t^2 \Rightarrow t = \sqrt{3/2}\) seconds, ie the constraint is that the particle hasn't already past \(Q\) before the collision. The collision will occur when \(s = 6t - \frac12 6 t^2\) and \(s =3 - \frac12 4 (t+T-1)^2\) coincide, ie: \begin{align*} && 6t - 3t^2 &= 3 - 2(t+T-1)^2 \\ && 0 &= 3 -2(T-1)^2 -(4(T-1)+6)t + t^2 \\ && 0 &= 3 -2(T-1)^2 -(4T+2)t + t^2 \\ \Rightarrow && t &= \frac{4T+2 \pm \sqrt{(4T+2)^2 - 4(3-2(T-1)^2)}}{2} \\ &&&= \frac{4T+2 \pm \sqrt{24T^2}}{2} \\ &&&= 2T + 1 \pm \sqrt{6} T \\ &&&= (2 \pm \sqrt{6})T + 1 \end{align*} we must take the smaller root, ie \((2-\sqrt{6})T + 1\). In the case the collision occurs exactly at the start, the particle \(P\) has traveled \(6\) meters, against a force of \(4\) newtons of friction, ie work done is \(24\) Joules.

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2004 Paper 2 Q11
D: 1600.0 B: 1500.0

The maximum power that can be developed by the engine of train \(A\), of mass \(m\), when travelling at speed \(v\) is \(Pv^{3/2}\,\), where \(P\) is a constant. The maximum power that can be developed by the engine of train \(B\), of mass \(2m\), when travelling at speed \(v\) is \(2Pv^{3/2}.\) For both \(A\) and \(B\) resistance to motion is equal to \(kv\), where \(k\) is a constant. For \(t\le0\), the engines are crawling along at very low equal speeds. At \(t = 0\,\), both drivers switch on full power and at time \(t\) the speeds of \(A\) and \(B\) are \(v_{\vphantom{\dot A}\!A}\) and \(v_{\vphantom{\dot B}\hspace{-1pt}B},\) respectively.

  1. Show that \[ v_{\vphantom{\dot A}\!A} = \frac{P^2 \left(1-\e^{-kt/2m}\right)^2}{k^2} \] and write down the corresponding result for \(v_{\vphantom{\dot B}B}\).
  2. Find \(v_{\vphantom{\dot B}A}\) and \(v_{\vphantom{\dot B}B}\) when \(9 v_{\vphantom{\dot B}A} =4v_{\vphantom{\dot B}B}\;\). %Show that %\(1 < v_{\vphantom{\dot B}\hspace{-1pt}B} /v_{\vphantom{\dot A}\!A} < 4\) for \(t>0\,\).
  3. Both engines are switched off when \(9 v_{\vphantom{\dot B}A} =4v_{\vphantom{\dot B}B}\,\). Show that thereafter \(k^2 v_{\vphantom{\dot B}B}^2 = 4 P^2 v_{\vphantom{\dot B}A}\,\).

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2004 Paper 2 Q12
D: 1600.0 B: 1516.0

Sketch the graph, for \(x \ge 0\,\), of $$ y = kx\e^{-ax^2} \;, $$ where \(a\) and \(k\) are positive constants. The random variable \(X\) has probability density function \(\f(x)\) given by \begin{equation*} \f(x)= \begin{cases} kx\e^{-ax^2} & \text{for \(0 \le x \le 1\)}\\[3pt] 0 & \text{otherwise}. \end{cases} \end{equation*} Show that \(\ds k=\frac{2a}{1-\e^{-a}}\) and find the mode \(m\) in terms of \(a\,\), distinguishing between the cases \(a < \frac12\) and \(a > \frac12\,\). Find the median \(h\) in terms of \(a\)\, and show that \(h > m\) if \(a > -\ln\left(2\e^{-1/2} - 1\right).\) Show that, \(-\ln\left(2\e^{-1/2}-1\right)> \frac12 \,\). Show also that, if \(a > -\ln\left(2\e^{-1/2} - 1\right) \,\), then $$ P(X > m \;\vert\; X < h) = {{2\e^{-1/2}-\e^{-a}-1} \over 1-\e^{-a}}\;. $$

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2004 Paper 2 Q13
D: 1600.0 B: 1500.0

A bag contains \(b\) balls, \(r\) of them red and the rest white. In a game the player must remove balls one at a time from the bag (without replacement). She may remove as many balls as she wishes, but if she removes any red ball, she loses and gets no reward at all. If she does not remove a red ball, she is rewarded with \pounds 1 for each white ball she has removed. If she removes \(n\) white balls on her first \(n\) draws, calculate her expected gain on the next draw and show that %her expected total reward would be the same as before it is zero if \(\ds n = {b-r \over r+1}\,\). Hence, or otherwise, show that she will maximise her expected total reward if she aims to remove \(n\) balls, where \[ n = \mbox{ the integer part of } \ds {b + 1 \over r + 1}\;. \] With this value of \(n\), show that in the case \(r=1\) and \(b\) even, her expected total reward is \(\pounds {1 \over 4}b\,\), and find her expected total reward in the case \(r=1\) and \(b\) odd.

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2004 Paper 2 Q14
D: 1600.0 B: 1484.0

Explain why, if \(\mathrm{A, B}\) and \(\mathrm{C}\) are three events, \[ \mathrm{P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) +P(A \cap B \cap C)}, \] where \(\mathrm{P(X)}\) denotes the probability of event \(\mathrm{X}\). A cook makes three plum puddings for Christmas. He stirs \(r\) silver sixpences thoroughly into the pudding mixture before dividing it into three equal portions. Find an expression for the probability that each pudding contains at least one sixpence. Show that the cook must stir 6 or more sixpences into the mixture if there is to be less than \({1 \over 3}\) chance that at least one of the puddings contains no sixpence. Given that the cook stirs 6 sixpences into the mixture and that each pudding contains at least one sixpence, find the probability that there are two sixpences in each pudding.

Solution:

TikZ diagram
When we add everything in \(A\),\(B\), \(C\) we overcount the overlaps. When we remove the overlaps we remove the centre section too many times, so we have to add it back on in the end. Let \(X_i\) be the probability that the \(i\)th pudding contains a sixpence. \begin{align*} && \mathbb{P}(X_1^c \cup X_2^c \cup X_3^c) &=\mathbb{P}(X_1^c \cap X_2^c \cap X_3^c) + \mathbb{P}(X_1^c)+\mathbb{P}(X_2^c)-\mathbb{P}(X_3^c)+\\ &&&\quad\quad-\mathbb{P}(X_1^c \cap X_2^c )-\mathbb{P}( X_2^c \cap X_3^c)-\mathbb{P}(X_1^c \cap X_3^c) \\ &&&= 0 + (\tfrac23)^r+ (\tfrac23)^r+ (\tfrac23)^r + \\ &&&\quad\quad - (\tfrac13)^r- (\tfrac13)^r- (\tfrac13)^r \\ &&&= \frac{3\cdot2^r-3}{3^{r}} \\ \Rightarrow && \mathbb{P}(\text{all contain a sixpence}) &= 1 - \frac{3\cdot2^r-3}{3^{r}} \\ &&&= \frac{3^r-3\cdot2^r+3}{3^r} \end{align*} When \(r = 5\) we have \(\frac{3 \cdot 32-3}{3^5} = \frac{31}{81} > \frac13\) When \(r = 6\) we have \(\frac{3 \cdot 64-3}{3^6} = \frac{7}{27} < \frac13\) Therefore, the chef must stir in at least \(6\). \begin{align*} && \mathbb{P}(\text{two in each}|\text{at least 1 in each}) &= \frac{ \mathbb{P}(\text{two in each} \cap \text{at least 1 in each}) }{ \mathbb{P}(\text{at least 1 in each}) } \\ &&&= \frac{ \mathbb{P}(\text{two in each}) }{ \mathbb{P}(\text{at least 1 in each}) } \\ &&&= \frac{90/3^6}{20/27} \\ &&&= \frac{1/3}{2} = \frac16 \end{align*}

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2004 Paper 3 Q1
D: 1700.0 B: 1603.9

Show that \[ \int_0^a \frac{\sinh x}{2\cosh^2 x -1} \, \mathrm{d} x = \frac{1}{2 \sqrt{2}} \ln \l \frac{\sqrt{2}\cosh a -1}{\sqrt{2}\cosh a +1}\r + \frac{1}{2 \sqrt{2}} \ln \l \frac{\sqrt{2}+1}{\sqrt{2}-1}\r \] and find \[ \int_0^a \frac{\cosh x}{1+2\sinh^2 x} \, \mathrm{d} x \, . \] Hence show that \[ \int_0^\infty \frac{\cosh x - \sinh x}{1+2\sinh^2 x} \, \mathrm{d} x = \frac{\pi}{2\sqrt{2}} - \frac{1}{2 \sqrt{2}} \ln \l \frac{\sqrt{2}+1}{\sqrt{2}-1}\r \, . \] By substituting \(u = \e^x\) in this result, or otherwise, find \[ \int_1^\infty \frac{1}{1+u^4} \, \mathrm{d} u \, . \]

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2004 Paper 3 Q2
D: 1700.0 B: 1516.0

The equation of a curve is \(y=\f ( x )\) where \[ \f ( x ) = x-4-\frac{16 \l 2x+1 \r^2}{x^2 \l x - 4 \r} \;. \]

  1. Write down the equations of the vertical and oblique asymptotes to the curve and show that the oblique asymptote is a tangent to the curve.
  2. Show that the equation \(\f ( x ) =0\) has a double root.
  3. Sketch the curve.

Solution:

  1. Clearly \(x = 0\) and \(x = 4\) are vertical asymptotes. Notice that \(\frac{16 \l 2x+1 \r^2}{x^2 \l x - 4 \r}\) tends to \(0\) as \(x \to \infty\). Therefore the oblique asymptote is \(y = x-4\).
  2. \begin{align*} && 0 &= \frac{x^2(x-4)^2-4^2(2x+1)^2}{x^2(x-4)} \\ &&&= \frac{(x(x-4)-4(2x+1))(x(x-4)+4(2x+1))}{x^2(x-4)} \\ &&&= \frac{(x^2-12x-4)(x^2+4x+4)}{x^2(x-4)}\\ &&&= \frac{(x^2-12x-4)(x+2)^2}{x^2(x-4)} \end{align*} Therefore \(f(x) = 0\) has a double root at \(x = -2\). Notice it also has roots at \(6 \pm 2\sqrt{10}\)
  3. TikZ diagram

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