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1997 Paper 3 Q6
D: 1700.0 B: 1516.0

Suppose that \(y_n\) satisfies the equations \[(1-x^2)\frac{{\rm d}^2y_n}{{\rm d}x^2}-x\frac{{\rm d}y_n}{{\rm d}x}+n^2y_n=0,\] \[y_n(1)=1,\quad y_n(x)=(-1)^ny_n(-x).\] If \(x=\cos\theta\), show that \[\frac{{\rm d}^2y_n}{{\rm d}\theta^2}+n^2y_n=0,\] and hence obtain \(y_n\) as a function of \(\theta\). Deduce that for \(|x|\leqslant1\) \[y_0=1,\quad y_1=x,\] \[y_{n+1}-2xy_n+y_{n-1}=0.\]

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1997 Paper 3 Q7
D: 1700.0 B: 1516.0

For each positive integer \(n\), let \begin{align*} a_n&=\frac1{n+1}+\frac1{(n+1)(n+2)}+\frac1{(n+1)(n+2)(n+3)}+\cdots;\\ b_n&=\frac1{n+1}+\frac1{(n+1)^2}+\frac1{(n+1)^3}+\cdots. \end{align*}

  1. Evaluate \(b_n\).
  2. Show that \(0
  3. Deduce that \(a_n=n!{\rm e}-[n!{\rm e}]\) (where \([x]\) is the integer part of \(x\)).
  4. Hence show that \(\mathrm{e}\) is irrational.

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1997 Paper 3 Q8
D: 1700.0 B: 1484.0

Let \(R_{\alpha}\) be the \(2\times2\) matrix that represents a rotation through the angle \(\alpha\) and let $$A=\begin{pmatrix}a&b\\b&c\end{pmatrix}.$$

  1. Find in terms of \(a\), \(b\) and \(c\) an angle \(\alpha\) such that \(R_{-\alpha}AR_{\alpha}\) is a diagonal matrix (i.e. has the value zero in top-right and bottom-left positions).
  2. Find values of \(a\), \(b\) and \(c\) such that the equation of the ellipse \[x^2+(y+2x\cot2\theta)^2=1\qquad(0 < \theta < \tfrac{1}{4}\pi)\] can be expressed in the form \[\begin{pmatrix}x&y\end{pmatrix}A\begin{pmatrix}x\\y\end{pmatrix}=1.\] Show that, for this \(A\), \(R_{-\alpha}AR_{\alpha}\) is diagonal if \(\alpha=\theta\). Express the non--zero elements of this matrix in terms of \(\theta\).
  3. Deduce, or show otherwise, that the minimum and maximum distances from the centre to the circumference of this ellipse are \(\tan\theta\) and \(\cot\theta\).

Solution: \begin{questionparts} \item \begin{align*} R_{-\alpha}AR_{\alpha} &= \begin{pmatrix} \cos \alpha & \sin\alpha \\ -\sin \alpha & \cos \alpha \end{pmatrix}\begin{pmatrix} a & b \\ b & c \end{pmatrix} \begin{pmatrix} \cos \alpha & -\sin\alpha \\ \sin \alpha & \cos \alpha \end{pmatrix} \\ &= \begin{pmatrix} \cos \alpha & \sin\alpha \\ -\sin \alpha & \cos \alpha \end{pmatrix} \begin{pmatrix} a\cos \alpha + b \sin \alpha & -a\sin\alpha + b \cos\alpha \\ b\cos\alpha + c \sin\alpha & c\cos\alpha-b\sin\alpha \end{pmatrix} \\ &= \begin{pmatrix} a\cos^2\alpha+2b\sin\alpha\cos\alpha+c\sin^2\alpha & -a\sin\alpha\cos \alpha+b\cos^2\alpha +c\sin\alpha\cos\alpha-b\sin^2 \alpha\\ (c-a)\sin\alpha\cos \alpha +b(\cos^2\alpha-\sin^2 \alpha) & a\sin^2 \alpha -2b\sin\alpha\cos\alpha+c\cos^2\alpha \end{pmatrix} \\ &= \begin{pmatrix} * & \frac{c-a}{2}\sin2\alpha+b \cos 2\alpha\\\frac{c-a}{2}\sin2\alpha+b \cos 2\alpha & * \end{pmatrix} \end{align*} Therefore this will be diagonal if \(\tan 2\alpha = \frac{2b}{a-c} \Rightarrow \alpha = \frac12 \tan^{-1} \l \frac{2b}{a-c} \r\) \item \begin{align*} x^2+(y+2x\cot2\theta)^2 &= x^2(1 + 4\cot^22\theta) + 4\cot2\theta xy + y^2 \\ &= \begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix} 1 + 4\cot^22\theta & 2\cot 2\theta \\ 2\cot 2\theta & 1 \end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} \end{align*} Plugging this \(\mathbf{A}\) in our result from before we discover \begin{align*} \frac12 \tan^{-1} \l \frac{2b}{a-c} \r &= \frac12 \tan^{-1} \l \frac{4\cot 2\theta}{1 + 4\cot^22\theta-1} \r \\ &= \frac12 \tan^{-1} \l \tan 2 \theta \r \\ &= \theta \end{align*} Therefore, the matrix will be: \begin{align*} & \textrm{diag}\begin{pmatrix} (1+4\cot^2 2\theta)\cos^2 \theta + 4\cot2\theta \sin\theta\cos\theta + \sin^2 \theta \\ (1+4\cot^2 2\theta)\sin^2 \theta - 4\cot2\theta \sin\theta\cos\theta + \cos^2 \theta \end{pmatrix} \\ =& \textrm{diag}\begin{pmatrix} \cos^2\theta + \frac{\cos^2 2\theta}{\sin^2 \theta} + 2\cos 2\theta + \sin^2 \theta \\ \sin^2\theta + \frac{\cos^2 2\theta}{\cos^2 \theta} - 2\cos 2\theta + \cos^2 \theta \end{pmatrix} \\ =& \textrm{diag}\begin{pmatrix} 1 + \cos 2\theta \l \frac{\cos2\theta}{\sin^2 \theta} + 2\r \\ 1 + \cos 2\theta \l \frac{\cos2\theta}{\cos^2 \theta} - 2\r \\ \end{pmatrix} \\ =& \textrm{diag}\begin{pmatrix} 1 + \cos 2\theta \l \frac{\cos^2 \theta + \sin^2 \theta}{\sin^2 \theta}\r \\ 1 -\cos 2\theta \l \frac{-\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta}\r \\ \end{pmatrix} \\ =& \textrm{diag}\begin{pmatrix} 1 + (\cos^2\theta - \sin^2 \theta) \cosec^2 \theta \\ 1 - (\cos^2\theta - \sin^2 \theta) \sec^2 \theta \\ \end{pmatrix} \\ =& \textrm{diag}\begin{pmatrix} \cot^2 \theta \\ \tan^2 \theta \\ \end{pmatrix} \\ \end{align*} Therefore this is a rotation of an ellipse with equation: \((\cot \theta x)^2 + (\tan \theta y)^2 = 1\), ie the shortest side and longest side are \(\cot \theta\) and \(\tan \theta\) respectively, but we know since \(0 < \theta < \tfrac{1}{4}\pi\) the shortest will be \(\tan \theta\) and the longest \(\cot \theta\).

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1997 Paper 3 Q9
D: 1700.0 B: 1500.0

A uniform rigid rod \(BC\) is suspended from a fixed point \(A\) by light stretched springs \(AB,AC\). The springs are of different natural lengths but the ratio of tension to extension is the same constant \(\kappa\) for each. The rod is not hanging vertically. Show that the ratio of the lengths of the stretched springs is equal to the ratio of the natural lengths of the unstretched springs.

Solution:

TikZ diagram
By moments or "centre of mass" or whatever argument you choose, the centre of mass is directly below \(A\). \begin{align*} N2:&& 0 &= \frac{1}{|AC|}\binom{-l\cos \theta}{h-l \sin \theta} T_{AC} + \frac{1}{|AB|} \binom{l \cos \theta}{h+l \sin \theta}T_{AB} + \binom{0}{-1}mg \\ \Rightarrow && \frac{T_{AC}}{AC} &= \frac{T_{AB}}{AB} \\ \Rightarrow && \frac{\kappa(AC-l_{AC})}{AC} &= \frac{\kappa(BC-l_{BC})}{BC} \\ \Rightarrow && \frac{l_{AC}}{AC} &= \frac{l_{BC}}{BC} \\ \Rightarrow && \frac{l_{AC}}{l_{BC}} &= \frac{AC}{BC} \end{align*}

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1997 Paper 3 Q10
D: 1700.0 B: 1484.0

By pressing a finger down on it, a uniform spherical marble of radius \(a\) is made to slide along a horizontal table top with an initial linear velocity \(v_0\) and an initial {\em backward} angular velocity \(\omega_0\) about the horizontal axis perpendicular to \(v_0\). The frictional force between the marble and the table is constant (independent of speed). For what value of \(v_0/(a\omega_0)\) does the marble

  1. slide to a complete stop,
  2. come to a stop and then roll back towards its initial position with linear speed \(v_0/7\).

Solution:

TikZ diagram
If the frictional force is \(F\), then: \begin{align*} L = I\ddot{\theta} && -Fa &= \frac{2}{5}ma^2 \dot{\omega} \\ \text{N2}(\rightarrow) && -F &= m\dot{v} \\ \\ \Rightarrow && \l \frac{2}{5}ma \dot{\omega} - \dot{v} \r &= 0 \\ \Rightarrow && \frac{2}{5}a \omega - v &= c \\ \end{align*}
  1. If the ball completely stops, then \(\omega = v = 0 \Rightarrow \frac{2}{5}a \omega_0 - v_0 = 0 \Rightarrow \frac{v_0}{a \omega_0} = \frac25\).
  2. If the ball rolls backwards with linear speed \(v_0/7\), \(v = - \frac{v_0}{7}\) and \(a \omega = \frac{v_0}{7}\), \begin{align*} && \frac{2}{5}a \omega_0 - v_0 &= \frac{2}{5} \frac{v_0}{7} + \frac{v_0}{7} \\ && &= \frac{1}{5} v_0 \\ \Rightarrow && \frac{v_0}{a \omega_0} &= \frac{1}{3} \end{align*}

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1997 Paper 3 Q11
D: 1700.0 B: 1500.0

\(\,\)

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A heavy symmetrical bell and clapper can both swung freely in a vertical plane about a point \(O\) on a horizontal beam at the apex of the bell. The mass of the bell is \(M\) and its moment of inertia about the beam is \(Mk^{2}\). Its centre of mass, \(G\), is a distance \(h\) from \(O\). The clapper may be regarded as a small heavy ball on a light rod of length \(l\). Initially the bell is held with its axis vertical and its mouth above the beam. The clapper ball rests against the side of the bell, with the rod making an angle \(\beta\) with the axis. The bell is then released. Show that, at the moment when the clapper and bell separate, the clapper rod makes an angle \(\alpha\) with the upwards vertical, where \[ \cot\alpha=\cot\beta-\frac{k^{2}}{hl}\mathrm{cosec}\beta. \]

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1997 Paper 3 Q12
D: 1700.0 B: 1500.0

  1. I toss a biased coin which has a probability \(p\) of landing heads and a probability \(q=1-p\) of landing tails. Let \(K\) be the number of tosses required to obtain the first head and let \[ \mathrm{G}(s)=\sum_{k=1}^{\infty}\mathrm{P}(K=k)s^{k}. \] Show that \[ \mathrm{G}(s)=\frac{ps}{1-qs} \] and hence find the expectation and variance of \(K\).
  2. I sample cards at random with replacement from a normal pack of \(52\). Let \(N\) be the total number of draws I make in order to sample every card at least once. By expressing \(N\) as a sum \(N=N_{1}+N_{2}+\cdots+N_{52}\) of random variables, or otherwise, find the expectation of \(N\). Estimate the numerical value of this expectation, using the approximations \(\mathrm{e}\approx2.7\) and \(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\approx0.5+\ln n\) if \(n\) is large.

Solution:

  1. Let \(N_i\) be the number of draws between the \((i-1)\)th new card and the \(i\)th new card. (Where \(N_1 = 1\)0 then \(N_i \sim K\) with \(p = \frac{53-i}{52}\)). Therefore \begin{align*} \E[N] &= \E[N_1 + \cdots + N_{52}] \\ &= \E[N_1] + \cdots + \E[N_i] + \cdots + \E[N_{52}] \\ &= 1 + \frac{52}{51} + \cdots + \frac{52}{53-k} + \cdots + \frac{52}{1} \\ &= 52 \left (1 + \frac{1}{2} + \cdots + \frac{1}{52} \right) \\ &= 52 \cdot \left ( 1 + \ln 52 \right) \end{align*} Notice that \(2.7 \times 2.7 = 7.29\) and \(7.3 \times 7.3 \approx 53.3\) so \(\ln 52 \approx 4\) and so our number is \(\approx 52 \cdot 4.5 =234\). [The correct answer actual number is 235.9782]

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1997 Paper 3 Q13
D: 1700.0 B: 1500.0

Let \(X\) and \(Y\) be independent standard normal random variables: the probability density function, \(\f\), of each is therefore given by \[ \f(x)=\left(2\pi\right)^{-\frac{1}{2}}\e^{-\frac{1}{2}x^{2}}. \]

  1. Find the moment generating function \(\mathrm{E}(\e^{\theta X})\) of \(X\).
  2. Find the moment generating function of \(aX+bY\) and hence obtain the condition on \(a\) and \(b\) which ensures that \(aX+bY\) has the same distribution as \(X\) and \(Y\).
  3. Let \(Z=\e^{\mu+\sigma X}\). Show that \[ \mathrm{E}(Z^{\theta})=\e^{\mu\theta+\frac{1}{2}\sigma^{2}\theta^{2}}, \] and hence find the expectation and variance of \(Z\).

Solution:

  1. \(\,\) \begin{align*} && \E[e^{\theta X}] &= \int_{-\infty}^{\infty} e^{\theta x} \frac{1}{\sqrt{2\pi}} e^{-\frac12 x^2 } \d x\\ &&&= \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac12 x^2+\theta x} \d x\\ &&&= \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac12 (x^2-2\theta x)} \d x\\ &&&= \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac12 (x-\theta )^2+\frac12\theta^2 } \d x\\ &&&= e^{\frac12\theta^2 }\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac12 (x-\theta )^2 } \d x\\ &&&=e^{\frac12\theta^2 } \end{align*}
  2. \begin{align*} && M_{aX+bY} (\theta) &= \mathbb{E}[e^{\theta (aX+bY)}] \\ &&&= e^{\frac12(a\theta)^2} \cdot e^{\frac12(b\theta)^2} \\ &&&= e^{\frac12(a^2+b^2)\theta^2} \end{align*} Therefore we need \(a^2+b^2 = 1\)
  3. \(\,\) \begin{align*} && \E[Z^\theta] &= \E[e^{\mu \theta + \sigma \theta X}] \\ &&&= e^{\mu \theta}e^{\frac12 \sigma^2 \theta^2} \\ &&&=e^{\mu \theta + \frac12 \sigma^2 \theta^2} \\ \end{align*} \begin{align*} \mathbb{E}(Z) &= \mathbb{E}[Z^1] \\ &= e^{\mu + \frac12 \sigma^2} \\ \var[Z] &= \E[Z^2] - \left ( \E[Z] \right)^2 \\ &= e^{2 \mu+ 2\sigma^2} - e^{2\mu + \sigma^2} \\ &= e^{2\mu+\sigma^2} \left (e^{\sigma^2}-1 \right) \end{align*} [NB: This is the lognormal distribution]

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1997 Paper 3 Q14
D: 1700.0 B: 1516.0

An industrial process produces rectangular plates of mean length \(\mu_{1}\) and mean breadth \(\mu_{2}\). The length and breadth vary independently with non-zero standard deviations \(\sigma_{1}\) and \(\sigma_{2}\) respectively. Find the means and standard deviations of the perimeter and of the area of the plates. Show that the perimeter and area are not independent.

Solution: Let \(L \sim N(\mu_1, \sigma_1^2)\), \(B \sim N(\mu_2, \sigma_2)^2\), so \begin{align*} && \mathbb{E}(\text{perimeter}) &= \E(2(L+B)) \\ &&&= 2\E[L]+2\E[B] \\ &&&= 2(\mu_1+\mu_2) \\ &&\var[\text{perimeter}] &= \E\left [ (2(L+B))^2 \right] - \left ( \E[2(L+B)] \right)^2 \\ &&&= 4\E[L^2+2LB+B^2] - 4(\mu_1+\mu_2)^2 \\ &&&= 4(\sigma_1^2+\mu_1^2+2\mu_1\mu_2+\sigma_2^2+\mu_2^2) - 4(\mu_1+\mu_2)^2\\ &&&= 4(\sigma_1^2+\sigma_2^2) \\ &&\text{sd}[\text{perimeter}] &= 2\sqrt{\sigma_1^2+\sigma_2^2} \\ \\ && \E[\text{area}] &= \E[LB] \\ &&&= \E[L]\E[B] \\ &&&= \mu_1\mu_2 \\ && \var[\text{area}] &= \E[(LB)^2] - \left (\E[LB] \right)^2 \\ &&&= \E[L^2]\E[B^2]-\mu_1^2\mu_2^2 \\ &&&= (\mu_1^2+\sigma_1^2)(\mu_2^2+\sigma_2^2) -\mu_1^2\mu_2^2 \\ &&&= \sigma_1^2\mu_2^2 + \sigma_2^2\mu_1^2 + \sigma_1^2\sigma_2^2\\ && \text{sd}(\text{area}) &= \sqrt{\sigma_1^2\mu_2^2 + \sigma_2^2\mu_1^2 + \sigma_1^2\sigma_2^2} \\ \\ && \E[\text{perimeter} \cdot \text{area}] &= \E[2(L+B)LB] \\ &&&= 2\E[L^2]\E[B] + 2\E[L]\E[B^2] \\ &&&= 2(\sigma_1^2+\mu_1^2)\mu_2 + 2(\sigma_2^2+\mu_2^2)\mu_1 \\ && \E[\text{perimeter}] \E[\text{area}] &= 2(\mu_1+\mu_2) \cdot \mu_1\mu_2 \end{align*} Since the latter does not depend on \(\sigma_i\) but the former does they cannot be equal in general, therefore they cannot be independent. [See also STEP 2006 Paper 3 Q14]

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1996 Paper 1 Q1
D: 1484.0 B: 1500.0

A cylindrical biscuit tin has volume \(V\) and surface area \(S\) (including the ends). Show that the minimum possible surface area for a given value of \(V\) is \(S=3(2\pi V^{2})^{1/3}.\) For this value of \(S\) show that the volume of the largest sphere which can fit inside the tin is \(\frac{2}{3}V\), and find the volume of the smallest sphere into which the tin fits.

Solution: Suppose we have height \(h\) and radius \(r\), then: \(V = \pi r^2 h\) and \(S = 2\pi r^2 + 2\pi r h\). \(h = \frac{V}{\pi r^2}\), so \begin{align*} S &= 2 \pi r^2 + 2 \pi r\frac{V}{\pi r^2} \\ &= 2\pi r^2 +V \frac1{r}+V \frac1{r} \\ &\underbrace{ \geq }_{\text{AM-GM}} 3 \sqrt[3]{2\pi r^2 \frac{V^2}{r^2} } = 3 (2 \pi V^2)^{1/3} \end{align*} Equality holds when \(r = \sqrt[3]{\frac{V}{2 \pi}}, h = \frac{V}{\pi (V/2\pi)^{2/3}} = \sqrt[3]{\frac{4V}{\pi}}\) Since \(h > r\) the sphere has a maximum radius of \(r\) and so it's largest volume is \(\frac43 \pi r^3 = \frac43 \pi \frac{V}{2 \pi} = \frac23 V\).

TikZ diagram
The radius of the sphere is \(\sqrt{\left (\frac{r}{2} \right)^2 + \left (\frac{h}{2} \right)^2 } = \frac12 \sqrt{r^2+h^2}\) \begin{align*} V_{sphere} &= \frac43 \pi (r^2+h^2)^{3/2} \\ &= \frac43 \pi \left (\left( \frac{V}{2 \pi} \right)^{2/3}+\left( \frac{4V}{ \pi} \right)^{2/3} \right)^{3/2} \\ &= \frac43 \pi \frac{V}{ \pi} \left ( 2^{-2/3}+4^{2/3}\right)^{3/2} \\ &= \frac 43 V \left ( \frac{1+4}{2^{2/3}} \right)^{3/2} \\ &= \frac43 \frac{5^{3/2}}{2} V \\ &= \frac{2 \cdot \sqrt{125}}{3} V \end{align*}

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