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2014 Paper 3 Q5
D: 1700.0 B: 1500.0

A quadrilateral drawn in the complex plane has vertices \(A\), \(B\), \(C\) and~\(D\), labelled anticlockwise. These vertices are represented, respectively, by the complex numbers \(a\), \(b\), \(c\) and~\(d\). Show that \(ABCD\) is a parallelogram (defined as a quadrilateral in which opposite sides are parallel and equal in length) if and only if \(a+c =b+d\,\). Show further that, in this case, \(ABCD\) is a square if and only if \({\rm i}(a-c)=b-d\). Let \(PQRS\) be a quadrilateral in the complex plane, with vertices labelled anticlockwise, the internal angles of which are all less than~\(180^\circ\). Squares with centres \(X\), \(Y\), \(Z\) and~\(T\) are constructed externally to the quadrilateral on the sides \(PQ\), \(QR\), \(RS\) and~\(SP\), respectively.

  1. If \(P\) and \(Q\) are represented by the complex numbers \(p\) and~\(q\), respectively, show that \(X\) can be represented by \[ \tfrac 12 \big( p(1+{\rm i} ) + q (1-{\rm i})\big) \,. \]
  2. Show that \(XY\!ZT\) is a square if and only if \(PQRS\) is a parallelogram.

2014 Paper 3 Q6
D: 1700.0 B: 1516.0

Starting from the result that \[ \.h(t) >0\ \mathrm{for}\ 0< t < x \Longrightarrow \int_0^x \.h(t)\ud t > 0 \,, \] show that, if \(\.f''(t) > 0\) for \(0 < t < x_0\) and \(\.f(0)=\.f'(0) =0\), then \(\.f(t)>0\) for \(0 < t < x_0\).

  1. Show that, for \(0 < x < \frac12\pi\), \[ \cos x \cosh x <1 \,. \]
  2. Show that, for \(0 < x < \frac12\pi\), \[ \frac 1 {\cosh x} < \frac {\sin x} x < \frac x {\sinh x} \,. \] %
  3. Show that, for \(0 < x < \frac12\pi\), \(\tanh x < \tan x\).

2014 Paper 3 Q7
D: 1700.0 B: 1484.0

The four distinct points \(P_i\) (\(i=1\), \(2\), \(3\), \(4\)) are the vertices, labelled anticlockwise, of a cyclic quadrilateral. The lines \(P_1P_3\) and \(P_2P_4\) intersect at \(Q\).

  1. By considering the triangles \(P_1QP_4\) and \(P_2QP_3\) show that \((P_1Q)( QP_3) = (P_2Q) (QP_4)\,\).
  2. Let \(\+p_i\) be the position vector of the point \(P_i\) (\(i=1\), \(2\), \(3\), \(4\)). Show that there exist numbers \(a_i\), not all zero, such that \begin{equation} \sum\limits_{i=1}^4 a_i =0 \qquad\text{and}\qquad \sum\limits_{i=1}^4 a_i \+p_i ={\bf 0} \,. \tag{\(*\)} \end{equation}
  3. Let \(a_i\) (\(i=1\),~\(2\), \(3\),~\(4\)) be any numbers, not all zero, that satisfy~\((*)\). Show that \(a_1+a_3\ne 0\) and that the lines \(P_1P_3\) and \(P_2P_4\) intersect at the point with position vector \[ \frac{a_1 \+p_1 + a_3 \+p_3}{a_1+a_3} \,. \] Deduce that \(a_1a_3 (P_1P_3)^2 = a_2a_4 (P_2P_4)^2\,\).

2014 Paper 3 Q8
D: 1700.0 B: 1516.0

The numbers \(\.f(r)\) satisfy \(\.f(r)>\.f(r+1)\) for \(r=1\), \(2\), \dots. Show that, for any non-negative integer \(n\), \[ k^n(k-1) \, \.f(k^{n+1}) \le \sum_{r=k^n}^{k^{n+1}-1}\.f(r) \le k^n(k-1)\, \.f(k^n)\, \] where \(k\) is an integer greater than 1.

  1. By taking \(\.f(r) = 1/r\), show that \[ \frac{N+1}2 \le \sum_{r=1}^{2^{N+1}-1} \frac1r \le N+1 \,. \] Deduce that the sum \(\sum\limits_{r=1}^\infty \frac1r\) does not converge.
  2. By taking \(\.f(r)= 1/r^3\), show that \[ \sum_{r=1}^\infty \frac1 {r^3} \le 1 \tfrac 13 \,. \]
  3. Let \(S(n)\) be the set of positive integers less than \(n\) which do not have a \(2\) in their decimal representation and let \(\sigma(n)\) be the sum of the reciprocals of the numbers in \(S(n)\), so for example \(\sigma(5) = 1+\frac13+\frac14\). Show that \(S(1000)\) contains \(9^3-1\) distinct numbers. Show that \(\sigma (n) < 80\) for all \(n\).

2014 Paper 3 Q9
D: 1700.0 B: 1500.0

A particle of mass \(m\) is projected with velocity \(\+ u\). It is acted upon by the force \(m\+g\) due to gravity and by a resistive force \(-mk \+v\), where \(\+v\) is its velocity and \(k\) is a positive constant. Given that, at time \(t\) after projection, its position \(\+r\) relative to the point of projection is given by \[ \+r = \frac{kt -1 +\.e^{-kt}} {k^2} \, \+g + \frac{ 1-\.e^{-kt}}{k} \, \+u \,, \] find an expression for \(\+v\) in terms of \(k\), \(t\), \(\+g\) and \(\+u\). Verify that the equation of motion and the initial conditions are satisfied. Let \(\+u = u\cos\alpha \, \+i + u \sin\alpha \, \+j\) and $\+g = -g\, \+j\(, where \)0<\alpha<90^\circ\(, and let \)T$ be the time after projection at which \(\+r \,.\, \+j =0\). Show that \[ uk \sin\alpha = \left(\frac{kT}{1-\.e^{-kT}} -1\right)g\,. \] Let \(\beta\) be the acute angle between \(\+v\) and \(\+i\) at time \(T\). Show that \[ \tan\beta = \frac{(\.e^{kT}-1)g}{uk\cos\alpha}-\tan\alpha \,. \] Show further that \(\tan\beta >\tan\alpha\) (you may assume that \(\sinh kT >kT\)) and deduce that~\(\beta >\alpha\).

2014 Paper 3 Q10
D: 1700.0 B: 1473.3

Two particles \(X\) and \(Y\), of equal mass \(m\), lie on a smooth horizontal table and are connected by a light elastic spring of natural length \(a\) and modulus of elasticity \(\lambda\). Two more springs, identical to the first, connect \(X\) to a point \(P\) on the table and \(Y\) to a point \(Q\) on the table. The distance between \(P\) and \(Q\) is \(3a\). Initially, the particles are held so that \(XP=a\), \(YQ= \frac12 a\,\), and \(PXYQ\) is a straight line. The particles are then released. At time \(t\), the particle \(X\) is a distance \(a+x\) from \(P\) and the particle \(Y\) is a distance \(a+y\) from \(Q\). Show that \[ m \frac{\.d ^2 x}{\.d t^2} = -\frac\lambda a (2x+y) \] and find a similar expression involving \(\dfrac{\.d^2 y}{\.d t^2}\). Deduce that \[ x-y = A\cos \omega t +B \sin\omega t \] where \(A\) and \(B\) are constants to be determined and \(ma\omega^2=\lambda\). Find a similar expression for \(x+y\). Show that \(Y\) will never return to its initial position.

2014 Paper 3 Q11
D: 1700.0 B: 1484.0

A particle \(P\) of mass \(m\) is connected by two light inextensible strings to two fixed points \(A\) and \(B\), with \(A\) vertically above \(B\). The string \(AP\) has length \(x\). The particle is rotating about the vertical through \(A\) and \(B\) with angular velocity \(\omega\), and both strings are taut. Angles \(PAB\) and \(PBA\) are \(\alpha\) and \(\beta\), respectively. Find the tensions \(T_A\) and \(T_B\) in the strings \(AP\) and \(BP\) (respectively), and hence show that \(\omega^2 x\cos\alpha \ge g\). Consider now the case that \(\omega^2 x\cos\alpha = g\). Given that \(AB=h\) and \(BP=d\), where \(h>d\), show that \(h\cos\alpha \ge \sqrt{h^2-d^2}\). Show further that \[ mg < T_A \le \frac{mgh}{\sqrt{h^2-d^2}\,}\,. \] Describe the geometry of the strings when \(T_A\) attains its upper bound.


Solution:

TikZ diagram
\begin{align*} \text{N2}(\uparrow): && T_A \cos \alpha - T_B \cos\alpha - mg &= 0 \\ \Rightarrow && T_A \cos \alpha - T_B \cos\beta &= mg \\ \text{N2}(\leftarrow, \text{radially}): && T_A \sin \alpha + T_B \sin \beta &= m x \sin \alpha \omega^2 \\ \Rightarrow && T_A(\cos \alpha \sin \beta+\sin \alpha \cos \beta) &= mg \sin \beta + mx \sin \alpha \omega^2 \cos \beta \\ \Rightarrow && T_A &=\frac{mg\sin \beta + m x \sin \alpha \omega^2 \cos \beta }{\sin(\alpha + \beta)} \\ \Rightarrow && T_B(\sin \beta \cos \alpha- \cos \beta \sin \alpha)&= mx \sin \alpha \omega^2 \cos \alpha -mg \sin \alpha \\ \Rightarrow && T_B &= \frac{m x \sin \alpha \omega^2 \cos \alpha - mg \sin \alpha}{\sin(\beta - \alpha)} \\ &&&= \frac{m \sin \alpha(\omega^2 \cos\alpha - g)}{\sin (\beta - \alpha)} \end{align*} Since \(T_B \geq 0 \Rightarrow \omega^2 \cos\alpha - g \geq 0\) as required.
TikZ diagram
\(\sqrt{h^2-d^2}\) is the length of the final side on the dashed right angle triangle with hypotenuse \(AB\). \(h \cos \alpha\) will be clearly longer as the angle \(\alpha\) will be smaller and so \(\cos \alpha\) will be larger. When \(\omega^2 x \cos \alpha = g\) we must have \(T_B = 0\). \(T_A\cos \alpha = mg \Rightarrow T_A > mg\) since \(\alpha \neq 0\). \(T_A = \frac{mg}{\cos \alpha} \leq \frac{mgh}{\sqrt{h^2-d^2}}\) \(T_A\) will attain it's upper bound when \(\angle APB\) is a right angle.

2014 Paper 3 Q12
D: 1700.0 B: 1500.0

The random variable \(X\) has probability density function \(f(x)\) (which you may assume is differentiable) and cumulative distribution function \(F(x)\) where \(-\infty < x < \infty \). The random variable \(Y\) is defined by \(Y= \e^X\). You may assume throughout this question that \(X\) and \(Y\) have unique modes.

  1. Find the median value \(y_m\) of \(Y\) in terms of the median value \(x_m\) of \(X\).
  2. Show that the probability density function of \(Y\) is \(f(\ln y)/y\), and deduce that the mode \(\lambda\) of \(Y\) satisfies \(\f'(\ln \lambda) = \f(\ln \lambda)\).
  3. Suppose now that \(X \sim {\rm N} (\mu,\sigma^2)\), so that \[ f(x) = \frac{1}{\sigma \sqrt{2\pi}\,} \e^{-(x-\mu)^2/(2\sigma^2)} \,. \] Explain why \[\frac{1}{\sigma \sqrt{2\pi}\,} \int_{-\infty}^{\infty}\e^{-(x-\mu-\sigma^2)^2/(2\sigma^2)} \d x = 1 \] and hence show that \( \E(Y) = \e ^{\mu+\frac12\sigma^2}\).
  4. Show that, when \(X \sim {\rm N} (\mu,\sigma^2)\), \[ \lambda < y_m < \E(Y)\,. \]


Solution:

  1. \begin{align*} && \frac12 &= \mathbb{P}(X \leq x_m) \\ \Leftrightarrow && \frac12 &= \mathbb{P}(e^X \leq e^{x_m} = y_m) \end{align*} Therefore the median is \(y_m = e^{x_m}\)
  2. \begin{align*} && \mathbb{P}(Y \leq y) &= \mathbb{P}(e^X \leq y) \\ &&&= \mathbb{P}(X \leq \ln y) \\ &&&= F(\ln y) \\ \Rightarrow && f_Y(y) &= f(\ln y)/y \\ \\ && f'_Y(y) &= \frac{f'(\ln y) - f(\ln y)}{y^2} \end{align*} Therefore since the mode satisfies \(f'_Y = 0\) we must have \(f'(\ln \lambda ) = f(\ln \lambda)\)
  3. This is the integral of the pdf of \(N(\mu + \sigma^2, \sigma^2)\) and therefore is clearly \(1\). \begin{align*} && \E[Y] &= \int_{-\infty}^{\infty} e^x \cdot \frac{1}{\sqrt{2\pi \sigma^2}} e^{-(x-\mu)^2/(2\sigma^2)} \d x \\ &&&= \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp (x - (x-\mu)^2/(2\sigma^2)) \d x\\ &&&= \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp ((2x \sigma^2- (x-\mu)^2)/(2\sigma^2)) \d x\\ &&&= \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp (-(x-\mu-\sigma^2)^2+2\mu \sigma^2-\sigma^4)/(2\sigma^2)) \d x\\ &&&= \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp (-(x-\mu+\sigma^2)^2)/(2\sigma^2)+\mu +\frac12\sigma^2) \d x\\ &&&= \e^{\mu +\frac12\sigma^2}\frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp (-(x-\mu-\sigma^2)^2)/(2\sigma^2)) \d x\\ &&&= \e^{\mu +\frac12\sigma^2} \end{align*}
  4. Notice that \(y_m = e^\mu < e^{\mu + \tfrac12 \sigma^2} = \E[Y]\), so it suffices to prove that \(\lambda < e^{\mu}\) Notice that \(f'(x) - f(x) = f(x)[-(x-\mu)/\sigma^2 - 1]\) and therefore \(\ln y - \mu = -\sigma^2\) so \(\lambda = e^{\mu - \sigma^2}\) which is clearly less than \(e^{\mu}\) as required.

2014 Paper 3 Q13
D: 1700.0 B: 1500.0

I play a game which has repeated rounds. Before the first round, my score is \(0\). Each round can have three outcomes:

  1. my score is unchanged and the game ends;
  2. my score is unchanged and I continue to the next round;
  3. my score is increased by one and I continue to the next round.
The probabilities of these outcomes are \(a\), \(b\) and \(c\), respectively (the same in each round), where \(a+b+c=1\) and \(abc\ne0\). The random variable \(N\) represents my score at the end of a randomly chosen game. Let \(G(t)\) be the probability generating function of \(N\).
  1. Suppose in the first round, the game ends. Show that the probability generating function conditional on this happening is 1.
  2. Suppose in the first round, the game continues to the next round with no change in score. Show that the probability generating function conditional on this happening is \(G(t)\).
  3. By comparing the coefficients of \(t^n\), show that $ G(t) = a + bG(t) + ctG(t)\,. $ Deduce that, for \(n\ge0\), \[ P(N=n) = \frac{ac^n}{(1-b)^{n+1}}\,. \]
  4. Show further that, for \(n\ge0\), \[ P(N=n) = \frac{\mu^n}{(1+\mu)^{n+1}}\,, \] where \(\mu=\E(N)\).


Solution:

  1. If the game ends in the first round then the score is exactly \(0\) and the pgf is \(1\cdot x^0 = 1\)
  2. If the game moves onto the next round with no change in the first round then it's as if nothing happened, therefore the pgf is the original pgf \(G(t)\)
  3. If the game moves into the next round with the score increased by one, then the pgf is \(tG(t)\) since all the scores are increased by \(1\). Therefore \begin{align*} && G(t) &= \E[t^N] \\ &&&= \E[\E[t^N | \text{first round}]] \\ &&&= a + bG(t) + ctG(t) \\ \Rightarrow && G(t)(1-b-ct) = a \\ \Rightarrow && G(t) &= \frac{a}{(1-b)-ct} \\ &&&= \frac{a}{(1-b)} \frac{1}{1- \left(\frac{c}{1-b}\right)t} \\ &&&= \sum_{n=0}^\infty \frac{a}{1-b} \frac{c^n}{(1-b)^n} t^n\\ &&&= \sum_{n=0}^{\infty} \frac{ac^n}{(1-b)^{n+1}}t^n \end{align*} Therefore by comparing coefficients, \(\mathbb{P}(N=n) = \frac{ac^n}{(1-b)^{n+1}}\)
  4. \(\,\) \begin{align*} && \E[N] &= G'(1) \\ &&&= \frac{ac}{((1-b)-c)^2} \\ &&&= \frac{ac}{a^2} = \frac{c}{a} \\ \\ && \frac{\mu^n}{(1+\mu)^{n+1}} &= \frac{c^na^{-n}}{(a+c)^{n+1}a^{-n-1}} \\ &&&= \frac{ac^n}{(a+c)^{n+1}}\\ &&&= \frac{ac^n}{(1-b)^{n+1}}\\ &&&= \mathbb{P}(N=n) \end{align*} as required