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2013 Paper 3 Q5
D: 1700.0 B: 1487.0

In this question, you may assume that, if \(a\), \(b\) and \(c\) are positive integers such that \(a\) and \(b\) are coprime and \(a\) divides \(bc\), then \(a\) divides \(c\). (Two positive integers are said to be coprime if their highest common factor is 1.)

  1. Suppose that there are positive integers \(p\), \(q\), \(n\) and \(N\) such that \(p\) and \(q\) are coprime and \(q^nN=p^n\). Show that \(N=kp^n\) for some positive integer \(k\) and deduce the value of \(q\). Hence prove that, for any positive integers \(n\) and \(N\), \(\sqrt[n]N\) is either a positive integer or irrational.
  2. Suppose that there are positive integers \(a\), \(b\), \(c\) and \(d\) such that \(a\) and \(b\) are coprime and \(c\) and \(d\) are coprime, and \(a^ad^b = b^a c^b \,\). Prove that \(d^b = b^a\) and deduce that, if \(p\) is a prime factor of \(d\), then \(p\) is also a prime factor of \(b\). If \(p^m\) and \(p^n\) are the highest powers of the prime number \(p\) that divide \(d\) and \(b\), respectively, express \(b\) in terms of \(a\), \(m\) and \(n\) and hence show that \(p^n\le n\). Deduce the value of \(b\). (You may assume that if \(x > 0\) and \(y\ge2\) then \(y^x > x\).) Hence prove that, if \(r\) is a positive rational number such that \(r^r\) is rational, then \(r\) is a positive integer.

2013 Paper 3 Q6
D: 1700.0 B: 1500.0

Let \(z\) and \(w\) be complex numbers. Use a diagram to show that \(\vert z-w \vert \le \vert z\vert + \vert w \vert\,.\) For any complex numbers \(z\) and \(w\), \(E\) is defined by \[ E = zw^* + z^*w +2 \vert zw \vert\,. \]

  1. Show that \(\vert z-w\vert^2 = \left( \vert z \vert + \vert w\vert\right)^2 -E\,\), and deduce that \(E\) is real and non-negative.
  2. Show that \(\vert 1-zw^*\vert^2 = \left ( 1 +\vert zw \vert \right)^2 -E\,\).
Hence show that, if both \(\vert z \vert >1\) and \(\vert w \vert >1\), then \[ \frac {\vert z-w\vert} {\vert 1-zw^*\vert } \le \frac{\vert z \vert +\vert w\vert }{1+\vert z w \vert}\,. \] Does this inequality also hold if both \(\vert z \vert <1\) and \(\vert w \vert <1\)?

2013 Paper 3 Q7
D: 1700.0 B: 1500.0

  1. Let \(y(x)\) be a solution of the differential equation \( \dfrac {\d^2 y}{\d x^2}+y^3=0\) with \(y = 1\) and \(\dfrac{\d y}{\d x} =0\) at \(x=0\), and let \[ {\rm E} (x)= \left ( \frac {\d y}{\d x}\right)^{\!\!2} + \tfrac 12 y^4\,. \] Show by differentiation that \({\rm E}\) is constant and deduce that \( \vert y(x) \vert \le 1\) for all \(x\).
  2. Let \(v(x)\) be a solution of the differential equation $ \dfrac{\d^2 v}{\d x^2} + x \dfrac {\d v}{\d x} +\sinh v =0 $ with \(v = \ln 3\) and \(\dfrac{\d v}{\d x} =0\) at \(x=0\), and let \[ {\rm E} (x)= \left ( \frac {\d v}{\d x}\right)^{\!\!2} + 2 \cosh v\,. \] Show that \(\dfrac{\d{\rm E}}{\d x}\le 0\) for \(x\ge0\) and deduce that \(\cosh v(x) \le \frac53\) for \(x\ge0\).
  3. Let \(w(x)\) be a solution of the differential equation \[ \frac{\d^2 w}{\d x^2} + (5\cosh x - 4 \sinh x -3) \frac{\d w}{\d x} + (w\cosh w + 2 \sinh w) =0 \] with \(\dfrac{\d w }{\d x}=\dfrac 1 { \sqrt 2 }\) and \(w=0\) at \(x=0\). Show that \(\cosh w(x) \le \frac54\) for \(x\ge0\).

2013 Paper 3 Q8
D: 1700.0 B: 1484.0

Evaluate \(\displaystyle \sum_{r=0}^{n-1} \e^{2i(\alpha + r\pi/n)}\) where \(\alpha\) is a fixed angle and \(n\ge2\). The fixed point \(O\) is a distance \(d\) from a fixed line \(D\). For any point \(P\), let \(s\) be the distance from \(P\) to \(D\) and let \(r\) be the distance from \(P\) to \(O\). Write down an expression for \(s\) in terms of \(d\), \(r\) and the angle \(\theta\), where \(\theta\) is as shown in the diagram below.

TikZ diagram
The curve \(E\) shown in the diagram is such that, for any point \(P\) on \(E\), the relation \(r = k s\) holds, where \(k\) is a fixed number with \(0< k <1\). Each of the \(n\) lines \(L_1\), \(\ldots\,\), \(L_n\) passes through \(O\) and the angle between adjacent lines is \(\frac \pi n\). The line \(L_j\) (\(j=1\), \(\ldots\,\), \(n\)) intersects \(E\) in two points forming a chord of length \(l_j\). Show that, for \(n\ge2\), \[ \sum_{j=1}^n \frac 1 {l_j} = \frac {(2-k^2)n} {4kd}\,. \]


Solution: \begin{align*} \sum_{r=0}^{n-1} \e^{2i(\alpha + r\pi/n)} &= e^{2i\alpha} \sum_{r=0}^{n-1} \left (\e^{2i\pi/n} \right)^r \\ &= e^{2i\alpha} \frac{1-\left (\e^{2i\pi/n} \right)^n}{1-\e^{2i\pi/n} } \\ &= 0 \end{align*} \(d = s + r \cos \theta\) ie \(s = d - r \cos \theta\) Therefore \(d = \frac{r}{k} + r \cos \theta \Rightarrow r = \frac{kd}{1+k \cos \theta}\). The \(l_j\) will come from \(r(\alpha + \frac{j \pi}{n} )+r(\alpha + \pi + \frac{j \pi}{n} )\) \begin{align*} && l_j &= r(\alpha + \frac{(j-1) \pi}{n} )+r(\alpha + \pi + \frac{(j-1) \pi}{n} ) \\ &&&= \frac{kd}{1+k \cos \left ( \alpha + \frac{(j-1) \pi}{n}\right)}+\frac{kd}{1+k \cos \left ( \alpha+\pi+ \frac{(j-1) \pi}{n}\right)}\\ &&&= \frac{kd}{1+k \cos \left ( \alpha + \frac{(j-1) \pi}{n}\right)}+\frac{kd}{1-k \cos \left ( \alpha+ \frac{(j-1) \pi}{n}\right)}\\ &&&= \frac{2kd}{1-k^2 \cos^2 \left ( \alpha + \frac{(j-1) \pi}{n}\right)}\\ \Rightarrow && \sum_{j=1}^n \frac 1 {l_j} &= \sum_{j=0}^{n-1} \frac{1-k^2 \cos^2 \left ( \alpha + \frac{j \pi}{n}\right)}{2kd} \\ &&&= \frac{n}{2kd}-\frac{k^2}{2kd} \sum_{j=0}^{n-1} \cos^2 \left ( \alpha + \frac{j \pi}{n}\right) \\ &&&= \frac{n}{2kd}-\frac{k^2}{2kd} \sum_{j=0}^{n-1} \frac{1+ \cos \left ( 2\alpha + \frac{2j \pi}{n}\right)}{2} \\ &&&= \frac{n}{2kd}-\frac{nk^2}{2kd}-\frac{k^2}{4kd} \sum_{j=0}^{n-1}\cos \left ( 2\alpha + \frac{2j \pi}{n}\right) \\ &&&= \frac{n}{2kd}-\frac{nk^2}{2kd}-\frac{k^2}{4kd} \underbrace{\textrm{Re} \left ( \sum_{j=0}^{n-1}e^{ 2i(\alpha + \frac{j \pi}{n})} \right)}_{=0} \\ &&&= \frac{n}{2kd} - \frac{nk^2}{4kd} \\ &&&= \frac{n(2-k^2)}{4kd} \end{align*}

2013 Paper 3 Q9
D: 1700.0 B: 1484.8

A sphere of radius \(R\) and uniform density \(\rho_{\text{s}}\) is floating in a large tank of liquid of uniform density \(\rho\). Given that the centre of the sphere is a distance \(x\) above the level of the liquid, where \(x < R\), show that the volume of liquid displaced is \[ \frac \pi 3 (2R^3-3R^2x +x^3)\,. \] The sphere is acted upon by two forces only: its weight and an upward force equal in magnitude to the weight of the liquid it has displaced. Show that \[ 4 R^3\rho_{\text{s}} (g+\ddot x) = (2R^3 -3R^2x +x^3)\rho g\,. \] Given that the sphere is in equilibrium when \(x=\frac12 R\), find \(\rho_{\text{s}}\) in terms of \(\rho\). Find, in terms of \(R\) and \(g\), the period of small oscillations about this equilibrium position.

2013 Paper 3 Q10
D: 1700.0 B: 1500.0

A uniform rod \(AB\) has mass \(M\) and length \(2a\). The point \(P\) lies on the rod a distance \(a-x\) from~\(A\). Show that the moment of inertia of the rod about an axis through \(P\) and perpendicular to the rod is \[ \tfrac13 M(a^2 +3x^2)\,. \] The rod is free to rotate, in a horizontal plane, about a fixed vertical axis through \(P\). Initially the rod is at rest. The end \(B\) is struck by a particle of mass \(m\) moving horizontally with speed \(u\) in a direction perpendicular to the rod. The coefficient of restitution between the rod and the particle is \(e\). Show that the angular velocity of the rod immediately after impact is \[ \frac{3mu(1+e)(a+x)}{M(a^2+3x^2) +3m(a+x)^2}\,. \] In the case \(m=2M\), find the value of \(x\) for which the angular velocity is greatest and show that this angular velocity is \(u(1+e)/a\,\).

2013 Paper 3 Q11
D: 1700.0 B: 1484.0

An equilateral triangle, comprising three light rods each of length \(\sqrt3a\), has a particle of mass \(m\) attached to each of its vertices. The triangle is suspended horizontally from a point vertically above its centre by three identical springs, so that the springs and rods form a tetrahedron. Each spring has natural length \(a\) and modulus of elasticity \(kmg\), and is light. Show that when the springs make an angle \(\theta\) with the horizontal the tension in each spring is \[ \frac{ kmg(1-\cos\theta)}{\cos\theta}\,. \] Given that the triangle is in equilibrium when \(\theta = \frac16 \pi\), show that \(k=4\sqrt3 +6\). The triangle is released from rest from the position at which \(\theta=\frac13\pi\). Show that when it passes through the equilibrium position its speed \(V\) satisfies \[ V^2 = \frac{4ag}3(6+\sqrt3)\,. \]

2013 Paper 3 Q12
D: 1700.0 B: 1500.0

A list consists only of letters \(A\) and \(B\) arranged in a row. In the list, there are \(a\) letter \(A\)s and \(b\) letter \(B\)s, where \(a\ge2\) and \(b\ge2\), and \(a+b=n\). Each possible ordering of the letters is equally probable. The random variable \(X_1\) is defined by \[ X_1 = \begin{cases} 1 & \text{if the first letter in the row is \(A\)};\\ 0 & \text{otherwise.} \end{cases} \] The random variables \(X_k\) (\(2 \le k \le n\)) are defined by \[ X_k = \begin{cases} 1 & \text{if the \((k-1)\)th letter is \(B\) and the \(k\)th is \(A\)};\\ 0 & \text{otherwise.} \end{cases} \] The random variable \(S\) is defined by \(S = \sum\limits_ {i=1}^n X_i\,\).

  1. Find expressions for \(\E(X_i)\), distinguishing between the cases \(i=1\) and \(i\ne1\), and show that \(\E(S)= \dfrac{a(b+1)}n\,\).
  2. Show that:
    1. [ for \(j\ge3\), \(\E(X_1X_j) = \dfrac{a(a-1)b}{n(n-1)(n-2)}\,\);
    2. $\ds \sum\limits_{i=2}^{n-2} \bigg( \sum\limits_{j=i+2}^n \E(X_iX_j)\bigg) = \dfrac{a(a-1)b(b-1)}{2n(n-1)}\,\(;
    3. \)\var(S) = \dfrac {a(a-1)b(b+1)}{n^2(n-1)}\,$.

2013 Paper 3 Q13
D: 1700.0 B: 1484.0

  1. The continuous random variable \(X\) satisfies \(0\le X\le 1\), and has probability density function \(\f(x)\) and cumulative distribution function \(\F(x)\). The greatest value of \(\f(x)\) is \(M\), so that \(0\le \f(x) \le M\).
    1. Show that \(0\le \F(x) \le Mx\) for \(0\le x\le1\).
    2. For any function \(\g(x)\), show that \[ \int_0^1 2 \g(x) \F(x) \f(x) \d x = \g(1) - \int_0^1 \g'(x) \big( \F(x)\big)^2 \d x \,. \]
  2. The continuous random variable \(Y\) satisfies \(0\le Y\le 1\), and has probability density function \(k \F(y) \f(y)\), where \(\f\) and \(\F\) are as above.
    1. Determine the value of the constant \(k\).
    2. Show that \[ 1+ \frac{nM}{n+1}\mu_{n+1} - \frac{nM}{n+1} \le \E(Y^n) \le 2M\mu_{n+1}\,, \] where \(\mu_{n+1} = \E(X^{n+1})\) and \(n\ge0\).
    3. Hence show that, for \(n\ge 1\), \[ \mu _n \ge \frac{n}{(n+1)M} -\frac{n-1}{n+1} \,.\]


Solution:

    1. \(\,\) \begin{align*} && 0 &\leq f(t) &\leq M \\ \Rightarrow && \int_0^x 0 \d t &\leq \int_0^x f(t) \d t & \leq \int_0^x M \d x \\ \Rightarrow && 0 &\leq F(x) &\leq Mx \end{align*}
    2. \(\,\) \begin{align*} && \int_0^1 2g(x)F(x)f(x) \d x &= \left [ g(x) F(x)^2 \right] - \int_0^1 g'(x) \left ( F(x)\right)^2 \d x \\ &&&= g(1) - \int_0^1 g'(x) \left ( F(x)\right)^2 \d x \end{align*}
    1. \(\,\) \begin{align*} && 1 &= \int_0^1 kF(y)f(y) \d y \\ &&&= k\left [ \frac12 F(y)^2\right]_0^1 \\ &&&= \frac{k}{2} \\ \Rightarrow && k &= 2 \end{align*}
    2. \(\,\) \begin{align*} \E[Y^n] &= \int_0^1 y^n 2F(y)f(y) \d y \\ &\geq \int_0^1 y^n 2My f(y) \d y \\ &= 2M\int_0^1 y^{n+1} f(y) \d y \\ &= 2M \E[X^{n+1}] = 2M\mu_{n+1} \\ \\ \E[Y^n] &= \int_0^1 y^n 2F(y)f(y) \d y \\ &= 1 - \int_0^1 ny^{n-1} F(y)^2 \d y \\ &\geq 1 - \int_0^1 ny^{n-1}My F(y) \d y \\ &= 1 - M\int_0^1 ny^n F(y) \d y \\ &= 1 - M[\frac{n}{n+1}y^{n+1} F(y)]_0^1 + M\int_0^1\frac{n}{n+1} y^{n+1} f(y) \d y \\ &= 1 - \frac{nM}{n+1} + \frac{nM}{n+1} \mu_{n+1} \end{align*}
    3. Since \(\E[Y^{n-1}] \geq 0\) we must have \begin{align*} && 2M\mu_n \geq 1 + \frac{(n-1)M}{n}\mu_n - \frac{(n-1)M}{n} \\ \Rightarrow && \mu_n \left (2M + \frac{(n-1)M}{n} \right) \geq 1 - \frac{(n-1)M}{n} \\ \Rightarrow && \mu_n \frac{3Mn-M}{n} & \geq \frac{n-(n-1)M}{n} \\ \Rightarrow && \mu_n & \geq \frac{n-(n-1)M}{3Mn-M} \end{align*}