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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*}