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1993 Paper 3 Q1
D: 1700.0 B: 1500.0

The curve \(P\) has the parametric equations $$ x= \sin\theta, \quad y=\cos2\theta \qquad\hbox{ for }-\pi/2 \le \theta \le \pi/2. $$ Show that \(P\) is part of the parabola \(y=1-2x^2\) and sketch \(P\). Show that the length of \(P\) is \(\surd (17) + {1\over 4} \sinh^{-1}4\). Obtain the volume of the solid enclosed when \(P\) is rotated through \(2\pi\) radians about the line \(y=-1\).


Solution: First notice that \(y = \cos 2 \theta = 1 - 2\sin^2 \theta = 1- 2x^2\), therefore \(P\) is lies on that parabola.

TikZ diagram
The arc length is \begin{align*} && s &= \int_{-\pi/2}^{\pi/2} \sqrt{\left ( \frac{\d x}{\d \theta} \right)^2+\left ( \frac{\d y}{\d \theta} \right)^2} \d \theta\\ && &= \int_{-\pi/2}^{\pi/2} \sqrt{\cos^2 \theta+16 \sin^2 \theta \cos^2 \theta } \d \theta\\ && &= \int_{-\pi/2}^{\pi/2} \cos \theta\sqrt{1+16 \sin^2 \theta} \d \theta\\ u = \sin \theta, \d u = \cos \theta \d \theta && &= \int_{u=-1}^{u=1} \sqrt{1+16 u^2} \d u\\ 4u = \sinh v, 4\d u = \cosh v: && &= \int_{v=-\sinh^{-1} 4}^{v=\sinh^{-1} 4} \sqrt{1+\sinh^2 v} \tfrac14\cosh v \d v\\ && &= \frac14 \int_{-\sinh^{-1} 4}^{\sinh^{-1} 4} \cosh^2 v \d v\\ && &= \frac18 \int_{-\sinh^{-1} 4}^{\sinh^{-1} 4} (1 + \cosh 2v) \d v\\ && &= \frac14 \sinh^{-1} 4 + \frac18\left [ \frac12\sinh 2v \right]_{-\sinh^{-1} 4}^{\sinh^{-1} 4}\\ && &= \frac14 \sinh^{-1} 4 + \frac18\left [ \sinh v \sqrt{1 + \sinh^2 v} \right]_{-\sinh^{-1} 4}^{\sinh^{-1} 4}\\ && &= \frac14 \sinh^{-1} 4 + \left (\frac18 \cdot 4 \sqrt{17} \right) - \left (\frac18 \cdot (-4) \sqrt{17} \right)\\ && &= \frac14 \sinh^{-1} 4 + \sqrt{17}\\ \end{align*} The volume of revolution is \begin{align*} && V &=\pi \int_{-1}^1 (2-2x^2)^2 \d x \\ &&&= \pi \left [4x-\frac83x^3+\frac45x^5 \right]_{-1}^1 \\ &&&= \pi \left ( 8-\frac{16}3+\frac85 \right) \\ &&&= \frac{64}{15}\pi \end{align*}

1993 Paper 3 Q2
D: 1700.0 B: 1500.0

The curve \(C\) has the equation \(x^3+y^3 = 3xy\).

  1. Show that there is no point of inflection on \(C\). You may assume that the origin is not a point of inflection.
  2. The part of \(C\) which lies in the first quadrant is a closed loop touching the axes at the origin. By converting to polar coordinates, or otherwise, evaluate the area of this loop.

1993 Paper 3 Q3
D: 1700.0 B: 1516.0

The matrices \(\mathbf{A},\mathbf{B}\) and \(\mathbf{M}\) are given by \[ \mathbf{A}=\begin{pmatrix}a & 0 & 0\\ b & c & 0\\ d & e & f \end{pmatrix},\quad\mathbf{B}=\begin{pmatrix}1 & p & q\\ 0 & 1 & r\\ 0 & 0 & 1 \end{pmatrix},\quad\mathbf{M}=\begin{pmatrix}1 & 3 & 2\\ 4 & 13 & 5\\ 3 & 8 & 7 \end{pmatrix}, \] where \(a,b,\ldots,r\) are real numbers. Given that \(\mathbf{M=AB},\) show that \(a=1,b=4,c=1,d=3,e=1,f=-2,p=3,q=2\) and \(r=-3\) gives the unique solution for \(\mathbf{A}\) and \(\mathbf{B}.\) Evaluate \(\mathbf{A}^{-1}\) and \(\mathbf{B}^{-1},\) Hence, or otherwise, solve the simultaneous equations \begin{alignat*}{1} x+3y+2z & =7\\ 4x+13y+5z & =18\\ 3x+8y+7z & =25. \end{alignat*}


Solution: \begin{align*} && \begin{pmatrix}1 & 3 & 2\\ 4 & 13 & 5\\ 3 & 8 & 7 \end{pmatrix} &= \begin{pmatrix}a & 0 & 0\\ b & c & 0\\ d & e & f \end{pmatrix}\begin{pmatrix}1 & p & q\\ 0 & 1 & r\\ 0 & 0 & 1 \end{pmatrix} \\ &&&= \begin{pmatrix} a & ap & aq \\ b & pb + c & qb + cr\\ d & pd + e & qd + er +f \end{pmatrix} \\ \Rightarrow && a,b,d,p,q&=1,4,3,3,2\\ &&&= \begin{pmatrix} 1 & 3 & 2 \\ 4 & 12 + c & 8+ cr\\ 3 & 9 + e & 6 + er +f \end{pmatrix} \\ \Rightarrow && c, e&=1,-1\\ &&&= \begin{pmatrix} 1 & 3 & 2 \\ 4 & 13 & 8+ r\\ 3 & 8 & 6 -r +f \end{pmatrix} \\ \Rightarrow && r, f &= -3, -2 \end{align*} \begin{align*} \mathbf{A}^{-1} &= \begin{pmatrix} 1 & 0 & 0 \\ 4 & 1 & 0\\ 3 & -1 & -2 \end{pmatrix}^{-1} \\ &=\frac{1}{-2} \begin{pmatrix} -2 & 0 & 0 \\ 8 & -2 & 0\\ -7 & 1 & 1 \end{pmatrix} \\ \\ \mathbf{B}^{-1} &= \begin{pmatrix} 1 & 3 & 2 \\ 0 & 1 & -3\\ 0 & 0 & 1 \end{pmatrix}^{-1} \\ &= \begin{pmatrix} 1 & -3 & -11 \\ 0 & 1 & 3\\ 0 & 0 & 1 \end{pmatrix} \\ \end{align*} We want to solve \(\mathbf{M}\mathbf{v} = \begin{pmatrix} 7 \\ 18 \\ 25 \end{pmatrix}\), ie \begin{align*} \mathbf{v} &= \mathbf{M}^{-1}\begin{pmatrix} 7 \\ 18 \\ 25 \end{pmatrix} \\ &= \mathbf{B}^{-1} \mathbf{A}^{-1}\begin{pmatrix} 7 \\ 18 \\ 25 \end{pmatrix} \\ &= \frac{1}{-2}\mathbf{B}^{-1} \begin{pmatrix} -2 & 0 & 0 \\ 8 & -2 & 0\\ -7 & 1 & 1 \end{pmatrix} \begin{pmatrix} 7 \\ 18 \\ 25 \end{pmatrix} \\ &= \frac{1}{-2}\mathbf{B}^{-1} \begin{pmatrix} -14 \\ 20 \\ -6 \end{pmatrix} \\ &= \begin{pmatrix} 1 & -3 & -11 \\ 0 & 1 & 3\\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 7 \\ -10 \\ 3 \end{pmatrix} \\ &= \begin{pmatrix} 4\\ -1 \\ 3 \end{pmatrix} \end{align*} This algorithm is called the "LU-decomposition"

1993 Paper 3 Q4
D: 1700.0 B: 1500.0

Sum the following infinite series.

  1. \[ 1 + \frac13 \bigg({\frac12}\bigg)^2 +\frac15\bigg(\frac12\bigg)^4 + \cdots + \frac{1}{2n+1} \bigg(\frac12\bigg)^{2n} + \cdots \] .
  2. \[ 2 -x -x^3 +2x^4 - \cdots + 2x^{4k} - x^{4k+1} - x^{4k+3} +\cdots \] where \(|x| < 1\).
  3. \[ \sum _{r=2}^\infty {r\, 2^{r-2} \over 3^{r-1} } \].
  4. \[ \sum_{r=2}^\infty {2 \over r(r^2-1) } \].


Solution:

  1. \begin{align*} && \sum_{i=0}^{\infty} x^{2i+1}&= \frac{x}{1-x^2} \\ \Rightarrow &&&=\frac12 \left ( \frac{1}{1-x} - \frac{1}{1+x} \right) \\ \underbrace{\Rightarrow}_{\int} && \sum_{i=0}^{\infty} \frac{1}{2i+1} x^{2i+2} &= \frac12 \left ( -\ln (1-x) - \ln(1+x) \right) \\ \underbrace{\Rightarrow}_{x = 1/2} && \sum_{i=0}^\infty \frac{1}{2i+1} \left (\frac12 \right)^{2i+2} &= -\frac12 \ln \frac12 - \frac12 \ln \frac32 \\ &&&= -\frac12 \ln \frac34 \\ &&\frac14\sum_{i=0}^\infty \frac{1}{2i+1} \left (\frac12 \right)^{2i}&= \frac12 \ln \frac43 \\ \Rightarrow&& S &= 2 \ln \frac43 \end{align*}
  2. \begin{align*} \sum_{k=0}^{\infty} \left (2x^{4k} - x^{4k+1} - x^{4k+3} \right) &= \sum_{k=0}^{\infty} \left (2 - x^{1} - x^{3} \right) x^{4k} \\ &= \frac{2-x-x^3}{1-x^4} \\ &= \frac{(1-x)(2+x+x^2)}{(1-x)(1+x+x^2+x^3)} \\ &= \frac{2+x+x^2}{1+x+x^2+x^3} \end{align*}
  3. \begin{align*} && \frac{1}{(1-x)^2} &= \sum_{r=0}^{\infty} r x^{r-1} \\ \Rightarrow && 9 &= \sum_{r=1}^{\infty} r \left ( \frac23 \right)^{r-1} \\ \Rightarrow && \sum_{r=2}^{\infty} r \left ( \frac{2^{r-2}}{3^{r-1}} \right) &= \frac12 \left ( 9 - 1 \right) \\ &&&= 4 \end{align*}
  4. \begin{align*} && \frac{2}{r(r^2-1)} &= \frac{1}{r-1} - \frac{2}{r} + \frac{1}{r+1} \\ \Rightarrow && \sum_{r=2}^{\infty} \left ( \frac{1}{r-1} - \frac{2}{r} + \frac{1}{r+1} \right) &= \sum_{r=2}^{\infty} \left ( \frac{1}{r-1} - \frac{1}{r} - \frac{1}{r} + \frac{1}{r+1} \right) \\ &&&= \sum_{r=2}^{\infty} \left ( \frac{1}{r-1} - \frac{1}{r} \right)-\sum_{r=2}^{\infty} \left ( \frac{1}{r} - \frac{1}{r+1} \right) \\ &&&= 1 - \frac12 \\ &&&= \frac12 \end{align*}

1993 Paper 3 Q5
D: 1730.5 B: 1466.6

The set \(S\) consists of ordered pairs of complex numbers \((z_1,z_2)\) and a binary operation \(\circ\) on \(S\) is defined by $$ (z_1,z_2)\circ(w_1,w_2)= (z_1w_1-z_2w^*_2, \; z_1w_2+z_2w^*_1). $$ Show that the operation \(\circ\) is associative and determine whether it is commutative. Evaluate \((z,0)\circ(w,0)\), \((z,0)\circ(0,w)\), \((0,z)\circ(w,0)\) and \((0,z)\circ(0,w)\). The set \(S_1\) is the subset of \(S\) consisting of \(A\), \(B\), \(\ldots\,\), \(H\), where \(A=(1,0)\), \(B=(0,1)\), \(C=(i,0)\), \(D=(0,i)\), \(E=(-1,0)\), \(F=(0,-1)\), \(G=(-i,0)\) and \(H=(0,-i)\). Show that \(S_1\) is closed under \(\circ\) and that it has an identity element. Determine the inverse and order of each element of \(S_1\). Show that \(S_1\) is a group under \(\circ\). \hfil\break [You are not required to compute the multiplication table in full.] Show that \(\{A,B,E,F\}\) is a subgroup of \(S_1\) and determine whether it is isomorphic to the group generated by the \(2\times2\) matrix $\begin{pmatrix}0 & 1\\ -1 & 0 \end{pmatrix}$ under matrix multiplication.

1993 Paper 3 Q6
D: 1700.0 B: 1484.0

The point in the Argand diagram representing the complex number \(z\) lies on the circle with centre \(K\) and radius \(r\), where \(K\) represents the complex number \(k\). Show that $$ zz^* -kz^* -k^*z +kk^* -r^2 =0. $$ The points \(P\), \(Q_1\) and \(Q_2\) represent the complex numbers \(z\), \(w_1\) and \(w_2\) respectively. The point \(P\) lies on the circle with \(OA\) as diameter, where \(O\) and \(A\) represent \(0\) and \(2i\) respectively. Given that \(w_1=z/(z-1)\), find the equation of the locus \(L\) of \(Q_1\) in terms of \(w_1\) and describe the geometrical form of \(L\). Given that \(w_2=z^*\), show that the locus of \(Q_2\) is also \(L\). Determine the positions of \(P\) for which \(Q_1\) coincides with \(Q_2\).

1993 Paper 3 Q7
D: 1700.0 B: 1516.0

The real numbers \(x\) and \(y\) satisfy the simultaneous equations $$ \sinh (2x) = \cosh y \qquad\hbox{and}\qquad \sinh(2y) = 2 \cosh x. $$ Show that \(\sinh^2 y\) is a root of the equation $$ 4t^3 + 4t^2 -4t -1=0 $$ and demonstrate that this gives at most one valid solution for \(y\). Show that the relevant value of \(t\) lies between \(0.7\) and \(0.8\), and use an iterative process to find \(t\) to 6 decimal places. Find \(y\) and hence find \(x\), checking your answers and stating the final answers to four decimal places.


Solution: Let \(t = \sinh^2 y\), then \begin{align*} && \sinh(2x) &= \cosh y \tag{1}\\ && \sinh(2y) &= 2 \cosh x \tag{2} \\ \\ && \cosh(2x) &= 2 \cosh^2 x -1 \\ (2): &&&= \frac12 \sinh^2(2y) -1 \\ && 1 &= \left (\frac12 \sinh^2(2y) -1 \right)^2 - \cosh^2 y \\ &&&= \frac14 \sinh^4(2y)-\sinh^2(2y)+1-\cosh^2 y \\ \Rightarrow && 0 &= \frac14 (\cosh^2 (2y)-1)^2- (\cosh^2 (2y)-1) - \cosh^2 y \\ &&&= \frac14 \left ( \left (1+2\sinh^2 y \right)^2-1 \right)^2 -\left ( \left (1+2\sinh^2 y \right)^2 -1\right) - (1 + \sinh^2 y ) \\ &&&= \frac14 \left ( 1 + 4t+4t^2 -1\right)^2 - \left ( 1+4t+4t^2-1\right) - (1 + t) \\ &&&= \frac14 (4t + 4t^2)^2 - (4t+4t^2)-1-t \\ &&&= 4(t+t^2)^2 - 4t^2-5t-1 \\ &&&= 4t^4+8t^3+4t^2-4t^2-5t-1 \\ &&& = 4t^4+8t^3-5t-1 \\ &&&= (t+1)(4t^3+4t^2-4t-1) \end{align*} Since \(\sinh^2 y\) is positive, we must be a root of the second cubic. Let \(f(t) = 4t^3+4t^2-4t-1\), then \(f(0) = -1\) and \(f'(t) = 12t^2+8t-4 = 4(t+1)(3t-1)\), so we have turning points at \(-1\) and \(\frac13\). Since \(f(-1) = 3 > 0\) and \(f(0) < 0\) we must have exactly one root larger than zero. Therefore there is a unique root. \(f(0.7) = -0.468 < 0\) \(f(0.8) = 0.408 > 0\) since \(f\) is continuous and changes sign, the root must fall in the interval \((0.7, 0.8)\). Let \(t_{n+1} = t_n - \frac{f(t_n)}{f'(t_n)}\), and \(t_0 = 0.75\), then \begin{align*} t_0 &= 0.75 \\ t_1 &= 0.7571428571 \\ t_2 &= 0.7570684728 \\ t_3 &= 0.7570684647 \end{align*} So \(t \approx 0.757068\), \(\sinh y \approx 0.870097\), \(y \approx 0.786474\), \(x \approx 0.546965\)

1993 Paper 3 Q8
D: 1700.0 B: 1484.0

A square pyramid has its base vertices at the points \(A\) \((a,0,0)\), \(B\) \((0,a,0)\), \(C\) \((-a,0,0)\) and \(D\) \((0,-a,0)\), and its vertex at \(E\) \((0,0,a)\). The point \(P\) lies on \(AE\) with \(x\)-coordinate \(\lambda a\), where \(0<\lambda<1\), and the point \(Q\) lies on \(CE\) with \(x\)-coordinate \(-\mu a\), where \(0<\mu<1\). The plane \(BPQ\) cuts \(DE\) at \(R\) and the \(y\)-coordinate of \(R\) is \(-\gamma a\). Prove that $$ \gamma = {\lambda \mu \over \lambda + \mu - \lambda \mu}. $$ Show that the quadrilateral \(BPRQ\) cannot be a parallelogram.

1993 Paper 3 Q9
D: 1700.0 B: 1485.7

For the real numbers \(a_1\), \(a_2\), \(a_3\), \(\ldots\),

  1. prove that \(a_1^2+a_2^2 \ge 2a_1a_2\),
  2. prove that \(a_1^2+a_2^2 +a_3^2 \ge a_2a_3 + a_3a_1 +a_1a_2\),
  3. prove that $3(a_1^2+a_2^2 +a_3^2 +a_4^2) \ge 2(a_1a_2+a_1a_3 + a_1a_4 +a_2a_3 + a_2a_4 +a_3a_4)\(,
  4. state and prove a generalisation of (iii) to the case of \)n$ real numbers,
  5. prove that $$ \left(\sum_{i=1}^n a_i \right)^2 \ge {2n\over n-1} \sum_{i,j} a_ia_j, $$ where the latter sum is taken over all pairs \((i,j)\) with $1\le i

1993 Paper 3 Q10
D: 1700.0 B: 1484.0

The transformation \(T\) of the point \(P\) in the \(x\),\(y\) plane to the point \(P'\) is constructed as follows: \hfil\break Lines are drawn through \(P\) parallel to the lines \(y=mx\) and \(y=-mx\) to cut the line \(y=kx\) at \(Q\) and \(R\) respectively, \(m\) and \(k\) being given constants. \(P'\) is the fourth vertex of the parallelogram \(PQP'R\). Show that if \(P\) is \((x_1,y_1)\) then \(Q\) is $$ \left( {mx_1-y_1 \over m-k}, {k(mx_1-y_1)\over m-k}\right). $$ Obtain the coordinates of \(P'\) in terms of \(x_1\), \(y_1\), \(m\) and \(k\), and express \(T\) as a matrix transformation. Show that areas are transformed under \(T\) into areas of the same magnitude.

1993 Paper 3 Q11
D: 1700.0 B: 1500.0

In this question, all gravitational forces are to be neglected. A rigid frame is constructed from 12 equal uniform rods, each of length \(a\) and mass \(m,\) forming the edges of a cube. Three of the edges are \(OA,OB\) and \(OC,\) and the vertices opposite \(O,A,B\) and \(C\) are \(O',A',B'\) and \(C'\) respectively. Forces act along the lines as follows, in the directions indicated by the order of the letters: \begin{alignat*}{3} 2mg\mbox{ along }OA, & \qquad & mg\mbox{ along }AC', & \qquad & \sqrt{2}mg\mbox{ along }O'A,\\ \sqrt{2}mg\mbox{ along }OA', & & 2mg\mbox{ along }C'B, & & mg\mbox{ along }A'C. \end{alignat*}

  1. The frame is freely pivoted at \(O\). Show that the direction of the line about which it will start to rotate is $\begin{pmatrix}1\\ 1\\ 2 \end{pmatrix}$ with respect to axes along \(OA\), \(OB\) and \(OC\) respectively.
  2. Show that the moment of inertia of the rod \(OA\) about the axis \(OO'\) is \(2ma^2/9\) and about a parallel axis through its mid-point is \(ma^2/18\). Hence find the moment of inertia of \(B'C\) about \(OO'\) and show that the moment of inertia of the frame about \(OO'\) is \(14ma^2/3\). If the frame is freely pivoted about the line \(OO'\) and the forces continue to act along the specified lines, find the initial angular acceleration of the frame.

1993 Paper 3 Q12
D: 1700.0 B: 1432.3

\(ABCD\) is a horizontal line with \(AB=CD=a\) and \(BC=6a\). There are fixed smooth pegs at \(B\) and \(C\). A uniform string of natural length \(2a\) and modulus of elasticity \(kmg\) is stretched from \(A\) to \(D\), passing over the pegs at \(B\) and \(C\). A particle of mass \(m\) is attached to the midpoint \(P\) of the string. When the system is in equilibrium, \(P\) is a distance \(a/4\) below \(BC\). Evaluate \(k\). The particle is pulled down to a point \(Q\), which is at a distance \(pa\) below the mid-point of \(BC\), and is released from rest. \(P\) rises to a point \(R\), which is at a distance \(3a\) above \(BC\). Show that \(2p^2-p-17=0\). Show also that the tension in the strings is less when the particle is at \(R\) than when the particle is at \(Q\).

1993 Paper 3 Q13
D: 1700.0 B: 1484.0

\(\ \)\vspace{-1cm} \noindent

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A uniform circular disc with radius \(a\), mass \(4m\) and centre \(O\) is freely mounted on a fixed horizontal axis which is perpendicular to its plane and passes through \(O\). A uniform heavy chain \(PS\) of length \((4+\pi)a\), mass \((4+\pi)m\) and negligible thickness is hung over the rim of the disc as shown in the diagram: \(Q\) and \(R\) are the points of the chain at the same level as \(O\). The contact between the chain and the rim of the disc is sufficiently rough to prevent slipping. Initially, the system is at rest with \(PQ=RS =2a\). A particle of mass \(m\) is attached to the chain at \(P\) and the system is released. By considering the energy of the system, show that when \(P\) has descended a distance \(x\), its speed \(v\) is given by $$ (\pi+7)av^2 = 2g(x^2+ax). $$ By considering the part \(PQ\) of the chain as a body of variable mass, show that when \(S\) reaches \(R\) the tension in the chain at \(Q\) is $$ {5\pi -2 \over \pi +7} mg. $$

1993 Paper 3 Q14
D: 1700.0 B: 1484.0

A particle rests at a point \(A\) on a horizontal table and is joined to a point \(O\) on the table by a taut inextensible string of length \(c\). The particle is projected vertically upwards at a speed \(64\surd(6gc)\). It next strikes the table at a point \(B\) and rebounds. The coefficient of restitution for any impact between the particle and the table is \({1\over 2}\). After rebounding at \(B\), the particle will rebound alternately at \(A\) and \(B\) until the string becomes slack. Show that when the string becomes slack the particle is at height \(c/2\) above the table. Determine whether the first rebound between \(A\) and \(B\) is nearer to \(A\) or to \(B\).


Solution: \begin{align*} \text{N2}(radially): && T + mg \cos \theta &= m\frac{v^2}{r} \\ \Rightarrow && v^2-gc \cos \theta &\geq 0 \\ \text{COE}: && \frac12 m u^2 &= \frac12mv^2 + mgc\cos \theta \\ \Rightarrow && u^2 &= v^2 + 2gc\cos \theta \\ && u^2 &\geq gc \cos \theta+2gc\cos \theta \\ \Rightarrow && u^2 &\geq 3gc\cos \theta \end{align*} Therefore it will complete bounces with the string taught if it leaves the table with \(u^2 \geq 3gc\). After \(6\) bounces it will leave the table with speed \(\sqrt{6gc} > \sqrt{3gc}\) and after \(7\) bounces it will leave the table with speed \(\sqrt{\frac{3}{2} gc} < \sqrt{3 gc}\). When it leaves the table with speed \(\sqrt{\tfrac32 gc}\), the string will go slack when \begin{align*} && \tfrac32 gc &= 3gc \cos \theta \\ \Rightarrow && \cos \theta &= \frac{1}{2} \end{align*} ie at a height \(c\cos \theta = \frac12c\) above the table. Once the string goes slack, the particle travels under circular motion, \begin{align*} && u^2 &= \frac12 gc \\ \Rightarrow && u_\rightarrow &= \sqrt{\tfrac12 gc} \cos \theta \\ && u_{\uparrow} &= \sqrt{\tfrac12 gc} \sin \theta \\ \Rightarrow && s &= ut - \tfrac12 gt^2 \\ \Rightarrow && -\frac{c}{2} &= \sqrt{\tfrac12 gc} \frac{\sqrt{3}}{2} t - \tfrac12 g t^2 \\ \Rightarrow && t &= \sqrt{\frac{c}{g}} \left ( \frac{\sqrt{6}+\sqrt{22}}{4} \right) \\ \Rightarrow && s_{\rightarrow} &= \tfrac12 \sqrt{\tfrac12 gc} \sqrt{\frac{c}{g}} \left ( \frac{\sqrt{6}+\sqrt{22}}{4} \right) - \frac{\sqrt{3}}{2}c \\ &&&= \left ( \frac{\sqrt{11}-3\sqrt{3}}{8} \right)c \\ \end{align*} We need to establish whether this position is positive or negative (ie if we cross the centre line. Clearly \(\sqrt{11} < 3\sqrt{3}\) so we haven't crossed the centre line and we land closer to where we took off. Since it's the 7th take off, this is closer to \(B\).

1993 Paper 3 Q15
D: 1700.0 B: 1501.5

The probability of throwing a head with a certain coin is \(p\) and the probability of throwing a tail is \(q=1-p\). The coin is thrown until at least two heads and at least two tails have been thrown; this happens when the coin has been thrown \(N\) times. Write down an expression for the probability that \(N=n\). Show that the expectation of \(N\) is $$ 2\bigg({1\over pq} -1-pq\bigg). $$


Solution: This can either occur via \(N-2\) heads and \(1\) tail in the first \(N-1\) flips, followed by a tail, or \(N-2\) tails and \(1\) head in the first \(N-1\) flips, followed by another head, ie \begin{align*} \mathbb{P}(N = n) &= \underbrace{\binom{n-1}{1}}_{\text{ways to choose when the first tail occurs}}p^{n-2}q^2 + \underbrace{\binom{n-1}{1}}_{\text{ways to choose when the first head occurs}}q^{n-2}p^2 \\ &= (n-1)p^2q^2(p^{n-4}+q^{n-4}) \\ \\ \mathbb{E}(N) &= \sum_{n=4}^{\infty} n \cdot \mathbb{P}(N = n) \\ &= \sum_{n=4}^{\infty} n \cdot (n-1)p^2q^2(p^{n-4}+q^{n-4}) \\ &= \sum_{n=4}^{\infty} n \cdot (n-1)(p^{n-2}q^2+q^{n-2}p^2) \\ &= q^2\sum_{n=4}^{\infty} n(n-1)p^{n-2}+p^2\sum_{n=4}^{\infty} n(n-1)q^{n-2} \\ &= q^2\left ( \sum_{n=2}^{\infty} n(n-1)p^{n-2} -2 \cdot 1 - 3 \cdot 2 \cdot p\right)+p^2\left ( \sum_{n=2}^{\infty} n(n-1)q^{n-2} - 2-6q\right) \\ &= q^2\left ( 2(1-p)^{-3} -2 - 6 p\right)+p^2\left ( 2(1-q)^{-3} - 2-6q\right) \\ &= q^2\left ( 2q^{-3} -2 - 6 p\right)+p^2\left ( 2p^{-3} - 2-6q\right) \\ &= \frac{2}{q} - 2q^2 - 6pq^2+\frac{2}{p} -2p^2-6p^2q \\ &= \frac{2}{q}+\frac2p - 2(p^2+q^2) - 6pq \\ &= \frac{2}{pq} - 2((p+q)^2-2pq) - 6pq \\ &= \frac{2}{pq} - 2 -2pq \\ &= 2 \left (\frac1{pq} - 1 - pq \right) \end{align*}