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1994 Paper 2 Q14
D: 1600.0 B: 1502.2

When Septimus Moneybags throws darts at a dart board they are certain to end on the board (a disc of radius \(a\)) but, it must be admitted, otherwise are uniformly randomly distributed over the board.

  1. Show that the distance \(R\) that his shot lands from the centre of the board is a random variable with variance \(a^{2}/18.\)
  2. At a charity fete he can buy \(m\) throws for \(\pounds(12+m)\), but he must choose \(m\) before he starts to throw. If at least one of his throws lands with \(a/\sqrt{10}\) of the centre he wins back \(\pounds 12\). In order to show that a good sport he is, he is determined to play but, being a careful man, he wishes to choose \(m\) so as to minimise his expected loss. What values of \(m\) should he choose?


Solution:

  1. \(\,\) \begin{align*} && \mathbb{P}(R < d) &= \frac{\pi d^2}{\pi a^2} \\ &&&= \frac{d^2}{a^2} \\ \Rightarrow && f_R(d) &= \frac{2d}{a^2}\\ \\ && \E[R] &= \int_0^a x \cdot f_R(x) \d x \\ &&&= \int_0^a \frac{2x^2}{a^2} \d x \\ &&&= \frac{2a}{3} \\ \\ && \E[R^2] &= \int_0^a x^2 \cdot f_R(x) \d x \\ &&&= \int_0^a \frac{2x^3}{a^2} \d x \\ &&&= \frac{a^2}{2} \\ \Rightarrow && \var[R] &= \frac{a^2}2 - \frac{4a^2}{9} \\ &&7= \frac{a^2}{18} \end{align*}
  2. Let \(p = \mathbb{P}(R < \frac{a}{\sqrt{10}}) = \frac{a^2}{10a^2} = \frac{1}{10}\) be the probability of hitting the target on each throw. His expected loss is \((12+m)p^m + m(1-p^m) = 12p^m + m\). \begin{array}{c|c} m & \text{expected loss} \\ \hline 0 & 12 \\ 1 & \frac{12}{10} + 1 \approx 2.2 \\ 2 & \frac{12}{100} + 2 \approx 2.12 \\ \end{array} If he takes more than \(2\) throws it will definitely cost more than \(3\), therefore he should take exactly \(2\) throws.

1994 Paper 3 Q1
D: 1700.0 B: 1516.0

Calculate \[ \int_{0}^{x}\mathrm{sech}\, t\,\mathrm{d}t. \] Find the reduction formula involving \(I_{n}\) and \(I_{n-2}\), where \[ I_{n}=\int_{0}^{x}\mathrm{sech}^{n}t\,\mathrm{d}t \] and, hence or otherwise, find \(I_{5}\) and \(I_{6}.\)


Solution: \begin{align*} && \int_0^x \mathrm{sech}\, t \d t &= \int_0^x \frac{2}{e^t+e^{-t}} \d t \\ &&&= \int_0^x \frac{2e^t}{e^{2t}+1} \d t \\ &&&= \left [2 \arctan e^t \right ]_0^x \\ &&&= 2\tan^{-1}e^x- \frac{\pi}{2} \end{align*} \begin{align*} && I_n &= \int_0^x \mathrm{sech}^n\, t \d t \\ &&&= \int_0^x \mathrm{sech}^{n-2}\, t \mathrm{sech}^2\, t \d t \\ &&&= \left [ \mathrm{sech}^{n-2}\, t \cdot \tanh t\right ]_0^x - \int_0^x (n-2) \mathrm{sech}^{n-3} \, t \cdot (- \tanh t \mathrm{sech}\, t) \tanh t \d t \\ &&&= \mathrm{sech}^{n-2}\, x \cdot \tanh x+(n-2)\int_0^x \mathrm{sech}^{n-2} t \tanh^2 t \d t \\ &&&= \mathrm{sech}^{n-2}\, x \cdot \tanh x+(n-2)\int_0^x \mathrm{sech}^{n-2} t (1-\mathrm{sech}^2 \, t) \d t \\ &&&= \mathrm{sech}^{n-2}\, x \cdot \tanh x+(n-2)I_{n-2}-(n-2)I_n \\ \Rightarrow && (n-1)I_n &= \mathrm{sech}^{n-2}\, x \cdot \tanh x+(n-2)I_{n-2} \\ \Rightarrow && I_n &= \frac{1}{n-1} \left ( \mathrm{sech}^{n-2}\, x \cdot \tanh x+(n-2)I_{n-2} \right) \\ \end{align*} \begin{align*} I_1 &= 2\tan^{-1}e^x- \frac{\pi}{2} \\ I_3 &= \frac12 \left ( \mathrm{sech}\, x \cdot \tanh x+ 2\tan^{-1}e^x- \frac{\pi}{2}\right) \\ &= \frac12 \mathrm{sech}\, x \cdot \tanh x+ \tan^{-1}e^x- \frac{\pi}{4} \\ I_5 &= \frac14 \left (\mathrm{sech}^3\, x \cdot \tanh x + 3 \left ( \frac12 \mathrm{sech}\, x \cdot \tanh x+ \tan^{-1}e^x- \frac{\pi}{4} \right) \right) \\ &= \frac14 \mathrm{sech}^3\, x \cdot \tanh x +\frac38 \mathrm{sech}\, x \cdot \tanh x+\frac34 \tan^{-1}e^x- \frac{3\pi}{16} \\ \\ I_2 &= \tanh x \\ I_4 &= \frac13 \left ( \mathrm{sech}^2 x\tanh x +2\tanh x\right) \\ &= \frac13 \mathrm{sech}^2 x\tanh x +\frac23 \tanh x \\ I_6 &= \frac15 \left ( \mathrm{sech}^4 x \tanh x+4\left ( \frac13 \mathrm{sech}^2 x\tanh x +\frac23 \tanh x \right) \right) \\ &= \frac15 \mathrm{sech}^4 x \tanh x + \frac4{15}\mathrm{sech}^2 x\tanh x + \frac{8}{15} \tanh x \end{align*}

1994 Paper 3 Q2
D: 1700.0 B: 1516.0

  1. By setting \(y=x+x^{-1},\) find the solutions of \[ x^{4}+10x^{3}+26x^{2}+10x+1=0. \]
  2. Solve \[ x^{4}+x^{3}-10x^{2}-4x+16=0. \]


Solution:

  1. \begin{align*} && x^{4}+10x^{3}+26x^{2}+10x+1 &= 0 \\ \Leftrightarrow && x^2 + 10x + 26 + 10x^{-1} + x^{-2} &= 0 \\ \Leftrightarrow && (x^2 + x^{-2} + 2) + 10(x+x^{-1}) + 24 &= 0 \\ \Leftrightarrow && y^2 + 10y + 24 &= 0 \tag{\(y = x + x^{-1}\)} \\ \Leftrightarrow && (y+6)(y+4) &= 0 \\ \Leftrightarrow && \begin{cases} x+x^{-1} = -4 \\ x+x^{-1} = -6 \\ \end{cases} \\ \Leftrightarrow && \begin{cases} x^2+4x+1 = 0 \\ x^2+6x+1 = 0 \\ \end{cases} \\ \Leftrightarrow && \boxed{\begin{cases} x = -2 \pm \sqrt{3} \\ x = -3 \pm 2\sqrt{2} \\ \end{cases}} \\ \end{align*}
  2. \begin{align*} && x^{4}+x^{3}-10x^{2}-4x+16=0 &= 0 \\ \Leftrightarrow && x^2 + x - 10 - 4x^{-1} + 4x^{-2} &= 0 \\ \Leftrightarrow && (x^2+4x^{-2} - 4) + (x - 4x^{-1}) - 6 &= 0 \\ \Leftrightarrow && (x^2+4x^{-2} - 4) + (x - 4x^{-1}) - 6 &= 0 \\ \Leftrightarrow && z^2 + z - 6 &= 0 \tag{\(z = x -2x^{-1}\)} \\ \Leftrightarrow && (z+3)(z-2) &= 0 \\ \Leftrightarrow && \begin{cases} x-2x^{-1} = -3 \\ x-2x^{-1} = 2 \\ \end{cases} \\ \Leftrightarrow && \begin{cases} x^2+3x-2 = 0 \\ x^2-2x-2 = 0 \\ \end{cases} \\ \Leftrightarrow && \boxed{\begin{cases} x = \frac{-3 \pm \sqrt{17}}{2} \\ x = 1 \pm \sqrt{3} \\ \end{cases}} \\ \end{align*}

1994 Paper 3 Q3
D: 1700.0 B: 1516.0

Describe geometrically the possible intersections of a plane with a sphere. Let \(P_{1}\) and \(P_{2}\) be the planes with equations \begin{alignat*}{1} 3x-y-1 & =0,\\ x-y+1 & =0, \end{alignat*} respectively, and let \(S_{1}\) and \(S_{2}\) be the spheres with equations \begin{alignat*}{1} x^{2}+y^{2}+z^{2} & =7,\\ x^{2}+y^{2}+z^{2}-6y-4z+10 & =0, \end{alignat*} respectively. Let \(C_{1}\) be the intersection of \(P_{1}\) and \(S_{1},\) let \(C_{2}\) be the intersection of \(P_{2}\) and \(S_{2}\) and let \(L\) be the intersection of \(P_{1}\) and \(P_{2}.\) Find the points where \(L\) meets each of \(S_{1}\) and \(S_{2}.\) Determine, giving your reasons, whether the circles \(C_{1}\) and \(C_{2}\) are linked.

1994 Paper 3 Q4
D: 1700.0 B: 1484.7

Find the two solutions of the differential equation \[ \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}=4y \] which pass through the point \((a,b^{2}),\) where \(b\neq0.\) Find two distinct points \((a_{1},1)\) and \((a_{2},1)\) such that one of the solutions through each of them also passes through the origin. Show that the graphs of these two solutions coincide and sketch their common graph, together with the other solutions through \((a_{1},1)\) and \((a_{2},1)\). Now sketch sufficient members of the family of solutions (for varying \(a\) and \(b\)) to indicate the general behaviour. Use your sketch to identify a common tangent, and comment briefly on its relevance to the differential equation.

1994 Paper 3 Q5
D: 1700.0 B: 1486.8

The function \(\mathrm{f}\) is given by \(\mathrm{f}(x)=\sin^{-1}x\) for \(-1 < x < 1.\) Prove that \[ (1-x^{2})\mathrm{f}''(x)-x\mathrm{f}'(x)=0. \] Prove also that \[ (1-x^{2})\mathrm{f}^{(n+2)}(x)-(2n+1)x\mathrm{f}^{(n+1)}(x)-n^{2}\mathrm{f}^{(n)}(x)=0, \] for all \(n>0\), where \(\mathrm{f}^{(n)}\) denotes the \(n\)th derivative of \(\mathrm{f}\). Hence express \(\mathrm{f}(x)\) as a Maclaurin series. The function \(\mathrm{g}\) is given by \[ \mathrm{g}(x)=\ln\sqrt{\frac{1+x}{1-x}}, \] for \(-1 < x < 1.\) Write down a power series expression for \(\mathrm{g}(x),\) and show that the coefficient of \(x^{2n+1}\) is greater than that in the expansion of \(\mathrm{f},\) for each \(n > 0\).

1994 Paper 3 Q6
D: 1700.0 B: 1501.5

The four points \(A,B,C,D\) in the Argand diagram (complex plane) correspond to the complex numbers \(a,b,c,d\) respectively. The point \(P_{1}\) is mapped to \(P_{2}\) by rotating about \(A\) through \(\pi/2\) radians. Then \(P_{2}\) is mapped to \(P_{3}\) by rotating about \(B\) through \(\pi/2\) radians, \(P_{3}\) is mapped to \(P_{4}\) by rotating about \(C\) through \(\pi/2\) radians and \(P_{4}\) is mapped to \(P_{5}\) by rotating about \(D\) through \(\pi/2\) radians, each rotation being in the positive sense. If \(z_{i}\) is the complex number corresponding to \(P_{i},\) find \(z_{5}\) in terms of \(a,b,c,d\) and \(z_{1}.\) Show that \(P_{5}\) will coincide with \(P_{1},\) irrespective of the choice of the latter if, and only if \[a-c=\mathrm{i}(b-d)\] and interpret this condition geometrically. The points \(A,B\) and \(C\) are now chosen to be distinct points on the unit circle and the angle of rotation is changed to \(\theta,\) where \(\theta\neq0,\) on each occasion. Find the necessary and sufficient condition on \(\theta\) and the points \(A,B\) and \(C\) for \(P_{4}\) always to coincide with \(P_{1}.\)

1994 Paper 3 Q7
D: 1679.5 B: 1503.1

Let \(S_{3}\) be the group of permutations of three objects and \(Z_{6}\) be the group of integers under addition modulo 6. List all the elements of each group, stating the order of each element. State, with reasons, whether \(S_{3}\) is isomorphic with \(Z_{6}.\) Let \(C_{6}\) be the group of 6th roots of unity. That is, \(C_{6}=\{1,\alpha,\alpha^{2},\alpha^{3},\alpha^{4},\alpha^{5}\}\) where \(\alpha=\mathrm{e}^{\mathrm{i}\pi/3}\) and the group operation is complex multiplication. Prove that \(C_{6}\) is isomorphic with \(Z_{6}.\) Is there any (multiplicative or additive) subgroup of the complex numbers which is isomorphic with \(S_{3}\)? Give a reason for your answer.


Solution: \(S_3 \) $\begin{array}{c | c |c |c |c |c |c |} \text{elements} & e & (12) & (13) & (23) & (123) & (132) \\ \text{order} & 1 & 2 & 2 & 2 & 3 & 3 \\ \end{array}$ \(\mathbb{Z}_6\) $\begin{array}{c | c |c |c |c |c |c |} \text{elements} & 0 & 1 & 2 & 3 & 4 & 5 \\ \text{order} & 1 & 6 & 3 & 2 & 3 & 6 \\ \end{array}$ \(S_3\) is not isomorphic to \(\mathbb{Z}_6\) since \(\mathbb{Z}_6\) has two elements of order \(6\) but \(S_3\) has none. Consider the map \(f : \mathbb{Z}_6 \to C_6\) with \(i \mapsto \alpha^i\). This is an isomorphism, since \(i + j \mapsto \alpha^{i+j} = \alpha^i\alpha^j\) \(S_3\) is non-abelian, since \((12)(123) = (23) \neq (13) = (123)(12)\) but multiplication and addition of complex numbers is commutative.

1994 Paper 3 Q8
D: 1700.0 B: 1516.0

Let \(a,b,c,d,p,q,r\) and \(s\) be real numbers. By considering the determinant of the matrix product \[ \begin{pmatrix}z_{1} & z_{2}\\ -z_{2}^{*} & z_{1}^{*} \end{pmatrix}\begin{pmatrix}z_{3} & z_{4}\\ -z_{4}^{*} & z_{3}^{*} \end{pmatrix}, \] where \(z_{1},z_{2},z_{3}\) and \(z_{4}\) are suitably chosen complex numbers, find expressions \(L_{1},L_{2},L_{3}\) and \(L_{4},\) each of which is linear in \(a,b,c\) and \(d\) and also linear in \(p,q,r\) and \(s,\) such that \[ (a^{2}+b^{2}+c^{2}+d^{2})(p^{2}+q^{2}+r^{2}+s^{2})=L_{1}^{2}+L_{2}^{2}+L_{3}^{2}+L_{4}^{2}. \]


Solution: Supppose \(z_1 = a+ib, z_2 = c+id, z_3 = p+iq, z_4 = r+is\) then: \begin{align*} && \det \left (\begin{pmatrix}z_{1} & z_{2}\\ -z_{2}^{*} & z_{1}^{*} \end{pmatrix}\begin{pmatrix}z_{3} & z_{4}\\ -z_{4}^{*} & z_{3}^{*} \end{pmatrix} \right) &= \det \begin{pmatrix}z_{1} & z_{2}\\ -z_{2}^{*} & z_{1}^{*} \end{pmatrix}\det\begin{pmatrix}z_{3} & z_{4}\\ -z_{4}^{*} & z_{3}^{*} \end{pmatrix} \\ && \det \begin{pmatrix}z_{1}z_3-z_2z_4^* & z_1z_4+z_2z_3^*\\ -z_2^*z_3-z_1^*z_4*& -z_2^*z_4+z_{1}^*z_3^* \end{pmatrix}&= (z_1z_1^*+z_2z_2^*)(z_3z_3^*+z_4z_4^*) \\ && |z_{1}z_3-z_2z_4^*|^2+|z_1z_4+z_2z_3^*|^2&= (a^2+b^2+c^2+d^2)(p^2+q^2+r^2+s^2) \\ && L_1^2 + L_2^2+L_3^2+L_4^2 &= \ldots \end{align*}

1994 Paper 3 Q9
D: 1700.0 B: 1500.0

A smooth, axially symmetric bowl has its vertical cross-sections determined by \(s=2\sqrt{ky},\) where \(s\) is the arc-length measured from its lowest point \(V\), and \(y\) is the height above \(V\). A particle is released from rest at a point on the surface at a height \(h\) above \(V\). Explain why \[ \left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)^{2}+2gy \] is constant. Show that the time for the particle to reach \(V\) is \[ \pi\sqrt{\frac{k}{2g}}. \] Two elastic particles of mass \(m\) and \(\alpha m,\) where \(\alpha<1,\) are released simultaneously from opposite sides of the bowl at heights \(\alpha^{2}h\) and \(h\) respectively. If the coefficient of restitution between the particles is \(\alpha,\) describe the subsequent motion.