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1967 Paper 1 Q1
D: 1500.0 B: 1500.0

Show that, for every positive integer \(n\), the number \(n^9 - n\) is divisible by 30 and that, for every odd positive integer \(n\), \(n^9 - n\) is divisible by 480.

1967 Paper 1 Q2
D: 1500.0 B: 1500.0

Let $$p(x) = 8x^4 - 8x^2 + 1.$$ Given that \(\cos 4\theta = p(\cos \theta)\), sketch the graph of \(y = p(x)\) as \(x\) ranges from \(-1\) to \(+1\). Now suppose that $$f(x) = 8x^4 + a_3x^3 + a_2x^2 + a_1x + a_0$$ is a polynomial such that \(-1 \leq f(x) \leq 1\) whenever \(-1 \leq x \leq 1\). What conclusion do you draw from a consideration of the number of roots of \(f(x) - p(x)\)? Show that, whatever the values of the real numbers \(b_0, b_1, b_2\) and \(b_3\), there exists an \(x\) such that \(0 \leq x \leq 4\) and such that $$|x^4 + b_3x^3 + b_2x^2 + b_1x + b_0| \geq 2.$$

1967 Paper 1 Q3
D: 1500.0 B: 1500.0

The sequences \(x_1, x_2, x_3, \ldots\) and \(y_1, y_2, y_3, \ldots\) are connected by the simultaneous equations \begin{align} x_{n+1} - x_n + y_n &= 0\\ x_n + y_{n+1} - y_n &= 1 \end{align} \((n \geq 1)\). It is given that \(x_1 = y_1 = 1\); find \(x_n\) and \(y_n\) for all \(n > 1\).

1967 Paper 1 Q4
D: 1500.0 B: 1500.0

Six chairs are equally spaced around a circular table at which three married couples are to have a meal. There are, of course, 6! possible seating arrangements. (Assuming here rotations are counted as separate arrangements)

  1. [(i)] In how many of these arrangements would each man sit by his wife?
  2. [(ii)] For how many arrangements is it true that no man would sit by his wife?
  3. [(iii)] How many arrangements are there in which exactly one man would sit by his wife?


Solution:

  1. If we label the chairs \(1, 2, \ldots, 6\) then we can either have couples in \(\{1,2\}, \{3,4\}, \{5, 6\}\) or in \(\{1, 6\}, \{2, 3\}, \{4, 5\}\). There are \(3!\) ways to arrange the couples into the pairs and 2^3 ways to order them. We can do this for each configuration, therefore there are \(2^4 \cdot 3! = 96\) arrangements.
  2. Let \(P_i\) be the set of arrangements where the \(i\)th couple sit together. Then \(|P_i|\) can be counted as pick the pair of chairs for the couple \(6\) ways. Pick the order \(2\) ways. The remaining foursome can go in \(4!\) ways, ie \(12 \cdot 4! = 288\). \(|P_i \cap P_j|\) can be counted by picking a set of chairs for the first couple (\(6\) ways), picking a set of chairs for the second couple (\(3\) ways), permuting each couple (\(2\) ways each) and then placing the remaining couple, \(2\) ways, so \(6 \cdot 3 \cdot 2^2 \cdot 2 = 144\) We have already calculated \(|P_1 \cap P_2 \cap P_3| = 96\). Therefore the ways there can be no couples together is \(720 - 3 \cdot 288 + 3 \cdot 144 - 96 = 192\)
  3. If exactly one person sits by their wife, we want: \begin{align*} |P_1 \cup P_2 \cup P_3| &= 3\sum |P_i| - 2 \sum |P_i \cap P_j| + 3|P_1 \cap P_2 \cap P_3| \\ &= 3 \cdot 288 - 2 \cdot 3 \cdot 144 + 3\cdot96 \\ &= 288 \end{align*}

1967 Paper 1 Q5
D: 1500.0 B: 1500.0

The complex numbers \(\alpha, \beta, \gamma, \delta\) are all non-zero and are also such that $$s_1 = s_5 = s_8 = 0,$$ where \(s_n\) (\(n = 1, 2, 3, \ldots\)) is defined by the equation $$s_n = \alpha^n + \beta^n + \gamma^n + \delta^n.$$ Prove that \(s_3 = 0\).

1967 Paper 1 Q6
D: 1500.0 B: 1500.0

The vertices \(A_1, A_2, A_3, A_4, A_5\) of a regular pentagon lie on a circle of unit radius with centre at the point \(O\). \(A_1\) is the mid-point of \(OP\). Prove that

  1. [(i)] \(PA_1 \cdot PA_2 \cdot PA_3 \cdot PA_4 \cdot PA_5 = 31\);
  2. [(ii)] \(\sum_{r=1}^{5} (PA_r)^2 = 25\);
  3. [(iii)] \(A_1A_2 \cdot A_1A_3 \cdot A_1A_4 \cdot A_1A_5 = 5\).

1967 Paper 1 Q7
D: 1500.0 B: 1500.0

\(OA\), \(OB\), \(OC\) are three lines through the point \(O\). The angles \(BOC\), \(COA\) and \(AOB\) are, respectively, \(\alpha\), \(\beta\) and \(\gamma\). Calculate \(\cos^2\theta\), where \(\theta\) is the angle between the line \(OA\) and the plane \(BOC\).

1967 Paper 1 Q8
D: 1500.0 B: 1500.0

Points \(X\) and \(Y\) are chosen, on the perpendiculars (produced if necessary) from the vertices \(A\) and \(B\) of a triangle \(ABC\) to the opposite sides, so that \(AX = BC\) and \(BY = AC\). Prove that \(XOY\) is a right-angled isosceles triangle.

1967 Paper 1 Q9
D: 1500.0 B: 1500.0

\(ABC\) is an isosceles triangle, with \(AB = AC\), \(I\) is the centre of the inscribed circle. \(S, I_1\) is the centre of the circle \(S_1\) touching \(BC\) internally and \(AB\), \(AC\) externally. Prove that the circle \(S'\) on \(I_1\) as diameter touches \(AB\), \(AC\) at \(B\) and \(C\), and that if \(S'\) meets \(S\) in \(P, Q\) and \(S_1\) in \(P_1, Q_1\), then \(I_1P\), \(I_1Q\) touch \(S\) and \(IP_1\), \(IQ_1\) touch \(S_1\).

1967 Paper 1 Q10
D: 1500.0 B: 1500.0

Two circles \(C_1, C_2\) of radii \(r_1\) and \(r_2\), each touch the parabola \(y^2 = 4ax\) in two points. Show that the centres of the circles lie on the axis of the parabola, and that, if \(C_1\) and \(C_2\) touch each other, then the difference between their radii is \(4a\).