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1960 Paper 1 Q101
D: 1500.0 B: 1500.0

Sketch the curve \(x^2 = (y-k)^2(y-2k)\), where \(x\), \(y\) are real variables and \(k\) is constant, in the three cases (i) \(k < 0\), (ii) \(k = 0\), (iii) \(k > 0\). Describe the nature of the singularity in each case.

1960 Paper 1 Q102
D: 1500.0 B: 1500.0

Establish a condition on the coefficients \(p\), \(q\), \(r\) for the equation \(x^3 + 3px^2 + 3qx + r = 0\) to have a repeated root. Show that if this condition is satisfied and \(q \neq p^2\), then the repeated root is $$\frac{pq - r}{2(q - p^2)}.$$ Find an expression for the third root.

1960 Paper 1 Q103
D: 1500.0 B: 1500.0

Solve the simultaneous recurrence relations \begin{align} x_{n+1} &= x_n + y_n,\\ y_{n+1} &= 4x_n - 2y_n, \end{align} with \(x_0 = 5\), \(y_0 = 0\).

1960 Paper 1 Q104
D: 1500.0 B: 1460.7

Sum the series $$\sum_1^q \frac{1}{n(n+1)}.$$ Prove that $$\frac{1}{p+1} - \frac{1}{q+1} < \sum_{p+1}^q \frac{1}{n^3} < \frac{1}{p} - \frac{1}{q}.$$ If \(S\) denotes the sum in the middle, and \(A\) denotes the average of the two bounds, prove that $$\frac{1}{3}\left\{\frac{1}{p(p+1)(p+2)} - \frac{1}{q(q+1)(q+2)}\right\} < A - S < \frac{1}{3}\left\{\frac{1}{(p-1)p(p+1)} - \frac{1}{(q-1)q(q+1)}\right\}.$$

1960 Paper 1 Q105
D: 1500.0 B: 1500.0

Let \(p\) be a prime greater than 3. Assume the theorem that if \(0 < n < p\) then there are integers \(a\), \(b\) such that \(1 = an + bp\). Prove that if \(1 < n < p-1\) then \(a \neq n\). Hence or otherwise prove that \((p-2)! - 1\) is divisible by \(p\).

1960 Paper 1 Q106
D: 1500.0 B: 1500.0

The circle \(A\) is contained inside the circle \(B\). Let \(L\), \(L'\) be the limit points of the coaxal system of circles containing \(A\), \(B\); suppose that \(L\) lies outside \(B\). Let \(\alpha\), \(\beta\) be the angles between \(LL'\) and the tangents from \(L\) to \(A\), \(B\) respectively. Prove that there is a sequence of circles \(C_1\), \(C_2\), \(\ldots\), \(C_n\), each \(C_i\) touching \(A\), \(B\) and \(C_{i-1}\) and \(C_1\) touching \(C_n\), provided that $$\frac{\sin\alpha}{\sin\beta} = \frac{1-\sin(\pi/n)}{1+\sin(\pi/n)}.$$

1960 Paper 1 Q107
D: 1500.0 B: 1500.0

In each of the following cases either prove the statement true, or give a counter-example to show it is false.

  1. [(i)] The diagonals of a parallelogram are perpendicular and bisect each other.
  2. [(ii)] If \(x\), \(y\) are the sides and \(a\), \(b\) the diagonals of a parallelogram, then $$2(x^2 + y^2) = a^2 + b^2.$$
  3. [(iii)] If in a hexagon opposite sides are equal and parallel, then the three diagonals are concurrent and bisect each other.
  4. [(iv)] If in a hexagon opposite sides are equal and parallel, with lengths \(x\), \(y\), \(z\), and if \(a\), \(b\), \(c\) are the diagonals, then $$3(x^2 + y^2 + z^2) = a^2 + b^2 + c^2.$$

1960 Paper 1 Q108
D: 1500.0 B: 1500.0

Tangents \(TP\), \(TP'\) are drawn to an ellipse whose foci are \(F\), \(F'\). Prove that the angles \(FTP\), \(F'TP'\) are equal.

1960 Paper 1 Q109
D: 1500.0 B: 1500.0

The parabola \(y^2 = 4ax\) is parametrised by \((at^2, 2at)\) where \(t\) is variable. If the normal at \(P\) meets the parabola again in \(Q\), find the relation between the parameters at \(P\) and \(Q\). Let \(P_1 P_2\) be a chord through the focus, and let the normals at \(P_1\), \(P_2\) meet the parabola again in \(Q_1\), \(Q_2\) respectively. Let \(R\), \(S\) be the midpoints of \(P_1 P_2\), \(Q_1 Q_2\) respectively. Find the coordinates of \(S\) in terms of the coordinates of \(R\).

1960 Paper 1 Q110
D: 1500.0 B: 1500.0

Four points \(P_1\), \(P_2\), \(P_3\), \(P_4\) are not coplanar. The line through \(P_1\) and \(P_3\) is denoted by \(l_{13}\). A plane \(\pi\) cuts \(l_{13}\) in \(Q_{13}\). Prove that \(Q_{13}\), \(Q_{24}\), \(Q_{14}\) are collinear. Draw the figure in the plane \(\pi\) that is the intersection of \(\pi\) with the edges and faces of the tetrahedron \(P_1 P_2 P_3 P_4\). Describe the effect on the figure if \(\pi\), \(P_1\), \(P_3\) are kept fixed, and \(P_2\), \(P_4\) are moved to different positions on \(l_{24}\), \(l_{24}\), respectively.