Problems

Filters
Clear Filters
1959 Paper 2 Q108
D: 1500.0 B: 1500.0

Sketch the curve \(C\) whose equation in polar coordinates is $$r^2 = a^2\cos 2\theta,$$ where \(a > 0\) and it is understood that \(r\) may take negative as well as positive values. Show that the perimeter \(s\) of \(C\) is given by $$s = 4a \int_0^{\pi/4} \frac{d\theta}{\sqrt{\cos 2\theta}}.$$ By means of the substitution \(t = \tan^4\theta\), or otherwise, express \(s\) in terms of the function \(B(p, q)\) defined by $$B(p, q) = \int_0^1 t^{p-1}(1-t)^{q-1} dt \quad (p > 0, q > 0).$$

1959 Paper 2 Q109
D: 1500.0 B: 1500.0

The functions \(u = u(x, y)\) and \(v = v(x, y)\) satisfy the equations $$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x},$$ identically. By means of the substitution \(x = X + Y\), \(y = X - Y\), or otherwise, prove that \(u + v\) is a function of \(x + y\), and \(u - v\) is a function of \(x - y\). What can be said about \(f(x, y)\) if it satisfies the equation $$\frac{\partial^2 f}{\partial x^2} = \frac{\partial^2 f}{\partial y^2}$$ identically? Illustrate your conclusion in the cases

  1. [(i)] \(f(x, y) = \sin x \cos y\), (ii) \(f(x, y) = xy\).
(It may be assumed that all partial derivatives involved are continuous, and that $$\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x}.)$$

1959 Paper 2 Q110
D: 1500.0 B: 1500.0

Verify that the differential equation $$y'' = (x^2 - 1)y,$$ where dashes denote differentiations with respect to \(x\), is satisfied by \(y = e^{-\frac{1}{2}x^2}\). By writing \(y = ue^{-\frac{1}{2}x^2}\) and forming the differential equation for \(u\), or otherwise, obtain an expression for the general solution of the equation in \(y\). Show that the solution for which \(y = a\) and \(y' = b\) when \(x = 0\) may be written in the form $$y = ae^{-\frac{1}{2}x^2} + b \int_0^x e^{-\frac{1}{2}t^2} dt.$$

1959 Paper 2 Q201
D: 1500.0 B: 1500.0

In a certain examination the possible marks were integers from 0 to 100; for each such integer there was at least one candidate who obtained that mark. In a second examination, taken by the same candidates, and with the same possible marks, it was found that for each pair of candidates the mark obtained by one was greater than that obtained by the other only if the same had been true in the first examination. Prove that at least one candidate obtained the same mark in both examinations. If a third examination is taken, and the second sentence of the above paragraph remains true when 'third' and 'second' are substituted for 'second' and 'first' respectively, is it necessarily true—

  1. [(a)] that some candidate obtained the same mark in the second and third examinations;
  2. [(b)] that some candidate obtained the same mark in the first and third;
  3. [(c)] that some candidate obtained the same mark in all three examinations?
Justify your answer in each case, either by a proof or by constructing an example in disproof.

1959 Paper 2 Q202
D: 1500.0 B: 1500.0

Prove that there cannot exist four (real or complex) numbers, all different, such that the square of each of them is equal to the sum of the cubes of the other three. Prove also that there is one (and only one) set of four distinct numbers such that the square of each is equal to the sum of the fourth powers of the other three. Find the equation whose roots are these numbers.

1959 Paper 2 Q203
D: 1500.0 B: 1500.0

Let \(p_n\) be the number of ways in which a collection of \(n\) dissimilar objects may be divided into parts, each of which contains at least one object, no regard being paid to order. By convention, the 'division' of the collection into a single part, namely the whole collection, is to be included in this number; also \(p_0\) is to be taken as 1. Prove that \[p_{n+1} = \sum_{m=0}^n \binom{n}{m} p_{n-m},\] where \[\binom{n}{m} = \frac{n!}{m!(n-m)!}\] (0! being taken as 1), and hence, or otherwise, prove that \(p_n/n!\) is the coefficient of \(x^n\) in the expansion in ascending powers of \(x\) of \[e^{x-1}.\]

1959 Paper 2 Q204
D: 1500.0 B: 1500.0

Given that any solution of the differential equation \[u'' + u = 0\] (where a dash denotes differentiation with respect to \(x\)) must be of the form \(A \sin x + B \cos x\), where \(A\) and \(B\) are constants, and that the function \(f(x)\) satisfies the differential equation \[y'' + \frac{2}{x} y' + y = 0,\] calculate \(\int_a^b xf(x) dx\) in terms of \(f(a)\) and \(f(b)\), indicating any special cases that may arise. Calculate \(\int_a^b x^2f(x) dx\) in terms of \(f'(a)\) and \(f'(b)\).

1959 Paper 2 Q205
D: 1500.0 B: 1500.0

Nine distinct points, not all collinear, are such that the line joining any two of them passes through a third. Prove that

  1. [(a)] no four of the points are collinear;
  2. [(b)] the points define exactly twelve lines, each passing through three of the points;
  3. [(c)] through each of the nine points there pass exactly four of the twelve lines.
[It is not possible to give an example of this configuration in which all the points are real.]

1959 Paper 2 Q206
D: 1500.0 B: 1500.0

'The centre of a circle that touches each of two given circles must lie on a certain hyperbola, whose centre is the mid-point of the straight line joining the centres of the given circles.' Examine the above statement; if you find it true, give a proof of it; if not, state and prove a suitably modified version.

1959 Paper 2 Q207
D: 1500.0 B: 1500.0

Enunciate the principle of virtual work. A light lever \(AOB\) of length \(2a\) can turn freely about its midpoint \(O\), and hangs from the end \(A\). A light rod \(BC\) of length \(h\) is smoothly jointed to \(AB\) and to a weight \(W\) and is restricted by a frictionless constraint to move in the vertical through \(O\). Investigate the positions of equilibrium according as \(a > b\), \(a = b\), or \(a < b\). Show also for the case \(a < b\) that if turning at \(B\) is resisted by a constant frictional couple, there is a position of limiting equilibrium in which the lever makes an angle \(\eta\) with the vertical given by the equation \[4Wa^2 F \sin \theta \cos^2 \theta = (b^2 - a^2)(Wa \sin \theta - F)^2.\]