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1966 Paper 4 Q1
D: 1500.0 B: 1500.0

A \emph{plane convex set} is a set of points in a plane such that any point of the line-segment joining any 2 points of the set also belongs to the set. 4 plane convex sets \(S_1, S_2, S_3, S_4\) are such that \(S_c, S_2, S_3\) have a common point \(P_1\); \(S_3, S_4, S_1\) have a common point \(P_2\); \(S_4, S_1, S_2\) have a common point \(P_3\); and \(S_1, S_2, S_3\) have a common point \(P_4\). By considering separately the three cases:

  1. where one of the points \(P_i\) falls inside (or on the boundary of) the triangle formed by the other three;
  2. where the points \(P_i\) form the vertices of a convex quadrilateral;
  3. where the points \(P_i\) are collinear;
and showing that these cases are exhaustive, prove that \(S_1, S_2, S_3, S_4\) all have at least one point in common.

1966 Paper 4 Q2
D: 1500.0 B: 1500.0

A \emph{semi-group} is a set of elements \(a, b, c, \ldots\) endowed with an operation, multiplication, denoted by juxtaposition (thus `\(a\) times \(b\)' is written \(ab\)) such that the product of any pair of elements is in the set and multiplication is associative, that is \((ab)c = a(bc)\), but not necessarily commutative (\(ab\) is not necessarily equal to \(ba\)). It is given that a certain semi-group possesses a right identity \(i\) (that is, \(ai = a\) for all elements \(a\)), and that every element \(a\) has a right inverse (that is, there exists an element \(a'\) such that \(aa' = i\)). By considering \(aa'a\) where \(aa' = i\), \(a'a = i\), or otherwise, prove (in either order):

  1. that \(i\) is also a left identity (that is, \(ia = a\) for all \(a\));
  2. that the right inverse \(a'\) (as defined above) is also a left inverse (that is, \(a'a = i\)).

1966 Paper 4 Q3
D: 1500.0 B: 1500.0

Four cards, the aces of hearts, diamonds, spades and clubs are well shuffled, and then dealt two to player \(A\), the other two to player \(B\). \(A\) is then asked whether at least one of his two cards is red. He replies in the affirmative. In the light of this information we wish to calculate the probability that he holds both the red aces. Consider the argument: `We know that \(A\) has one red ace; without loss of generality we may suppose that it is the heart ace. Among the other three cards there is no reason why one more than another should be the diamond ace; one only out of three equally likely possibilities gives \(A\) both the red cards; the required chance is thus 1 in 3.' Criticize this argument, and produce a correct argument and answer.

1966 Paper 4 Q4
D: 1500.0 B: 1500.0

It is given that $$f_n(x) = \sin x + \frac{1}{2}\sin 2x + \frac{1}{3}\sin 3x + \ldots + \left(\frac{1}{n}\right)\sin nx$$ For each integer \(n = 1, 2, 3, \ldots\) If \(x_0\) is any minimum of \(f_n(x)\) in the range \(0 < x < \pi\), prove that \(\sin x_0 < 0\), and hence that \(\sin x_0\) and \(\sin(n + \frac{1}{2})x_0 = \sin \frac{1}{2}x_0\). Deduce, by using mathematical induction on \(n\), that \(f_n(x)\) can never take negative or zero values in the range \(0 < x < \pi\), for any \(n \geq 1\).

1966 Paper 4 Q5
D: 1500.0 B: 1476.7

In the complex polynomial equation $$z^n + a_{n-1}z^{n-1} + a_{n-2}z^{n-2} + \ldots + a_2z^2 + a_1z + 1 = 0,$$ it is given that the complex numbers \(a_{n-1}, a_{n-2}, \ldots, a_2, a_1\) satisfy $$|a_{n-1}| \leq 1, \quad |a_{n-2}| \leq 1, \quad \ldots, \quad |a_2| \leq 1, \quad |a_1| \leq 1.$$ Show that any root of the equation in the complex plane must lie in the annular region \(\frac{1}{2} < |z| < 2\).

1966 Paper 4 Q6
D: 1500.0 B: 1500.0

A hill \(\frac{1}{2}\) mile high is in the shape of a spherical cap, with a horizontal circular rim, the radius of the sphere being 1 mile. A man walks up from a point of the rim to the peak at a steady speed of 3 miles per hour but never ascending at a gradient of more than \(\sin^{-1}(\frac{1}{3})\). Find the minimum time the walk can take him, and sketch roughly a possible minimum path (as seen from above); does it necessarily have no sharp corners?

1966 Paper 4 Q7
D: 1500.0 B: 1500.0

\(p, n\) are positive integers with \(p\) a prime (\(\geq 2\)). Prove that the highest power of \(p\) that divides \(n!\) is exactly $$\left\lfloor\frac{n}{p}\right\rfloor + \left\lfloor\frac{n}{p^2}\right\rfloor + \left\lfloor\frac{n}{p^3}\right\rfloor + \ldots,$$ where \(\lfloor x \rfloor\) denotes the greatest integer not greater than \(x\). Find the highest power of 12 that divides \(120!\)

1966 Paper 4 Q8
D: 1500.0 B: 1500.0

A particle is attached to the end of a light string which passes through a fixed ring. Initially the particle is moving in a horizontal circle, the string making an angle with the vertical. The string is then drawn upwards slowly through the ring until the distance of the particle from it has been halved. Assuming angular momentum is conserved, show that the string now makes an angle \(\alpha'\) with the vertical, where $$\frac{\sin^4 \alpha'}{\cos \alpha'} = 8 \frac{\sin^4 \alpha}{\cos \alpha}.$$

1966 Paper 4 Q9
D: 1500.0 B: 1500.0

A circle of radius \(a\) lies inside a circle of radius \(2a\) and touches it. The two circles lie in the boundary of a uniform lamina which is free to rotate in a vertical plane about a fixed horizontal axis through a point \(P\) on the line of centres. Show that, as \(P\) is varied, the minimum period of oscillations of small amplitude is $$2\pi \left(\frac{a}{g}\right)^{\frac{1}{2}} \left(\frac{74}{9}\right)^{\frac{1}{2}}.$$

1966 Paper 4 Q10
D: 1500.0 B: 1500.0

A light inextensible string \(AB\) of length \(l\) carries a small ring \(A\) at one end and a bob \(B\) at the other end. The ring can slide on a fixed horizontal wire, and initially the system hangs in equilibrium with \(AB\) vertical. At time \(t = 0\) a variable force begins to act on the ring, constraining it to move along the wire in such a way that its displacement is \(a \sin \omega t\). Find the components of acceleration of the bob along and perpendicular to \(AB\) and show that the angle \(\theta\) that \(AB\) makes with the downward vertical satisfies the differential equation $$\frac{d^2\theta}{dt^2} + \frac{g}{l}\sin \theta = \frac{a\omega^2}{l}\sin \omega t \cos \theta.$$ Assuming that \(\theta\) always remains small, calculate it as a function of time.