Show that the differential equation \[x^3y'' + (x - 2)(xy' - y) = 0\] has a solution proportional to \(x^\alpha\) for some \(\alpha\). By making the substitution \(y = x^\tau\), or otherwise, find the general solution of this equation.
The cubic equation \[x^3 + ax^2 + bx + c = 0\] has roots \(\alpha, \beta, \gamma\). Find a cubic with roots \(\alpha^3, \beta^3, \gamma^3\), its coefficients being expressed in terms of \(a, b\) and \(c\).
By looking at the coefficient of \(x^n\) in \((1 + x)^{2n}\) in two different ways, or otherwise, show that \[\sum_{r=0}^{n} \left(\frac{1}{r!(n-r)!}\right)^2 = \frac{(2n)!}{(n!)^2}.\] By applying the theorem of the arithmetic and geometric means deduce that \[\left(\frac{(n!)^2(n+1)}{(2n)!}\right)^{(n+1)/4} \leq 1!2!3! \ldots n!.\]
(i) Show that there are 18 four figure numbers containing at least three successive sevens. How many five figure numbers are there with the same property? (ii) What is the last figure (i.e. unit place) of \(7^{1000}\) when written out in full? How many figures are there in the expansion?
Given two sets \(A\) and \(B\), we define the symmetric difference \[A\triangle B = (A \cap B^c) \cup (A^c \cap B)\] (where for any set \(C\), \(C^c\) denotes its complement) Show that (i) \(A\triangle \emptyset = \emptyset\triangle A = A\) (where \(\emptyset\) is the empty set), (ii) \(A\triangle A = \emptyset\), (iii) the operator \(\triangle\) is associative (i.e. \(A\triangle(B\triangle C) = (A\triangle B)\triangle C\) for any three sets \(A, B, C\)), (iv) \(x \in A_1\triangle A_2\triangle \ldots \triangle A_n\) (\(n \geq 2\)) if and only if \(x \in A_j\) for an odd number of \(j\)'s.
Suppose that \(x\) and \(y\) are real and satisfy the equations \begin{align*} 2x^3\cos 3y + 2x^2\cos 2y + x\cos y &= -\frac14\\ 2x^3\sin 3y + 2x^2\sin 2y + x\sin y &= 0 \end{align*} Show that \(x^2 = \frac{1}{4}\) and find the possible values of \(y\).
Solution: Note that \(2(e^{yi}x)^3+2(xe^{yi})^2 + (xe^{iy}) = -4\) Let \(z = xe^{iy}\) in which case \(2z^3 + 2z^2+ z + \frac14 =0\), but then \begin{align*} 0 &= 1 + 4z+8x^2+8z^3 \\ &= (1+2z)(1+2z+4z^2) \end{align*} Therefore the roots are \(z = -\frac12\) (and so \(x^2 =\frac14\)) or the roots are complex conjugates whose product is \(\frac14\) so we also have \(x^2 =\frac12\). Solving \(4z^2 + 2z + 1 = 0 \Rightarrow z = \frac{-2 \pm \sqrt{4-16}}{8} = \frac{-2 \pm 2i\sqrt{3}}{8} = \frac12 \left (\frac{-1 \pm i\sqrt{3}}{2} \right)\) so all solutions are \begin{align*} (x,y) = \left (-\frac12, 2n \pi \right),\left (\frac12, (2n+1) \pi \right), \left (\frac12, \frac{2 \pi}{3}+ n \pi \right), \left (-\frac12, \frac{2 \pi}{3} + \frac{\pi}{2} +n \pi \right), \end{align*}
Two Oxford undergraduates, Algy and Berty, resolve to duel with champagne corks at twenty paces. Each shot that Algy fires has a probability 1/4 of hitting Berty and each shot that Berty fires has a probability 1/5 of hitting Algy. The supply of corks is unlimited. What are Algy's chances of escaping without being hit if (i) they fire alternately starting with Algy until one of them is hit? (ii) they fire alternately starting with Berty until one of them is hit? (iii) they fire simultaneously and repeatedly until one or both is hit?
(i) Juggins and Muggins throw two fair dice each. What is the probability that Juggins' total score is strictly greater than that of Muggins? (ii) Villages A, B, C and D are linked by power transmission lines between A and B, A and C, B and C, B and D, C and D. The generator plant is at A. During a severe storm the probability that any particular line will be brought down by the weather is \(p\) (independent of any other). What is the probability that it will be possible to supply D with power after the storm?
A mirror has the form of the ellipse \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.\] Light rays are emitted, in the plane of the ellipse, from the focus \((ae, 0)\) and are reflected at the mirror. Show that they all pass through the other focus, \((-ae, 0)\).
Zarg's Law of space combat says that the rate of destruction of each side's battle cruisers is a constant \(k\) times the number of cruisers on the other side. In a battle there are always so many cruisers that their numbers can be treated as continuous variables. The Args start a battle with \(A_0\) cruisers, and the Bryds with \(B_0\) cruisers, with \(A_0 > B_0\). Show that when all the Bryd cruisers are destroyed there are \((A_0^2 - B_0^2)^{1/2}\) Arg cruisers left. However, before the battle the Bryds discover that \(A_0 < B_0\sqrt{2}\), and that they can split the Arg force into two parts. They can fight one part first and the other part later. How should they split the Arg force in order to win with the smallest losses? What happens if \(A_0 > B_0\sqrt{2}\)?