A particle is projected under the influence of gravity from a point \(O\) on a level plane in such a way that, when its horizontal distance from \(O\) is \(c\), its height is \(h\). It then lands on the plane at a distance \(c+d\) from \(O\). Show that the angle of projection \(\alpha\) satisfies \[ \tan\alpha=\frac{h(c+d)}{cd} \] and that the speed of projection \(v\) satisfies \[ v^{2}=\frac{g}{2}\left(\frac{cd}{h}+\frac{(c+d)^{2}h}{cd}\right)\,. \]
An examiner has to assign a mark between 1 and \(m\) inclusive to each of \(n\) examination scripts (\(n\leqslant m\)). He does this randomly, but never assigns the same mark twice. If \(K\) is the highest mark that he assigns, explain why \[ \mathrm{P}(K=k)=\left.\binom{k-1}{n-1}\right/\binom{m}{n} \] for \(n\leqslant k\leqslant m,\) and deduce that \[ \sum_{k=n}^{m}\binom{k-1}{n-1}=\binom{m}{n}\,. \] Find the expected value of \(K\).
Solution: If the highest mark is \(k\), then there are \(n-1\) remaining marks to give, and they have to be chosen from the numbers \(1, 2, \ldots, k-1\), ie in \(\binom{k-1}{n-1}\) ways. There are \(n\) numbers to be chosen from \(1, 2, \ldots, m\) in total, therefore \(\displaystyle \mathbb{P}(K=k) = \left.\binom{k-1}{n-1} \right/ \binom{m}{n}\) Since \(K\) can take any of the values \(n, \cdots, m\), we must have \begin{align*} && 1 &= \sum_{k=n}^m \mathbb{P}(K=k) \\ &&&= \sum_{k=n}^m \left.\binom{k-1}{n-1} \right/ \binom{m}{n} \\ \Rightarrow && \binom{m}{n} &= \sum_{k=n}^m \binom{k-1}{n-1} \\ \\ && \mathbb{E}(K) &= \sum_{k=n}^m k \cdot \mathbb{P}(K=k) \\ &&&= \sum_{k=n}^m k \cdot \left.\binom{k-1}{n-1} \right/ \binom{m}{n} \\ &&&= n\binom{m}{n}^{-1} \sum_{k=n}^m \frac{k}{n} \cdot \binom{k-1}{n-1} \\ &&&= n\binom{m}{n}^{-1} \sum_{k=n}^m \binom{k}{n} \\ &&&= n\binom{m}{n}^{-1} \sum_{k=n+1}^{m+1} \binom{k-1}{n+1-1} \\ &&&= n\binom{m}{n}^{-1} \binom{m+1}{n+1} \\ &&&= n \cdot \frac{m+1}{n+1} \end{align*}
I have a Penny Black stamp which I want to sell to my friend Jim, but we cannot agree a price. So I put the stamp under one of two cups, jumble them up, and let Jim guess which one it is under. If he guesses correctly, I add a third cup, jumble them up, and let Jim guess correctly, adding another cup each time. The price he pays for the stamp is \(\pounds N,\) where \(N\) is the number of cups present when Jim fails to guess correctly. Find \(\mathrm{P}(N=k)\). Show that \(\mathrm{E}(N)=\mathrm{e}\) and calculate \(\mathrm{Var}(N).\)
Solution: \begin{align*} && \mathbb{P}(N = k) &= \mathbb{P}(\text{guesses }k-1\text{ correctly then 1 wrong})\\ &&&= \frac12 \cdot \frac{1}{3} \cdots \frac{1}{k-1} \frac{k-1}{k} \\ &&&= \frac{k-1}{k!} \\ &&\mathbb{E}(N) &= \sum_{k=2}^\infty k \cdot \mathbb{P}(N=k) \\ &&&= \sum_{k=2}^{\infty} \frac{k(k-1)}{k!} \\ &&&= \sum_{k=0}^{\infty} \frac{1}{k!} = e \\ && \textrm{Var}(N) &= \mathbb{E}(N^2) - \mathbb{E}(N)^2 \\ && \mathbb{E}(N^2) &= \sum_{k=2}^{\infty} k^2 \mathbb{P}(N=k) \\ &&&= \sum_{k=2}^{\infty} \frac{k^2(k-1)}{k!} \\ &&&= \sum_{k=0}^{\infty} \frac{k+2}{k!} \\ &&&= \sum_{k=0}^{\infty} \frac{1}{k!} + 2 \sum_{k=0}^{\infty} \frac{1}{k!} = 3e \\ \Rightarrow && \textrm{Var}(N) &= 3e-e^2 \end{align*}
A biased coin, with a probability \(p\) of coming up heads and a probability \(q=1-p\) of coming up tails, is tossed repeatedly. Let \(A\) be the event that the first run of \(r\) successive heads occurs before the first run of \(s\) successive tails. If \(H\) is the even that on the first toss the coin comes up heads and \(T\) is the event that it comes up tails, show that \begin{alignat*}{1} \mathrm{P}(A|H) & =p^{\alpha}+(1-p^{\alpha})\mathrm{P}(A|T),\\ \mathrm{P}(A|T) & =(1-q^{\beta})\mathrm{P}(A|H), \end{alignat*} where \(\alpha\) and \(\beta\) are to be determined. Use these two equations to find \(\mathrm{P}(A|H),\) \(\mathrm{P}(A|T),\) and hence \(\mathrm{P}(A).\)
Solution:
Consider the system of equations \begin{alignat*}{1} 2yz+zx-5xy & =2\\ yz-zx+2xy & =1\\ yz-2zx+6xy & =3 \end{alignat*} Show that \[xyz=\pm 6\] and find the possible values of \(x\), \(y\) and \(z\).
Solution: Consider the linear \(3\times 3\) system in \(yz, zx, xy\), then \begin{align*} \left(\begin{array}{ccc|c} 2 & 1 & -5 & 2 \\ 1 & -1 & 2 & 1 \\ 1 & -2 & 6 & 3 \\ \end{array}\right) \\ \left(\begin{array}{ccc|c} 1 & -1 & 2 & 1 \\ 1 & -2 & 6 & 3 \\ 2 & 1 & -5 & 2 \\ \end{array}\right) \\ \left(\begin{array}{ccc|c} 1 & -1 & 2 & 1 \\ 0 & -1 & 4 & 2 \\ 0 & 3 & -9 & 0 \\ \end{array}\right) \\ \left(\begin{array}{ccc|c} 1 & -1 & 2 & 1 \\ 0 & -1 & 4 & 2 \\ 0 & 0 & 3 & 6 \\ \end{array}\right) \\ \end{align*} Therefore \(yz = 2, zx = 6, xy = 3 \Rightarrow (xyz)^2 = 36 \Rightarrow xyz = \pm 6\). If \(xyz = 6, x = 3, y = 1, z = 2\), if \(xyz = -6, x = -3, y = -1, z = -2\)
The Fibonacci numbers \(F_{n}\) are defined by the conditions \(F_{0}=0\), \(F_{1}=1\) and \[F_{n+1}=F_{n}+F_{n-1}\] for all \(n\geqslant 1\). Show that \(F_{2}=1\), \(F_{3}=2\), \(F_{4}=3\) and compute \(F_{5}\), \(F_{6}\) and~\(F_{7}\). Compute \(F_{n+1}F_{n-1}-F_{n}^{2}\) for a few values of \(n\); guess a general formula and prove it by induction, or otherwise. By induction on \(k\), or otherwise, show that \[F_{n+k}=F_{k}F_{n+1}+F_{k-1}F_{n}\] for all positive integers \(n\) and \(k\).
Show that \(\cos 4u=8\cos^{4}u-8\cos^{2}u+1\). If \[ I=\int_{-1}^{1} \frac{1}{\vphantom{{\big(}^2}\; \surd(1+x)+\surd(1-x)+2\; }\;{\rm d}x ,\] show, by using the change of variable \(x=\cos t\), that \[ I= \int_0^\pi \frac{\sin t}{4\cos^{2}\left(\frac{t}{4}-\frac{\pi}{8}\right)}\,{\rm d}t.\] By using the further change of variable \(u=\frac{t}{4}-\frac{\pi}{8}\), or otherwise, show that \[I=4\surd{2}-\pi-2.\] \noindent[You may assume that \(\tan\frac{\pi}{8}=\surd{2}-1\).]
If $$ z^{4}+z^{3}+z^{2}+z+1=0\tag{*} $$ and \(u=z+z^{-1}\), find the possible values of \(u\). Hence find the possible values of \(z\). [Do not try to simplify your answers.] Show that, if \(z\) satisfies \((*)\), then \[z^{5}-1=0.\] Hence write the solutions of \((*)\) in the form \(z=r(\cos\theta+i\sin\theta)\) for suitable real \(r\) and \(\theta\). Deduce that \[\sin\frac{2\pi}{5}=\frac{\surd(10+2\surd 5)}{4} \ \ \hbox{and}\ \ \cos\frac{2\pi}{5}=\frac{-1+\surd 5}{4}.\]
Solution: \begin{align*} && 0 &= z^4+z^3+z^2+z+1 \\ \Rightarrow && 0 &= z^2+z+1+z^{-1}+z^{-2} \tag{\(z \neq 0\)} \\ &&&= \left ( z+z^{-1} \right)^2-2 + z+z^{-1} + 1 \\ &&&= u^2+u-1 \\ \Rightarrow && u &= \frac{-1 \pm \sqrt{5}}{2} \\ \Rightarrow && z+z^{-1} &= \frac{-1 \pm \sqrt{5}}{2} \\ \Rightarrow && 0 &= z^2-\left ( \frac{-1 \pm \sqrt{5}}{2}\right)z+1 \\ \Rightarrow && z &= \frac{\left ( \frac{-1 \pm \sqrt{5}}{2}\right) \pm \sqrt{\left ( \frac{-1 \pm \sqrt{5}}{2}\right)^2-4}}{2} \\ &&&= \frac{\left ( \frac{-1 \pm \sqrt{5}}{2}\right) \pm \sqrt{\frac{1+5\mp2\sqrt{5}-16}{4}}}{2} \\ &&&= \frac{\left ( \frac{-1 \pm \sqrt{5}}{2}\right) \pm \sqrt{\frac{-10\mp2\sqrt{5}-16}{4}}}{2} \\ &&&= \frac{-1\pm\sqrt{5}}{4} \pm i\frac{\sqrt{10\pm 2\sqrt{5}}}{4} \end{align*} Since \(z^4+z^3+z^2+z+1 = 0\) we can multiply both sides by \(z-1\) to obtain \(z^5-1 = 0\). Therefore if \(z = r(\cos \theta + i \sin \theta)\) we see that \(z^5 = 1 \Rightarrow r^5 (\cos 5 \theta + i \sin 5 \theta) = 1 \Rightarrow r = 1, 5 \theta = 2n \pi\) ie \(z = \cos \frac{2n\pi}{5} + i\sin \frac{2n \pi}{5}\). We are looking for a solution in the first quadrant, therefore \(\cos \frac{2\pi}{5} = \frac{-1 + \sqrt{5}}4\) and \(\sin \frac{2\pi}{5} = \frac{\sqrt{10+2\sqrt{5}}}{4}\)
A {\sl proper factor} of a positive integer \(N\) is an integer \(M\), with \(M\ne 1\) and \(M\ne N\), which divides \(N\) without remainder. Show that \(12\) has \(4\) proper factors and \(16\) has \(3\). Suppose that \(N\) has the prime factorisation \[N=p_{1}^{m_{1}}p_{2}^{m_{2}}\dots p_{r}^{m_{r}},\] where \(p_{1}\), \(p_{2}\), \dots, \(p_{r}\) are distinct primes and \(m_{1}\), \(m_{2}\), \dots, \(m_{r}\) are positive integers. How many proper factors does \(N\) have and why? Find: