\documentclass[a4, 11pt]{report} \pagestyle{myheadings} \markboth{}{Paper III, 1987 \ \ \ \ \ \today } \RequirePackage{amssymb} \RequirePackage{amsmath} \RequirePackage{graphicx} \RequirePackage{color} \RequirePackage[flushleft]{paralist}[2013/06/09] \RequirePackage{geometry} \geometry{% a4paper, lmargin=2cm, rmargin=2.5cm, tmargin=3.5cm, bmargin=2.5cm, footskip=12pt, headheight=24pt} \newcommand{\comment}[1]{{\bf Comment} {\it #1}} %\renewcommand{\comment}[1]{} \newcommand{\bluecomment}[1]{{\color{blue}#1}} %\renewcommand{\comment}[1]{} \newcommand{\redcomment}[1]{{\color{red}#1}} \usepackage{epsfig} \usepackage{pstricks-add} \usepackage{tgheros} %% changes sans-serif font to TeX Gyre Heros (tex-gyre) \renewcommand{\familydefault}{\sfdefault} %% changes font to sans-serif %\usepackage{sfmath} %%%% this makes equation sans-serif %\input RexFigs \setlength{\parskip}{10pt} \setlength{\parindent}{0pt} \newlength{\qspace} \setlength{\qspace}{20pt} \newcounter{qnumber} \setcounter{qnumber}{0} \newenvironment{question}% {\vspace{\qspace} \begin{enumerate}[\bfseries 1\quad][10]% \setcounter{enumi}{\value{qnumber}}% \item% } { \end{enumerate} \filbreak \stepcounter{qnumber} } \newenvironment{questionparts}[1][1]% { \begin{enumerate}[\bfseries (i)]% \setcounter{enumii}{#1} \addtocounter{enumii}{-1} \setlength{\itemsep}{5mm} \setlength{\parskip}{8pt} } { \end{enumerate} } \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\Var}{Var} \def\d{{\rm d}} \def\e{{\rm e}} \def\g{{\rm g}} \def\h{{\rm h}} \def\f{{\rm f}} \def\p{{\rm p}} \def\s{{\rm s}} \def\t{{\rm t}} \def\A{{\rm A}} \def\B{{\rm B}} \def\E{{\rm E}} \def\F{{\rm F}} \def\G{{\rm G}} \def\H{{\rm H}} \def\P{{\rm P}} \def\bb{\mathbf b} \def \bc{\mathbf c} \def\bx {\mathbf x} \def\bn {\mathbf n} \newcommand{\low}{^{\vphantom{()}}} %%%%% to lower suffices: $X\low_1$ etc \newcommand{\subone}{ {\vphantom{\dot A}1}} \newcommand{\subtwo}{ {\vphantom{\dot A}2}} \def\le{\leqslant} \def\ge{\geqslant} \def\var{{\rm Var}\,} \newcommand{\ds}{\displaystyle} \newcommand{\ts}{\textstyle} \begin{document} \setcounter{page}{2} \section*{Section A: \ \ \ Pure Mathematics} %%%%%%%%%%Q1 \begin{question} Find the set of positive integers $n$ for which $n$ does not divide $(n-1)!.$ Justify your answer. [Note that small values of $n$ may require special consideration.] \end{question} %%%%%%%%%%Q2 \begin{question} Let ${\displaystyle I_{m,n}=\int\cos^{m}x\sin nx\,\mathrm{d}x,}$ where $m$ and $n$ are non-negative integers. Prove that for $m,n\geqslant1,$ \[ (m+n)I_{m,n}=-\cos^{m}\cos nx+mI_{m-1,n-1}. \] \begin{questionparts} \item Show that ${\displaystyle \int_{0}^{\pi}\cos^{m}x\sin nx\,\mathrm{d}x=0}$ whenever $m,n$ are both even or both odd. \item Evaluate ${\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{2}x\sin3x\,\mathrm{d}x.}$ \end{questionparts} \end{question} %%%%%%%%% Q3 \begin{question} \begin{questionparts} \item If $z=x+\mathrm{i}y,$ with $x,y$ real, show that \[ \left|x\right|\cos\alpha+\left|y\right|\sin\alpha\leqslant\left|z\right|\leqslant\left|x\right|+\left|y\right| \] for all real $\alpha.$ \item By considering $(5-\mathrm{i})^{4}(1+\mathrm{i}),$ show that \[ \frac{\pi}{4}=4\tan^{-1}\left(\frac{1}{5}\right)-\tan^{-1}\left(\frac{1}{239}\right). \] Prove similarly that \[ \frac{\pi}{4}=3\tan^{-1}\left(\frac{1}{4}\right)+\tan^{-1}\left(\frac{1}{20}\right)+\tan^{-1}\left(\frac{1}{1985}\right). \] \end{questionparts} \end{question} %%%%%% Q4 \begin{question}$\ $ \begin{center} 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The bottom of $B$ is one unit lower than the bottom of $A$ and they are connected by a thin rubber tube with a tap in it. The tap is closed and $A$ is filled with water to a depth of 4 units. The tap is then closed. When the water comes to rest, both surfaces are at a height $h$ above the bottom of $B$, as shown in the diagram. Show that $h$ satisfies the equation \[ h^{2}-3h+\sinh h=15. \] \end{question} %%%%%%%%% Q5 \begin{question} A secret message consists of the numbers $1,3,7,23,24,37,39,43,43,43,45,47$ arranged in some order as $a_{1},a_{2},\ldots,a_{12}.$ The message is encoded as $b_{1},b_{2},\ldots,b_{12}$ with $0\leqslant b_{j}\leqslant49$ and \begin{alignat*}{1} b_{2j} & \equiv a_{2j}+n_{0}+j\pmod{50},\\ b_{2j+1} & \equiv a_{2j+1}+n_{1}+j\pmod{50}, \end{alignat*} for some integers $n_{0}$ and $n_{1}.$ If the coded message is $35,27,2,36,15,35,8,40,40,37,24,48,$ find the original message, explaining your method carefully. \end{question} %%%%%%%%% Q6 \begin{question} The functions $x(t)$ and $y(t)$ satisfy the simultaneous differential equations \begin{alignat*}{1} \dfrac{\mathrm{d}x}{\mathrm{d}t}+2x-5y & =0\\ \frac{\mathrm{d}y}{\mathrm{d}t}+ax-2y & =2\cos t, \end{alignat*} subject to $x=0,$ $\dfrac{\mathrm{d}y}{\mathrm{d}t}=0$ at $t=0.$ Solve these equations for $x$ and $y$ in the case when $a=1$. Without solving the equations explicitly, state briefly how the form of the solutions for $x$ and $y$ if $a>1$ would differ from the form when $a=1.$ \end{question} %%%%%%%%% Q7 \begin{question} Prove that \[ \tan^{-1}t=t-\frac{t^{3}}{3}+\frac{t^{5}}{5}-\cdots+\frac{(-1)^{n}t^{2n+1}}{2n+1}+(-1)^{n+1}\int_{0}^{t}\frac{x^{2n+2}}{1+x^{2}}\,\mathrm{d}x. \] Hence show that, if $0\leqslant t\leqslant1,$ then \[ \frac{t^{2n+3}}{2(2n+3)}\leqslant\left|\tan^{-1}t-\sum_{r=0}^{n}\frac{(-1)^{r}t^{2r+1}}{2r+1}\right|\leqslant\frac{t^{2n+3}}{2n+3}. \] Show that, as $n\rightarrow\infty,$ \[ 4\sum_{r=0}^{n}\frac{(-1)^{r}}{(2r+1)}\rightarrow\pi, \] but that the error in approximating $\pi$ by ${\displaystyle 4\sum_{r=0}^{n}\frac{(-1)^{r}}{(2r+1)}}$ is at least $10^{-2}$ if $n$ is less than or equal to $98$. \end{question} %%%%%%%%% Q8 \begin{question} Show that, if the lengths of the diagonals of a parallelogram are specified, then the parallogram has maximum area when the diagonals are perpendicular. Show also that the area of a parallelogram is less than or equal to half the square of the length of its longer diagonal. The set $A$ of points $(x,y)$ is given by \begin{alignat*}{1} \left|a_{1}x+b_{1}y-c_{1}\right| & \leqslant\delta,\\ \left|a_{2}x+b_{2}y-c_{2}\right| & \leqslant\delta, \end{alignat*} with $a_{1}b_{2}\neq a_{2}b_{1}.$ Sketch this set and show that it is possible to find $(x_{1},y_{1}),(x_{2},y_{2})\in A$ with \[ (x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}\geqslant\frac{8\delta^{2}}{\left|a_{1}b_{2}-a_{2}b_{1}\right|}. \] \end{question} %%%%%%%%% Q9 \begin{question} Let $(G,*)$ and $(H,\circ)$ be two groups and $G\times H$ be the set of ordered pairs $(g,h)$ with $g\in G$ and $h\in H.$ A multiplication on $G\times H$ is defined by \[ (g_{1},h_{1})(g_{2},h_{2})=(g_{1}*g_{2},h_{1}\circ h_{2}) \] for all $g_{1},g_{2}\in G$ and $h_{1},h_{2}\in H$. Show that, with this multiplication, $G\times H$ is a group. State whether the following are true or false and prove your answers. \begin{enumerate} \item[\bf (i)] $G\times H$ is abelian if and only if both $G$ and $H$ are abelian. \item[\bf (ii)] $G\times H$ contains a subgroup isomorphic to $G$. \item[\bf (iii)] $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ is isomorphic to $\mathbb{Z}_{4}.$ \item[\bf (iv)] $S_{2}\times S_{3}$ is isomorphic to $S_{6}.$ \end{enumerate} {[}$\mathbb{Z}_{n}$ is the cyclic group of order $n$, and $S_{n}$ is the permutation group on $n$ objects.{]} \end{question} %%%%%%%%%% 10 \begin{question} The \textit{Bernoulli polynomials} $P_{n}(x)$, where $n$ is a non-negative integer, are defined by $P_{0}(x)=1$ and, for $n\geqslant1$, \[ \frac{\mathrm{d}P_{n}}{\mathrm{d}x}=nP_{n-1}(x),\qquad\int_{0}^{1}P_{n}(x)\,\mathrm{d}x=0 \] Show by induction or otherwise, that \[ P_{n}(x+1)-P_{n}(x)=nx^{n-1},\quad\mbox{ for }n\geqslant1. \] Deduce that \[ n\sum_{m=0}^{k}m^{n-1}=P_{n}(k+1)-P_{n}(0) \] Hence show that ${\displaystyle \sum_{m=0}^{1000}m^{3}=(500500)^{2}}$ \end{question} \newpage \section*{Section B: \ \ \ Mechanics} %%%%%%%%%% Q11 \begin{question} A woman stands in a field at a distance of $a\,\mathrm{m}$ from the straight bank of a river which flows with negligible speed. She sees her frightened child clinging to a tree stump standing in the river $b\,\mathrm{m}$ downstream from where she stands and $c\,\mathrm{m}$ from the bank. She runs at a speed of $u\,\mathrm{ms}^{-1}$ and swims at $v\,\mathrm{ms}^{-1}$ in straight lines. Find an equation to be satisfied by $x,$ where $x\,\mathrm{m}$ is the distance upstream from the stump at which she should enter the river if she is to reach the child in the shortest possible time. Suppose now that the river flows with speed $v$ ms$^{-1}$ and the stump remains fixed. Show that, in this case, $x$ must satisfy the equation \[ 2vx^{2}(b-x)=u(x^{2}-c^{2})[a^{2}+(b-x)^{2}]^{\frac{1}{2}}. \] For this second case, draw sketches of the woman's path for the three possibilities $b>c,$ $b=c$ and $b