Problems

2008 Paper 3 Q1

D: 1700.0 B: 1516.0

Find all values of $a$, $b$, $x$ and $y$ that satisfy the simultaneous equations \begin{alignat*}{3} a&+b & &=1 &\\ ax&+by & &= \tfrac13& \\ ax^2&+by^2& &=\tfrac15& \\ ax^3 &+by^3& &=\tfrac17\,.& \end{alignat*} \noindent{\bf [} {\bf Hint}: you may wish to start by multiplying the second equation by $x+y$. {\bf ]}

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2008 Paper 3 Q2

D: 1700.0 B: 1555.2

Let $S_k(n) \equiv \sum\limits_{r=0}^n r^k\,$, where $k$ is a positive integer, so that \[ S_1(n) \equiv \tfrac12 n(n+1) \text{ and } S_2(n) \equiv \tfrac16 n(n+1)(2n+1)\,. \]

By considering $\sum\limits_{r=0}^n \left[ (r+1)^k-r^k\right]\, $, show that \[ kS_{k-1}(n)=(n+1)^k -(n+1) - \binom{k}{2} S_{k-2}(n) - \binom {k}{3} S_{k-3}(n) - \cdots - \binom{k}{k-1} S_{1}(n) \;. \tag{$*$} \] Obtain simplified expressions for $S_3(n)$ and $S_4(n)$.
Explain, using $(*)$, why $S_k(n)$ is a polynomial of degree $k+1$ in $n$. Show that in this polynomial the constant term is zero and the sum of the coefficients is 1.

Solution:

\begin{align*} &&(n+1)^k &= \sum_{r=0}^n \left [ (r+1)^k - r^k \right] \\ &&&= \sum_{r=0}^n \left [ \left ( \binom{k}{0}r^k+\binom{k}1r^{k-1} + \binom{k}{2}r^{k-2} + \cdots + \binom{k}{k} 1 \right) - r^k\right] \\ &&&= \sum_{r=0}^n \left ( \binom{k}1r^{k-1} + \binom{k}{2}r^{k-2} + \cdots + \binom{k}{k} 1 \right) \\ &&&=k \sum_{r=0}^n r^{k-1} + \binom{k}{2}\sum_{r=0}^nr^{k-2} + \cdots + \binom{k}{k} \sum_{r=0}^n 1 \\ &&&= kS_{k-1}(n) + \binom{k}2 S_{k-2}(n) + \cdots +\binom{k}{k-1}S_1(n) + (n+1) \\ \Rightarrow && k S_{k-1}(n) &= (n+1)^k -(n+1) -\binom{k}2 S_{k-2}(n) - \cdots -\binom{k}{k-1}S_1(n) \\ && 4S_3(n) &= (n+1)^4-(n+1) - \binom{4}{2} \frac{n(n+1)(2n+1)}{6} - \binom{4}{3} \frac{n(n+1)}{2} \\ &&&= (n+1) \left ( (n+1)^3-1 - n(2n+1)-2n \right) \\ &&&= (n+1) \left ( n^3+3n^2+3n+1-1 - 2n^2-3n \right) \\ &&&= (n+1) \left ( n^3+n^2 \right) \\ &&&= n^2(n+1)^2 \\ \Rightarrow && S_3(n) &= \frac{n^2(n+1)^2}{4} \\ \\ &&5S_4(n) &=(n+1)^5-(n+1) - \binom{5}{2} \frac{n^2(n+1)^2}4 - \binom{5}{3} \frac{n(n+1)(2n+1)}{6} - \binom{5}{4} \frac{n(n+1)}{2} \\ &&&= (n+1) \left ((n+1)^4 - 1-\frac{5n^2(n+1)}{2} - \frac{5n(2n+1)}{3} -\frac{5n}{2}\right)\\ &&&= \frac{n+1}{6} \left (6(n+1)^4-6-15n^2(n+1)-10n(2n+1)-15n \right) \\ &&&= \frac{n+1}{6} \left (6n^4+24n^3+36n^2+24n+6 -6-15n^3-15n^2-20n^2-10n-15n\right) \\ &&&= \frac{n+1}{6} \left (6n^4+9n^3+n^2-n\right) \\ &&&= \frac{(n+1)n(2n+1)(3n^2+3n-1)}{6} \\ \Rightarrow && S_4(n) &= \frac{(n+1)n(2n+1)(3n^2+3n-1)}{30} \end{align*}
Proceeding by induction, since $S_k(n)$ is a polynomial of degree $k+1$ for small $k$, we can see that \[ (k+1)S_k(n) = \underbrace{(n+1)^{k+1}}_{\text{poly deg }=k+1} - \underbrace{(n+1)}_{\text{poly deg}=1} - \underbrace{\binom{k+1}{2}S_{k-1}(n)}_{\text{poly deg}=k} - \underbrace{\cdots}_{\text{polys deg}< k} - \underbrace{\binom{k+1}{k} S_1(n)}_{\text{poly deg}=1}\] therefore $S_k(n)$ is a polynomial of degree $k+1$ (in fact with leading coefficient $\frac{1}{k+1}$. Since $S_k(0) = \sum_{r=0}^{0} r^k = 0$ there is no constant term, and since $S_k(1) = \sum_{r=0}^1 r^k = 1$ the sum of the coefficients is $1$

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2008 Paper 3 Q3

D: 1700.0 B: 1484.8

The point $P(a\cos\theta\,,\, b\sin\theta)$, where $a>b>0$, lies on the ellipse \[\dfrac {x^2}{a^2} + \dfrac {y^2}{b^2}=1\,.\] The point $S(-ea\,,\,0)$, where $b^2=a^2(1-e^2)\,$, is a focus of the ellipse. The point $N$ is the foot of the perpendicular from the origin, $O$, to the tangent to the ellipse at $P$. The lines $SP$ and $ON$ intersect at $T$. Show that the $y$-coordinate of $T$ is \[\dfrac{b\sin\theta}{1+e\cos\theta}\,.\] Show that $T$ lies on the circle with centre $S$ and radius $a$.

Solution: Find the gradient of the tangent of the ellipse at $P$: \begin{align*} && \frac{2x}{a^2} + \frac{2y}{b^2} \frac{\d y}{\d x} &= 0 \\ \Rightarrow && \frac{\d y}{\d x} &= - \frac{2xb^2}{2ya^2} \\ &&&=- \frac{a \cos \theta b^2}{b \sin \theta a^2} \\ &&&=-\frac{b}{a} \cot \theta \end{align*} Therefore the gradient of $ON$ is $\frac{a}{b} \tan \theta$. \begin{align*} && y &= \frac{a}{b} \tan \theta x \\ && \frac{y-0}{x-(-ea)} &= \frac{b\sin \theta-0}{a\cos \theta -(-ea)} \\ && y &= \frac{b \sin \theta}{a(e+\cos \theta)}(x+ea) \\ \Rightarrow && y &= \frac{b \sin \theta}{a(\cos \theta+e)}\frac{b}{a} \cot \theta y+ \frac{eb \sin \theta}{\cos \theta + e} \\ &&&= \frac{b^2 \cos \theta}{a^2(\cos \theta +e)}y + \frac{eb \sin \theta}{\cos \theta + e} \\ \Rightarrow && (\cos \theta+e)y &= (1-e^2)\cos \theta y +eb \sin \theta\\ && e(1+e\cos \theta)y &= eb \sin \theta \\ \Rightarrow && y &= \frac{b \sin \theta}{1+e\cos \theta} \\ && x &= \frac{b \sin \theta}{1+e\cos \theta} \frac{b}{a} \cot \theta \\ &&&= \frac{b^2 \cos \theta}{a(1+e\cos \theta)} \end{align*} Therefore $\displaystyle T\left (\frac{b^2 \cos \theta}{a(1+e\cos \theta)}, \frac{b \sin \theta}{1+e\cos \theta} \right)$. Finally, we can look at the distance of $T$ from $S$ \begin{align*} && d^2 &= \left (\frac{b^2 \cos \theta}{a(1+e\cos \theta)}-(-ea) \right)^2 + \left (\frac{b \sin \theta}{1+e\cos \theta} -0\right)^2 \\ &&&= \frac{\left (b^2 \cos \theta+ea^2(1+e\cos\theta)\right)^2 + \left ( ab \sin \theta\right)^2}{a^2(1+e\cos \theta)^2} \\ &&&= \frac{b^4\cos^2\theta+e^2a^4(1+e\cos\theta)^2+2ea^2b^2(1+e\cos\theta)+a^2b^2\sin^2\theta}{a^2(1+e\cos\theta)^2} \\ &&&= \frac{a^4(1-e^2)^2\cos^2\theta+e^2a^4(1+e\cos\theta)^2+2ea^2a^2(1-e^2)(1+e\cos\theta)+a^4(1-e^2)\sin^2\theta}{a^2(1+e\cos\theta)^2} \\ &&&= a^2 \left ( \frac{(1-e^2)^2\cos^2\theta+e^2(1+e\cos\theta)^2+2e(1-e^2)(1+e\cos\theta)+(1-e^2)(1-\cos^2\theta)}{(1+e\cos\theta)^2} \right) \\ &&&= a^2 \left ( \frac{e^2(1+e\cos\theta)^2+(1-e^2)((1-e^2)\cos^2\theta+2e(1+e\cos\theta)+(1-\cos^2\theta))}{(1+e\cos\theta)^2} \right) \\ &&&= a^2 \left ( \frac{e^2(1+e\cos\theta)^2+(1-e^2)(1+e\cos\theta)^2}{(1+e\cos\theta)^2} \right) \\ &&&= a^2 \end{align*} Therefore a circle radius $a$ centre $S$.

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2008 Paper 3 Q4

D: 1700.0 B: 1516.0

Show, with the aid of a sketch, that $y> \tanh (y/2)$ for $y>0$ and deduce that \begin{equation} \textrm{arcosh} x > \dfrac{x-1}{\sqrt{x^2-1}} \text{ for } x>1. \tag{$*$} \end{equation}
By integrating $(*)$, show that $\textrm{arcosh} x > 2 \dfrac{{x-1}}{\sqrt{x^2-1}} $ for $x>1$.
Show that $\textrm{arcosh} x >3 \dfrac{\sqrt{x^2-1}}{{x+2}} $ for $x>1$.

[Note: $\textrm{arcosh} x $ is another notation for $\cosh^{-1}x$.]

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2008 Paper 3 Q5

D: 1700.0 B: 1499.3

The functions ${\rm T}_n(x)$, for $n=0$, 1, 2, $\ldots\,$, satisfy the recurrence relation \[ {\rm T}_{n+1}(x) -2x {\rm T}_n(x) + {\rm T}_{n-1}(x) =0\, \ \ \ \ \ \ \ (n\ge1). \tag{$*$} \] Show by induction that \[ \left({\rm T}_n(x)\right)^2 - {\rm T}_{n-1}(x) {\rm T}_{n+1}(x) = \f(x)\,, \] where $\f(x) = \left({\rm T}_1(x)\right)^2 - {\rm T}_0(x){\rm T}_2(x)\,$. In the case $\f(x)\equiv 0$, determine (with proof) an expression for ${\rm T}_n(x)$ in terms of ${\rm T}_0(x)$ (assumed to be non-zero) and ${\rm r}(x)$, where ${\rm r}(x) = {\rm T}_1(x)/ {\rm T}_0(x)$. Find the two possible expressions for ${\rm r}(x)$ in terms of $x$. %Conjecture (without proof) the general form of the solution of $(*)$.

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2008 Paper 3 Q6

D: 1700.0 B: 1500.0

In this question, $p$ denotes $\dfrac{\d y}{\d x}\,$.

Given that \[ y=p^2 +2 xp\,, \] show by differentiating with respect to $x$ that \[ \frac{\d x}{\d p} = -2 - \frac {2x} p . \] Hence show that $x = -\frac23p +Ap^{-2}\,,$ where $A$ is an arbitrary constant. Find $y$ in terms of $x$ if $p=-3$ when $x=2$.
Given instead that \[ y=2xp +p \ln p\,,\] and that $p=1$ when $x=-\frac14$, show that $x=-\frac12 \ln p - \frac14\,$ and find $y$ in terms of $x$.

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2008 Paper 3 Q7

D: 1700.0 B: 1500.0

The points $A$, $B$ and $C$ in the Argand diagram are the vertices of an equilateral triangle described anticlockwise. Show that the complex numbers $a$, $b$ and $c$ representing $A$, $B$ and $C$ satisfy \[2c= (a+b) +\mathrm{i}\sqrt3(b-a).\] Find a similar relation in the case that $A$, $B$ and $C$ are the vertices of an equilateral triangle described clockwise.

The quadrilateral $DEFG$ lies in the Argand diagram. Show that points $P$, $Q$, $R$ and $S$ can be chosen so that $PDE$, $QEF$, $RFG$ and $SGD$ are equilateral triangles and $PQRS$ is a parallelogram.
The triangle $LMN$ lies in the Argand diagram. Show that the centroids $U$, $V$ and $W$ of the equilateral triangles drawn externally on the sides of $LMN$ are the vertices of an equilateral triangle. \noindent [{\bf Note:} The {\em centroid} of a triangle with vertices represented by the complex numbers $x$,~$y$ and~$z$ is the point represented by $\frac13(x+y+z)\,$.]

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2008 Paper 3 Q8

D: 1700.0 B: 1500.0

The coefficients in the series \[ S= \tfrac13 x + \tfrac 16 x^2 + \tfrac1{12} x^3 + \cdots + a_rx^r + \cdots \] satisfy a recurrence relation of the form $a_{r+1} + p a_r =0$. Write down the value of $p$. By considering $(1+px)S$, find an expression for the sum to infinity of $S$ (assuming that it exists). Find also an expression for the sum of the first $n+1$ terms of $S$.
The coefficients in the series \[ T=2 + 8x +18x^2+37 x^3 +\cdots + a_rx^r + \cdots \] satisfy a recurrence relation of the form $a_{r+2}+pa_{r+1} +qa_r=0$. Find an expression for the sum to infinity of $T$ (assuming that it exists). By expressing $T$ in partial fractions, or otherwise, find an expression for the sum of the first $n+1$ terms of $T$.

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2008 Paper 3 Q9

D: 1700.0 B: 1484.0

A particle of mass $m$ is initially at rest on a rough horizontal surface. The particle experiences a force $mg\sin \pi t$, where $t$ is time, acting in a fixed horizontal direction. The coefficient of friction between the particle and the surface is $\mu$. Given that the particle starts to move first at $t=T_0$, state the relation between $T_0$ and $\mu$.

For $\mu = \mu_0$, the particle comes to rest for the first time at $t=1$. Sketch the acceleration-time graph for $0\le t \le 1$. Show that \[ 1+\left(1-\mu_0^2\right)^{\frac12} -\mu_0\pi +\mu_0 \arcsin \mu_0 =0\,. \]
For $\mu=\mu_0$ sketch the acceleration-time graph for $0\le t\le 3$. Describe the motion of the particle in this case and in the case $\mu=0$.

\noindent[{\bf Note:} $\arcsin x$ is another notation for $\sin^{-1}x$.\ ]

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2008 Paper 3 Q10

D: 1700.0 B: 1484.0

A long string consists of $n$ short light strings joined together, each of natural length $\ell$ and modulus of elasticity $\lambda$. It hangs vertically at rest, suspended from one end. Each of the short strings has a particle of mass $m$ attached to its lower end. The short strings are numbered $1$ to $n$, the $n$th short string being at the top. By considering the tension in the $r$th short string, determine the length of the long string. Find also the elastic energy stored in the long string. A uniform heavy rope of mass $M$ and natural length $L_0$ has modulus of elasticity $\lambda$. The rope hangs vertically at rest, suspended from one end. Show that the length, $L$, of the rope is given by \[ L=L_0\biggl(1+ \frac{Mg}{2\lambda}\biggr), \] and find an expression in terms of $L$, $L_0$ and $\lambda$ for the elastic energy stored in the rope.

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