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1994 Paper 2 Q1
In this question we consider only positive, non-zero integers written out in the usual (decimal) way. We say, for example, that 207 ends in 7 and that 5310 ends in 1 followed by 0. Show that, if \(n\) does not end in 5 or an even number, then there exists \(m\) such that \(n\times m\) ends in 1. Show that, given any \(n\), we can find \(m\) such that \(n\times m\) ends either in 1 or in 1 followed by one or more zeros. Show that, given any \(n\) which ends in 1 or in 1 followed by one or more zeros, we can find \(m\) such that \(n\times m\) contains all the digits \(0,1,2,\ldots,9\).
Solution: \begin{array}{c|c} \text{ends in} & \text{multiply by} \\ \hline 1 & 1 \\ 3 & 7 \\ 7 & 3 \\ 9 & 9 \end{array} If if \(n = 2^a \cdot 5^b \cdot c\) where \(c\) has no factors of \(2\) and \(5\) then we can multiply by \(2^b \cdot 5^a\) to obtain \(c\) followed by \(0\)s. Since \(c\) is neither even, nor a multiple of \(5\), by the earlier part of the question we can find a multiple such that it ends in \(1\). Suppose it is a \(k\) digit number, the consider Now consider \(1\underbrace{00\cdots0}_{k\text{ digits}}2\underbrace{00\cdots0}_{k\text{ digits}}\cdots 8\underbrace{00\cdots0}_{k\text{ digits}}9\cdot 0\), then clearly each section will end in the leading digit (ie all digits from \(1\) to \(9\)) and also end with a \(0\)
1994 Paper 2 Q2
If \(\mathrm{Q}\) is a polynomial, \(m\) is an integer, \(m\geqslant1\) and \(\mathrm{P}(x)=(x-a)^{m}\mathrm{Q}(x),\) show that \[ \mathrm{P}'(x)=(x-a)^{m-1}\mathrm{R}(x) \] where \(\mathrm{R}\) is a polynomial. Explain why \(\mathrm{P}^{(r)}(a)=0\) whenever \(1\leqslant r\leqslant m-1\). (\(\mathrm{P}^{(r)}\) is the \(r\)th derivative of \(\mathrm{P}.\)) If \[ \mathrm{P}_{n}(x)=\frac{\mathrm{d}^{n}}{\mathrm{d}x^{n}}(x^{2}-1)^{n} \] for \(n\geqslant1\) show that \(\mathrm{P}_{n}\) is a polynomial of degree \(n\). By repeated integration by parts, or otherwise, show that, if \(n-1\geqslant m\geqslant0,\) \[ \int_{-1}^{1}x^{m}\mathrm{P}_{n}(x)\,\mathrm{d}x=0 \] and find the value of \[ \int_{-1}^{1}x^{n}\mathrm{P}_{n}(x)\,\mathrm{d}x. \] {[}Hint. \textit{You may use the formula \[ \int_{0}^{\frac{\pi}{2}}\cos^{2n+1}t\,\mathrm{d}t=\frac{(2^{2n})(n!)^{2}}{(2n+1)!} \] without proof if you need it. However some ways of doing this question do not use this formula.}{]}
Solution: \begin{align*} && P(x) &= (x-a)^mQ(x) \\ \Rightarrow && P'(x) &= m(x-a)^{m-1}Q(x) + (x-a)^mQ'(x) \\ &&&= (x-a)^{m-1}(\underbrace{mQ(x) + (x-a)Q'(x)}_{\text{a polynomial}}) \\ &&&= (x-a)^{m-1}R(x) \end{align*} Therefore \(P^{(r)}(a) = 0\) for \(1 \leq r \leq m-1\) since each time we differentiate we will have a factor of \((x-a)^{m-r}\) which is zero when we evaluate at \(x = a\). If \(P_n(x) = \frac{\d^n}{\d x^n}(x^2-1)^n\) then we are differentiating a degree \(2n\) polynomial \(n\) times. Each time we differentiate we reduce the degree by \(1\), therefore the degree of \(P_n\) is \(n\). \begin{align*} && \int_{-1}^1 x^mP_n(x) \d x &= \left [x^m \underbrace{\frac{\d^{n-1}}{\d x^{n-1}}\left ( (x-1)^{n} (x+1)^{n} \right)}_{\text{has a factor of }x-1\text{ and }x+1}\right]_{-1}^1 - \int_{-1}^1 mx^{m-1}\frac{\d^{n-1}}{\d x^{n-1}}\left ( (x-1)^{n} (x+1)^{n} \right) \d x\\ &&&= 0 - \int_{-1}^1 mx^{m-1}\frac{\d^{n-1}}{\d x^{n-1}}\left ( (x-1)^{n} (x+1)^{n} \right) \d x\\ &&&= -\left [mx^{m-1} \underbrace{\frac{\d^{n-2}}{\d x^{n-2}}\left ( (x-1)^{n} (x+1)^{n} \right)}_{\text{has a factor of }x-1\text{ and }x+1}\right]_{-1}^1+ \int_{-1}^1 m(m-1)x^{m-2}\frac{\d^{n-2}}{\d x^{n-2}}\left ( (x-1)^{n} (x+1)^{n} \right) \d x\\ &&&= m(m-1)\int_{-1}^1 x^{m-2}\frac{\d^{n-2}}{\d x^{n-2}}\left ( (x-1)^{n} (x+1)^{n} \right) \d x\\ &&& \cdots \\ &&&= (-1)^m m!\int_{-1}^1 \frac{\d^{n-m}}{\d x^{n-m}} \left ( (x-1)^{n} (x+1)^{n} \right) \d x\\ &&&= 0 \end{align*} If \(n = m\), we have \begin{align*} && \int_{-1}^1 x^n P_n(x) \d x&= (-1)^nn! \int_{-1}^1 (x^2-1)^n \d x \\ && &= (-1)^{2n}n! \cdot 2\int_{0}^1 (1-x^2)^n \d x \\ x = \sin \theta, \d x = \cos \theta \d \theta: &&&= 2 \cdot n!\int_{0}^{\pi/2} \cos^{2n} \theta \cdot \cos \theta \d \theta \\ &&&= 2 \cdot n!\int_{0}^{\pi/2} \cos^{2n+1} \theta \d \theta \\ &&&= 2 \cdot n!\frac{(2^{2n})(n!)^{2}}{(2n+1)!} \\ &&&= \frac{(2^{2n+1})(n!)^{3}}{(2n+1)!} \\ \end{align*}
1994 Paper 2 Q3
The function \(\mathrm{f}\) satisfies \(\mathrm{f}(0)=1\) and \[ \mathrm{f}(x-y)=\mathrm{f}(x)\mathrm{f}(y)-\mathrm{f}(a-x)\mathrm{f}(a+y) \] for some fixed number \(a\) and all \(x\) and \(y\). Without making any further assumptions about the nature of the function show that \(\mathrm{f}(a)=0\). Show that, for all \(t\),
- \(\mathrm{f}(t)=\mathrm{f}(-t)\),
- \(\mathrm{f}(2a)=-1\),
- \(\mathrm{f}(2a-t)=-\mathrm{f}(t)\),
- \(\mathrm{f}(4a+t)=\mathrm{f}(t)\).
Solution: Let \(P(x,y)\) be the statement that the functional equation holds, then: \begin{align*} P(0,0): && f(0) &= f(0)f(0)-f(a)f(a) \\ \Rightarrow && 1 &= 1 - f(a)^2 \\ \Rightarrow && f(a)^2 &= 0 \\ \Rightarrow && f(a) &= 0 \end{align*}
- \begin{align*} P(0,t): && f(-t) &= f(0)f(t) - f(a)f(a-t) \\ \Rightarrow && f(-t) &= f(t) - 0 \\ \Rightarrow && f(t) &= f(-t) \end{align*}
- \begin{align*} P(a,a): && f(0) &= f(a)f(a)-f(0)f(2a) \\ \Rightarrow && 1 &= 0 - f(2a) \\ \Rightarrow && f(2a) &= -1 \end{align*}
- \begin{align*} P(2a,t): && f(2a-t) &= f(2a)f(t) - f(-a)f(a+t) \\ \Rightarrow && f(2a-t) &= -f(t)-f(a)f(a+t) \\ &&&= -f(t)-0 \\ \Rightarrow && f(2a-t) &= -f(t) \end{align*}
- \begin{align*} && f(4a+t) &= f(2a-(-2a-t)) \\ &&&=-f(2a+t) \\ &&&=-f(2a-(-t)) \\ &&&=f(-t) \\ &&&=f(t) \end{align*}
1994 Paper 2 Q4
By considering the area of the region defined in terms of Cartesian coordinates \((x,y)\) by \[ \{(x,y):\ x^{2}+y^{2}=1,\ 0\leqslant y,\ 0\leqslant x\leqslant c\}, \] show that \[ \int_{0}^{c}(1-x^{2})^{\frac{1}{2}}\,\mathrm{d}x=\tfrac{1}{2}[c(1-c^{2})^{\frac{1}{2}}+\sin^{-1}c], \] if \(0 < c\leqslant1.\) Show that the area of the region defined by \[ \left\{ (x,y):\ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,\ 0\leqslant y,\ 0\leqslant x\leqslant c\right\} , \] is \[ \frac{ab}{2}\left[\frac{c}{a}\left(1-\frac{c^{2}}{a^{2}}\right)^{\frac{1}{2}}+\sin^{-1}\left(\frac{c}{a}\right)\right], \] if \(0 < c\leqslant a\) and \(0 < b.\) Suppose that \(0 < b\leqslant a.\) Show that the area of intersection \(E\cap F\) of the two regions defined by \[ E=\left\{ (x,y):\ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\leqslant1\right\} \qquad\mbox{ and }\qquad F=\left\{ (x,y):\ \frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}\leqslant1\right\} \] is \[ 4ab\sin^{-1}\left(\frac{b}{\sqrt{a^{2}+b^{2}}}\right). \]
1994 Paper 2 Q5
- Show that the equation \[ (x-1)^{4}+(x+1)^{4}=c \] has exactly two real roots if \(c>2,\) one root if \(c=2\) and no roots if \(c<2\).
- How many real roots does the equation \(\left(x-3\right)^{4}+\left(x-1\right)^{4}=c\) have?
- How many real roots does the equation \(\left|x-3\right|+\left|x-1\right|=c\) have?
- How many real roots does the equation \(\left(x-3\right)^{3}+\left(x-1\right)^{3}=c\) have?
Solution:
- \(\,\) \begin{align*} && c &= (x-1)^4+(x+1)^4 \\ &&&= 2x^4+12x^2+2 \\ \Rightarrow && 0 &= (x^2+6)^2-\frac{c}{2} - 35 \\ \Rightarrow && \underbrace{x^2+6}_{\geq 6} &= \pm \sqrt{35 + \frac{c}{2}}\\ \end{align*} Therefore there are two solutions if \(c > 2\), one solution if \(c = 2\) and no solutions otherwise.
- \(\,\) This equation is the same equation if \(y = x-2\), ie there are two solutions if \(c > 2\), one solution if \(c = 2\) and no solutions otherwise.
- Rewriting as \(|x-1|+|x+1| = c\) we have For \(x < -1\): \(1-x-1-x = -2x\) For \(-1 \leq x \leq 1\): \(1-x+x+1 = 2\) For \(x > 1\): \(x-1+x+1 = 2x\) Therefore there are infinitely many solutions if \(c = 2\) (the interval \([-3,-1]\)), two solutions if \(c > 2\) and none otherwise.
- Rewriting as \((x-1)^3+(x+1)^3\) we have \(x^3+6x = c\). Notice that \(3x^2+6 > 0\) so the function is increasing, ie there is one solution for all \(c\)
1994 Paper 2 Q6
Prove by induction, or otherwise, that, if \(0<\theta<\pi\), \[ \frac{1}{2}\tan\frac{\theta}{2}+\frac{1}{2^{2}}\tan\frac{\theta}{2^{2}}+\cdots+\frac{1}{2^{n}}\tan\frac{\theta}{2^{n}}=\frac{1}{2^{n}}\cot\frac{\theta}{2^{n}}-\cot\theta. \] Deduce that \[ \sum_{r=1}^{\infty}\frac{1}{2^{r}}\tan\frac{\theta}{2^{r}}=\frac{1}{\theta}-\cot\theta. \]
Solution: Claim: \(\displaystyle \sum_{r=1}^n \frac1{2^r} \tan \tfrac{\theta}{2^r} = \frac1{2^n}\cot \tfrac{\theta}{2^n} - \cot \theta\) Proof: (By Induction) Base case: \(n = 1\) \begin{align*} && LHS &= \sum_{r=1}^1 \frac1{2^r} \tan \frac{\theta}{2^r} \\ &&&= \frac1{2} \tan \frac{\theta}{2}\\ \\ && RHS &= \frac12 \cot \frac{\theta}{2} - \cot \theta \\ &&&= \frac12 \frac{1}{\tan \frac{\theta}{2}} - \frac{1-\tan^2 \frac{\theta}{2}}{2 \tan \frac{\theta}{2}} \\ &&&= \frac{1}{2} \tan \frac{\theta}{2} = LHS \end{align*} Therefore our base case is true. Assume our statement is true for some \(n=k\), then consider \(n = k+1\), ie \begin{align*} \sum_{r=1}^{k+1} \frac1{2^r} \tan \tfrac{\theta}{2^r} &= \sum_{r=1}^{k} \frac1{2^r} \tan \tfrac{\theta}{2^r} + \frac1{2^{k+1}} \tan \frac{\theta}{2^{k+1}} \\ &= \frac{1}{2^k} \cot \frac{\theta}{2^k} - \cot \theta + \frac{1}{2^{k+1}}\tan \frac{\theta}{2^{k+1}} \\ &= \frac{1}{2^{k+1}} \left (2 \cot \frac{\theta}{2^k} +\tan \frac{\theta}{2^{k+1}} \right) - \cot \theta \\ &= \frac{1}{2^{k+1}} \left (2\frac{1-\tan^2 \frac{\theta}{2^{k+1}}}{2 \tan \frac{\theta}{2^{k+1}}} + \tan \frac{\theta}{2^{k+1}} \right) - \cot \theta \\ &= \frac{1}{2^{k+1}} \cot \frac{\theta}{2^{k+1}} - \cot \theta \\ \end{align*} Therefore, since as \(x \to 0, x\cot x \to 1\) or \(x \cot \theta x \to \frac{1}{\theta}\) \begin{align*} \sum_{r=1}^{\infty}\frac{1}{2^{r}}\tan\frac{\theta}{2^{r}} &= \lim_{k\to \infty} \sum_{r=1}^{k}\frac{1}{2^{r}}\tan\frac{\theta}{2^{r}} \\ &= \lim_{k\to \infty} \left ( \frac{1}{2^{k+1}} \cot \frac{\theta}{2^{k+1}} - \cot \theta \right) \\ &= \lim_{k\to \infty}\frac{1}{2^{k+1}} \cot \frac{\theta}{2^{k+1}} - \cot \theta \\ &= \lim_{k\to \infty}\frac{1}{2^{k+1}} \cot \frac{\theta}{2^{k+1}} - \cot \theta \\ &= \lim_{k\to \infty}\frac{1}{2^{k+1}} \cot \frac{\theta}{2^{k+1}} - \cot \theta \\ &= \frac{1}{\theta} - \cot \theta \end{align*}
1994 Paper 2 Q7
Show that the equation \[ ax^{2}+ay^{2}+2gx+2fy+c=0 \] where \(a>0\) and \(f^{2}+g^{2}>ac\) represents a circle in Cartesian coordinates and find its centre. The smooth and level parade ground of the First Ruritanian Infantry Division is ornamented by two tall vertical flagpoles of heights \(h_{1}\) and \(h_{2}\) a distance \(d\) apart. As part of an initiative test a soldier has to march in such a way that he keeps the angles of elevation of the tops of the two flagpoles equal to one another. Show that if the two flagpoles are of different heights he will march in a circle. What happens if the two flagpoles have the same height? To celebrate the King's birthday a third flagpole is added. Soldiers are then assigned to each of the three different pairs of flagpoles and are told to march in such a way that they always keep the tops of their two assigned flagpoles at equal angles of elevation to one another. Show that, if the three flagpoles have different heights \(h_{1},h_{2}\) and \(h_{3}\) and the circles in which the soldiers march have centres of \((x_{ij},y_{ij})\) (for the flagpoles of height \(h_{i}\) and \(h_{j}\)) relative to Cartesian coordinates fixed in the parade ground, then the \(x_{ij}\) satisfy \[ h_{3}^{2}\left(h_{1}^{2}-h_{2}^{2}\right)x_{12}+h_{1}^{2}\left(h_{2}^{2}-h_{3}^{2}\right)x_{23}+h_{2}^{2}\left(h_{3}^{2}-h_{1}^{2}\right)x_{31}=0, \] and the same equation connects the \(y_{ij}\). Deduce that the three centres lie in a straight line.
1994 Paper 2 Q8
`24 Hour Spares' stocks a small, widely used and cheap component. Every \(T\) hours \(X\) units arrive by lorry from the wholesaler, for which the owner pays a total \(\pounds (a+qX)\). It costs the owner \(\pounds b\) per hour to store one unit. If she has the units in stock she expects to sell \(r\) units per hour at \(\pounds(p+q)\) per unit. The other running costs of her business remain at \(\pounds c\) pounds an hour irrespective of whether she has stock or not. (All of the quantities \(T,X,a,b,r,q,p\) and \(c\) are greater than 0.) Explain why she should take \(X\leqslant rT\). Given that the process may be assumed continuous (the items are very small and she sells many each hour), sketch \(S(t)\) the amount of stock remaining as a function of \(t\) the time from the last delivery. Compute the total profit over each period of \(T\) hours. Show that, if \(T\) is fixed with \(T\geqslant p/b\), the business can be made profitable if \[ p^{2}>2\frac{(a+cT)b}{r}. \]
1994 Paper 2 Q9
A light rod of length \(2a\) is hung from a point \(O\) by two light inextensible strings \(OA\) and \(OB\) each of length \(b\) and each fixed at \(O\). A particle of mass \(m\) is attached to the end \(A\) and a particle of mass \(2m\) is attached to the end \(B.\) Show that, in equilibrium, the angle \(\theta\) that the rod makes the horizontal satisfies the equation \[ \tan\theta=\frac{a}{3\sqrt{b^{2}-a^{2}}}. \] Express the tension in the string \(AO\) in terms of \(m,g,a\) and \(b\).
Solution:
1994 Paper 2 Q10
A truck is towing a trailer of mass \(m\) across level ground by means of an elastic rope of natural length \(l\) whose modulus of elasticity is \(\lambda.\) At first the rope is slack and the trailer stationary. The truck then accelerates until the rope becomes taut and thereafter the truck travels in a straight line at a constant speed \(u\). Assuming that the effect of friction on the trailer is negligible, show that the trailer will collide with the truck at a time \[ \pi\left(\frac{lm}{\lambda}\right)^{\frac{1}{2}}+\frac{l}{u} \] after the rope first becomes taut.