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2006 Paper 1 Q11
Particles \(A_1\), \(A_2\), \(A_3\), \(\ldots\), \(A_n\) (where \(n\ge 2\)) lie at rest in that order in a smooth straight horizontal trough. The mass of \(A_{n-1}\) is \(m\) and the mass of \(A_n\) is \(\lambda m\), where \(\lambda>1\). Another particle, \(A_0\), of mass \(m\), slides along the trough with speed \(u\) towards the particles and collides with \(A_1\). Momentum and energy are conserved in all collisions.
- Show that it is not possible for there to be exactly one particle moving after all collisions have taken place.
- Show that it is not possible for \(A_{n-1}\) and \(A_n\) to be the only particles moving after all collisions have taken place.
- Show that it is not possible for \(A_{n-2}\), \(A_{n-1}\) and \(A_n\) to be the only particles moving after all collisions have taken place.
- Given that there are exactly two particles moving after all collisions have taken place, find the speeds of these particles in terms of \(u\) and \(\lambda\).
2006 Paper 1 Q12
Oxtown and Camville are connected by three roads, which are at risk of being blocked by flooding. On two of the three roads there are two sections which may be blocked. On the third road there is only one section which may be blocked. The probability that each section is blocked is \(p\). Each section is blocked independently of the other four sections. Show that the probability that Oxtown is cut off from Camville is \(p^3 \l 2-p \r^2\). I want to travel from Oxtown to Camville. I choose one of the three roads at random and find that my road is not blocked. Find the probability that I would not have reached Camville if I had chosen either of the other two roads. You should factorise your answer as fully as possible. Comment briefly on the value of this probability in the limit \(p\to1\).
2006 Paper 1 Q13
A very generous shop-owner is hiding small diamonds in chocolate bars. Each diamond is hidden independently of any other diamond, and on average there is one diamond per kilogram of chocolate.
- I go to the shop and roll a fair six-sided die once. I decide that if I roll a score of \(N\), I will buy \(100N\) grams of chocolate. Show that the probability that I will have no diamonds is \[ \frac{\e^{-0.1}}{ 6} \l \frac{1 - \e^{-0.6} }{ 1 - \e^{-0.1}} \r \] Show also that the expected number of diamonds I find is 0.35.
- Instead, I decide to roll a fair six-sided die repeatedly until I score a 6. If I roll my first 6 on my \(T\)th throw, I will buy \(100T\) grams of chocolate. Show that the probability that I will have no diamonds is \[ \frac{\e^{-0.1}}{ 6 - 5\e^{-0.1}} \] Calculate also the expected number of diamonds that I find. (You may find it useful to consider the the binomial expansion of \(\l 1 - x \r^{-2}\).)
Solution: Not that the number of diamonds per kilogram is \(1\) so we are assuming it is \(Po(M)\) where \(M\) is the mass in kg. In particular \(\E[X] = M\) and \(\mathbb{P}(X = 0) = e^{-M}\)
- \(\,\) \begin{align*} && \mathbb{P}(\text{no diamonds}) &= \sum_{n=1}^6\mathbb{P}(\text{no diamonds and roll }n) \\ &&&= \sum_{n=1}^6 \tfrac16 e^{-\frac{n}{10}} \\ &&&= \frac{e^{-0.1}}6 \left ( \frac{1-e^{-0.6}}{1-e^{-0.1}}\right) \\ && \E[\text{diamonds}] &= \sum_{n=1}^6 \E(\text{diamonds}|N = n)\mathbb{P}(N = n) \\ &&&= \sum_{n=1}^6 0.1n \cdot \frac16 \\ &&&= 0.1 \cdot \frac{7}{2} = 0.35 \end{align*}
- \(\mathbb{P}(T = k) = \left ( \frac56 \right)^{t-1} \frac16\), so \begin{align*} && \mathbb{P}(\text{no diamonds}) &= \sum_{n=1}^\infty\mathbb{P}(\text{no diamonds and }T=n) \\ &&&= \sum_{n=1}^\infty e^{-0.1n} \left ( \frac56 \right)^{n-1} \frac16 \\ &&&= \frac{e^{-0.1}}{6} \frac1{1- \frac56 e^{-0.1}} \\ &&&= \frac{e^{-0.1}}{6 - 5e^{-0.1}} \\ \\ && \E[\text{diamonds}] &= \sum_{n=1}^\infty \E(\text{diamonds}|T = n)\mathbb{P}(T = n) \\ &&&= \sum_{n=1}^\infty 0.1n \cdot \left ( \frac56 \right)^{n-1} \frac16 \\ &&&= \frac{0.1}{6} \sum_{n=1}^\infty n \cdot \left ( \frac56 \right)^{n-1} \\ &&&= \frac{1}{60} \frac{1}{(1- \tfrac56)^2} \\ &&&= \frac{6}{10} = \frac35 \end{align*}
2006 Paper 1 Q14
- A bag of sweets contains one red sweet and \(n\) blue sweets. I take a sweet from the bag, note its colour, return it to the bag, then shake the bag. I repeat this until the sweet I take is the red one. Find an expression for the probability that I take the red sweet on the \(r\)th attempt. What value of \(n\) maximises this probability?
- Instead, I take sweets from the bag, without replacing them in the bag, until I take the red sweet. Find an expression for the probability that I take the red sweet on the \(r\)th attempt. What value of \(n\) maximises this probability?
Solution:
- This is the probability of having the sequence \(\underbrace{BB\cdots B}_{r-1 \text{ times}}R\) which has probability \(\displaystyle \left ( \frac{n}{n+1} \right)^{r-1}\frac{1}{n+1}\). Maximising this, is equivalent to maximising \(\log\) of it, ie \begin{align*} && y &= (r-1) \ln n - r \ln (n+1) \\ \Rightarrow && \frac{\d y}{\d n} &= \frac{r-1}{n} - \frac{r}{n+1} \\ &&&= \frac{(r-1)(n+1)-rn}{n(n+1)} \\ &&&= \frac{r-n-1}{n(n+1)} \end{align*} Therefore this is maximised when \(n = r-1\)
2006 Paper 2 Q1
The sequence of real numbers \(u_1\), \(u_2\), \(u_3\), \(\ldots\) is defined by \begin{equation*} u_1=2 \,, \qquad\text{and} \qquad u_{n+1} = k - \frac{36}{u_n} \quad \text{for } n\ge1, \tag{\(*\)} \end{equation*} where \(k\) is a constant.
- Determine the values of \(k\) for which the sequence \((*)\) is: (a) constant; (b) periodic with period 2; (c) periodic with period 4.
- In the case \(k=37\), show that \(u_n\ge 2\) for all \(n\). Given that in this case the sequence \((*)\) converges to a limit \(\ell\), find the value of \(\ell\).
2006 Paper 2 Q2
Using the series \[ \e^x = 1 + x +\frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}+\cdots\,, \] show that \(\e>\frac83\). Show that \(n!>2^n\) for \(n\ge4\) and hence show that \(\e<\frac {67}{24}\). Show that the curve with equation \[ y= 3\e^{2x} +14 \ln (\tfrac43-x)\,, \qquad {x<\tfrac43} \] has a minimum turning point between \(x=\frac12\) and \(x=1\) and give a sketch to show the shape of the curve.
Solution: \begin{align*} && e &= 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots \\ &&&> 1 + 1+ \frac12 + \frac16 \\ &&&= \frac{12+3+1}{6} = \frac83 \end{align*} \(4! = 24 > 16 = 2^4\), notice that \(n! = \underbrace{n \cdot (n-1) \cdots 5}_{>2^{n-4}} \cdot \underbrace{4!}_{>2^4} >2^n\). \begin{align*} && e &= 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots \\ &&&< \frac83 + \frac{1}{2^4} + \frac{1}{2^5} + \cdots \\ &&&= \frac83 + \frac{1}{2^4} \frac{1}{1-\tfrac12} \\ &&&= \frac83 + \frac1{8} \\ &&&= \frac{67}{24} \end{align*} \begin{align*} && y &= 3e^{2x} +14 \ln(\tfrac43-x) \\ && y' &= 6e^{2x} - \frac{14}{\tfrac43-x} \\ && y'(\tfrac12) &= 6e - \frac{14}{\tfrac43-\tfrac12} \\ &&&= 6e -\tfrac{84}{5} = 6(e-\tfrac{14}5) < 0 \\ && y'(1) &= 6e^2 - \frac{14}{\tfrac43-1} \\ &&&= 6e^2 - 42 = 6(e^2-7) \\ &&&> 6(\tfrac{64}{9} - 7) > 0 \end{align*} Therefore \(y'\) changes from negative (decreasing) to positive (increasing) in our range, and therefore there is a minima in this range.
2006 Paper 2 Q3
- Show that $\displaystyle \big( 5 + \sqrt {24}\;\big)^4 + \frac{1 }{\big(5 + \sqrt {24}\;\big)^4} \ $ is an integer. Show also that \[\displaystyle 0.1 < \frac{1}{ 5 + \sqrt {24}} <\frac 2 {19}< 0.11\,.\] Hence determine, with clear reasoning, the value of \(\l 5 + \sqrt {24}\r^4\) correct to four decimal places.
- If \(N\) is an integer greater than 1, show that \(( N + \sqrt {N^2 - 1} \,) ^k\), where \(k\) is a positive integer, differs from the integer nearest to it by less than \(\big( 2N - \frac12 \big)^{-k}\).
Solution:
- First notice that \(\frac{1}{5+\sqrt{24}} = \frac{5-\sqrt{24}}{25-24} = 5 - \sqrt{24}\), hence \begin{align*} && ( 5 + \sqrt {24})^4 + \frac{1 }{(5 + \sqrt {24})^4} &= ( 5 + \sqrt {24})^4 + ( 5 - \sqrt {24})^4 \\ \end{align*} where clearly all terms including \(\sqrt{24}\) will cancel out, therefore it is an integer. \begin{align*} && 5 + \sqrt{24} &< 5 + 5 = 10 \\ \Rightarrow && \frac{1}{5+\sqrt{24}}& > \frac{1}{10} = 0.1 \\ && 2(5 + \sqrt{24}) &=10 + \sqrt{96} > 19 \\ \Rightarrow && \frac{1}{5+\sqrt{24}} & < \frac{2}{19} < \frac{2}{18} = \frac19 = 0.11111\ldots < 0.11 \end{align*} Therefore, \(10^{-4} < (5+\sqrt{24})^4 < 0.11^{-4} = 0.00014641\) \begin{align*} && (5+\sqrt{24})^4 + (5-\sqrt{24})^4 &= 2(5^4+6\cdot5^2\cdot24+24^2) \\ &&&= 2\cdot (625 + 3600+576) \\ &&&= 9602 \\ \Rightarrow && (5+\sqrt{24})^4 &= 9602 - \epsilon, \epsilon \in (0.0001, 0.00014641) \\ \Rightarrow && (5+\sqrt{24})^4 &\in (9601.999854, ,9601.9999) \\ \Rightarrow && (5+\sqrt{24})^4 &= 9601.9998 \, (4 \text{ d.p.}) \end{align*}
- Notice that \((N+\sqrt{N^2-1})^{k}+(N-\sqrt{N^2-1})^{k}\) is an integer for the same reason as before (sum of conjugates). Notice also that \(\frac{1}{N+\sqrt{N^2-1}} = N - \sqrt{N^2-1}\) and that so it sufficies to show that \begin{align*} && N + \sqrt{N^2-1} &> 2N-\tfrac12 \\ \Leftrightarrow && \sqrt{N^2-1} &> N - \tfrac12 \\ \Leftrightarrow && N^2-1 &> N^2-N+1\\ \Leftrightarrow && N &> \tfrac32\\ \end{align*} Which is true since \(N > 1\) and \(N\) is an integer.
2006 Paper 2 Q4
By making the substitution \(x=\pi-t\,\), show that \[ \! \int_0^\pi x\f(\sin x) \d x = \tfrac12 \pi \! \int_0^\pi \f(\sin x) \d x\,, \] where \(\f(\sin x)\) is a given function of \(\sin x\). Evaluate the following integrals:
- \(\displaystyle \int_0^\pi \frac {x \sin x}{3+\sin^2 x}\,\d x\,\);
- $\displaystyle \int_0^{2\pi} \frac {x \sin x}{3+\sin^2 x}\,\d x\,\(;
- \)\displaystyle \int_{0}^{\pi} \frac {x \big\vert\sin 2x\big\vert}{3+\sin^2 x}\,\d x\,$.
2006 Paper 2 Q5
The notation \({\lfloor } x \rfloor\) denotes the greatest integer less than or equal to the real number \(x\). Thus, for example, \(\lfloor \pi\rfloor =3\,\), \(\lfloor 18\rfloor =18\,\) and \(\lfloor-4.2\rfloor = -5\,\).
- Two curves are given by \(y= x^2+3x-1\) and \(y=x^2 +3\lfloor x\rfloor -1\,\). Sketch the curves, for \(1\le x \le 3\,\), on the same axes. Find the area between the two curves for \(1\le x \le n\), where \(n\) is a positive integer.
- Two curves are given by \(y= x^2+3x-1\) and \(y=\lfloor x\rfloor ^2+3\lfloor x\rfloor -1\,\). Sketch the curves, for \(1\le x \le 3\,\), on the same axes. Show that the area between the two curves for \(1\le x \le n\), where \(n\) is a positive integer, is \[ \tfrac 16 (n-1)(3n+11)\,. \]
Solution:
- \(\,\)
The difference between the curves is \(3x - 3\lfloor x \rfloor\), which has area \(\frac32\) for each step. Therefore the area between the curves from \(1 \leq x \leq n\) is \(\frac32 (n-1)\)
- \(\,\)
The area between the curves is \(x^2 - \lfloor x \rfloor ^2 + 3(x - \lfloor x \rfloor)\). Looking at \begin{align*} && A &= \int_1^n \left ( x^2 - \lfloor x \rfloor ^2 \right )\d x \\ &&&= \frac{n^3-1^3}{3} - \sum_{k=1}^{n-1} k^2 \\ &&&= \frac{(n-1)(n^2+n+1)}{3} - \frac{(n-1)n(2n-1)}{6} \\ &&&= \frac{(n-1) \left (2n^2+2n+2-2n^2+n \right)}{6} \\ &&&= \frac{(n-1)(3n+2)}{6} \end{align*} Therefore the total area is \(\frac{(n-1)(3n+2)}{6}+\frac32(n-1) = \frac{(n-1)}{6}\left ( 3n+2+9\right) =\frac{(n-1)(3n+11)}{6}\)
2006 Paper 2 Q6
By considering a suitable scalar product, prove that \[ (ax+by+cz)^2 \le (a^2+b^2+c^2)(x^2+y^2+z^2) \] for any real numbers \(a\), \(b\), \(c\), \(x\), \(y\) and \(z\). Deduce a necessary and sufficient condition on \(a\), \(b\), \(c\), \(x\), \(y\) and \(z\) for the following equation to hold: \[ (ax+by+cz)^2 = (a^2+b^2+c^2)(x^2+y^2+z^2) \,. \]
- Show that \((x+2y+2z)^2 \le 9(x^2+y^2+z^2)\) for all real numbers \(x\), \(y\) and \(z\).
- Find real numbers \(p\), \(q\) and \(r\) that satisfy both \[ p^2+4q^2+9r^2 = 729 \text{ and } 8p+8q+3r = 243\,. \]
Solution: Consider \(\begin{pmatrix} a \\ b \\ c \end{pmatrix}\), \(\begin{pmatrix} x \\ y \\ z \end{pmatrix}\), then we know that \begin{align*} && \begin{pmatrix} a \\ b \\ c \end{pmatrix} \cdot \begin{pmatrix} x \\ y \\ z \end{pmatrix} &= \sqrt{a^2+b^2+c^2} \sqrt{x^2+y^2+z^2} \cos \theta \\ \Rightarrow && (ax+by+cz)^2 &= (a^2+b^2+c^2)(x^2+y^2+z^2) \cos^2 \theta \\ &&&\leq (a^2+b^2+c^2)(x^2+y^2+z^2) \end{align*} For equality to hold, we must have that the vectors are parallel, ie \(\begin{pmatrix} a \\ b \\ c \end{pmatrix} = \lambda \begin{pmatrix} x \\ y \\ z \end{pmatrix}\)
- By applying our inequality from the first part with \(a=1, b = 2, c=2\) we have \((x+2y+2z)^2 \leq (1+2^2+2^2)(x^2+y^2+z^2) = 9(x^2+y^2+z^2)\)
- Since \begin{align*} && (p^2+(2q)^2+(3r)^2)\left (8^2 +4^2+1^2 \right) &\geq (8p+8q+3r)^2 \\ \Leftrightarrow && 729 \cdot 81 &\geq 243^2 \\ &&3^6 \cdot 3^4 &\geq 3^{10} \end{align*} Therefore we must be in the equality case, ie \(p = 8\lambda, 2q = 4\lambda, 3r = \lambda\) as well as \(64\lambda + 16\lambda +\lambda = 243 \Rightarrow 81\lambda = 243 \Rightarrow \lambda = 3\) so we have \[ (p,q,r) = \left (24, 6, 1 \right) \]