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2003 Paper 2 Q7
Show that, if \(n>0\,\), then $$ \int_{e^{1/n}}^\infty\,{{\ln x} \over {x^{n+1}}}\,\d x = {2 \over {n^2\e}}\;. $$ You may assume that \(\ds \frac{\ln x} x \to 0\;\) as \(x\to\infty\,\). Explain why, if \(1 < a < b\,\), then $$ \int_b^\infty\,{{\ln x} \over {x^{n+1}}}\,\d x < \int_a^\infty\,{{\ln x} \over {x^{n+1}}}\,\d x\;. $$ Deduce that $$ \sum_{n=1}^{N}{1 \over n^2} < {\e \over 2}\int_{\e^{1/N}}^{\infty} \left({1-x^{-N}} \over {x^2-x}\right) \ln x\,\d x\;, $$ where \(N\,\) is any integer greater than \(1\).
2003 Paper 2 Q8
It is given that \(y\) satisfies $$ {{\d y} \over { \d t}} + k\left({{t^2-3t+2} \over {t+1}}\right)y = 0\;, $$ where \(k\) is a constant, and \(y=A \) when \(t=0\,\), where \(A\) is a positive constant. Find \(y\) in terms of \(t\,\), \(k\) and \(A\,\). Show that \(y\) has two stationary values whose ratio is \((3/2)^{6k}\e^{-5{k}/2}.\) Describe the behaviour of \(y\) as \(t \to +\infty\) for the case where \(k> 0\) and for the case where \(k<0\,.\) In separate diagrams, sketch the graph of \(y\) for \(t>0\) for each of these cases.
Solution: \begin{align*} && \frac{\d y}{\d t} &= - k \left (\frac{t^2-3t+2}{t+1} \right) y \\ \Rightarrow && \int \frac1y \d y &= -k\int \left (t-4 + \frac{6}{t+1}\right) \d t \\ \Rightarrow && \ln y &= -k \left ( \frac12 t^2 -4t + 6\ln (t+1) \right) + C \\ (t,y) = (0,A): && \ln A &=C \\ \Rightarrow && \ln y &= -k \left ( \frac12 t^2 -4t + 6\ln (t+1) \right) + \ln A \\ && \ln \left ( \frac{y}{A}(t+1)^{6k} \right) &= -k \l \frac12 t^2 - 4t \r \\ \Rightarrow && y &= A\frac{\exp \l -k(\frac12 t^2-4t)\r}{(t+1)^{6k}} \end{align*} \(y\) wil have stationary values when \(\frac{\d y}{\d t} = 0\), ie \begin{align*} k \left (\frac{t^2-3t+2}{t+1} \right) y &= 0 \\ k \left ( \frac{(t-2)(t-1)}{t+1} \right) y &= 0 \end{align*} ie when \(y = 0, t = 1, t =2\). Clearly \(y = 0\) is not a solution, so \(y\) has the values: \begin{align*} t = 1: && y &= A\frac{\exp \l -k(\frac12 -4)\r}{(2)^{6k}} \\ &&&= A \frac{e^{7/2 k}}{2^{6k}} \\ t = 2: && y &= A\frac{\exp \l -k(2 -8)\r}{(3)^{6k}} \\ &&&= A \frac{e^{6 k}}{3^{6k}} \\ \text{ratio}: && \frac{e^{7/2k}}{2^{6k}} \cdot \frac{3^{6k}}{e^{6k}} &= (3/2)^{6k} e^{-5k/2} \end{align*} If \(k > 0\) as \(t \to \infty\) \(y \to 0\) since the \(e^{-kt^2/2}\) term dominates everything. If \(k < 0\) as \(t \to \infty\) \(y \to \infty\) as since the \(e^{-kt^2}\) term also dominates but now it growing to infinity faster than everything else.
2003 Paper 2 Q9
\(AB\) is a uniform rod of weight \(W\,\). The point \(C\) on \(AB\) is such that \(AC>CB\,\). The rod is in contact with a rough horizontal floor at \(A\,\) and with a cylinder at \(C\,\). The cylinder is fixed to the floor with its axis horizontal. The rod makes an angle \({\alpha}\) with the horizontal and lies in a vertical plane perpendicular to the axis of the cylinder. The coefficient of friction between the rod and the floor is \(\tan \lambda_1\) and the coefficient of friction between the rod and the cylinder is \(\tan \lambda_2\,\). Show that if friction is limiting both at \(A\) and at \(C\), and \({\alpha} \ne {\lambda}_2 - {\lambda}_1\,\), then the frictional force acting on the rod at \(A\) has magnitude $$ \frac{ W\sin {\lambda}_1 \, \sin({\alpha}-{\lambda}_2)} {\sin ({\alpha}+{\lambda}_1-{\lambda}_2)} \;.$$ %and that %$$ %p=\frac{\cos{\alpha} \, \sin({\alpha}+{\lambda}_1-{\lambda}_2)} %{2\cos{\lambda}_1 \, \sin {\lambda}_2}\;. %$$
2003 Paper 2 Q10
A bead \(B\) of mass \(m\) can slide along a rough horizontal wire. A light inextensible string of length \(2\ell\) has one end attached to a fixed point \(A\) of the wire and the other to \(B\,\). A particle \(P\) of mass \(3m\) is attached to the mid-point of the string and \(B\) is held at a distance \(\ell\) from~\(A\,\). The bead is released from rest. Let \(a_1\) and \(a_2\) be the magnitudes of the horizontal and vertical components of the initial acceleration of \(P\,\). Show by considering the motion of \(P\) relative to \(A\,\), or otherwise, that \(a_1= \sqrt 3 a_2\,\). Show also that the magnitude of the initial acceleration of \(B\) is \(2a_1\,\). Given that the frictional force opposing the motion of \(B\) is equal to \(({\sqrt{3}}/6)R\), where \(R\) is the normal reaction between \(B\) and the wire, show that the magnitude of the initial acceleration of \(P\) is~\(g/18\,\).
2003 Paper 2 Q11
A particle \(P_1\) is projected with speed \(V\) at an angle of elevation \({\alpha}\,\,\,( > 45^{\circ})\,,\,\,\,\) from a point in a horizontal plane. Find \(T_1\), the flight time of \(P_1\), in terms of \({\alpha}, V \hbox{ and } g\,\). Show that the time after projection at which the direction of motion of \(P_1\) first makes an angle of \(45^{\circ}\) with the horizontal is \(\frac12 (1-\cot \alpha)T_1\,\). A particle \(P_2\) is projected under the same conditions. When the direction of the motion of \(P_2\) first makes an angle of \(45^{\circ}\) with the horizontal, the speed of \(P_2\) is instantaneously doubled. If \(T_2\) is the total flight time of \(P_2\), show that $$ \frac{2T_2}{T_1} = 1+\cot{\alpha} +\sqrt{1+3\cot^2{\alpha}} \;. $$
2003 Paper 2 Q12
The life of a certain species of elementary particles can be described as follows. Each particle has a life time of \(T\) seconds, after which it disintegrates into \(X\) particles of the same species, where \(X\) is a random variable with binomial distribution \(\mathrm{B}(2,p)\,\). A population of these particles starts with the creation of a single such particle at \(t=0\,\). Let \(X_n\) be the number of particles in existence in the time interval \(nT < t < (n+1)T\,\), where \(n=1\,\), \(2\,\), \(\ldots\). Show that \(\P(X_1=2 \mbox { and } X_2=2) = 6p^4q^2\;\), where \(q=1-p\,\). Find the possible values of \(p\) if it is known that \(\P(X_1=2 \vert X_2=2) =9/25\,\). Explain briefly why \(\E(X_n) =2p\E(X_{n-1})\) and hence determine \(\E(X_n)\) in terms of \(p\). Show that for one of the values of \(p\) found above \(\lim_{n \to \infty}\E(X_n) = 0\) and that for the other \(\lim_{n \to \infty}\E(X_n) = + \infty\,\).
Solution: Notice that we can see the total number generated as \(X_n \sim B(2X_{n-1},p)\), since a Binomial is a sum of independent Bernoullis, and there are two Bernoullis per particle. \begin{align*} && \mathbb{P}(X_1=2 \mbox { and } X_2=2) &= \underbrace{p^2}_{\text{two generated in first iteration}} \cdot \underbrace{\binom{4}{2}p^2q^2}_{\text{two generated from the first two}} \\ &&&= 6p^4q^2 \end{align*} \begin{align*} && \mathbb{P})(X_1 = 2 |X_2 = 2) &= \frac{ \mathbb{P}(X_1=2 \mbox { and } X_2=2) }{ \mathbb{P}( X_2=2) } \\ &&&= \frac{6p^4q^2}{6p^4q^2+2pq \cdot p^2} \\ &&&= \frac{3pq}{3pq+1} \\ \Rightarrow && \frac{9}{25} &= \frac{3pq}{3pq+1} \\ \Rightarrow && 27pq + 9 &= 75pq \\ \Rightarrow && 9 &= 48pq \\ \Rightarrow && pq &= \frac{3}{16} \\ \Rightarrow && 0 &= p^2 - p + \frac3{16} \\ \Rightarrow && p &= \frac14, \frac34 \end{align*} By the same reasoning about the Bernoullis, we must have \(\E[X_n] = \E[\E[X_n | X_{n-1}]] = \E[2pX_{n-1}] = 2p \E[X_{n-1}]\) therefore \(\E[X_n] = (2p)^n\). If \(p = \frac14\) then \(\E[X_n] = \frac1{2^n} \to 0\) If \(p = \frac34\) then \(\E[X_n] = \left(\frac32 \right)^n \to \infty\)
2003 Paper 2 Q13
The random variable \(X\) takes the values \(k=1\), \(2\), \(3\), \(\dotsc\), and has probability distribution $$ \P(X=k)= A{{{\lambda}^k\e^{-{\lambda}}} \over {k!}}\,, $$ where \(\lambda \) is a positive constant. Show that \(A = (1-\e^{-\lambda})^{-1}\,\). Find the mean \({\mu}\) in terms of \({\lambda}\) and show that $$ \var(X) = {\mu}(1-{\mu}+{\lambda})\;. $$ Deduce that \({\lambda} < {\mu} < 1+{\lambda}\,\). Use a normal approximation to find the value of \(P(X={\lambda})\) in the case where \({\lambda}=100\,\), giving your answer to 2 decimal places.
Solution: Let \(Y \sim Po(\lambda)\) \begin{align*} && 1 &= \sum_{k=1}^\infty \mathbb{P}(X = k ) \\ &&&= \sum_{k=1}^\infty A \frac{\lambda^k e^{-\lambda}}{k!}\\ &&&= Ae^{-\lambda} \sum_{k=1}^{\infty} \frac{\lambda^k e^{-\lambda}}{k!} \\ &&&= Ae^{-\lambda} \left (e^{\lambda}-1 \right) \\ \Rightarrow && A &= (1-e^{-\lambda})^{-1} \\ \\ && \E[X] &= \sum_{k=1}^{\infty} k \cdot \mathbb{P}(X=k) \\ &&&= A\sum_{k=1}^{\infty} k \frac{\lambda^k e^{-\lambda}}{k!} \\ &&&= A\E[Y] = A\lambda = \lambda(1-e^{-\lambda})^{-1} \\ \\ && \var[X] &= \E[X^2] - (\E[X])^2 \\ &&&= A\sum_{k=1}^{\infty} k^2 \frac{\lambda^k e^{-\lambda}}{k!} - \mu^2 \\ &&&= A\E[Y^2] - \mu^2 \\ &&&= A(\var[Y]+\lambda^2) - \mu^2 \\ &&&= A(\lambda + \lambda^2) - \mu^2 \\ &&&= A\lambda(1+\lambda) - \mu^2 \\ &&&= \mu(1+\lambda - \mu) \end{align*} Since \(A > 1\) we must have \(\mu > \lambda\) and since \(\var[X] > 0\) we must have \(1 + \lambda > \mu\) as required. If \(\lambda = 100\), then \(A \approx 1\) and \(P(X=\lambda) \approx P(Y = \lambda)\) and \(Y \approx N(\lambda, \lambda)\) so the value is approximately \(\displaystyle \int_{-\frac12}^{\frac12} \frac{1}{\sqrt{2\pi \lambda}} e^{-\frac{x^2}{2\lambda}} \d x \approx \frac{1}{\sqrt{200\pi}} = \frac{1}{\sqrt{630.\ldots}} \approx \frac{1}{25} = 0.04 \)
2003 Paper 2 Q14
The probability of throwing a 6 with a biased die is \(p\,\). It is known that \(p\) is equal to one or other of the numbers \(A\) and \(B\) where \(0 < A < B < 1 \,\). Accordingly the following statistical test of the hypothesis \(H_0: \,p=B\) against the alternative hypothesis \(H_1: \,p=A\) is performed. The die is thrown repeatedly until a 6 is obtained. Then if \(X\) is the total number of throws, \(H_0\) is accepted if \(X \le M\,\), where \(M\) is a given positive integer; otherwise \(H_1\) is accepted. Let \({\alpha}\) be the probability that \(H_1\) is accepted if \(H_0\) is true, and let \({\beta}\) be the probability that \(H_0\) is accepted if \(H_1\) is true. Show that \({\beta} = 1- {\alpha}^K,\) where \(K\) is independent of \(M\) and is to be determined in terms of \(A\) and \(B\,\). Sketch the graph of \({\beta}\) against \({\alpha}\,\).
Solution: \(X \sim Geo(p)\). \(\alpha = \mathbb{P}(X > M | p = B) = (1-B)^{M}\) \(\beta = \mathbb{P}(X \leq M | p = A) = 1 - \mathbb{P}(X > M | p = A) = 1 - (1-A)^{M}\) \begin{align*} \ln \alpha &= M \ln(1-B) \\ \ln (1-\beta) &= M \ln(1-A) \\ \frac{\ln \alpha}{\ln (1-\beta)} &= \frac{\ln(1-B)}{\ln(1-A)} \\ \ln(1-\beta) &= \ln \alpha \frac{\ln (1-A)}{\ln(1-B)} \\ \beta &= 1- \alpha^{ \frac{\ln (1-A)}{\ln(1-B)} } \end{align*} and \(K = \frac{\ln (1-A)}{\ln(1-B)} \) Since \(0 < A < B < 1\) we must have that \(0 < 1 - B < 1-A < 1\) and \(\ln(1-B) < \ln(1-A) < 0\) so \(0 < K < 1\)
2003 Paper 3 Q1
Given that \(x+a>0\) and \(x+b>0\,\), and that \(b>a\,\), show that \[ \frac{\mathrm{d} \ }{\mathrm{d} x} \arcsin \left ( \frac{x + a }{ \ x + b} \right) = \frac{ \sqrt{\;b - a\;}} {( x + b ) \sqrt{ a + b + 2x} \ \ } \] and find $\displaystyle \frac{\mathrm{d} \ }{ \mathrm{d} x} \; \mathrm{arcosh} \left ( \frac{x + b }{ \ x + a} \right)$. Hence, or otherwise, integrate, for \(x > -1\,\),
- \[ \int \frac{1}{ ( x + 1) \sqrt{x + 3} } \mathrm{d} x \]
- \[ \int \frac{1} {( x + 3 ) \sqrt{x + 1} } \mathrm{d} x \] .
Solution: \begin{align*} \frac{\mathrm{d} \ }{\mathrm{d} x} \arcsin \left ( \frac{x + a }{ \ x + b} \right ) &= \frac{1}{\sqrt{1-\left ( \frac{x + a }{ \ x + b} \right )^2}} \left ( \frac{b-a}{(x+b)^2} \right) \\ &= \frac{b-a}{(x+b)\sqrt{(x+b)^2-(x+a)^2}} \\ &= \frac{b-a}{(x+b)\sqrt{(b-a)(2x+b+a)}} \\ &= \frac{\sqrt{b-a}}{(x+b)\sqrt{a+b+2x}} \\ \\ \frac{\mathrm{d} \ }{ \mathrm{d} x} \; \mathrm{arcosh} \left ( \frac{x + b }{ \ x + a} \right) &= \frac{1}{\sqrt{\left ( \frac{x + b }{ \ x + a} \right)^2-1}} \left ( -\frac{b-a}{(x+a)^2} \right) \\ &= -\frac{b-a}{(x+a)\sqrt{(x+b)^2-(x+a)^2}} \\ &= -\frac{b-a}{(x+a)\sqrt{(b-a)(a+b+2x)}} \\ &= -\frac{\sqrt{b-a}}{(x+a)\sqrt{a+b+2x}} \end{align*}
- \begin{align*} \int \frac{1}{ ( x + 1) \sqrt{x + 3} } \mathrm{d} x &= \int \frac{1}{(x+1)\sqrt{\frac12 (2x+6)}} \d x\\ &= \int \frac{\sqrt{2}}{(x+1)\sqrt{2x+1+5}} \d x \\ &= \frac{\sqrt{2}}{2}\int \frac{\sqrt{5-1}}{(x+1)\sqrt{2x+1+5}} \d x \\ &= - \frac{\sqrt{2}}{2}\textrm{arcosh} \left ( \frac{x+5}{x+1} \right) + C \end{align*}
- \begin{align*} \int \frac{1}{(x+3)\sqrt{x+1}} \d x &= \int \frac{1}{(x+3)\sqrt{\tfrac12(2x+2)}} \d x + C \\ &= \int \frac{\sqrt{3-1}}{(x+3)\sqrt{2x+3-1}} \d x \\ &= \textrm{arcsin} \left ( \frac{x-1}{x+3} \right) \end{align*}
2003 Paper 3 Q2
Show that $\ds ^{2r} \! {\rm C}_r =\frac{1\times3\times\dots\times (2r-1)}{r!} \, \times 2^r \;, $ for \(r\ge1\,\).
- Give the first four terms of the binomial series for \(\l 1 - p \r^{-\frac12}\). By choosing a suitable value for \(p\) in this series, or otherwise, show that $$ \displaystyle \sum_{r=0}^\infty \frac{ {\vphantom {\A}}^{2r} \! {\rm C}_r }{ 8^r} = \sqrt 2 \; .$$
- Show that $$ \displaystyle \sum_{r=0}^\infty \frac{\l 2r + 1 \r \; {\vphantom{A}}^{2r} \! {\rm C} _r }{ 5^r} =\big( \sqrt 5\big)^3 \;. $$
Solution: \begin{align*} \binom{2r}{r} &= \frac{(2r)!}{r!r!} \\ &= \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdots (2r-1)(2r)}{r! r!} \\ &= \frac{1 \cdot 3 \cdot 5 \cdots (2r-1) \cdot (2 \cdot 1) \cdot (2 \cdot 2) \cdots (2 \cdot r)}{r!}{r!} \\ &= \frac{1\cdot 3 \cdots (2r-1) \cdot 2^r \cdot 1 \cdot 2 \cdots r}{r!r!} \\ &= \frac{1\cdot 3 \cdots (2r-1) \cdot 2^r \cdot r!}{r!r!} \\ &= \frac{1\cdot 3 \cdots (2r-1)}{r!} \cdot 2^r \end{align*} which is what we wanted to show
- \begin{align*} (1 - p)^{-\frac12} &= 1 + \left ( -\frac12 \right )(-p) + \frac{1}{2!} \left (-\frac12 \right )\left (-\frac32 \right )(-p)^2 + \ldots \\ & \quad \quad \quad \cdots +\frac{1}{3!} \left (-\frac12 \right )\left (-\frac32 \right )\left (-\frac52 \right )(-p)^3 + O(p^4) \\ &= \boxed{1 + \frac{1}{2}p + \frac{3}{8}p^2 + \frac{5}{16}p^3} + O(p^4) \end{align*} More generally: \begin{align*} \binom{-\frac{1}{2}}{k} &=\frac{(-\frac{1}{2})\cdot(-\frac{1}{2} -1)\cdots(-\frac12 -k+1)}{k!} \\ &= \frac{(-1)(-3)(-5)\cdots(-(2k-1))}{k!2^k} \\ &= \frac{(-1)^k(1)(3)(5)\cdots((2k-1))}{k!2^k} \\ &= (-1)^k \frac{1}{4^k}\binom{2k}{k} \\ \end{align*} Therefore, \begin{align*} \sqrt{2} = \left (1-\frac12 \right)^{-\frac12} &= \sum_{r=0}^{\infty} \binom{-\frac12}{r} \left (-\frac12 \right )^r \tag{\(\frac12 < 1\) so series is valid} \\ &= \sum_{r=0}^{\infty} (-1)^r \frac{1}{4^r}\binom{2r}{r} \left (-\frac12 \right )^r \\ &= \sum_{r=0}^{\infty} \frac{1}{8^r}\binom{2r}{r} \end{align*}, which is what we wanted to show.
- \begin{align*} p(1-p^2)^{-\frac12} &= \sum_{r=0}^{\infty} \binom{-\frac12}{r} \left (-p^2 \right )^rp \\ &= \sum_{r=0}^{\infty} \frac{1}{4^r}\binom{2r}{r} p^{2r+1} \end{align*} Differentiating with respect to \(p\), \begin{align*} (1-p^2)^{-\frac12} +p^2(1-p^2)^{-\frac32} &= \sum_{r=0}^{\infty} \frac{1}{4^r}(2r+1)\binom{2r}{r} p^{2r} \\ (1-p^2)^{-\frac32} &= \sum_{r=0}^{\infty} \frac{1}{4^r}(2r+1)\binom{2r}{r} p^{2r} \end{align*} Letting \(p = \frac{2}{\sqrt{5}}\), and \(|\frac2{\sqrt{5}}| < 1\) we have \begin{align*} \left (1-\frac45 \right )^{-\frac32} &= \sum_{r=0}^{\infty} \frac{1}{5^r}(2r+1)\binom{2r}{r} \\ (\sqrt{5})^3 &= \sum_{r=0}^{\infty} \frac{1}{5^r}(2r+1)\binom{2r}{r} \end{align*} (Alternative) \begin{align*} (\sqrt5)^3 &= \left ( \frac{1}{5} \right )^{-\frac32} \\ &= \left ( 1- \frac{4}{5} \right )^{-\frac32} \\ &= \sum_{r=0}^{\infty} \binom{-\frac32}{r} \left (-\frac45 \right)^r \\ &= \sum_{r=0}^{\infty} \binom{-\frac12}{r} \frac{-\frac32-(r-1)}{-\frac12} \left (-\frac45 \right)^r \\ &= \sum_{r=0}^{\infty} \binom{-\frac12}{r} (2r+1) \left (-\frac45 \right)^r \\ &= \sum_{r=0}^{\infty} (-1)^r \frac{1}{4^r}\binom{2r}{r} (2r+1) \left (-\frac45 \right)^r \\ &= \sum_{r=0}^{\infty}(2r+1)\binom{2r}{r} \left (\frac15 \right)^r \\ (\sqrt{5})^3 &= \sum_{r=0}^{\infty}\frac{1}{5^r}(2r+1)\binom{2r}{r} \\ \end{align*}