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2008 Paper 1 Q11
A straight uniform rod has mass \(m\). Its ends \(P_1\) and \(P_2\) are attached to small light rings that are constrained to move on a rough rigid circular wire with centre \(O\) fixed in a vertical plane, and the angle \(P_1OP_2\) is a right angle. The rod rests with \(P_1\) lower than \(P_2\), and with both ends lower than \(O\). The coefficient of friction between each of the rings and the wire is \(\mu\). Given that the rod is in limiting equilibrium (i.e. \ on the point of slipping at both ends), show that \[ \tan \alpha = \frac{1-2\mu -\mu^2}{1+2\mu -\mu^2}\,, \] where \(\alpha\) is the angle between \(P_1O\) and the vertical (\(0<\alpha<45^\circ\)). Let \(\theta\) be the acute angle between the rod and the horizontal. Show that \(\theta =2\lambda\), where \(\lambda \) is defined by \(\tan \lambda= \mu\) and \(0<\lambda<22.5^\circ\).
2008 Paper 1 Q12
In this question, you may use without proof the results: \[ \sum_{r=1}^n r = \tfrac12 n(n+1) \qquad\text{and}\qquad \sum_{r=1}^n r^2 = \tfrac1 6 n(n+1)(2n+1)\,. \] The independent random variables \(X_1\) and \(X_2\) each take values \(1\), \(2\), \(\ldots\), \(N\), each value being equally likely. The random variable \(X\) is defined by \[ X= \begin{cases} X_1 & \text { if } X_1\ge X_2\\ X_2 & \text { if } X_2\ge X_1\;. \end{cases} \]
- Show that \(\P(X=r) = \dfrac{2r-1}{N^2}\,\) for \(r=1\), \(2\), \(\ldots\), \(N\).
- Find an expression for the expectation, \(\mu\), of \(X\) and show that \(\mu=67.165\) in the case \(N=100\).
- The median, \(m\), of \(X\) is defined to be the integer such that \(\P(X\ge m) \ge \frac 12\) and \(\P(X\le m)\ge \frac12\). Find an expression for \(m\) in terms of \(N\) and give an explicit value for \(m\) in the case \(N=100\).
- Show that when \(N\) is very large, \[ \frac \mu m \approx \frac {2\sqrt2}3\,. \]
Solution: \begin{align*} \P(X = r) &= \P(X_1 = r, X_2 \leq r) + \P(X_2 = r, X_1 < r) \\ &= \P(X_1 = r) \P(X_2 \leq r) + \P(X_2 = r)\P( X_1 < r) \\ &= \frac{1}{N} \frac{r}{N} + \frac{1}{N} \frac{r-1}{N} \\ &= \frac{2r-1}{N^2} \end{align*} \begin{align*} \E(X) &= \sum_{r=1}^N r \P(X = r) \\ &= \sum_{r=1}^N \frac{2r^2 - r}{N^2} \\ &= \frac{1}{N^2} \l \frac{N(N+1)(2N+1)}{3} - \frac{N(N+1)}{2} \r \\ &= \frac{N+1}{N} \l \frac{4N-1}{6} \r \end{align*} When \(N = 100\), this is equal to \(\frac{101 \cdot 399}{6 \cdot 100} = \frac{101 \cdot 133}{200} = 67.165\) \begin{align*} &&\frac12 &\leq \P(X \leq m) \\ &&&=\sum_{r=1}^m \P(X=r) \\ &&&=\sum_{r=1}^m \frac{2r-1}{N^2} \\ &&&= \frac{1}{N^2} \l m(m+1) - m \r \\ &&&= \frac{m^2}{N^2} \\ \Rightarrow && m^2 &\geq \frac{N^2}{2} \\ \Rightarrow && m &\geq \frac{N}{\sqrt{2}} \\ \Rightarrow && m &= \left \lceil \frac{N}{\sqrt{2}} \right \rceil \end{align*} When \(N = 100\), \(100/\sqrt{2} = \sqrt{2}50\). \(\sqrt{2} > 1.4 \Rightarrow 50\sqrt{2} > 70\) \(\sqrt{2} < 1.42 \Rightarrow 50 \sqrt{2} < 71\), therefore \(\displaystyle \left \lceil \frac{100}{\sqrt{2}} \right \rceil = 71\) \begin{align*} \lim_{N \to \infty} \frac{\frac{(N+1)(4N-1)}{6N}}{ \left \lceil\frac{N}{\sqrt{2}} \right \rceil} &= \lim_{N \to \infty} \frac{\sqrt{2}}{3}\l \frac{4N^2 +3N - 1}{2N^2} \r \tag{since the floor will be irrelevant}\\ &= \lim_{N \to \infty} \frac{\sqrt{2}}{3}\l 2 + \frac{3}{2N} - \frac{1}{N^2} \r \\ &= \lim_{N \to \infty} \frac{2\sqrt{2}}{3} \end{align*}
2008 Paper 1 Q13
Three married couples sit down at a round table at which there are six chairs. All of the possible seating arrangements of the six people are equally likely.
- Show that the probability that each husband sits next to his wife is \(\frac{2}{15}\).
- Find the probability that exactly two husbands sit next to their wives.
- Find the probability that no husband sits next to his wife.
2008 Paper 2 Q1
A sequence of points \((x_1,y_1)\), \((x_2,y_2)\), \(\ldots\) in the cartesian plane is generated by first choosing \((x_1,y_1)\) then applying the rule, for \(n=1\), \(2\), \(\ldots\), \[ (x_{n+1}, y_{n+1}) = (x_n^2-y_n^2 +a, \; 2x_ny_n+b+2)\,, \] where \(a\) and \(b\) are given real constants.
- In the case \(a=1\) and \(b=-1\), find the values of \((x_1,y_1)\) for which the sequence is constant.
- Given that \((x_1,y_1) = (-1,1)\), find the values of \(a\) and \(b\) for which the sequence has period 2.
2008 Paper 2 Q2
Let \(a_n\) be the coefficient of \(x^n\) in the series expansion, in ascending powers of \(x\), of \[\displaystyle \frac{1+x}{(1-x)^2(1+x^2)} \,, \] where \(\vert x \vert <1\,\). Show, using partial fractions, that either \(a_n =n+1\) or \(a_n = n+2\) according to the value of \(n\). Hence find a decimal approximation, to nine significant figures, for the fraction \( \displaystyle \frac{11\,000}{8181}\). \newline [You are not required to justify the accuracy of your approximation.]
Solution: \begin{align*} && \frac{1+x}{(1-x)^2(1+x^2)} &= \frac{A}{1-x} + \frac{B}{(1-x)^2} + \frac{Cx+D}{1+x^2} \\ \Rightarrow && 1 + x &= A(1-x)(1+x^2) + B(1+x^2) + Cx(1-x)^2 + D(1-x)^2 \\ \Rightarrow && 2 &= 2B \tag{\(x = 1\)} \\ \Rightarrow && 1 &= B \\ \Rightarrow && 1 &= A+B+D \tag{\(x = 0\)}\\ \Rightarrow && A &= -D \\ \Rightarrow && 0 &= 4A+2B-4C+4D \tag{\(x = -1\)}\\ \Rightarrow && C &= \frac12\\ \Rightarrow && 3 &= -5A+5B+2C+D \tag{\(x=2\)} \\ \Rightarrow && 3 &= -6A+6 \\ \Rightarrow && A,D &=-\frac12,\frac12 \\ \Rightarrow && \frac{1+x}{(1-x)^2(1+x^2)} &= \frac{1}{(1-x)^2} +\frac{1}{2(1-x)}+ \frac{x-1}{2(1+x^2)} \\ &&&=\sum_{k=0}^{\infty}(k+1)x^k + \sum_{k=0}^{\infty}\frac12 x^k + \sum_{k=0}^{\infty}\frac12 (-1)^kx^{2k+1} - \sum_{k=0}^{\infty}\frac12 (-1)^kx^{2k} \end{align*} Therefore the coefficient of \(x^n\) is \(n+1\) or \(n+2\) depending on whether the coefficients from the final series add constructively \(n \equiv 1, 2 \pmod{4}\) or destructively. \begin{align*} \frac{11\, 000}{8181} &= \frac{(10+1) \cdot 1000}{(10-1)^2 \cdot (100+1)} \\ &= \frac{(1+\frac{1}{10})}{(1-\frac{1}{10})^2 \cdot (1+\frac1{10})} \\ &= 1 + \frac3{10} + \frac4{10^2} + \frac{4}{10^3}+\frac{5}{10^4} + \frac{7}{10^5} + \frac{8}{10^6} + \cdots \\ & \quad \quad \cdots + \frac{8}{10^7} + \frac{9}{10^8} + \frac{11}{10^9} + \frac{12}{10^{10}} + \cdots \\ &= 1.34457890 + \frac{12}{10^{10}} + \cdots \end{align*} \begin{align*} && \sum_{k=m}^{\infty} (k+2)x^k &= x^m \sum_{k=0}^{\infty} (k+m+2)x^{k} \\ && &= \frac{x^k}{(1-x)^2} + \frac{(m+2)x^k}{1-x} \\ \Rightarrow && |\sum_{k=m}^{\infty} a_k \left ( \frac1{10} \right )^k | &\leq \frac{1}{10^m}\left ( \frac{1}{(1-\frac1{10})^2} + \frac{m+2}{1-\frac1{10}} \right) \\ &&&= \frac{1}{10^{m-1}} \left ( \frac{9m+28}{81}\right ) \end{align*} Therefore for this will be less than \(10^{-9}\), when \(m = 11\), so our approximation is valid to 9sf
2008 Paper 2 Q3
- Find the coordinates of the turning points of the curve \(y=27x^3-27x^2+4\). Sketch the curve and deduce that \(x^2(1-x)\le 4/27\) for all \(x\ge0\,\). Given that each of the numbers \(a\), \(b\) and \(c\) lies between \(0\) and \(1\), prove by contradiction that at least one of the numbers \(bc(1-a)\), \(ca(1-b)\) and \(ab(1-c)\) is less than or equal to \(4/27\).
- Given that each of the numbers \(p\) and \(q\) lies between \(0\) and \(1\), prove that at least one of the numbers \(p(1-q)\) and \(q(1-p)\) is less than or equal to \(1/4\).
Solution:
- \(\,\)
\begin{align*}
&& y & = 27x^3 - 27x^2 + 4 \\
\Rightarrow && \frac{\d y}{\d x} &= 81x^2 - 54x \\
\Rightarrow && x &= 0, \frac23 \\
\Rightarrow && (x,y) &= (0, 4), \left (\frac23, 0 \right)
\end{align*}
Since \(f(x) \geq 0\) for \(x \geq 0\) we must have \(27x^2(1-x) \leq 4 \Rightarrow x^2(1-x) \leq \frac{4}{27}\) Suppose for contradiction that \(bc(1-a) > \frac{4}{27}, ca(1-b) > \frac{4}{27}, ab(1-c) > \frac{4}{27}\) then taking the product we see \begin{align*} && \left ( \frac{4}{27} \right)^3 &< bc(1-a) \cdot ca(1-b) \cdot ab(1-c) \\ &&&= a^2(1-c) \cdot b^2(1-b) \cdot c^2(1-c) \leq \left ( \frac{4}{27}\right)^3 \end{align*} which is a contradiction.
- Notice that \(f(x) = x(1-x)\) has a turning point at \((\frac12, \frac14)\), and so \(f(x) \leq \frac14\). Suppose for contradiction that both \(p(1-q)\) and \(q(1-p)\) are larger than \(1/4\) \begin{align*} && \left ( \frac14 \right)^2 &< p(1-q) \cdot q(1-p) \\ &&&= p(1-p) \cdot q(1-q) \\ &&&\leq \left ( \frac14 \right)^2 \end{align*} which is a contradiction.
2008 Paper 2 Q4
A curve is given by \[x^2+y^2 +2axy = 1,\] where \(a\) is a constant satisfying \(0 < a < 1\). Show that the gradient of the curve at the point~\(P\) with coordinates \((x,y)\) is \[\displaystyle - \frac {x+ay}{ax+y}\,,\] provided \(ax+y \ne0\). Show that \(\theta\), the acute angle between \(OP\) and the normal to the curve at \(P\), satisfies \[ \tan\theta = a\vert y^2-x^2\vert\;. \] Show further that, if \(\ \displaystyle \frac{\d \theta}{\d x}=0\) at \(P\), then:
- \(a(x^2+y^2)+2xy=0\,\);
- \((1+a)(x^2+y^2+2xy)=1\,\);
- \(\displaystyle \tan\theta = \frac a{\sqrt{1-a^2}}\,\).
2008 Paper 2 Q5
Evaluate the integrals \[\int_0^{\frac{1}{2}\pi} \frac{\sin 2x}{1+\sin^2x} \d x \text{ and } \int_0^{\frac{1}{2}\pi} \frac{\sin x}{1+\sin^2x} \d x\] Show, using the binomial expansion, that \((1+\sqrt2\,)^5<99\). Show also that \(\sqrt 2 > 1.4\). Deduce that \(2^{\sqrt2} > 1+ \sqrt2\,\). Use this result to determine which of the above integrals is greater.
Solution: \begin{align*} && I &= \int_0^{\frac{1}{2}\pi} \frac{\sin 2x}{1+\sin^2x} \d x \\ &&&= \int_0^{\frac{1}{2}\pi} \frac{2 \sin x \cos x}{1+\sin^2x} \d x \\ &&&= \left [\ln (1 + \sin^2 x) \right]_0^{\pi/2} \\ &&&= \ln 2 \\ \\ && J &= \int_0^{\frac{1}{2}\pi} \frac{\sin x}{1+\sin^2x} \d x \\ &&&= \int_0^{\frac{1}{2}\pi} \frac{\sin x}{2-\cos^2x} \d x \\ &&&= \frac{1}{2\sqrt{2}}\int_0^{\frac{1}{2}\pi} \left ( \frac{\sin x}{\sqrt{2}-\cos x}+ \frac{\sin x}{\sqrt{2}+\cos x} \right) \d x \\ &&&= \frac{1}{2\sqrt{2}} \left [\ln (\sqrt{2}-\cos x) - \ln (\sqrt{2}+\cos x) \right]_0^{\pi/2} \\ &&&= \frac{1}{2\sqrt{2}} \left (-\ln(\sqrt{2}-1)+\ln(\sqrt{2}+1) \right) \\ &&&= \frac1{2\sqrt{2}} \ln \left (\frac{\sqrt{2}+1}{\sqrt{2}-1} \right)\\ &&&= \frac1{\sqrt{2}} \ln (\sqrt{2}+1) \end{align*} \begin{align*} && (1+\sqrt{2})^5 + (1-\sqrt{2})^5 &= 2(1+10\cdot2+5\cdot2^2) \\ &&&= 82 \\ && |(1-\sqrt{2})^5| & < 1 \\ && (1+\sqrt{2})^5 &< 83 < 99 \\ \\ && 1.4^2 &= 1.96 \\ &&&< 2 \\ \Rightarrow && 1.4 &<\sqrt{2} \\ \\ \Rightarrow && 2^{\sqrt{2}} &> 2^{1.4} \\ &&&=2^{7/5} \\ &&&= {128}^{1/5} \\ &&&>99^{1/5} \\ &&&>1+\sqrt{2} \end{align*} \begin{align*} && \ln 2 & > \frac{1}{\sqrt{2}} \ln(\sqrt{2}+1) \\ \Leftrightarrow && \sqrt{2} \ln 2 &> \ln(\sqrt{2}+1) \\ \Leftrightarrow && 2^{\sqrt{2}} &> 1+\sqrt{2} \end{align*} which we have already shown, so the first integral is larger.
2008 Paper 2 Q6
A curve has the equation \(y=\f(x)\), where \[ \f(x) = \cos \Big( 2x+ \frac \pi 3\Big) + \sin \Big ( \frac{3x}2 - \frac \pi 4\Big). \]
- Find the period of \(\f(x)\).
- Determine all values of \(x\) in the interval \(-\pi\le x \le \pi\) for which \(\f(x)=0\). Find a value of \(x\) in this interval at which the curve touches the \(x\)-axis without crossing it.
- Find the value or values of \(x\) in the interval \(0\le x \le 2\pi\) for which \(\f(x)=2\,\).
Solution: \begin{align*} && f(x) &= \cos \left( 2x+ \frac \pi 3\right) + \sin \left ( \frac{3x}2 - \frac\pi 4\right) \\ &&&= \cos \left( 2x+ \frac \pi 3\right) + \cos\left (\frac{\pi}{2} - \left ( \frac{3x}2 - \frac\pi 4\right) \right)\\ &&&= \cos \left( 2x+ \frac \pi 3\right) + \cos\left (\frac{3\pi}{4} - \frac{3x}2 \right)\\ &&&= 2 \cos \left (\frac{2x+ \frac \pi 3+\frac{3\pi}{4} - \frac{3x}2}{2} \right) \cos \left ( \frac{\left (2x+ \frac \pi 3 \right) - \left (\frac{3\pi}{4} - \frac{3x}2 \right)}{2} \right)\\ &&&= 2 \cos \left (\frac{\frac{x}{2}+ \frac {13\pi}{12}}{2} \right) \cos \left ( \frac{\frac{7x}{2}- \frac {5\pi}{12}}{2} \right)\\ &&&= 2 \cos \left (\frac{x}{4}+ \frac {13\pi}{24} \right) \cos \left ( \frac{7x}{4}- \frac {5\pi}{24} \right)\\ \end{align*}
- The period of \(f\) will be the LCM of \(\frac{2\pi}{\pi}\) and \(\frac{2\pi}{\frac32} = \frac{4\pi}{3}\) which is \(4\pi\). (This is also clear from the factorised form).
- \(f(x) = 0\) means one of those two factors is zero, ie \begin{align*} \text{first factor}: && 0 &= \cos \left (\frac{x}{4}+ \frac {13\pi}{24} \right) \\ &&n\pi + \frac{\pi}{2}&= \frac{x}{4}+ \frac {13\pi}{24} \\ \Rightarrow && x &= 4n\pi - \frac{\pi}{6} \\ \Rightarrow && x &= -\frac{\pi}{6} \\ \\ \text{second factor}: && 0 &= \cos \left ( \frac{7x}{4}- \frac {5\pi}{24} \right) \\ && n\pi + \frac{\pi}{2} &= \frac{7x}{4}- \frac {5\pi}{24} \\ \Rightarrow && 7x &= 4n\pi + \frac{17}{6}\pi \\ \Rightarrow && x &= \frac{4n}7\pi + \frac{17}{42}\pi \\ \Rightarrow && x &= -\frac{31}{42} \pi, -\frac16\pi, \frac{17}{42}\pi, \frac{41}{42}\pi \end{align*} Therefore all solutions are \(-\frac{31}{42} \pi, -\frac16\pi, \frac{17}{42}\pi, \frac{41}{42}\pi\) We can see that \(-\frac{\pi}{6}\) is a repeated root, therefore it touches the axis and does not cross.
- \(f(x) = 2\) requires both factors to be \(1\) or \(-1\). \begin{align*} \text{first factor}: && \pm1 &= \cos \left (\frac{x}{4}+ \frac {13\pi}{24} \right) \\ &&n\pi &= \frac{x}{4}+ \frac {13\pi}{24} \\ \Rightarrow && x &= 4n\pi - \frac{13\pi}{6} \\ \Rightarrow && x &= \frac{11}{6}\pi \\ \end{align*} We only need to test this value, where it's \(-1\), so we look at \( \cos \left ( \frac{77\pi}{24}- \frac {5\pi}{24} \right) = \cos (3\pi) = -1\), so the only value is \(\frac{11}{6}\pi\)
2008 Paper 2 Q7
- By writing \(y=u{(1+x^2)\vphantom{\dot A}}^{\frac12}\), where \(u\) is a function of \(x\), find the solution of the equation \[ \frac 1 y \frac{\d y} {\d x} = xy + \frac x {1+x^2} \] for which \(y=1\) when \(x=0\).
- Find the solution of the equation \[ \frac 1 y \frac{\d y} {\d x} = x^2y + \frac {x^2 } {1+x^3} \] for which \(y=1\) when \(x=0\).
- Give, without proof, a conjecture for the solution of the equation \[ \frac 1 y \frac{\d y} {\d x} = x^{n-1}y + \frac {x^{n-1} } {1+x^n} \] for which \(y=1\) when \(x=0\), where \(n\) is an integer greater than 1.