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2002 Paper 3 Q1
D: 1700.0 B: 1500.0

Find the area of the region between the curve \(\displaystyle y = {\ln x \over x}\,\) and the \(x\)-axis, for \(1 \le x \le a\). What happens to this area as \(a\) tends to infinity? Find the volume of the solid obtained when the region between the curve \(\displaystyle y = {\ln x \over x}\,\) and the \(x\)-axis, for \(1 \le x\le a\), is rotated through \(2 \pi\) radians about the \(x\)-axis. What happens to this volume as \(a\) tends to infinity?

Solution: \begin{align*} && \int_1^a \frac{\ln x}{x} \d x &= \left [ \ln x \cdot \ln x\right ]_1^a - \int_1^a \frac{\ln x}{x} \d x \\ \Rightarrow && \int_1^a \frac{\ln x}{x} \d x &= \frac12 \left ( \ln a \right) ^2 \\ && \int_1^\infty \frac{\ln x}{x} \d x &= \lim_{a \to \infty} \frac12 (\ln a)^2 \\ &&&= \infty \end{align*} \begin{align*} && \pi \int_1^a \left ( \frac{\ln x}{x} \right)^2 \d x &= \pi \int_{u=0}^{u=\ln a} \left ( \frac{u}{e^u} \right)^2 e^u \d u \\ &&&= \pi \int_0^{\ln a} u^2 e^{-u} \d u \\ &&&= \pi \left [-u^2e^{-u} \right]_0^{\ln a} +\pi \int_0^{\ln a} 2u e^{-u} \d u \\ &&&= -\frac{\pi}{a} (\ln a)^2 + \pi \left [-2u e^{-u} \right]_0^{\ln a} + \pi \int_0^{\ln a} e^{-u} \d u \\ &&&= -\frac{\pi}{a} (\ln a)^2- \frac{2 \pi}{a}\ln a+\pi \left (1 - \frac{1}{a} \right) \\ \\ && \pi \int_1^{\infty} \left ( \frac{\ln x}{x} \right)^2 \d x &= \lim_{a \to \infty} \left ( -\frac{\pi}{a} (\ln a)^2- \frac{2 \pi}{a}\ln a+\pi \left (1 - \frac{1}{a} \right) \right) \\ &&&= \pi \end{align*}

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2002 Paper 3 Q2
D: 1700.0 B: 1500.0

Prove that \(\displaystyle \arctan a + \arctan b = \arctan \l {a + b \over 1-ab} \r\,\) when \(0 < a < 1\) and \(0 < b < 1\,\). Prove by induction that, for \(n \ge 1\,\), \[ \sum_{r = 1}^n \arctan \l {1 \over r^2 + r + 1} \r = \arctan \l {n \over n+2} \r \] and hence find \[ \sum_{r = 1}^\infty \arctan \l {1 \over r^2 + r + 1} \r\,. \] Hence prove that \[ \sum_{r = 1}^\infty \arctan \l {1 \over r^2 - r + 1} \r = {\pi \over 2}\,. \]

Solution: \begin{align*} && \arctan a &\in (0, \tfrac{\pi}{4}) \\ && \arctan b &\in (0, \tfrac{\pi}{4}) \\ \Rightarrow && \arctan a+\arctan b &\in (0, \tfrac{\pi}{2}) \\ && \tan \left ( \arctan a+\arctan b \right) &= \frac{\tan \arctan a + \tan \arctan b}{1 - \tan \arctan a \tan \arctan b} \\ &&&= \frac{a+b}{1-ab} \in (0, \infty) \\ \Rightarrow && \arctan \left ( \frac{a+b}{1-ab} \right) &\in (0, \tfrac{\pi}{2}) \\ \Rightarrow && \arctan a + \arctan b &= \arctan \left ( \frac{a+b}{1-ab} \right) \end{align*} Claim: \(\displaystyle \sum_{r = 1}^n \arctan \l {1 \over r^2 + r + 1} \r = \arctan \l {n \over n+2} \r\) Proof: (By Induction): Base case (\(n=1\)): \begin{align*} && LHS &= \sum_{r=1}^1 \arctan \left ( \frac{1}{r^2+r+1} \right) \\ &&&= \arctan \left ( \frac{1}{3} \right) \\ && RHS &= \arctan \left ( \frac{1}{1+2} \right)\\ &&&= \arctan \left ( \frac{1}{3} \right) = LHS \end{align*} Inductive step, suppose true for \(n = k\), ie \begin{align*} && \sum_{r = 1}^k \arctan \l {1 \over r^2 + r + 1} \r &= \arctan \l {k \over k+2} \r \\ \Rightarrow && \sum_{r = 1}^{k+1} \arctan \l {1 \over r^2 + r + 1} \r &= \sum_{r = 1}^k \arctan \l {1 \over r^2 + r + 1} \r+ \arctan \left ( \frac{1}{(k+1)^2+(k+1)+1} \right) \\ &&&= \arctan \l {k \over k+2} \r+\arctan \left ( \frac{1}{(k+1)^2+(k+1)+1} \right) \\ &&&= \arctan \left ( \frac{{k \over k+2}+\frac{1}{(k+1)^2+(k+1)+1} }{1-\frac{k}{k+2}\frac{1}{(k+1)^2+(k+1)+1} } \right) \\ &&&= \arctan \left ( \frac{k((k+1)^2+k+1+k)+(k+2) }{(k+2)((k+1)^2+(k+1)+1)-k} \right) \\ &&&= \arctan \left ( \frac{k^3+3k^2+4k+2 }{k^3+5k^2+8k+6} \right) \\ &&&= \arctan \left ( \frac{(k+1)(k^2+2k+2) }{(k+3)(k^2+2k+2)} \right) \\ &&&= \arctan \left ( \frac{k+1 }{(k+1)+2} \right) \\ \end{align*} Therefore it is true for \(n = k+1\), therefore it is true for all \(n \geq 1\) by the principle of mathematical induction. \begin{align*} && S &= \lim_{n \to \infty} \sum_{r = 1}^n \arctan \l {1 \over r^2 + r + 1} \r \\ &&&= \lim_{n \to \infty} \arctan \l \frac{n}{n+2} \r \\ &&&= \lim_{n \to \infty} \arctan \l \frac{1}{1+2/n} \r \\ &&&=\arctan\l \lim_{n \to \infty} \frac{1}{1+2/n} \r \\ &&&= \frac{\pi}{4} \end{align*} \begin{align*} && \sum_{r = 1}^\infty \arctan \l {1 \over r^2 - r + 1} \r &= \sum_{r = 0}^\infty \arctan \left( \frac{1}{ (r+1)^2 - (r+1) + 1} \right) \\ &&&= \sum_{r = 0}^\infty \arctan \left( \frac{1}{ r^2+r+1} \right) \\ &&&= \arctan \l \frac{1}{0^2+0+1} \r + \frac{\pi}{4} \\ &&&= \frac{\pi}{2} \end{align*}

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2002 Paper 3 Q3
D: 1700.0 B: 1516.0

Let \[\f(x) = a \sqrt{x} - \sqrt{x - b}\;, \] where \(x\ge b >0\) and \(a>1\,\). Sketch the graph of \(\f(x)\,\). Hence show that the equation \(\f(x) = c\), where \(c>0\), has no solution when \(c^2 < b \l a^2 - 1 \r\,\). Find conditions on \(c^2\) in terms of \(a\) and \(b\) for the equation to have exactly one or exactly two solutions. Solve the equations

  1. \(3 \sqrt{x} - \sqrt{x - 2} = 4\, ,\)
  2. \(3 \sqrt{x} - \sqrt{x - 3} = 5\;\).

Solution: \begin{align*} && f'(x) &= \frac12 ax^{-1/2}-\frac12(x-b)^{-1/2} \\ \Rightarrow f'(x) = 0: && 0 &= \frac{a\sqrt{x-b}-\sqrt{x}}{\sqrt{x(x-b)}} \\ \Rightarrow && x &= a^2(x-b)\\ \Rightarrow && x &= \frac{a^2b}{a^2-1} \\ && f(x) &= a^2 \sqrt{\frac{b}{a^2-1}} - \sqrt{\frac{a^2b}{a^2-1}-b} \\ &&&= a^2 \sqrt{\frac{b}{a^2-1}} - \sqrt{\frac{b}{a^2-1}} \\ &&&= \sqrt{b(a^2-1)} \end{align*}

TikZ diagram
If \(c\) is below the turning point, ie \(c^2 < b(a^2-1)\) there is no solution. If \(c^2 = b(a^2-1)\) there is exactly one solution. If \(b(a^2-1) < c^2 < (f(b))^2 = a^2b\) then there are two solutions, otherwise there is exactly one solution.
  1. \(c^2 = 16\), \(2 \cdot (3^2-1) = 16\), so we should have exactly one solution at \(x = \frac{3^2 \cdot 2}{3^2 -1 } = \frac{9}{4}\)
  2. \(c^2 = 25\) and \(3 \cdot (3^2 - 1) = 24, 3 \cdot (3^2) = 27\), so we look for two solutions. \begin{align*} && 5 & = 3 \sqrt{x} - \sqrt{x-3} \\ \Rightarrow && 25 &= 9x+x-3-6\sqrt{x(x-3)} \\ \Rightarrow && 3\sqrt{x(x-3)} &= 5x-14 \\ \Rightarrow && 9x(x-3) &= 25x^2-140x+196 \\ \Rightarrow && 0 &= 16x^2-113x+196 \\ &&&= (x-4)(16x-49) \\ \Rightarrow && x &= 4, \frac{49}{16} \end{align*}

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2002 Paper 3 Q4
D: 1700.0 B: 1490.1

Show that if \(x\) and \(y\) are positive and \(x^3 + x^2 = y^3 - y^2\) then \(x < y\,\). Show further that if \(0 < x \le y - 1\), then \(x^3 + x^2 < y^3 - y^2\). Prove that there does not exist a pair of {\sl positive} integers such that the difference of their cubes is equal to the sum of their squares. Find all the pairs of integers such that the difference of their cubes is equal to the sum of their squares.

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2002 Paper 3 Q5
D: 1700.0 B: 1516.0

Give a condition that must be satisfied by \(p\), \(q\) and \(r\) for it to be possible to write the quadratic polynomial \(px^2 + qx + r\) in the form \(p \l x + h \r^2\), for some \(h\). Obtain an equation, which you need not simplify, that must be satisfied by \(t\) if it is possible to write \[ \l x^2 + \textstyle{{1 \over 2}} bx + t \r^2 - \l x^4 + bx^3 + cx^2 +dx +e \r \] in the form \(k \l x + h \r^2\), for some \(k\) and \(h\). Hence, or otherwise, write \(x^4 + 6x^3 + 9x^2 -2x -7\) as a product of two quadratic factors.

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2002 Paper 3 Q6
D: 1700.0 B: 1484.9

Find all the solution curves of the differential equation \[ y^4 \l {\mathrm{d}y \over \mathrm{d}x }\r^{\! \! 4} = \l y^2 - 1 \r^2 \] that pass through either of the points

  1. \(\l 0, \, \frac{1}{2}\sqrt3 \r\),
  2. \(\l 0, \, \frac{1}{2}\sqrt5 \r\).
Show also that \(y = 1\) and \(y = -1\) are solutions of the differential equation. Sketch all these solution curves on a single set of axes.

Solution: \begin{align*} && y^4 \left (\frac{\d y}{\d x} \right)^4 &= (y^2 - 1)^2 \\ \Rightarrow && y^2 \left (\frac{\d y}{\d x} \right)^2 &= |y^2 - 1| \\ && y \left (\frac{\d y}{\d x} \right) &= \pm \sqrt{|y^2-1|} \\ \Rightarrow &&\int \frac{y}{\sqrt{|y^2-1|}} \d y &= \int \pm 1 \d x \\ \Rightarrow && \pm \sqrt{|y^2-1|} &= \pm x + C \\ \end{align*}

  1. Since \(y^2 < 1\), our solution curve should be of the from \(-\sqrt{1-y^2} = \pm x + C\) Plugging in \((0, \tfrac12 \sqrt{3})\), we obtain \(-\tfrac12 = C\), therefore our solution curves are \(\pm x = \frac12 - \sqrt{1-y^2}\)
  2. Since \(y^2 > 1\), our solution curve should be of the from \(\sqrt{y^2-1} = \pm x + C\) Plugging in \((0, \tfrac12 \sqrt{5})\), we obtain \(\tfrac12 = C\), therefore our solution curves are \(\pm x = \sqrt{y^2-1}-\frac12\)
Clearly if \(y = \pm 1\), \(y'=0\) and the equation is satisfied.
TikZ diagram

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2002 Paper 3 Q7
D: 1700.0 B: 1484.0

Given that \(\alpha\) and \(\beta\) are acute angles, show that \(\alpha + \beta = \tfrac{1}{2}\pi\) if and only if \(\cos^2 \alpha + \cos^2 \beta = 1\). In the \(x\)--\(y\) plane, the point \(A\) has coordinates \((0,s)\) and the point \(C\) has coordinates \((s,0)\), where \(s>0\). The point \(B\) lies in the first quadrant (\(x>0\), \(y>0\)). The lengths of \(AB\), \(OB\) and \(CB\) are respectively \(a\), \(b\) and \(c\). Show that \[ (s^2 +b^2 - a^2)^2 + (s^2 +b^2 -c^2)^2 = 4s^2b^2 \] and hence that \[ (2s^2 -a^2-c^2)^2 + (2b^2 -a^2-c^2)^2 =4a^2c^2\;. \] Deduce that $$ \l a - c \r^2 \le 2b^2 \le \l a + c \r^2\;. $$ %Show, %by considering the case \(a=1+\surd2\,\), \(b=c=1\,\), % that the condition \(\l \ast \r\,\) %is not sufficient to ensure that \(B\) lies in the first quadrant.

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2002 Paper 3 Q8
D: 1700.0 B: 1469.7

Four complex numbers \(u_1\), \(u_2\), \(u_3\) and \(u_4\) have unit modulus, and arguments \(\theta_1\), \(\theta_2\), \(\theta_3\) and \(\theta_4\), respectively, with \(-\pi < \theta_1 < \theta_2 < \theta_3 < \theta_4 < \pi\). Show that \[ \arg \l u_1 - u_2 \r = \tfrac{1}{2} \l \theta_1 + \theta_2 -\pi \r + 2n\pi \] where \(n = 0 \hspace{4 pt} \mbox{or} \hspace{4 pt} 1\,\). Deduce that \[ \arg \l \l u_1 - u_2 \r \l u_4 - u_3 \r \r = \arg \l \l u_1 - u_4 \r \l u_3 - u_2 \r \r + 2n\pi \] for some integer \(n\). Prove that \[ | \l u_1 - u_2 \r \l u_4 - u_3 \r | + | \l u_1 - u_4 \r \l u_3 - u_2 \r | = | \l u_1 - u_3 \r \l u_4 - u_2 \r |\;. \]

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2002 Paper 3 Q9
D: 1700.0 B: 1500.0

A tall container made of light material of negligible thickness has the form of a prism, with a square base of area \(a^2\). It contains a volume \(ka^3\) of fluid of uniform density. The container is held so that it stands on a rough plane, which is inclined at angle \(\theta\) to the horizontal, with two of the edges of the base of the container horizontal. In the case \(k > \frac12 \tan\theta\), show that the centre of mass of the fluid is at a distance \(x\) from the lower side of the container and at a distance \(y\) from the base of the container, where \[ \frac x a = \frac12 - \frac {\tan\theta}{12k}\;, \ \ \ \ \ \ \frac y a = \frac k 2 + \frac{\tan^2\theta}{24k}\;. \] Determine the corresponding coordinates in the case \(k < \frac12 \tan\theta\). The container is now released. Given that \(k < \frac12\), show that the container will topple if \(\theta >45^\circ\).

Solution:

TikZ diagram
The fluid can be divided into a cuboid parallel to the slope and a right-angled triangle. If the height of the water on the longer side is \(\ell a\), then we have \(ka^3 = (\ell a - a\tan \theta)a^2 + \frac12 a^3\tan \theta \Rightarrow \ell = k + \frac12 \tan \theta\) This is acceptable when \(k > \frac12 \tan \theta\). The centre of mass of the cuboid will be \((\frac{a}{2}, \frac12 (k - \frac12 \tan \theta))\) and of the triangle will be \((\frac13 a, \frac13 \tan \theta + (k - \frac12 \tan \theta) )\) Therefore we have: \begin{align*} && \text{COM} && \text{mass} \\ \text{cuboid} && (\frac{a}{2}, \frac{a}2 (k - \frac12 \tan \theta)) && a^3(k - \frac12 \tan \theta) \\ \text{triangle} && (\frac13 a, \frac{a}3 \tan \theta + a(k - \frac12 \tan \theta) ) && a^3\frac12 \tan \theta \\ \text{whole system} && (x, y) && a^3k \end{align*} Therefore \begin{align*} && a^3k x &= \frac{a}{2} \cdot a^3(k - \frac12 \tan \theta) + \frac13 a \cdot a^3\frac12 \tan \theta \\ &&&= a^4 \frac{k}{2} - \frac{1}{12}a^4 \tan \theta \\ \Rightarrow && \frac{x}{a} &= \frac12 - \frac{\tan \theta}{12 k} \\ \\ && a^3k y &= \frac{a}2 (k - \frac12 \tan \theta) \cdot a^3(k - \frac12 \tan \theta) + \\ &&& \qquad\qquad \cdots + \l \frac{a}3 \tan \theta + a(k - \frac12 \tan \theta) \r \cdot a^3\frac12 \tan \theta \\ &&&= \frac{a^4k^2}{2} -\frac{a^4k \tan \theta}{2} + \frac{a^4 \tan^2 \theta}{8} - \frac{a^4 \tan^2 \theta}{12} + \frac{a^4k \tan \theta}{2} \\ \Rightarrow && \frac{y}{a} &= \frac{k}2 + \frac{\tan^2 \theta}{24k} \end{align*}
TikZ diagram
If the water only fills up a prism, it's sides must be \(b\) and \(b\tan \theta \), therefore the volume is \(\frac12 ab^2 \tan \theta = ka^3 \Rightarrow b = a\sqrt{\frac{2k}{\tan \theta}}\) The centre of mass will be at \(\l \frac13 a\sqrt{\frac{2k}{\tan \theta}}, a\sqrt{2k \tan \theta}\r\) The container will topple if the centre of mass is outside the base, ie if the centre of mass \((x,y)\) lies above the line \(y = \tan (90^\circ- x) = \frac{1}{\tan \theta} x\). If \(\theta > 45^\circ\) then \(\tan \theta > 1\) and so we are in the \(\frac12 \tan \theta > \frac12 > k\) and so we are in the second case. \begin{align*} \frac{y}{x} &= \frac{\frac 13 a\sqrt{2k \tan \theta}}{\frac13 a\sqrt{\frac{2k}{\tan \theta}}} \\ &= \tan \theta \end{align*} \(\tan \theta > \frac{1}{\tan \theta} \Leftrightarrow \tan \theta > 1 \Leftrightarrow \theta > 45^\circ\).

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2002 Paper 3 Q10
D: 1700.0 B: 1516.0

A light hollow cylinder of radius \(a\) can rotate freely about its axis of symmetry, which is fixed and horizontal. A particle of mass \(m\) is fixed to the cylinder, and a second particle, also of mass \(m\), moves on the rough inside surface of the cylinder. Initially, the cylinder is at rest, with the fixed particle on the same horizontal level as its axis and the second particle at rest vertically below this axis. The system is then released. Show that, if \(\theta\) is the angle through which the cylinder has rotated, then \[ \ddot{\theta} = {g \over 2a} \l \cos \theta - \sin \theta \r \,, \] provided that the second particle does not slip. Given that the coefficient of friction is \( (3 + \sqrt{3})/6\), show that the second particle starts to slip when the cylinder has rotated through \(60^\circ\).

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