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2005 Paper 3 Q1
D: 1700.0 B: 1500.0

Show that \(\sin A = \cos B\) if and only if \(A = (4n+1)\frac{\pi}{2}\pm B\) for some integer \(n\). Show also that \(\big\vert\sin x \pm \cos x \big\vert \le \sqrt{2}\) for all values of \(x\) and deduce that there are no solutions to the equation \(\sin\left( \sin x \right) = \cos \left( \cos x \right)\). Sketch, on the same axes, the graphs of \(y= \sin \left( \sin x \right)\) and \(y = \cos \left( \cos x \right)\). Sketch, not on the previous axes, the graph of \(y= \sin \left(2 \sin x \right)\).


Solution: \begin{align*} && \sin A &= \cos B \\ \Leftrightarrow && 0 &= \sin A - \cos B \\ &&&= \sin A - \sin ( \frac{\pi}{2} - B) \\ &&&= 2 \sin \left ( \frac{A + B - \frac{\pi}{2}}{2} \right) \cos \left (\frac{A - B + \frac\pi2}{2} \right) \\ \Leftrightarrow && n \pi &= \frac{A+B - \frac{\pi}{2}}{2}, n\pi + \frac{\pi}{2} = \frac{A-B+\frac{\pi}{2}}{2} \\ \Leftrightarrow && A \pm B &= 2n\pi + \frac{\pi}{2} \\ &&&= (4n+1) \frac{\pi}{2} \end{align*} \begin{align*} |\sin x \pm \cos x| &= | \sqrt{2} \sin(x \pm \frac{\pi}{4} )| \\ & \leq \sqrt{2} \end{align*} Therefore if \(\sin(\sin x) = \cos (\cos x)\) we must have that \(|\sin x \pm \cos x| = |(4n+1) \frac{\pi}{2}| \geq \frac{\pi}{2} > 1.5 > \sqrt{2}\) contradiction.

TikZ diagram
TikZ diagram

2005 Paper 3 Q2
D: 1700.0 B: 1502.0

Find the general solution of the differential equation \(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{xy}{x^2+a^2}\;\), where \(a\ne0\,\), and show that it can be written in the form \(\displaystyle y^2(x^2+a^2)= c^2\,\), where \(c\) is an arbitrary constant. Sketch this curve. Find an expression for \(\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} (x^2+y^2)\) and show that \[ \frac{\mathrm{d^2}}{\mathrm{d}x^2} (x^2+y^2) = 2\left(1 -\frac {c^2}{(x^2+a^2)^2} \right) + \frac{8c^2x^2}{(x^2+a^2)^3}\;. \]

  1. Show that, if \(0 < c < a^2\), the points on the curve whose distance from the origin is least are \(\displaystyle \l 0\,,\;\pm \frac{c}{a}\r\).
  2. If \(c > a^2\), determine the points on the curve whose distance from the origin is least.


Solution: \begin{align*} && \frac{\d y}{\d x} &= - \frac{xy}{x^2+a^2} \\ \Rightarrow && \int \frac{1}{y} \d y &= \int -\frac{x}{x^2+a^2} \d x \\ && \ln |y| &= -\frac12 \ln |x^2 + a^2| + C \\ \Rightarrow && C' &= \ln y^2 + \ln (x^2+a^2) \\ \Rightarrow && c^2 &= y^2(x^2+a^2) \end{align*} (where the final constant \(c^2\) can be taken as a square since it is clearly positive).

TikZ diagram
\begin{align*} && \frac{\d }{\d x} \left (x^2 + y^2 \right) &= 2x - \frac{2xy^2}{x^2+a^2} \\ &&&=2x - \frac{2x c^2}{(x^2+a^2)^2} \\ &&&= 2x \left ( 1 - \frac{c^2}{(x^2+a^2)^2}\right) \\ \\ && \frac{\d ^2}{\d x^2} \left (x^2 + y^2 \right) &= \frac{\d }{\d x} \left (2x \left ( 1 - \frac{c^2}{(x^2+a^2)^2}\right) \right) \\ &&&= 2 \left ( 1 - \frac{c^2}{(x^2+a^2)^2}\right) + 2x \left (\frac{2c^2 \cdot 2x}{(x^2+a^2)^3} \right) \\ &&&= 2 \left ( 1 - \frac{c^2}{(x^2+a^2)^2}\right) + \frac{8x^2c^2 }{(x^2+a^2)^3} \\ \end{align*}
  1. The shortest distance from the origin will have the first derivative as \(0\), ie \(x = 0\) or \(x^2 + a^2 = c\), but if \(c < a^2\) this can only occur for \(x = 0\), so the closest to the origin is \((0, \pm \frac{c}{a})\)
  2. If \(c > a^2\) then we can have \(x = 0\) or \(x = \pm \sqrt{c-a^2}\). Looking at the second derivative, when \(x = 0\) we have \(2(1- \frac{c^2}{a^4}) < 0\) which is a local maximum. When \(x = \pm\sqrt{c-a^2}\) we have \(8(c-a^2)c^2/c^3 > 0\) which is the minimum, therefore the points are \((\pm \sqrt{c-a^2}, c)\)
TikZ diagram

2005 Paper 3 Q3
D: 1700.0 B: 1500.0

Let \(\f(x)=x^2+px+q\) and \(\g(x)=x^2+rx+s\,\). Find an expression for \(\f ( \g (x))\) and hence find a necessary and sufficient condition on \(a\), \(b\) and \(c\) for it to be possible to write the quartic expression \(x^4+ax^3+bx^2+cx+d\) in the form \(\f ( \g (x))\), for some choice of values of \(p\), \(q\), \(r\) and~\(s\). Show further that this condition holds if and only if it is possible to write the quartic expression \(x^4+ax^3+bx^2+cx+d\) in the form \((x^2+vx+w)^2-k\), for some choice of values of \(v\), \(w\) and \(k\). Find the roots of the quartic equation \(x^4-4x^3+10x^2-12x+4=0\,\).

2005 Paper 3 Q4
D: 1700.0 B: 1457.9

The sequence \(u_n\) (\(n= 1, 2, \ldots\)) satisfies the recurrence relation \[ u_{n+2}= \frac{u_{n+1}}{u_n}(ku_n-u_{n+1}) \] where \(k\) is a constant. If \(u_1=a\) and \(u_2=b\,\), where \(a\) and \(b\) are non-zero and \(b \ne ka\,\), prove by induction that \[ u_{2n}=\Big(\frac b a \Big) u_{2n-1} \] \[ u_{2n+1}= c u_{2n} \] for \(n \ge 1\), where \(c\) is a constant to be found in terms of \(k\), \(a\) and \(b\). Hence express \(u_{2n}\) and \(u_{2n-1}\) in terms of \(a\), \(b\), \(c\) and \(n\). Find conditions on \(a\), \(b\) and \(k\) in the three cases:

  1. the sequence \(u_n\) is geometric;
  2. \(u_n\) has period 2;
  3. the sequence \(u_n\) has period 4.

2005 Paper 3 Q5
D: 1700.0 B: 1502.1

Let \(P\) be the point on the curve \(y=ax^2+bx+c\) (where \(a\) is non-zero) at which the gradient is \(m\). Show that the equation of the tangent at \(P\) is \[ y-mx=c-\frac{(m-b)^2}{4a}\;. \] Show that the curves \(y=a_1 x^2+b_1 x+c_1\) and \(y=a_2 x^2+b_2 x+c_2\) (where \(a_1\) and \(a_2\) are non-zero) have a common tangent with gradient \(m\) if and only if \[ (a_2 -a_1 )m^2 + 2(a_1 b_2-a_2 b_1)m + 4a_1 a_2(c_2-c_1)+ a_2 b_1^2-a_1 b_2 ^2=0\;. \] Show that, in the case \(a_1 \ne a_2 \,\), the two curves have exactly one common tangent if and only if they touch each other. In the case \(a_1 =a_2\,\), find a necessary and sufficient condition for the two curves to have exactly one common tangent.


Solution: \begin{align*} && y' &= 2ax+b \\ \Rightarrow && m &= 2ax_t+b \\ \Rightarrow && x_t &= \frac{m-b}{2a} \end{align*} Therefore we must have \begin{align*} mx_t &= 2ax_t^2+bx_t \\ y - mx &= ax_t^2+bx_t+c - mx_t \\ &= ax_t^2+bx_t+c - (2ax_t^2+bx_t) \\ &= c - ax_t^2 \\ &= c-a\left (\frac{m-b}{2a} \right)^2 \\ &= c - \frac{(m-b)^2}{4a} \end{align*} They will have a common tangent if and only if the constant terms are equal, ie \begin{align*} && c_1 - \frac{(m-b_1)^2}{4a_1} &= c_2 - \frac{(m-b_2)^2}{4a_2} \\ \Leftrightarrow && (c_1-c_2) &= \frac{(m-b_1)^2}{4a_1} -\frac{(m-b_2)^2}{4a_2} \\ \Leftrightarrow && 4a_1a_2(c_1-c_2) &= a_2(m-b_1)^2-a_1(m-b_2)^2 \\ &&&= (a_2-a_1)m^2+2(a_1b_2-a_2b_1)m+a_2b_1^2-a_1b_2^2 \end{align*} as required. Treating this as a polynomial in \(m\), we can see that the two curves will have exactly one common tangent iff \(\Delta = 0\), ie: \begin{align*} && 0 &= \Delta \\ &&&= (2(a_1b_2-a_2b_1))^2 - 4 (a_2-a_1)(4a_1 a_2(c_2-c_1)+ a_2 b_1^2-a_1 b_2 ^2) \\ &&&= 4a_1^2b_2^2-8a_1a_2b_1b_2+4a_2b_1^2 - 4a_2^2b_1^2-4a_1^2b_2^2 + 4a_1a_2(b_1^2+b_2^2)-16(a_2-a_1)a_1a_2(c_2-c_1) \\ &&&=-8a_1a_2b_1b_2+4a_1a_2(b_1^2+b_2^2)-16(a_2-a_1)a_1a_2(c_2-c_1) \\ &&&=a_1a_2(4(b_1-b_2)^2-16(a_2-a_1)(c_2-c_1)) \\ &&&= 4a_1a_2((b_2-b_1)^2 - 4(a_2-a_1)(c_2-c_1) \end{align*} But this is just the discriminant of the difference, ie equivalent to the two parabolas just touching. (Assuming \(a_1-a_2 \neq 0\) and we do end up with a quadratic). If \(a_1 = a_2 = a\) then we need exactly one solution to \(2a(b_1-b_2)m +4a^2(c_2-c_1)+a(b_1^2-b_2^2) = 0\), ie \(b_1 \neq b_2\).

2005 Paper 3 Q6
D: 1700.0 B: 1516.0

In this question, you may use without proof the results \[ 4 \cosh^3 y - 3 \cosh y = \cosh (3y) \ \ \ \ \text{and} \ \ \ \ \mathrm{arcosh} \, y = \ln ( y+\sqrt{y^2-1}). \] \noindent[ {\bf Note: } \(\mathrm{arcosh}y\) is another notation for \(\cosh^{-1}y\,\)] Show that the equation \(x^3 - 3a^2x = 2a^3 \cosh T\) is satisfied by \( 2a \cosh \l \frac13 T \r\) and hence that, if \(c^2\ge b^3>0\), one of the roots of the equation \(x^3-3bx=2c\) is \(\ds u+\frac{b}{u}\), where \(u = (c+\sqrt{c^2-b^3})^{\frac13}\;\). Show that the other two roots of the equation \(x^3-3bx=2c\) are the roots of the quadratic equation \[\ds x^2 + \Big( u+\frac{b}{u}\Big) x + u^2+\frac{b^2}{u^2}-b=0\, ,\] and find these roots in terms of \(u\), \(b\) and \(\omega\), where \(\omega = \frac{1}{2}(-1 + \mathrm{i}\sqrt{3})\). Solve completely the equation \(x^3-6x=6\,\).

2005 Paper 3 Q7
D: 1700.0 B: 1471.4

Show that if \(\displaystyle \int\frac1{u \, \f(u)}\; \d u = \F(u) + c\;\), then \(\displaystyle \int\frac{m}{x \, \f(x^m)} \;\d x = \F(x^m) + c\;\), where \(m\ne0\). Find:

  1. \(\displaystyle\int\frac1{x^n-x} \, \d x\,\);
  2. \(\displaystyle\int\frac1 {\sqrt{x^n+x^2}}\, \d x\,\).


Solution: \begin{align*} u = x^m, \d u = m x^{m-1} && \int \frac{m}{x f(x^m)} \d x &= \int \frac{m x^{m-1}}{uf(u)} \d x \\ &&&= \int \frac{1}{u f(u)} \d u \\ &&&= F(u) + c \\ &&&= F(x^m) + c \end{align*}

  1. \begin{align*} && \int \frac{1}{u(u-1)} \d u &= \int \left ( \frac{1}{u-1}-\frac{1}{u} \right ) \d u \\ &&&= \ln \left ( \frac{u-1}{u} \right) + c \\ &&&= \ln \left ( 1 - \frac{1}{u} \right) + c \\ && \int \frac{1}{x^n - x} \d x &= \int \frac{1}{x (x^{n-1}-1)} \d x \\ f(u) = u - 1: && &= \frac{1}{n-1} \ln \left ( 1 - \frac{1}{x^{n-1}} \right) + c \end{align*}
  2. \begin{align*} v = \sqrt{u+1}, \d v = \tfrac12 (u+1)^{-1/2} \d u && \int \frac{1}{u\sqrt{u+1}} \d u &= \int \frac{1}{(v^2-1)} (u+1)^{-1/2} \d u \\ &&&= \int \frac{2}{v^2-1} \d v \\ &&&=\ln \frac{1-v}{1+v} + c \\ &&&= \ln \left (\frac{1-\sqrt{u+1}}{1+\sqrt{u+1}} \right)+ c \\ f(u) = \sqrt{x+1}:&& \int \frac{1}{\sqrt{x^n + x^2}} \d x &= \int \frac{1}{x\sqrt{x^{n-2}+1}} \d x \\ &&&= \frac{1}{n-2} \ln \left ( \frac{1-\sqrt{x^{n-2}+1}}{1+\sqrt{x^{n-2}+1}} \right)+c \end{align*}

2005 Paper 3 Q8
D: 1700.0 B: 1484.0

In this question, \(a\) and \(c\) are distinct non-zero complex numbers. The complex conjugate of any complex number \(z\) is denoted by \(z^*\). Show that \[ |a - c|^2 = aa^* + cc^* -ac^* - ca^* \] and hence prove that the triangle \(OAC\) in the Argand diagram, whose vertices are represented by \(0\), \(a\) and \(c\) respectively, is right angled at \(A\) if and only if \(2aa^* = ac^*+ca^*\,\). Points \(P\) and \(P'\) in the Argand diagram are represented by the complex numbers \(ab\) and \(\ds \frac{a}{b^*}\,\), where \(b\) is a non-zero complex number. A circle in the Argand diagram has centre \(C\) and passes through the point \(A\), and is such that \(OA\) is a tangent to the circle. Show that the point \(P\) lies on the circle if and only if the point \(P'\) lies on the circle. Conversely, show that if the points represented by the complex numbers \(ab\) and \(\ds \frac{a}{b^*}\), for some non-zero complex number \(b\) with \(bb^* \ne 1\,\), both lie on a circle centre \(C\) in the Argand diagram which passes through \(A\), then \(OA\) is a tangent to the circle.

2005 Paper 3 Q9
D: 1700.0 B: 1484.0

Two particles, A and B, move without friction along a horizontal line which is perpendicular to a vertical wall. The coefficient of restitution between the two particles is \(e\) and the coefficient of restitution between particle B and the wall is also \(e\), where \( 0< e < 1\). The mass of particle~A is \(4em\) (with \(m > 0\)), and the mass of particle B is \((1-e)^2m\)\,. Initially, A is moving towards the wall with speed \((1-e)v\) (where \(v > 0\)) and B is moving away from the wall and towards A with speed \(2ev\). The two particles collide at a distance \(d\) from the wall. Find the speeds of A and B after the collision. When B strikes the wall, it rebounds along the same line. Show that a second collision will take place, at a distance \(de\) from the wall. Deduce that further collisions will take place. Find the distance from the wall at which the \(n\)th collision takes place, and show that the times between successive collisions are equal.

2005 Paper 3 Q10
D: 1700.0 B: 1486.8

Two thin discs, each of radius \(r\) and mass \(m\), are held on a rough horizontal surface with their centres a distance \(6r\) apart. A thin light elastic band, of natural length \(2\pi r\) and modulus \(\dfrac{\pi mg}{12}\), is wrapped once round the discs, its straight sections being parallel. The contact between the elastic band and the discs is smooth. The coefficient of static friction between each disc and the horizontal surface is \(\mu\), and each disc experiences a force due to friction equal to \(\mu mg\) when it is sliding. The discs are released simultaneously. If the discs collide, they rebound and a half of their total kinetic energy is lost in the collision.

  1. Show that the discs start sliding, but come to rest before colliding, if and only if \mbox{\(\frac23 <\mu <1\)}.
  2. Show that, if the discs collide at least once, their total kinetic energy just before the first collision is \(\frac43 mgr(2-3\mu)\).
  3. Show that if \(\frac 4 9 > \mu^2 >\frac{5}{27}\) the discs come to rest exactly once after the first collision.