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2010 Paper 2 Q1
D: 1600.0 B: 1516.0

Let \(P\) be a given point on a given curve \(C\). The {\em osculating circle} to \(C\) at \(P\) is defined to be the circle that satisfies the following two conditions at \(P\): it touches \(C\); and the rate of change of its gradient is equal to the rate of change of the gradient of \(C\). Find the centre and radius of the osculating circle to the curve \(y=1-x+\tan x\) at the point on the curve with \(x\)-coordinate \(\frac14 \pi\).

2010 Paper 2 Q2
D: 1600.0 B: 1506.8

Prove that \[ \cos 3x = 4 \cos^3 x - 3 \cos x \,. \] Find and prove a similar result for \(\sin 3x\) in terms of \(\sin x\).

  1. Let \[ {\rm I}(\alpha) = \int_0^\alpha \big(7\sin x - 8 \sin^3 x\big) \d x\,. \] Show that \[ {\rm I}(\alpha) = -\tfrac 8 3 c^3 + c +\tfrac5 3\,, \] where \(c = \cos \alpha\). Write down one value of \(c\) for which \({\rm I}(\alpha) =0\).
  2. Useless Eustace believes that \[ \int \sin^n x \, \d x =\dfrac {\sin^{n+1}x}{n+1}\, \] for \(n=1, \ 2, \ 3, \ldots\, \). Show that Eustace would obtain the correct value of \({\rm I}(\beta)\,\), where \(\cos \beta= -\frac16\). Find all values of \(\alpha\) for which he would obtain the correct value of \({\rm I}(\alpha)\).


Solution: \begin{align*} \cos 3x &\equiv \cos (2x + x) \\ &\equiv \cos 2x \cos x - \sin 2x \sin x \\ &\equiv (2\cos^2 x - 1) \cos x - 2 \sin x \cos x \sin x \\ &\equiv 2 \cos^3 x - \cos x - 2\cos x (\sin^2 x) \\ &\equiv 2 \cos^3 x - \cos x - 2\cos x (1- \cos^2 x) \\ &\equiv 4\cos^3 x - 3\cos x \end{align*} Similarly, \begin{align*} \sin 3x &\equiv \sin (2x + x) \\ &\equiv \sin 2x \cos x + \cos 2x \sin x \\ &\equiv 2 \sin x \cos x \cos x + (1-2\sin^2 x) \sin x \\ &\equiv 2 \sin x (1-\sin^2 x) + \sin x - 2 \sin^3 x \\ &\equiv 3 \sin x -4 \sin ^3 x \end{align*}

  1. \begin{align*} I(\alpha) &= \int_0^{\alpha} (7 \sin x - 8 \sin^3 x) \d x \\ &= \int_0^{\alpha} (7 \sin x - (6\sin x-2 \sin 3x) ) \d x \\ &= \int_0^{\alpha} (\sin x +2 \sin 3x ) \d x \\ &= -\cos \alpha - \frac23 \cos 3\alpha +1+\frac23 \\ &= -c - \frac23 (4c^3-3c) + \frac53 \\ &= -\frac83 c^3 +c + \frac53 \end{align*} as required. When \(c = -1\) this value is \(0\). Eustace will obtain the value \(\frac{7}{2} \sin^2 \beta - 2 \sin^4 \beta = \frac72 (1-\cos^2 \beta) - 2(1-\cos^2 \beta)^2 = \frac32 + \frac12\cos^2 \beta -2\cos^4 \beta\) So if \(\cos \beta = -\frac16\) he will obtain \(\frac32 + \frac{1}{2\cdot36} - \frac{2}{6^4}\) and he should obtain \(\frac{8}{3} \frac{1}{6^3} - \frac{1}{6} + \frac{5}{3}\) which are equal. We want to find all roots of: \begin{align*} && \frac32 + \frac12 c^2 - 2c^4 &= -\frac83 c^3+ c + \frac53 \\ \Rightarrow && 0 &=2c^4-\frac83c^3-\frac12 c^2+c +\frac{1}{6} \\ &&&= 12c^4-16c^3-3c^2+6c+1\\ &&&= (6c+1)(2c^3-3c^2+1) \\ &&&= (6c+1)(2c+1)(c-1)^2 \end{align*} Therefore \(\cos \alpha = - \frac16, -\frac12, 1\) will give the correct answers.

2010 Paper 2 Q3
D: 1600.0 B: 1500.0

The first four terms of a sequence are given by \(F_0=0\), \(F_1=1\), \(F_2=1\) and \(F_3=2\). The general term is given by \[ F_n= a\lambda^n+b\mu^n\,, \tag{\(*\)} \] where \(a\), \(b\), \(\lambda\) and \(\mu\) are independent of \(n\), and \(a\) is positive.

  1. Show that \(\lambda^2 +\lambda\mu+ \mu^2 = 2\), and find the values of \(\lambda\), \(\mu\), \(a\) and \(b\).
  2. Use \((*)\) to evaluate \(F_6\).
  3. Evaluate \(\displaystyle \sum_{n=0}^\infty \frac{F_n}{2^{n+1}}\,.\)


Solution:

  1. \(\,\) \begin{align*} && 0 &= a+b \tag{1}\\ && 1 &= a\lambda -a\mu \tag{2} \\ && 1 &= a\lambda^2 -a\mu^2 \tag{3} \\ && 2 &= a\lambda^3 - a\mu^3 \tag{4} \\ (4) \div (2): && 2 & = \lambda^2+\lambda \mu + \mu^2 \\ (3) \div (2): && 1 &= \lambda + \mu \\ \Rightarrow && 2 &= \lambda^2 + \lambda(1-\lambda) + (1-\lambda)^2 \\ &&&= \lambda^2-\lambda+1\\ \Rightarrow && \lambda, \mu &= \frac{1 \pm \sqrt{5}}{2} \\ \Rightarrow && a &= \frac{1}{\lambda - \mu} = \frac{1}{\sqrt{5}} \\ \Rightarrow && b &= -\frac{1}{\sqrt{5}} \end{align*} (NB: This is Binet's formula)
  2. \(\,\) \begin{align*} F_6 &= \frac{1}{\sqrt{5}} \left ( \left ( \frac{1 +\sqrt{5}}{2} \right)^6- \left ( \frac{1 -\sqrt{5}}{2} \right)^6 \right) \\ &= \frac{1}{2^6 \sqrt{5}} \left ( (1+\sqrt{5})^6-(1-\sqrt{5})^6 \right) \\ &= \frac{1}{2^5 \cdot \sqrt{5}} \left (6 \sqrt{5} +\binom{6}{3} (\sqrt{5})^3+\binom{6}{5}(\sqrt{5})^5 \right)\\ &= \frac{1}{2^5} \left (6 +20\cdot 5+6\cdot 5^2 \right)\\ &= \frac{1}{2^5} 256 = 2^3 = 8 \end{align*} (way more painful than just computing it by adding terms!)
  3. \(\,\) \begin{align*} && \sum_{n=0}^{\infty} \frac{F_n}{2^{n+1}} &= \sum_{n=0}^{\infty} \frac{a\lambda^n + b\mu^n}{2^{n+1}} \\ &&&= \frac12 \left ( \frac{a}{1-\frac{\lambda}2} + \frac{b}{1-\frac{\mu}2} \right) \\ &&&= \frac12 \left ( \frac{2a}{2-\lambda} + \frac{2b}{2-\mu}\right) \\ &&&= \frac{2a}{4-2\lambda} + \frac{2b}{4-2\mu}\\ &&&= \frac{2a}{4-(1+\sqrt{5})} - \frac{2a}{4-(1-\sqrt{5})} \\ &&&= \frac{2}{3\sqrt{5}-5} - \frac{2}{3\sqrt{5}+5} \\ &&&= \frac{6\sqrt{5}+10-6\sqrt{5}+10}{45-25} \\ &&&= 1 \end{align*}

2010 Paper 2 Q4
D: 1600.0 B: 1516.0

  1. Let \[ I=\int_0^a \frac {\f(x)}{\f(x)+\f(a-x)} \, \d x\,. \] Use a substitution to show that \[ I = \int_0^a \frac {\f(a-x)}{\f(x)+\f(a-x)} \, \d x\, \] and hence evaluate \(I\) in terms of \(a\). Use this result to evaluate the integrals \[ \int_0^1 \frac{\ln (x+1)}{\ln (2+x-x^2)}\, \d x \text{ \ \ \ \ \ \ and \ \ \ \ \ } \int_0^{\frac\pi 2} \frac{\sin x } {\sin(x+\frac \pi 4 )} \, \d x \,. \]
  2. Evaluate \[ \int_{\frac12}^2 \frac {\sin x}{x \big(\sin x + \sin \frac 1 x\big)} \, \d x\,. \]

2010 Paper 2 Q5
D: 1600.0 B: 1500.0

The points \(A\) and \(B\) have position vectors \(\bf i +j+k\) and \(5{\bf i} - {\bf j} -{\bf k}\), respectively, relative to the origin \(O\). Find \(\cos2\alpha\), where \(2\alpha\) is the angle \(\angle AOB\).

  1. The line \(L _1\) has equation \({\bf r} =\lambda(m{\bf i}+n {\bf j} + p{\bf k})\). Given that \(L _1\) is inclined equally to \(OA\) and to \(OB\), determine a relationship between \(m\), \(n\) and~\(p\). Find also values of \(m\), \(n\) and~\(p\) for which \(L _1\) is the angle bisector of \(\angle AOB\).
  2. The line \(L _2\) has equation \({\bf r} =\mu(u{\bf i}+v {\bf j} + w{\bf k})\). Given that \( L _2\) is inclined at an angle \(\alpha\) to \(OA\), where \(2\alpha = \angle AOB\), determine a relationship between \(u\), \(v\) and \(w\). Hence describe the surface with Cartesian equation \(x^2+y^2+z^2 =2(yz+zx+xy)\).

2010 Paper 2 Q6
D: 1600.0 B: 1516.0

Each edge of the tetrahedron \(ABCD\) has unit length. The face \(ABC\) is horizontal, and \(P\) is the point in \(ABC\) that is vertically below \(D\).

  1. Find the length of \(PD\).
  2. Show that the cosine of the angle between adjacent faces of the tetrahedron is \(1/3\).
  3. Find the radius of the largest sphere that can fit inside the tetrahedron.


Solution:

  1. \(D\) must be above the centre (of any kind) of the equilateral triangle \(ABC\). Therefore it is a distance \(\frac23 \frac{\sqrt{3}}2 = \frac{\sqrt{3}}3\) from \(A\). \(D\) is \(1\) from \(A\), therefore by Pythagoras \(PD = \sqrt{1-\frac13} = \sqrt{\frac23}\)
  2. We can place \(D\) at \(\langle 0,0,\sqrt{\frac23}\rangle\) and \(A'\) (the midpoint of \(BC\)) at \(\langle-\frac{\sqrt{3}}{6},0,0 \rangle\) and we find: \begin{align*} && \cos \theta &= \frac{(\mathbf{a}'-\mathbf{d})\cdot \mathbf{a}'}{|\mathbf{a}'-\mathbf{d}|| \mathbf{a}'|} \\ &&&= \frac{|\mathbf{a}'|}{|\mathbf{a}'-\mathbf{d}|} \\ &&&= \frac{\frac{\sqrt{3}}{6}}{\sqrt{\frac23+\frac{3}{36}}} = \frac13 \end{align*}
  3. We have
    TikZ diagram
    And therefore we must have \(\tan \frac{\cos^{-1} \frac13}{2} = \frac{r}{\frac{\sqrt{3}}{6}}\) therefore \begin{align*} && r &= \frac{\sqrt{3}}{6} \tan \left (\frac{\cos^{-1} \frac13}{2} \right) \\ &&&= \frac{\sqrt{3}}6 \sqrt{\frac{1-\cos(\cos^{-1}\frac13)}{1+\cos(\cos^{-1}\frac13)}} \\ &&&= \frac{\sqrt{3}}6 \sqrt{\frac{\frac23}{\frac43}} \\ &&&= \frac{\sqrt{6}}{12} \end{align*}

2010 Paper 2 Q7
D: 1600.0 B: 1484.0

  1. By considering the positions of its turning points, show that the curve with equation \[ y=x^3-3qx-q(1+q)\,, \] where \(q>0\) and \(q\ne1\), crosses the \(x\)-axis once only.
  2. Given that \(x\) satisfies the cubic equation \[ x^3-3qx-q(1+q)=0\,, \] and that \[ x=u+q/u\,, \] obtain a quadratic equation satisfied by \(u^3\). Hence find the real root of the cubic equation in the case \(q>0\), \(q\ne1\).
  3. The quadratic equation \[ t^2 -pt +q =0\, \] has roots \(\alpha \) and \(\beta\). Show that \[ \alpha^3+\beta^3 = p^3 -3qp\,. \] It is given that one of these roots is the square of the other. By considering the expression \((\alpha^2 -\beta)(\beta^2-\alpha)\), find a relationship between \(p\) and \(q\). Given further that \(q>0\), \(q\ne1\) and \(p\) is real, determine the value of \(p\) in terms of \(q\).

2010 Paper 2 Q8
D: 1600.0 B: 1498.7

The curves \(C_1\) and \(C_2\) are defined by \[ y= \e^{-x} \ \ \ (x>0) \text{ \ \ \ and \ \ \ } y= \e^{-x}\sin x \ \ \ (x>0), \] respectively. Sketch roughly \(C_1\) and \(C_2\) on the same diagram. Let \(x_n\) denote the \(x\)-coordinate of the \(n\)th point of contact between the two curves, where \(0 < x_1 < x_2 < \cdots\), and let \(A_n\) denote the area of the region enclosed by the two curves between \(x_n\) and \(x_{n+1}\). Show that \[ A_n = \tfrac12(\e^{2\pi}-1) \e^{-(4n+1)\pi/2} \] and hence find \(\displaystyle \sum_{n=1}^\infty A_n\).

2010 Paper 2 Q9
D: 1600.0 B: 1500.0

Two points \(A\) and \(B\) lie on horizontal ground. A particle \(P_1\) is projected from \(A\) towards \(B\) at an acute angle of elevation \(\alpha\) and simultaneously a particle \(P_2\) is projected from \(B\) towards \(A\) at an acute angle of elevation \(\beta\). Given that the two particles collide in the air a horizontal distance \(b\) from \(B\), and that the collision occurs after \(P_1\) has attained its maximum height \(h\), show that \[ 2h \cot\beta < b < 4h \cot\beta \hphantom{\,,} \] and \[ 2h \cot\alpha < a < 4h \cot\alpha \,, \] where \(a\) is the horizontal distance from \(A\) to the point of collision.

2010 Paper 2 Q10
D: 1600.0 B: 1516.0

  1. In an experiment, a particle \(A\) of mass \(m\) is at rest on a smooth horizontal table. A particle \(B\) of mass \(bm\), where \(b >1\), is projected along the table directly towards \(A\) with speed \(u\). The collision is perfectly elastic. Find an expression for the speed of \(A\) after the collision in terms of \(b\) and \(u\), and show that, irrespective of the relative masses of the particles, \(A\) cannot be made to move at twice the initial speed of \(B\).
  2. In a second experiment, a particle \(B_1\) is projected along the table directly towards \(A\) with speed \(u\). This time, particles \(B_2\), \(B_3\), \(\ldots\,\), \(B_n\) are at rest in order on the line between \(B_1\) and \(A\). The mass of \(B_i\) (\(i=1\), \(2\), \(\ldots\,\), \(n\)) is \(\lambda^{n+1-i}m\), where \(\lambda>1\). All collisions are perfectly elastic. Show that, by choosing \(n\) sufficiently large, there is no upper limit on the speed at which \(A\) can be made to move. In the case \(\lambda=4\), determine the least value of \(n\) for which \(A\) moves at more than \(20u\). You may use the approximation \(\log_{10}2 \approx 0.30103\).


Solution:

  1. TikZ diagram
    Since the collision is perfectly elastic, the speed of approach and separation are equal, ie \(v_B = v_A - u\) \begin{align*} \text{COM}: && bmu &= bm(v_A - u) + mv_A \\ \Rightarrow && (b+1)v_A &= 2bu \\ \Rightarrow && v_A &= \frac{2b}{b+1} u \end{align*} Since \(0 < \frac{b}{b+1} < 1\), the largest \(0 < v_A < 2u\)
  2. After the first collision with each \(B_i\) we will have \(\displaystyle v_{i+1} = \frac{2\lambda}{\lambda + 1}v_i\), ie \(\displaystyle v_{i+1} = \left (\frac{2\lambda}{\lambda + 1} \right)^i u\) and so \(\displaystyle v_A = \left (\frac{2\lambda}{\lambda + 1} \right)^n u\) which can be arbitrarily large. Suppose \(\lambda = 4\), then \begin{align*} && 20u &< v_A \\ &&&= \left (\frac{8}{5} \right)^n u \\ \Rightarrow && \log_{10} 20 < n \log_{10}(16/10) \\ && \log_{10} 2 + 1 < n 4\log_{10} 2 - n \\ \Rightarrow && n &> \frac{ \log_{10} 2 + 1}{ 4\log_{10} 2 - 1} \\ &&&\approx \frac{0.30103+1}{4 \times 0.30103 -1}\\ &&&= \frac{1.30103}{0.20412} \\ &&&>6 \end{align*} So \(n =7\) is the smallest possible