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2015 Paper 2 Q3
D: 1600.0 B: 1483.4

Three rods have lengths \(a\), \(b\) and \(c\), where \(a< b< c\). The three rods can be made into a triangle (possibly of zero area) if \(a+b\ge c\). Let \(T_{n}\) be the number of triangles that can be made with three rods chosen from \(n\) rods of lengths \(1\), \(2\), \(3\), \(\ldots\) , \(n\) (where \(n\ge3\)). Show that \(T_8-T_7 = 2+4+6\) and evaluate \(T_8 -T_6\). Write down expressions for \(T_{2m}-T_{2m-1}\) and \(T_{2m} - T_{2m-2}\). Prove by induction that \(T_{2m}=\frac 16 m (m-1)(4m+1)\,\), and find the corresponding result for an odd number of rods.


Solution: Every \(T_7\) triangle is valid, so we are interested in new triangles which have \(8\) has a longest side. We can have: \begin{array}{c|c|c} \text{longest} & \text{middle} & \text{shortest} \\ \hline 8 & 7 & 1-6 \\ 8 & 6 & 2-5 \\ 8 & 5 & 3-4 \end{array} which is \(6+4+2\) extra triangles. The new ones excluding all the sixes are: \begin{array}{c|c|c} \text{longest} & \text{middle} & \text{shortest} \\ \hline 8 & 7 & 1-6 \\ 8 & 6 & 2-5 \\ 8 & 5 & 3-4 \\ 7 & 6 & 1-5 \\ 7 & 5 & 2-4 \\ 7 & 4 & 3 \\ \end{array} Ie \(2+4+6 + 1 + 3+5\) \(T_{2m}-T_{2m-1} = 2 \frac{(m-1)m}{2} = m(m-1)\) and \(T_{2m}-T_{2m-2} = \frac{(2m-2)(2m-1)}{2}\) \(T_4 = 3\) (\(1,2,3\), \(1,3,4\), \(2,3,4\)) and \(\frac16 \cdot 2 \cdot 1 \cdot 9 = 3\) so the base case holds. Suppose it's true for some \(m = k\), then \begin{align*} && T_{2(k+1)} &= T_{2k} + \frac{2m(2m+1)}{2} \\ &&&= \frac{m(m-1)(4m+1)}{6} + \frac{6m(2m+1)}{6}\\ &&&= \frac{m(4m^2-3m-1+12m+6)}{6} \\ &&&= \frac{m(4m^2+9m+5)}{6}\\ &&&= \frac{m(4m+5)(m+1)}{6}\\ &&&= \frac{(m+1-1)(4(m+1)+5)(m+1)}{6}\\ \end{align*} as required, therefore it is true by induction. For odd numbers, we can see that \(T_{2m-1} = \frac{m(m-1)(4m+1)}{6} - m(m-1) = \frac{m(m-1)(4m-5)}{6}\)

2015 Paper 2 Q4
D: 1600.0 B: 1516.0

  1. The continuous function \(\f\) is defined by \[ \tan \f(x) = x \ \ \ \ \ (-\infty < x <\infty) \] and \(\f(0)=\pi\). Sketch the curve \(y=\f(x)\)\,.
  2. The continuous function \(\g\) is defined by \[ \tan \g(x) = \frac x {1+x^2} \ \ \ \ \ \ (-\infty < x <\infty) \] and \(\g(0)=\pi\). Sketch the curves \(y= \dfrac x {1+x^2} \ \) and \(y=\g(x)\)\,.
  3. The continuous function \(\h \) is defined by \(\h (0)=\pi\) and \[ \tan \h (x)= \frac x {1-x^2}\, \ \ \ \ \ (x \ne \pm 1) \,. \] (The values of \(\h (x)\) at \(x=\pm1\) are such that \(\h (x)\) is continuous at these points.) Sketch the curves \(y= \dfrac x {1-x^2} \ \) and \(y=\h (x)\). %
  4. The continuous functions \(\h_1\) and \(\h_2\) are % defined by: \(\h_1(0)=\h_2(0)=\pi \), %\[ %\tan \h_1(x) = \frac {x+x^4} {1+x^2+x^4} %\ \ \ \ \ \text{and} \ \ \ \ \ \ %\tan \h_2(x) = \frac {4x-x^3} {1-x^4} %\,. %\] %for values of \(x\) at which the right hand sides are defined. %Find \(\lim\limits_{x\to\infty}\h_1(x)\) and \(\lim\limits_{x\to\infty}\h_2(x)\,\).

2015 Paper 2 Q5
D: 1600.0 B: 1484.9

In this question, the \(\mathrm{arctan}\) function satisfies \(0\le \arctan x <\frac12 \pi\) for \(x\ge0\,\).

  1. Let \[ S_n= \sum_{m=1}^n \arctan \left(\frac1 {2m^2}\right) \,, \] for \(n=1, 2, 3, \ldots\) . Prove by induction that \[ \tan S_n = \frac n {n+1} \,. \] Prove also that \[ S_n = \arctan \frac n {n+1} \,. \]
  2. In a triangle \(ABC\), the lengths of the sides \(AB\) and \(BC\) are \(4n^2\) and \(4n^4-1\), respectively, and the angle at \(B\) is a right angle. Let \(\angle BCA = 2\alpha_n\). Show that \[ \sum_{n=1}^\infty \alpha_n = \tfrac14\pi \,. \]


Solution:

  1. Claim: \(\tan S_n = \frac n {n+1}\) Proof: (By Induction) Base case: (\(n=1\)): \begin{align*} && \tan \left ( \sum_{m=1}^1 \arctan \left ( \frac{1}{2m^2} \right) \right) &= \tan \left ( \arctan \left ( \frac{1}{2} \right) \right) \\ &&&= \frac12 = \frac{1}{1+1} \end{align*} Therefore the base case is true. Inductive step: Suppose our statement is true for some \(n = k\), ie \begin{align*} && \frac{k}{k+1} &= \tan \left ( \sum_{m=1}^k \arctan \left ( \frac{1}{2m^2} \right) \right) \\ \Rightarrow && \tan S_{k+1} &= \tan \left ( \sum_{m=1}^k \arctan \left ( \frac{1}{2m^2} \right) + \arctan \left ( \frac{1}{2 (k+1)^2} \right) \right) \\ &&&= \frac{\tan S_k + \tan \left ( \arctan \left ( \frac{1}{2 (k+1)^2} \right) \right)}{1-\tan S_k \tan \left ( \arctan \left ( \frac{1}{2 (k+1)^2} \right) \right)} \\ &&&= \frac{\frac{k}{k+1} + \frac{1}{2(k+1)^2}}{1-\frac{k}{k+1} \frac{1}{2(k+1)^2}} \\ &&&= \frac{2k(k+1)^2+(k+1)}{2(k+1)^3-k} \\ &&&= \frac{k+1}{(k+1)+1} \end{align*} Therefore it is true for \(n=k+1\). Conclusion: Therefore by the principle of mathematical induction since our statement is true for \(n=1\) and if it is true for \(n=k\) it is true for \(n=k+1\) it is true for all \(n\geq1\) Since \(S_n < \frac12 \pi\) for all \(n\), we must have \(\arctan \frac{n}{n+1} = S_n\)
  2. \(\tan (2\alpha_n) = \frac{4n^2}{4n^4-1} = \frac{2n^2+2n^2}{(2n^2)(2n^2)-1} = \frac{\frac{1}{2n^2}+\frac{1}{2n^2}}{1-\frac{1}{2n^2}\frac{1}{2n^2}} \Rightarrow \tan (\alpha_n) = \arctan \frac{1}{2n^2}\). In particular \(\displaystyle \sum_{n=1}^{N} \alpha_n = \arctan \frac{n}{n+1} \Rightarrow \sum_{n=1}^{\infty} \alpha_n \to \arctan 1 = \frac{\pi}{4} \)

2015 Paper 2 Q6
D: 1600.0 B: 1504.2

  1. Show that \[ \mathrm{sec}^2\left(\tfrac14\pi-\tfrac12 x\right)=\frac{2}{1+\sin x} \,. \] Hence integrate \(\dfrac{1}{1+\sin x}\) with respect to \(x\).
  2. By means of the substitution \(y=\pi -x\), show that \[ \int_0^\pi x \f (\sin x)\, \d x = \frac \pi 2 \int_0^\pi \f(\sin x) \, \d x ,\] where \(\mathrm{f}\) is any function for which these integrals exist. Hence evaluate \[ \int_0^\pi \frac x {1+\sin x} \, \d x \,. \]
  3. Evaluate \[ \int_0^\pi\frac{ 2x^3 -3\pi x^2}{(1+\sin x)^2}\, \d x .\]

2015 Paper 2 Q7
D: 1600.0 B: 1484.0

A circle \(C\) is said to be {\em bisected} by a curve \(X\) if \(X\) meets \(C\) in exactly two points and these points are diametrically opposite each other on \(C\).

  1. Let \(C\) be the circle of radius \(a\) in the \(x\)-\(y\) plane with centre at the origin. Show, by giving its equation, that it is possible to find a circle of given radius \(r\) that bisects \(C\) provided \(r>a\). Show that no circle of radius \(r\) bisects \(C\) if \(r\le a\,\).
  2. Let \(C_1\) and \(C_2\) be circles with centres at \((-d,0)\) and \((d,0)\) and radii \(a_1\) and \(a_2\), respectively, where \(d>a_1\) and \(d>a_2\). Let \(D\) be a circle of radius~\(r\) that bisects both \(C_1\) and \(C_2\). Show that the \(x\)-coordinate of the centre of \(D\) is \(\dfrac{a_2^2 - a_1^2}{4d}\). Obtain an expression in terms of \(d\), \(r\), \(a_1\) and \(a_2\) for the \(y\)-coordinate of the centre of~\(D\), and deduce that \(r\) must satisfy \[ 16r^2d^2 \ge \big (4d^2 +(a_2-a_1)^2\big) \, \big (4d^2 +(a_2+a_1)^2\big) \,. \]

2015 Paper 2 Q8
D: 1600.0 B: 1500.0

\noindent

\psset{xunit=1.0cm,yunit=1.0cm,algebraic=true,dimen=middle,dotstyle=o,dotsize=3pt 0,linewidth=0.3pt,arrowsize=3pt 2,arrowinset=0.25} \begin{pspicture*}(-2.94,-1.87)(7.07,3.86) \pscircle(0,1){1.25} \pscircle(3,0){0.55} \rput[tl](5.33,-0.41){\(P\)} \psline(-2.44,-0.03)(6.18,-0.85) \psline(-2.04,3.71)(6.55,-1.48) \rput[tl](-0.18,1.1){\(C_1\)} \rput[tl](2.85 ,0.15){\(C_2\)} \rput[tl](-0.65,3.29){\(L'\)} \rput[tl](-1.5,-0.34){\(L\)} \end{pspicture*}
The diagram above shows two non-overlapping circles \(C_1\) and \(C_2\) of different sizes. The lines \(L\) and \(L'\) are the two common tangents to \(C_1\) and \(C_2\) such that the two circles lie on the same side of each of the tangents. The lines \(L\) and \(L'\) intersect at the point \(P\) which is called the {\em focus} of \(C_1\) and \(C_2\).
  1. Let \({\bf x}_1\) and \({\bf x}_2\) be the position vectors of the centres of \(C_1\) and \(C_2\), respectively. Show that the position vector of \(P\) is \[ \frac{r_1 {\bf x}_2- r_2 {\bf x}_1}{r_1-r_2} \,, \] where \(r_1\) and \(r_2\) are the radii of \(C_1\) and \(C_2\), respectively.
  2. The circle \(C_3\) does not overlap either \(C_1\) or \(C_2\) and its radius, \(r_3\), satisfies \(r_1 \ne r_3 \ne r_2\). The focus of \(C_1\) and \(C_3\) is \(Q\), and the focus of \(C_2\) and \(C_3\) is \(R\). Show that \(P\), \(Q\) and~\(R\) lie on the same straight line.
  3. Find a condition on \(r_1\), \(r_2\) and \(r_3\) for \(Q\) to lie half-way between \(P\) and \(R\).

2015 Paper 2 Q9
D: 1600.0 B: 1484.0

An equilateral triangle \(ABC\) is made of three light rods each of length \(a\). It is free to rotate in a vertical plane about a horizontal axis through \(A\). Particles of mass \(3m\) and \(5m\) are attached to \(B\) and \(C\) respectively. Initially, the system hangs in equilibrium with \(BC\) below \(A\).

  1. Show that, initially, the angle \(\theta\) that \(BC\) makes with the horizontal is given by \(\sin\theta = \frac17\).
  2. The triangle receives an impulse that imparts a speed \(v\) to the particle \(B\). Find the minimum speed \(v_0\) such that the system will perform complete rotations if \(v>v_0\).


Solution:

TikZ diagram
  1. The sine rule tells us: \begin{align*} && \frac{\frac58 a}{\sin(30^\circ + \theta)} &= \frac{a}{\sin(90^{\circ}-\theta)} \\ \Rightarrow &&\frac58 \cos \theta &= \frac12 \cos \theta+ \frac{\sqrt{3}}2 \sin \theta \\ \Rightarrow && \frac{1}{4\sqrt{3}} &= \tan \theta \\ \Rightarrow && \sin \theta &= \sqrt{\frac{1}{48+1}} = \frac17 \end{align*}
  2. \(\,\) \begin{align*} && \text{initial energy} &= \frac12(5m)v^2 + \frac12 (3m)v^2 - 3m \cdot g \cdot a \cos(30^{\circ}+\theta) -5m \cdot g \cdot a\cos(30^\circ - \theta) \\ &&&= 4m v^2 - amg(4\sqrt{3} \cos \theta + \sin \theta) \\ &&&= 4mv^2 - 7amg \\ && \text{energy at top} &= \frac12 m v_{top}^2 + 7amg \end{align*} We need this equation to be positive for all values of \(v_{top} \geq 0\), so \(4mv^2 \geq 14amg \Rightarrow v_0 = \sqrt{\frac{7ag}2}\)

2015 Paper 2 Q10
D: 1600.0 B: 1516.0

A particle of mass \(m\) is pulled along the floor of a room in a straight line by a light string which is pulled at constant speed \(V\) through a hole in the ceiling. The floor is smooth and horizontal, and the height of the room is \(h\). Find, in terms of \(V\) and \(\theta\), the speed of the particle when the string makes an angle of \(\theta\) with the vertical (and the particle is still in contact with the floor). Find also the acceleration, in terms of \(V\), \(h\) and \(\theta\). Find the tension in the string and hence show that the particle will leave the floor when \[ \tan^4\theta = \frac{V^2}{gh}\,. \]

2015 Paper 2 Q11
D: 1600.0 B: 1484.0

Three particles, \(A\), \(B\) and \(C\), each of mass \(m\), lie on a smooth horizontal table. Particles \(A\) and \(C\) are attached to the two ends of a light inextensible string of length \(2a\) and particle~\(B\) is attached to the midpoint of the string. Initially, \(A\), \(B\) and \(C\) are at rest at points \((0,a)\), \((0,0)\) and \((0,-a)\), respectively. An impulse is delivered to \(B\), imparting to it a speed \(u\) in the positive \(x\) direction. The string remains taut throughout the subsequent motion.

\psset{xunit=1.0cm,yunit=1.0cm,algebraic=true,dimen=middle,dotstyle=o,dotsize=3pt 0,linewidth=0.5pt,arrowsize=3pt 2,arrowinset=0.25} \begin{pspicture*}(-2.18,-3.26)(6.26,3.18) \psline[linewidth=2pt](2.,1.5)(3.,0.) \psline[linewidth=2pt](3.,0.)(2.,-1.5) \psline(-2.,0.)(6.,0.) \psline(0.,3.)(0.,-3.) \parametricplot{2.1587989303424644}{3.1415926535897936}{0.7*cos(t)+3.|0.7*sin(t)+0.} \rput[tl](-0.25,2.96){\(y\)} \rput[tl](5.76,-0.1){\(x\)} \rput[tl](2.5,0.38){\(\theta\)} \rput[tl](2.1,1.94){\(A\)} \rput[tl](3.2,0.4){\(B\)} \rput[tl](2.16,-1.46){\(C\)} \begin{scriptsize} \psdots[dotsize=8pt 0,dotstyle=*](2.,1.5) \psdots[dotsize=8pt 0,dotstyle=*](3.,0.) \psdots[dotsize=8pt 0,dotstyle=*](2.,-1.5) \end{scriptsize} \end{pspicture*}
  1. At time \(t\), the angle between the \(x\)-axis and the string joining \(A\) and \(B\) is \(\theta\), as shown in the diagram, and \(B\) is at \((x,0)\). Write down the coordinates of \(A\) in terms of \(x,a\) and \(\theta\). Given that the velocity of \(B\) is \((v,0)\), show that the velocity of \(A\) is \((\dot x + a\sin\theta \,\dot \theta\,,\, a\cos\theta\, \dot\theta)\), where the dot denotes differentiation with respect to time.
  2. Show that, before particles \(A\) and \(C\) first collide, \[ 3\dot x + 2a \dot\theta \sin\theta =v \text{ \ \ \ \ \ \ and \ \ \ \ \ \ } \dot \theta^2 = \frac{v^2}{a^2(3-2\sin^2\theta)} \,. \]
  3. When \(A\) and \(C\) collide, the collision is elastic (no energy is lost). At what value of \(\theta\) does the second collision between particles \(A\) and \(C\) occur? (You should justify your answer.)
  4. When \(v=0\), what are the possible values of \(\theta\)? Is \(v =0\) whenever \(\theta\) takes these values?

2015 Paper 2 Q12
D: 1600.0 B: 1500.0

Four players \(A\), \(B\), \(C\) and \(D\) play a coin-tossing game with a fair coin. Each player chooses a sequence of heads and tails, as follows: Player A: HHT; Player B: THH; Player C: TTH; Player D: HTT. The coin is then tossed until one of these sequences occurs, in which case the corresponding player is the winner.

  1. Show that, if only \(A\) and \(B\) play, then \(A\) has a probability of \(\frac14\) of winning.
  2. If all four players play together, find the probabilities of each one winning.
  3. Only \(B\) and \(C\) play. What is the probability of \(C\) winning if the first two tosses are TT? Let the probabilities of \(C\) winning if the first two tosses are HT, TH and HH be \(p\), \(q\) and \(r\), respectively. Show that \(p=\frac12 +\frac12q\). Find the probability that \(C\) wins.


Solution:

  1. The only way \(A\) can win is if the sequence starts HH, if it does not start like this, then the only way HHT can appear is after a sequence of THH...H, but then THH has already appeared and \(B\) has won. Therefore the probability is \(\frac14\)
  2. If HH appears before TT then either \(A\) or \(B\) will win. If HH appears first, then \(A\) has a \(\frac14\) probability of winning. So \(A\): \(\frac18\), \(B:\), \(\frac38\), \(C:\), \(\frac18\), \(D: \frac38\)
  3. If the first two tosses are TT then \(C\) will win. If the first two tosses are HT, then either the next toss is T and \(C\) wins, or the next toss is H, and it's as if we started TH. ie \(p = \frac12 + \frac12 q\). If the first two tosses are TH, then either the next toss is H and \(C\) losses or the next toss is T and it's like starting HT. So \(q = \frac12 p\). Therefore \(p = \frac12 + \frac14p \Rightarrow p = \frac13\) If the first two tosses are HH, then eventually a T appears, and it's the same as starting HT. Therefore the probability \(C\) wins is: \(\frac14 + \frac14 \cdot \frac13 + \frac14 \cdot \frac16 + \frac14 \cdot \frac13 = \frac{11}{24}\)