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2014 Paper 3 Q1
D: 1700.0 B: 1542.2

Let \(a\), \(b\) and \(c\) be real numbers such that \(a+b+c=0\) and let \[(1+ax)(1+bx)(1+cx) = 1+qx^2 +rx^3\,\] for all real \(x\). Show that \(q = bc+ca+ab\) and \(r= abc\).

  1. Show that the coefficient of \(x^n\) in the series expansion (in ascending powers of \(x\)) of \(\ln (1+qx^2+rx^3)\) is \((-1)^{n+1} S_n\) where \[S_n = \frac{a^n+b^n+c^n}{n} \,, \ \ \ \ \ \ \ \ (n\ge1).\]
  2. Find, in terms of \(q\) and \(r\), the coefficients of \(x^2\), \(x^3\) and \(x^5\) in the series expansion (in ascending powers of \(x\)) of \(\ln (1+qx^2+rx^3)\) and hence show that \(S_2S_3 =S_5\).
  3. Show that \(S_2S_5 =S_7\).
  4. Give a proof of, or find a counterexample to, the claim that \(S_2S_7=S_9\).

Solution: \begin{align*} (1+ax)(1+bx)(1+cx) &= (1+(a+b)x+abx^2)(1+cx) \\ &= 1+(a+b+c)x+(ab+bc+ca)x^2+abcx^3 \end{align*} Therefore by comparing coefficients, \(q = bc + ca + ab\) and \(r = abc\) as required.

  1. \begin{align*} \ln (1+qx^2 + rx^3) &= \ln(1+ax) + \ln(1+bx)+\ln(1+cx) \\ &= -\sum_{n=1}^{\infty} \frac{(-ax)^n}{n}-\sum_{n=1}^{\infty} \frac{(-bx)^n}{n}-\sum_{n=1}^{\infty} \frac{(-cx)^n}{n} \\ &= \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(a^n+b^n+c^n)}{n} x^n \\ &= \sum_{n=1}^{\infty} (-1)^{n+1} S_n x^n \\ \end{align*}
  2. \begin{align*} \ln (1 + qx^2 + rx^3) &= (qx^2+rx^3) -\frac{(qx^2+rx^3)^2}{2} + O(x^6) \\ &= qx^2 + rx^3 - \frac12 q^2 x^4 - qr x^5 + O(x^6) \\ \end{align*} Comparing coefficients we see that \(S_2 = -q\) and \(S_3 = r\), we also must have \(S_5 = -qr = S_2S_3\) as required.
  3. \begin{align*} \ln (1 + qx^2 + rx^3) &= (qx^2+rx^3) -\frac{(qx^2+rx^3)^2}{2} +\frac{(qx^2+rx^3)^3}{3}+ O(x^8) \\ &= qx^2 + rx^3 - \frac12 q^2 x^4 - qr x^5 + \frac12 rx^6 + \frac13 q^3 x^6 + q^2r x^7 + O(x^8) \\ &= qx^2 + rx^3 - \frac12 q^2 x^4 - qr x^5 + \left ( \frac12 r+ \frac13 q^3 \right)x^6 + q^2r x^7 \end{align*} Comparing coefficients we see that \(S_2 = -q\) and \(S_5 =-qr\), we also must have \(S_7 = q^2r = S_2S_5\) as required.
  4. Let \(a = b = 1, c = -2\), then \(S_2 = \frac{1^2+1^2 + (-2)^2}{2} = 3, S_7 = \frac{1^2+1^2+(-2)^7}{7} = -18, S_9 = \frac{1^1+1^2+(-2)^9}{9} = \frac{2 - 512}{9} \neq 3 \cdot (-18)\)

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2014 Paper 3 Q2
D: 1700.0 B: 1513.2

  1. Show, by means of the substitution \(u=\cosh x\,\), that \[ \int \frac{\sinh x}{\cosh 2x} \d x = \frac 1{2\sqrt2} \ln \left\vert \frac{\sqrt2 \cosh x - 1}{\sqrt2 \cosh x + 1 } \right\vert + C \,.\]
  2. Use a similar substitution to find an expression for \[ \int \frac{\cosh x}{\cosh 2x} \d x \,.\]
  3. Using parts (i) and (ii) above, show that \[ \int_0^1 \frac 1{1+u^4} \d u = \frac{\pi + 2\ln(\sqrt2 +1)}{4\sqrt2}\,. \]

Solution:

  1. \begin{align*} && \int \frac{\sinh x}{\cosh 2x} \d x &= \int \frac{\sinh x}{2\cosh^2 x -1} \d x \\ u = \cosh x, \d u = \sinh x \d x &&&= \int \frac{1}{2u^2 -1} \d u \\ &&&= \int\frac12 \left ( \frac{1}{\sqrt{2}u-1}-\frac{1}{\sqrt{2}u+1} \right) \d u \\ &&&= \frac1{2\sqrt{2}} \left (\ln (\sqrt{2}u-1) - \ln(\sqrt{2}u+1) \right) + C \\ &&&= \frac{1}{2\sqrt{2}} \ln \left ( \frac{\sqrt{2}\cosh x -1}{\sqrt{2}\cosh x +1} \right) + C \end{align*}
  2. \begin{align*} && \int \frac{\cosh x}{\cosh 2x} \d x &= \int \frac{\cosh x}{1+2\sinh^2 x} \d x \\ u = \sinh x && &= \int \frac{1}{1+2u^2} \d u \\ &&&=\frac{1}{\sqrt{2}} \tan^{-1} (\sqrt{2}u) + C \\ &&&= \frac{1}{\sqrt{2}} \tan^{-1}(\sqrt{2}\sinh x) + C \end{align*}
  3. \begin{align*} u = e^x : && \int_0^1 \frac{1}{1+u^4} \d u &= \int_{x=-\infty}^{x=0} \frac{1}{1+e^{4x}}e^{x} \d x \\ &&&= \int_{-\infty}^{0} \frac{e^{-x}}{e^{2x}+e^{-2x}} \d x \\ &&&= \int_{-\infty}^{0} \frac{\cosh x - \sinh x}{2\cosh 2x } \d x \\ &&&= \frac12 \int_{-\infty}^{0} \frac{\cosh x}{\cosh 2x} \d x - \frac12 \int_{-\infty}^{0} \frac{\sinh x}{\cosh 2x} \\ &&&= \frac12 \left [\frac{1}{\sqrt{2}} \tan^{-1}(\sqrt{2}\sinh x) \right]_{-\infty}^{0}-\frac12 \left [ \frac{1}{2\sqrt{2}}\ln \left ( \frac{\sqrt{2}\cosh x -1}{\sqrt{2}\cosh x +1} \right) \right]_{-\infty}^{0} \\ &&&= 0 - \frac1{2\sqrt{2}} \frac{-\pi}{2}-\left (\frac1{4\sqrt{2}} \ln \left (\frac{\sqrt{2}-1}{\sqrt{2}+1} \right) - 0 \right) \\ &&&= \frac{\pi - \ln((\sqrt{2}-1)^2)}{4\sqrt{2}} \\ &&&= \frac{\pi + 2 \ln(1+\sqrt{2})}{4\sqrt{2}} \end{align*}

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2014 Paper 3 Q3
D: 1700.0 B: 1500.0

  1. The line \(L\) has equation \(y=mx+c\), where \(m>0\) and \(c>0\). Show that, in the case \(mc>a>0\), the shortest distance between \(L\) and the parabola \(y^2=4ax\) is \[ \frac{mc-a}{m\sqrt{m^2+1}}\,.\] What is the shortest distance in the case that \(mc\le a\)?
  2. Find the shortest distance between the point \((p,0)\), where \(p>0\), and the parabola \(y^2=4ax\), where \(a>0\), in the different cases that arise according to the value of \(p/a\). [\textit{You may wish to use the parametric coordinates \((at^2, 2at)\) of points on the parabola.}] Hence find the shortest distance between the circle $(x-p)^2 + y^2 =b^2\(, where \)p>0\( and \)b>0\(, and the parabola \)y^2=4ax\(, where \)a>0$, in the different cases that arise according to the values of \(p\), \(a\) and~\(b\).

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2014 Paper 3 Q4
D: 1700.0 B: 1500.0

  1. Let \[ I = \int_0^1 \bigl((y')^2 -y^2\bigr)\d x \qquad\text{and}\qquad I_1=\int_0^1 (y'+y\tan x)^2 \d x \,, \] where \(y\) is a given function of \(x\) satisfying \(y=0\) at \(x=1\). Show that \(I-I_1=0\) and deduce that \(I\ge0\). Show further that \(I=0\) only if \(y=0\) for all \(x\) (\(0\le x \le 1\)).
  2. Let \[ J = \int_0^1 \bigl((y')^2 -a^2y^2\bigr)\d x \,, \] where \(a\) is a given positive constant and \(y\) is a given function of \(x\), not identically zero, satisfying \(y=0\) at \(x=1\). By considering an integral of the form \[ \int_0^1 (y'+ay\tan bx)^2 \d x \,, \] where \(b\) is suitably chosen, show that \(J\ge0\). You should state the range of values of \(a\), in the form \(a < k\), for which your proof is valid. In the case \(a=k\), find a function \(y\) (not everywhere zero) such that \(J=0\).

Solution:

  1. \begin{align*} && I - I_1 &= \int_0^1 \left ( \left ( y' \right)^2 - y^2 \right) \d x - \int_0^1 \left ( y' + y \tan x \right)^2 \d x\\ &&&= \int_0^1 \left ( \left ( y' \right)^2 - y^2 \right) - \left ( y' + y \tan x \right)^2 \d x\\ &&&= \int_0^1 \left (-y^2-2yy' \tan x - y^2 \tan^2 x \right) \d x\\ &&&= \int_0^1 \left (-2yy' \tan x - y^2(1+ \tan^2 x )\right) \d x\\ &&&= \int_0^1 \left (-2yy' \tan x - y^2 \sec^2 x\right) \d x\\ &&&= \int_0^1 -\frac{\d}{\d x} \left (y^2 \tan x \right) \d x\\ &&&= \left [-y^2 \tan x \right]_0^1 \\ &&&= 0 \\ \\ \Rightarrow && I &= I_1 = \int_0^1 \left ( y' + y \tan x \right)^2 d x \geq 0 \end{align*} The only way \(I_0 = 0\) is is \(y' + y \tan x =0\), so \begin{align*} && \frac{\d y}{\d x} &= - y \tan x \\ \Rightarrow && \int \frac{1}{y} &= \int -\tan x \d x \\ \Rightarrow && \ln |y| &= \ln |\cos x| + C \\ \Rightarrow && y &= A \cos x \\ \Rightarrow && A &= 0 \Rightarrow y = 0 \end{align*}
  2. Let \(J_1 = \int_0^1 (y'+ay\tan ax)^2 \d x\), then \begin{align*} && J-J_1 &= \int_0^1 \left ( \left ( y' \right)^2 - a^2y^2 \right) - \left ( y' + ya \tan ax \right)^2 \d x\\ &&&= \int_0^1 \left (-a^2y^2-2yy' a \tan a x-y^2a^2 \tan^2 ax \right) \d x \\ &&&= \int_0^1 \left (-2yy' a \tan ax - a^2y^2(1+\tan^2 ax) \right) \d x \\ &&&= \int_0^1 \left (-2yy' a \tan ax - a^2y^2\sec^2 ax \right) \d x \\ &&&= \left [ - a y^2 \tan a x \right]_0^1 = 0 \end{align*} This is true if \(a < \frac{\pi}{2}\), since otherwise we might care about the order of the zero for \(y\) at \(x = 1\). Consider \(y = \cos \frac{\pi}{2} x\), then \(y' = -\frac{\pi}{2} \sin^2\frac{\pi}{2} x\) and \begin{align*} && \int_0^1 \frac{\pi^2}{4} \left (\sin^2 \frac{\pi}{2}x - \cos^2 \frac{\pi}{2} x \right) \d x &= -\frac{\pi^2}{4} \int_0^1 \cos(\pi x) \d x \\ &&&= 0 \end{align*}

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2014 Paper 3 Q5
D: 1700.0 B: 1500.0

A quadrilateral drawn in the complex plane has vertices \(A\), \(B\), \(C\) and~\(D\), labelled anticlockwise. These vertices are represented, respectively, by the complex numbers \(a\), \(b\), \(c\) and~\(d\). Show that \(ABCD\) is a parallelogram (defined as a quadrilateral in which opposite sides are parallel and equal in length) if and only if \(a+c =b+d\,\). Show further that, in this case, \(ABCD\) is a square if and only if \({\rm i}(a-c)=b-d\). Let \(PQRS\) be a quadrilateral in the complex plane, with vertices labelled anticlockwise, the internal angles of which are all less than~\(180^\circ\). Squares with centres \(X\), \(Y\), \(Z\) and~\(T\) are constructed externally to the quadrilateral on the sides \(PQ\), \(QR\), \(RS\) and~\(SP\), respectively.

  1. If \(P\) and \(Q\) are represented by the complex numbers \(p\) and~\(q\), respectively, show that \(X\) can be represented by \[ \tfrac 12 \big( p(1+{\rm i} ) + q (1-{\rm i})\big) \,. \]
  2. Show that \(XY\!ZT\) is a square if and only if \(PQRS\) is a parallelogram.

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2014 Paper 3 Q6
D: 1700.0 B: 1516.0

Starting from the result that \[ \.h(t) >0\ \mathrm{for}\ 0< t < x \Longrightarrow \int_0^x \.h(t)\ud t > 0 \,, \] show that, if \(\.f''(t) > 0\) for \(0 < t < x_0\) and \(\.f(0)=\.f'(0) =0\), then \(\.f(t)>0\) for \(0 < t < x_0\).

  1. Show that, for \(0 < x < \frac12\pi\), \[ \cos x \cosh x <1 \,. \]
  2. Show that, for \(0 < x < \frac12\pi\), \[ \frac 1 {\cosh x} < \frac {\sin x} x < \frac x {\sinh x} \,. \] %
  3. Show that, for \(0 < x < \frac12\pi\), \(\tanh x < \tan x\).

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2014 Paper 3 Q7
D: 1700.0 B: 1484.0

The four distinct points \(P_i\) (\(i=1\), \(2\), \(3\), \(4\)) are the vertices, labelled anticlockwise, of a cyclic quadrilateral. The lines \(P_1P_3\) and \(P_2P_4\) intersect at \(Q\).

  1. By considering the triangles \(P_1QP_4\) and \(P_2QP_3\) show that \((P_1Q)( QP_3) = (P_2Q) (QP_4)\,\).
  2. Let \(\+p_i\) be the position vector of the point \(P_i\) (\(i=1\), \(2\), \(3\), \(4\)). Show that there exist numbers \(a_i\), not all zero, such that \begin{equation} \sum\limits_{i=1}^4 a_i =0 \qquad\text{and}\qquad \sum\limits_{i=1}^4 a_i \+p_i ={\bf 0} \,. \tag{\(*\)} \end{equation}
  3. Let \(a_i\) (\(i=1\),~\(2\), \(3\),~\(4\)) be any numbers, not all zero, that satisfy~\((*)\). Show that \(a_1+a_3\ne 0\) and that the lines \(P_1P_3\) and \(P_2P_4\) intersect at the point with position vector \[ \frac{a_1 \+p_1 + a_3 \+p_3}{a_1+a_3} \,. \] Deduce that \(a_1a_3 (P_1P_3)^2 = a_2a_4 (P_2P_4)^2\,\).

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2014 Paper 3 Q8
D: 1700.0 B: 1516.0

The numbers \(\.f(r)\) satisfy \(\.f(r)>\.f(r+1)\) for \(r=1\), \(2\), \dots. Show that, for any non-negative integer \(n\), \[ k^n(k-1) \, \.f(k^{n+1}) \le \sum_{r=k^n}^{k^{n+1}-1}\.f(r) \le k^n(k-1)\, \.f(k^n)\, \] where \(k\) is an integer greater than 1.

  1. By taking \(\.f(r) = 1/r\), show that \[ \frac{N+1}2 \le \sum_{r=1}^{2^{N+1}-1} \frac1r \le N+1 \,. \] Deduce that the sum \(\sum\limits_{r=1}^\infty \frac1r\) does not converge.
  2. By taking \(\.f(r)= 1/r^3\), show that \[ \sum_{r=1}^\infty \frac1 {r^3} \le 1 \tfrac 13 \,. \]
  3. Let \(S(n)\) be the set of positive integers less than \(n\) which do not have a \(2\) in their decimal representation and let \(\sigma(n)\) be the sum of the reciprocals of the numbers in \(S(n)\), so for example \(\sigma(5) = 1+\frac13+\frac14\). Show that \(S(1000)\) contains \(9^3-1\) distinct numbers. Show that \(\sigma (n) < 80\) for all \(n\).

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2014 Paper 3 Q9
D: 1700.0 B: 1500.0

A particle of mass \(m\) is projected with velocity \(\+ u\). It is acted upon by the force \(m\+g\) due to gravity and by a resistive force \(-mk \+v\), where \(\+v\) is its velocity and \(k\) is a positive constant. Given that, at time \(t\) after projection, its position \(\+r\) relative to the point of projection is given by \[ \+r = \frac{kt -1 +\.e^{-kt}} {k^2} \, \+g + \frac{ 1-\.e^{-kt}}{k} \, \+u \,, \] find an expression for \(\+v\) in terms of \(k\), \(t\), \(\+g\) and \(\+u\). Verify that the equation of motion and the initial conditions are satisfied. Let \(\+u = u\cos\alpha \, \+i + u \sin\alpha \, \+j\) and $\+g = -g\, \+j\(, where \)0<\alpha<90^\circ\(, and let \)T$ be the time after projection at which \(\+r \,.\, \+j =0\). Show that \[ uk \sin\alpha = \left(\frac{kT}{1-\.e^{-kT}} -1\right)g\,. \] Let \(\beta\) be the acute angle between \(\+v\) and \(\+i\) at time \(T\). Show that \[ \tan\beta = \frac{(\.e^{kT}-1)g}{uk\cos\alpha}-\tan\alpha \,. \] Show further that \(\tan\beta >\tan\alpha\) (you may assume that \(\sinh kT >kT\)) and deduce that~\(\beta >\alpha\).

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2014 Paper 3 Q10
D: 1700.0 B: 1473.3

Two particles \(X\) and \(Y\), of equal mass \(m\), lie on a smooth horizontal table and are connected by a light elastic spring of natural length \(a\) and modulus of elasticity \(\lambda\). Two more springs, identical to the first, connect \(X\) to a point \(P\) on the table and \(Y\) to a point \(Q\) on the table. The distance between \(P\) and \(Q\) is \(3a\). Initially, the particles are held so that \(XP=a\), \(YQ= \frac12 a\,\), and \(PXYQ\) is a straight line. The particles are then released. At time \(t\), the particle \(X\) is a distance \(a+x\) from \(P\) and the particle \(Y\) is a distance \(a+y\) from \(Q\). Show that \[ m \frac{\.d ^2 x}{\.d t^2} = -\frac\lambda a (2x+y) \] and find a similar expression involving \(\dfrac{\.d^2 y}{\.d t^2}\). Deduce that \[ x-y = A\cos \omega t +B \sin\omega t \] where \(A\) and \(B\) are constants to be determined and \(ma\omega^2=\lambda\). Find a similar expression for \(x+y\). Show that \(Y\) will never return to its initial position.

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