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2018 Paper 2 Q1
D: 1600.0 B: 1516.0

Show that, if \(k\) is a root of the quartic equation \[ x^4 + ax^3 + bx^2 + ax + 1 = 0\,, \tag{\(*\)} \] then \(k^{-1}\) is a root. You are now given that \(a\) and \(b\) in \((*)\) are both real and are such that the roots are all real.

  1. Write down all the values of \(a\) and \(b\) for which \((*)\) has only one distinct root.
  2. Given that \((*)\) has exactly three distinct roots, show that either \(b=2a-2\) or \(b=-2a-2\,\).
  3. Solve \((*)\) in the case \(b= 2 a -2\,\), giving your solutions in terms of \(a\).
Given that \(a\) and \(b\) are both real and that the roots of \((*)\) are all real, find necessary and sufficient conditions, in terms of \(a\) and \(b\), for \((*)\) to have exactly three distinct real roots.

Solution: Let \(f(x) = x^4 + ax^3 + bx^2 + ax + 1\), and suppose \(f(k) = 0\). Since \(f(0) = 1\), \(k \neq 0\), therefore we can talk about \(k^{-1}\). \begin{align*} && f(k^{-1}) &= k^{-4} + ak^{-3} + bk^{-2} + ak^{-1} + 1 \\ &&&= k^{-4}(1 + ak + bk^2 + ak^3 + k^4) \\ &&&= k^{-4}(k^4+ak^3+bk^2+ak+1) \\ &&&= k^{-4}f(k) = 0 \end{align*} Therefore \(k^{-1}\) is also a root of \(f\)

  1. If \(f\) has only on distinct root, we must have \(f(x) = (x+k)^4\) therefore \(k = k^{-1} \Rightarrow k^2 = 1 \Rightarrow k = \pm1\), or \(a = 4, b = 6\) or \(a = -4, b = 6\)
  2. If \(f\) has exactly three distinct roots then one of the roots must be a repeated \(1\) or \(-1\), ie \(0 = f(1) = 1 + a + b + a + 1 = 2 + b +2a \Rightarrow b = -2a-2\) or \(0 = f(-1) = 1 -a + b -a + 1 \Rightarrow b = 2a - 2\)
  3. If \(b = 2a-2\), we have \begin{align*} && f(x) &= 1 + ax + (2a-2)x^2 + ax^3 + x^4 \\ &&&= (x^2+2x+1)(1+(a-2)x+x^2) \\ \Rightarrow && x &= \frac{2-a \pm \sqrt{(a-2)^2 - 4}}{2} \\ &&&= \frac{2-a \pm \sqrt{a^2-4a}}{2} \end{align*}
\(f\) has exactly three distinct real roots iff \(b = \pm 2a - 2\) and \(b \neq 6\)

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2018 Paper 2 Q2
D: 1600.0 B: 1516.0

A function \(\f(x)\) is said to be {\em concave} for \(a< x < b\) if \[ \ t\,\f(x_1) +(1-t)\,\f(x_2) \le \f\big(tx_1+ (1-t)x_2\big) \, ,\] for \(a< x_1 < b\,\), \ \(a< x_2< b\) and \(0\le t \le 1\,\). Illustrate this definition by means of a sketch, showing the chord joining the points \(\big(x_1, \f(x_1)\big) \) and \(\big(x_2, \f(x_2)\big) \), in the case \(x_1

  • By choosing \(t\), \(x_1\) and \(x_2\) suitably, show that, if \(\f(x)\) is concave for \(a< x < b\,\), then \[ \f\Big(\frac{u+ v+w}3\Big) \ge \frac{ \f(u) +\f(v) +\f(w)}3 \, ,\] for \(a< u < b\,\), \(a< v < b\,\) and \(a< w < b\,\).
  • Show that, if \(A\), \(B\) and \(C\) are the angles of a triangle, then \[ \sin A +\sin B + \sin C \le \frac{3\sqrt3}2 \,. \]
  • By considering \(\ln (\sin x)\), show that, if \(A\), \(B\) and \(C\) are the angles of a triangle, then \[ \sin A \times \sin B \times \sin C \le \frac {3 \sqrt 3} 8 \,. \]

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    2018 Paper 2 Q3
    D: 1600.0 B: 1529.7

    1. Let \[ \f(x) = \frac 1 {1+\tan x} \] for \(0\le x < \frac12\pi\,\). Show that \(\f'(x)= -\dfrac{1}{1+\sin 2x}\) and hence find the range of \(\f'(x)\). Sketch the curve \(y=\f(x)\).
    2. The function \(\g(x)\) is continuous for \(-1\le x \le 1\,\). Show that the curve \(y=\g(x)\) has rotational symmetry of order 2 about the point \((a,b)\) on the curve if and only if \[ \g(x) + \g(2a-x) = 2b\,. \] Given that the curve \(y=\g(x)\) passes through the origin and has rotational symmetry of order 2 about the origin, write down the value of \[\displaystyle \int_{-1}^1 \g(x)\,\d x\,. \]
    3. Show that the curve \(y=\dfrac{1}{1+\tan^kx}\,\), where \(k\) is a positive constant and \(0 < x < \frac12\pi\,\), \\[3mm] has rotational symmetry of order~2 about a certain point (which you should specify) and evaluate \[ \int_{\frac16\pi}^{\frac13\pi} \frac 1 {1+\tan^kx} \, \d x \,. \]

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    2018 Paper 2 Q4
    D: 1600.0 B: 1484.0

    In this question, you may use the following identity without proof: \[ \cos A + \cos B = 2\cos\tfrac12(A+B) \, \cos \tfrac12(A-B) \;. \]

    1. Given that \(0\le x \le 2\pi\), find all the values of \(x\) that satisfy the equation \[ \cos x + 3\cos 2x + 3\cos 3 x + \cos 4x= 0 \,. \]
    2. Given that \(0\le x \le \pi\) and \(0\le y \le \pi\) and that \[ \cos (x+y) + \cos (x-y) -\cos2x = 1 \,, \] show that either \(x=y\) or \(x\) takes one specific value which you should find.
    3. Given that \(0\le x \le \pi\) and \(0\le y \le \pi\,\), find the values of \(x\) and \(y\) that satisfy the equation \[ \cos x + \cos y -\cos (x+y) = \tfrac32 \,. \]

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    2018 Paper 2 Q5
    D: 1600.0 B: 1505.3

    In this question, you should ignore issues of convergence.

    1. Write down the binomial expansion, for \(\vert x \vert<1\,\), of \(\;\dfrac{1}{1+x}\,\) and deduce that %. By considering %$ %\displaystyle \int \frac 1 {1+x} \, \d x %\,, %$ %show that \[ \displaystyle \ln (1+x) = -\sum_{n=1}^\infty \frac {(-x)^n}n \, \] for \(\vert x \vert <1 \,\).
    2. Write down the series expansion in powers of \(x\) of \(\displaystyle \e^{-ax}\,\). Use this expansion to show that \[ \int_0^\infty \frac {\left(1- \e^{-ax}\right)\e^{-x}}x \,\d x = \ln(1+a) \ \ \ \ \ \ \ (\vert a \vert <1)\,. \]
    3. Deduce the value of \[ \int_0^1 \frac{x^p - x^q}{\ln x} \, \d x \ \ \ \ \ \ (\vert p\vert <1, \ \vert q\vert <1) \,. \]

    Solution:

    1. \begin{align*} && \frac1{1+x} &= 1 - x+ x^2 - x^3+ \cdots \\ \Rightarrow && \int_0^x \frac{1}{1+t} \d t &= \int_0^x \sum_{n=0}^{\infty} (-t)^n \d t \\ &&&= \left [\sum_{n=0}^{\infty} -\frac{(-t)^{n+1}}{n+1} \right]_0^x \\ \Rightarrow &&\ln(1+x)&=- \sum_{n=1}^\infty \frac{(-x)^n}{n} \end{align*}
    2. \begin{align*} && e^{-ax} &= \sum_{n=0}^\infty \frac{(-a)^n}{n!} x^n \\ \Rightarrow && \int_0^{\infty} \frac{1}{x} \left (1-e^{-ax} \right)e^{-x} \d x &= \int_0^{\infty} \frac{1}{x} \left (-\sum_{n=1}^\infty \frac{(-a)^n}{n!}x^n \right)e^{-x} \d x \\ &&&= -\int_0^{\infty} \sum_{n=1}^\infty \frac{(-a)^n}{n!} x^{n-1} e^{-x} \d x \\ &&&= -\sum_{n=1}^\infty \frac{(-a)^n}{n!} \int_0^{\infty} x^{n-1} e^{-x} \d x \\ &&&= -\sum_{n=1}^\infty \frac{(-a)^n}{n!} (n-1)! \\ &&&= -\sum_{n=1}^\infty \frac{(-a)^n}{n} \\ &&&= \ln (1+a) \end{align*}
    3. \begin{align*} && \int_0^1 \frac{x^p - x^q}{\ln x} \, \d x &= \int_0^1 \frac{x^p(1 - x^{q-p})}{\ln x} \, \d x \\ e^{-u} = x, \d x = -e^{-u} \d u: &&&=\int_{u=\infty}^{0} \frac{e^{-pu}-e^{-qu}}{-u} (-e^{-u})\d u \\ &&&= \int_0^\infty \frac{e^{-u}(e^{-qu}-e^{-pu})}{u} \d u \\ &&&= \int_0^\infty \frac{e^{-(1+q)u}(1-e^{-(p-q)u})}{u} \d u \\ v = (1+q)u, \d v = (1+q) \d u: &&&=\int_0^{\infty} \frac{e^{-v}(1-e^{-\left(\frac{p-q}{1+q}\right)v}}{v}\d v \\ &&&= \ln \left(1 + \frac{p-q}{1+q} \right) \\ &&&= \ln \left ( \frac{1+p}{1+q} \right) \end{align*}

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    2018 Paper 2 Q6
    D: 1600.0 B: 1484.7

    1. Find all pairs of positive integers \((n,p)\), where \(p\) is a prime number, that satisfy \[ n!+ 5 =p \,. \]
    2. In this part of the question you may use the following two theorems:
      1. For \(n\ge 7\), \(1! \times 3! \times \cdots \times (2n-1)! > (4n)!\,\).
      2. For every positive integer \(n\), there is a prime number between \(2n\) and \(4n\).
      Find all pairs of positive integers \((n,m)\) that satisfy \[ 1! \times 3! \times \cdots \times (2n-1)! = m! \,. \]

    Solution:

    1. Let \(n! + 5 = p\). If \(n \geq 5\) then \(5\) divides the LHS and so must also divide the RHS. Since \(n!+5 > 5\) this means the RHS cannot be prime. Therefore consider \(n = 1, 2, 3, 4\). \begin{align*} n = 1: && 1! + 5 = 6 &&\text{ nope} \\ n=2: && 2! + 5 = 7 && \checkmark \\ n=3: && 3! + 5 = 11 && \checkmark \\ n=4: && 4! + 5 = 29 && \checkmark \end{align*} Therefore the solutions are \((2,7), (3,11), (4,29)\).
    2. Suppose \(1! \times 3! \times \cdots \times (2n-1)! = m!\). If \(n \geq 7\) then \(m! > (4n)!\) (by the first theorem) in particular \(m > 4n\). Therefore (by the second theorem) the RHS is divisible by some prime which cannot divide the LHS. Therefore consider \(n = 1,2,3,4,5,6\) \begin{align*} n = 1: && 1! = 1 = 1! && \checkmark \\ n = 2: && 1! \times 3! = 6 = 3! && \checkmark \\ n = 3: && 1! \times 3! \times 5! = 6! && \checkmark \\ n = 4: && 1! \times 3! \times 5! \times 7! = 6! \times 7! = 10! && \checkmark \\ n = 5: && 1! \times 3! \times 5! \times 7! \times 9! = 10! 9! > 11! && \text{would need a factor of } 11\text{ so no} \\ n = 6: && 1! \times 3! \times 5! \times 7! \times 9! \times 11! = 10! 11! 9! > 13! && \text{would need a factor of } 13\text{ so no} \\ \end{align*} Therefore all solutions are \((1,1), (2,3), (3,6), (4,10)\)

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    2018 Paper 2 Q7
    D: 1600.0 B: 1500.0

    The points \(O\), \(A\) and \(B\) are the vertices of an acute-angled triangle. The points \(M\) and \(N\) lie on the sides \(OA\) and \(OB\) respectively, and the lines \(AN\) and \(BM\) intersect at \(Q\). The position vector of \(A\) with respect to \(O\) is \(\bf a\), and the position vectors of the other points are labelled similarly. Given that \(\vert MQ \vert = \mu \vert QB\vert \), and that \(\vert NQ \vert = \nu \vert QA\vert \), where \(\mu\) and \(\nu\) are positive and \(\mu \nu <1\), show that \[ {\bf m} = \frac {(1+\mu)\nu}{1+\nu} \, {\bf a} \,. \] The point \(L\) lies on the side \(OB\), and \(\vert OL \vert = \lambda \vert OB \vert \,\). Given that \(ML\) is parallel to \(AN\), express~\(\lambda\) in terms of \(\mu\) and \(\nu\). What is the geometrical significance of the condition \(\mu\nu<1\,\)?

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    2018 Paper 2 Q8
    D: 1600.0 B: 1484.0

    1. Use the substitution \(v= \sqrt y\) to solve the differential equation \[ \frac{\d y}{\d t} = \alpha y^{\frac12} - \beta y \ \ \ \ \ \ \ \ \ \ (y\ge0, \ \ t\ge0) \,, \] where \(\alpha\) and \(\beta\) are positive constants. Find the non-constant solution \(y_1(x)\) that satisfies \(y_1(0)=0\,\).
    2. Solve the differential equation \[ \frac{\d y}{\d t} = \alpha y^{\frac23} - \beta y \ \ \ \ \ \ \ \ \ \ (y\ge0, \ \ t\ge0) \,, \] where \(\alpha\) and \(\beta\) are positive constants. Find the non-constant solution \(y_2(x)\) that satisfies \(y_2(0)=0\,\).
    3. In the case \(\alpha=\beta\), sketch \(y_1(x)\) and \(y_2(x)\) on the same axes, indicating clearly which is \(y_1(x)\) and which is \(y_2(x)\). You should explain how you determined the positions of the curves relative to each other.

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    2018 Paper 2 Q9
    D: 1600.0 B: 1485.5

    Two small beads, \(A\) and \(B\), of the same mass, are threaded onto a vertical wire on which they slide without friction, and which is fixed to the ground at \(P\). They are released simultaneously from rest, \(A\) from a height of \(8h\) above \(P\) and \(B\) from a height of \(17h\) above \(P\). When \(A\) reaches the ground for the first time, it is moving with speed \( V\). It then rebounds with coefficient of restitution \(\frac{1}{2}\) and subsequently collides with \(B\) at height \(H\) above \(P\). Show that \(H= \frac{15}8h\) and find, in terms of \(g\) and \(h\), the speeds \(u_A\) and \(u_B\) of the two beads just before the collision. When \(A\) reaches the ground for the second time, it is again moving with speed \( V\). Determine the coefficient of restitution between the two beads.

    Solution: \begin{align*} && v^2 &= u^2 +2as \\ \Rightarrow && V^2 &= 2 g \cdot (8h)\\ \Rightarrow && V &=4\sqrt{hg}\\ \end{align*} When the first particle collides with the ground, the second particle is at \(9h\) traveling with speed \(V\), the first particle is at \(0\) traveling (upwards) with speed \(\tfrac12 V\). For a collision we need: \begin{align*} && \underbrace{\frac12 V t- \frac12 g t^2}_{\text{position of A}} &= \underbrace{9h - Vt - \frac12 gt^2}_{\text{position of B}} \\ \Rightarrow && \frac32Vt &= 9h \\ \Rightarrow && t &= \frac{6h}{V} \\ \\ && \underbrace{\frac12 V t- \frac12 g t^2}_{\text{position of A}} &= \frac12 V \frac{6h}{V} - \frac12 g t^2 \\ &&&= 3h - \frac12 g\frac{36h^2}{16hg} \\ &&&= 3h - \frac{9}{8}h \\ &&&= \frac{15}{8}h \end{align*} Just before the collision, \(A\) will be moving with velocity (taking upwards as positive) \begin{align*} && u_A &= \frac12 V-gt \\ &&&= 2\sqrt{hg}-g \frac{6h}{V} \\ &&&= 2\sqrt{hg} - g \frac{6h}{4\sqrt{hg}} \\ &&&= 2\sqrt{hg}-\frac32\sqrt{hg} \\ &&&= \frac12 \sqrt{hg} \end{align*} Similarly, for \(B\). \begin{align*} && u_B &= -V -gt \\ &&&= -4\sqrt{hg} - \frac32\sqrt{hg} \\ &&&= -\frac{11}{2}\sqrt{hg} \end{align*} Considering \(A\), to figure out \(v_A\). \begin{align*} && v^2 &= u^2 + 2as \\ && V^2 &= v_A^2 + 2g\frac{15}{8}h \\ && 16hg &= v_A^2 + \frac{15}{4}gh \\ \Rightarrow && v_A^2 &= \frac{49}{4}gh \\ \Rightarrow && v_A &= -\frac{7}{2}\sqrt{gh} \end{align*}

    TikZ diagram
    To keep things clean, lets use units of \(\sqrt{hg}\) so we don't need to focus on that for now: \begin{align*} \text{COM}: && \frac12 - \frac{11}{2} &= -\frac{7}{2}+v_B \\ \Rightarrow && v_B& =-\frac{3}{2} \\ \text{NEL}: && e &= \frac{2}{6} = \frac13 \end{align*}

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    2018 Paper 2 Q10
    D: 1600.0 B: 1500.0

    A uniform elastic string lies on a smooth horizontal table. One end of the string is attached to a fixed peg, and the other end is pulled at constant speed \(u\). At time \(t=0\), the string is taut and its length is \(a\). Obtain an expression for the speed, at time \(t\), of the point on the string which is a distance \(x\) from the peg at time~\(t\). An ant walks along the string starting at \(t=0\) at the peg. The ant walks at constant speed~\(v\) along the string (so that its speed relative to the peg is the sum of \(v\) and the speed of the point on the string beneath the ant). At time \(t\), the ant is a distance \(x\) from the peg. Write down a first order differential equation for \(x\), and verify that \[ \frac{\d }{\d t} \left( \frac x {a+ut}\right) = \frac v {a+ut} \,. \] Show that the time \(T\) taken for the ant to reach the end of the string is given by \[uT = a(\e^k-1)\,,\] where \(k=u/v\). On reaching the end of the string, the ant turns round and walks back to the peg. Find in terms of \(T\) and \(k\) the time taken for the journey back.

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