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2017 Paper 1 Q1
D: 1500.0 B: 1484.0

  1. Use the substitution \(u= x\sin x +\cos x\) to find \[ \int \frac{x }{x\tan x +1 } \, \d x \,. \] Find by means of a similar substitution, or otherwise, \[ \int \frac{x }{x\cot x -1 } \, \d x \,. \]
  2. Use a substitution to find \[ \int \frac{x\sec^2 x \, \tan x}{x\sec^2 x -\tan x} \,\d x \, \] and \[ \int \frac{x\sin x \cos x}{(x-\sin x \cos x)^2} \, \d x \,. \]


Solution:

  1. \(\,\) \begin{align*} && I &= \int \frac{x}{x \tan x + 1} \d x \\ &&&= \int \frac{x \cos x}{x \sin x + \cos x} \d x \\ u = x \sin x + \cos x , \d u = x \cos x \d x: &&&= \int \frac{\d u}{u} \\ &&&= \ln u + C \\ &&&= \ln (x \sin x + \cos x) + C \\ \\ && J &= \int \frac{x}{x \cot x - 1} \d x \\ &&&= \int \frac{x \sin x }{x \cos x - \sin x} \d x \\ u = x \cos x - \sin x, \d u = x \sin x \d x: &&&= \int \frac{1}{u} \d u \\ &&&= \ln u + K \\ &&&= \ln (x \cos x -\sin x) + K \end{align*}
  2. \(\,\) \begin{align*} && I &= \int \frac{x \sec^2 x \tan x}{x \sec^2 x - \tan x} \d x \\ u = x\sec^2 x-\tan x, \d u = 2x \sec^2 x \tan x&&&= \frac12 \int \frac{1}{u} \d u \\ &&&= \frac12 \ln (x \sec^2 x - \tan x) + C \\ \\ && J &= \int \frac{x \sin x \cos x}{(x - \sin x \cos x)^2} \d x \\ u = x \sec^2 x -\tan x, \d u=2x \frac{\sin x}{\cos^3 x} &&&= \int \frac{x \sin x \cos x}{\cos^4x(x\sec^2 x -\tan x)^2} \d x \\ &&&= \frac12 \int \frac{1}{u^2} \d u \\ &&&= -\frac12u^{-1} + K \\ &&&= \frac{1}{2(\tan x - x \sec^2 x)} + K \end{align*}

2017 Paper 1 Q2
D: 1484.0 B: 1500.1

  1. The inequality \(\dfrac 1 t \le 1\) holds for \(t\ge1\). By integrating both sides of this inequality over the interval \(1\le t \le x\), show that \[ \ln x \le x-1 \tag{\(*\)} \] for \(x \ge 1\). Show similarly that \((*)\) also holds for \(0 < x \le 1\).
  2. Starting from the inequality \(\dfrac{1}{t^2} \le \dfrac1 t \) for \(t \ge 1\), show that \[ \ln x \ge 1-\frac{1}{x} \tag{\(**\)} \] for \(x > 0\).
  3. Show, by integrating (\(*\)) and (\(**\)), that \[ \frac{2}{ y+1} \le \frac{\ln y}{ y-1} \le \frac{ y+1}{2 y} \] for \( y > 0\) and \( y\ne1\).


Solution:

  1. \(\,\) \begin{align*} (x \geq 1): && \int_1^x \frac{1}{t} \d t &\leq \int_1^x 1 \d t \\ \Rightarrow && \ln x - \ln 1 &\leq x - 1 \\ \Rightarrow && \ln x & \leq x - 1 \\ \\ (0 < x \leq 1):&& \int_x^1 1\d t &\leq \int_x^1 \frac{1}{t} \d t \\ \Rightarrow&& 1- x &\leq \ln 1 - \ln x \\ \Rightarrow&& \ln x &\leq x - 1 \end{align*}
  2. \(\,\) \begin{align*} (x \geq 1): && \int_1^x \frac{1}{t^2} \d t &\leq \int_1^x \frac{1}{t} \d t \\ \Rightarrow && -\frac1x+1 &\leq \ln x - \ln 1 \\ \Rightarrow && 1 - \frac1x &\leq \ln x \\ \\ (0 < x \leq 1): && \int_x^1 \frac{1}{t} \d t &\leq \int_x^1 \frac{1}{t^2} \d t \\ \Rightarrow && \ln 1 - \ln x & \leq -1 + \frac{1}{x} \\ \Rightarrow && 1 - \frac1x &\leq \ln x \\ \end{align*}
  3. \(\,\) \begin{align*} (1 < y): && \int_1^y \left (1 - \frac1{x} \right)\d x &\leq \int_1^y \ln x \d x \\ \Rightarrow && \left [x - \ln x \right]_1^y & \leq \left [ x \ln x - x\right]_1^y \\ \Rightarrow && y - \ln y - 1 &\leq y \ln y - y +1 \\ \Rightarrow && 2y-2 & \leq (y+1) \ln y \\ \Rightarrow && \frac{2}{y+1} & \leq \frac{\ln y}{y-1} \\ (0 < y < 1): && \int_y^1 \left (1 - \frac1{x} \right)\d x &\leq \int_y^1 \ln x \d x \\ \Rightarrow && \left [x - \ln x \right]_y^1 & \leq \left [ x \ln x - x\right]_y^1 \\ \Rightarrow && 1 - (y - \ln y) &\leq -1-(y \ln y-y) \\ \Rightarrow && 2-2y &\leq -(y+1)\ln y \\ \Rightarrow && \frac{2}{y+1} &\leq \frac{-\ln y}{1-y} \tag{\(1-y > 0\)} \\ \Rightarrow && \frac{2}{y+1} &\leq \frac{\ln y}{y-1} \\ \\ (1 < y): && \int_1^y \ln x \d x &\leq \int_1^y (x-1) \d x \\ \Rightarrow && \left [x \ln x -x \right]_1^y &\leq \left[ \frac12 x^2 - x \right]_1^y\\ \Rightarrow && y \ln y - y +1 &\leq \frac12y^2 - y+\frac12 \\ \Rightarrow && y \ln y &\leq \frac12 \left (y^2-1 \right) \\ \Rightarrow && \frac{\ln y}{y-1} &\leq \frac{y+1}{2y} \\ \\ (0 < y < 1) && \int_y^1 \ln x \d x &\leq \int_y^1 (x-1) \d x \\ \Rightarrow && \left [x \ln x -x \right]_y^1&\leq \left[ \frac12 x^2 - x \right]_y^1\\ \Rightarrow && -1-(y \ln y - y +1) &\leq-\frac12 - \left ( \frac12y^2 - y\right)\\ \Rightarrow && \frac12 \left (y^2-1 \right) &\leq y \ln y \\ \Rightarrow && \frac{\ln y}{y-1} & \leq \frac{y+1}{2y} \tag{\(y-1 < 0\)} \end{align*}

2017 Paper 1 Q3
D: 1500.0 B: 1500.0

The points \(P(ap^2, 2ap)\) and \(Q(aq^2, 2aq)\), where \(p>0\) and \(q<0\), lie on the curve \(C\) with equation $$y^2= 4ax\,, $$ where \(a>0\,\). Show that the equation of the tangent to \(C\) at \(P\) is $$y= \frac 1 p \, x +ap\,.$$ The tangents to the curve at \(P\) and at \(Q \) meet at \(R\). These tangents meet the \(y\)-axis at \(S\) and \(T\) respectively, and \(O\) is the origin. Prove that the area of triangle \(OPQ\) is twice the area of triangle \(RST\).

2017 Paper 1 Q4
D: 1500.0 B: 1516.0

  1. Let \(r\) be a real number with \(\vert r \vert<1\) and let \[ S = \sum_{n=0}^\infty r^n\,. \] You may assume without proof that \(S = \ds \frac{1}{1-r}\, \). Let \(p= 1 + r +r^2\). Sketch the graph of the function \(1+r+r^2\) and deduce that \(\frac{3}{4} \le p < 3\,\). Show that, if \(1 < p < 3\), then the value of \( p\) determines \(r\), and hence \(S\), uniquely. Show also that, if \(\frac{3}{4} < p < 1\), then there are two possible values of \(S\) and these values satisfy the equation \((3- p)S^2-3S+1=0\).
  2. Let \(r\) be a real number with \(\vert r \vert<1\) and let \[ T =\sum_{n=1}^\infty nr^{n-1} \,. \] You may assume without proof that $ T = \ds \dfrac{1}{(1-r)^2}\,. $ Let \( q= 1+2r+3r^2\). Find the set of values of \( q\) that determine \(T\) uniquely. Find the set of values of \( q\) for which \(T\) has two possible values. Find also a quadratic equation, with coefficients depending on \( q\), that is satisfied by these two values.

2017 Paper 1 Q5
D: 1500.0 B: 1456.4

A circle of radius \(a\) is centred at the origin \(O\). A rectangle \(PQRS\) lies in the minor sector \(OMN\) of this circle where \(M\) is \((a,0)\) and \(N\) is \((a \cos \beta, a \sin \beta)\), and \(\beta\) is a constant with \(0 < \beta < \frac{\pi}{2}\,\). Vertex \(P\) lies on the positive \(x\)-axis at \((x,0)\); vertex \(Q\) lies on \(ON\); vertex \(R\) lies on the arc of the circle between \(M\) and \(N\); and vertex \(S\) lies on the positive \(x\)-axis at \((s,0)\). Show that the area \(A\) of the rectangle can be written in the form \[ A= x(s-x)\tan\beta \,. \] Obtain an expression for \(s\) in terms of \(a\), \(x\) and \(\beta\), and use it to show that \[ \frac{\d A}{\d x} = (s-2x) \tan \beta - \frac {x^2} s \tan^3\beta \,. \] Deduce that the greatest possible area of rectangle \(PQRS\) occurs when \(s= x(1+\sec\beta)\) and show that this greatest area is \(\tfrac12 a^2 \tan \frac12 \beta\,\). Show also that this greatest area occurs when \(\angle ROS = \frac12\beta\,\).

2017 Paper 1 Q6
D: 1516.0 B: 1484.0

In this question, you may assume that, if a continuous function takes both positive and negative values in an interval, then it takes the value \(0\) at some point in that interval.

  1. The function \(\f\) is continuous and \(\f(x)\) is non-zero for some value of \(x\) in the interval \(0\le x \le 1\). Prove by contradiction, or otherwise, that if \[ \int_0^1 \f(x) \d x = 0\,, \] then \(\f(x)\) takes both positive and negative values in the interval \(0\le x\le 1\).
  2. The function \(\g\) is continuous and \[ \int_0^1 \g(x) \, \d x = 1\,, \quad \int_0^1 x\g(x) \, \d x = \alpha\, , \quad \int_0^1 x^2\g(x) \, \d x = \alpha^2\,. \tag{\(*\)} \] Show, by considering \[ \int_0^1 (x - \alpha)^2 \g(x) \, \d x \,, \] that \(\g(x)=0\) for some value of \(x\) in the interval \(0\le x\le 1\). Find a function of the form \(\g(x) = a+bx\) that satisfies the conditions \((*)\) and verify that \(\g(x)=0\) for some value of \(x\) in the interval \(0\le x \le 1\).
  3. The function \(\h\) has a continuous derivative \(\h'\) and \[ \h(0) = 0\,, \quad \h(1) = 1\,, \quad \int_0^1 \h(x) \, \d x = \beta\,, \quad \int_0^1 x \h(x) \, \d x = \ts \frac{1}{2} \ds \beta (2 - \beta) \,. \] Use the result in part \bf (ii) \rm to show that \(\h^\prime(x)=0\) for some value of \(x\) in the interval \(0\le x\le 1\).

2017 Paper 1 Q7
D: 1500.0 B: 1516.0

The triangle \(ABC\) has side lengths \(\left| BC \right| = a\), \(\left| CA \right| = b\) and \(\left| AB \right| = c\). Equilateral triangles \(BXC\), \; \(CY\!A\) \hspace{0.0mm} and \(AZB\) are erected on the sides of the triangle \(ABC\), with~\(X\) on the other side of \(BC\) from \(A\), and similarly for \(Y\) and \(Z\). Points \(L\), \(M\) and \(N\) are the centres of rotational symmetry of triangles \(BXC\), \(CY\!A\) and \(AZB\) respectively.

  1. Show that \(| CM| = \dfrac {\ b} {\sqrt3} \,\) and write down the corresponding expression for \(| CL|\).
  2. Use the cosine rule to show that \[ 6 \left| LM \right|^2 = a^2+b^2+c^2 + 4\sqrt3 \, \Delta \,, \] where \(\Delta\) is the area of triangle \(ABC\). Deduce that \(LMN\) is an equilateral triangle. Show further that the areas of triangles \(LMN\) and \(ABC\) are equal if and only if \[ a^2+b^2 +c^2 = 4\sqrt3 \, \Delta \,. \]
  3. Show that the conditions \[ (a -b)^2 = -2ab \big( 1 -\cos(C-60^\circ)\big) \,\] and \[ a^2+b^2 +c^2 = 4\sqrt3 \, \Delta \] are equivalent. Deduce that the areas of triangles \(LMN\) and \(ABC\) are equal if and only if \(ABC\) is equilateral.

2017 Paper 1 Q8
D: 1500.0 B: 1516.0

Two sequences are defined by \(a_1 = 1\) and \(b_1 = 2\) and, for \(n \ge 1\), \begin{equation*} \begin{split} a_{n+1} & = a_n+ 2b_n \,, \\ b_{n+1} & = 2a_n + 5b_n \,. \end{split} \end{equation*} Prove by induction that, for all \(n \ge 1\), \[ a_n^2+2a_nb_n - b_n^2 = 1 \,. \tag{\(*\)}\]

  1. Let \(c_n = \dfrac{a_n}{b_n}\)\,. Show that \(b_n \ge 2 \times 5^{n-1}\) and use \((*)\) to show that \[ c_n \to \sqrt 2 -1 \text{ as } n\to\infty\,. \]
  2. Show also that \(c_n > \sqrt2 -1\) and hence that $\dfrac2 {c_n+1}<\sqrt2

2017 Paper 1 Q9
D: 1484.0 B: 1484.0

A particle is projected at speed \(u\) from a point \(O\) on a horizontal plane. It passes through a fixed point \(P\) which is at a horizontal distance \(d\) from \(O\) and at a height \(d \tan \beta\) above the plane, where \(d>0\) and \(\beta \) is an acute angle. The angle of projection \(\alpha\) is chosen so that \(u\) is as small as possible.

  1. Show that \(u^2 = gd \tan \alpha\) and \(2\alpha = \beta + 90^\circ\,\).
  2. At what angle to the horizontal is the particle travelling when it passes through~\(P\)? Express your answer in terms of \(\alpha\) in its simplest form.

2017 Paper 1 Q10
D: 1500.0 B: 1484.0

Particles \(P_1\), \(P_2\), \(\ldots\) are at rest on the \(x\)-axis, and the \(x\)-coordinate of \(P_n\) is \(n\). The mass of \(P_n\) is \(\lambda^nm\). Particle \(P\), of mass \(m\), is projected from the origin at speed \(u\) towards \(P_1\). A series of collisions takes place, and the coefficient of restitution at each collision is \(e\), where \(0 < e <1\). The speed of \(P_n\) immediately after its first collision is \(u_n\) and the speed of \(P_n\) immediately after its second collision is \(v_n\). No external forces act on the particles.

  1. Show that \(u_1=\dfrac{1+e}{1+\lambda}\, u\) and find expressions for \(u_n\) and \(v_n\) in terms of \(e\), \(\lambda\), \(u\) and \(n\).
  2. Show that, if \(e > \lambda\), then each particle (except \(P\)) is involved in exactly two collisions.
  3. Describe what happens if \(e=\lambda\) and show that, in this case, the fraction of the initial kinetic energy lost approaches \(e\) as the number of collisions increases.
  4. Describe what happens if \(\lambda e=1\). What fraction of the initial kinetic energy is \mbox{eventually} lost in this case?


Solution:

  1. TikZ diagram
    \begin{align*} \text{COM}: && mu &= mv + \lambda m u_1 \\ \Rightarrow && u &= v + \lambda u_1 \tag{1} \\ \text{NEL}: && e &= \frac{u_1-v}{u} \\ \Rightarrow && eu &= u_1 - v \tag{2} \\ (1)+(2) && (1+e)u &= (1+\lambda) u_1 \\ \Rightarrow && u_1 &= \frac{1+e}{1+\lambda}u \\ && v &= u_1 - eu \\ &&&= \frac{1+e - (1+\lambda)e}{1+\lambda} u \\ &&&= \frac{1-\lambda e}{1+\lambda}u \end{align*} Note that subsequent (first (and second)) are the same as these, therefore: \begin{align*} u_n &= \left ( \frac{1+e}{1+\lambda} \right)^n u \\ v_n &= \frac{1-\lambda e}{1+\lambda } u_n \\ &= \frac{1-\lambda e}{1+\lambda } \left ( \frac{1+e}{1+\lambda} \right)^n u \end{align*}
  2. If \(e > \lambda\) then \((1-\lambda e) > 1-e^2 > 0\) and \begin{align*} \frac{v_{n+1}}{v_n} &= \frac{1+e}{1+\lambda} > 1 \end{align*} So the particles are moving away from each other - hence no more collisions.
  3. If \(e = \lambda\) then \(u_n = u\) and \(v_n = (1-\lambda)u\) so all the particles end up moving at the same speed. \begin{align*} \text{initial k.e.} &= \frac12 m u^2 \\ \text{final k.e.} &= \frac12 m((1-e)u)^2 + \sum_{n = 1}^{\infty} \frac12 \lambda^n m ((1-e)u)^2 \\ &= \frac12mu^2(1-e)^2 \left ( \sum_{n=0}^{\infty} e^n \right) \tag{\(e = \lambda\)} \\ &= \frac12 mu^2(1-e)^2 \frac{1}{1-e} \\ &= \frac12m u^2 (1-e) \\ \text{change in k.e.} &= \frac12 m u^2 - \frac12m u^2 (1-e) \\ &= e\frac12m u^2 \end{align*} Ie the total energy lost approaches a fraction of \(e\).
  4. If \(\lambda e = 1\), after the second collision the particle will be stationary. ie \begin{align*} \text{initial k.e.} &= \frac12 m u^2 \\ \text{k.e. after }n\text{ collisions} &= \frac12 \lambda^n m \left (\left ( \frac{1+e}{1+\lambda} \right)^n u \right)^2\\ &= \frac12 \lambda^n m \left ( \frac{1+\frac1{\lambda}}{1+\lambda} \right)^{2n} u&2\\ &= \frac12 \lambda^n m \left ( \frac{1+\frac1{\lambda}}{1+\lambda} \right)^{2n} u\\ &= \frac12 \lambda^n m \left ( \frac{1}{\lambda} \right)^{2n} u\\ &= \frac12 m \lambda^{-n} u\\ &\to 0 \end{align*} Eventually we lose all the kinetic energy.