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2007 Paper 2 Q2
D: 1600.0 B: 1500.0

A curve has equation \(y=2x^3-bx^2+cx\). It has a maximum point at \((p,m)\) and a minimum point at \((q,n)\) where \(p>0\) and \(n>0\). Let \(R\) be the region enclosed by the curve, the line \(x=p\) and the line \(y=n\).

  1. Express \(b\) and \(c\) in terms of \(p\) and \(q\).
  2. Sketch the curve. Mark on your sketch the point of inflection and shade the region \(R\). Describe the symmetry of the curve.
  3. Show that \(m-n=(q-p)^3\).
  4. Show that the area of \(R\) is \(\frac12 (q-p)^4\).


Solution:

  1. \(\,\) \begin{align*} && y &= 2x^3-bx^2+cx \\ \Rightarrow && y' &= 6x^2-2bx+c \end{align*} We must have \(p, q\) are the roots of this equation, ie \(\frac13b = p+q, \frac16c = pq\)
  2. The point of inflection will be at \(\frac{b}6\) The graph will have rotational symmetry of \(180^{\circ}\) about the point of inflection.
    TikZ diagram
  3. \begin{align*} && m-n &= 2(p^3-q^3)-b(p^2-q^2)+c(p-q) \\ &&&= (p-q)(2(p^2+qp+q^2)-b(p+q)+c) \\ &&&= (p-q)(2(p^2+qp+q^2)-3(p+q)^2+6pq) \\ &&&= (p-q)(-p^2-q^2+2pq) \\ &&&= (q-p)^3 \end{align*}
  4. The area of \(R\) is \begin{align*} A &= \frac12 bh \\ &= \frac12 (q-p)(m-n) = \frac12(q-p)^4 \end{align*} as required.

2007 Paper 2 Q7
D: 1600.0 B: 1516.0

A function \(\f(x)\) is said to be concave on some interval if \(\f''(x)<0\) in that interval. Show that \(\sin x\) is concave for \(0< x < \pi\) and that \(\ln x\) is concave for \(x > 0\). Let \(\f(x)\) be concave on a given interval and let \(x_1\), \(x_2\), \(\ldots\), \(x_n\) lie in the interval. Jensen's inequality states that \[ \frac1 n \sum_{k=1}^n\f(x_k) \le \f \bigg (\frac1 n \sum_{k=1}^n x_k\bigg) \] and that equality holds if and only if \(x_1=x_2= \cdots =x_n\). You may use this result without proving it.

  1. Given that \(A\), \(B\) and \(C\) are angles of a triangle, show that \[ \sin A + \sin B + \sin C \le \frac{3\sqrt3}2 \,. \]
  2. By choosing a suitable function \(\f\), prove that \[ \sqrt[n]{t_1t_2\cdots t_n}\; \le \; \frac{t_1+t_2+\cdots+t_n}n \] for any positive integer \(n\) and for any positive numbers \(t_1\), \(t_2\), \(\ldots\), \(t_n\). Hence:
    1. show that \(x^4+y^4+z^4 +16 \ge 8xyz\), where \(x\), \(y\) and \(z\) are any positive numbers;
    2. find the minimum value of \(x^5+y^5+z^5 -5xyz\), where \(x\), \(y\) and \(z\) are any positive numbers.


Solution: \begin{align*} && f(x) &= \sin x \\ \Rightarrow && f''(x) &= -\sin x \end{align*} which is clearly negative on \((0,\pi)\) since \(\sin\) is positive on this interval. \begin{align*} && f(x) &= \ln x \\ \Rightarrow && f''(x) &= -1/x^2 \end{align*} which is clearly negative for \(x > 0\)

  1. Since \(A,B,C\) are angles in a triangle, we must have \(0 < A,B,C< \pi\) and so we can apply Jensen with \(f = \sin\) to obtain: \begin{align*} &&\frac13( \sin A + \sin B + \sin C) &\leq \sin \left ( \frac{A+B+C}{3}\right) \\ &&&= \sin \frac{\pi}{3} = \frac{\sqrt{3}}2 \\ \Rightarrow && \sin A + \sin B + \sin C &\leq\frac{3\sqrt{3}}2 \end{align*}
  2. Suppose \(f(x) = \ln x\), then applying Jensen on the positive numbers \(t_1, \ldots, t_n\) we obtain \begin{align*} && \frac1n \left ( \sum_{i=1}^n \ln t_n \right) &\leq \ln \left ( \frac1n\sum_{i=1}^n t_n \right) \\ \Rightarrow && \frac1n \ln\left (\prod_{i=1} t_n\right)&\leq \ln \left ( \frac1n\sum_{i=1}^n t_n \right) \\ \Rightarrow && \ln\left (\left (\prod_{i=1} t_n\right)^{1/n}\right)&\leq \ln \left ( \frac1n\sum_{i=1}^n t_n \right) \\ \Rightarrow && \left (\prod_{i=1} t_n\right)^{1/n}&\leq\frac1n\sum_{i=1}^n t_n \\ \Rightarrow && \sqrt[n]{t_1t_2 \cdots t_n}&\leq\frac1n(t_1 + t_2 + \cdots + t_n) \tag{AM-GM}\\ \end{align*}
    1. Applying AM-GM with \(t_1 = x^4, t_2 = y^4, t_3 = z^4, t_4 = 2^4\) we have \begin{align*} && \frac{x^4+y^4+z^4+16}{4} & \geq \sqrt[4]{x^4y^4z^42^4} \\ \Rightarrow && x^4+y^4+z^4+16 &\geq 8xyz \end{align*}
    2. Applying AM-GM with \(t_1 = x^5, t_2 = y^5, t_3 = z^5, t_4 = 1^5, t_5 = 1^5\) we have \begin{align*} && \frac{x^5+y^5+z^5+1+1}{5} & \geq \sqrt[5]{x^5y^5z^5} \\ \Rightarrow && x^5+y^5+z^5+2 &\geq 5xyz \\ \Rightarrow && x^5+y^5+z^5 - 5xyz &\geq -2 \end{align*} Therefore the minimum is \(-2\)

2007 Paper 2 Q9
D: 1600.0 B: 1500.0

A solid right circular cone, of mass \(M\), has semi-vertical angle \(\alpha\) and smooth surfaces. It stands with its base on a smooth horizontal table. A particle of mass \(m\) is projected so that it strikes the curved surface of the cone at speed \(u\). The coefficient of restitution between the particle and the cone is \(e\). The impact has no rotational effect on the cone and the cone has no vertical velocity after the impact.

  1. The particle strikes the cone in the direction of the normal at the point of impact. Explain why the trajectory of the particle immediately after the impact is parallel to the normal to the surface of the cone. Find an expression, in terms of \(M\), \(m\), \(\alpha\), \(e\) and \(u\), for the speed at which the cone slides along the table immediately after impact.
  2. If instead the particle falls vertically onto the cone, show that the speed \(w\) at which the cone slides along the table immediately after impact is given by \[ w= \frac{mu(1+e)\sin\alpha\cos\alpha}{M+m\cos^2\alpha}\,. \] Show also that the value of \(\alpha\) for which \(w\) is greatest is given by \[ \cos \alpha = \sqrt{ \frac{M}{2M+m}}\ . \]

2007 Paper 2 Q11
D: 1600.0 B: 1500.0

{\sl In this question take the acceleration due to gravity to be \(10\,{\rm m \,s}^{-2}\) and neglect air resistance.} The point \(O\) lies in a horizontal field. The point \(B\) lies \(50\,\)m east of \(O\). A particle is projected from \(B\) at speed \(25\,{\rm m\,s}^{-1}\) at an angle \(\arctan \frac12\) above the horizontal and in a direction that makes an angle \(60^\circ\) with \(OB\); it passes to the north of \(O\).

  1. Taking unit vectors \(\mathbf i\), \(\mathbf j\) and \(\mathbf k\) in the directions east, north and vertically upwards, respectively, find the position vector of the particle relative to \(O\) at time \(t\)~seconds after the particle was projected, and show that its distance from \(O\) is \[ 5(t^2- \sqrt5 t +10)\, {\rm m}. \] When this distance is shortest, the particle is at point \(P\). Find the position vector of \(P\) and its horizontal bearing from \(O\).
  2. Show that the particle reaches its maximum height at \(P\).
  3. When the particle is at \(P\), a marksman fires a bullet from \(O\) directly at \(P\). The initial speed of the bullet is \(350\,{\rm m\,s}^{-1}\). Ignoring the effect of gravity on the bullet show that, when it passes through \(P\), the distance between \(P\) and the particle is approximately~\(3\,\)m.

2007 Paper 3 Q10
D: 1700.0 B: 1500.7

A particle is projected from a point on a plane that is inclined at an angle~\(\phi\) to the horizontal. The position of the particle at time \(t\) after it is projected is \((x,y)\), where \((0,0)\) is the point of projection, \(x\) measures distance up the line of greatest slope and \(y\) measures perpendicular distance from the plane. Initially, the velocity of the particle is given by \((\dot x, \dot y) = (V\cos\theta, V\sin\theta)\), where \(V>0\) and \(\phi+\theta<\pi/2\,\). Write down expressions for \(x\) and \(y\). The particle bounces on the plane and returns along the same path to the point of projection. Show that \[2\tan\phi\tan\theta =1\] and that \[ R= \frac{V^2\cos^2\theta}{2g\sin\phi}\,, \] where \(R\) is the range along the plane. Show further that \[ \frac{2V^2}{gR} = 3\sin\phi + {\rm cosec}\,\phi \] and deduce that the largest possible value of \(R\) is \(V^2/ (\sqrt{3}\,g)\,\).

2006 Paper 1 Q2
D: 1516.0 B: 1500.0

A small goat is tethered by a rope to a point at ground level on a side of a square barn which stands in a large horizontal field of grass. The sides of the barn are of length \(2a\) and the rope is of length \(4a\). Let \(A\) be the area of the grass that the goat can graze. Prove that \(A\le14\pi a^2\) and determine the minimum value of \(A\).


Solution:

TikZ diagram
The areas are \(8\pi a^2 + \frac14 \pi (4a-x)^2 + \frac14 \pi (2a-x)^2 + \frac14\pi(2a+x)^2+\frac14 \pi x^2\) ie \begin{align*} A &= \frac{\pi}{4} \left ( x^2 \left (1 + 1 + 1 + 1 \right) + x \left (4a-4a-8a \right)+\left (32a^2+16a^2+4a^2+4a^2 \right)\right) \\ &= \frac{\pi}{4} \left (4x^2-8ax+56a^2 \right) \\ &= \pi(x^2-2ax+14a^2) \\ &= \pi ((x-a)^2+13a^2) \end{align*} Since \(x \in [0, 2a]\) we have \(13\pi a^2 \leq A \leq 14 \pi a^2\)

2006 Paper 1 Q4
D: 1500.0 B: 1514.2

By sketching on the same axes the graphs of \(y=\sin x\) and \(y=x\), show that, for \(x>0\):

  1. \(x>\sin x\,\);
  2. \(\dfrac {\sin x} {x} \approx 1\) for small \(x\).
A regular polygon has \(n\) sides, and perimeter \(P\). Show that the area of the polygon is \[ \displaystyle \frac{P^2} { {4n \tan \l\dfrac{ \pi} { n} \r}} \;. \] Show by differentiation (treating \(n\) as a continuous variable) that the area of the polygon increases as \(n\) increases with \(P\) fixed. Show also that, for large \(n\), the ratio of the area of the polygon to the area of the smallest circle which can be drawn around the polygon is approximately \(1\).

2006 Paper 1 Q9
D: 1500.0 B: 1484.0

A block of mass \(4\,\)kg is at rest on a smooth, horizontal table. A smooth pulley \(P\) is fixed to one edge of the table and a smooth pulley \(Q\) is fixed to the opposite edge. The two pulleys and the block lie in a straight line. Two horizontal strings are attached to the block. One string runs over pulley \(P\); a particle of mass \(x\,\)kg hangs at the end of this string. The other string runs over pulley \(Q\); a particle of mass \(y\,\)kg hangs at the end of this string, where \(x > y\) and \(x + y = 6\,\). The system is released from rest with the strings taut. When the \(4\,\)kg block has moved a distance \(d\), the string connecting it to the particle of mass \(x\,\)kg is cut. Show that the time taken by the block from the start of the motion until it first returns to rest (assuming that it does not reach the edge of the table) is \(\sqrt{d/(5g)\,} \,\f(y)\), where \[ \f(y)= \frac{10}{ \sqrt{6-2y}}+ \left(1 + \frac{4}{ y} \right) \sqrt {6 -2y}. \] Calculate the value of \(y\) for which \(\f'(y)=0\).

2006 Paper 1 Q10
D: 1484.0 B: 1500.0

A particle \(P\) is projected in the \(x\)-\(y\) plane, where the \(y\)-axis is vertical and the \(x\)-axis is horizontal. The particle is projected with speed \(V\) from the origin at an angle of \(45 ^\circ\) above the positive \(x\)-axis. Determine the equation of the trajectory of \(P\). The point of projection (the origin) is on the floor of a barn. The roof of the barn is given by the equation \(y= x \tan \alpha +b\,\), where \(b>0\) and \(\alpha\) is an acute angle. Show that, if the particle just touches the roof, then \(V(-1+ \tan\alpha) =-2 \sqrt{bg}\); you should justify the choice of the negative root. If this condition is satisfied, find, in terms of \(\alpha\), \(V\) and \(g\), the time after projection at which touching takes place. A particle \(Q\) can slide along a smooth rail fixed, in the \(x\)-\(y\) plane, to the under-side of the roof. It is projected from the point \((0,b)\) with speed \(U\) at the same time as \(P\) is projected from the origin. Given that the particles just touch in the course of their motions, show that \[ 2 \sqrt 2 \, U \cos \alpha = V \big(2 + \sin\alpha\cos\alpha -\sin^2\alpha) . \]

2006 Paper 1 Q14
D: 1500.0 B: 1502.6

  1. A bag of sweets contains one red sweet and \(n\) blue sweets. I take a sweet from the bag, note its colour, return it to the bag, then shake the bag. I repeat this until the sweet I take is the red one. Find an expression for the probability that I take the red sweet on the \(r\)th attempt. What value of \(n\) maximises this probability?
  2. Instead, I take sweets from the bag, without replacing them in the bag, until I take the red sweet. Find an expression for the probability that I take the red sweet on the \(r\)th attempt. What value of \(n\) maximises this probability?


Solution:

  1. This is the probability of having the sequence \(\underbrace{BB\cdots B}_{r-1 \text{ times}}R\) which has probability \(\displaystyle \left ( \frac{n}{n+1} \right)^{r-1}\frac{1}{n+1}\). Maximising this, is equivalent to maximising \(\log\) of it, ie \begin{align*} && y &= (r-1) \ln n - r \ln (n+1) \\ \Rightarrow && \frac{\d y}{\d n} &= \frac{r-1}{n} - \frac{r}{n+1} \\ &&&= \frac{(r-1)(n+1)-rn}{n(n+1)} \\ &&&= \frac{r-n-1}{n(n+1)} \end{align*} Therefore this is maximised when \(n = r-1\)

2006 Paper 2 Q11
D: 1600.0 B: 1484.0

A projectile of unit mass is fired in a northerly direction from a point on a horizontal plain at speed \(u\) and an angle \(\theta\) above the horizontal. It lands at a point \(A\) on the plain. In flight, the projectile experiences two forces: gravity, of magnitude \(g\); and a horizontal force of constant magnitude \(f\) due to a wind blowing from North to South. Derive an expression, in terms of \(u\), \(g\), \(f\) and \(\theta\) for the distance \(OA\).

  1. Determine the angle \(\alpha\) such that, for all \(\theta>\alpha\), the wind starts to blow the projectile back towards \(O\) before it lands at \(A\).
  2. An identical projectile, which experiences the same forces, is fired from \(O\) in a northerly direction at speed \(u\) and angle \(45^\circ\) above the horizontal and lands at a point \(B\) on the plain. Given that \(\theta\) is chosen to maximise \(OA\), show that \[ \frac{OB}{OA} = \frac{ g-f}{\; \sqrt{g^2+f^2\;}- f \;\;}\;. \] Describe carefully the motion of the second projectile when \(f=g\).

2006 Paper 3 Q12
D: 1700.0 B: 1500.0

Fifty times a year, 1024 tourists disembark from a cruise liner at a port. From there they must travel to the city centre either by bus or by taxi. Tourists are equally likely to be directed to the bus station or to the taxi rank. Each bus of the bus company holds 32 passengers, and the company currently runs 15 buses. The company makes a profit of \(\pounds\)1 for each passenger carried. It carries as many passengers as it can, with any excess being (eventually) transported by taxi. Show that the largest annual licence fee, in pounds, that the company should consider paying to be allowed to run an extra bus is approximately \[ 1600 \Phi(2) - \frac{800}{\sqrt{2\pi}}\big(1- \e^{-2}\big)\,, \] where \(\displaystyle \Phi(x) =\dfrac1{\sqrt{2\pi}} \int_{-\infty}^x \e^{-\frac12t^2}\d t\,\). You should not consider continuity corrections.


Solution: The the number of people being directed towards the buses (each cruise) is \(X \sim B(1024, \tfrac12) \approx N(512, 256) \approx 16Z + 512\). Therefore without an extra bus, the expected profit is \(\mathbb{E}[\min(X, 15 \times 32)]\). With the extra bus, the extra profit is \(\mathbb{E}[\min(X, 16 \times 32)]\), therefore the expected extra profit is: \(\mathbb{E}[\min(X, 16 \times 32)]-\mathbb{E}[\min(X, 15 \times 32)] = \mathbb{E}[\min(X, 16 \times 32)-\min(X, 15 \times 32)] \) \begin{align*} \text{Expected extra profit} &= \mathbb{E}[\min(X, 16 \times 32)-\min(X, 15 \times 32)] \\ &= \mathbb{E}[\min(16Z+512, 16 \times 32)-\min(16Z+512, 15 \times 32)] \\ &= 16\mathbb{E}[\min(Z+32, 32)-\min(Z+32, 30)] \\ &=16\int_{-\infty}^{\infty} \left (\min(Z+32, 32)-\min(Z+32, 30) \right)p_Z(z) \d z \\ &= 16 \left ( \int_{-2}^{0} (z+32-30) p_Z(z) \d z + \int_0^\infty (32-30)p_Z(z) \d z \right) \\ &= 16 \left ( \int_{-2}^{0} (z+2) p_Z(z) \d z + \int_0^\infty 2p_Z(z) \d z \right) \\ &= 16 \left ( \int_{-2}^{0} zp_Z(z) \d z + 2\int_{-2}^\infty p_Z(z) \d z \right) \\ &= 16 \left ( \int_{-2}^{0} z \frac{1}{\sqrt{2\pi}} e^{-\frac12 z^2} \d z + 2(1-\Phi(2)) \right) \\ &= 32(1-\Phi(2)) + \frac{16}{\sqrt{2\pi}} \left [ -e^{-\frac12z^2} \right]_{-2}^0 \\ &= 32(1-\Phi(2)) - \frac{16}{\sqrt{2\pi}} \left ( 1-e^{-2}\right) \end{align*} Across \(50\) different runs, this profit is \[ 1600(1-\Phi(2)) - \frac{800}{\sqrt{2\pi}} \left ( 1-e^{-2}\right) \]

2005 Paper 1 Q11
D: 1500.0 B: 1485.7

A particle moves so that \({\bf r}\), its displacement from a fixed origin at time \(t\), is given by \[{\bf r} = \l \sin{2t} \r {\bf i} + \l 2\cos t \r \bf{j}\,,\] where \(0 \le t < 2\pi\).

  1. Show that the particle passes through the origin exactly twice.
  2. Determine the times when the velocity of the particle is perpendicular to its displacement.
  3. Show that, when the particle is not at the origin, its velocity is never parallel to its displacement.
  4. Determine the maximum distance of the particle from the origin, and sketch the path of the particle.


Solution:

  1. It is at the origin when both \(\sin 2t\) and \(\cos t = 0\), but this \(\sin 2t = 2 \sin t \cos t\) so this happens precisely when \(\cos t = 0\), ie when \(t = \frac{\pi}{2}, \frac{3\pi}{2}\)
  2. \(\,\) \begin{align*} && \dot{\mathbf{r}} &= 2 \cos 2t \mathbf{i} - 2 \sin t \mathbf{j} \\ && \mathbf{r} \cdot \dot{\mathbf{r}} &= 2\cos 2t \sin 2t - 2 \sin t 2 \cos t \\ &&&= \sin 2t \left (2\cos 2t - 2 \right) \end{align*} Therefore they are perpendicular when \(\sin 2t = 0 \Rightarrow t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\) and when \(\cos 2t = 1 \Rightarrow 2t = 0, 2\pi, 4\pi \Rightarrow t = 0, \pi, 2\pi\), therefore all solutions are \( t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\)
  3. For \(\mathbf{r}\) and \(\dot{\mathbf{r}}\) to be parallel, we would need \begin{align*} && \frac{2 \cos 2t}{\sin 2t} &= \frac{-2 \sin t}{2 \cos t}\\ && 2 \cos 2t \cos t &= - \sin t \sin 2t \\ && 0 &= 2\cos t (\cos 2t + \sin ^2 t) \\ &&&= 2 \cos t (\cos^2 t) \\ &&&= 2 \cos^3 t \end{align*} Therefore the only time we can be parallel is when \(\cos t = 0\), which is when we are at the origin.
  4. \(\frac{\d }{\d t} (\mathbf{r} \cdot \mathbf{r}) = 2 \mathbf{r} \cdot \mathbf{\dot{r}}\) so we should check the values when velocity and displacement are perpendicular, ie \( t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\) which have values \(\mathbf{r} = \binom{0}{2}, \binom{0}{0}, \binom{0}{-2}, \binom{0}{0}, \binom{0}{2}\). Therefore the maximum distance is \(2\).
    TikZ diagram

2005 Paper 2 Q5
D: 1600.0 B: 1470.1

The angle \(A\) of triangle \(ABC\) is a right angle and the sides \(BC\), \(CA\) and \(AB\) are of lengths \(a\), \(b\) and \(c\), respectively. Each side of the triangle is tangent to the circle \(S_1\) which is of radius \(r\). Show that \(2r = b+c-a\). Each vertex of the triangle lies on the circle~\(S_2\). The ratio of the area of the region between~\(S_1\) and the triangle to the area of \(S_2\) is denoted by \(R\,\). Show that $$ \pi R = -(\pi-1)q^2 + 2\pi q -(\pi+1) \;, $$ where \(q=\dfrac{b+c}a\,\). Deduce that $$ R\le \frac1 {\pi( \pi - 1)} \;. $$

2005 Paper 2 Q9
D: 1600.0 B: 1484.0

Two particles, \(A\) and \(B\), of masses \(m\) and \(2m\), respectively, are placed on a line of greatest slope, \(\ell\), of a rough inclined plane which makes an angle of \(30^{\circ}\) with the horizontal. The coefficient of friction between \(A\) and the plane is \(\frac16\sqrt{3}\) and the coefficient of friction between \(B\) and the plane is \(\frac13 \sqrt{3}\). The particles are at rest with \(B\) higher up \(\ell\) than \(A\) and are connected by a light inextensible string which is taut. A force \(P\) is applied to \(B\).

  1. Show that the least magnitude of \(P\) for which the two particles move upwards along \(\ell\) is \(\frac{11}8 \sqrt{3}\, mg\) and give, in this case, the direction in which \(P\) acts.
  2. Find the least magnitude of \(P\) for which the particles do not slip downwards along~\(\ell\).