Problems

Filters
Clear Filters
2014 Paper 1 Q7
D: 1516.0 B: 1500.0

In the triangle \(OAB\), the point \(D\) divides the side \(BO\) in the ratio \(r:1\) (so that \(BD = rDO\)), and the point \(E\) divides the side \(OA\) in the ratio \(s:1\) (so that \(OE =s EA\)), where \(r\) and \(s\) are both positive.

  1. The lines \(AD\) and \(BE\) intersect at \(G\). Show that \[ \+g= \frac{rs}{1+r+rs} \, \+a + \frac 1 {1+r+rs} \, \+b \,, \] where \+a, \+b and \+g are the position vectors with respect to \(O\) of \(A\), \(B\) and \(G\), respectively.
  2. The line through \(G\) and~\(O\) meets \(AB\) at \(F\). Given that \(F\) divides \(AB\) in the ratio \(t:1\), find an expression for~\(t\) in terms of \(r\) and~\(s\).

2014 Paper 1 Q8
D: 1500.0 B: 1484.0

Let \(L_a\) denote the line joining the points \((a,0)\) and \((0, 1-a)\), where \(0< a < 1\). The line~\(L_b\) is defined similarly.

  1. Determine the point of intersection of \(L_a\) and \(L_b\), where \(a\ne b\).
  2. Show that this point of intersection, in the limit as $b\to a\(, lies on the curve~\)C$ given by \[ y=(1-\sqrt x)^2\, \ \ \ \ (0< x < 1)\,. \]
  3. Show that every tangent to~\(C\) is of the form~\(L_a\) for some~\(a\).

2014 Paper 1 Q9
D: 1516.0 B: 1500.0

A particle of mass \(m\) is projected due east at speed \(U\) from a point on horizontal ground at an angle \(\theta\) above the horizontal, where $0<\theta< 90^\circ\(. In addition to the gravitational force \)mg$, it experiences a horizontal force of magnitude \(mkg\), where \(k\) is a positive constant, acting due west in the plane of motion of the particle. Determine expressions in terms of \(U\), \(\theta\) and~\(g\) for the time, \(T_H\), at which the particle reaches its greatest height and the time, \(T_L \), at which it lands. Let \(T = U\cos\theta /(kg)\). By considering the relative magnitudes of \(T_H\), \(T_L \) and \(T\), or otherwise, sketch the trajectory of the particle in the cases \(k\tan\theta<\frac12\), \ \ \(\frac12 < k\tan\theta<1\), and \(k\tan\theta>1\). What happens when \(k\tan\theta =1\)?

2014 Paper 1 Q10
D: 1500.0 B: 1500.0

  1. A uniform spherical ball of mass \(M\) and radius \(R\) is released from rest with its centre a distance \(H+R\) above horizontal ground. The coefficient of restitution between the ball and the ground is \(e\). Show that, after bouncing, the centre of the ball reaches a height \(R+He^2\) above the ground.
  2. A second uniform spherical ball, of mass \(m\) and radius \(r\), is now released from rest together with the first ball (whose centre is again a distance \(H+R\) above the ground when it is released). The two balls are initially one on top of the other, with the second ball (of mass \(m\)) above the first. The two balls separate slightly during their fall, with their centres remaining in the same vertical line, so that they collide immediately after the first ball has bounced on the ground. The coefficient of restitution between the balls is also \(e\). The centre of the second ball attains a height \(h\) above the ground. Given that \(R=0.2\), \(r=0.05\), \(H=1.8\), \(h=4.5\) and \(e=\frac23\), determine the value of \(M/m\).


Solution:

  1. The story for first ball (before it collides with the second ball) will be exactly the same, ie it rebounds with speed \(eV\). For the second ball, it will also have fallen a distance \(H\) and will be travelling with the same speed \(V\). Their speed of approach therefore will be \((1+e)V\), and the speed of separating therefore must be \(e(1+e)V\) Given the centre of the second ball reaches a height of \(h\) (from a position of height) \(2R+r\), we must have: \begin{align*} && v^2 &= u^2 + 2as \\ && 0 &= w^2 - 2g(h - 2R-r) \\ \Rightarrow && w^2 &= 2g(h-2R-r) \end{align*} Taking upwards to be positive, then we have:
    TikZ diagram
    \begin{align*} \text{COM}: && MeV + m(-V) &= M(w-e(1+e)V) + mw \\ \Rightarrow && V(Me-m)+e(1+e)MV &= w(M+m) \\ \Rightarrow && w &= \frac{2Me+e^2M-m}{M+m} V \\ \Rightarrow && w^2 &= \left ( \frac{M/m(2e+e^2)-1}{M/m+1}\right)^2 V^2 \\ \Rightarrow && 2g(h-2R-r) &= \left ( \frac{M/m(2e+e^2)-1}{M/m+1}\right)^2 2gH \\ \Rightarrow && \left ( \frac{M/m(2e+e^2)-1}{M/m+1}\right)^2 &= \frac{h-2R-r}{H} \\ &&&= \frac{4.5-0.4-0.05}{1.8} \\ &&&= \frac{9}{4} \\ \Rightarrow && \frac{M/m(\frac43+\frac49)-1}{M/m+1} &= \frac32 \\ \Rightarrow && \frac{16}9M/m-1 &= \frac32 M/m+\frac32 \\ \Rightarrow && \frac{5}{18}M/m &= \frac{5}{2} \\ \Rightarrow && M/m &= 9 \end{align*}

2014 Paper 1 Q11
D: 1500.0 B: 1500.0

The diagrams below show two separate systems of particles, strings and pulleys. In both systems, the pulleys are smooth and light, the strings are light and inextensible, the particles move vertically and the pulleys labelled with \(P\) are fixed. The masses of the particles are as indicated on the diagrams.

\psset{xunit=0.7cm,yunit=0.7cm,algebraic=true,dimen=middle,dotstyle=o,dotsize=3pt 0,linewidth=0.3pt,arrowsize=3pt 2,arrowinset=0.25} \begin{pspicture*}(-0.6,-5.31)(12.33,7.94) \pspolygon(1,-3)(1,-2)(0,-2)(0,-3) \pspolygon(4.01,-1)(4.01,0)(3.01,0)(3.01,-1) \pspolygon(10.3,-1)(10.3,0)(9.3,0)(9.3,-1) \pspolygon(11.76,-3)(11.76,-2)(10.76,-2)(10.76,-3) \pspolygon(8,0)(8,1)(7,1)(7,0) \pscircle(2,6){1.05} \pscircle(9,6){1.05} \psline(3.5,6)(3.49,0) \psline(0.5,6.01)(0.5,-2) \psline(1,-3)(1,-2) \psline(1,-2)(0,-2) \psline(0,-2)(0,-3) \psline(0,-3)(1,-3) \psline(4.01,-1)(4.01,0) \psline(4.01,0)(3.01,0) \psline(3.01,0)(3.01,-1) \psline(3.01,-1)(4.01,-1) \psline(7.5,5.99)(7.51,1.01) \psline(10.5,6)(10.48,3.98) \pscircle(10.5,3.25){0.51} \psline(9.77,3.26)(9.77,-0.01) \psline(11.23,3.26)(11.23,-2) \psline(10.3,-1)(10.3,0) \psline(10.3,0)(9.3,0) \psline(9.3,0)(9.3,-1) \psline(9.3,-1)(10.3,-1) \psline(11.76,-3)(11.76,-2) \psline(11.76,-2)(10.76,-2) \psline(10.76,-2)(10.76,-3) \psline(10.76,-3)(11.76,-3) \psline(8,0)(8,1) \psline(8,1)(7,1) \psline(7,1)(7,0) \psline(7,0)(8,0) \rput[tl](0.16,-2.33){\(M\)} \rput[tl](3.2,-0.4){\(m\)} \rput[tl](7.18,0.67){\(M\)} \rput[tl](9.46,-0.4){\(m_1\)} \rput[tl](10.91,-2.4){\(m_2\)} \rput[tl](1.82,6.14){\(P\)} \rput[tl](8.84,6.14){\(P\)} \rput[tl](10.22,3.48){\(P_1\)} \rput[tl](1.06,-3.72){\(\text{System I}\)} \rput[tl](8.11,-3.68){\(\text{System II}\)} \end{pspicture*}
  1. For system I show that the acceleration, \(a_1\), of the particle of mass~\(M\), measured in the downwards direction, is given by \[ a_1= \frac{M-m}{M+m} \, g \,, \] where \(g\) is the acceleration due to gravity. Give an expression for the force on the pulley due to the tension in the string.
  2. For system II show that the acceleration, \(a_2\), of the particle of mass \(M\), measured in the downwards direction, is given by \[ a_2= \frac{ M - 4\mu}{M+4\mu}\,g \,, \] where \(\mu = \dfrac{m_1m_2}{m_1+m_2}\). In the case \(m= m_1+m_2\), show that \(a_1= a_2\) if and only if \(m_1=m_2\).

2014 Paper 1 Q12
D: 1484.0 B: 1441.7

A game in a casino is played with a fair coin and an unbiased cubical die whose faces are labelled \(1, 1, 1, 2, 2\) and \(3.\) In each round of the game, the die is rolled once and the coin is tossed once. The outcome of the round is a random variable \(X\). The value, \(x\), of \(X\) is determined as follows. If the result of the toss is heads then \(x= \vert ks -1\vert\), and if the result of the toss is tails then \(x=\vert k-s\vert\), where \(s\) is the number on the die and \(k\) is a given number. Show that \(\mathbb{E}(X^2) = k +13(k-1)^2 /6\). Given that both \(\mathbb{E}(X^2)\) and \(\mathbb{E}(X)\) are positive integers, and that \(k\) is a single-digit positive integer, determine the value of \(k\), and write down the probability distribution of \(X\). A gambler pays \(\pounds 1\) to play the game, which consists of two rounds. The gambler is paid:

  • \(\pounds w\), where \(w\) is an integer, if the sum of the outcomes of the two rounds exceeds \(25\);
  • \(\pounds 1\) if the sum of the outcomes equals \(25\);
  • nothing if the sum of the outcomes is less that \(25\).
Find, in terms of \(w\), an expression for the amount the gambler expects to be paid in a game, and deduce the maximum possible value of \(w\), given that the casino's owners choose \(w\) so that the game is in their favour.


Solution: \begin{align*} && \mathbb{E}(X^2) &= \frac12 \left (\frac16 \left ( 3(k -1)^2+2(2k-1)^2+(3k-1)^2 \right) +\frac16 \left ( 3(k -1)^2+2(k-2)^2+(k-3)^2 \right) \right) \\ &&&= \frac12 \left (\frac16 \left (20k^2-20k+6 \right) + \frac16 \left ( 6k^2-20k+20\right) \right) \\ &&&= \frac1{12} \left (26k^2-40k+ 26\right) \\ &&&= \frac{13}{6} (k^2+1) - \frac{10}{3}k \\ &&&= \frac{13}{6}(k-1)^2+k \end{align*} Since \(k\) a single digit positive number and \(\mathbb{E}(X^2)\) is an integer, \(6 \mid k-1 \Rightarrow k = 1, 7\). \begin{align*} \mathbb{E}(X | k=1) &= \frac12 \left (\frac16 \left ( 2+2 \right) +\frac16 \left ( 2+2 \right) \right) = \frac23 \not \in \mathbb{Z}\\ \mathbb{E}(X | k=7) &= \frac12 \left (\frac16 \left ( 3\cdot6+2\cdot13+20 \right) +\frac16 \left ( 3\cdot6+2\cdot5+4 \right) \right) = 8 \end{align*} Therefore \(k = 7\) The probability distribution is \begin{align*} && \mathbb{P}(X=4) = \frac1{12} \\ && \mathbb{P}(X=5) = \frac1{6} \\ && \mathbb{P}(X=6) = \frac12 \\ && \mathbb{P}(X=13) = \frac1{6} \\ && \mathbb{P}(X=20)= \frac1{12} \\ \end{align*} The only ways to score more than \(25\) are: \(20+6, 20+13, 20+20, 13+13\) The only ways to score exactly \(25\) are \(20+5\) \begin{align*} \mathbb{P}(>25) &= \frac1{12} \cdot\left(2\cdot \frac12+2\cdot\frac16+\frac1{12}\right) + \frac{1}{6^2} \\ &= \frac{7}{48} \\ \mathbb{P}(=25) &= \frac{2}{12 \cdot 6} = \frac{1}{36} \\ \\ \mathbb{E}(\text{payout}) &= \frac{7}{48}w + \frac{1}{36} = \frac{21w+4}{144} \end{align*} The casino needs \(\frac{21w+4}{144} < 1 \Rightarrow 21w< 140 \Rightarrow w < \frac{20}{3}\)

2014 Paper 1 Q13
D: 1500.0 B: 1483.3

A continuous random variable \(X\) has a triangular distribution, which means that it has a probability density function of the form \[ \f(x) = \begin{cases} \g(x) & \text{for \(a< x \le c\)} \\ \h(x) & \text{for \(c\le x < b\)} \\ 0 & \text{otherwise,} \end{cases} \] where \(\g(x)\) is an increasing linear function with \(\g(a)=0\), \(\h(x)\) is a decreasing linear function with \(\h(b) =0\), and \(\g(c)=\h(c)\). Show that \(\g(x) = \dfrac{2(x-a)}{(b-a)(c-a)}\) and find a similar expression for \(\h(x)\).

  1. Show that the mean of the distribution is \(\frac13(a+b+c)\).
  2. Find the median of the distribution in the different cases that arise.


Solution: Since \(\int f(x) \, dx = 1\), and \(f(x)\) is a triangle with base \(b-a\), it must have height \(\frac{2}{b-a}\) in order to have the desired area. Since \(g(a) = 0, g(c) = \frac{2}{b-a}\), \(g(x) = A(x-a)\) and \(\frac{2}{b-a} = A (c-a) \Rightarrow g(x) = \frac{2(x-a)}{(b-a)(c-a)}\) as required. Similarly, \(h(x) = B(x-b)\) and \(\frac{2}{b-a} = B(c-b) \Rightarrow h(x) = \frac{2(b-x)}{(b-a)(b-c)}\) The mean of the distribution will be: \begin{align*} \int_a^b xf(x) \, dx &= \int_a^c xg(x) \, dx + \int_c^b xh(x) \, dx \\ &= \frac{2}{(b-a)(c-a)} \int_a^c x(x-a) dx + \frac{2}{(b-a)(b-c)} \int_c^b x(b-x) \, dx \\ &= \frac{2}{(b-a)} \l \frac{1}{c-a} \left [ \frac{x^3}{3} - a\frac{x^2}{2} \right ]_a^c + \frac{1}{b-c} \left [ b\frac{x^2}{2} - \frac{x^3}{3} \right ]_c^b\r \\ &= \frac{2}{(b-a)} \l \frac{1}{c-a} \l \frac{c^3}{3} - a\frac{c^2}{2} - \frac{a^3}{3} + \frac{a^3}{2} \r + \frac{1}{b-c} \l \frac{b^3}{2} - \frac{b^3}{3} - \frac{bc^2}{2} + \frac{c^3}{3} \r \r \\ &= \frac{2}{(b-a)} \l \l \frac{c^2+ac+a^2}{3} - \frac{a(a+c)}{2} \r +\l \frac{b(b+c)}{2} - \frac{b^2+bc+c^2}{3} \r\r \\ &= \frac{2}{(b-a)} \l \frac{2c^2+2ac+2a^2}{6} - \frac{3a^2+3ac}{6} + \frac{3b^2+3bc}{6} - \frac{2b^2+2bc+2c^2}{6} \r \\ &= \frac{2}{(b-a)} \l \frac{-a^2+b^2-ac+bc}{6} \r \\ &= \frac{a+b+c}{3} \\ \end{align*} The median \(M\) satisfies: \begin{align*} && \int_a^M f(x) \, dx &= \frac12 \\ \end{align*} The left hand triangle will have area: \(\frac{c-a}{b-a}\) which will be \(\geq \frac12\) if \(c \geq \frac{a+b}{2}\). In this case we need \begin{align*} && \frac{(M-a)^2}{(b-a)(c-a)} &= \frac12 \\ \Rightarrow && M &= a + \sqrt{\frac12 (b-a)(c-a)} \end{align*} Otherwise, we need: \begin{align*} && \frac{(b-M)^2}{(b-a)(b-c)} &= \frac12 \\ \Rightarrow && M &= b - \sqrt{\frac12 (b-a)(b-c)} \end{align*} These are consistent, if \(c = \frac{b+a}{2}\)

2014 Paper 2 Q1
D: 1600.0 B: 1500.0

In the triangle \(ABC\), the base \(AB\) is of length 1 unit and the angles at~\(A\) and~\(B\) are \(\alpha\) and~\(\beta\) respectively, where \(0<\alpha\le\beta\). The points \(P\) and~\(Q\) lie on the sides \(AC\) and \(BC\) respectively, with \(AP=PQ=QB=x\). The line \(PQ\) makes an angle of~\(\theta\) with the line through~\(P\) parallel to~\(AB\).

  1. Show that \(x\cos\theta = 1- x\cos\alpha - x\cos\beta\), and obtain an expression for \(x\sin\theta\) in terms of \(x\), \(\alpha\) and~\(\beta\). Hence show that \begin{equation} \label{eq:2*} \bigl(1+2\cos(\alpha+\beta)\bigr)x^2 - 2(\cos\alpha + \cos\beta)x + 1 = 0\,. \tag{\(*\)} \end{equation} Show that \((*)\) is also satisfied if \(P\) and \(Q\) lie on \(AC\) produced and \(BC\) produced, respectively. [By definition, \(P\) lies on \(AC\) produced if \(P\) lies on the line through \(A\) and~\(C\) and the points are in the order \(A\), \(C\), \(P\)\,.]
  2. State the condition on \(\alpha\) and \(\beta\) for \((*)\) to be linear in \(x\). If this condition does not hold (but the condition \(0<\alpha \le \beta\) still holds), show that \((*)\) has distinct real roots.
  3. Find the possible values of~\(x\) in the two cases (a) \(\alpha = \beta = 45^\circ\) and (b) \(\alpha = 30^\circ\), \(\beta = 90^\circ\), and illustrate each case with a sketch.

2014 Paper 2 Q2
D: 1600.0 B: 1500.0

This question concerns the inequality \begin{equation} \label{eq6:*} \int_0^\pi \bigl( \.f(x) \bigr)^2 \ud x \le \int_0^\pi \bigl( \.f'(x)\bigr)^2 \ud x\,.\tag{\(*\)} \end{equation}

  1. Show that \((*)\) is satisfied in the case \(\.f(x)=\sin nx\), where \(n\)~is a positive integer. Show by means of counterexamples that \((*)\) is not necessarily satisfied if either \(\.f(0) \ne 0\) or \(\.f(\pi)\ne0\).
  2. You may now assume that \((*)\) is satisfied for any (differentiable) function~\(\.f\) for which \(\.f(0)=\.f(\pi)=0\). By setting \(\.f(x) = ax^2 + bx +c\), where \(a\), \(b\) and \(c\) are suitably chosen, show that \(\pi^2\le 10\). By setting \(\.f(x) = p \sin \frac12 x + q\cos \frac12 x +r\), where \(p\), \(q\) and \(r\) are suitably chosen, obtain another inequality for \(\pi\). Which of these inequalities leads to a better estimate for \(\pi^2\,\)?

2014 Paper 2 Q3
D: 1600.0 B: 1516.0

  1. Show, geometrically or otherwise, that the shortest distance between the origin and the line \(y= mx+c\), where \(c\ge0\), is \(c(m^2+1)^{-\frac12}\).
  2. The curve \(C\) lies in the \(x\)-\(y\) plane. Let the line \(L\) be tangent to~\(C\) at a point~\(P\) on~\(C\), and let \(a\)~be the shortest distance between the origin and \(L\). The curve~\(C\) has the property that the distance~\(a\) is the same for all points~\(P\) on~\(C\). Let \(P\) be the point on \(C\) with coordinates \((x,y(x))\). Given that the tangent to \(C\) at \(P\) is not vertical, show that \begin{equation} \label{eq:8*} (y-xy')^2 = a^2\big (1+(y')^2 \big) \,. \tag{\(*\)} \end{equation} By first differentiating \((*)\) with respect to \(x\), show that either \(y= mx \pm a(1+m^2)^{\frac12}\) for some \(m\) or \(x^2+y^2 =a^2\).
  3. Now suppose that \(C\) (as defined above) is a continuous curve for \(-\infty < x < \infty\), consisting of the arc of a circle and two straight lines. Sketch an example of such a curve which has a non-vertical tangent at each point.