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2014 Paper 2 Q4
D: 1600.0 B: 1500.0

  1. By using the substitution \(u=1/x\), show that for \(b>0\) \[ \int_{1/b}^b \frac{x \ln x}{(a^2+x^2)(a^2x^2+1)} \d x =0 \,. \]
  2. By using the substitution \(u=1/x\), show that for \(b>0\), \[ \int_{1/b}^b \frac{\arctan x}{x} \d x = \frac{\pi \ln b} 2\,. \]
  3. By using the result \( \displaystyle \int_0^\infty \frac 1 {a^2+x^2} \d x = \frac {\pi}{2 a} \) (where \(a > 0\)),and a substitution of the form \(u=k/x\), for suitable \(k\), show that \[ \int_0^\infty \frac 1 {(a^2+x^2)^2} \d x = \frac {\pi}{4a^3 } \, \ \ \ \ \ \ (a > 0). \]


Solution:

  1. \begin{align*} && I &= \int_{1/b}^b \frac{x \ln x}{(a^2+x^2)(a^2x^2+1} \d x \\ u = 1/x, \d u = -1/x^2 \d x: &&&= \int_{u=b}^{u=1/b} \frac{1/u \ln(1/u)}{(a^2+u^{-2})(a^2u^{-2}+1)} (- \frac{1}{u^2}) \d u \\ &&&= \int_{1/b}^b \frac{-u\ln u}{(a^2u^2+1)(a^2+u^2)} \d u \\ &&&= -I \\ \Rightarrow && I &= 0 \end{align*}
  2. \(\,\) \begin{align*} && I &= \int_{1/b}^b \frac{\arctan x}{x} \d x \\ u = 1/x, \d x = -1/u^2 \d u: &&&= \int_{u=b}^{u=1/b} \frac{\arctan \frac1u}{\frac1u} \frac{-1}{u^2} \d u \\ &&&= \int_{1/b}^b \frac{\arctan \frac1u}{u} \d u \\ \Rightarrow && 2I &= \int_{1/b}^b \frac{\arctan x + \arctan \frac1x}{x} \d x \\ &&&= \int_{1/b}^b \frac{\frac{\pi}2}{x} \d x \\ &&&= \pi \ln b \\ \Rightarrow && I &= \frac{\pi}{2} \ln b \end{align*}
  3. \(\,\) \begin{align*} && I_a &= \int_0^{\infty} \frac{1}{(a^2+x^2)^2} \d x \\ u = a/x, \d x = -a/u^2 \d u:&&&= \int_{u=0}^{u=\infty} \frac{1}{\left (a^2+\frac{a^2}{u^2} \right)^2} \frac{a}{u^2} \d u \\ &&&= \frac1{a^3}\int_0^{\infty} \frac{1}{(u+1/u)^2} \d u \\ &&&= \frac{1}{a^3} \int_0^{\infty} \frac{u^2}{(u^2+1)^2} \d u \\ &&&= \frac{1}{a^3} \int_0^{\infty} \frac{u^2+1-1}{(u^2+1)^2} \d u \\ &&&= \frac{1}{a^3} \int_0^{\infty} \frac{1}{(u^2+1)} - \frac{1}{(u^2+1)^2} \d u \\ &&&= \frac1{a^3} \frac{\pi}{2} - \frac{1}{a^3} I_1 \\ \Rightarrow && 2I_1 &= \frac{\pi}{2} \\ \Rightarrow && I_1 &= \frac{\pi}{4} \\ \Rightarrow && I_a &= \frac{\pi}{4a^3} \end{align*}

2014 Paper 2 Q5
D: 1600.0 B: 1486.1

Given that \(y=xu\), where \(u\) is a function of \(x\), write down an expression for \(\dfrac {\d y}{\d x}\).

  1. Use the substitution \(y=xu\) to solve \[ \frac {\d y}{\d x} = \frac {2y+x}{y-2x} \] given that the solution curve passes through the point \((1,1)\). Give your answer in the form of a quadratic in \(x\) and \(y\).
  2. Using the substitutions \(x=X+a\) and \(y=Y+b\) for appropriate values of \(a\) and \(b\), or otherwise, solve \[ \frac {\d y}{\d x} = \frac {x-2y-4} {2x+y-3}\,, \] given that the solution curve passes through the point \((1,1)\).


Solution: \begin{align*} \frac{\d y}{\d x} &= \frac{\d }{\d x} \l y \r \\ &= \frac{\d }{\d x} \l xu \r \\ &\underbrace{=}_{\text{product rule}} \frac{\d}{\d x} \l x \r u + x \frac{\d}{\d x} \l u \r \\ &= u + x \frac{\d u}{\d x} \end{align*} \begin{questionparts} \item \begin{align*} && \frac{\d y}{\d x} &= \frac{2y + x}{y - 2x} \\ && u + x \frac{\d u}{\d x} &= \frac{2u + 1}{u - 2} \\ && x \frac{\d u}{\d x} &= \frac{2u-1-u^2+2u}{u-2} \\ \Rightarrow && \int \frac{2-u}{u^2-4u+1} \d u &= \int \frac{1}{x} \d x \\ && \int \frac{2-u}{(u-2)^2-5} \d u &= \int \frac1x \d x \\ && -\frac12\ln| (u-2)^2 - 5| &= \ln x + C \\ (x,y) = (1,1): && - \ln 2 &= C \\ \Rightarrow && \ln x^2 &= \ln 4 - \ln |5 - (u-2)^2| \\ \Rightarrow && x^2 &= \frac{4}{5- (u-2)^2} \\ \Rightarrow && 4 & = x^2(5 - (\frac{y}{x} - 2)^2) \\ &&&= 5x^2 - (y-2x)^2 \\ &&&= x^2+4xy-y^2 \end{align*} \item It would be convienient if \(x-2y -4 = X-2Y\) and \(2x+y-3 = 2X+Y\), ie \(a-2b = 4\) and \(2a+b = 3\), ie \(a = 2, b = -1\). Now our differential equation is: \begin{align*} && \frac{\d Y}{\d X} &= \frac{X - 2Y}{2X+Y} \\ && \frac{\d X}{\d Y} &= \frac{2X + Y}{X-2Y} \end{align*} This is the same differential equation we have already solved, just with the roles of \(x\) and \(y\) interchanged with \(Y\) and \(X\) and with the point \((0,3)\) being on the curve, ie: \(Y^2 + 4XY-X^2 = c\) and \(c = 9\), therefore our equation is: \[ (y-1)^2 + 4(y-1)(x+2)-(x+2)^2 = 9\]

2014 Paper 2 Q6
D: 1600.0 B: 1484.2

By simplifying \(\sin(r+\frac12)x - \sin(r-\frac12)x\) or otherwise show that, for \(\sin\frac12 x \ne0\), \[ \cos x + \cos 2x +\cdots + \cos nx = \frac{\sin(n+\frac12)x - \sin\frac12 x}{2\sin\frac12x}\,. \] The functions \(\.S_n\), for \(n=1\), \(2\), \dots, are defined by \[ \.S_n(x) = \sum_{r=1}^n \frac 1 r \sin rx \qquad (0\le x \le \pi). \]

  1. Find the stationary points of \(\.S_2(x)\) for \(0\le x\le\pi\), and sketch this function.
  2. Show that if \(\.S_n(x)\) has a stationary point at \(x=x_0\), where \(0< x_0 < \pi\), then \[ \sin nx_0 = (1-\cos nx_0) \tan\tfrac12 x_0 \] and hence that \(\.S_n(x_0) \ge \.S_{n-1}(x_0)\). Deduce that if \(\.S_{n-1}(x)>0\) for all \(x\) in the interval \(00\) for all \(x\) in this interval.
  3. Prove that \(\.S_n(x)\ge0\) for \(n\ge1\) and \(0\le x\le\pi\).

2014 Paper 2 Q7
D: 1600.0 B: 1486.9

  1. The function \(\f\) is defined by \(\f(x)= |x-a| + |x-b| \), where \(a < b\). Sketch the graph of \(\f(x)\), giving the gradient in each of the regions \(x < a\), \(a < x < b\) and \(x > b\). Sketch on the same diagram the graph of \(\g(x)\), where \(\g(x)= |2x-a-b|\). What shape is the quadrilateral with vertices \((a,0)\), \((b,0)\), \((b,\f(b))\) and \((a, \f(a))\)?
  2. Show graphically that the equation \[ |x-a| + |x-b| = |x-c|\,, \] where \(a < b\), has \(0\), \(1\) or \(2\) solutions, stating the relationship of \(c\) to \(a\) and \(b\) in each case.
  3. For the equation \[ |x-a| + |x-b| = |x-c|+|x-d|\,, \] where \(a < b\), \(c < d\) and \(d-c < b-a\), determine the number of solutions in the various cases that arise, stating the relationship between \(a\), \(b\), \(c\) and \(d\) in each case.


Solution:

  1. \(\,\)
    TikZ diagram
    \((a,0)\), \((b,0)\), \((b,\f(b))\) and \((a, \f(a))\) forms a rectangle.
  2. There are no solutions if \(a < c < b\):
    TikZ diagram
    There is one solution if \(a=c\) or \(a = b\)
    TikZ diagram
    And there are two solution if \(c \not \in [a,b]\)
    TikZ diagram
    There is exactly one solution unless....
    TikZ diagram
    ... there are infinitely many solutions when the gradients line up perfectly, ie when \(a+b=c+d\)
    TikZ diagram

2014 Paper 2 Q8
D: 1600.0 B: 1486.3

For positive integers \(n\), \(a\) and \(b\), the integer \(c_r\) (\(0\le r\le n\)) is defined to be the coefficient of~\(x^r\) in the expansion in powers of \(x\) of \((a+bx)^n\). Write down an expression for \(c_r\) in terms of \(r\), \(n\), \(a\) and~\(b\). For given \(n\), \(a\) and \(b\), let \(m\)~denote a value of~\(r\) for which \(c_r\)~is greatest (that is, \(c_m \ge c_r\) for \(0\le r\le n\)). Show that \[ \frac{b(n+1)}{a+b} - 1 \le m \le \frac {b(n+1)}{a+b} \,. \] Deduce that \(m\) is either a unique integer or one of two consecutive integers. Let \(\.G(n,a,b)\) denote the unique value of~\(m\) (if there is one) or the larger of the two possible values of~\(m\).

  1. Evaluate \(\.G(9,1,3)\) and \(\.G(9,2,3)\).
  2. For any positive integer \(k\), find \(\.G(2k,a,a)\) and \(\.G(2k-1,a,a)\) in terms of~\(k\).
  3. For fixed \(n\) and \(b\), determine a value of~\(a\) for which \(\.G(n,a,b)\) is greatest.
  4. For fixed~\(n\), find the greatest possible value of \(\.G(n,1,b)\). For which values of~\(b\) is this greatest value achieved?

2014 Paper 2 Q9
D: 1600.0 B: 1484.0

A uniform rectangular lamina \(ABCD\) rests in equilibrium in a vertical plane with the \(A\) in contact with a rough vertical wall. The plane of the lamina is perpendicular to the wall. It is supported by a light inextensible string attached to the side \(AB\) at a distance \(d\) from \(A\). The other end of the string is attached to a point on the wall above \(A\) where it makes an acute angle \(\theta\) with the downwards vertical. The side \(AB\) makes an acute angle \(\phi\) with the upwards vertical at \(A\). The sides \(BC\) and \(AB\) have lengths \(2a\) and \(2b\) respectively. The coefficient of friction between the lamina and the wall is \(\mu\).

  1. Show that, when the lamina is in limiting equilibrium with the frictional force acting upwards, \begin{equation} d\sin(\theta +\phi) = (\cos\theta +\mu \sin\theta)(a\cos\phi +b\sin\phi)\,. \tag{\(*\)} \end{equation}
  2. How should \((*)\) be modified if the lamina is in limiting equilibrium with the frictional force acting downwards?
  3. Find a condition on \(d\), in terms of \(a\), \(b\), \(\tan\theta\) and \(\tan\phi\), which is necessary and sufficient for the frictional force to act upwards. Show that this condition cannot be satisfied if \(b(2\tan\theta+ \tan \phi) < a\).


Solution:

TikZ diagram
  1. \begin{align*} \text{N2}(\uparrow): && T \cos \theta + F -W &= 0 \\ && W &= T\cos \theta + \mu R \tag{1} \\ \text{N2}(\rightarrow): && R-T\sin \theta &= 0 \\ && R &= T \sin \theta \tag{2}\\ \\ (1)+(2): && W&=(\cos \theta + \mu \sin \theta)T \tag{3} \\ \overset{\curvearrowright}{A}: && 0 &= W(b\sin \phi + a \cos \phi) - Td\sin(\phi+\theta) \tag{4} \\ \\ (3)+(4): && 0 &= (\cos \theta + \mu \sin \theta)(b\sin \phi + a \cos \phi)-d\sin(\phi+\theta) \\ \Rightarrow && d\sin(\phi+\theta) &= (\cos \theta + \mu \sin \theta)(b\sin \phi + a \cos \phi) \end{align*} as required.
  2. If \(F\) is operating downwards, it's equivalent to \(-\mu\), ie: \[d\sin(\phi+\theta) = (\cos \theta - \mu \sin \theta)(b\sin \phi + a \cos \phi)\]
  3. For the frictional force to be acting upwards, we need \begin{align*} && d\sin(\phi+\theta) &\geq \cos \theta(b\sin \phi + a \cos \phi) \\ \Rightarrow && d &\geq \frac{\cos \theta(b\sin \phi + a \cos \phi)}{\sin(\phi + \theta)} \\ &&&= \frac{\cos \theta(b\sin \phi + a \cos \phi)}{\sin\phi \cos\theta+\cos\phi\sin \theta)}\\ &&&= \frac{(b\sin \phi + a \cos \phi)}{\sin\phi+\cos \phi \tan \theta)}\\ &&&= \frac{a+b\tan \phi}{\tan\theta+\tan\phi }\\ \end{align*} We know that \(d < 2b\), so \begin{align*} && 2b &>\frac{a+b\tan \phi}{\tan\theta+\tan\phi }\\ \Rightarrow && 2b \tan \theta + 2b \tan \phi &> a + b \tan \phi \\ \Rightarrow &&b(2 \tan \theta + \tan \phi) &> a\\ \end{align*} Therefore we will have problems if the inequality is reversed!

2014 Paper 2 Q10
D: 1600.0 B: 1484.0

A particle is projected from a point \(O\) on horizontal ground with initial speed \(u\) and at an angle of \(\theta\) above the ground. The motion takes place in the \(x\)-\(y\) plane, where the \(x\)-axis is horizontal, the \(y\)-axis is vertical and the origin is \(O\). Obtain the Cartesian equation of the particle's trajectory in terms of \(u\), \(g\) and~\(\lambda\), where \(\lambda=\tan\theta\). Now consider the trajectories for different values of \(\theta\) with \(u\)~fixed. Show that for a given value of~\(x\), the coordinate~\(y\) can take all values up to a maximum value,~\(Y\), which you should determine as a function of \(x\), \(u\) and~\(g\). Sketch a graph of \(Y\) against \(x\) and indicate on your graph the set of points that can be reached by a particle projected from \(O\) with speed \(u\). Hence find the furthest distance from \(O\) that can be achieved by such a projectile.

2014 Paper 2 Q11
D: 1600.0 B: 1504.7

A small smooth ring \(R\) of mass \(m\) is free to slide on a fixed smooth horizontal rail. A light inextensible string of length~\(L\) is attached to one end,~\(O\), of the rail. The string passes through the ring, and a particle~\(P\) of mass~\(km\) (where \(k>0\)) is attached to its other end; this part of the string hangs at an acute angle \(\alpha\) to the vertical and it is given that \(\alpha\) is constant in the motion. Let \(x\) be the distance between \(O\) and the ring. Taking the \(y\)-axis to be vertically upwards, write down the Cartesian coordinates of~\(P\) relative to~\(O\) in terms of \(x\), \(L\) and~\(\alpha\).

  1. By considering the vertical component of the equation of motion of \(P\), show that \[ km\ddot x \cos\alpha = T \cos\alpha - kmg\,, \] where \(T\) is the tension in the string. Obtain two similar equations relating to the horizontal components of the equations of motion of \(P\) and \(R\).
  2. Show that \(\dfrac {\sin\alpha}{(1-\sin\alpha)^2_{\vphantom|}} = k\), and deduce, by means of a sketch or otherwise, that motion with \(\alpha\) constant is possible for all values of~\(k\).
  3. Show that \(\ddot x = -g\tan\alpha\,\).

2014 Paper 2 Q12
D: 1600.0 B: 1484.8

The lifetime of a fly (measured in hours) is given by the continuous random variable~\(T\) with probability density function \(\.f(t)\) and cumulative distribution function \(\.F(t)\). The \emph{hazard function}, \(\.h(t)\), is defined, for \(\.F(t)<1\), by \[ \.h(t) = \frac{\.f(t)}{1-\.F(t)}\,. \]

  1. Given that the fly lives to at least time \(t\), show that the probability of its dying within the following \(\delta t\) is approximately \(\.h (t) \, \delta t\) for small values of \(\delta t\).
  2. Find the hazard function in the case \(\.F(t) = t/a\) for \(0< t < a\). Sketch \(\.f(t)\) and \(\.h(t)\) in this case.
  3. The random variable \(T\) is distributed on the interval $t> a\(, where \)a>0\(, and its hazard function is \)t^{-1}$. Determine the probability density function for \(T\).
  4. Show that \(\.h(t)\) is constant for \(t>b\) and zero otherwise if and only if \(\.f(t) =k\.e^{-k(t-b)}\) for \(t>b\), where \(k\)~is a positive constant.
  5. The random variable \(T\) is distributed on the interval \(t> 0\) and its hazard function is given by \[ \.h(t) = \left(\frac{\lambda}{\theta^\lambda}\right)t^{\lambda-1}\,, \] where \(\lambda\) and \(\theta\) are positive constants. Find the probability density function for \(T\).

2014 Paper 2 Q13
D: 1600.0 B: 1469.5

A random number generator prints out a sequence of integers \(I_1, I_2, I_3, \dots\). Each integer is independently equally likely to be any one of \(1, 2, \dots, n\), where \(n\) is fixed. The random variable \(X\) takes the value \(r\), where \(I_r\) is the first integer which is a repeat of some earlier integer. Write down an expression for \(\mathbb{P}(X=4)\).

  1. Find an expression for \(\mathbb{P}(X=r)\), where \(2\le r\le n+1\). Hence show that, for any positive integer \(n\), \[ \frac 1n + \left(1-\frac1n\right) \frac 2 n + \left(1-\frac1n\right)\left(1-\frac2n\right) \frac3 n + \cdots \ = \ 1 \,. \]
  2. Write down an expression for \(\mathbb{E}(X)\). (You do not need to simplify it.)
  3. Write down an expression for \(\mathbb{P}(X\ge k)\).
  4. Show that, for any discrete random variable \(Y\) taking the values \(1, 2, \dots, N\), \[ \mathbb{E}(Y) = \sum_{k=1}^N \mathbb{P}(Y\ge k)\,. \] Hence show that, for any positive integer \(n\), \[ \left(1-\frac{1^2}n\right) + \left(1-\frac1n\right)\left(1-\frac{2^2}n\right) + \left(1-\frac1n\right)\left(1-\frac{2}n\right)\left(1-\frac{3^2}n\right) + \cdots \ = \ 0. \]


Solution: \begin{align*} && \mathbb{P}(X > 4) &= 1 \cdot \frac{n-1}{n} \cdot \frac{n-2}{n} \cdot \frac{n-3}{n} \\ && \mathbb{P}(X > 3) &= 1 \cdot \frac{n-1}{n} \cdot \frac{n-2}{n} \\ \Rightarrow && \mathbb{P}(X =4) &= \mathbb{P}(X > 3) - \mathbb{P}(X > 4) \\ &&&= \frac{(n-1)(n-2)}{n^2} \left (1 - \frac{n-3}{n} \right) \\ &&&= \frac{3(n-1)(n-2)}{n^3} \end{align*}

  1. Notice that \begin{align*} && \mathbb{P}(X > r) &= \frac{n-1}{n} \cdots \frac{n-r+1}{n} \\ \Rightarrow && \mathbb{P}(X = r) &= \frac{n-1}{n} \cdots \frac{n-r+2}{n} \left (1 - \frac{n-r+1}{n} \right) \\ &&&= \frac{(n-1)\cdots(n-r+2)(r-1)}{n^{r-1}} \\ &&&= \left (1 - \frac{1}n \right)\left (1 - \frac{2}{n} \right) \cdots \left (1 - \frac{r-2}{n} \right) \frac{r-1}{n} \\ \Rightarrow && 1 &= \sum \mathbb{P}(X = r) \\ &&&= \sum_{r=2}^{n+1} \mathbb{P}(X = r) \\ &&&= \frac 1n + \left(1-\frac1n\right) \frac 2 n + \left(1-\frac1n\right)\left(1-\frac2n\right) \frac3 n + \cdots \end{align*}
  2. \(\,\) \begin{align*} \mathbb{E}(X) &= \sum_{r=2}^{n+1} r\cdot\mathbb{P}(X = r) \\ &= \frac 2n + \left(1-\frac1n\right) \frac {2\cdot3} n + \left(1-\frac1n\right)\left(1-\frac2n\right) \frac{3\cdot4} n + \cdots \end{align*}
  3. \(\displaystyle \mathbb{P}(X \geq k) = \frac{n-1}{n} \cdots \frac{n-r+2}{n}\)
  4. \(\,\) \begin{align*} && \mathbb{E}(Y) &= \sum_{r=1}^N r \cdot \mathbb{P}(Y = r) \\ &&&= \sum_{r=1}^N \sum_{j=1}^r \mathbb{P}(Y = r) \\ &&&= \sum_{j=1}^N \sum_{r=j}^N \mathbb{P}(Y=r) \\ &&&= \sum_{j=1}^N \mathbb{P}(Y \geq j) \end{align*} Let \(P_k = \left(1-\frac1n\right)\left(1-\frac2n\right) \cdots \left(1-\frac1n\right)\left(1-\frac{k}n\right) \) \begin{align*} && \mathbb{E}(X) &= P_1 \frac{1 \cdot 2 }{n} + P_2 \cdot \frac{2 \cdot 3}{n} + \cdots + P_k \cdot \frac{k(k+1)}{n} + \cdots \\ && &= \sum_{k=1}^{n} \frac{k^2}{n}P_k + \sum_{k=1}^{n} \frac{k}{n}P_k \\ && \text{Using the identity } & \frac{k}{n}P_k = \frac{k}{n} \prod_{i=1}^{k-1} \left(1-\frac{i}{n}\right) = P_k - P_{k+1}: \\ && \sum_{k=1}^{n} \frac{k}{n}P_k &= (P_1 - P_2) + (P_2 - P_3) + \cdots + (P_n - P_{n+1}) \\ && &= P_1 - P_{n+1} = 1 - 0 = 1 \\ \\ \Rightarrow && \mathbb{E}(X) &= \sum_{k=1}^{n} \frac{k^2}{n}P_k + 1 \\ && &= \mathbb{P}(X \geq 1) + \mathbb{P}(X \geq 2) + \mathbb{P}(X \geq 3) + \cdots \\ && &= 1 + P_1 + P_2 + P_3 + \cdots \\ && &= 1 + \sum_{k=1}^{n} P_k \\ \\ \Rightarrow && 1 + \sum_{k=1}^{n} P_k &= \sum_{k=1}^{n} \frac{k^2}{n}P_k + 1 \\ \Rightarrow && \sum_{k=1}^{n} P_k &= \sum_{k=1}^{n} \frac{k^2}{n}P_k \\ \Rightarrow && 0 &= \sum_{k=1}^{n} P_k \left( 1 - \frac{k^2}{n} \right) \end{align*}