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2009 Paper 1 Q12
D: 1500.0 B: 1501.5

Prove that, for any real numbers \(x\) and \(y\), \(x^2+y^2\ge2xy\,\).

  1. Carol has two bags of sweets. The first bag contains \(a\) red sweets and \(b\) blue sweets, whereas the second bag contains \(b\) red sweets and \(a\) blue sweets, where \(a\) and \(b\) are positive integers. Carol shakes the bags and picks one sweet from each bag without looking. Prove that the probability that the sweets are of the same colour cannot exceed the probability that they are of different colours.
  2. Simon has three bags of sweets. The first bag contains \(a\) red sweets, \(b\) white sweets and \(c\) yellow sweets, where \(a\), \(b\) and \(c\) are positive integers. The second bag contains \(b\) red sweets, \(c\) white sweets and \(a\) yellow sweets. The third bag contains \(c\) red sweets, \(a\) white sweets and \(b\) yellow sweets. Simon shakes the bags and picks one sweet from each bag without looking. Show that the probability that exactly two of the sweets are of the same colour is \[ \frac {3(a^2b+b^2c+c^2a+ab^2 + bc^2 +ca^2)}{(a+b+c)^3}\,, \] and find the probability that the sweets are all of the same colour. Deduce that the probability that exactly two of the sweets are of the same colour is at least 6 times the probability that the sweets are all of the same colour.

2009 Paper 1 Q13
D: 1500.0 B: 1504.1

I seat \(n\) boys and \(3\) girls in a line at random, so that each order of the \(n+3\) children is as likely to occur as any other. Let \(K\) be the maximum number of consecutive girls in the line so, for example, \(K=1\) if there is at least one boy between each pair of girls.

  1. Find \(\P(K=3)\).
  2. Show that \[\P(K=1)= \frac{n(n-1)}{(n+2)(n+3)}\,. \]
  3. Find \(\E(K)\).


Solution:

  1. If all the girls are say together there are \(n+1\) ways to place the block of 3 girls. There are \(\binom{n+3}{3}\) ways to choose where to place the girls in total, therefore: \begin{align*} && \mathbb{P}(K =3) &= \frac{n+1}{\binom{n+3}3} \\ &&&= \frac{6(n+1)}{(n+3)(n+2)(n+1)} \\ &&&= \frac{6}{(n+3)(n+2)} \end{align*}
  2. If \(K= 1\) then all of the girls are separated. We can place three girls and two boys separating them, then we are allocating \(N-2\) boys to \(4\) gaps, ie \(\binom{N-2+3}{3} = \binom{N+1}{3}\). \begin{align*} && \mathbb{P}(K=3) &= \frac{\binom{n+1}{3}}{\binom{n+3}{3}} \\ &&&= \frac{(n+1)n(n-1)}{(n+3)(n+2)(n+1)} \\ &&&= \frac{n(n-1)}{(n+3)(n+2)} \end{align*}
  3. \(\,\) \begin{align*} \mathbb{E}(K) &= \sum_{k=1}^3 k \mathbb{P}(K=k) \\ &= \frac{6}{(n+3)(n+2)} + 2 \left (1 - \frac{6}{(n+3)(n+2)} - \frac{n(n-1)}{(n+3)(n+2)} \right) + 3\frac{n(n-1)}{(n+3)(n+2)} \\ &= 2+\frac{6-12+n(n-1)}{(n+3)(n+2)} \\ &= 2 + \frac{n^2-n-6}{(n+2)(n+3)}\\ &= 2 + \frac{(n-3)(n+2)}{(n+2)(n+3)} \\ &= 2 + \frac{n-3}{n+3} \\ &= \frac{2n}{n+3} \end{align*}

2009 Paper 2 Q1
D: 1600.0 B: 1516.0

Two curves have equations \(\; x^4+y^4=u\;\) and \(\; xy = v\;\), where \(u\) and \(v\) are positive constants. State the equations of the lines of symmetry of each curve. The curves intersect at the distinct points \(A\), \(B\), \(C\) and \(D\) (taken anticlockwise from \(A\)). The coordinates of \(A\) are \((\alpha,\beta)\), where \(\alpha > \beta > 0\). Write down, in terms of \(\alpha\) and \(\beta\), the coordinates of \(B\), \(C\) and \(D\). Show that the quadrilateral \(ABCD\) is a rectangle and find its area in terms of \(u\) and \(v\) only. Verify that, for the case \(u=81\) and \(v=4\), the area is \(14\).


Solution: The curve \(x^4 + y^4 = u\) has lines of symmetry:

  • \(y = 0\)
  • \(x = 0\)
  • \(y = x\)
  • \(y = -x\)
The curve \(xy = v\) has lines of symmetry:
  • \(y = x\)
  • \(y = -x\)
TikZ diagram
The points are \(A = (\alpha, \beta), B = (\beta, \alpha), C = (-\alpha, -\beta), D = (-\beta, -\alpha)\) \(AD\) has gradient \(\frac{\beta+\alpha}{\alpha+\beta} = 1\), \(BC\) has the same gradient. \(AB\) has gradient \(\frac{\alpha-\beta}{\beta-\alpha} = -1\), as does \(CD\). Therefore it has two sets of perpendicular and parallel sides, hence a rectangle. The area is \(|AD||AB| = \sqrt{2(\alpha+\beta)^2}\sqrt{2(\alpha-\beta)^2} = 2(\alpha^2-\beta^2)\) The squared area is \(4(\alpha^4+\beta^4 - 2 \alpha^2\beta^2) = 4(u - 2v^2)\) ie the area is \(2\sqrt{u-2v^2}\) When \(u = 81, v = 4\) we have the area is \(2 \sqrt{81 - 2 \cdot 16} = 14\) as required.

2009 Paper 2 Q2
D: 1600.0 B: 1500.0

The curve \(C\) has equation \[ y= a^{\sin (\pi \e^ x)}\,, \] where \(a>1\).

  1. Find the coordinates of the stationary points on \(C\).
  2. Use the approximations \(\e^t \approx 1+t\) and \(\sin t \approx t\) (both valid for small values of \(t\)) to show that \[ y\approx 1-\pi x \ln a \; \] for small values of \(x\).
  3. Sketch \(C\).
  4. By approximating \(C\) by means of straight lines joining consecutive stationary points, show that the area between \(C\) and the \(x\)-axis between the \(k\)th and \((k+1)\)th maxima is approximately \[ \Big( \frac {a^2+1}{2a} \Big) \ln \Big ( 1+ \big( k-\tfrac34)^{-1} \Big)\,. \]


Solution:

  1. \(\,\) \begin{align*} && y & = a^{\sin(\pi e^x)} \\ \Rightarrow && \frac{\d y}{\d x} &= a^{\sin(\pi e^x)} \cdot ( \ln a) \cdot (\cos (\pi e^x)) \cdot \pi e^x \\ \frac{\d y}{\d x} = 0: && 0 &= \cos(\pi e^x) \\ \Rightarrow && \pi e^x &= \left ( \frac{2n+1}{2} \right) \pi \\ \Rightarrow && x &= \ln \left ( \frac{2n+1}{2} \right) \\ && y &= a^{(-1)^n} \\ &&(x,y) &= \left (\ln \left ( \frac{2n+1}{2} \right), a^{(-1)^n} \right) \end{align*}
  2. \(\,\) \begin{align*} && y &= a^{\sin(\pi e^x)} \\ &&&= e^{\ln a \cdot \sin(\pi e^x)} \\ &&&\approx e^{\ln a \cdot \sin(\pi (1+x))} \\ &&&\approx e^{-\ln a \cdot \sin(\pi x)} \\ &&&\approx e^{-\ln a \cdot \pi x} \\ &&&\approx 1-( \pi\ln a) x \end{align*}
  3. TikZ diagram
  4. The \(k\)th maxima is at \(\ln \left ( \frac{4(k-1)+1}{2}\right)\) and \(a\) ,and the \((k+1)\)th is at \(\ln \left ( \frac{4k+1}{2}\right)\). They have a minima between at \(\ln \left ( \frac{4k-3}{2}\right)\). \begin{align*} && \text{Area} &\approx \frac12 \left (\ln \left ( \frac{4k-1}{2}\right)- \ln \left ( \frac{4k-3}{2}\right)\right) \left ( a + \frac1a \right) + \frac12 \left ( \ln \left ( \frac{4k+1}{2}\right)-\ln \left ( \frac{4k-1}{2}\right)\right) \left ( a + \frac1a \right) \\ &&&= \frac{a^2+1}{2a} \ln \left (\frac{4k+1}{4k-3} \right) \\ &&&= \frac{a^2+1}{2a} \ln \left (1 + \frac{4}{4k-3} \right) \\ &&&= \frac{a^2+1}{2a} \ln \left (1 + \frac{1}{k-\tfrac34} \right) \\ &&&= \frac{a^2+1}{2a} \ln \left (1 + (k-\tfrac34)^{-1} \right) \\ \end{align*}

2009 Paper 2 Q3
D: 1600.0 B: 1500.0

Prove that \[ \tan \left ( \tfrac14 \pi -\tfrac12 x \right)\equiv \sec x -\tan x\,. \tag{\(*\)} \]

  1. Use \((*)\) to find the value of \(\tan\frac18\pi\,\). Hence show that \[ \tan \tfrac{11}{24} \pi = \frac{\sqrt3 + \sqrt2 -1}{\sqrt3 -\sqrt6+1}\;. \]
  2. Show that \[ \frac{\sqrt3 + \sqrt2 -1}{\sqrt3 -\sqrt6+1}= 2+\sqrt2+\sqrt3+\sqrt6\,. \]
  3. Use \((*)\) to show that \[ \tan \tfrac1{48}\pi = \sqrt{16+10\sqrt2+8\sqrt3 +6\sqrt6 \ }-2-\sqrt2-\sqrt3-\sqrt6\,. \]


Solution: \begin{align*} && \tan \left ( \tfrac14 \pi -\tfrac12 x \right) &\equiv \frac{\tan \tfrac{\pi}{4} - \tan \tfrac12 x}{1 + \tan \tfrac{\pi}{4} \tan \tfrac12 x} \\ &&&= \frac{1-\tan \tfrac12 x}{1+\tan \tfrac12 x} \\ \\ && \sec x - \tan x &= \frac{1+t^2}{1-t^2} - \frac{2t}{1-t^2} \\ &&&= \frac{(1-t)^2}{(1-t)(1+t)} \\ &&&= \frac{1-t}{1+t} \end{align*} Therefore both sides are equal to the same thing.

  1. \(\tan \tfrac18 \pi = \tan(\tfrac14 \pi - \tfrac12 \tfrac14\pi) = \sec \tfrac14 \pi - \tan \tfrac14 \pi = \sqrt{2} - 1\) \begin{align*} && \tan \tfrac{11}{24} \pi &= \tan (\tfrac13 \pi +\tfrac18 \pi) \\ &&&= \frac{\tan \tfrac13 \pi +\tan \tfrac18 \pi}{1-\tan \tfrac13 \pi \tan \tfrac18 \pi} \\ &&&= \frac{\sqrt{3} + \sqrt{2} - 1}{1 - \sqrt{3}(\sqrt{2}-1)} \\ &&&= \frac{\sqrt{3} + \sqrt{2} - 1}{1 +\sqrt{3}-\sqrt{6}} \\ \end{align*}
  2. \(\,\) \begin{align*} && (\sqrt{3}-\sqrt{6}+1)(2+\sqrt{2}+\sqrt{3}+\sqrt{6}) &= (2\sqrt{3}+\sqrt{6}+3+3\sqrt{2}) + \\ &&&\quad+(-2\sqrt{6}-2\sqrt{3}-3\sqrt{2}-6) + \\ &&&\quad+(2+\sqrt{2}+\sqrt{3}+\sqrt{6}) \\ &&&= \sqrt{3}+\sqrt{2}-1 \end{align*}
  3. \(\,\) \begin{align*} && \tan \tfrac{1}{48} \pi &= \tan (\tfrac14\pi - \tfrac{11}{48} \pi) \\ &&&= \sec \tfrac{11}{24} \pi - \tan \tfrac{11}{24} \pi \\ &&&= \sqrt{1+\tan^2 \tfrac{11}{24}\pi} - \tan \tfrac{11}{24} \pi \\ &&&= \sqrt{1 + (2+\sqrt{2}+\sqrt{3}+\sqrt{6})^2} - (2+\sqrt{2}+\sqrt{3}+\sqrt{6}) \\ &&&= \sqrt{16+10\sqrt{2} + 8\sqrt{3}+6\sqrt{6}} - 2 - \sqrt2 - \sqrt3-\sqrt6 \end{align*}

2009 Paper 2 Q4
D: 1600.0 B: 1500.0

The polynomial \(\p(x)\) is of degree 9 and \(\p(x)-1\) is exactly divisible by \((x-1)^5\).

  1. Find the value of \(\p(1)\).
  2. Show that \(\p'(x)\) is exactly divisible by \((x-1)^4\).
  3. Given also that \(\p(x)+1\) is exactly divisible by \((x+1)^5\), find \(\p(x)\).


Solution: \(p(x) = q(x)(x-1)^5 + 1\) where \(q(x)\) has degree \(4\).

  1. \(p(1) = q(1)(1-1)^5 + 1 = 1\).
  2. \(p'(x) = q'(x)(x-1)^5 + 5(x-1)^4q(x) + 0 = (x-1)^4((x-1)q'(x) + 5q(x))\) so \(p'(x)\) is divisible by \((x-1)^4\)
  3. \(p(x)+1\) divisible by \((x+1)^5\) means that \(p(-1) = -1\) and \(p'(x)\) is divisible by \((x+1)^4\). Since \(p'(x)\) is degree \(8\) it must be \(c(x+1)^4(x-1)^4 = c(x^2 - 1)^4\). Expanding and integrating, we get \(p(x) = c(\frac{1}{9}x^9 -\frac{4}{7}x^7 + \frac{6}{5}x^5 - \frac{4}{3}x^3 + x) + d\). When \(x = 1\) we get \(c \frac{128}{315} + d = 1\) and when \(x = -1\) we get \(-c \frac{128}{315} + d = -1\) so \(2d = 0 \Rightarrow d = 0, c = \frac{315}{128}\) and \[ p(x) =\frac{315}{128} \l \frac{1}{9}x^9 -\frac{4}{7}x^7 + \frac{6}{5}x^5 - \frac{4}{3}x^3 + x\r \]

2009 Paper 2 Q5
D: 1600.0 B: 1500.0

Expand and simplify \((\sqrt{x-1}+1)^2\,\).

  1. Evaluate \[ \int_{5}^{10} \frac{ \sqrt{x+2\sqrt{x-1} \;} + \sqrt{x-2\sqrt{x-1} \;} } {\sqrt{x-1}} \,\d x\;. \]
  2. Find the total area between the curve \[ y= \frac{\sqrt{x-2\sqrt{x-1}\;}}{\sqrt{x-1}\;} \] and the \(x\)-axis between the points \(x=\frac54\) and \(x=10\).
  3. Evaluate \[ \int_{\frac54}^{10} \frac{ \sqrt{x+2\sqrt{x-1}\;} + \sqrt{x-2\sqrt{x+1}+2 \;} } {\sqrt{x^2-1} } \;\d x\;. \]

2009 Paper 2 Q6
D: 1600.0 B: 1516.0

The Fibonacci sequence \(F_1\), \(F_2\), \(F_3\), \(\ldots\) is defined by \(F_1=1\), \(F_2= 1\) and \[ F_{n+1} = F_n+F_{n-1} \qquad\qquad (n\ge 2). \] Write down the values of \(F_3\), \(F_4\), \(\ldots\), \(F_{10}\). Let \(\displaystyle S=\sum_{i=1}^\infty \dfrac1 {F_i}\,\).

  1. Show that \(\displaystyle \frac 1{F_i} > \frac1{2F_{i-1}}\,\) for \(i\ge4\) and deduce that \(S > 3\,\). Show also that \(S < 3\frac23\,\).
  2. Show further that \(3.2 < S < 3.5\,\).


Solution: \begin{array}{c|r} n & F_n \\ \hline 1 & 1 \\ 2 & 1 \\ 3 & 2 \\ 4 & 3 \\ 5 & 5 \\ 6 & 8 \\ 7 & 13 \\ 8 & 21 \\ 9 & 34 \\ 10 & 55 \end{array} \begin{questionparts} \item Claim: \(\frac1{F_i} > \frac1{2F_{i-1}}\) for \(i \geq 4\). Proof: Since \(F_i = F_{i-1}+F_{i-2}\) and \(F_i > 1\) for \(i \geq 1\) we have \(F_i > F_{i-1}\) for \(i \geq 3\). In particular we have \(F_i = F_{i-1}+F_{i-2} < 2F_{i-1}\) for \(i -1 \geq 3\) or \(i \geq 4\). Therefore \(\frac{1}{F_i} > \frac1{2F_{i-1}}\)

2009 Paper 2 Q7
D: 1600.0 B: 1500.0

Let \(y= (x-a)^n \e^{bx} \sqrt{1+x^2}\,\), where \(n\) and \(a\) are constants and \(b\) is a non-zero constant. Show that \[ \frac{\d y}{\d x} = \frac{(x-a)^{n-1} \e^{bx} \q(x)}{\sqrt{1+x^2}}\,, \] where \(\q(x)\) is a cubic polynomial. Using this result, determine:

  1. $\displaystyle \int \frac {(x-4)^{14} \e^{4x}(4x^3-1)} {\sqrt{1+x^2\;}} \, \d x\,;\(
  2. \)\displaystyle \int \frac{(x-1)^{21}\e^{12x}(12x^4-x^2-11)} {\sqrt{1+x^2\;}}\,\d x\,;\(
  3. \)\displaystyle \int \frac{(x-2)^{6}\e^{4x}(4x^4+x^3-2)} {\sqrt{1+x^2\;}}\,\d x\,.$

2009 Paper 2 Q8
D: 1600.0 B: 1484.8

The non-collinear points \(A\), \(B\) and \(C\) have position vectors \(\bf a\), \(\bf b\) and \(\bf c\), respectively. The points \(P\) and \(Q\) have position vectors \(\bf p\) and \(\bf q\), respectively, given by \[ {\bf p}= \lambda {\bf a} +(1-\lambda){\bf b} \text{ \ \ \ and \ \ \ } {\bf q}= \mu {\bf a} +(1-\mu){\bf c} \] where \(0<\lambda<1\) and \(\mu>1\). Draw a diagram showing \(A\), \(B\), \(C\), \(P\) and \(Q\). Given that \(CQ\times BP = AB\times AC\), find \(\mu\) in terms of \(\lambda\), and show that, for all values of \(\lambda\), the the line \(PQ\) passes through the fixed point \(D\), with position vector \({\bf d}\) given by \({\bf d= -a +b +c}\,\). What can be said about the quadrilateral \(ABDC\)?

2009 Paper 2 Q9
D: 1600.0 B: 1484.0

  1. A uniform lamina \(OXYZ\) is in the shape of the trapezium shown in the diagram. It is right-angled at \(O\) and \(Z\), and \(OX\) is parallel to \(YZ\). The lengths of the sides are given by \(OX=9\,\)cm, \(XY=41\,\)cm, \(YZ=18\,\)cm and \(ZO=40\,\)cm. Show that its centre of mass is a distance \(7\,\)cm from the edge \(OZ\).
    TikZ diagram
  2. The diagram shows a tank with no lid made of thin sheet metal. The base \(OXUT\), the back \(OTWZ\) and the front \(XUVY\) are rectangular, and each end is a trapezium as in part (i). The width of the tank is \(d\,\)cm.
    TikZ diagram
    Show that the centre of mass of the tank, when empty, is a distance \[ \frac {3(140+11d)}{5(12+d)}\,\text{cm} \] from the back of the tank. The tank is then filled with a liquid. The mass per unit volume of this liquid is \(k\) times the mass per unit area of the sheet metal. In the case \(d=20\), find an expression for the distance of the centre of mass of the filled tank from the back of the tank.


Solution:

  1. TikZ diagram
    \begin{array}{c|c|c|c} & OXX'Z & XX'Y & OXYZ \\ \hline \text{Area} & 360 & 180 & 540\\ \text{COM} & \binom{4.5}{20} & \binom{12}{\frac{80}{3}} & \binom{\overline{x}}{\overline{y}} \end{array} \begin{align*} && 2 \binom{3}{20} + \binom{12}{\frac{80}{3}} &= 3 \binom{\overline{x}}{\overline{y}} \\ \Rightarrow && \binom{\overline{x}}{\overline{y}} &= \frac13 \binom{21}{\frac{200}{3}} \\ &&&= \binom{7}{\frac{200}{9}} \end{align*} ie, the centre of mass is \(7\text{ cm}\) from \(OZ\)
  2. \begin{align*} && \underbrace{540 \cdot 7}_{OXYZ} + \underbrace{540 \cdot 7}_{TUVW} + \underbrace{40d\cdot 0}_{OTWZ} + \underbrace{9d\cdot 4.5}_{OXUT} + \underbrace{41d \cdot 13.5}_{XUVY} &= (540+540+40d+9d+41d) \overline{x} \\ \Rightarrow && \overline{x} &= \frac{540\cdot 14 + 50d \cdot 4.5 + 41d \cdot 9}{1080 + 90d} \\ &&&= \frac{90 \cdot 84 + 225d + 369d}{1080+90d} \\ &&&= \frac{90 \cdot 84 + 594d}{1080+90d} \\ &&&= \frac{54(140+11d)}{90(12+d)} \\ &&&= \frac{3(140+11d)}{5(12+d)} \end{align*} The volume of the prizm is \(540d\), it's center of mass is \(7\). For the tank, it COM is \(\frac{3(140+11\cdot20)}{5(12+20)} = \frac{27}4\) and area is \(2880\) Therefore for the combined shape we have: \begin{align*} && 540dk \cdot 7 + 2880 \cdot \frac{27}{4} &= (540 \cdot20 k+2880) \overline{x} \\ \Rightarrow && \overline{x} &= \frac{720(150k+27)}{720(15k + 4)} \\ &&&= \frac{3(50k+9)}{15k+4} \end{align*} \begin{align*} && \end{align*}

2009 Paper 2 Q10
D: 1600.0 B: 1500.0

\(\,\)

\psset{xunit=1.5cm,yunit=1.5cm,algebraic=true,dotstyle=o,dotsize=3pt 0,linewidth=0.3pt,arrowsize=3pt 2,arrowinset=0.25} \begin{pspicture*}(-2.73,-2.6)(3.4,1.82) \psline{->}(-2.73,0)(2.5,0) \psline{->}(0,-2.2)(0,1.5) \rput[tl](2.55,0.05){\(x\)} \rput[tl](-0.05,1.75){\(y\)} \rput[tl](-2.09,-0.3){\(P_1\)} \rput[tl](-1.11,-0.3){\(P_2\)} \rput[tl](-0.55,-0.6){\(P_3\)} \rput[tl](-0.55,-1.6){\(P_4\)} \rput[tl](0.07,0.3){\(O\)} \psline{->}(-2.1,0.4)(-1.6,0.4) \psline{->}(0.4,-1.9)(0.4,-1.4) \rput[tl](-1.55,0.45){\(u\)} \rput[tl](0.32,-1.2){\(u\)} \begin{scriptsize} \psdots[dotsize=18pt 0,dotstyle=*](-1,0) \psdots[dotsize=18pt 0,dotstyle=*](-2,0) \psdots[dotsize=18pt 0,dotstyle=*](0,-0.7) \psdots[dotsize=18pt 0,dotstyle=*](0,-1.7) \end{scriptsize} \end{pspicture*}
Four particles \(P_1\), \(P_2\), \(P_3\) and \(P_4\), of masses \(m_1\), \(m_2\), \(m_3\) and \(m_4\), respectively, are arranged on smooth horizontal axes as shown in the diagram. Initially, \(P_2\) and \(P_3\) are stationary, and both \(P_1\) and \(P_4\) are moving towards \(O\) with speed \(u\). Then \(P_1\) and \(P_2\) collide, at the same moment as \(P_4\) and \(P_3\) collide. Subsequently, \(P_2\) and \(P_3\) collide at \(O\), as do \(P_1\) and \(P_4\) some time later. The coefficient of restitution between each pair of particles is~\(e\), and \(e>0\). Show that initially \(P_2\) and \(P_3\) are equidistant from \(O\).

2009 Paper 2 Q11
D: 1600.0 B: 1516.0

A train consists of an engine and \(n\) trucks. It is travelling along a straight horizontal section of track. The mass of the engine and of each truck is \(M\). The resistance to motion of the engine and of each truck is \(R\), which is constant. The maximum power at which the engine can work is \(P\). Obtain an expression for the acceleration of the train when its speed is \(v\) and the engine is working at maximum power. The train starts from rest with the engine working at maximum power. Obtain an expression for the time \(T\) taken to reach a given speed \(V\), and show that this speed is only achievable if \[ P>(n+1)RV\,. \]

  1. In the case when \((n+1) RV/P\) is small, use the approximation \(\ln (1-x) \approx -x -\frac12 x^2\) (valid for small \( x \)) to obtain the approximation \[ PT\approx \tfrac 12 (n+1) MV^2\, \] and interpret this result.
  2. In the general case, the distance moved from rest in time \(T\) is \(X\). {\em Write down}, with explanation, an equation relating \(P\), \(T\), \(X\), \(M\), \(V\), \(R\) and \(n\) and hence show that \[ X= \frac{2PT - (n+1)MV^2}{2(n+1)R} \,. \]

2009 Paper 2 Q12
D: 1600.0 B: 1496.6

A continuous random variable \(X\) has probability density function given by \[ \f(x) = \begin{cases} 0 & \mbox{for } x<0 \\ k\e^{-2 x^2} & \mbox{for } 0\le x< \infty \;,\\ \end{cases} \] where \(k\) is a constant.

  1. Sketch the graph of \(\f(x)\).
  2. Find the value of \(k\).
  3. Determine \(\E(X)\) and \(\var(X)\).
  4. Use statistical tables to find, to three significant figures, the median value of \(X\).


Solution:

  1. \par
    TikZ diagram
  2. Let \(Y \sim N(0,\frac14)\), then: \begin{align*} &&\int_0^\infty \frac{1}{\sqrt{2 \pi \cdot \frac14}} e^{-2x^2} \, dx &= \frac12\\ \Rightarrow && \int_0^\infty e^{-2x^2} &= \frac{\sqrt{\pi}}{2 \sqrt{2}} \\ \Rightarrow && k &= \boxed{\frac{2\sqrt{2}}{\sqrt{\pi}}} \end{align*}
  3. \begin{align*} \mathbb{E}[X] &= \int_0^\infty x f(x) \, dx \\ &= \frac{2\sqrt{2}}{\sqrt{\pi}}\int_0^\infty x e^{-2x^2}\, dx \\ &= \frac{2\sqrt{2}}{\sqrt{\pi}} \left [-\frac{1}{4}e^{-2x^2} \right]_0^\infty \\ &= \frac{1}{\sqrt{2\pi}} \\ \end{align*} In order to calculate \(\mathbb{E}(X^2)\) it is useful to consider the related computation \(\mathbb{E}(Y^2)\). In fact, by symmetry, these will be the same values. Therefore \(\mathbb{E}(X^2) = \mathbb{E}(Y^2) = \mathrm{Var}(Y) = \frac{1}{4}\) (since \(\mathbb{E}(Y) = 0\)). Therefore \(\mathrm{Var}(Y) = \mathbb{E}(Y^2) - \mathbb{E}(Y)^2 = \frac14 - \frac{1}{2\pi}\)
  4. \begin{align*} && \mathbb{P}(X < x) &= \frac12 \\ \Leftrightarrow && 2\mathbb{P}(0 \leq Y < x) &= \frac12 \\ \Leftrightarrow && 2\l \mathbb{P}(Y < x) - \frac12 \r &= \frac12 \\ \Leftrightarrow && \mathbb{P}(Y < x)&= \frac34 \\ \Leftrightarrow && \mathbb{P}(\frac{Y-0}{1/2} < \frac{x}{1/2})&= \frac34 \\ \Leftrightarrow && \mathbb{P}(Z < \frac{x}{1/2})&= \frac34 \\ \Leftrightarrow && \Phi(2x)&= \frac34 \\ \Leftrightarrow && 2x &= 0.6744895\cdots \\ \Leftrightarrow && x &= 0.3372\cdots \\ \Leftrightarrow && &= 0.337 \, (3 \text{sf}) \\ \end{align*}

2009 Paper 2 Q13
D: 1600.0 B: 1500.0

Satellites are launched using two different types of rocket: the Andover and the Basingstoke. The Andover has four engines and the Basingstoke has six. Each engine has a probability~\(p\) of failing during any given launch. After the launch, the rockets are retrieved and repaired by replacing some or all of the engines. The cost of replacing each engine is \(K\). For the Andover, if more than one engine fails, all four engines are replaced. Otherwise, only the failed engine (if there is one) is replaced. Show that the expected repair cost for a single launch using the Andover is \[ 4Kp(1+q+q^2-2q^3) \ \ \ \ \ \ \ \ \ \ \ \ \ (q=1-p) \tag{*} \] For the Basingstoke, if more than two engines fail, all six engines are replaced. Otherwise only the failed engines (if there are any) are replaced. Find, in a form similar to \((*)\), the expected repair cost for a single launch using the Basingstoke. Find the values of \(p\) for which the expected repair cost for the Andover is \(\frac23\) of the expected repair cost for the Basingstoke.