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2009 Paper 1 Q11
D: 1500.0 B: 1500.0

Two particles move on a smooth horizontal table and collide. The masses of the particles are \(m\) and \(M\). Their velocities before the collision are \(u{\bf i}\) and \(v{\bf i}\,\), respectively, where \(\bf i\) is a unit vector and \(u>v\). Their velocities after the collision are \(p{\bf i}\) and \(q{\bf i}\,\), respectively. The coefficient of restitution between the two particles is \(e\), where \(e<1\).

  1. Show that the loss of kinetic energy due to the collision is \[ \tfrac12 m (u-p)(u-v)(1-e)\,, \] and deduce that \(u\ge p\).
  2. Given that each particle loses the same (non-zero) amount of kinetic energy in the collision, show that \[ u+v+p+q=0\,, \] and that, if \(m\ne M\), \[ e= \frac{(M+3m)u + (3M+m)v}{(M-m)(u-v)}\,. \]


Solution:

  1. \begin{align*} \text{COM}: && mu + Mv &= mp + Mq \\ \Rightarrow && m(u-p) &= M(q-v) \\ \text{NEL}: && q-p &= e(u-v) \\ && q +ev &= p+eu \\ && \Delta \text{ k.e.} &= \frac12 m u^2 + \frac12 M v^2 -\frac12 m p^2 - \frac12 M q^2 \\ &&&= \frac12m (u^2 - p^2)+\frac12M(v^2-q^2) \\ &&&= \frac12m (u^2 - p^2)+\frac12M(v-q)(v+q) \\ &&&= \frac12m(u^2-p^2) - \frac12 m(u-p)(v+q) \\ &&&= \frac12 m(u-p) \left ( u+p-v-q\right) \\ &&&= \frac12 m(u-p) \left (u-v+(p-q)\right) \\ &&&= \frac12 m(u-p) \left (u-v-e(u-v)\right) \\ &&&= \frac12m(u-p)(u-v)(1-e) \end{align*} Since the loss in energy is positive, and \(m\), \(u-v\) and \(1-e\) are all positive, so is \(u-p\), ie \(u \geq p\)
  2. \begin{align*} && \frac12 m u^2 - \frac12mp^2 &= \frac12Mv^2 - \frac12Mq^2 \\ && \frac12 m(u-p)(u+p) &= \frac12 M (v-q)(v+q) \\ && \frac12 m (u-p)(u+p) &= -\frac12 m(u-p)(v+q) \\ \Rightarrow && u+p+v+q &= 0 \end{align*} \begin{align*} && p+q &= -(u+v)\\ &&mp+Mq &= mu+Mv \\ \Rightarrow && (M-m)q &= mu+Mv+mu+mv\\ \Rightarrow && q &= \frac{(M+m)v+2mu}{M-m} \\ \Rightarrow && (m-M)p &= mu+Mv+Mu+Mv \\ \Rightarrow && p &= -\frac{(M+m)u+2Mv}{M-m} \\ \\ && e &= \frac{q-p}{u-v} \\ &&&= \frac{(M+m)v+2mu+(M+m)u+2Mv}{(u-v)(M-m)} \\ &&&= \frac{(3M+m)v+(3m+M)u}{(u-v)(M-m)} \end{align*}

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 \[ \Big( \frac {a^2+1}{2a} \Big) \ln \Big ( 1+ \big( k-\tfrac34)^{-1} \Big)\,. \]

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 \[ \tan \tfrac1{48}\pi = \sqrt{16+10\sqrt2+8\sqrt3 +6\sqrt6 \ }-2-\sqrt2-\sqrt3-\sqrt6\,. \]

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 \(\displaystyle S=\sum_{i=1}^\infty \dfrac1 {F_i}\,\). \begin{questionparts} \item Show that \(\displaystyle \frac 1{F_i} > \frac1{2F_{i-1}}\,\) for \(i\ge4\) and deduce that \(S > 3\,\). Show also that \(S < 3\frac23\,\). \item Show further that \(3.2 < S < 3.5\,\). \end{questionpart}


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\,.$