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2010 Paper 2 Q2
D: 1600.0 B: 1506.8
integration trigonometric triple angle formula cos 3x sin 3x polynomial equation cubic definite integral proof

Prove that \[ \cos 3x = 4 \cos^3 x - 3 \cos x \,. \] Find and prove a similar result for \(\sin 3x\) in terms of \(\sin x\).

  1. Let \[ {\rm I}(\alpha) = \int_0^\alpha \big(7\sin x - 8 \sin^3 x\big) \d x\,. \] Show that \[ {\rm I}(\alpha) = -\tfrac 8 3 c^3 + c +\tfrac5 3\,, \] where \(c = \cos \alpha\). Write down one value of \(c\) for which \({\rm I}(\alpha) =0\).
  2. Useless Eustace believes that \[ \int \sin^n x \, \d x =\dfrac {\sin^{n+1}x}{n+1}\, \] for \(n=1, \ 2, \ 3, \ldots\, \). Show that Eustace would obtain the correct value of \({\rm I}(\beta)\,\), where \(\cos \beta= -\frac16\). Find all values of \(\alpha\) for which he would obtain the correct value of \({\rm I}(\alpha)\).

Solution: \begin{align*} \cos 3x &\equiv \cos (2x + x) \\ &\equiv \cos 2x \cos x - \sin 2x \sin x \\ &\equiv (2\cos^2 x - 1) \cos x - 2 \sin x \cos x \sin x \\ &\equiv 2 \cos^3 x - \cos x - 2\cos x (\sin^2 x) \\ &\equiv 2 \cos^3 x - \cos x - 2\cos x (1- \cos^2 x) \\ &\equiv 4\cos^3 x - 3\cos x \end{align*} Similarly, \begin{align*} \sin 3x &\equiv \sin (2x + x) \\ &\equiv \sin 2x \cos x + \cos 2x \sin x \\ &\equiv 2 \sin x \cos x \cos x + (1-2\sin^2 x) \sin x \\ &\equiv 2 \sin x (1-\sin^2 x) + \sin x - 2 \sin^3 x \\ &\equiv 3 \sin x -4 \sin ^3 x \end{align*}

  1. \begin{align*} I(\alpha) &= \int_0^{\alpha} (7 \sin x - 8 \sin^3 x) \d x \\ &= \int_0^{\alpha} (7 \sin x - (6\sin x-2 \sin 3x) ) \d x \\ &= \int_0^{\alpha} (\sin x +2 \sin 3x ) \d x \\ &= -\cos \alpha - \frac23 \cos 3\alpha +1+\frac23 \\ &= -c - \frac23 (4c^3-3c) + \frac53 \\ &= -\frac83 c^3 +c + \frac53 \end{align*} as required. When \(c = -1\) this value is \(0\). Eustace will obtain the value \(\frac{7}{2} \sin^2 \beta - 2 \sin^4 \beta = \frac72 (1-\cos^2 \beta) - 2(1-\cos^2 \beta)^2 = \frac32 + \frac12\cos^2 \beta -2\cos^4 \beta\) So if \(\cos \beta = -\frac16\) he will obtain \(\frac32 + \frac{1}{2\cdot36} - \frac{2}{6^4}\) and he should obtain \(\frac{8}{3} \frac{1}{6^3} - \frac{1}{6} + \frac{5}{3}\) which are equal. We want to find all roots of: \begin{align*} && \frac32 + \frac12 c^2 - 2c^4 &= -\frac83 c^3+ c + \frac53 \\ \Rightarrow && 0 &=2c^4-\frac83c^3-\frac12 c^2+c +\frac{1}{6} \\ &&&= 12c^4-16c^3-3c^2+6c+1\\ &&&= (6c+1)(2c^3-3c^2+1) \\ &&&= (6c+1)(2c+1)(c-1)^2 \end{align*} Therefore \(\cos \alpha = - \frac16, -\frac12, 1\) will give the correct answers.

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2009 Paper 1 Q2
D: 1500.0 B: 1500.0
implicit differentiation tangent-normal cubic curve polynomial equation Diophantine equation algebraic manipulation

A curve has the equation \[ y^3 = x^3 +a^3+b^3\,, \] where \(a\) and \(b\) are positive constants. Show that the tangent to the curve at the point \((-a,b)\) is \[ b^2y-a^2x = a^3+b^3\,. \] In the case \(a=1\) and \(b=2\), show that the \(x\)-coordinates of the points where the tangent meets the curve satisfy \[ 7x^3 -3x^2 -27x-17 =0\,. \] Hence find positive integers \(p\), \(q\), \(r\) and \(s\) such that \[ p^3 = q^3 +r^3 +s^3\,. \]

Solution: \begin{align*} && y^3 &= x^3 + a^3 + b^3 \\ \Rightarrow && 3y^2 \frac{\d y}{\d x} &= 3x^2 \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{x^2}{y^2} \end{align*} Therefore the tangent at the point \((-a,b)\) has gradient \(\frac{a^2}{b^2}\), ie \begin{align*} && \frac{y-b}{x+a} &= \frac{a^2}{b^2} \\ \Rightarrow && b^2y - b^3 &= a^2 x + a^3 \\ \Rightarrow && b^2 y-a^2 x &= a^3 + b^3 \end{align*} Notice that tangent will be, \(4y-x = 9\) so substituting this we obtain: \begin{align*} && \left (\frac{9+x}{4} \right)^3 &= x^3 + 9 \\ \Rightarrow && 9^3 + 3 \cdot 9^2 x + 3 \cdot 9x^2 + x^3 &= 64x^3 + 64 \cdot 9 \\ \Rightarrow && 9 \cdot (9^2 - 8^2) + 9 \cdot (3 \cdot 9) + 9 \cdot 3x^2 -9 \cdot 7x^3 &= 0 \\ \Rightarrow && 7x^3-3x^2-27x-17 &= 0 \\ \Rightarrow && (x+1)^2(7x-17) &= 0 \tag{repeated root since tangent} \end{align*} So we have another point on the curve \(y^3 = x^3 + 2^3 + 1^3\), namely \((\frac{17}7, \frac{17+9 \cdot 7}{28}) = (\frac{17}7, \frac{20}{7})\), so \begin{align*} 20^3 &= 17^3 + 14^3 + 7^3 \end{align*}

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2009 Paper 2 Q13
D: 1600.0 B: 1500.0
probability tree diagram binomial distribution expected value polynomial equation engine failure cost optimisation

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.

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2002 Paper 2 Q2
D: 1600.0 B: 1484.6
quadratics roots of polynomials substitution reciprocal polynomial quartic polynomial equation algebraic manipulation

Show that setting \(z - z^{-1}=w\) in the quartic equation \[ z^4 +5z^3 +4z^2 -5z +1=0 \] results in the quadratic equation \(w^2+5w+6=0\). Hence solve the above quartic equation. Solve similarly the equation \[ 2z^8 -3z^7-12z^6 +12z^5 +22z^4-12z^3 -12 z^2 +3z +2=0 \;. \]

Solution: \begin{align*} && 0 &= z^4 +5z^3 +4z^2 -5z +1 \\ &&0 &= z^2 + z^{-2} + 5(z-z^{-1}) + 4 \\ &&&= (z-z^{-1})^2+2+5(z-z^{-1})+4 \\ &&&= w^2 + 5w + 6 \\ &&&= (w+3)(w+2) \\ \Rightarrow && 0 &= z-z^{-1}+3 \\ \Rightarrow && 0 &= z^2+3z-1 \\ \Rightarrow && z &= \frac{-3 \pm \sqrt{3^2+4}}{2} = \frac{-3 \pm \sqrt{13}}{2} \\ \Rightarrow && 0 &= z-z^{-1}+2 \\ \Rightarrow && 0 &= z^2+2z-1 \\ \Rightarrow && z &= \frac{-2 \pm \sqrt{2^2+4}}{2} = - 1 \pm \sqrt{2} \\ \end{align*} \begin{align*} &&0 &= 2z^8 -3z^7-12z^6 +12z^5 +22z^4-12z^3 -12 z^2 +3z +2 \\ && 0 &= 2(z^4+z^{-4}) - 3(z^3-z^{-3})-12(z^2+z^{-2})+12(z-z^{-1})+22 \\ &&&= 2\left ((z-z^{-1})^4+4(z^2+z^{-2})-6\right)-3 \left ((z-z^{-1})^3+3(z-z^{-1}) \right)-12 \left ((z-z^{-1})^2+2 \right)+12(z-z^{-1})+22 \\ &&&= 2(w^4+4(w^2+2)-6)-3w^3-9w-12w^2-24+12w+22 \\ &&&= 2 w^4-3w^3-4w^2+3w+2 \\ \Rightarrow && 0 &= 2(w^2+w^{-2})-3(w-w^{-1})-4 \\ &&&= 2((w-w^{-1})^2+2)-3(w-w^{-1})-4 \\ &&&= 2x^2-3x \\ &&&= x(2x-3) \\ \Rightarrow && 0 &= w -w^{-1} \\ \Rightarrow && w &= \pm 1 \\ \Rightarrow && \pm 1 &= z-z^{-1} \\ \Rightarrow && 0 &= z^2 \mp z-1 \\ \Rightarrow && z &= \frac{\pm 1 \pm \sqrt{5}}{2} \\ \Rightarrow && \frac32 &= w-w^{-1} \\ \Rightarrow && 0 &= 2w^2-3w -2 \\ &&&= (2w+1)(w-2) \\ \Rightarrow && 2 &= z-z^{-1} \\ \Rightarrow && 0 &= z^2-2z-1 \\ \Rightarrow && z &= 1 \pm \sqrt{2} \\ \Rightarrow && -\frac12 &= z-z^{-1} \\ \Rightarrow && 0 &= 2z^2+z-2 \\ \Rightarrow && z &= \frac{-1 \pm \sqrt{17}}{4} \\ \Rightarrow && z &\in \left \{ \frac{\pm 1 \pm \sqrt{5}}{2}, 1 \pm \sqrt{2}, \frac{-1 \pm \sqrt{17}}{4} \right \} \end{align*}

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