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2004 Paper 1 Q1
D: 1500.0 B: 1516.0

  1. Express \(\left(3+2\sqrt{5} \, \right)^3\) in the form \(a+b\sqrt{5}\) where \(a\) and \(b\) are integers.
  2. Find the positive integers \(c\) and \(d\) such that \(\sqrt[3]{99-70\sqrt{2}\;}\) = \(c - d\sqrt{2} \,\).
  3. Find the two real solutions of \(x^6 - 198 x^3 + 1 = 0 \,\).

Solution:

  1. \begin{align*} (3+2\sqrt{5})^3 &= 3^3 + 3 \cdot 3^2 \cdot 2\sqrt{5} + 3 \cdot 3 \cdot (2 \sqrt{5})^2 + (2\sqrt{5})^3 \\ &= 27 + 180 + (54+40)\sqrt{5} \\ &= 207 + 94\sqrt{5} \end{align*}
  2. \begin{align*} && (c-d\sqrt{2})^3 &= c^3+6cd -(3c^2d+2d^3)\sqrt{2} \\ \Rightarrow && 99 &= c(c^2+6d^2) \\ && 70 &= d(3c^2+2d^2) \\ \Rightarrow && c & \mid 99, d \mid 70 \\ && c &= 3, d = 2 \end{align*}
  3. \begin{align*} && 0 &= x^6 - 198x^3 + 1 \\ \Rightarrow && 0 &= (x^3-99)^2+1-99^2 \\ \Rightarrow && x^3 &= 99 \pm \sqrt{99^2-1} \\ &&&= 99 \pm 10 \sqrt{98} \\ &&&= 99 \pm 70 \sqrt{2} \\ \Rightarrow && x &= 3 \pm 2 \sqrt{2} \end{align*}

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2004 Paper 1 Q2
D: 1484.0 B: 1499.3

The square bracket notation \(\boldsymbol{[} x\boldsymbol{]}\) means the greatest integer less than or equal to \(x\,\). For example, \(\boldsymbol{[}\pi\boldsymbol{]} = 3\,\), \(\boldsymbol{[}\sqrt{24}\,\boldsymbol{]} = 4\,\) and \(\boldsymbol{[}5\boldsymbol{]}=5\,\).

  1. Sketch the graph of \(y = \sqrt{\boldsymbol{[}x\boldsymbol{]}}\) and show that \[ \displaystyle \int^a_0 \sqrt{\boldsymbol{[}x\boldsymbol{]}} \; \mathrm{d}x = \sum^{a-1}_{r=0} \sqrt{r} \] when \(a\) is a positive integer.
  2. Show that $\displaystyle \int^{a}_0 2_{\vphantom{A}}^{\pmb{\boldsymbol {[} } x \pmb{ \boldsymbol{]}} }\; \mathrm{d}x = 2^{a}-1$ when \(a\) is a positive integer.
  3. Determine an expression for $\displaystyle \int^{a}_0 2_{\vphantom{\dot A}}^{\pmb{\boldsymbol{[} }x \pmb{ \boldsymbol{]}} } \; \mathrm{d}x\( when \)a$ is positive but not an integer.

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2004 Paper 1 Q3
D: 1500.0 B: 1500.0

  1. Show that \(x-3\) is a factor of \begin{equation} x^3-5x^2+2x^2y+xy^2-8xy-3y^2+6x+6y \;. \tag{\(*\)} \end{equation} Express (\( * \)) in the form \((x-3)(x+ay+b)(x+cy+d)\) where \(a\), \(b\), \(c\) and \(d\) are integers to be determined.
  2. Factorise \(6y^3-y^2-21y+2x^2+12x-4xy+x^2y-5xy^2+10\) into three linear factors.

Solution:

  1. Let \(f(x,y) = x^3-5x^2+2x^2y+xy^2-8xy-3y^2+6x+6y\), then \begin{align*} f(3,y) &= 27 - 5 \cdot 9 +18y + 3y^2-24y-3y^2+18 + 6y \\ &= 0 \end{align*}, therefore \(x-3\) is a factor of \(f(x,y)\). \begin{align*} f(x,y) &= x^3-5x^2+6x+y(2x^2-8x+6) + y^2(x-3) \\ &= (x-3)(x^2-2x)+y(x-3)(2x-2)+y^2(x-3) \\ &= (x-3)(x^2-2x+2y(x-1)+y^2) \\ &= (x-3)(x+y)(x+y-2) \end{align*}
  2. Let \(g(x,y) = 6y^3-y^2-21y+2x^2+12x-4xy+x^2y-5xy^2+10\), notice that \(g(x,-2) = 0\), so \(y+2\) is a factor, \begin{align*} g(x,y) &= 6y^3-y^2-21y+2x^2+12x-4xy+x^2y-5xy^2+10 \\ &= x^2(2+y) + x(12-4y-5y^2) + 6y^3-y^2-21y+10 \\ &= x^2(y+2) + x(y+2)(6-5y) + (y+2)(6y^2-13y+5) \\ &= (y+2)(x^2+(6-5y)x+(6y^2-13y+5)) \\ &= (y+2)(x-2y +1)(x-3y+5) \end{align*}

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2004 Paper 1 Q4
D: 1516.0 B: 1484.0

Differentiate \(\sec {t}\) with respect to \(t\).

  1. Use the substitution \(x=\sec t\) to show that $\displaystyle \int^2_{\sqrt 2} \frac{1}{ x^3\sqrt {x^2-1} } \; \mathrm{d}x =\frac{\sqrt 3 - 2}{8} + \frac {\pi}{24} \;.$
  2. Determine $\displaystyle \int \frac{1} {( x+2) \sqrt {(x+1)(x+3)} } \; \mathrm{d}x \;$.
  3. Determine $\displaystyle \int \frac {1} {(x+2) \sqrt {x^2+4x-5} } \; \mathrm{d}x \;$.

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2004 Paper 1 Q5
D: 1484.0 B: 1500.0

The positive integers can be split into five distinct arithmetic progressions, as shown: \begin{align*} A&: \ \ 1, \ 6, \ 11, \ 16, \ ... \\ B&: \ \ 2, \ 7, \ 12, \ 17, \ ...\\ C&: \ \ 3, \ 8, \ 13, \ 18, \ ... \\ D&: \ \ 4, \ 9, \ 14, \ 19, \ ... \\ E&: \ \ 5, 10, \ 15, \ 20, \ ... \end{align*} Write down an expression for the value of the general term in each of the five progressions. Hence prove that the sum of any term in \(B\) and any term in \(C\) is a term in \(E\). Prove also that the square of every term in \(B\) is a term in \(D\). State and prove a similar claim about the square of every term in \(C\).

  1. Prove that there are no positive integers \(x\) and \(y\) such that \[ x^2+5y=243\,723 \,. \]
  2. Prove also that there are no positive integers \(x\) and \(y\) such that \[ x^4+2y^4=26\,081\,974 \,. \]

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2004 Paper 1 Q6
D: 1484.0 B: 1500.0

The three points \(A\), \(B\) and \(C\) have coordinates \(\l p_1 \, , \; q_1 \r\), \(\l p_2 \, , \; q_2 \r\) and \(\l p_3 \, , \; q_3 \r\,\), respectively. Find the point of intersection of the line joining \(A\) to the midpoint of \(BC\), and the line joining~\(B\) to the midpoint of \(AC\). Verify that this point lies on the line joining \(C\) to the midpoint of~\(AB\). The point \(H\) has coordinates \(\l p_1 + p_2 + p_3 \, , \; q_1 + q_2 + q_3 \r\,\). Show that if the line \(AH\) intersects the line \(BC\) at right angles, then \(p_2^2 + q_2^2 = p_3^2 + q_3^2\,\), and write down a similar result if the line \(BH\) intersects the line \(AC\) at right angles. Deduce that if \(AH\) is perpendicular to \(BC\) and also \(BH\) is perpendicular to \(AC\), then \(CH\) is perpendicular to \(AB\).

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2004 Paper 1 Q7
D: 1500.0 B: 1500.0

  1. The function \(\f(x)\) is defined for \(\vert x \vert < \frac15\) by \[ \f(x) = \sum_{n=0}^\infty a_n x^n\;, \] where \(a_0=2\), \(a_1=7\) and \(a_n =7a_{n-1} - 10a_{n-2}\) for \(n\ge{2}\,\). Simplify \(\f(x) - 7x\f(x) + 10x^2\f(x)\,\), and hence show that \(\displaystyle\f(x) = {1\over 1-2x} + {1 \over 1-5x} \;\). Hence show that \(a_n=2^n + 5^n\,\).
  2. The function \(\g(x)\) is defined for \(\vert x \vert < \frac13\) by \[ \g(x) = \sum_{n=0}^\infty b_n x^n \;, \] where \(b_0=5\,\), \(b_1 =10 \,\), \(b_2=40\,\), \(b_3=100\) and \(b_n = pb_{n-1} + qb_{n-2}\) for \(n\ge{2}\,\). Obtain an expression for \(\g(x)\) as the sum of two algebraic fractions and determine \(b_n\) in terms of \(n\).

Solution:

  1. \begin{align*} && f(x) -7xf(x)+10x^2f(x) &= \sum_{n=0}^\infty a_n x^n - 7x \sum_{n=0}^{\infty} a_n x^n + 10x^2 \sum_{n=0}^{\infty} a_nx^n \\ &&&= \sum_{n=2}^\infty (a_n-7a_{n-1}+10a_{n-2})x^n + a_0+a_1x-7a_0x \\ &&&= 0 + 2-7x \\ \\ \Rightarrow && f(x) &= \frac{2-7x}{1-7x+10x^2} \\ &&&= \frac{2-7x}{(1-5x)(1-2x)} \\ &&&= \frac{1}{1-2x} + \frac{1}{1-5x} \\ &&&= \sum_{n=0}^{\infty} (2^n + 5^n)x^n \end{align*} Therefore \(a_n = 2^n +5^n\)
  2. \(\,\) \begin{align*} && 40 &= 10p + 5 q \\ && 100 &= 40p+10q \\ && 10 &= 4p + q \\ \Rightarrow && (p,q) &= (1,6) \\ \\ && g(x) -xg(x)-6x^2g(x) &= 5+5x \\ \Rightarrow && g(x) &= \frac{5+5x}{1-x-6x^2} \\ &&&= \frac{5+5x}{(1-3x)(1+2x)} \\ &&&= \frac{4}{1-3x} + \frac{1}{1+2x} \\ &&&= \sum_{n=0}^{\infty} (4 \cdot 3^n + (-2)^n)x^n \\ \Rightarrow && b_n &= 4 \cdot 3^n + (-2)^n \end{align*}

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2004 Paper 1 Q8
D: 1500.0 B: 1547.8

A sequence \(t_0\), \(t_1\), \(t_2\), \(...\) is said to be {\sl strictly increasing} if \(t_{n+1} > t_n\) for all \(n\ge{0}\,\).

  1. The terms of the sequence \(x_0\,\), \(x_1\,\), \(x_2\,\), \(\ldots\) satisfy $$ \ds x_{n+1}=\frac{x_n^2 +6}{5} $$ for \(n\ge{0}\,\). Prove that if \(x_0 > 3\) then the sequence is strictly increasing.
  2. The terms of the sequence \(y_0\,\), \(y_1\,\), \(y_2\,\), \(\ldots\) satisfy $$ \ds y_{n+1}= 5-\frac 6 {y_n} $$ for \(n\ge{0}\,\). Prove that if \(2 < y_0 < 3\) then the sequence is strictly increasing but that \(y_n<3\) for all \(n\,\).

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2004 Paper 1 Q9
D: 1484.0 B: 1489.3

A particle is projected over level ground with a speed \(u\) at an angle \(\theta\) above the horizontal. Derive an expression for the greatest height of the particle in terms of \(u\), \(\theta\) and \(g\). A particle is projected from the floor of a horizontal tunnel of height \({9\over 10}d\). Point \(P\) is \({1\over 2}d\) metres vertically and \(d\) metres horizontally along the tunnel from the point of projection. The particle passes through point \(P\) and lands inside the tunnel without hitting the roof. Show that \[ \arctan \textstyle {3 \over 5} < \theta < \arctan \, 3 \;. \]

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2004 Paper 1 Q10
D: 1516.0 B: 1516.0

A particle is travelling in a straight line. It accelerates from its initial velocity \(u\) to velocity \(v\), where \(v > \vert u \vert > 0\,\), travelling a distance \(d_1\) with uniform acceleration of magnitude \(3a\,\). It then comes to rest after travelling a further distance \(d_2\,\) with uniform deceleration of magnitude \(a\,\). Show that

  1. if \(u>0\) then \(3d_1 < d_2\,\);
  2. if \(u<0\) then \(d_2 < 3d_1 < 2d_2\,\).
Show also that the average speed of the particle (that is, the total distance travelled divided by the total time) is greater in the case \(u>0\) than in the case \(u<0\,\). \noindent {\bf Note:} In this question \(d_1\) and \(d_2\) are distances travelled by the particle which are not the same, in the second case, as displacements from the starting point.

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