Edexcel FP1 (Further Pure Mathematics 1) 2013 June

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Question 1 3 marks
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The complex numbers \(z\) and \(w\) are given by $$z = 8 + 3\text{i}, \quad w = -2\text{i}$$ Express in the form \(a + b\text{i}\), where \(a\) and \(b\) are real constants,
  1. \(z - w\), [1]
  2. \(zw\). [2]
Question 2 6 marks
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  1. \(\mathbf{A} = \begin{pmatrix} 2k + 1 & k \\ -3 & -5 \end{pmatrix}\), where \(k\) is a constant Given that $$\mathbf{B} = \mathbf{A} + 3\mathbf{I}$$ where \(\mathbf{I}\) is the \(2 \times 2\) identity matrix, find
    1. \(\mathbf{B}\) in terms of \(k\), [2]
    2. the value of \(k\) for which \(\mathbf{B}\) is singular. [2]
  2. Given that $$\mathbf{C} = \begin{pmatrix} 2 \\ -3 \\ 4 \end{pmatrix}, \quad \mathbf{D} = (2 \quad -1 \quad 5)$$ and $$\mathbf{E} = \mathbf{CD}$$ find \(\mathbf{E}\). [2]
Question 3 10 marks
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$$f(x) = \frac{1}{2}x^4 - x^3 + x - 3$$
  1. Show that the equation \(f(x) = 0\) has a root \(\alpha\) between \(x = 2\) and \(x = 2.5\) [2]
  2. Starting with the interval \([2, 2.5]\) use interval bisection twice to find an interval of width \(0.125\) which contains \(\alpha\). [3]
The equation \(f(x) = 0\) has a root \(\beta\) in the interval \([-2, -1]\).
  1. Taking \(-1.5\) as a first approximation to \(\beta\), apply the Newton-Raphson process once to \(f(x)\) to obtain a second approximation to \(\beta\). Give your answer to 2 decimal places. [5]
Question 4 6 marks
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$$f(x) = (4x^2 + 9)(x^2 - 2x + 5)$$
  1. Find the four roots of \(f(x) = 0\) [4]
  2. Show the four roots of \(f(x) = 0\) on a single Argand diagram. [2]
Question 5 8 marks
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\includegraphics{figure_1} Figure 1 shows a rectangular hyperbola \(H\) with parametric equations $$x = 3t, \quad y = \frac{3}{t}, \quad t \neq 0$$ The line \(L\) with equation \(6y = 4x - 15\) intersects \(H\) at the point \(P\) and at the point \(Q\) as shown in Figure 1.
  1. Show that \(L\) intersects \(H\) where \(4t^2 - 5t - 6 = 0\) [3]
  2. Hence, or otherwise, find the coordinates of points \(P\) and \(Q\). [5]
Question 6 7 marks
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$$\mathbf{A} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \quad \mathbf{B} = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$$ The transformation represented by \(\mathbf{B}\) followed by the transformation represented by \(\mathbf{A}\) is equivalent to the transformation represented by \(\mathbf{P}\).
  1. Find the matrix \(\mathbf{P}\). [2]
Triangle \(T\) is transformed to the triangle \(T'\) by the transformation represented by \(\mathbf{P}\). Given that the area of triangle \(T'\) is 24 square units,
  1. find the area of triangle \(T\). [3]
Triangle \(T'\) is transformed to the original triangle \(T\) by the matrix represented by \(\mathbf{Q}\).
  1. Find the matrix \(\mathbf{Q}\). [2]
Question 7 8 marks
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The parabola \(C\) has equation \(y^2 = 4ax\), where \(a\) is a positive constant. The point \(P(at^2, 2at)\) is a general point on \(C\).
  1. Show that the equation of the tangent to \(C\) at \(P(at^2, 2at)\) is $$ty = x + at^2$$ [4]
The tangent to \(C\) at \(P\) meets the \(y\)-axis at a point \(Q\).
  1. Find the coordinates of \(Q\). [1]
Given that the point \(S\) is the focus of \(C\),
  1. show that \(PQ\) is perpendicular to \(SQ\). [3]
Question 8 10 marks
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  1. Prove by induction, that for \(n \in \mathbb{Z}^+\), $$\sum_{r=1}^{n} r(2r - 1) = \frac{1}{6}n(n + 1)(4n - 1)$$ [6]
  2. Hence, show that $$\sum_{r=n+1}^{2n} r(2r - 1) = \frac{1}{3}n(an^2 + bn + c)$$ where \(a\), \(b\) and \(c\) are integers to be found. [4]
Question 9 9 marks
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The complex number \(w\) is given by $$w = 10 - 5\text{i}$$
  1. Find \(|w|\). [1]
  2. Find \(\arg w\), giving your answer in radians to 2 decimal places. [2]
The complex numbers \(z\) and \(w\) satisfy the equation $$(2 + \text{i})(z + 3\text{i}) = w$$
  1. Use algebra to find \(z\), giving your answer in the form \(a + b\text{i}\), where \(a\) and \(b\) are real numbers. [4]
Given that $$\arg(\lambda + 9\text{i} + w) = \frac{\pi}{4}$$ where \(\lambda\) is a real constant,
  1. find the value of \(\lambda\). [2]
Question 10 8 marks
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  1. Use the standard results for \(\sum_{r=1}^{n} r^3\) and \(\sum_{r=1}^{n} r\) to evaluate $$\sum_{r=1}^{24} (r^3 - 4r)$$ [2]
  2. Use the standard results for \(\sum_{r=1}^{n} r^2\) and \(\sum_{r=1}^{n} r\) to show that $$\sum_{r=0}^{n} (r^2 - 2r + 2n + 1) = \frac{1}{6}(n + 1)(n + a)(bn + c)$$ for all integers \(n \geqslant 0\), where \(a\), \(b\) and \(c\) are constant integers to be found. [6]