AQA C4 (Core Mathematics 4) 2016 June

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Question 1 11 marks
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  1. Express \(\frac{19x - 3}{(1 + 2x)(3 - 4x)}\) in the form \(\frac{A}{1 + 2x} + \frac{B}{3 - 4x}\). [3 marks]
    1. Find the binomial expansion of \(\frac{19x - 3}{(1 + 2x)(3 - 4x)}\) up to and including the term in \(x^2\). [7 marks]
    2. State the range of values of \(x\) for which this expansion is valid. [1 mark]
Question 2 5 marks
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By forming and solving a suitable quadratic equation, find the solutions of the equation $$3 \cos 2\theta - 5 \cos \theta + 2 = 0$$ in the interval \(0° < \theta < 360°\), giving your answers to the nearest \(0.1°\). [5 marks]
Question 3 8 marks
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  1. Express \(\frac{3 + 13x - 6x^2}{2x - 3}\) in the form \(Ax + B + \frac{C}{2x - 3}\). [4 marks]
  2. Show that \(\int_3^6 \frac{3 + 13x - 6x^2}{2x - 3} \, dx = p + q \ln 3\), where \(p\) and \(q\) are rational numbers. [4 marks]
Question 4 7 marks
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The mass of radioactive atoms in a substance can be modelled by the equation $$m = m_0 k^t$$ where \(m_0\) grams is the initial mass, \(m\) grams is the mass after \(t\) days and \(k\) is a constant. The value of \(k\) differs from one substance to another.
    1. A sample of radioactive iodine reduced in mass from 24 grams to 12 grams in 8 days. Show that the value of the constant \(k\) for this substance is 0.917004, correct to six decimal places. [1 mark]
    2. A similar sample of radioactive iodine reduced in mass to 1 gram after 60 days. Calculate the initial mass of this sample, giving your answer to the nearest gram. [2 marks]
  1. The half-life of a radioactive substance is the time it takes for a mass of \(m_0\) to reduce to a mass of \(\frac{1}{2}m_0\). A sample of radioactive vanadium reduced in mass from exactly 10 grams to 8.106 grams in 100 days. Find the half-life of radioactive vanadium, giving your answer to the nearest day. [4 marks]
Question 5 10 marks
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It is given that \(\sin A = \frac{\sqrt{5}}{3}\) and \(\sin B = \frac{1}{\sqrt{5}}\), where the angles \(A\) and \(B\) are both acute.
    1. Show that the exact value of \(\cos B = \frac{2}{\sqrt{5}}\). [1 mark]
    2. Hence show that the exact value of \(\sin 2B\) is \(\frac{4}{5}\). [2 marks]
    1. Show that the exact value of \(\sin(A - B)\) can be written as \(p(5 - \sqrt{5})\), where \(p\) is a rational number. [4 marks]
    2. Find the exact value of \(\cos(A - B)\) in the form \(r + s\sqrt{5}\), where \(r\) and \(s\) are rational numbers. [3 marks]
Question 6 15 marks
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The line \(l_1\) passes through the point \(A(0, 6, 9)\) and the point \(B(4, -6, -11)\). The line \(l_2\) has equation \(\mathbf{r} = \begin{bmatrix} -1 \\ 5 \\ -2 \end{bmatrix} + \lambda \begin{bmatrix} 3 \\ -5 \\ 1 \end{bmatrix}\).
  1. The acute angle between the lines \(l_1\) and \(l_2\) is \(\theta\). Find the value of \(\cos \theta\) as a fraction in its lowest terms. [5 marks]
  2. Show that the lines \(l_1\) and \(l_2\) intersect and find the coordinates of the point of intersection. [5 marks]
  3. The points \(C\) and \(D\) lie on line \(l_2\) such that \(ACBD\) is a parallelogram. \includegraphics{figure_6} The length of \(AB\) is three times the length of \(CD\). Find the coordinates of the points \(C\) and \(D\). [5 marks]
Question 7 9 marks
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A curve \(C\) is defined by the parametric equations $$x = \frac{4 - e^{-6t}}{4}, \quad y = \frac{e^{3t}}{3t}, \quad t \neq 0$$
  1. Find the exact value of \(\frac{dy}{dx}\) at the point on \(C\) where \(t = \frac{2}{3}\). [5 marks]
  2. Show that \(x = \frac{4 - e^{-6t}}{4}\) can be rearranged into the form \(e^{3t} = \frac{e}{2\sqrt{(1-x)}}\). [2 marks]
  3. Hence find the Cartesian equation of \(C\), giving your answer in the form $$y = \frac{e}{f(x)[1 - \ln(f(x))]}$$ [2 marks]
Question 8 10 marks
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It is given that \(\theta = \tan^{-1}\left(\frac{3x}{2}\right)\).
  1. By writing \(\theta = \tan^{-1}\left(\frac{3x}{2}\right)\) as \(2\tan\theta = 3x\), use implicit differentiation to show that $$\frac{d\theta}{dx} = \frac{k}{4 + 9x^2}$$, where \(k\) is an integer. [3 marks]
  2. Hence solve the differential equation $$9y(4 + 9x^2)\frac{dy}{dx} = \cosec 3y$$ given that \(x = 0\) when \(y = \frac{\pi}{3}\). Give your answer in the form \(\mathbf{g}(y) = \mathbf{h}(x)\). [7 marks]