AQA C4 (Core Mathematics 4) 2012 January

Question 1
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1
  1. Express \(\frac { 2 x + 3 } { 4 x ^ { 2 } - 1 }\) in the form \(\frac { A } { 2 x - 1 } + \frac { B } { 2 x + 1 }\), where \(A\) and \(B\) are integers. (3 marks)
  2. Express \(\frac { 12 x ^ { 3 } - 7 x - 6 } { 4 x ^ { 2 } - 1 }\) in the form \(C x + \frac { D ( 2 x + 3 ) } { 4 x ^ { 2 } - 1 }\), where \(C\) and \(D\) are integers.
    (3 marks)
  3. Evaluate \(\int _ { 1 } ^ { 2 } \frac { 12 x ^ { 3 } - 7 x - 6 } { 4 x ^ { 2 } - 1 } \mathrm {~d} x\), giving your answer in the form \(p + \ln q\), where \(p\) and \(q\) are rational numbers.
    (5 marks)
Question 2
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2 Angle \(\alpha\) is acute and \(\cos \alpha = \frac { 3 } { 5 }\). Angle \(\beta\) is obtuse and \(\sin \beta = \frac { 1 } { 2 }\).
    1. Find the value of \(\tan \alpha\) as a fraction.
      (1 mark)
    2. Find the value of \(\tan \beta\) in surd form.
  1. Hence show that \(\tan ( \alpha + \beta ) = \frac { m \sqrt { 3 } - n } { n \sqrt { 3 } + m }\), where \(m\) and \(n\) are integers.
    (3 marks)
Question 3
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3
  1. Find the binomial expansion of \(( 1 + 6 x ) ^ { \frac { 2 } { 3 } }\) up to and including the term in \(x ^ { 2 }\).
    (2 marks)
  2. Find the binomial expansion of \(( 8 + 6 x ) ^ { \frac { 2 } { 3 } }\) up to and including the term in \(x ^ { 2 }\).
    (3 marks)
  3. Use your answer from part (b) to find an estimate for \(\sqrt [ 3 ] { 100 }\) in the form \(\frac { a } { b }\), where \(a\) and \(b\) are integers.
    (2 marks)
Question 4
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4 A scientist is testing models for the growth and decay of colonies of bacteria. For a particular colony, which is growing, the model is \(P = A \mathrm { e } ^ { \frac { 1 } { 8 } t }\), where \(P\) is the number of bacteria after a time \(t\) minutes and \(A\) is a constant.
  1. This growing colony consists initially of 500 bacteria. Calculate the number of bacteria, according to the model, after one hour. Give your answer to the nearest thousand.
  2. For a second colony, which is decaying, the model is \(Q = 500000 \mathrm { e } ^ { - \frac { 1 } { 8 } t }\), where \(Q\) is the number of bacteria after a time \(t\) minutes. Initially, the growing colony has 500 bacteria and, at the same time, the decaying colony has 500000 bacteria.
    1. Find the time at which the populations of the two colonies will be equal, giving your answer to the nearest 0.1 of a minute.
    2. The population of the growing colony will exceed that of the decaying colony by 45000 bacteria at time \(T\) minutes. Show that $$\left( \mathrm { e } ^ { \frac { 1 } { 8 } T } \right) ^ { 2 } - 90 \mathrm { e } ^ { \frac { 1 } { 8 } T } - 1000 = 0$$ and hence find the value of \(T\), giving your answer to one decimal place.
      (4 marks)
Question 5
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5 A curve is defined by the parametric equations $$x = 8 t ^ { 2 } - t , \quad y = \frac { 3 } { t }$$
  1. Show that the cartesian equation of the curve can be written as \(x y ^ { 2 } + 3 y = k\), stating the value of the integer \(k\).
    (2 marks)
    1. Find an equation of the tangent to the curve at the point \(P\), where \(t = \frac { 1 } { 4 }\).
    2. Verify that the tangent at \(P\) intersects the curve when \(x = \frac { 3 } { 2 }\).
Question 6
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6
  1. Use the Factor Theorem to show that \(4 x - 3\) is a factor of $$16 x ^ { 3 } + 11 x - 15$$
  2. Given that \(x = \cos \theta\), show that the equation $$27 \cos \theta \cos 2 \theta + 19 \sin \theta \sin 2 \theta - 15 = 0$$ can be written in the form $$16 x ^ { 3 } + 11 x - 15 = 0$$
  3. Hence show that the only solutions of the equation $$27 \cos \theta \cos 2 \theta + 19 \sin \theta \sin 2 \theta - 15 = 0$$ are given by \(\cos \theta = \frac { 3 } { 4 }\).
Question 7
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7 Solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = y ^ { 2 } x \sin 3 x$$ given that \(y = 1\) when \(x = \frac { \pi } { 6 }\). Give your answer in the form \(y = \frac { 9 } { \mathrm { f } ( x ) }\).
Question 8
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8 The points \(A\) and \(B\) have coordinates \(( 4 , - 2,3 )\) and \(( 2,0 , - 1 )\) respectively. The line \(l\) passes through \(A\) and has equation \(\mathbf { r } = \left[ \begin{array} { r } 4
- 2
3 \end{array} \right] + \lambda \left[ \begin{array} { r } 1
5
- 2 \end{array} \right]\).
    1. Find the vector \(\overrightarrow { A B }\).
    2. Find the acute angle between \(A B\) and the line \(l\), giving your answer to the nearest degree.
  1. The point \(C\) lies on the line \(l\) such that the angle \(A B C\) is a right angle. Given that \(A B C D\) is a rectangle, find the coordinates of the point \(D\).