AQA C4 (Core Mathematics 4) 2016 June

1

1. Express \(\frac { 19 x - 3 } { ( 1 + 2 x ) ( 3 - 4 x ) }\) in the form \(\frac { A } { 1 + 2 x } + \frac { B } { 3 - 4 x }\).
2. 1. Find the binomial expansion of \(\frac { 19 x - 3 } { ( 1 + 2 x ) ( 3 - 4 x ) }\) up to and including the term in \(x ^ { 2 }\).
   2. State the range of values of \(x\) for which this expansion is valid.
      [0pt] [1 mark]

3

1. Express \(\frac { 3 + 13 x - 6 x ^ { 2 } } { 2 x - 3 }\) in the form \(A x + B + \frac { C } { 2 x - 3 }\).
2. Show that \(\int _ { 3 } ^ { 6 } \frac { 3 + 13 x - 6 x ^ { 2 } } { 2 x - 3 } \mathrm {~d} x = p + q \ln 3\), where \(p\) and \(q\) are rational numbers.
   [0pt] [4 marks]

4 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. 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.
   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. 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]
   \includegraphics[max width=\textwidth, alt={}]{c42685e9-bfa4-48d4-8abb-13e88a4b765e-08_1182_1707_1525_153}

5 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. 1. Show that the exact value of \(\cos B = \frac { 2 } { \sqrt { 5 } }\).
   2. Hence show that the exact value of \(\sin 2 B\) is \(\frac { 4 } { 5 }\).
2. 1. Show that the exact value of \(\sin ( A - B )\) can be written as \(p ( 5 - \sqrt { 5 } )\), where \(p\) is a rational number.
   2. Find the exact value of \(\cos ( A - B )\) in the form \(r + s \sqrt { 5 }\), where \(r\) and \(s\) are rational numbers.

6 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 \(\mathbf { r } = \left[ \begin{array} { r } - 1
5
- 2 \end{array} \right] + \lambda \left[ \begin{array} { r } 3
- 5
1 \end{array} \right]\).

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.
2. Show that the lines \(l _ { 1 }\) and \(l _ { 2 }\) intersect and find the coordinates of the point of intersection.
3. The points \(C\) and \(D\) lie on line \(l _ { 2 }\) such that \(A C B D\) is a parallelogram.
   \includegraphics[max width=\textwidth, alt={}, center]{c42685e9-bfa4-48d4-8abb-13e88a4b765e-12_392_949_1018_548} The length of \(A B\) is three times the length of \(C D\).
   Find the coordinates of the points \(C\) and \(D\).
   [0pt] [5 marks] \(7 \quad\) A curve \(C\) is defined by the parametric equations $$x = \frac { 4 - \mathrm { e } ^ { 2 - 6 t } } { 4 } , \quad y = \frac { \mathrm { e } ^ { 3 t } } { 3 t } , \quad t \neq 0$$

1. Find the exact value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) at the point on \(C\) where \(t = \frac { 2 } { 3 }\).
2. Show that \(x = \frac { 4 - \mathrm { e } ^ { 2 - 6 t } } { 4 }\) can be rearranged into the form \(\mathrm { e } ^ { 3 t } = \frac { \mathrm { e } } { 2 \sqrt { ( 1 - x ) } }\).
3. Hence find the Cartesian equation of \(C\), giving your answer in the form $$y = \frac { \mathrm { e } } { \mathrm { f } ( x ) [ 1 - \ln ( \mathrm { f } ( x ) ) ] }$$

8 It is given that \(\theta = \tan ^ { - 1 } \left( \frac { 3 x } { 2 } \right)\).

1. By writing \(\theta = \tan ^ { - 1 } \left( \frac { 3 x } { 2 } \right)\) as \(2 \tan \theta = 3 x\), use implicit differentiation to show that \(\frac { \mathrm { d } \theta } { \mathrm { d } x } = \frac { k } { 4 + 9 x ^ { 2 } }\), where \(k\) is an integer.
   [0pt] [3 marks]
2. Hence solve the differential equation $$9 y \left( 4 + 9 x ^ { 2 } \right) \frac { \mathrm { d } y } { \mathrm {~d} x } = \operatorname { cosec } 3 y$$ given that \(x = 0\) when \(y = \frac { \pi } { 3 }\). Give your answer in the form \(\mathrm { g } ( y ) = \mathrm { h } ( x )\).
   [0pt] [7 marks]