AQA C4 (Core Mathematics 4) 2014 June

1 A curve is defined by the parametric equations \(x = \frac { t ^ { 2 } } { 2 } + 1 , y = \frac { 4 } { t } - 1\).

1. Find the gradient at the point on the curve where \(t = 2\).
2. Find a Cartesian equation of the curve.
   \includegraphics[max width=\textwidth, alt={}]{9f03a5f3-7fea-4fb7-b3bd-b4c0cdf662a2-02_1730_1709_977_153}

2

1. Given that \(\frac { 4 x ^ { 3 } - 2 x ^ { 2 } + 16 x - 3 } { 2 x ^ { 2 } - x + 2 }\) can be expressed as \(A x + \frac { B ( 4 x - 1 ) } { 2 x ^ { 2 } - x + 2 }\), find the values of the constants \(A\) and \(B\).
2. The gradient of a curve is given by $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 x ^ { 3 } - 2 x ^ { 2 } + 16 x - 3 } { 2 x ^ { 2 } - x + 2 }$$ The point \(( - 1,2 )\) lies on the curve. Find the equation of the curve.
   [0pt] [4 marks]

3

1. Find the binomial expansion of \(( 1 - 4 x ) ^ { \frac { 1 } { 4 } }\) up to and including the term in \(x ^ { 2 }\).
   [0pt] [2 marks]
2. Find the binomial expansion of \(( 2 + 3 x ) ^ { - 3 }\) up to and including the term in \(x ^ { 2 }\).
3. Hence find the binomial expansion of \(\frac { ( 1 - 4 x ) ^ { \frac { 1 } { 4 } } } { ( 2 + 3 x ) ^ { 3 } }\) up to and including the term in \(x ^ { 2 }\).
   [0pt] [2 marks]

4 A painting was valued on 1 April 2001 at \(\pounds 5000\).
The value of this painting is modelled by $$V = A p ^ { t }$$ where \(\pounds V\) is the value \(t\) years after 1 April 2001, and \(A\) and \(p\) are constants.

1. Write down the value of \(A\).
2. According to the model, the value of this painting on 1 April 2011 was \(\pounds 25000\). Using this model:
   1. show that \(p ^ { 10 } = 5\);
   2. use logarithms to find the year in which the painting will be valued at \(\pounds 75000\).
3. A painting by another artist was valued at \(\pounds 2500\) on 1 April 1991. The value of this painting is modelled by $$W = 2500 q ^ { t }$$ where \(\pounds W\) is the value \(t\) years after 1 April 1991, and \(q\) is a constant.
   1. Show that, according to the two models, the value of the two paintings will be the same \(T\) years after 1 April 1991, $$\text { where } T = \frac { \ln \left( \frac { 5 } { 2 } \right) } { \ln \left( \frac { p } { q } \right) }$$
   2. Given that \(p = 1.029 q\), find the year in which the two paintings will have the same value.
      [0pt] [1 mark]

5

1. 1. Express \(3 \sin x + 4 \cos x\) in the form \(R \sin ( x + \alpha )\) where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\), giving your value of \(\alpha\) to the nearest \(0.1 ^ { \circ }\).
   2. Hence solve the equation \(3 \sin 2 \theta + 4 \cos 2 \theta = 5\) in the interval \(0 ^ { \circ } < \theta < 360 ^ { \circ }\), giving your solutions to the nearest \(0.1 ^ { \circ }\).
2. 1. Show that the equation \(\tan 2 \theta \tan \theta = 2\) can be written as \(2 \tan ^ { 2 } \theta = 1\).
   2. Hence solve the equation \(\tan 2 \theta \tan \theta = 2\) in the interval \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\), giving your solutions to the nearest \(0.1 ^ { \circ }\).
3. 1. Use the Factor Theorem to show that \(2 x - 1\) is a factor of \(8 x ^ { 3 } - 4 x + 1\).
   2. Show that \(4 \cos 2 \theta \cos \theta + 1\) can be written as \(8 x ^ { 3 } - 4 x + 1\) where \(x = \cos \theta\).
   3. Given that \(\theta = 72 ^ { \circ }\) is a solution of \(4 \cos 2 \theta \cos \theta + 1 = 0\), use the results from parts (c)(i) and (c)(ii) to show that the exact value of \(\cos 72 ^ { \circ }\) is \(\frac { ( \sqrt { 5 } - 1 ) } { p }\) where \(p\) is an integer.
      [0pt] [3 marks]

6 The line \(l _ { 1 }\) has equation \(\mathbf { r } = \left[ \begin{array} { r } 4
- 5
3 \end{array} \right] + \lambda \left[ \begin{array} { r } - 1
3
1 \end{array} \right]\).
The line \(l _ { 2 }\) has equation \(\mathbf { r } = \left[ \begin{array} { r } 7
- 8
6 \end{array} \right] + \mu \left[ \begin{array} { r } 2
- 3
1 \end{array} \right]\).
The point \(P\) lies on \(l _ { 1 }\) where \(\lambda = - 1\). The point \(Q\) lies on \(l _ { 2 }\) where \(\mu = 2\).

1. Show that the vector \(\overrightarrow { P Q }\) is parallel to \(\left[ \begin{array} { r } 1
   - 1
   1 \end{array} \right]\).
2. The lines \(l _ { 1 }\) and \(l _ { 2 }\) intersect at the point \(R ( 3 , b , c )\).
   1. Show that \(b = - 2\) and find the value of \(c\).
   2. The point \(S\) lies on a line through \(P\) that is parallel to \(l _ { 2 }\). The line \(R S\) is perpendicular to the line \(P Q\).
      \includegraphics[max width=\textwidth, alt={}, center]{9f03a5f3-7fea-4fb7-b3bd-b4c0cdf662a2-16_887_1159_1320_443} Find the coordinates of \(S\).
      \(7 \quad\) A curve has equation \(\cos 2 y + y \mathrm { e } ^ { 3 x } = 2 \pi\).
      The point \(A \left( \ln 2 , \frac { \pi } { 4 } \right)\) lies on this curve.

1. 1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
   2. Hence find the exact value of the gradient of the curve at \(A\).
2. The normal at \(A\) crosses the \(y\)-axis at the point \(B\). Find the exact value of the \(y\)-coordinate of \(B\).
   [0pt] [2 marks]

8

1. Express \(\frac { 16 x } { ( 1 - 3 x ) ( 1 + x ) ^ { 2 } }\) in the form \(\frac { A } { 1 - 3 x } + \frac { B } { 1 + x } + \frac { C } { ( 1 + x ) ^ { 2 } }\).
2. Solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 16 x \mathrm { e } ^ { 2 y } } { ( 1 - 3 x ) ( 1 + x ) ^ { 2 } }$$ where \(y = 0\) when \(x = 0\).
   Give your answer in the form \(\mathrm { f } ( y ) = \mathrm { g } ( x )\).
   [0pt] [7 marks]