AQA C4 (Core Mathematics 4) 2011 June

1 The polynomial \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = 4 x ^ { 3 } - 13 x + 6\).

1. Find \(\mathrm { f } ( - 2 )\).
2. Use the Factor Theorem to show that \(2 x - 3\) is a factor of \(\mathrm { f } ( x )\).
3. Simplify \(\frac { 2 x ^ { 2 } + x - 6 } { \mathrm { f } ( x ) }\).

2 The average weekly pay of a footballer at a certain club was \(\pounds 80\) on 1 August 1960. By 1 August 1985, this had risen to \(\pounds 2000\). The average weekly pay of a footballer at this club can be modelled by the equation $$P = A k ^ { t }$$ where \(\pounds P\) is the average weekly pay \(t\) years after 1 August 1960, and \(A\) and \(k\) are constants.

1. 1. Write down the value of \(A\).
   2. Show that the value of \(k\) is 1.137411 , correct to six decimal places.
2. Use this model to predict the year in which, on 1 August, the average weekly pay of a footballer at this club will first exceed \(\pounds 100000\).

3

1. 1. Find the binomial expansion of \(( 1 - x ) ^ { \frac { 1 } { 3 } }\) up to and including the term in \(x ^ { 2 }\).
   2. Hence, or otherwise, show that $$( 125 - 27 x ) ^ { \frac { 1 } { 3 } } \approx 5 + \frac { m } { 25 } x + \frac { n } { 3125 } x ^ { 2 }$$ for small values of \(x\), stating the values of the integers \(m\) and \(n\).
2. Use your result from part (a)(ii) to find an approximate value of \(\sqrt [ 3 ] { 119 }\), giving your answer to five decimal places.
   (2 marks)

4

1. A curve is defined by the parametric equations \(x = 3 \cos 2 \theta , y = 2 \cos \theta\).
   1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { k \cos \theta }\), where \(k\) is an integer.
   2. Find an equation of the normal to the curve at the point where \(\theta = \frac { \pi } { 3 }\).
2. Find the exact value of \(\int _ { - \frac { \pi } { 4 } } ^ { \frac { \pi } { 4 } } \sin ^ { 2 } x \mathrm {~d} x\).

5 The points \(A\) and \(B\) have coordinates \(( 5,1 , - 2 )\) and \(( 4 , - 1,3 )\) respectively.
The line \(l\) has equation \(\mathbf { r } = \left[ \begin{array} { r } - 8
5
- 6 \end{array} \right] + \mu \left[ \begin{array} { r } 5
0
- 2 \end{array} \right]\).

1. Find a vector equation of the line that passes through \(A\) and \(B\).
2. 1. Show that the line that passes through \(A\) and \(B\) intersects the line \(l\), and find the coordinates of the point of intersection, \(P\).
   2. The point \(C\) lies on \(l\) such that triangle \(P B C\) has a right angle at \(B\). Find the coordinates of \(C\).

6 A curve is defined by the equation \(2 y + \mathrm { e } ^ { 2 x } y ^ { 2 } = x ^ { 2 } + C\), where \(C\) is a constant. The point \(P \left( 1 , \frac { 1 } { \mathrm { e } } \right)\) lies on the curve.

1. Find the exact value of \(C\).
2. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).
3. Verify that \(P \left( 1 , \frac { 1 } { \mathrm { e } } \right)\) is a stationary point on the curve.

7 A giant snowball is melting. The snowball can be modelled as a sphere whose surface area is decreasing at a constant rate with respect to time. The surface area of the sphere is \(A \mathrm {~cm} ^ { 2 }\) at time \(t\) days after it begins to melt.

1. Write down a differential equation in terms of the variables \(A\) and \(t\) and a constant \(k\), where \(k > 0\), to model the melting snowball.
2. 1. Initially, the radius of the snowball is 60 cm , and 9 days later, the radius has halved. Show that \(A = 1200 \pi ( 12 - t )\).
      (You may assume that the surface area of a sphere is given by \(A = 4 \pi r ^ { 2 }\), where \(r\) is the radius.)
   2. Use this model to find the number of days that it takes the snowball to melt completely.

8

1. Express \(\frac { 1 } { ( 3 - 2 x ) ( 1 - x ) ^ { 2 } }\) in the form \(\frac { A } { 3 - 2 x } + \frac { B } { 1 - x } + \frac { C } { ( 1 - x ) ^ { 2 } }\).
   (4 marks)
2. Solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 \sqrt { y } } { ( 3 - 2 x ) ( 1 - x ) ^ { 2 } }$$ where \(y = 0\) when \(x = 0\), expressing your answer in the form $$y ^ { p } = q \ln [ \mathrm { f } ( x ) ] + \frac { x } { 1 - x }$$ where \(p\) and \(q\) are constants.