AQA C3 (Core Mathematics 3) 2014 June

1 Use Simpson's rule, with five ordinates (four strips), to calculate an estimate for $$\int _ { 0 } ^ { \pi } x ^ { \frac { 1 } { 2 } } \sin x d x$$ Give your answer to four significant figures.
[0pt] [4 marks]

2 A curve has equation \(y = 2 \ln ( 2 \mathrm { e } - x )\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Find an equation of the normal to the curve \(y = 2 \ln ( 2 \mathrm { e } - x )\) at the point on the curve where \(x = \mathrm { e }\).
   [0pt] [4 marks]
3. The curve \(y = 2 \ln ( 2 \mathrm { e } - x )\) intersects the line \(y = x\) at a single point, where \(x = \alpha\).
   1. Show that \(\alpha\) lies between 1 and 3 .
   2. Use the recurrence relation $$x _ { n + 1 } = 2 \ln \left( 2 \mathrm { e } - x _ { n } \right)$$ with \(x _ { 1 } = 1\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
   3. Figure 1, on the opposite page, shows a sketch of parts of the graphs of \(y = 2 \ln ( 2 \mathrm { e } - x )\) and \(y = x\), and the position of \(x _ { 1 }\). On Figure 1, draw a cobweb or staircase diagram to show how convergence takes place, indicating the positions of \(x _ { 2 }\) and \(x _ { 3 }\) on the \(x\)-axis.
      [0pt] [2 marks] \section*{(c)(iii)} \begin{figure}[h]
      \captionsetup{labelformat=empty} \caption{Figure 1} \includegraphics[alt={},max width=\textwidth]{57412ec0-ad97-4418-8ba8-93f1f7d8aac1-05_864_1284_1802_386}
      \end{figure}

3

1. 1. Differentiate \(\left( x ^ { 2 } + 1 \right) ^ { \frac { 5 } { 2 } }\) with respect to \(x\).
   2. Given that \(y = \mathrm { e } ^ { 2 x } \left( x ^ { 2 } + 1 \right) ^ { \frac { 5 } { 2 } }\), find the value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) when \(x = 0\).
2. A curve has equation \(y = \frac { 4 x - 3 } { x ^ { 2 } + 1 }\). Use the quotient rule to find the \(x\)-coordinates of the stationary points of the curve.
   [0pt] [5 marks]
   \includegraphics[max width=\textwidth, alt={}]{57412ec0-ad97-4418-8ba8-93f1f7d8aac1-06_1855_1709_852_153}

4 The sketch shows part of the curve with equation \(y = \mathrm { f } ( x )\).
\includegraphics[max width=\textwidth, alt={}, center]{57412ec0-ad97-4418-8ba8-93f1f7d8aac1-08_536_1054_367_539}

1. On Figure 2 below, sketch the curve with equation \(y = - | \mathrm { f } ( x ) |\).
2. On Figure 3 on the page opposite, sketch the curve with equation \(y = \mathrm { f } ( | 2 x | )\).
3. 1. Describe a sequence of two geometrical transformations that maps the graph of \(y = \mathrm { f } ( x )\) onto the graph of \(y = \mathrm { f } ( 2 x + 2 )\).
   2. Find the coordinates of the image of the point \(P ( 4 , - 3 )\) under the sequence of transformations given in part (c)(i). \begin{figure}[h]
      \captionsetup{labelformat=empty} \caption{Figure 3} \includegraphics[alt={},max width=\textwidth]{57412ec0-ad97-4418-8ba8-93f1f7d8aac1-09_778_1032_424_529}
      \end{figure}

5 The functions \(f\) and \(g\) are defined with their respective domains by $$\begin{array} { l l } \mathrm { f } ( x ) = x ^ { 2 } - 6 x + 5 , & \text { for } x \geqslant 3
\mathrm {~g} ( x ) = | x - 6 | , & \text { for all real values of } x \end{array}$$

1. Find the range of f.
2. The inverse of f is \(\mathrm { f } ^ { - 1 }\). Find \(\mathrm { f } ^ { - 1 } ( x )\). Give your answer in its simplest form.
3. 1. Find \(\mathrm { gf } ( x )\).
   2. Solve the equation \(\operatorname { gf } ( x ) = 6\).

6

1. By using integration by parts twice, find $$\int x ^ { 2 } \sin 2 x d x$$
2. A curve has equation \(y = x \sqrt { \sin 2 x }\), for \(0 \leqslant x \leqslant \frac { \pi } { 2 }\). The region bounded by the curve and the \(x\)-axis is rotated through \(2 \pi\) radians about the \(x\)-axis to generate a solid. Find the exact value of the volume of the solid generated.
   [0pt] [3 marks]

7 Use the substitution \(u = 3 - x ^ { 3 }\) to find the exact value of \(\int _ { 0 } ^ { 1 } \frac { x ^ { 5 } } { 3 - x ^ { 3 } } \mathrm {~d} x\).
[0pt] [6 marks]

8

1. Show that the expression \(\frac { 1 - \sin x } { \cos x } + \frac { \cos x } { 1 - \sin x }\) can be written as \(2 \sec x\).
   [0pt] [4 marks]
2. Hence solve the equation $$\frac { 1 - \sin x } { \cos x } + \frac { \cos x } { 1 - \sin x } = \tan ^ { 2 } x - 2$$ giving the values of \(x\) to the nearest degree in the interval \(0 ^ { \circ } \leqslant x < 360 ^ { \circ }\).
   [0pt] [6 marks]
3. Hence solve the equation $$\frac { 1 - \sin \left( 2 \theta - 30 ^ { \circ } \right) } { \cos \left( 2 \theta - 30 ^ { \circ } \right) } + \frac { \cos \left( 2 \theta - 30 ^ { \circ } \right) } { 1 - \sin \left( 2 \theta - 30 ^ { \circ } \right) } = \tan ^ { 2 } \left( 2 \theta - 30 ^ { \circ } \right) - 2$$ giving the values of \(\theta\) to the nearest degree in the interval \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).
   [0pt] [2 marks]
   \includegraphics[max width=\textwidth, alt={}]{57412ec0-ad97-4418-8ba8-93f1f7d8aac1-16_1517_1709_1190_153}
   \includegraphics[max width=\textwidth, alt={}]{57412ec0-ad97-4418-8ba8-93f1f7d8aac1-20_2489_1730_221_139}