AQA C3 (Core Mathematics 3) 2006 January

1

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) when \(y = \tan 3 x\).
   (2 marks)
2. Given that \(y = \frac { 3 x + 1 } { 2 x + 1 }\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { ( 2 x + 1 ) ^ { 2 } }\).
   (3 marks)

2 Use Simpson's rule with 5 ordinates (4 strips) to find an approximation to $$\int _ { 1 } ^ { 3 } \frac { 1 } { \sqrt { 1 + x ^ { 3 } } } \mathrm {~d} x$$ giving your answer to three significant figures.

3

1. 1. Given that \(\mathrm { f } ( x ) = x ^ { 4 } + 2 x\), find \(\mathrm { f } ^ { \prime } ( x )\).
   2. Hence, or otherwise, find \(\int \frac { 2 x ^ { 3 } + 1 } { x ^ { 4 } + 2 x } \mathrm {~d} x\).
2. 1. Use the substitution \(u = 2 x + 1\) to show that $$\int x \sqrt { 2 x + 1 } \mathrm {~d} x = \frac { 1 } { 4 } \int \left( u ^ { \frac { 3 } { 2 } } - u ^ { \frac { 1 } { 2 } } \right) \mathrm { d } u$$
   2. Hence show that \(\int _ { 0 } ^ { 4 } x \sqrt { 2 x + 1 } \mathrm {~d} x = 19.9\) correct to three significant figures.

4 It is given that \(2 \operatorname { cosec } ^ { 2 } x = 5 - 5 \cot x\).

1. Show that the equation \(2 \operatorname { cosec } ^ { 2 } x = 5 - 5 \cot x\) can be written in the form $$2 \cot ^ { 2 } x + 5 \cot x - 3 = 0$$
2. Hence show that \(\tan x = 2\) or \(\tan x = - \frac { 1 } { 3 }\).
3. Hence, or otherwise, solve the equation \(2 \operatorname { cosec } ^ { 2 } x = 5 - 5 \cot x\), giving all values of \(x\) in radians to one decimal place in the interval \(- \pi < x \leqslant \pi\).

5 The diagram shows part of the graph of \(y = \mathrm { e } ^ { 2 x } - 9\). The graph cuts the coordinate axes at \(( 0 , a )\) and \(( b , 0 )\).
\includegraphics[max width=\textwidth, alt={}, center]{908f530c-076d-47b1-90dd-38dbfe44f898-03_826_924_477_541}

1. State the value of \(a\), and show that \(b = \ln 3\).
2. Show that \(y ^ { 2 } = \mathrm { e } ^ { 4 x } - 18 \mathrm { e } ^ { 2 x } + 81\).
3. The shaded region \(R\) is rotated through \(360 ^ { \circ }\) about the \(x\)-axis. Find the volume of the solid formed, giving your answer in the form \(\pi ( p \ln 3 + q )\), where \(p\) and \(q\) are integers.
4. Sketch the curve with equation \(y = \left| \mathrm { e } ^ { 2 x } - 9 \right|\) for \(x \geqslant 0\).

6 [Figure 1, printed on the insert, is provided for use in this question.]
The curve \(y = x ^ { 3 } + 4 x - 3\) intersects the \(x\)-axis at the point \(A\) where \(x = \alpha\).

1. Show that \(\alpha\) lies between 0.5 and 1.0.
2. Show that the equation \(x ^ { 3 } + 4 x - 3 = 0\) can be rearranged into the form \(x = \frac { 3 - x ^ { 3 } } { 4 }\).
   (1 mark)
3. 1. Use the iteration \(x _ { n + 1 } = \frac { 3 - x _ { n } { } ^ { 3 } } { 4 }\) with \(x _ { 1 } = 0.5\) to find \(x _ { 3 }\), giving your answer to two decimal places.
      (3 marks)
   2. The sketch on Figure 1 shows parts of the graphs of \(y = \frac { 3 - x ^ { 3 } } { 4 }\) 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.
      (3 marks)

7

1. The sketch shows the graph of \(y = \sin ^ { - 1 } x\).
   \includegraphics[max width=\textwidth, alt={}, center]{908f530c-076d-47b1-90dd-38dbfe44f898-05_835_834_447_587} Write down the coordinates of the points \(P\) and \(Q\), the end-points of the graph.
2. Sketch the graph of \(y = - \sin ^ { - 1 } ( x - 1 )\).

8 The functions \(f\) and \(g\) are defined with their respective domains by $$\begin{array} { l l } \mathrm { f } ( x ) = x ^ { 2 } & \text { for all real values of } x
\mathrm {~g} ( x ) = \frac { 1 } { x + 2 } & \text { for real values of } x , x \neq - 2 \end{array}$$

1. State the range of f.
2. 1. Find fg(x).
   2. Solve the equation \(\operatorname { fg } ( x ) = 4\).
3. 1. Explain why the function f does not have an inverse.
   2. The inverse of g is \(\mathrm { g } ^ { - 1 }\). Find \(\mathrm { g } ^ { - 1 } ( x )\).

9

1. Given that \(y = x ^ { - 2 } \ln x\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 - 2 \ln x } { x ^ { 3 } }\).
2. Using integration by parts, find \(\int x ^ { - 2 } \ln x \mathrm {~d} x\).
3. The sketch shows the graph of \(y = x ^ { - 2 } \ln x\).
   \includegraphics[max width=\textwidth, alt={}, center]{908f530c-076d-47b1-90dd-38dbfe44f898-06_604_1045_687_536}
   1. Using the answer to part (a), find, in terms of e, the \(x\)-coordinate of the stationary point \(A\).
   2. The region \(R\) is bounded by the curve, the \(x\)-axis and the line \(x = 5\). Using your answer to part (b), show that the area of \(R\) is $$\frac { 1 } { 5 } ( 4 - \ln 5 )$$