AQA C3 (Core Mathematics 3) 2013 January

1

1. Show that the equation \(x ^ { 3 } - 6 x + 1 = 0\) has a root \(\alpha\), where \(2 < \alpha < 3\).
2. Show that the equation \(x ^ { 3 } - 6 x + 1 = 0\) can be rearranged into the form $$x ^ { 2 } = 6 - \frac { 1 } { x }$$ (1 mark)
3. Use the recurrence relation \(x _ { n + 1 } = \sqrt { 6 - \frac { 1 } { x _ { n } } }\), with \(x _ { 1 } = 2.5\), to find the value of \(x _ { 3 }\), giving your answer to four significant figures.
   (2 marks)

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   \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\)

2

1. Use Simpson's rule, with five ordinates (four strips), to calculate an estimate for $$\int _ { 0 } ^ { 4 } \frac { x } { x ^ { 2 } + 2 } \mathrm {~d} x$$ Give your answer to four significant figures.
2. Show that the exact value of \(\int _ { 0 } ^ { 4 } \frac { x } { x ^ { 2 } + 2 } \mathrm {~d} x\) is \(\ln k\), where \(k\) is an integer. (5 marks)

3

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) when $$y = \mathrm { e } ^ { 3 x } + \ln x$$
2. 1. Given that \(u = \frac { \sin x } { 1 + \cos x }\), show that \(\frac { \mathrm { d } u } { \mathrm {~d} x } = \frac { 1 } { 1 + \cos x }\).
   2. Hence show that if \(y = \ln \left( \frac { \sin x } { 1 + \cos x } \right)\), then \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \operatorname { cosec } x\).

4 The diagram shows a sketch of the curve with equation \(y = \mathrm { f } ( x )\).
\includegraphics[max width=\textwidth, alt={}, center]{b8614dd6-2197-40c3-a673-5bef3e3653a5-5_629_1113_370_461}

1. On the axes below, sketch the curve with equation \(y = | \mathrm { f } ( x ) |\).
2. Describe a sequence of two geometrical transformations that maps the graph of \(y = \mathrm { f } ( x )\) onto the graph of \(y = \mathrm { f } ( 2 x - 1 )\).

5 The function f is defined by $$\mathrm { f } ( x ) = \frac { x ^ { 2 } - 4 } { 3 } , \text { for real values of } x , \text { where } \boldsymbol { x } \leqslant \mathbf { 0 }$$

1. State the range of f.
2. The inverse of f is \(\mathrm { f } ^ { - 1 }\).
   1. Write down the domain of \(\mathrm { f } ^ { - 1 }\).
   2. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\).
3. The function g is defined by $$\mathrm { g } ( x ) = \ln | 3 x - 1 | , \quad \text { for real values of } x , \text { where } x \neq \frac { 1 } { 3 }$$ The curve with equation \(y = \mathrm { g } ( x )\) is sketched below.
   \includegraphics[max width=\textwidth, alt={}, center]{b8614dd6-2197-40c3-a673-5bef3e3653a5-6_469_819_1254_612}
   1. The curve \(y = \mathrm { g } ( x )\) intersects the \(x\)-axis at the origin and at the point \(P\). Find the \(x\)-coordinate of \(P\).
   2. State whether the function \(g\) has an inverse. Give a reason for your answer.
   3. Show that \(\operatorname { gf } ( x ) = \ln \left| x ^ { 2 } - k \right|\), stating the value of the constant \(k\).
   4. Solve the equation \(\mathrm { gf } ( x ) = 0\).

6

1. Show that $$\frac { \sec ^ { 2 } x } { ( \sec x + 1 ) ( \sec x - 1 ) }$$ can be written as \(\operatorname { cosec } ^ { 2 } x\).
2. Hence solve the equation $$\frac { \sec ^ { 2 } x } { ( \sec x + 1 ) ( \sec x - 1 ) } = \operatorname { cosec } x + 3$$ giving the values of \(x\) to the nearest degree in the interval \(- 180 ^ { \circ } < x < 180 ^ { \circ }\).
3. Hence solve the equation $$\frac { \sec ^ { 2 } \left( 2 \theta - 60 ^ { \circ } \right) } { \left( \sec \left( 2 \theta - 60 ^ { \circ } \right) + 1 \right) \left( \sec \left( 2 \theta - 60 ^ { \circ } \right) - 1 \right) } = \operatorname { cosec } \left( 2 \theta - 60 ^ { \circ } \right) + 3$$ giving the values of \(\theta\) to the nearest degree in the interval \(0 ^ { \circ } < \theta < 90 ^ { \circ }\).

7 A curve has equation \(y = 4 x \cos 2 x\).

1. Find an exact equation of the tangent to the curve at the point on the curve where $$x = \frac { \pi } { 4 }$$
2. The region shaded on the diagram below is bounded by the curve \(y = 4 x \cos 2 x\) and the \(x\)-axis from \(x = 0\) to \(x = \frac { \pi } { 4 }\).
   \includegraphics[max width=\textwidth, alt={}, center]{b8614dd6-2197-40c3-a673-5bef3e3653a5-8_487_878_740_591} By using integration by parts, find the exact value of the area of the shaded region.
   (5 marks)
   \includegraphics[max width=\textwidth, alt={}]{b8614dd6-2197-40c3-a673-5bef3e3653a5-8_1275_1717_1432_150}

8

1. Show that $$\int _ { 0 } ^ { \ln 2 } \mathrm { e } ^ { 1 - 2 x } \mathrm {~d} x = \frac { 3 } { 8 } \mathrm { e }$$
2. Use the substitution \(u = \tan x\) to find the exact value of $$\int _ { 0 } ^ { \frac { \pi } { 4 } } \sec ^ { 4 } x \sqrt { \tan x } d x$$ (8 marks)