Edexcel C3 (Core Mathematics 3) 2012 June

1. Express

$$\frac { 2 ( 3 x + 2 ) } { 9 x ^ { 2 } - 4 } - \frac { 2 } { 3 x + 1 }$$ as a single fraction in its simplest form.

2. $$f ( x ) = x ^ { 3 } + 3 x ^ { 2 } + 4 x - 12$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written as $$x = \sqrt { } \left( \frac { 4 ( 3 - x ) } { ( 3 + x ) } \right) , \quad x \neq - 3$$ The equation \(x ^ { 3 } + 3 x ^ { 2 } + 4 x - 12 = 0\) has a single root which is between 1 and 2
2. Use the iteration formula $$x _ { n + 1 } = \sqrt { } \left( \frac { 4 \left( 3 - x _ { n } \right) } { \left( 3 + x _ { n } \right) } \right) , n \geqslant 0$$ with \(x _ { 0 } = 1\) to find, to 2 decimal places, the value of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\). The root of \(\mathrm { f } ( x ) = 0\) is \(\alpha\).
3. By choosing a suitable interval, prove that \(\alpha = 1.272\) to 3 decimal places.

3. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of the curve \(C\) which has equation $$y = \mathrm { e } ^ { x \sqrt { 3 } } \sin 3 x , \quad - \frac { \pi } { 3 } \leqslant x \leqslant \frac { \pi } { 3 }$$

1. Find the \(x\) coordinate of the turning point \(P\) on \(C\), for which \(x > 0\) Give your answer as a multiple of \(\pi\).
2. Find an equation of the normal to \(C\) at the point where \(x = 0\)

4. \begin{figure}[h] \end{figure} Figure 2 shows part of the curve with equation \(y = \mathrm { f } ( x )\)
The curve passes through the points \(P ( - 1.5,0 )\) and \(Q ( 0,5 )\) as shown.
On separate diagrams, sketch the curve with equation

1. \(y = | f ( x ) |\)
2. \(y = \mathrm { f } ( | x | )\)
3. \(y = 2 f ( 3 x )\) Indicate clearly on each sketch the coordinates of the points at which the curve crosses or meets the axes.

1. (a) Express \(4 \operatorname { cosec } ^ { 2 } 2 \theta - \operatorname { cosec } ^ { 2 } \theta\) in terms of \(\sin \theta\) and \(\cos \theta\).
   (b) Hence show that

$$4 \operatorname { cosec } ^ { 2 } 2 \theta - \operatorname { cosec } ^ { 2 } \theta = \sec ^ { 2 } \theta$$ (c) Hence or otherwise solve, for \(0 < \theta < \pi\), $$4 \operatorname { cosec } ^ { 2 } 2 \theta - \operatorname { cosec } ^ { 2 } \theta = 4$$ giving your answers in terms of \(\pi\).

6. The functions \(f\) and \(g\) are defined by $$\begin{aligned} & \mathrm { f } : x \mapsto \mathrm { e } ^ { x } + 2 , \quad x \in \mathbb { R }
& \mathrm {~g} : x \mapsto \ln x , \quad x > 0 \end{aligned}$$

1. State the range of f.
2. Find \(\mathrm { fg } ( x )\), giving your answer in its simplest form.
3. Find the exact value of \(x\) for which \(\mathrm { f } ( 2 x + 3 ) = 6\)
4. Find \(\mathrm { f } ^ { - 1 }\), the inverse function of f , stating its domain.
5. On the same axes sketch the curves with equation \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\), giving the coordinates of all the points where the curves cross the axes.

1. Differentiate with respect to \(x\),
   1. \(x ^ { \frac { 1 } { 2 } } \ln ( 3 x )\)
   2. \(\frac { 1 - 10 x } { ( 2 x - 1 ) ^ { 5 } }\), giving your answer in its simplest form.
2. Given that \(x = 3 \tan 2 y\) find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\).

$$f ( x ) = 7 \cos 2 x - 24 \sin 2 x$$ Given that \(\mathrm { f } ( x ) = R \cos ( 2 x + \alpha )\), where \(R > 0\) and \(0 < \alpha < 90 ^ { \circ }\),

1. find the value of \(R\) and the value of \(\alpha\).
2. Hence solve the equation $$7 \cos 2 x - 24 \sin 2 x = 12.5$$ for \(0 \leqslant x < 180 ^ { \circ }\), giving your answers to 1 decimal place.
3. Express \(14 \cos ^ { 2 } x - 48 \sin x \cos x\) in the form \(a \cos 2 x + b \sin 2 x + c\), where \(a , b\), and \(c\) are constants to be found.
4. Hence, using your answers to parts (a) and (c), deduce the maximum value of $$14 \cos ^ { 2 } x - 48 \sin x \cos x$$