Edexcel C3 (Core Mathematics 3) 2006 January

1. \begin{figure}[h] \end{figure} Figure 1 shows the graph of \(y = \mathrm { f } ( x ) , - 5 \leqslant x \leqslant 5\).
The point \(M ( 2,4 )\) is the maximum turning point of the graph.
Sketch, on separate diagrams, the graphs of

1. \(y = \mathrm { f } ( x ) + 3\),
2. \(y = | \mathrm { f } ( x ) |\),
3. \(y = \mathrm { f } ( | x | )\). Show on each graph the coordinates of any maximum turning points.

1. Express

$$\frac { 2 x ^ { 2 } + 3 x } { ( 2 x + 3 ) ( x - 2 ) } - \frac { 6 } { x ^ { 2 } - x - 2 }$$ as a single fraction in its simplest form.

3. The point \(P\) lies on the curve with equation \(y = \ln \left( \frac { 1 } { 3 } x \right)\). The \(x\)-coordinate of \(P\) is 3 . Find an equation of the normal to the curve at the point \(P\) in the form \(y = a x + b\), where \(a\) and \(b\) are constants.
(5)

4. (a) Differentiate with respect to \(x\)

1. \(x ^ { 2 } \mathrm { e } ^ { 3 x + 2 }\),
2. \(\frac { \cos \left( 2 x ^ { 3 } \right) } { 3 x }\).
   (b) Given that \(x = 4 \sin ( 2 y + 6 )\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\).

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

1. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written as $$x = \sqrt { \left( \frac { 2 } { x } + \frac { 1 } { 2 } \right) }$$ The equation \(2 x ^ { 3 } - x - 4 = 0\) has a root between 1.35 and 1.4.
2. Use the iteration formula $$x _ { n + 1 } = \sqrt { } \left( \frac { 2 } { x _ { n } } + \frac { 1 } { 2 } \right)$$ with \(x _ { 0 } = 1.35\), to find, to 2 decimal places, the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\). The only real root of \(\mathrm { f } ( x ) = 0\) is \(\alpha\).
3. By choosing a suitable interval, prove that \(\alpha = 1.392\), to 3 decimal places.

6. $$f ( x ) = 12 \cos x - 4 \sin x$$ Given that \(\mathrm { f } ( x ) = R \cos ( x + \alpha )\), where \(R \geqslant 0\) and \(0 \leqslant \alpha \leqslant 90 ^ { \circ }\),

1. find the value of \(R\) and the value of \(\alpha\).
2. Hence solve the equation $$12 \cos x - 4 \sin x = 7$$ for \(0 \leqslant x < 360 ^ { \circ }\), giving your answers to one decimal place.
3. 1. Write down the minimum value of \(12 \cos x - 4 \sin x\).
   2. Find, to 2 decimal places, the smallest positive value of \(x\) for which this minimum value occurs.
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7. (a) Show that

1. \(\frac { \cos 2 x } { \cos x + \sin x } \equiv \cos x - \sin x , \quad x \neq \left( n - \frac { 1 } { 4 } \right) \pi , n \in \mathbb { Z }\),
2. \(\frac { 1 } { 2 } ( \cos 2 x - \sin 2 x ) \equiv \cos ^ { 2 } x - \cos x \sin x - \frac { 1 } { 2 }\).
   (b) Hence, or otherwise, show that the equation $$\cos \theta \left( \frac { \cos 2 \theta } { \cos \theta + \sin \theta } \right) = \frac { 1 } { 2 }$$ can be written as $$\sin 2 \theta = \cos 2 \theta$$ (c) Solve, for \(0 \leqslant \theta < 2 \pi\), $$\sin 2 \theta = \cos 2 \theta$$ giving your answers in terms of \(\pi\).

8. The functions \(f\) and \(g\) are defined by $$\begin{array} { l l } \mathrm { f } : x \rightarrow 2 x + \ln 2 , & x \in \mathbb { R } ,
\mathrm {~g} : x \rightarrow \mathrm { e } ^ { 2 x } , & x \in \mathbb { R } . \end{array}$$

1. Prove that the composite function gf is $$\operatorname { gf } : x \rightarrow 4 \mathrm { e } ^ { 4 x } , \quad x \in \mathbb { R }$$
2. In the space provided on page 19, sketch the curve with equation \(y = \operatorname { gf } ( x )\), and show the coordinates of the point where the curve cuts the \(y\)-axis.
3. Write down the range of gf.
4. Find the value of \(x\) for which \(\frac { \mathrm { d } } { \mathrm { d } x } [ \operatorname { gf } ( x ) ] = 3\), giving your answer to 3 significant figures.