Edexcel C3 (Core Mathematics 3)

1. Express

$$\frac { 3 x ^ { 2 } } { \left( 2 x ^ { 2 } + 7 x + 6 \right) } \times \frac { 7 ( 3 + 2 x ) } { 3 x ^ { 5 } }$$ as a single fraction in its simplest form.

2. The function \(f\) is defined by $$\mathrm { f } : x \mapsto 2 x , \quad x \in \mathbb { R }$$

1. Find \(\mathrm { f } ^ { - 1 } ( x )\) and state the domain of \(\mathrm { f } ^ { - 1 }\). The function g is defined by $$\mathrm { g } : x \mapsto 3 x ^ { 2 } + 2 , \quad x \in \mathbb { R }$$
2. Find \(\mathrm { gf } ^ { - 1 } ( x )\).
3. State the range of \(\mathrm { gf } ^ { - 1 } ( x )\).

3. Find the exact solutions of

1. \(\mathrm { e } ^ { 2 x + 3 } = 6\),
2. \(\ln ( 3 x + 2 ) = 4\).

4. Differentiate with respect to \(x\)

1. \(x ^ { 3 } \mathrm { e } ^ { 3 x }\),
2. \(\frac { 2 x } { \cos x }\),
3. \(\tan ^ { 2 } x\). Given that \(x = \cos y ^ { 2 }\),
4. find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(y\).

5. (a) Using the formulae $$\begin{gathered} \sin ( A \pm B ) = \sin A \cos B \pm \cos A \sin B
\cos ( A \pm B ) = \cos A \cos B \mp \sin A \sin B \end{gathered}$$ show that

1. \(\sin ( A + B ) - \sin ( A - B ) = 2 \cos A \sin B\),
2. \(\cos ( A - B ) - \cos ( A + B ) = 2 \sin A \sin B\).
   (b) Use the above results to show that $$\frac { \sin ( A + B ) - \sin ( A - B ) } { \cos ( A - B ) - \cos ( A + B ) } = \cot A$$ Using the result of part (b) and the exact values of \(\sin 60 ^ { \circ }\) and \(\cos 60 ^ { \circ }\),
   (c) find an exact value for \(\cot 75 ^ { \circ }\) in its simplest form.
   

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6. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation \(y = \mathrm { f } ( x ) , x \in \mathbb { R }\). The curve has a minimum point at \(( - 0.5 , - 2 )\) and a maximum point at \(( 0.4 , - 4 )\). The lines \(x = 1\), the \(x\)-axis and the \(y\)-axis are asymptotes of the curve, as shown in Fig. 1. On a separate diagram sketch the graphs of

1. \(y = | \mathrm { f } ( x ) |\),
2. \(y = \mathrm { f } ( x - 3 )\),
3. \(y = \mathrm { f } ( | x | )\). In each case show clearly
   1. the coordinates of any points at which the curve has a maximum or minimum point,
   2. how the curve approaches the asymptotes of the curve.
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7. (a) Sketch the curve with equation \(y = \ln x\).
(b) Show that the tangent to the curve with equation \(y = \ln x\) at the point ( \(\mathrm { e } , 1\) ) passes through the origin.
(c) Use your sketch to explain why the line \(y = m x\) cuts the curve \(y = \ln x\) between \(x = 1\) and \(x = \mathrm { e }\) if \(0 < m < \frac { 1 } { \mathrm { e } }\). Taking \(x _ { 0 } = 1.86\) and using the iteration \(x _ { n + 1 } = \mathrm { e } ^ { \frac { 1 } { 3 } x _ { n } }\),
(d) calculate \(x _ { 1 } , x _ { 2 } , x _ { 3 } , x _ { 4 }\) and \(x _ { 5 }\), giving your answer to \(x _ { 5 }\) to 3 decimal places. The root of \(\ln x - \frac { 1 } { 3 } x = 0\) is \(\alpha\).
(e) By considering the change of sign of \(\ln x - \frac { 1 } { 3 } x\) over a suitable interval, show that your answer for \(x _ { 5 }\) is an accurate estimate of \(\alpha\), correct to 3 decimal places.

1. In a particular circuit the current, \(I\) amperes, is given by

$$I = 4 \sin \theta - 3 \cos \theta , \quad \theta > 0$$ where \(\theta\) is an angle related to the voltage. Given that \(I = R \sin ( \theta - \alpha )\), where \(R > 0\) and \(0 \leqslant \alpha < 360 ^ { \circ }\),

1. find the value of \(R\), and the value of \(\alpha\) to 1 decimal place.
2. Hence solve the equation \(4 \sin \theta - 3 \cos \theta = 3\) to find the values of \(\theta\) between 0 and \(360 ^ { \circ }\).
3. Write down the greatest value for \(I\).
4. Find the value of \(\theta\) between 0 and \(360 ^ { \circ }\) at which the greatest value of \(I\) occurs.
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