Edexcel C3 (Core Mathematics 3) 2005 June

1. (a) Given that \(\sin ^ { 2 } \theta + \cos ^ { 2 } \theta \equiv 1\), show that \(1 + \tan ^ { 2 } \theta \equiv \sec ^ { 2 } \theta\).
   (b) Solve, for \(0 \leqslant \theta < 360 ^ { \circ }\), the equation

$$2 \tan ^ { 2 } \theta + \sec \theta = 1 ,$$ giving your answers to 1 decimal place.

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

1. \(3 \sin ^ { 2 } x + \sec 2 x\),
2. \(\{ x + \ln ( 2 x ) \} ^ { 3 }\). Given that \(y = \frac { 5 x ^ { 2 } - 10 x + 9 } { ( x - 1 ) ^ { 2 } } , \quad x \neq 1\),
   (b) show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { 8 } { ( x - 1 ) ^ { 3 } }\).

3. The function \(f\) is defined by $$f : x \rightarrow \frac { 5 x + 1 } { x ^ { 2 } + x - 2 } - \frac { 3 } { x + 2 } , x > 1$$

1. Show that \(\mathrm { f } ( x ) = \frac { 2 } { x - 1 } , x > 1\).
2. Find \(\mathrm { f } ^ { - 1 } ( x )\). The function \(g\) is defined by $$\mathrm { g } : x \rightarrow x ^ { 2 } + 5 , \quad x \in \mathbb { R }$$
3. Solve \(\operatorname { fg } ( x ) = \frac { 1 } { 4 }\).

4. $$\mathrm { f } ( x ) = 3 \mathrm { e } ^ { x } - \frac { 1 } { 2 } \ln x - 2 , \quad x > 0 .$$

1. Differentiate to find \(\mathrm { f } ^ { \prime } ( x )\). The curve with equation \(y = \mathrm { f } ( x )\) has a turning point at \(P\). The \(x\)-coordinate of \(P\) is \(\alpha\).
2. Show that \(\alpha = \frac { 1 } { 6 } \mathrm { e } ^ { - \alpha }\). The iterative formula $$x _ { n + 1 } = \frac { 1 } { 6 } \mathrm { e } ^ { - x _ { n } } , x _ { 0 } = 1$$ is used to find an approximate value for \(\alpha\).
3. Calculate the values of \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to 4 decimal places.
4. By considering the change of sign of \(\mathrm { f } ^ { \prime } ( x )\) in a suitable interval, prove that \(\alpha = 0.1443\) correct to 4 decimal places.

5. (a) Using the identity \(\cos ( A + B ) \equiv \cos A \cos B - \sin A \sin B\), prove that $$\cos 2 A \equiv 1 - 2 \sin ^ { 2 } A$$ (b) Show that $$2 \sin 2 \theta - 3 \cos 2 \theta - 3 \sin \theta + 3 \equiv \sin \theta ( 4 \cos \theta + 6 \sin \theta - 3 )$$ (c) Express \(4 \cos \theta + 6 \sin \theta\) in the form \(R \sin ( \theta + \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\).
(d) Hence, for \(0 \leqslant \theta < \pi\), solve $$2 \sin 2 \theta = 3 ( \cos 2 \theta + \sin \theta - 1 )$$ giving your answers in radians to 3 significant figures, where appropriate.

6. \begin{figure}[h] \end{figure} Figure 1 shows part of the graph of \(y = \mathrm { f } ( x ) , x \in \mathbb { R }\). The graph consists of two line segments that meet at the point \(( 1 , a ) , a < 0\). One line meets the \(x\)-axis at \(( 3,0 )\). The other line meets the \(x\)-axis at \(( - 1,0 )\) and the \(y\)-axis at \(( 0 , b ) , b < 0\). In separate diagrams, sketch the graph with equation

1. \(y = \mathrm { f } ( x + 1 )\),
2. \(y = \mathrm { f } ( | x | )\). Indicate clearly on each sketch the coordinates of any points of intersection with the axes. Given that \(\mathrm { f } ( x ) = | x - 1 | - 2\), find
3. the value of \(a\) and the value of \(b\),
4. the value of \(x\) for which \(\mathrm { f } ( x ) = 5 x\).

1. A particular species of orchid is being studied. The population \(p\) at time \(t\) years after the study started is assumed to be

$$p = \frac { 2800 a \mathrm { e } ^ { 0.2 t } } { 1 + a \mathrm { e } ^ { 0.2 t } } , \text { where } a \text { is a constant. }$$ Given that there were 300 orchids when the study started,

1. show that \(a = 0.12\),
2. use the equation with \(a = 0.12\) to predict the number of years before the population of orchids reaches 1850.
3. Show that \(p = \frac { 336 } { 0.12 + \mathrm { e } ^ { - 0.2 t } }\).
4. Hence show that the population cannot exceed 2800.