Edexcel C3 (Core Mathematics 3) 2011 January

1. (a) Express \(7 \cos x - 24 \sin x\) in the form \(R \cos ( x + \alpha )\) where \(R > 0\) and \(0 < \alpha < \frac { \pi } { 2 }\). Give the value of \(\alpha\) to 3 decimal places.
   (b) Hence write down the minimum value of \(7 \cos x - 24 \sin x\).
   (c) Solve, for \(0 \leqslant x < 2 \pi\), the equation

$$7 \cos x - 24 \sin x = 10$$ giving your answers to 2 decimal places.

2. (a) Express $$\frac { 4 x - 1 } { 2 ( x - 1 ) } - \frac { 3 } { 2 ( x - 1 ) ( 2 x - 1 ) }$$ as a single fraction in its simplest form. Given that $$f ( x ) = \frac { 4 x - 1 } { 2 ( x - 1 ) } - \frac { 3 } { 2 ( x - 1 ) ( 2 x - 1 ) } - 2 , \quad x > 1$$ (b) show that $$f ( x ) = \frac { 3 } { 2 x - 1 }$$ (c) Hence differentiate \(\mathrm { f } ( x )\) and find \(\mathrm { f } ^ { \prime } ( 2 )\).

1. Find all the solutions of

$$2 \cos 2 \theta = 1 - 2 \sin \theta$$ in the interval \(0 \leqslant \theta < 360 ^ { \circ }\).

4. Joan brings a cup of hot tea into a room and places the cup on a table. At time \(t\) minutes after Joan places the cup on the table, the temperature, \(\theta ^ { \circ } \mathrm { C }\), of the tea is modelled by the equation $$\theta = 20 + A \mathrm { e } ^ { - k t } ,$$ where \(A\) and \(k\) are positive constants. Given that the initial temperature of the tea was \(90 ^ { \circ } \mathrm { C }\),

1. find the value of \(A\). The tea takes 5 minutes to decrease in temperature from \(90 ^ { \circ } \mathrm { C }\) to \(55 ^ { \circ } \mathrm { C }\).
2. Show that \(k = \frac { 1 } { 5 } \ln 2\).
3. Find the rate at which the temperature of the tea is decreasing at the instant when \(t = 10\). Give your answer, in \({ } ^ { \circ } \mathrm { C }\) per minute, to 3 decimal places.

5. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation \(y = \mathrm { f } ( x )\), where $$\mathrm { f } ( x ) = ( 8 - x ) \ln x , \quad x > 0$$ The curve cuts the \(x\)-axis at the points \(A\) and \(B\) and has a maximum turning point at \(Q\), as shown in Figure 1.

1. Write down the coordinates of \(A\) and the coordinates of \(B\).
2. Find f'(x).
3. Show that the \(x\)-coordinate of \(Q\) lies between 3.5 and 3.6
4. Show that the \(x\)-coordinate of \(Q\) is the solution of $$x = \frac { 8 } { 1 + \ln x }$$ To find an approximation for the \(x\)-coordinate of \(Q\), the iteration formula $$x _ { n + 1 } = \frac { 8 } { 1 + \ln x _ { n } }$$ is used.
5. Taking \(x _ { 0 } = 3.55\), find the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\). Give your answers to 3 decimal places.

1. The function \(f\) is defined by

$$\mathrm { f } : x \mapsto \frac { 3 - 2 x } { x - 5 } , \quad x \in \mathbb { R } , x \neq 5$$

1. Find \(\mathrm { f } ^ { - 1 } ( x )\).
   (3) \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{3ff6824f-9fbf-4b5b-8bab-91332c549b36-10_901_1091_593_429} \captionsetup{labelformat=empty} \caption{Figure 2}
   \end{figure} The function g has domain \(- 1 \leqslant x \leqslant 8\), and is linear from \(( - 1 , - 9 )\) to \(( 2,0 )\) and from \(( 2,0 )\) to \(( 8,4 )\). Figure 2 shows a sketch of the graph of \(y = \mathrm { g } ( x )\).
2. Write down the range of g.
3. Find \(\operatorname { gg } ( 2 )\).
4. Find \(\mathrm { fg } ( 8 )\).
5. On separate diagrams, sketch the graph with equation
   1. \(y = | \mathrm { g } ( x ) |\),
   2. \(y = \mathrm { g } ^ { - 1 } ( x )\). Show on each sketch the coordinates of each point at which the graph meets or cuts the axes.
6. State the domain of the inverse function \(\mathrm { g } ^ { - 1 }\).

1. The curve \(C\) has equation

$$y = \frac { 3 + \sin 2 x } { 2 + \cos 2 x }$$

1. Show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 6 \sin 2 x + 4 \cos 2 x + 2 } { ( 2 + \cos 2 x ) ^ { 2 } }$$
2. Find an equation of the tangent to \(C\) at the point on \(C\) where \(x = \frac { \pi } { 2 }\). Write your answer in the form \(y = a x + b\), where \(a\) and \(b\) are exact constants.

8. (a) Given that $$\frac { \mathrm { d } } { \mathrm {~d} x } ( \cos x ) = - \sin x$$ show that \(\frac { \mathrm { d } } { \mathrm { d } x } ( \sec x ) = \sec x \tan x\). Given that $$x = \sec 2 y$$ (b) find \(\frac { \mathrm { d } x } { \mathrm {~d} y }\) in terms of \(y\).
(c) Hence find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\). \includegraphics[max width=\textwidth, alt={}, center]{3ff6824f-9fbf-4b5b-8bab-91332c549b36-14_102_93_2473_1804}