Edexcel C3 (Core Mathematics 3) 2014 January

1. $$f ( x ) = \sec x + 3 x - 2 , \quad - \frac { \pi } { 2 } < x < \frac { \pi } { 2 }$$

1. Show that there is a root of \(\mathrm { f } ( x ) = 0\) in the interval \([ 0.2,0.4 ]\)
2. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written in the form $$x = \frac { 2 } { 3 } - \frac { 1 } { 3 \cos x }$$ The solution of \(\mathrm { f } ( x ) = 0\) is \(\alpha\), where \(\alpha = 0.3\) to 1 decimal place.
3. Starting with \(x _ { 0 } = 0.3\), use the iterative formula $$x _ { n + 1 } = \frac { 2 } { 3 } - \frac { 1 } { 3 \cos x _ { n } }$$ to calculate the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 4 decimal places.
4. State the value of \(\alpha\) correct to 3 decimal places.

2. $$f ( x ) = \frac { 15 } { 3 x + 4 } - \frac { 2 x } { x - 1 } + \frac { 14 } { ( 3 x + 4 ) ( x - 1 ) } , \quad x > 1$$

1. Express \(\mathrm { f } ( x )\) as a single fraction in its simplest form.
2. Hence, or otherwise, find \(\mathrm { f } ^ { \prime } ( x )\), giving your answer as a single fraction in its simplest form.

1. (a) By writing \(\operatorname { cosec } x\) as \(\frac { 1 } { \sin x }\), show that

$$\frac { \mathrm { d } ( \operatorname { cosec } x ) } { \mathrm { d } x } = - \operatorname { cosec } x \cot x$$ Given that \(y = \mathrm { e } ^ { 3 x } \operatorname { cosec } 2 x , 0 < x < \frac { \pi } { 2 }\),
(b) find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). The curve with equation \(y = \mathrm { e } ^ { 3 x } \operatorname { cosec } 2 x , 0 < x < \frac { \pi } { 2 }\), has a single turning point.
(c) Show that the \(x\)-coordinate of this turning point is at \(x = \frac { 1 } { 2 } \arctan k\) where the value
of the constant \(k\) should be found. of the constant \(k\) should be found.

1. A pot of coffee is delivered to a meeting room at 11 am . At a time \(t\) minutes after 11 am the temperature, \(\theta ^ { \circ } \mathrm { C }\), of the coffee in the pot is given by the equation

$$\theta = A + 60 \mathrm { e } ^ { - k t }$$ where \(A\) and \(k\) are positive constants. Given also that the temperature of the coffee at 11 am is \(85 ^ { \circ } \mathrm { C }\) and that 15 minutes later it is \(58 ^ { \circ } \mathrm { C }\),

1. find the value of \(A\).
2. Show that \(k = \frac { 1 } { 15 } \ln \left( \frac { 20 } { 11 } \right)\)
3. Find, to the nearest minute, the time at which the temperature of the coffee reaches \(50 ^ { \circ } \mathrm { C }\).

5. \begin{figure}[h] \end{figure} The curve shown in Figure 1 has equation $$x = 3 \sin y + 3 \cos y , \quad - \frac { \pi } { 4 } < y < \frac { \pi } { 4 }$$

1. Express the equation of the curve in the form
   \(x = R \sin ( y + \alpha )\), where \(R\) and \(\alpha\) are constants, \(R > 0\) and \(0 < \alpha < \frac { \pi } { 2 }\)
2. Find the coordinates of the point on the curve where the value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) is \(\frac { 1 } { 2 }\). Give your answers to 3 decimal places.

1. Given that \(a\) and \(b\) are constants and that \(0 < a < b\),
   1. on separate diagrams, sketch the graph with equation
      1. \(y = | 2 x + a |\),
      2. \(y = | 2 x + a | - b\).
   Show on each sketch the coordinates of each point at which the graph crosses or meets the axes.
2. Solve, for \(x\), the equation $$| 2 x + a | - b = \frac { 1 } { 3 } x$$ giving any answers in terms of \(a\) and \(b\).

7. (i) (a) Prove that $$\cos 3 \theta \equiv 4 \cos ^ { 3 } \theta - 3 \cos \theta$$ (You may use the double angle formulae and the identity $$\cos ( A + B ) \equiv \cos A \cos B - \sin A \sin B )$$ (b) Hence solve the equation $$2 \cos 3 \theta + \cos 2 \theta + 1 = 0$$ giving answers in the interval \(0 \leqslant \theta \leqslant \pi\).
Solutions based entirely on graphical or numerical methods are not acceptable.
(ii) Given that \(\theta = \arcsin x\) and that \(0 < \theta < \frac { \pi } { 2 }\), show that $$\cot \theta = \frac { \sqrt { \left( 1 - x ^ { 2 } \right) } } { x } , \quad 0 < x < 1$$

8. The function \(f\) is defined by $$\mathrm { f } : x \rightarrow 3 - 2 \mathrm { e } ^ { - x } , \quad x \in \mathbb { R }$$

1. Find the inverse function, \(\mathrm { f } ^ { - 1 } ( x )\) and give its domain.
2. Solve the equation \(\mathrm { f } ^ { - 1 } ( x ) = \ln x\). The equation \(\mathrm { f } ( t ) = k \mathrm { e } ^ { t }\), where \(k\) is a positive constant, has exactly one real solution.
3. Find the value of \(k\).