Edexcel C3 (Core Mathematics 3)

1. A curve has the equation \(y = ( 3 x - 5 ) ^ { 3 }\).
   1. Find an equation for the tangent to the curve at the point \(P ( 2,1 )\).
   The tangent to the curve at the point \(Q\) is parallel to the tangent at \(P\).
2. Find the coordinates of \(Q\).

2. (a) Use the identities for \(\cos ( A + B )\) and \(\cos ( A - B )\) to prove that $$2 \cos A \cos B \equiv \cos ( A + B ) + \cos ( A - B ) .$$ (b) Hence, or otherwise, find in terms of \(\pi\) the solutions of the equation $$2 \cos \left( x + \frac { \pi } { 2 } \right) = \sec \left( x + \frac { \pi } { 6 } \right) ,$$ for \(x\) in the interval \(0 \leq x \leq \pi\).

3. Differentiate each of the following with respect to \(x\) and simplify your answers.

1. \(\quad \ln ( \cos x )\)
2. \(x ^ { 2 } \sin 3 x\)
3. \(\frac { 6 } { \sqrt { 2 x - 7 } }\)

4. (a) Express \(2 \sin x ^ { \circ } - 3 \cos x ^ { \circ }\) in the form \(R \sin ( x - \alpha ) ^ { \circ }\) where \(R > 0\) and \(0 < \alpha < 90\).
(b) Show that the equation $$\operatorname { cosec } x ^ { \circ } + 3 \cot x ^ { \circ } = 2$$ can be written in the form $$2 \sin x ^ { \circ } - 3 \cos x ^ { \circ } = 1 .$$ (c) Solve the equation $$\operatorname { cosec } x ^ { \circ } + 3 \cot x ^ { \circ } = 2 ,$$ for \(x\) in the interval \(0 \leq x \leq 360\), giving your answers to 1 decimal place.

5. (a) Show that \(( 2 x + 3 )\) is a factor of \(\left( 2 x ^ { 3 } - x ^ { 2 } + 4 x + 15 \right)\).
(b) Hence, simplify $$\frac { 2 x ^ { 2 } + x - 3 } { 2 x ^ { 3 } - x ^ { 2 } + 4 x + 15 } .$$ (c) Find the coordinates of the stationary points of the curve with equation $$y = \frac { 2 x ^ { 2 } + x - 3 } { 2 x ^ { 3 } - x ^ { 2 } + 4 x + 15 } .$$

1. The population in thousands, \(P\), of a town at time \(t\) years after \(1 ^ { \text {st } }\) January 1980 is modelled by the formula

$$P = 30 + 50 \mathrm { e } ^ { 0.002 t }$$ Use this model to estimate

1. the population of the town on \(1 { } ^ { \text {st } }\) January 2010,
2. the year in which the population first exceeds 84000 . The population in thousands, \(Q\), of another town is modelled by the formula $$Q = 26 + 50 \mathrm { e } ^ { 0.003 t }$$
3. Show that the value of \(t\) when \(P = Q\) is a solution of the equation $$t = 1000 \ln \left( 1 + 0.08 \mathrm { e } ^ { - 0.002 t } \right) .$$
4. Use the iteration formula $$t _ { n + 1 } = 1000 \ln \left( 1 + 0.08 \mathrm { e } ^ { - 0.002 t _ { n } } \right)$$ with \(t _ { 0 } = 50\) to find \(t _ { 1 } , t _ { 2 }\) and \(t _ { 3 }\) and hence, the year in which the populations of these two towns will be equal according to these models.

7. \begin{figure}[h] \end{figure} Figure 1 shows the graph of \(y = \mathrm { f } ( x )\) which meets the coordinate axes at the points \(( a , 0 )\) and \(( 0 , b )\), where \(a\) and \(b\) are constants.

1. Showing, in terms of \(a\) and \(b\), the coordinates of any points of intersection with the axes, sketch on separate diagrams the graphs of
   1. \(\quad y = \mathrm { f } ^ { - 1 } ( x )\),
   2. \(y = 2 \mathrm { f } ( 3 x )\). Given that $$\mathrm { f } ( x ) = 2 - \sqrt { x + 9 } , \quad x \in \mathbb { R } , \quad x \geq - 9 ,$$
2. find the values of \(a\) and \(b\),
3. find an expression for \(\mathrm { f } ^ { - 1 } ( x )\) and state its domain.