Edexcel FP2 (Further Pure Mathematics 2) 2007 June

1. Obtain the general solution of the differential equation

$$x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = \cos x , \quad x > 0$$ giving your answer in the form \(y = \mathrm { f } ( x )\).
(Total 8 marks)

2.
\includegraphics[max width=\textwidth, alt={}, center]{d6befd60-de40-41b6-8ae5-48656dbca40c-1_734_1228_888_479} The diagram above shows a sketch of the curve with equation $$y = \frac { x ^ { 2 } - 1 } { | x + 2 | } , \quad x \neq - 2$$ The curve crosses the \(x\)-axis at \(x = 1\) and \(x = - 1\) and the line \(x = - 2\) is an asymptote of the curve.

1. Use algebra to solve the equation \(\frac { x ^ { 2 } - 1 } { | x + 2 | } = 3 ( 1 - x )\).
2. Hence, or otherwise, find the set of values of \(x\) for which $$\frac { x ^ { 2 } - 1 } { | x + 2 | } < 3 ( 1 - x )$$ (Total 9 marks)

3. A scientist is modelling the amount of a chemical in the human bloodstream. The amount \(x\) of the chemical, measured in \(\mathrm { mg } l ^ { - 1 }\), at time \(t\) hours satisfies the differential equation $$2 x \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 6 \left( \frac { \mathrm { dx } } { \mathrm { dt } } \right) ^ { 2 } = x ^ { 2 } - 3 x ^ { 4 } , \quad x > 0$$

1. Show that the substitution \(\mathrm { y } = \frac { 1 } { x ^ { 2 } }\) transforms this differential equation into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + y = 3$$
2. Find the general solution of differential equation \(I\). Given that at time \(t = 0 , x = \frac { 1 } { 2 }\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 0\),
3. find an expression for \(x\) in terms of \(t\),
4. write down the maximum value of \(x\) as \(t\) varies.

4.
\includegraphics[max width=\textwidth, alt={}, center]{d6befd60-de40-41b6-8ae5-48656dbca40c-3_535_1027_276_577} The diagram above shows a sketch of the curve \(C\) with polar equation $$r = 4 \sin \theta \cos ^ { 2 } \theta , \quad 0 \leq \theta < \frac { \pi } { 2 }$$ The tangent to \(C\) at the point \(P\) is perpendicular to the initial line.

1. Show that \(P\) has polar coordinates \(\left( \frac { 3 } { 2 } , \frac { \pi } { 6 } \right)\). The point \(Q\) on \(C\) has polar coordinates \(\left( \sqrt { 2 } , \frac { \pi } { 4 } \right)\).
   The shaded region \(R\) is bounded by \(O P , O Q\) and \(C\), as shown in the diagram above.
2. Show that the area of \(R\) is given by $$\int _ { \frac { \pi } { 6 } } ^ { \frac { \pi } { 4 } } \left( \sin ^ { 2 } 2 \theta \cos 2 \theta + \frac { 1 } { 2 } - \frac { 1 } { 2 } \cos 4 \theta \right) \mathrm { d } \theta$$
3. Hence, or otherwise, find the area of \(R\), giving your answer in the form \(a + b \pi\), where \(a\) and \(b\) are rational numbers.
   (Total 14 marks)

5. Find the set of values of \(x\) for which $$\frac { x + 1 } { 2 x - 3 } < \frac { 1 } { x - 3 }$$

6. $$\frac { \mathrm { d } y } { \mathrm {~d} x } - y \tan x = 2 \sec ^ { 3 } x$$ Given that \(y = 3\) at \(x = 0\), find \(y\) in terms of \(x\)
(Total 7 marks)

7. For the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 3 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = 2 x ( x + 3 )$$ find the solution for which at \(x = 0 , \frac { \mathrm {~d} y } { \mathrm {~d} x } = 1\) and \(y = 1\).
(Total 12 marks)

8. (a) Sketch the curve \(C\) with polar equation $$r = 5 + \sqrt { 3 } \cos \theta , \quad 0 \leq \theta \leq 2 \pi$$ (b) Find the polar coordinates of the points where the tangents to \(C\) are parallel to the initial line \(\theta = 0\). Give your answers to 3 significant figures where appropriate.
(c) Using integration, find the area enclosed by the curve \(C\), giving your answer in terms of \(\pi\).

9. $$\frac { \mathrm { d } y } { \mathrm {~d} x } = y \mathrm { e } ^ { x ^ { 2 } } .$$ It is given that \(y = 0.2\) at \(x = 0\).

1. Use the approximation \(\frac { y _ { 1 } - y _ { 0 } } { h } \approx \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) _ { 0 }\), with \(h = 0.1\), to obtain an estimate of the value of \(y\) at \(x = 0.1\).
2. Use your answer to part (a) and the approximation \(\frac { y _ { 2 } - y _ { 0 } } { 2 h } \approx \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) _ { 1 }\), with \(h = 0.1\), to obtain an estimate of the value of \(y\) at \(x = 0.2\). Gives your answer to 4 decimal places.
   (Total 5 marks)

10. $$\left( 1 - x ^ { 2 } \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = 0$$ At \(x = 0 , y = 2\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - 1\).

1. Find the value of \(\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }\) at \(x = 0\).
2. Express \(y\) as a series in ascending powers of \(x\), up to and including the term in \(x ^ { 3 }\).
   (Total 7 marks)

11. (a) Given that \(z = \cos \theta + \mathrm { i } \sin \theta\), use de Moivre's theorem to show that $$z ^ { n } + \frac { 1 } { z ^ { n } } = 2 \cos n \theta$$ (b) Express \(32 \cos ^ { 6 } \theta\) in the form \(p \cos 6 \theta + q \cos 4 \theta + r \cos 2 \theta + \mathrm { s }\), where \(p , q , r\) and \(s\) are integers.
(c) Hence find the exact value of $$\int _ { 0 } ^ { \frac { \pi } { 3 } } \cos ^ { 6 } \theta \mathrm {~d} \theta$$

1. The transformation \(T\) from the \(z\)-plane, where \(z = x + \mathrm { i } y\), to the \(w\)-plane, where \(w = u + \mathrm { i } v\), is given by

$$w = \frac { z + \mathrm { i } } { \mathrm { z } } , \quad z \neq 0 .$$

1. The transformation \(T\) maps the points on the line with equation \(y = x\) in the \(z\)-plane, other than \(( 0,0 )\), to points on a line \(l\) in the \(w\)-plane. Find a cartesian equation of \(l\).
2. Show that the image, under \(T\), of the line with equation \(x + y + 1 = 0\) in the \(z\)-plane is a circle \(C\) in the \(w\)-plane, where \(C\) has cartesian equation $$u ^ { 2 } + v ^ { 2 } - u + v = 0$$
3. On the same Argand diagram, sketch \(l\) and \(C\).