Edexcel FP2 (Further Pure Mathematics 2) 2008 June

\begin{enumerate} \item Solve the differential equation \(\frac { \mathrm { d } y } { \mathrm {~d} x } - 3 y = x\) to obtain \(y\) as a function of \(x\). \item (a) Simplify the expression \(\frac { ( x + 3 ) ( x + 9 ) } { x - 1 } - ( 3 x - 5 )\), giving your answer in the form \(\frac { a ( x + b ) ( x + c ) } { x - 1 }\), where \(a , b\) and \(c\) are integers.
(b) Hence, or otherwise, solve the inequality \(\frac { ( x + 3 ) ( x + 9 ) } { x - 1 } > 3 x - 5 \quad\) (4)(Total \(\mathbf { 8 }\) marks) \item (a) Find the general solution of the differential equation \(3 \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} x } - 2 y = x ^ { 2 }\)
(b) Find the particular solution for which, at \(x = 0 , y = 2\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3\).(6)(Total 14 marks) \item The diagram above shows the curve \(C _ { 1 }\) which has polar equation \(\boldsymbol { r } = \boldsymbol { a } ( \mathbf { 3 } + \mathbf { 2 } \boldsymbol { \operatorname { c o s } } \boldsymbol { \theta } ) , 0 \leq \theta < 2 \pi\) and the circle \(C _ { 2 }\) with equation \(\boldsymbol { r } = \mathbf { 4 } \boldsymbol { a } , 0 \leq \theta < 2 \pi\), where \(a\) is a positive constant.

1. Find, in terms of \(a\), the polar coordinates of the points where the curve \(C _ { 1 }\) meets the circle \(C _ { 2 }\).(4) \end{enumerate} The regions enclosed by the curves \(C _ { 1 }\) and \(C _ { 2 }\) overlap and this common region \(R\) is shaded in the figure.
2. Find, in terms of \(a\), an exact expression for the area of the
   \includegraphics[max width=\textwidth, alt={}, center]{863ef52d-ae75-450c-9eab-8102804868f5-1_523_707_1262_1255}
   region \(R\).(8)
3. In a single diagram, copy the two curves in the diagram above and also sketch the curve \(C _ { 3 }\) with polar equation \(r = 2 a \cos \theta , 0 \leq \theta < 2 \pi\) Show clearly the coordinates of the points of intersection of \(\mathrm { C } _ { 1 } , \mathrm { C } _ { 2 }\) and \(\mathrm { C } _ { 3 }\) with the initial line, \(\theta = 0\).(3)(Total 15 marks)

5. (a) Find, in terms of \(k\), the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 3 x = k t + 5 , \text { where } k \text { is a constant and } t > 0 .$$ For large values of \(t\), this general solution may be approximated by a linear function.
(b) Given that \(k = 6\), find the equation of this linear function.(2)(Total 9 marks)

6. (a) Find, in the simplest surd form where appropriate, the exact values of \(x\) for which $$\frac { x } { 2 } + 3 = \left| \frac { 4 } { x } \right|$$ (b) Sketch, on the same axes, the line with equation \(y = \frac { x } { 2 } + 3\) and the graph of \(y = \left| \frac { 4 } { x } \right| , \quad x \neq 0\).
(c) Find the set of values of \(x\) for which \(\frac { x } { 2 } + 3 > \left| \frac { 4 } { x } \right|\).
(2)(Total 10 marks)

7. (a) Show that the substitution \(y = v x\) transforms the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x } { y } + \frac { 3 y } { x } , \quad x > 0 , \quad y > 0$$ into the differential equation $$x \frac { \mathrm {~d} v } { \mathrm {~d} x } = 2 v + \frac { 1 } { v } .$$ (b) By solving differential equation (II), find a general solution of differential equation (I) in the form \(y = \mathrm { f } ( x )\). Given that \(y = 3\) at \(x = 1\),
(c)find the particular solution of differential equation (I).(2)

8. The curve \(C\) shown in the diagram above has polar equation $$r = 4 ( 1 - \cos \theta ) , 0 \leq \theta \leq \frac { \pi } { 2 }$$ At the point \(P\) on \(C\), the tangent to \(C\) is parallel to the line \(\theta = \frac { \pi } { 2 }\).

1. Show that \(P\) has polar coordinates \(\left( 2 , \frac { \pi } { 3 } \right)\). The curve \(C\) meets the line \(\theta = \frac { \pi } { 2 }\) at the point \(A\). The tangent to \(C\) at the initial line at the point \(N\). The finite region \(R\), shown shaded in
   \includegraphics[max width=\textwidth, alt={}]{863ef52d-ae75-450c-9eab-8102804868f5-2_737_561_1395_1329} the diagram above, is bounded by the initial line, the line \(\theta = \frac { \pi } { 2 }\), the arc \(A P\) of \(C\) and the line \(P N\).
2. Calculate the exact area of \(R\).

9. $$\left( x ^ { 2 } + 1 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 2 y ^ { 2 } + ( 1 - 2 x ) \frac { \mathrm { d } y } { \mathrm {~d} x }$$

1. By differentiating equation (I) with respect to \(x\), show that $$\left( x ^ { 2 } + 1 \right) \frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } = ( 1 - 4 x ) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + ( 4 y - 2 ) \frac { \mathrm { d } y } { \mathrm {~d} x }$$ Given that \(y = 1\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 1\) at \(x = 0\),
2. find the series solution for \(y\), in ascending powers of \(x\), up to and including the term in \(x _ { 3 }\).(4)
3. Use your series to estimate the value of \(y\) at \(x = - 0.5\), giving your answer to two decimal places.(1)

10. The point \(P\) represents a complex number \(z\) on an Argand diagram such that $$| z - 3 | = 2 | z |$$

1. Show that, as \(z\) varies, the locus of \(P\) is a circle, and give the coordinates of the centre and the radius of the circle.(5) The point \(Q\) represents a complex number \(z\) on an Argand diagram such that $$| z + 3 | = | z - i \sqrt { } 3 |$$
2. Sketch, on the same Argand diagram, the locus of \(P\) and the locus of \(Q\) as \(z\) varies.(5)
3. On your diagram shade the region which satisfies $$| z - 3 | \geq 2 | z | \text { and } | z + 3 | \geq | z - i \sqrt { } 3 |$$

1. De Moivre's theorem states that \(\quad ( \cos \theta + \mathrm { i } \sin \theta ) ^ { n } = \cos n \theta + \mathrm { i } \sin n \theta\) for \(n \in \Re\)
   1. Use induction to prove de Moivre's theorem for \(n \in \mathbb { Z } ^ { + }\).
   2. Show that \(\cos 5 \theta = 16 \cos ^ { 5 } \theta - 20 \cos ^ { 3 } \theta + 5 \cos \theta\)
   3. Hence show that \(2 \cos \frac { \pi } { 10 }\) is a root of the equation
   $$x ^ { 4 } - 5 x ^ { 2 } + 5 = 0$$