Edexcel FP2 (Further Pure Mathematics 2) 2015 June

1. (a) Use algebra to find the set of values of \(x\) for which

$$x + 2 > \frac { 12 } { x + 3 }$$ (b) Hence, or otherwise, find the set of values of \(x\) for which $$x + 2 > \frac { 12 } { | x + 3 | }$$

2. $$z = - 2 + ( 2 \sqrt { 3 } ) \mathrm { i }$$

1. Find the modulus and the argument of \(z\). Using de Moivre's theorem,
2. find \(z ^ { 6 }\), simplifying your answer,
3. find the values of \(w\) such that \(w ^ { 4 } = z ^ { 3 }\), giving your answers in the form \(a + \mathrm { i } b\) where \(a , b \in \mathbb { R }\).

1. Find, in the form \(y = \mathrm { f } ( x )\), the general solution of the differential equation

$$\tan x \frac { \mathrm {~d} y } { \mathrm {~d} x } + y = 3 \cos 2 x \tan x , \quad 0 < x < \frac { \pi } { 2 }$$

4. (a) Show that $$r ^ { 2 } ( r + 1 ) ^ { 2 } - ( r - 1 ) ^ { 2 } r ^ { 2 } \equiv 4 r ^ { 3 }$$ Given that \(\sum _ { r = 1 } ^ { n } r = \frac { 1 } { 2 } n ( n + 1 )\)
(b) use the identity in (a) and the method of differences to show that $$\left( 1 ^ { 3 } + 2 ^ { 3 } + 3 ^ { 3 } + \ldots + n ^ { 3 } \right) = ( 1 + 2 + 3 + \ldots + n ) ^ { 2 }$$

1. A transformation \(T\) from the \(z\)-plane to the \(w\)-plane is given by

$$w = \frac { z } { z + 3 \mathrm { i } } , \quad z \neq - 3 \mathrm { i }$$ The circle with equation \(| z | = 2\) is mapped by \(T\) onto the curve \(C\).

1. 1. Show that \(C\) is a circle.
   2. Find the centre and radius of \(C\). The region \(| z | \leqslant 2\) in the \(z\)-plane is mapped by \(T\) onto the region \(R\) in the \(w\)-plane.
2. Shade the region \(R\) on an Argand diagram.

6. \begin{figure}[h] \end{figure} The curve \(C\), shown in Figure 1, has polar equation $$r = 3 a ( 1 + \cos \theta ) , \quad 0 \leqslant \theta < \pi$$ The tangent to \(C\) at the point \(A\) is parallel to the initial line.

1. Find the polar coordinates of \(A\). The finite region \(R\), shown shaded in Figure 1, is bounded by the curve \(C\), the initial line and the line \(O A\).
2. Use calculus to find the area of the shaded region \(R\), giving your answer in the form \(a ^ { 2 } ( p \pi + q \sqrt { 3 } )\), where \(p\) and \(q\) are rational numbers.

7. $$y = \tan ^ { 2 } x , \quad - \frac { \pi } { 2 } < x < \frac { \pi } { 2 }$$

1. Show that \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 6 \sec ^ { 4 } x - 4 \sec ^ { 2 } x\)
2. Hence show that \(\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } = 8 \sec ^ { 2 } x \tan x \left( A \sec ^ { 2 } x + B \right)\), where \(A\) and \(B\) are constants to be found.
3. Find the Taylor series expansion of \(\tan ^ { 2 } x\), in ascending powers of \(\left( x - \frac { \pi } { 3 } \right)\), up to and including the term in \(\left( x - \frac { \pi } { 3 } \right) ^ { 3 }\)

1. (a) Show that the transformation \(x = \mathrm { e } ^ { u }\) transforms the differential equation

$$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 7 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 16 y = 2 \ln x , \quad x > 0$$ into the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} u ^ { 2 } } - 8 \frac { \mathrm {~d} y } { \mathrm {~d} u } + 16 y = 2 u$$ (b) Find the general solution of the differential equation (II), expressing \(y\) as a function of \(u\).
(c) Hence obtain the general solution of the differential equation (I).