AQA FP3 (Further Pure Mathematics 3) 2010 June

1 The function \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = x + 3 + \sin y$$ and $$y ( 1 ) = 1$$

1. Use the Euler formula $$y _ { r + 1 } = y _ { r } + h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$ with \(h = 0.1\), to obtain an approximation to \(y ( 1.1 )\), giving your answer to four decimal places.
2. Use the formula $$y _ { r + 1 } = y _ { r - 1 } + 2 h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$ with your answer to part (a), to obtain an approximation to \(y ( 1.2 )\), giving your answer to three decimal places.

2

1. Find the value of the constant \(k\) for which \(k \sin 2 x\) is a particular integral of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + y = \sin 2 x$$
2. Hence find the general solution of this differential equation.

3

1. Explain why \(\int _ { 1 } ^ { \infty } 4 x \mathrm { e } ^ { - 4 x } \mathrm {~d} x\) is an improper integral.
2. Find \(\quad \int 4 x \mathrm { e } ^ { - 4 x } \mathrm {~d} x\).
3. Hence evaluate \(\int _ { 1 } ^ { \infty } 4 x \mathrm { e } ^ { - 4 x } \mathrm {~d} x\), showing the limiting process used.

4 By using an integrating factor, find the solution of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { 3 } { x } y = \left( x ^ { 4 } + 3 \right) ^ { \frac { 3 } { 2 } }$$ given that \(y = \frac { 1 } { 5 }\) when \(x = 1\).
(9 marks)

5

1. Write down the expansion of \(\cos 4 x\) in ascending powers of \(x\) up to and including the term in \(x ^ { 4 }\). Give your answer in its simplest form.
2. 1. Given that \(y = \ln \left( 2 - \mathrm { e } ^ { x } \right)\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x } , \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) and \(\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }\).
      (You may leave your expression for \(\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }\) unsimplified.)
   2. Hence, by using Maclaurin's theorem, show that the first three non-zero terms in the expansion, in ascending powers of \(x\), of \(\ln \left( 2 - \mathrm { e } ^ { x } \right)\) are $$- x - x ^ { 2 } - x ^ { 3 }$$
3. Find $$\lim _ { x \rightarrow 0 } \left[ \frac { x \ln \left( 2 - \mathrm { e } ^ { x } \right) } { 1 - \cos 4 x } \right]$$

6 The polar equation of a curve \(C _ { 1 }\) is $$r = 2 ( \cos \theta - \sin \theta ) , \quad 0 \leqslant \theta \leqslant 2 \pi$$

1. 1. Find the cartesian equation of \(C _ { 1 }\).
   2. Deduce that \(C _ { 1 }\) is a circle and find its radius and the cartesian coordinates of its centre.
2. The diagram shows the curve \(C _ { 2 }\) with polar equation $$r = 4 + \sin \theta , \quad 0 \leqslant \theta \leqslant 2 \pi$$ \includegraphics[max width=\textwidth, alt={}, center]{90a59b47-3799-46a2-b76b-ced5cc3e1aac-4_519_847_443_593}
   1. Find the area of the region that is bounded by \(C _ { 2 }\).
   2. Prove that the curves \(C _ { 1 }\) and \(C _ { 2 }\) do not intersect.
   3. Find the area of the region that is outside \(C _ { 1 }\) but inside \(C _ { 2 }\).

7

1. Given that \(x = t ^ { \frac { 1 } { 2 } } , x > 0 , t > 0\) and \(y\) is a function of \(x\), show that:
   1. \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 t ^ { \frac { 1 } { 2 } } \frac { \mathrm {~d} y } { \mathrm {~d} t }\);
      (2 marks)
   2. \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 4 t \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} t }\).
      (3 marks)
2. Hence show that the substitution \(x = t ^ { \frac { 1 } { 2 } }\) transforms the differential equation $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \left( 8 x ^ { 2 } + 1 \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + 12 x ^ { 3 } y = 12 x ^ { 5 }$$ into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 3 y = 3 t$$ (2 marks)
3. Hence find the general solution of the differential equation $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \left( 8 x ^ { 2 } + 1 \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + 12 x ^ { 3 } y = 12 x ^ { 5 }$$ giving your answer in the form \(y = \mathrm { f } ( x )\).