AQA FP3 (Further Pure Mathematics 3) 2011 January

1 The function \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = x + \sqrt { y }$$ and $$y ( 3 ) = 4$$ Use the improved Euler formula $$y _ { r + 1 } = y _ { r } + \frac { 1 } { 2 } \left( k _ { 1 } + k _ { 2 } \right)$$ where \(k _ { 1 } = h \mathrm { f } \left( x _ { r } , y _ { r } \right)\) and \(k _ { 2 } = h \mathrm { f } \left( x _ { r } + h , y _ { r } + k _ { 1 } \right)\) and \(h = 0.1\), to obtain an approximation to \(y ( 3.1 )\), giving your answer to three decimal places.

2

1. Find the values of the constants \(p\) and \(q\) for which \(p \sin x + q \cos x\) is a particular integral of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + 5 y = 13 \cos x$$
2. Hence find the general solution of this differential equation.

3 A curve \(C\) has polar equation \(r ( 1 + \cos \theta ) = 2\).

1. Find the cartesian equation of \(C\), giving your answer in the form \(y ^ { 2 } = \mathrm { f } ( x )\).
2. The straight line with polar equation \(4 r = 3 \sec \theta\) intersects the curve \(C\) at the points \(P\) and \(Q\). Find the length of \(P Q\).

4 By using an integrating factor, find the solution of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } - \frac { 2 } { x } y = 2 x ^ { 3 } \mathrm { e } ^ { 2 x }$$ given that \(y = \mathrm { e } ^ { 4 }\) when \(x = 2\). Give your answer in the form \(y = \mathrm { f } ( x )\).

5

1. Write \(\frac { 4 } { 4 x + 1 } - \frac { 3 } { 3 x + 2 }\) in the form \(\frac { C } { ( 4 x + 1 ) ( 3 x + 2 ) }\), where \(C\) is a constant.
   (l mark)
2. Evaluate the improper integral $$\int _ { 1 } ^ { \infty } \frac { 10 } { ( 4 x + 1 ) ( 3 x + 2 ) } d x$$ showing the limiting process used and giving your answer in the form \(\ln k\), where \(k\) is a constant.
   (6 marks)

6 The diagram shows a sketch of a curve \(C\).
\includegraphics[max width=\textwidth, alt={}, center]{8cb4b110-274e-47ec-a31b-ee8f84434a65-3_305_556_1078_721} The polar equation of the curve is $$r = 2 \sin 2 \theta \sqrt { \cos \theta } , \quad 0 \leqslant \theta \leqslant \frac { \pi } { 2 }$$ Show that the area of the region bounded by \(C\) is \(\frac { 16 } { 15 }\).

7

1. Write down the expansions in ascending powers of \(x\) up to and including the term in \(x ^ { 3 }\) of:
   1. \(\cos x + \sin x\);
   2. \(\quad \ln ( 1 + 3 x )\).
2. It is given that \(y = \mathrm { e } ^ { \tan x }\).
   1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and show that \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = ( 1 + \tan x ) ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x }\).
   2. Find the value of \(\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }\) when \(x = 0\).
   3. Hence, by using Maclaurin's theorem, show that the first four terms in the expansion, in ascending powers of \(x\), of \(\mathrm { e } ^ { \tan x }\) are $$1 + x + \frac { 1 } { 2 } x ^ { 2 } + \frac { 1 } { 2 } x ^ { 3 }$$
3. Find $$\lim _ { x \rightarrow 0 } \left[ \frac { \mathrm { e } ^ { \tan x } - ( \cos x + \sin x ) } { x \ln ( 1 + 3 x ) } \right]$$

8

1. Given that \(x = \mathrm { e } ^ { t }\) and that \(y\) is a function of \(x\), show that $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { \mathrm { d } y } { \mathrm {~d} t }$$
2. Hence show that the substitution \(x = \mathrm { e } ^ { t }\) transforms the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 3 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = 2 \ln x$$ into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 4 y = 2 t$$
3. Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 4 y = 2 t$$
4. Hence solve the differential equation \(x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 3 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = 2 \ln x\), given that \(y = \frac { 3 } { 2 }\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 2 }\) when \(x = 1\).
   (5 marks)