AQA FP3 (Further Pure Mathematics 3) 2013 January

1 It is given that \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = \sqrt { 2 x + y }$$ and $$y ( 3 ) = 5$$

1. Use the Euler formula $$y _ { r + 1 } = y _ { r } + h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$ with \(h = 0.2\), to obtain an approximation to \(y ( 3.2 )\), 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 \(y ( 3.4 )\), giving your answer to three decimal places.

2

1. Write down the expansion of \(\mathrm { e } ^ { 3 x }\) in ascending powers of \(x\) up to and including the term in \(x ^ { 2 }\).
2. Hence, or otherwise, find the term in \(x ^ { 2 }\) in the expansion, in ascending powers of \(x\), of \(\mathrm { e } ^ { 3 x } ( 1 + 2 x ) ^ { - \frac { 3 } { 2 } }\).
   (4 marks)

3 It is given that the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + y = 0$$ is \(y = \mathrm { e } ^ { x } ( A x + B )\). Hence find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + y = 6 \mathrm { e } ^ { x }$$

4

1. Explain why \(\int _ { 0 } ^ { 1 } x ^ { 4 } \ln x \mathrm {~d} x\) is an improper integral.
   (l mark)
2. Evaluate \(\int _ { 0 } ^ { 1 } x ^ { 4 } \ln x \mathrm {~d} x\), showing the limiting process used.
   (6 marks)

5

1. Show that \(\tan x\) is an integrating factor for the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { \sec ^ { 2 } x } { \tan x } y = \tan x$$ (2 marks)
2. Hence solve this differential equation, given that \(y = 3\) when \(x = \frac { \pi } { 4 }\).
   (6 marks)

6

1. It is given that \(y = \ln \left( \mathrm { e } ^ { 3 x } \cos x \right)\).
   1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 - \tan x\).
   2. Find \(\frac { \mathrm { d } ^ { 4 } y } { \mathrm {~d} x ^ { 4 } }\).
2. Hence use Maclaurin's theorem to show that the first three non-zero terms in the expansion, in ascending powers of \(x\), of \(\ln \left( \mathrm { e } ^ { 3 x } \cos x \right)\) are \(3 x - \frac { 1 } { 2 } x ^ { 2 } - \frac { 1 } { 12 } x ^ { 4 }\).
   (3 marks)
3. Write down the expansion of \(\ln ( 1 + p x )\), where \(p\) is a constant, in ascending powers of \(x\) up to and including the term in \(x ^ { 2 }\).
4. 1. Find the value of \(p\) for which \(\lim _ { x \rightarrow 0 } \left[ \frac { 1 } { x ^ { 2 } } \ln \left( \frac { \mathrm { e } ^ { 3 x } \cos x } { 1 + p x } \right) \right]\) exists.
   2. Hence find the value of \(\lim _ { x \rightarrow 0 } \left[ \frac { 1 } { x ^ { 2 } } \ln \left( \frac { \mathrm { e } ^ { 3 x } \cos x } { 1 + p x } \right) \right]\) when \(p\) takes the value found in part (d)(i).

7

1. Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 6 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 10 y = \mathrm { e } ^ { 2 t }$$ giving your answer in the form \(y = \mathrm { f } ( t )\).
2. Given that \(x = t ^ { \frac { 1 } { 2 } } , x > 0 , t > 0\) and \(y\) is a function of \(x\), show that $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 4 t \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} t }$$ (5 marks)
3. Hence show that the substitution \(x = t ^ { \frac { 1 } { 2 } }\) transforms the differential equation $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \left( 12 x ^ { 2 } + 1 \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + 40 x ^ { 3 } y = 4 x ^ { 3 } \mathrm { e } ^ { 2 x ^ { 2 } }$$ into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 6 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 10 y = \mathrm { e } ^ { 2 t }$$ (2 marks)
4. Hence write down the general solution of the differential equation $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \left( 12 x ^ { 2 } + 1 \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + 40 x ^ { 3 } y = 4 x ^ { 3 } \mathrm { e } ^ { 2 x ^ { 2 } }$$ (l mark)

8 The diagram shows a sketch of a curve.
\includegraphics[max width=\textwidth, alt={}, center]{f05737eb-adb1-4228-aebf-6b5c7f26a434-5_464_574_402_726} The polar equation of the curve is $$r = \sin 2 \theta \sqrt { \left( 2 + \frac { 1 } { 2 } \cos \theta \right) } , \quad 0 \leqslant \theta \leqslant \frac { \pi } { 2 }$$ The point \(P\) is the point of the curve at which \(\theta = \frac { \pi } { 3 }\). The perpendicular from \(P\) to the initial line meets the initial line at the point \(N\).

1. 1. Find the exact value of \(r\) when \(\theta = \frac { \pi } { 3 }\).
   2. Show that the polar equation of the line \(P N\) is \(r = \frac { 3 \sqrt { 3 } } { 8 } \sec \theta\).
   3. Find the area of triangle \(O N P\) in the form \(\frac { k \sqrt { 3 } } { 128 }\), where \(k\) is an integer.
2. 1. Using the substitution \(u = \sin \theta\), or otherwise, find \(\int \sin ^ { n } \theta \cos \theta \mathrm {~d} \theta\), where \(n \geqslant 2\).
   2. Find the area of the shaded region bounded by the line \(O P\) and the arc \(O P\) of the curve. Give your answer in the form \(a \pi + b \sqrt { 3 } + c\), where \(a , b\) and \(c\) are constants.
      (8 marks)