AQA FP3 (Further Pure Mathematics 3) 2014 June

1 It is given that \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = \frac { \ln ( x + y ) } { \ln y }$$ and $$y ( 6 ) = 3$$ 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 \(h = 0.4\), to obtain an approximation to \(y ( 6.4 )\), giving your answer to three decimal places.
[0pt] [5 marks]

2

1. Find the values of the constants \(a\), \(b\) and \(c\) for which \(a + b \sin 2 x + c \cos 2 x\) is a particular integral of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + 4 y = 20 - 20 \cos 2 x$$ [4 marks]
2. Hence find the solution of this differential equation, given that \(y = 4\) when \(x = 0\).
   [0pt] [4 marks]
   \includegraphics[max width=\textwidth, alt={}]{0eb3e96e-528c-4a99-b164-31cc865f0d68-04_1974_1709_733_153}

3 A curve has polar equation \(r ( 4 - 3 \cos \theta ) = 4\). Find its Cartesian equation in the form \(y ^ { 2 } = \mathrm { f } ( x )\).
[0pt] [4 marks]

4 Solve the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } - 3 y = 2 \mathrm { e } ^ { - x }$$ given that \(y \rightarrow 0\) as \(x \rightarrow \infty\) and that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - 3\) when \(x = 0\).
[0pt] [10 marks]
\includegraphics[max width=\textwidth, alt={}, center]{0eb3e96e-528c-4a99-b164-31cc865f0d68-08_46_145_829_159}

5

1. Find \(\int x \cos 8 x \mathrm {~d} x\).
2. Find \(\lim _ { x \rightarrow 0 } \left[ \frac { 1 } { x } \sin 2 x \right]\).
3. Explain why \(\int _ { 0 } ^ { \frac { \pi } { 4 } } \left( 2 \cot 2 x - \frac { 1 } { x } + x \cos 8 x \right) \mathrm { d } x\) is an improper integral.
4. Evaluate \(\int _ { 0 } ^ { \frac { \pi } { 4 } } \left( 2 \cot 2 x - \frac { 1 } { x } + x \cos 8 x \right) \mathrm { d } x\), showing the limiting process used. Give your answer as a single term.
   [0pt] [4 marks]

6

1. By using an integrating factor, find the general solution of the differential equation $$\frac { \mathrm { d } u } { \mathrm {~d} x } - \frac { 2 x } { x ^ { 2 } + 4 } u = 3 \left( x ^ { 2 } + 4 \right)$$ giving your answer in the form \(u = \mathrm { f } ( x )\).
   [0pt] [6 marks]
2. Show that the substitution \(u = x ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x }\) transforms the differential equation $$x ^ { 2 } \left( x ^ { 2 } + 4 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 8 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 3 \left( x ^ { 2 } + 4 \right) ^ { 2 }$$ into $$\frac { \mathrm { d } u } { \mathrm {~d} x } - \frac { 2 x } { x ^ { 2 } + 4 } u = 3 \left( x ^ { 2 } + 4 \right)$$
3. Hence, given that \(x > 0\), find the general solution of the differential equation $$x ^ { 2 } \left( x ^ { 2 } + 4 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 8 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 3 \left( x ^ { 2 } + 4 \right) ^ { 2 }$$ [2 marks]

7

1. It is given that \(y = \ln ( \cos x + \sin x )\).
   1. Show that \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = - \frac { 2 } { 1 + \sin 2 x }\).
   2. Find \(\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }\).
2. 1. Hence use Maclaurin's theorem to show that the first three non-zero terms in the expansion, in ascending powers of \(x\), of \(\ln ( \cos x + \sin x )\) are \(x - x ^ { 2 } + \frac { 2 } { 3 } x ^ { 3 }\).
   2. Write down the first three non-zero terms in the expansion, in ascending powers of \(x\), of \(\ln ( \cos x - \sin x )\).
3. Hence find the first three non-zero terms in the expansion, in ascending powers of \(x\), of \(\ln \left( \frac { \cos 2 x } { \mathrm { e } ^ { 3 x - 1 } } \right)\).
   [0pt] [4 marks]
   \includegraphics[max width=\textwidth, alt={}]{0eb3e96e-528c-4a99-b164-31cc865f0d68-17_2484_1707_221_153}
   \includegraphics[max width=\textwidth, alt={}]{0eb3e96e-528c-4a99-b164-31cc865f0d68-19_2484_1707_221_153}

8 The diagram shows a sketch of a curve \(C\), the pole \(O\) and the initial line. The curve \(C\) intersects the initial line at the point \(P\).
\includegraphics[max width=\textwidth, alt={}, center]{0eb3e96e-528c-4a99-b164-31cc865f0d68-20_432_949_402_525} The polar equation of \(C\) is \(r = \left( 1 - \tan ^ { 2 } \theta \right) \sec \theta , - \frac { \pi } { 4 } \leqslant \theta \leqslant \frac { \pi } { 4 }\).

1. Show that the area of the region bounded by the curve \(C\) is \(\frac { 8 } { 15 }\).
2. The curve whose polar equation is $$r = \frac { 1 } { 2 } \sec ^ { 3 } \theta , \quad - \frac { \pi } { 4 } \leqslant \theta \leqslant \frac { \pi } { 4 }$$ intersects \(C\) at the points \(A\) and \(B\).
   1. Find the polar coordinates of \(A\) and \(B\).
   2. Given that angle \(O A P =\) angle \(O B P = \alpha\), show that \(\tan \alpha = k \sqrt { 3 }\), where \(k\) is an integer.
   3. Using your value of \(k\) from part (b)(ii), state whether the point \(A\) lies inside or lies outside the circle whose diameter is \(O P\). Give a reason for your answer.
      [0pt] [1 mark]
      \includegraphics[max width=\textwidth, alt={}]{0eb3e96e-528c-4a99-b164-31cc865f0d68-21_2484_1707_221_153}
      \includegraphics[max width=\textwidth, alt={}]{0eb3e96e-528c-4a99-b164-31cc865f0d68-23_2484_1707_221_153}