AQA FP3 (Further Pure Mathematics 3) 2013 June

1 It is given that \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = ( x - y ) \sqrt { x + y }$$ and $$y ( 2 ) = 1$$ 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 \(h = 0.2\), to obtain an approximation to \(y ( 2.2 )\), giving your answer to three decimal places.

2 The Cartesian equation of a circle is \(( x + 8 ) ^ { 2 } + ( y - 6 ) ^ { 2 } = 100\).
Using the origin \(O\) as the pole and the positive \(x\)-axis as the initial line, find the polar equation of this circle, giving your answer in the form \(r = p \sin \theta + q \cos \theta\).
(4 marks)

3

1. Find the values of the constants \(a , b\) and \(c\) for which \(a + b x + c x \mathrm { e } ^ { - 3 x }\) is a particular integral of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } - 3 y = 3 x - 8 \mathrm { e } ^ { - 3 x }$$
2. Hence find the general solution of this differential equation.
3. Hence express \(y\) in terms of \(x\), given that \(y = 1\) when \(x = 0\) and that \(\frac { \mathrm { d } y } { \mathrm {~d} x } \rightarrow - 1\) as \(x \rightarrow \infty\).

4 Evaluate the improper integral $$\int _ { 0 } ^ { \infty } \left( \frac { 2 x } { x ^ { 2 } + 4 } - \frac { 4 } { 2 x + 3 } \right) \mathrm { d } x$$ showing the limiting process used and giving your answer in the form \(\ln k\), where \(k\) is a constant.

5

1. Differentiate \(\ln ( \ln x )\) with respect to \(x\).
2. 1. Show that \(\ln x\) is an integrating factor for the first-order differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { 1 } { x \ln x } y = 9 x ^ { 2 } , \quad x > 1$$
   2. Hence find the solution of this differential equation, given that \(y = 4 \mathrm { e } ^ { 3 }\) when \(x = \mathrm { e }\).
      (6 marks)

6 It is given that \(y = ( 4 + \sin x ) ^ { \frac { 1 } { 2 } }\).

1. Express \(y \frac { \mathrm {~d} y } { \mathrm {~d} x }\) in terms of \(\cos x\).
2. Find the value of \(\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }\) when \(x = 0\).
3. Hence, by using Maclaurin's theorem, find the first four terms in the expansion, in ascending powers of \(x\), of \(( 4 + \sin x ) ^ { \frac { 1 } { 2 } }\).
   (2 marks)

7 A differential equation is given by $$\sin ^ { 2 } x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 \sin x \cos x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = 2 \sin ^ { 4 } x \cos x , \quad 0 < x < \pi$$

1. Show that the substitution $$y = u \sin x$$ where \(u\) is a function of \(x\), transforms this differential equation into $$\frac { \mathrm { d } ^ { 2 } u } { \mathrm {~d} x ^ { 2 } } + u = \sin 2 x$$
2. Hence find the general solution of the differential equation $$\sin ^ { 2 } x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 \sin x \cos x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = 2 \sin ^ { 4 } x \cos x$$ giving your answer in the form \(y = \mathrm { f } ( x )\).
   (6 marks)

8 The diagram shows a sketch of a curve and a circle.
\includegraphics[max width=\textwidth, alt={}, center]{a2bc95fe-5588-4ff7-a8a3-0cd07df412c9-4_460_693_370_680} The polar equation of the curve is $$r = 3 + 2 \sin \theta , \quad 0 \leqslant \theta \leqslant 2 \pi$$ The circle, whose polar equation is \(r = 2\), intersects the curve at the points \(P\) and \(Q\), as shown in the diagram.

1. Find the polar coordinates of \(P\) and the polar coordinates of \(Q\).
2. A straight line, drawn from the point \(P\) through the pole \(O\), intersects the curve again at the point \(A\).
   1. Find the polar coordinates of \(A\).
   2. Find, in surd form, the length of \(A Q\).
   3. Hence, or otherwise, explain why the line \(A Q\) is a tangent to the circle \(r = 2\).
3. Find the area of the shaded region which lies inside the circle \(r = 2\) but outside the curve \(r = 3 + 2 \sin \theta\). Give your answer in the form \(\frac { 1 } { 6 } ( m \sqrt { 3 } + n \pi )\), where \(m\) and \(n\) are integers.