| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2005 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area of region with line boundary |
| Difficulty | Challenging +1.2 This is a standard Further Maths polar coordinates question requiring intersection points, distance calculation using polar formulas, and area integration. Part (a) involves algebraic manipulation to find intersections and apply the chord length formula. Part (b) requires setting up and evaluating a polar area integral with a line boundary—a routine FP2 technique but more involved than basic Pure maths questions. |
| Spec | 4.09a Polar coordinates: convert to/from cartesian4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4a(1+\cos\theta) = \frac{3a}{\cos\theta}\) or \(r = 4a\left(1 + \frac{3a}{r}\right)\) | M1 | |
| \(4\cos^2\theta + 4\cos\theta - 3 = 0\) or \(r^2 - 4ar - 12a^2 = 0\) | A1 | |
| \((2\cos\theta - 1)(2\cos\theta + 3) = 0\) or \((r-6a)(r+2a) = 0\) | M1 | |
| \(\cos\theta = \frac{1}{2},\ \theta = \frac{\pi}{3}\) or \(r = 6a\) | A1 | Note \(ON = 3a\) |
| \(PQ = 2 \times ON\tan\frac{\pi}{6} = 6\sqrt{3}a\) | cso M1 A1 | |
| or \(PQ = 2\times\sqrt{(6a)^2 - (3a)^2} = 2\sqrt{27a^2} = 6\sqrt{3}a\) | cso | Any complete equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2 \times \frac{1}{2}\int_0^{\pi/3} r^2\, d\theta = \int 16a^2(1+\cos\theta)^2\, d\theta\) | M1 | \(\int r^2\, d\theta\) |
| \(= \int\left(1 + 2\cos\theta + \frac{1}{2} + \frac{1}{2}\cos 2\theta\right)d\theta\) | M1 | \(\cos^2\theta \to \cos 2\theta\) |
| \(= \left[\frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta\right]\) | A1 | |
| \(= 16a^2\left[\frac{\pi}{2} + \sqrt{3} + \frac{\sqrt{3}}{8}\right]\ \left(= 2a^2[4\pi + 9\sqrt{3}] \approx 56.3a^2\right)\) | M1 A1 | Use of their \(\frac{\pi}{3}\) for M1 |
| Area of \(\triangle POQ = \frac{1}{2} \cdot 6\sqrt{3}\,a \times 3a\) or \(9a^2\sqrt{3}\) | B1 | |
| \(R = a^2(8\pi + 9\sqrt{3})\) | cao A1 | 7 marks total |
# Question 8:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4a(1+\cos\theta) = \frac{3a}{\cos\theta}$ or $r = 4a\left(1 + \frac{3a}{r}\right)$ | M1 | |
| $4\cos^2\theta + 4\cos\theta - 3 = 0$ or $r^2 - 4ar - 12a^2 = 0$ | A1 | |
| $(2\cos\theta - 1)(2\cos\theta + 3) = 0$ or $(r-6a)(r+2a) = 0$ | M1 | |
| $\cos\theta = \frac{1}{2},\ \theta = \frac{\pi}{3}$ or $r = 6a$ | A1 | Note $ON = 3a$ |
| $PQ = 2 \times ON\tan\frac{\pi}{6} = 6\sqrt{3}a$ | cso M1 A1 | |
| or $PQ = 2\times\sqrt{(6a)^2 - (3a)^2} = 2\sqrt{27a^2} = 6\sqrt{3}a$ | cso | Any complete equivalent |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2 \times \frac{1}{2}\int_0^{\pi/3} r^2\, d\theta = \int 16a^2(1+\cos\theta)^2\, d\theta$ | M1 | $\int r^2\, d\theta$ |
| $= \int\left(1 + 2\cos\theta + \frac{1}{2} + \frac{1}{2}\cos 2\theta\right)d\theta$ | M1 | $\cos^2\theta \to \cos 2\theta$ |
| $= \left[\frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta\right]$ | A1 | |
| $= 16a^2\left[\frac{\pi}{2} + \sqrt{3} + \frac{\sqrt{3}}{8}\right]\ \left(= 2a^2[4\pi + 9\sqrt{3}] \approx 56.3a^2\right)$ | M1 A1 | Use of their $\frac{\pi}{3}$ for M1 |
| Area of $\triangle POQ = \frac{1}{2} \cdot 6\sqrt{3}\,a \times 3a$ or $9a^2\sqrt{3}$ | B1 | |
| $R = a^2(8\pi + 9\sqrt{3})$ | cao A1 | 7 marks total |8. The curve $C$ which passes through $O$ has polar equation
$$r = 4 a ( 1 + \cos \theta ) , \quad - \pi < \theta \leq \pi .$$
The line $l$ has polar equation
$$r = 3 a \sec \theta , \quad - \frac { \pi } { 2 } < \theta < \frac { \pi } { 2 } .$$
The line $l$ cuts $C$ at the points $P$ and $Q$, as shown in the diagram.
\begin{enumerate}[label=(\alph*)]
\item Prove that $P Q = 6 \sqrt { } 3 a$.
The region $R$, shown shaded in the diagram, is bounded by $l$ and $C$.
\item Use calculus to find the exact area of $R$.\\
\includegraphics[max width=\textwidth, alt={}, center]{d9aa1f75-ef35-4bf0-85c2-dff36872ca46-2_714_778_1959_1153}
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2005 Q8 [13]}}