| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2022 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area between two polar curves |
| Difficulty | Challenging +1.3 This is a standard Further Maths polar coordinates question requiring sketching of polar curves and computing area between them using the formula A = ½∫(r₁² - r₂²)dθ. While it involves multiple steps (identifying the 6-petaled rose curve, finding intersection points, setting up and evaluating the integral with cos²(6θ) integration), these are all routine techniques for FM students. The 'show working' requirement and exact answer adds some rigour, but this remains a textbook application rather than requiring novel insight. |
| Spec | 4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Closed loop with "petals" containing circle of radius 1 | M1 | Allow any closed loop that oscillates, at least 4 petals, need not have correct number |
| Fully correct – 6 petals in roughly the right places, curvature need not be perfectly smooth | A1 | 6 petals in right places, maximum radius between 5 and 6 radius lines, minimum between 2 and 3 |
| Circle centre \(O\) radius 1 | B1 | Circle of radius 1 and centre \(O\) drawn |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(\frac{1}{2}\right)\int r^2\,d\theta = \left(\frac{1}{2}\right)\int\left(16-12\cos 6\theta+\frac{9}{4}\cos^2 6\theta\right)d\theta\) | M1 | Attempts to square \(r\), achieving a 3-term quadratic in \(\cos 6\theta\) |
| \(= \frac{1}{2}\int_0^{2\pi}\left(16-12\cos 6\theta+\frac{9}{8}(1+\cos 12\theta)\right)d\theta\) | M1 | Applies double angle formula to \(\cos^2\) term; accept \(\cos^2 6\theta \to \frac{1}{2}(\pm 1 \pm \cos 12\theta)\) |
| \(= \frac{1}{2}\left[16\theta - 2\sin 6\theta + \frac{9}{8}\left(\theta+\frac{1}{12}\sin 12\theta\right)\right]_0^{2\pi}\) | M1 A1 | M1: Achieves form \(\alpha\theta+\beta\sin 6\theta+\gamma\sin 12\theta\), \(\alpha,\beta,\gamma\neq 0\); A1: Correct integration (limits and \(\frac{1}{2}\) not needed); look for \(16\theta-2\sin 6\theta+\frac{9}{8}\left(\theta+\frac{1}{12}\sin 12\theta\right)\) |
| \(A_{outer} = \frac{1}{2}\int_0^{2\pi}r^2\,d\theta = \frac{1}{2}\left(32\pi - 0 + \frac{9}{8}(2\pi+0)-(0)\right)\) | dM1 | Depends on at least two previous M's; correct overall strategy for area in outer loop, correct limits, \(\frac{1}{2}\) present or implied |
| Area required is \(\frac{1}{2}\left(32\pi+\frac{9\pi}{4}\right)-\pi(1^2) = \ldots\) | B1 | Subtracts correct area of \(\pi\) for inner circle |
| \(= \dfrac{129}{8}\pi\) | A1 | cso; check sin terms disappear with limits |
# Question 4:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Closed loop with "petals" containing circle of radius 1 | M1 | Allow any closed loop that oscillates, at least 4 petals, need not have correct number |
| Fully correct – 6 petals in roughly the right places, curvature need not be perfectly smooth | A1 | 6 petals in right places, maximum radius between 5 and 6 radius lines, minimum between 2 and 3 |
| Circle centre $O$ radius 1 | B1 | Circle of radius 1 and centre $O$ drawn |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(\frac{1}{2}\right)\int r^2\,d\theta = \left(\frac{1}{2}\right)\int\left(16-12\cos 6\theta+\frac{9}{4}\cos^2 6\theta\right)d\theta$ | M1 | Attempts to square $r$, achieving a 3-term quadratic in $\cos 6\theta$ |
| $= \frac{1}{2}\int_0^{2\pi}\left(16-12\cos 6\theta+\frac{9}{8}(1+\cos 12\theta)\right)d\theta$ | M1 | Applies double angle formula to $\cos^2$ term; accept $\cos^2 6\theta \to \frac{1}{2}(\pm 1 \pm \cos 12\theta)$ |
| $= \frac{1}{2}\left[16\theta - 2\sin 6\theta + \frac{9}{8}\left(\theta+\frac{1}{12}\sin 12\theta\right)\right]_0^{2\pi}$ | M1 A1 | M1: Achieves form $\alpha\theta+\beta\sin 6\theta+\gamma\sin 12\theta$, $\alpha,\beta,\gamma\neq 0$; A1: Correct integration (limits and $\frac{1}{2}$ not needed); look for $16\theta-2\sin 6\theta+\frac{9}{8}\left(\theta+\frac{1}{12}\sin 12\theta\right)$ |
| $A_{outer} = \frac{1}{2}\int_0^{2\pi}r^2\,d\theta = \frac{1}{2}\left(32\pi - 0 + \frac{9}{8}(2\pi+0)-(0)\right)$ | dM1 | Depends on at least two previous M's; correct overall strategy for area in outer loop, correct limits, $\frac{1}{2}$ present or implied |
| Area required is $\frac{1}{2}\left(32\pi+\frac{9\pi}{4}\right)-\pi(1^2) = \ldots$ | B1 | Subtracts correct area of $\pi$ for inner circle |
| $= \dfrac{129}{8}\pi$ | A1 | cso; check sin terms disappear with limits |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0d458344-42cb-48d1-90b3-e071df8ea7bb-12_897_1040_205_534}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows part of the curve with polar equation
$$r = 4 - \frac { 3 } { 2 } \cos 6 \theta \quad 0 \leqslant \theta < 2 \pi$$
\begin{enumerate}[label=(\alph*)]
\item Sketch, on the polar grid in Figure 2,
\begin{enumerate}[label=(\roman*)]
\item the rest of the curve with equation
$$r = 4 - \frac { 3 } { 2 } \cos 6 \theta \quad 0 \leqslant \theta < 2 \pi$$
\item the polar curve with equation
$$r = 1$$
$$0 \leqslant \theta < 2 \pi$$
A spare copy of the grid is given on page 15.
In part (b) you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
\end{enumerate}\item Determine the exact area enclosed between the two curves defined in part (a).
Only use this grid if you need to redraw your answer to part (a)
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0d458344-42cb-48d1-90b3-e071df8ea7bb-15_901_1042_1651_532}
\captionsetup{labelformat=empty}
\caption{Copy of Figure 2}
\end{center}
\end{figure}
\end{enumerate}
\hfill \mbox{\textit{Edexcel F2 2022 Q4 [10]}}