| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Year | 2020 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area between two polar curves |
| Difficulty | Challenging +1.8 This is a challenging Further Maths polar coordinates problem requiring students to find intersection points by solving 1+sin θ = 3(1-sin θ), set up the correct area integral with appropriate limits, and evaluate ∫(sin²θ - cos 2θ) type integrals to reach the exact form p√3 - qπ. The multi-step nature, need for careful geometric reasoning about which curve is 'outer', and algebraic manipulation to match the given answer form make this significantly harder than average, though it follows a standard polar area template. |
| Spec | 4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(3(1-\sin\theta)=1+\sin\theta \Rightarrow \sin\theta = \frac{1}{2} \Rightarrow \theta = \ldots\) | M1 | Realises intersection angles required, solves \(C_1=C_2\) |
| \(\theta = \frac{\pi}{6}\ \left(\text{or}\ \frac{5\pi}{6}\right)\) | A1 | Correct value in radians |
| Use of \(\frac{1}{2}\int(1+\sin\theta)^2\,d\theta\) or \(\frac{1}{2}\int\{3(1-\sin\theta)\}^2\,d\theta\) | M1 | Evidence of selecting correct polar area formula on either curve |
| \(\left(\frac{1}{2}\right)\int\!\left[(1+\sin\theta)^2 - 9(1-\sin\theta)^2\right]d\theta\) \(= \left(\frac{1}{2}\right)\int\!\left[1+2\sin\theta+\sin^2\theta - 9+18\sin\theta-9\sin^2\theta\right]d\theta\) | M1, A1 | Fully expands both expressions for \(r^2\); correct expansions (may be unsimplified) |
| \(\int\sin^2\theta\,d\theta = \frac{1}{2}\int(1-\cos 2\theta)\,d\theta \Rightarrow\) \(\int\!\left[(1+\sin\theta)^2-9(1-\sin\theta)^2\right]d\theta = 2\sin 2\theta - 12\theta - 20\cos\theta\) | M1, A1 | Applies double angle identity to reach integrable form; correct integration. FYI: \(\int(1+\sin\theta)^2\,d\theta = \frac{3}{2}\theta - 2\cos\theta - \frac{1}{4}\sin 2\theta\); \(\int 9(1-\sin\theta)^2\,d\theta = \frac{27}{2}\theta + 18\cos\theta - \frac{9}{4}\sin 2\theta\) |
| \(A = \frac{1}{2}\int_{\pi/6}^{5\pi/6}\!\left[(1+\sin\theta)^2-9(1-\sin\theta)^2\right]d\theta\) or \(A = 2\times\frac{1}{2}\int_{\pi/6}^{\pi/2}\!\left[(1+\sin\theta)^2-9(1-\sin\theta)^2\right]d\theta\) | DM1 | Depends on all previous M marks. Correct strategy with appropriate limits; if limits \(\frac{\pi}{6}\) and \(\frac{\pi}{2}\) used, area must be doubled |
| \(= \frac{1}{2}\{(-\sqrt{3}-10\pi+10\sqrt{3})-(\sqrt{3}-2\pi-10\sqrt{3})\} = \ldots\) | ||
| \(= 9\sqrt{3} - 4\pi\) | A1 | Correct area |
## Question 3:
| Answer | Mark | Guidance |
|--------|------|----------|
| $3(1-\sin\theta)=1+\sin\theta \Rightarrow \sin\theta = \frac{1}{2} \Rightarrow \theta = \ldots$ | M1 | Realises intersection angles required, solves $C_1=C_2$ |
| $\theta = \frac{\pi}{6}\ \left(\text{or}\ \frac{5\pi}{6}\right)$ | A1 | Correct value in radians |
| Use of $\frac{1}{2}\int(1+\sin\theta)^2\,d\theta$ or $\frac{1}{2}\int\{3(1-\sin\theta)\}^2\,d\theta$ | M1 | Evidence of selecting correct polar area formula on either curve |
| $\left(\frac{1}{2}\right)\int\!\left[(1+\sin\theta)^2 - 9(1-\sin\theta)^2\right]d\theta$ $= \left(\frac{1}{2}\right)\int\!\left[1+2\sin\theta+\sin^2\theta - 9+18\sin\theta-9\sin^2\theta\right]d\theta$ | M1, A1 | Fully expands both expressions for $r^2$; correct expansions (may be unsimplified) |
| $\int\sin^2\theta\,d\theta = \frac{1}{2}\int(1-\cos 2\theta)\,d\theta \Rightarrow$ $\int\!\left[(1+\sin\theta)^2-9(1-\sin\theta)^2\right]d\theta = 2\sin 2\theta - 12\theta - 20\cos\theta$ | M1, A1 | Applies double angle identity to reach integrable form; correct integration. FYI: $\int(1+\sin\theta)^2\,d\theta = \frac{3}{2}\theta - 2\cos\theta - \frac{1}{4}\sin 2\theta$; $\int 9(1-\sin\theta)^2\,d\theta = \frac{27}{2}\theta + 18\cos\theta - \frac{9}{4}\sin 2\theta$ |
| $A = \frac{1}{2}\int_{\pi/6}^{5\pi/6}\!\left[(1+\sin\theta)^2-9(1-\sin\theta)^2\right]d\theta$ or $A = 2\times\frac{1}{2}\int_{\pi/6}^{\pi/2}\!\left[(1+\sin\theta)^2-9(1-\sin\theta)^2\right]d\theta$ | DM1 | Depends on all previous M marks. Correct strategy with appropriate limits; if limits $\frac{\pi}{6}$ and $\frac{\pi}{2}$ used, area must be doubled |
| $= \frac{1}{2}\{(-\sqrt{3}-10\pi+10\sqrt{3})-(\sqrt{3}-2\pi-10\sqrt{3})\} = \ldots$ | | |
| $= 9\sqrt{3} - 4\pi$ | A1 | Correct area |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{7458ec3b-1be1-4b46-893c-c7470d622e6e-08_549_908_246_790}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of two curves $C _ { 1 }$ and $C _ { 2 }$ with polar equations
$$\begin{array} { l l }
C _ { 1 } : r = ( 1 + \sin \theta ) & 0 \leqslant \theta < 2 \pi \\
C _ { 2 } : r = 3 ( 1 - \sin \theta ) & 0 \leqslant \theta < 2 \pi
\end{array}$$
The region $R$ lies inside $C _ { 1 }$ and outside $C _ { 2 }$ and is shown shaded in Figure 1.\\
Show that the area of $R$ is
$$p \sqrt { 3 } - q \pi$$
where $p$ and $q$ are integers to be determined.
\hfill \mbox{\textit{Edexcel CP1 2020 Q3 [9]}}