| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area of region with line boundary |
| Difficulty | Challenging +1.2 This question requires finding where r = 9/2, setting up a polar area integral with a line boundary (requiring subtraction of a rectangular region), and evaluating ∫(4 + cos 2θ)² dθ using standard trigonometric identities. While it involves multiple steps and careful geometric reasoning about the boundary, the techniques are all standard for Further Maths polar coordinates topics with no novel insights required. |
| Spec | 4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4+\cos2\theta=\frac{9}{2}\Rightarrow\theta=\ldots\) | M1 | Realises angle for \(A\) required and attempts to find it |
| \(\theta=\frac{\pi}{6}\) | A1 | Correct angle |
| \(\frac{1}{2}\int(4+\cos2\theta)^2\,d\theta=\frac{1}{2}\int(16+8\cos2\theta+\cos^22\theta)\,d\theta\) | M1 | Uses correct area formula and squares \(r\) to achieve 3TQ integrand in \(\cos2\theta\) |
| \(\cos^22\theta=\frac{1}{2}+\frac{1}{2}\cos4\theta\Rightarrow A=\frac{1}{2}\int\!\left(16+8\cos2\theta+\frac{1}{2}+\frac{1}{2}\cos4\theta\right)d\theta\) | M1 | Use of correct double angle identity to achieve suitable form for integration |
| \(=\frac{1}{2}\!\left[16\theta+4\sin2\theta+\frac{\sin4\theta}{8}+\frac{\theta}{2}\right]\) | A1 | Correct integration |
| Using limits \(0\) and \(\frac{\pi}{6}\): \(\frac{1}{2}\!\left[\frac{33\pi}{12}+2\sqrt{3}+\frac{\sqrt{3}}{16}-(0)\right]\) | M1 | Correct use of limits |
| Area of triangle \(=\frac{1}{2}(r\cos\theta)(r\sin\theta)=\frac{1}{2}\times\frac{81}{4}\times\frac{1}{2}\times\frac{\sqrt{3}}{2}\) | M1 | Identifies need to subtract area of triangle |
| Area of \(R=\frac{33\pi}{24}+\frac{33\sqrt{3}}{32}-\frac{81\sqrt{3}}{32}\) | M1 | Complete method for area of \(R\) |
| \(=\frac{11}{8}\pi-\frac{3\sqrt{3}}{2}\left(p=\frac{11}{8},\;q=-\frac{3}{2}\right)\) | A1 | Correct final answer |
## Question 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4+\cos2\theta=\frac{9}{2}\Rightarrow\theta=\ldots$ | M1 | Realises angle for $A$ required and attempts to find it |
| $\theta=\frac{\pi}{6}$ | A1 | Correct angle |
| $\frac{1}{2}\int(4+\cos2\theta)^2\,d\theta=\frac{1}{2}\int(16+8\cos2\theta+\cos^22\theta)\,d\theta$ | M1 | Uses correct area formula and squares $r$ to achieve 3TQ integrand in $\cos2\theta$ |
| $\cos^22\theta=\frac{1}{2}+\frac{1}{2}\cos4\theta\Rightarrow A=\frac{1}{2}\int\!\left(16+8\cos2\theta+\frac{1}{2}+\frac{1}{2}\cos4\theta\right)d\theta$ | M1 | Use of correct double angle identity to achieve suitable form for integration |
| $=\frac{1}{2}\!\left[16\theta+4\sin2\theta+\frac{\sin4\theta}{8}+\frac{\theta}{2}\right]$ | A1 | Correct integration |
| Using limits $0$ and $\frac{\pi}{6}$: $\frac{1}{2}\!\left[\frac{33\pi}{12}+2\sqrt{3}+\frac{\sqrt{3}}{16}-(0)\right]$ | M1 | Correct use of limits |
| Area of triangle $=\frac{1}{2}(r\cos\theta)(r\sin\theta)=\frac{1}{2}\times\frac{81}{4}\times\frac{1}{2}\times\frac{\sqrt{3}}{2}$ | M1 | Identifies need to subtract area of triangle |
| Area of $R=\frac{33\pi}{24}+\frac{33\sqrt{3}}{32}-\frac{81\sqrt{3}}{32}$ | M1 | Complete method for area of $R$ |
| $=\frac{11}{8}\pi-\frac{3\sqrt{3}}{2}\left(p=\frac{11}{8},\;q=-\frac{3}{2}\right)$ | A1 | Correct final answer |
**(9 marks)**
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3b36bdc3-a68d-4982-bf23-f780773df5cc-08_492_1063_214_502}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The curve $C$ shown in Figure 1 has polar equation
$$r = 4 + \cos 2 \theta \quad 0 \leqslant \theta \leqslant \frac { \pi } { 2 }$$
At the point $A$ on $C$, the value of $r$ is $\frac { 9 } { 2 }$\\
The point $N$ lies on the initial line and $A N$ is perpendicular to the initial line.\\
The finite region $R$, shown shaded in Figure 1, is bounded by the curve $C$, the initial line and the line $A N$.
Find the exact area of the shaded region $R$, giving your answer in the form $p \pi + q \sqrt { 3 }$ where $p$ and $q$ are rational numbers to be found.\\
\hfill \mbox{\textit{Edexcel CP1 Q4 [9]}}