| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area enclosed by polar curve |
| Difficulty | Standard +0.8 This question requires understanding polar coordinates, using boundary conditions to find a constant, then applying the polar area formula with integration of cos²(2θ). While the setup is moderately challenging (interpreting the geometry and establishing that r_max = 0.6 when θ=0), the integration itself is standard A-level technique. The multi-step nature and polar context place it above average but not exceptionally difficult for Further Maths students. |
| Spec | 4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(2(0.4 + a) = 1.2\) or \(0.4 + a = 0.6\) or \(0.4 + a\cos 0 = 0.6 \Rightarrow a = \ldots\) | M1 | 3.4 |
| \(a = 0.2\) | A1* | 1.1b - cso |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Area of rectangle is \(1.2 \times 0.6 (= 0.72)\) | B1 | 1.1b |
| Area enclosed by curve \(= \frac{1}{2}\int (0.4 + 0.2\cos 2\theta)^2 \, d\theta\) | M1 | 3.1a |
| \((0.4 + 0.2\cos 2\theta)^2 = 0.16 + 0.16\cos 2\theta + 0.04\cos^2 2\theta = 0.16 + 0.16\cos 2\theta + 0.04\left(\frac{\cos 4\theta + 1}{2}\right)\) | M1 | 2.1 |
| \(\frac{1}{2}\int(0.4 + 0.2\cos 2\theta)^2 \, d\theta = \frac{1}{2}\left[0.18\theta + 0.08\sin 2\theta + 0.005\sin 4\theta\right](+c)\) \(= 0.09\theta + 0.04\sin 2\theta + 0.0025\sin 4\theta (+c)\) | A1ft | 1.1b - o.e. |
| Area enclosed by curve \(= \left[0.09\theta + 0.04\sin 2\theta + 0.0025\sin 4\theta\right]_0^{2\pi}\) or \(= 2\left[0.09\theta + 0.04\sin 2\theta + 0.0025\sin 4\theta\right]_0^{\pi}\) or \(= 4\left[0.09\theta + 0.04\sin 2\theta + 0.0025\sin 4\theta\right]_0^{\pi/2}\) | dM1 | 3.1a |
| \(= \frac{9}{50}\pi\) or \(0.18\pi\ (= 0.5654\ldots)\) | A1 | 1.1b |
| Area of wood \(= 1.2 \times 0.6 - 0.18\pi\) | M1 | 1.1b |
| \(= \text{awrt } 0.155 \text{ (m}^2\text{)}\) | A1 | 1.1b |
# Question 3:
## Part (a)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| $2(0.4 + a) = 1.2$ or $0.4 + a = 0.6$ or $0.4 + a\cos 0 = 0.6 \Rightarrow a = \ldots$ | M1 | 3.4 |
| $a = 0.2$ | A1* | 1.1b - cso |
**(2 marks)**
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| Area of rectangle is $1.2 \times 0.6 (= 0.72)$ | B1 | 1.1b |
| Area enclosed by curve $= \frac{1}{2}\int (0.4 + 0.2\cos 2\theta)^2 \, d\theta$ | M1 | 3.1a |
| $(0.4 + 0.2\cos 2\theta)^2 = 0.16 + 0.16\cos 2\theta + 0.04\cos^2 2\theta = 0.16 + 0.16\cos 2\theta + 0.04\left(\frac{\cos 4\theta + 1}{2}\right)$ | M1 | 2.1 |
| $\frac{1}{2}\int(0.4 + 0.2\cos 2\theta)^2 \, d\theta = \frac{1}{2}\left[0.18\theta + 0.08\sin 2\theta + 0.005\sin 4\theta\right](+c)$ $= 0.09\theta + 0.04\sin 2\theta + 0.0025\sin 4\theta (+c)$ | A1ft | 1.1b - o.e. |
| Area enclosed by curve $= \left[0.09\theta + 0.04\sin 2\theta + 0.0025\sin 4\theta\right]_0^{2\pi}$ or $= 2\left[0.09\theta + 0.04\sin 2\theta + 0.0025\sin 4\theta\right]_0^{\pi}$ or $= 4\left[0.09\theta + 0.04\sin 2\theta + 0.0025\sin 4\theta\right]_0^{\pi/2}$ | dM1 | 3.1a |
| $= \frac{9}{50}\pi$ or $0.18\pi\ (= 0.5654\ldots)$ | A1 | 1.1b |
| Area of wood $= 1.2 \times 0.6 - 0.18\pi$ | M1 | 1.1b |
| $= \text{awrt } 0.155 \text{ (m}^2\text{)}$ | A1 | 1.1b |
**(8 marks) — Total: 10 marks**
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9f5761f9-15d0-499a-992a-c98539f2785c-10_508_874_244_609}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Diagram not to scale
Figure 1 shows the design for a table top in the shape of a rectangle $A B C D$. The length of the table, $A B$, is 1.2 m . The area inside the closed curve is made of glass and the surrounding area, shown shaded in Figure 1, is made of wood.
The perimeter of the glass is modelled by the curve with polar equation
$$r = 0.4 + a \cos 2 \theta \quad 0 \leqslant \theta < 2 \pi$$
where $a$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $a = 0.2$
Hence, given that $A D = 60 \mathrm {~cm}$,
\item find the area of the wooden part of the table top, giving your answer in $\mathrm { m } ^ { 2 }$ to 3 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel CP1 2019 Q3 [10]}}