| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2023 |
| Session | November |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area between two polar curves |
| Difficulty | Challenging +1.2 This is a standard Further Maths polar coordinates question requiring conversion between Cartesian and polar forms, finding intersections, sketching curves, and computing area between curves. While it involves multiple parts and requires careful setup of the area integral, all techniques are routine for Further Maths students with no novel insights needed. The algebra is straightforward and the curves (r=cos θ and r=sin 2θ) are standard examples. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x^2 - x + \frac{1}{4} + y^2 = \frac{1}{4} \Rightarrow r^2 - r\cos\theta + \frac{1}{4} = \frac{1}{4}\) | B1 | Uses \(x^2+y^2=r^2\) and \(x=r\cos\theta\) |
| \(r(r-\cos\theta)=0\) | M1 | Factorises |
| \([r\neq 0 \Rightarrow]\ r=\cos\theta\) | A1 | AG |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sin 2\theta = \cos\theta \Rightarrow 2\sin\theta\cos\theta = \cos\theta\) | M1 | Sets \(r\) values equal and uses \(\sin 2\theta = 2\sin\theta\cos\theta\) |
| \(\cos\theta \neq 0 \Rightarrow \sin\theta = \frac{1}{2}\) | A1 | \(\cos\theta \neq 0\) must be recognised |
| \(\left(\frac{1}{2}\sqrt{3}, \frac{1}{6}\pi\right)\) | A1 | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| [sketch] | B1 | Initial line drawn and one curve correct |
| [sketch] | B1 | Other curve correct |
| [sketch with P marked] | B1 | Intersection marked in correct position and both curves labelled |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2}\int_0^{\frac{1}{6}\pi}\sin^2 2\theta\, d\theta + \frac{1}{2}\int_{\frac{1}{6}\pi}^{\frac{1}{2}\pi}\cos^2\theta\, d\theta\) | M1 | Uses \(\frac{1}{2}\int r^2\, d\theta\) with correct limits |
| \(\frac{1}{2}\int_0^{\frac{1}{6}\pi}\sin^2 2\theta\, d\theta = \frac{1}{4}\int_0^{\frac{1}{6}\pi}1 - \cos 4\theta\, d\theta\) | M1 | Integrates \(\sin^2 2\theta\) using identity |
| \(= \frac{1}{4}\left[\theta - \frac{1}{4}\sin 4\theta\right]_0^{\frac{1}{6}\pi}\) | A1 | |
| \(\frac{1}{2}\int_{\frac{1}{6}\pi}^{\frac{1}{2}\pi}\cos^2\theta\, d\theta = \frac{1}{4}\int_{\frac{1}{6}\pi}^{\frac{1}{2}\pi}1 + \cos 2\theta\, d\theta\) | M1 | Integrates \(\cos^2\theta\) using identity |
| \(= \frac{1}{4}\left[\theta + \frac{1}{2}\sin 2\theta\right]_{\frac{1}{6}\pi}^{\frac{1}{2}\pi}\) | A1 | |
| \(\frac{1}{4}\left(\frac{1}{6}\pi - \frac{1}{8}\sqrt{3}\right) + \frac{1}{4}\left(\frac{1}{2}\pi - \frac{1}{6}\pi - \frac{1}{4}\sqrt{3}\right) = \frac{1}{8}\left(\pi - \frac{3}{4}\sqrt{3}\right)\) | A1 | |
| 6 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^2 - x + \frac{1}{4} + y^2 = \frac{1}{4} \Rightarrow r^2 - r\cos\theta + \frac{1}{4} = \frac{1}{4}$ | B1 | Uses $x^2+y^2=r^2$ and $x=r\cos\theta$ |
| $r(r-\cos\theta)=0$ | M1 | Factorises |
| $[r\neq 0 \Rightarrow]\ r=\cos\theta$ | A1 | AG |
| **Total: 3** | | |
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sin 2\theta = \cos\theta \Rightarrow 2\sin\theta\cos\theta = \cos\theta$ | M1 | Sets $r$ values equal and uses $\sin 2\theta = 2\sin\theta\cos\theta$ |
| $\cos\theta \neq 0 \Rightarrow \sin\theta = \frac{1}{2}$ | A1 | $\cos\theta \neq 0$ must be recognised |
| $\left(\frac{1}{2}\sqrt{3}, \frac{1}{6}\pi\right)$ | A1 | |
| | **3** | |
## Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| [sketch] | B1 | Initial line drawn and one curve correct |
| [sketch] | B1 | Other curve correct |
| [sketch with P marked] | B1 | Intersection marked in correct position and both curves labelled |
| | **3** | |
## Question 6(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}\int_0^{\frac{1}{6}\pi}\sin^2 2\theta\, d\theta + \frac{1}{2}\int_{\frac{1}{6}\pi}^{\frac{1}{2}\pi}\cos^2\theta\, d\theta$ | M1 | Uses $\frac{1}{2}\int r^2\, d\theta$ with correct limits |
| $\frac{1}{2}\int_0^{\frac{1}{6}\pi}\sin^2 2\theta\, d\theta = \frac{1}{4}\int_0^{\frac{1}{6}\pi}1 - \cos 4\theta\, d\theta$ | M1 | Integrates $\sin^2 2\theta$ using identity |
| $= \frac{1}{4}\left[\theta - \frac{1}{4}\sin 4\theta\right]_0^{\frac{1}{6}\pi}$ | A1 | |
| $\frac{1}{2}\int_{\frac{1}{6}\pi}^{\frac{1}{2}\pi}\cos^2\theta\, d\theta = \frac{1}{4}\int_{\frac{1}{6}\pi}^{\frac{1}{2}\pi}1 + \cos 2\theta\, d\theta$ | M1 | Integrates $\cos^2\theta$ using identity |
| $= \frac{1}{4}\left[\theta + \frac{1}{2}\sin 2\theta\right]_{\frac{1}{6}\pi}^{\frac{1}{2}\pi}$ | A1 | |
| $\frac{1}{4}\left(\frac{1}{6}\pi - \frac{1}{8}\sqrt{3}\right) + \frac{1}{4}\left(\frac{1}{2}\pi - \frac{1}{6}\pi - \frac{1}{4}\sqrt{3}\right) = \frac{1}{8}\left(\pi - \frac{3}{4}\sqrt{3}\right)$ | A1 | |
| | **6** | |6
\begin{enumerate}[label=(\alph*)]
\item Show that the curve with Cartesian equation
$$\left( x - \frac { 1 } { 2 } \right) ^ { 2 } + y ^ { 2 } = \frac { 1 } { 4 }$$
has polar equation $r = \cos \theta$.\\
The curves $C _ { 1 }$ and $C _ { 2 }$ have polar equations
$$r = \cos \theta \quad \text { and } \quad r = \sin 2 \theta$$
respectively, where $0 \leqslant \theta \leqslant \frac { 1 } { 2 } \pi$. The curves $C _ { 1 }$ and $C _ { 2 }$ intersect at the pole and at another point $P$.
\item Find the polar coordinates of $P$.
\item In a single diagram sketch $C _ { 1 }$ and $C _ { 2 }$, clearly identifying each curve, and mark the point $P$.
\item The region $R$ is enclosed by $C _ { 1 }$ and $C _ { 2 }$ and includes the line $O P$.
Find, in exact form, the area of $R$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2023 Q6 [15]}}