| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2024 |
| Session | November |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Show polar curve has Cartesian form |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths polar coordinates question requiring conversion to Cartesian form (standard technique), sketching, numerical verification of an intersection point, and area calculation using polar integration. While it involves several steps and Further Maths content, each part follows well-established procedures without requiring novel insight—the Cartesian conversion is routine, the area integral is a direct application of the polar area formula, and part (c) is just numerical verification. Slightly above average difficulty due to the multi-step nature and Further Maths context, but not exceptionally challenging. |
| Spec | 4.09a 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 |
| \(r = a\left(\frac{x}{r}+\frac{y}{r}\right) \Rightarrow r^2 = ax + ay\) | M1 | Uses \(x = r\cos\theta\) and \(y = r\sin\theta\) to eliminate \(\theta\) |
| \(x^2 - ax + y^2 - ay = 0\) | M1A1 | OE. Obtains Cartesian equation (using \(r^2 = x^2 + y^2\)) |
| \(\left(x-\frac{a}{2}\right)^2 + \left(y-\frac{a}{2}\right)^2 = \frac{a^2}{2}\), Centre \(\left(\frac{a}{2}, \frac{a}{2}\right)\) and radius \(\frac{a}{\sqrt{2}}\) | B1 | |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Circle sketch | B1 | Circle |
| Correct position, passing through \(O\) | B1 | Correct position, passing through \(O\) |
| \(a\sqrt{2}\) | B1 | States maximum distance or labels sketch |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\cos\phi + \sin\phi - \phi = 0\) | M1 | |
| \(\cos 1.25 + \sin 1.25 - 1.25 = 0.01\), \(\cos 1.26 + \sin 1.26 - 1.26 = -0.002\) | A1 | Shows sign change |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{a^2}{2}\int_0^{\phi}\theta^2\,d\theta + \frac{a^2}{2}\int_{\phi}^{\frac{3}{4}\pi}(\cos\theta+\sin\theta)^2\,d\theta\) | M1 | Uses area formula on both curves |
| A1 | Forms area of smaller region enclosed by \(C_1\) and \(C_2\) with correct limits | |
| \(\frac{a^2}{2}\int_0^{\phi}\theta^2\,d\theta + \frac{a^2}{2}\int_{\phi}^{\frac{3}{4}\pi}1+2\sin\theta\cos\theta\,d\theta\) | M1 | Applies relevant identities for circle integral to produce integrable form |
| \(\frac{a^2}{2}\left[\frac{1}{3}\theta^3\right]_0^{\phi} + \frac{a^2}{2}\left[\theta+\sin^2\theta\right]_{\phi}^{\frac{3}{4}\pi}\) | A1 | For correct integration of the circle part |
| \(\frac{a^2\phi^3}{6} + \frac{a^2}{2}\left(\frac{3}{4}\pi + \frac{1}{2} - \phi - \sin^2\phi\right) = \frac{a^2}{2}\left(\frac{\phi^3}{3}+\frac{3}{4}\pi - \phi + \frac{1}{2}\cos 2\phi\right)\) | A1 | Integrates spiral and substitutes correct limits. AG. |
| \(\frac{a^2}{2}\pi - \frac{a^2}{2}\left(\frac{\phi^3}{3}+\frac{3}{4}\pi-\phi+\frac{1}{2}\cos 2\phi\right)\) | M1 | Forms area of larger region enclosed by \(C_1\) and \(C_2\) |
| \(\frac{a^2}{2}\left(-\frac{\phi^3}{3}+\frac{1}{4}\pi+\phi-\frac{1}{2}\cos 2\phi\right)\) | A1 | |
| 7 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $r = a\left(\frac{x}{r}+\frac{y}{r}\right) \Rightarrow r^2 = ax + ay$ | M1 | Uses $x = r\cos\theta$ and $y = r\sin\theta$ to eliminate $\theta$ |
| $x^2 - ax + y^2 - ay = 0$ | M1A1 | OE. Obtains Cartesian equation (using $r^2 = x^2 + y^2$) |
| $\left(x-\frac{a}{2}\right)^2 + \left(y-\frac{a}{2}\right)^2 = \frac{a^2}{2}$, Centre $\left(\frac{a}{2}, \frac{a}{2}\right)$ and radius $\frac{a}{\sqrt{2}}$ | B1 | |
| | **4** | |
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Circle sketch | B1 | Circle |
| Correct position, passing through $O$ | B1 | Correct position, passing through $O$ |
| $a\sqrt{2}$ | B1 | States maximum distance or labels sketch |
| | **3** | |
## Question 7(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cos\phi + \sin\phi - \phi = 0$ | M1 | |
| $\cos 1.25 + \sin 1.25 - 1.25 = 0.01$, $\cos 1.26 + \sin 1.26 - 1.26 = -0.002$ | A1 | Shows sign change |
| | **2** | |
## Question 7(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{a^2}{2}\int_0^{\phi}\theta^2\,d\theta + \frac{a^2}{2}\int_{\phi}^{\frac{3}{4}\pi}(\cos\theta+\sin\theta)^2\,d\theta$ | M1 | Uses area formula on both curves |
| | A1 | Forms area of smaller region enclosed by $C_1$ and $C_2$ with correct limits |
| $\frac{a^2}{2}\int_0^{\phi}\theta^2\,d\theta + \frac{a^2}{2}\int_{\phi}^{\frac{3}{4}\pi}1+2\sin\theta\cos\theta\,d\theta$ | M1 | Applies relevant identities for circle integral to produce integrable form |
| $\frac{a^2}{2}\left[\frac{1}{3}\theta^3\right]_0^{\phi} + \frac{a^2}{2}\left[\theta+\sin^2\theta\right]_{\phi}^{\frac{3}{4}\pi}$ | A1 | For correct integration of the circle part |
| $\frac{a^2\phi^3}{6} + \frac{a^2}{2}\left(\frac{3}{4}\pi + \frac{1}{2} - \phi - \sin^2\phi\right) = \frac{a^2}{2}\left(\frac{\phi^3}{3}+\frac{3}{4}\pi - \phi + \frac{1}{2}\cos 2\phi\right)$ | A1 | Integrates spiral and substitutes correct limits. AG. |
| $\frac{a^2}{2}\pi - \frac{a^2}{2}\left(\frac{\phi^3}{3}+\frac{3}{4}\pi-\phi+\frac{1}{2}\cos 2\phi\right)$ | M1 | Forms area of larger region enclosed by $C_1$ and $C_2$ |
| $\frac{a^2}{2}\left(-\frac{\phi^3}{3}+\frac{1}{4}\pi+\phi-\frac{1}{2}\cos 2\phi\right)$ | A1 | |
| | **7** | |7 The curve $C _ { 1 }$ has polar equation $r = a ( \cos \theta + \sin \theta )$ for $- \frac { 1 } { 4 } \pi \leqslant \theta \leqslant \frac { 3 } { 4 } \pi$, where $a$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Find a Cartesian equation for $C _ { 1 }$ and show that it represents a circle, stating its radius and the Cartesian coordinates of its centre.
\item Sketch $C _ { 1 }$ and state the greatest distance of a point on $C _ { 1 }$ from the pole.\\
\includegraphics[max width=\textwidth, alt={}, center]{beb9c1f1-1676-4432-a42a-c418ff9f45d8-14_2721_40_107_2010}
The curve $C _ { 2 }$ with polar equation $r = a \theta$ intersects $C _ { 1 }$ at the pole and the point with polar coordinates $( a \phi , \phi )$.
\item Verify that $1.25 < \phi < 1.26$.
\item Show that the area of the smaller region enclosed by $C _ { 1 }$ and $C _ { 2 }$ is equal to
$$\frac { 1 } { 2 } a ^ { 2 } \left( \frac { 3 } { 4 } \pi + \frac { 1 } { 3 } \phi ^ { 3 } - \phi + \frac { 1 } { 2 } \cos 2 \phi \right)$$
and deduce, in terms of $a$ and $\phi$, the area of the larger region enclosed by $C _ { 1 }$ and $C _ { 2 }$.\\
If you use the following page to complete the answer to any question, the question number must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2024 Q7 [16]}}