| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Polar curve with substitution integral |
| Difficulty | Challenging +1.8 This is a substantial Further Maths polar coordinates question requiring curve sketching, area calculation with substitution (standard polar area formula with u-substitution), and implicit differentiation to find extrema followed by numerical verification. While multi-part and requiring several techniques, each component follows established methods without requiring novel insight—the substitution is given, and the derivative condition is provided to verify rather than derive independently. |
| Spec | 4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Correct shape of polar curve | B1 | Correct shape |
| Section \(\frac{1}{2}\pi \leqslant \theta \leqslant \pi\) correct | B1 | |
| \(\left(\sqrt{\pi\tan^{-1}\pi},\ 0\right) = (1.99,\ 0)\) | B1 | May be seen on their diagram |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(A = \frac{1}{2}\int_0^{\pi}(\pi-\theta)\tan^{-1}(\pi-\theta)\,\mathrm{d}\theta\) or \(A = -\frac{1}{2}\int_{\pi}^{0}(u)\tan^{-1}(u)\,\mathrm{d}u\) | M1 | Uses correct formula, or \(A = \frac{1}{2}\int_0^{\pi}(u)\tan^{-1}(u)\,\mathrm{d}u\) |
| \(\frac{1}{2}\left[\frac{1}{2}(u)^2\tan^{-1}(u)\right]_0^{\pi} - \frac{1}{4}\int_0^{\pi}\frac{u^2}{1+u^2}\,\mathrm{d}u\) | M1 A1 | Integrates by parts |
| \(\int_0^{\pi}\frac{u^2}{1+u^2}\,\mathrm{d}u = \int_0^{\pi}\frac{1+u^2-1}{1+u^2}\,\mathrm{d}u = \int_0^{\pi}1 - \frac{1}{1+u^2}\,\mathrm{d}u\) | M1 | Rearranges integral into form which can be integrated |
| \(\left[u - \tan^{-1}u\right]_0^{\pi}\) | A1 | |
| \(A = \frac{1}{2}\left[\frac{1}{2}(u)^2\tan^{-1}(u)\right]_0^{\pi} - \frac{1}{4}\left[u - \tan^{-1}(u)\right]_0^{\pi}\) | A1 | |
| \(A = \frac{1}{4}\pi^2\tan^{-1}\pi - \frac{1}{4}\left(\pi - \tan^{-1}\pi\right) = \frac{1}{4}(\pi^2+1)\tan^{-1}\pi - \frac{1}{4}\pi \approx 2.65\) | A1 | |
| Total | 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = \sqrt{(\pi-\theta)\tan^{-1}(\pi-\theta)}\sin\theta\) | B1 | Uses \(y = r\sin\theta\) |
| \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = \sqrt{(\pi-\theta)\tan^{-1}(\pi-\theta)}\cos\theta - \frac{(\pi-\theta)\left(1+(\pi-\theta)^2\right)^{-1}+\tan^{-1}(\pi-\theta)}{2\sqrt{(\pi-\theta)\tan^{-1}(\pi-\theta)}}\sin\theta\) | M1 A1 | Finds derivative |
| \([\theta\neq 0],\ \pi \Rightarrow 2(\pi-\theta)\tan^{-1}(\pi-\theta)\cot\theta - \frac{\pi-\theta}{1+(\pi-\theta)^2} - \tan^{-1}(\pi-\theta) = 0\) | A1 | Puts \(\frac{\mathrm{d}y}{\mathrm{d}\theta}=0\) and forms equation. AG |
| \(2(\pi-1.2)\tan^{-1}(\pi-1.2)\cot1.2 - \frac{\pi-1.2}{1+(\pi-1.2)^2} - \tan^{-1}(\pi-1.2) = 0.151\) and \(2(\pi-1.3)\tan^{-1}(\pi-1.3)\cot1.3 - \frac{\pi-1.3}{1+(\pi-1.3)^2} - \tan^{-1}(\pi-1.3) = -0.395\) | B1 | Shows sign change |
| Total | 5 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Correct shape of polar curve | B1 | Correct shape |
| Section $\frac{1}{2}\pi \leqslant \theta \leqslant \pi$ correct | B1 | |
| $\left(\sqrt{\pi\tan^{-1}\pi},\ 0\right) = (1.99,\ 0)$ | B1 | May be seen on their diagram |
| **Total** | **3** | |
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## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $A = \frac{1}{2}\int_0^{\pi}(\pi-\theta)\tan^{-1}(\pi-\theta)\,\mathrm{d}\theta$ or $A = -\frac{1}{2}\int_{\pi}^{0}(u)\tan^{-1}(u)\,\mathrm{d}u$ | M1 | Uses correct formula, or $A = \frac{1}{2}\int_0^{\pi}(u)\tan^{-1}(u)\,\mathrm{d}u$ |
| $\frac{1}{2}\left[\frac{1}{2}(u)^2\tan^{-1}(u)\right]_0^{\pi} - \frac{1}{4}\int_0^{\pi}\frac{u^2}{1+u^2}\,\mathrm{d}u$ | M1 A1 | Integrates by parts |
| $\int_0^{\pi}\frac{u^2}{1+u^2}\,\mathrm{d}u = \int_0^{\pi}\frac{1+u^2-1}{1+u^2}\,\mathrm{d}u = \int_0^{\pi}1 - \frac{1}{1+u^2}\,\mathrm{d}u$ | M1 | Rearranges integral into form which can be integrated |
| $\left[u - \tan^{-1}u\right]_0^{\pi}$ | A1 | |
| $A = \frac{1}{2}\left[\frac{1}{2}(u)^2\tan^{-1}(u)\right]_0^{\pi} - \frac{1}{4}\left[u - \tan^{-1}(u)\right]_0^{\pi}$ | A1 | |
| $A = \frac{1}{4}\pi^2\tan^{-1}\pi - \frac{1}{4}\left(\pi - \tan^{-1}\pi\right) = \frac{1}{4}(\pi^2+1)\tan^{-1}\pi - \frac{1}{4}\pi \approx 2.65$ | A1 | |
| **Total** | **7** | |
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## Question 7(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \sqrt{(\pi-\theta)\tan^{-1}(\pi-\theta)}\sin\theta$ | B1 | Uses $y = r\sin\theta$ |
| $\frac{\mathrm{d}y}{\mathrm{d}\theta} = \sqrt{(\pi-\theta)\tan^{-1}(\pi-\theta)}\cos\theta - \frac{(\pi-\theta)\left(1+(\pi-\theta)^2\right)^{-1}+\tan^{-1}(\pi-\theta)}{2\sqrt{(\pi-\theta)\tan^{-1}(\pi-\theta)}}\sin\theta$ | M1 A1 | Finds derivative |
| $[\theta\neq 0],\ \pi \Rightarrow 2(\pi-\theta)\tan^{-1}(\pi-\theta)\cot\theta - \frac{\pi-\theta}{1+(\pi-\theta)^2} - \tan^{-1}(\pi-\theta) = 0$ | A1 | Puts $\frac{\mathrm{d}y}{\mathrm{d}\theta}=0$ and forms equation. AG |
| $2(\pi-1.2)\tan^{-1}(\pi-1.2)\cot1.2 - \frac{\pi-1.2}{1+(\pi-1.2)^2} - \tan^{-1}(\pi-1.2) = 0.151$ and $2(\pi-1.3)\tan^{-1}(\pi-1.3)\cot1.3 - \frac{\pi-1.3}{1+(\pi-1.3)^2} - \tan^{-1}(\pi-1.3) = -0.395$ | B1 | Shows sign change |
| **Total** | **5** | |7 The curve $C$ has polar equation $r ^ { 2 } = ( \pi - \theta ) \tan ^ { - 1 } ( \pi - \theta )$, for $0 \leqslant \theta \leqslant \pi$.
\begin{enumerate}[label=(\alph*)]
\item Sketch $C$ and state the polar coordinates of the point of $C$ furthest from the pole.
\item Using the substitution $u = \pi - \theta$, or otherwise, find the area of the region enclosed by $C$ and the initial line.
\item Show that, at the point of $C$ furthest from the initial line,
$$2 ( \pi - \theta ) \tan ^ { - 1 } ( \pi - \theta ) \cot \theta - \frac { \pi - \theta } { 1 + ( \pi - \theta ) ^ { 2 } } - \tan ^ { - 1 } ( \pi - \theta ) = 0$$
and verify that this equation has a root for $\theta$ between 1.2 and 1.3.\\
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 [15]}}