| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Maximum/minimum distance from pole or line |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths polar coordinates question requiring optimization (finding maximum distance), solving transcendental equations, finding intersections, sketching, and computing areas. While it involves several techniques, each part follows standard procedures: part (a) uses differentiation of distance formula, part (b) solves r₁=r₂, and part (d) applies the standard polar area formula. The transcendental equation and exact form requirements add moderate challenge, but this is typical Further Maths fare without requiring exceptional insight. |
| 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 |
| \(x = \theta\cos^2\theta\) | B1 | |
| \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = -2\theta\cos\theta\sin\theta + \cos^2\theta = 0\) | M1 A1 | |
| \(\cos\theta \neq 0 \Rightarrow 2\theta\tan\theta - 1 = 0\) | A1 | |
| \(2(0.6)\tan 0.6 - 1 = -0.179\) and \(2(0.7)\tan 0.7 - 1 = 0.179\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\theta\cos\theta = \theta\sin\theta \Rightarrow \tan\theta = 1\) | M1 | |
| \(\left(\frac{1}{8}\pi\sqrt{2},\, \frac{1}{4}\pi\right)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Sketch showing initial line drawn and \(C_1\) correct | B1 | |
| \(C_2\) correct | B1 | |
| Intersection correct | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2}\int_0^{\frac{1}{4}\pi}\theta^2\left(\cos^2\theta - \sin^2\theta\right)\mathrm{d}\theta\) | M1 | |
| \(= \frac{1}{2}\int_0^{\frac{1}{4}\pi}\theta^2\cos 2\theta\,\mathrm{d}\theta\) | M1 | |
| \(= \frac{1}{2}\left(\left[\frac{1}{2}\theta^2\sin 2\theta\right]_0^{\frac{1}{4}\pi} - \int_0^{\frac{1}{4}\pi}\theta\sin 2\theta\,\mathrm{d}\theta\right)\) | M1 A1 | |
| \(= \frac{1}{2}\left(\left[\frac{1}{2}\theta^2\sin 2\theta\right]_0^{\frac{1}{4}\pi} - \left[-\frac{1}{2}\theta\cos 2\theta\right]_0^{\frac{1}{4}\pi} - \frac{1}{2}\int_0^{\frac{1}{4}\pi}\cos 2\theta\,\mathrm{d}\theta\right)\) | M1 | |
| \(= \frac{1}{4}\left[\theta^2\sin 2\theta + \theta\cos 2\theta - \frac{1}{2}\sin 2\theta\right]_0^{\frac{1}{4}\pi}\) | A1 | |
| \(= \frac{1}{64}\pi^2 - \frac{1}{8}\) | A1 |
## Question 7:
### Part 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = \theta\cos^2\theta$ | B1 | |
| $\frac{\mathrm{d}x}{\mathrm{d}\theta} = -2\theta\cos\theta\sin\theta + \cos^2\theta = 0$ | M1 A1 | |
| $\cos\theta \neq 0 \Rightarrow 2\theta\tan\theta - 1 = 0$ | A1 | |
| $2(0.6)\tan 0.6 - 1 = -0.179$ and $2(0.7)\tan 0.7 - 1 = 0.179$ | B1 | |
### Part 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\theta\cos\theta = \theta\sin\theta \Rightarrow \tan\theta = 1$ | M1 | |
| $\left(\frac{1}{8}\pi\sqrt{2},\, \frac{1}{4}\pi\right)$ | A1 | |
### Part 7(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Sketch showing initial line drawn and $C_1$ correct | B1 | |
| $C_2$ correct | B1 | |
| Intersection correct | B1 | |
### Part 7(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}\int_0^{\frac{1}{4}\pi}\theta^2\left(\cos^2\theta - \sin^2\theta\right)\mathrm{d}\theta$ | M1 | |
| $= \frac{1}{2}\int_0^{\frac{1}{4}\pi}\theta^2\cos 2\theta\,\mathrm{d}\theta$ | M1 | |
| $= \frac{1}{2}\left(\left[\frac{1}{2}\theta^2\sin 2\theta\right]_0^{\frac{1}{4}\pi} - \int_0^{\frac{1}{4}\pi}\theta\sin 2\theta\,\mathrm{d}\theta\right)$ | M1 A1 | |
| $= \frac{1}{2}\left(\left[\frac{1}{2}\theta^2\sin 2\theta\right]_0^{\frac{1}{4}\pi} - \left[-\frac{1}{2}\theta\cos 2\theta\right]_0^{\frac{1}{4}\pi} - \frac{1}{2}\int_0^{\frac{1}{4}\pi}\cos 2\theta\,\mathrm{d}\theta\right)$ | M1 | |
| $= \frac{1}{4}\left[\theta^2\sin 2\theta + \theta\cos 2\theta - \frac{1}{2}\sin 2\theta\right]_0^{\frac{1}{4}\pi}$ | A1 | |
| $= \frac{1}{64}\pi^2 - \frac{1}{8}$ | A1 | |7 The curve $C _ { 1 }$ has polar equation $r = \theta \cos \theta$, for $0 \leqslant \theta \leqslant \frac { 1 } { 2 } \pi$.
\begin{enumerate}[label=(\alph*)]
\item The point on $C _ { 1 }$ furthest from the line $\theta = \frac { 1 } { 2 } \pi$ is denoted by $P$. Show that, at $P$,
$$2 \theta \tan \theta - 1 = 0$$
and verify that this equation has a root between 0.6 and 0.7 .\\
The curve $C _ { 2 }$ has polar equation $r = \theta \sin \theta$, for $0 \leqslant \theta \leqslant \frac { 1 } { 2 } \pi$. The curves $C _ { 1 }$ and $C _ { 2 }$ intersect at the pole, denoted by $O$, and at another point $Q$.
\item Find the polar coordinates of $Q$, giving your answers in exact form.
\item Sketch $C _ { 1 }$ and $C _ { 2 }$ on the same diagram.
\item Find, in terms of $\pi$, the area of the region bounded by the arc $O Q$ of $C _ { 1 }$ and the arc $O Q$ of $C _ { 2 }$. [7]\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2020 Q7 [17]}}