| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2023 |
| Session | November |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Polar curve with exponential function |
| Difficulty | Challenging +1.8 This is a Further Maths polar coordinates question requiring curve sketching, area calculation using integration, and optimization with implicit differentiation. While it involves multiple techniques (exponential functions, polar area formula, and finding extrema by setting dy/dx = 0), each part follows standard Further Maths procedures without requiring particularly novel insights. The verification in part (c) is routine substitution. |
| 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 |
| Initial line drawn at \(\theta = 0\), correct shape with \(r\) strictly decreasing | B1 | Initial line drawn. Correct shape, \(r\) strictly decreasing. |
| Correct shape at extremities | B1 | Correct shape at extremities. |
| \(1 - e^{-\frac{1}{2}\pi}\) | B1 | May be seen on *their* diagram. |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2}\int_0^{\frac{1}{2}\pi}\left(e^{-\theta} - e^{-\frac{1}{2}\pi}\right)^2 d\theta\) | M1 | Uses correct formula with correct limits. |
| \(\frac{1}{2}\int_0^{\frac{1}{2}\pi} e^{-2\theta} - 2e^{-\theta - \frac{1}{2}\pi} + e^{-\pi}\, d\theta\) | A1 | |
| \(\frac{1}{2}\left[-\frac{1}{2}e^{-2\theta} + 2e^{-\theta-\frac{1}{2}\pi} + e^{-\pi}\theta\right]_0^{\frac{1}{2}\pi}\) | M1A1 | Integrates. |
| \(\frac{1}{2}\left(-\frac{1}{2}e^{-\pi} + 2e^{-\pi} + \frac{1}{2}\pi e^{-\pi} + \frac{1}{2} - 2e^{-\frac{1}{2}\pi}\right) = \frac{3}{4}e^{-\pi} + \frac{1}{4}\pi e^{-\pi} - e^{-\frac{1}{2}\pi} + \frac{1}{4}\) | A1 | |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = \left(e^{-\theta} - e^{-\frac{1}{2}\pi}\right)\sin\theta\) | B1 | Uses \(y = r\sin\theta\) |
| \(\frac{dy}{d\theta} = \left(e^{-\theta} - e^{-\frac{1}{2}\pi}\right)\cos\theta + \sin\theta\left(-e^{-\theta}\right) = 0\) | M1A1 | Sets derivative equal to zero. |
| \(\left[\theta \neq \frac{1}{2}\pi \Rightarrow\right] 1 + \left(\frac{-e^{-\theta}}{e^{-\theta} - e^{-\frac{1}{2}\pi}}\right)\tan\theta = 0 \Rightarrow 1 - e^{\theta - \frac{1}{2}\pi} - \tan\theta = 0\) | A1 | AG. |
| \(1 - e^{0.56 - \frac{1}{2}\pi} - \tan 0.56 = 0.00912\) and \(1 - e^{0.57 - \frac{1}{2}\pi} - \tan 0.57 = -0.00856\) | B1 | Shows sign change. |
| 5 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Initial line drawn at $\theta = 0$, correct shape with $r$ strictly decreasing | B1 | Initial line drawn. Correct shape, $r$ strictly decreasing. |
| Correct shape at extremities | B1 | Correct shape at extremities. |
| $1 - e^{-\frac{1}{2}\pi}$ | B1 | May be seen on *their* diagram. |
| | **3** | |
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}\int_0^{\frac{1}{2}\pi}\left(e^{-\theta} - e^{-\frac{1}{2}\pi}\right)^2 d\theta$ | M1 | Uses correct formula with correct limits. |
| $\frac{1}{2}\int_0^{\frac{1}{2}\pi} e^{-2\theta} - 2e^{-\theta - \frac{1}{2}\pi} + e^{-\pi}\, d\theta$ | A1 | |
| $\frac{1}{2}\left[-\frac{1}{2}e^{-2\theta} + 2e^{-\theta-\frac{1}{2}\pi} + e^{-\pi}\theta\right]_0^{\frac{1}{2}\pi}$ | M1A1 | Integrates. |
| $\frac{1}{2}\left(-\frac{1}{2}e^{-\pi} + 2e^{-\pi} + \frac{1}{2}\pi e^{-\pi} + \frac{1}{2} - 2e^{-\frac{1}{2}\pi}\right) = \frac{3}{4}e^{-\pi} + \frac{1}{4}\pi e^{-\pi} - e^{-\frac{1}{2}\pi} + \frac{1}{4}$ | A1 | |
| | **5** | |
## Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \left(e^{-\theta} - e^{-\frac{1}{2}\pi}\right)\sin\theta$ | B1 | Uses $y = r\sin\theta$ |
| $\frac{dy}{d\theta} = \left(e^{-\theta} - e^{-\frac{1}{2}\pi}\right)\cos\theta + \sin\theta\left(-e^{-\theta}\right) = 0$ | M1A1 | Sets derivative equal to zero. |
| $\left[\theta \neq \frac{1}{2}\pi \Rightarrow\right] 1 + \left(\frac{-e^{-\theta}}{e^{-\theta} - e^{-\frac{1}{2}\pi}}\right)\tan\theta = 0 \Rightarrow 1 - e^{\theta - \frac{1}{2}\pi} - \tan\theta = 0$ | A1 | AG. |
| $1 - e^{0.56 - \frac{1}{2}\pi} - \tan 0.56 = 0.00912$ and $1 - e^{0.57 - \frac{1}{2}\pi} - \tan 0.57 = -0.00856$ | B1 | Shows sign change. |
| | **5** | |6 The curve $C$ has polar equation $r = \mathrm { e } ^ { - \theta } - \mathrm { e } ^ { - \frac { 1 } { 2 } \pi }$, where $0 \leqslant \theta \leqslant \frac { 1 } { 2 } \pi$.
\begin{enumerate}[label=(\alph*)]
\item Sketch $C$ and state, in exact form, the greatest distance of a point on $C$ from the pole.
\item Find the exact value of the area of the region bounded by $C$ and the initial line.
\item Show that, at the point on $C$ furthest from the initial line,
$$1 - e ^ { \theta - \frac { 1 } { 2 } \pi } - \tan \theta = 0$$
and verify that this equation has a root between 0.56 and 0.57 .
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2023 Q6 [13]}}