| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area of region with line boundary |
| Difficulty | Challenging +1.3 This is a standard Further Maths polar coordinates question requiring a sketch and area calculation using the formula ½∫r²dθ. The integration involves partial fractions and logarithms but follows routine techniques. The 'show that' format provides a target answer, reducing problem-solving demand. Slightly above average difficulty due to the algebraic manipulation required, but well within expected Further Maths scope. |
| Spec | 4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| *(Polar curve sketch)* | B1 | Approximately correct curve passing through the pole \(O\), with correct domain. |
| B1 | \(r\) strictly increasing over the domain \(0\) to \(\frac{\pi}{2}\). | |
| B1 | Correct form at \(O\). Approx tangential to initial line at \(O\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2}\int_0^{\frac{1}{2}\pi}\left(\frac{1}{\pi-\theta}-\frac{1}{\pi}\right)^2 d\theta\) | M1 | Uses \(\frac{1}{2}\int r^2\, d\theta\) with correct limits |
| \(\frac{1}{2}\int_0^{\frac{1}{2}\pi} \frac{1}{(\pi-\theta)^2}-\frac{2}{\pi(\pi-\theta)}+\frac{1}{\pi^2}\, d\theta\) | M1 | Expands |
| \(\frac{1}{2}\left[\frac{1}{\pi-\theta}+\frac{2}{\pi}\ln(\pi-\theta)+\frac{\theta}{\pi^2}\right]_0^{\frac{1}{2}\pi}\) | M1 A1 | Integrates all terms to obtain correct form |
| \(\frac{1}{2}\left(\frac{2}{\pi}+\frac{2}{\pi}\ln\frac{\pi}{2}+\frac{1}{2\pi}-\left(\frac{1}{\pi}+\frac{2}{\pi}\ln\pi\right)\right)=\frac{1}{2}\left(\frac{3}{2\pi}+\frac{2}{\pi}\ln\frac{1}{2}\right)\) | M1 | Substitute limits into correct form and simplify log terms |
| \(=\dfrac{3-4\ln 2}{4\pi}\) | A1 | AG |
# Question 5:
## Part (a):
*(Polar curve sketch)* | **B1** | Approximately correct curve passing through the pole $O$, with correct domain.
| **B1** | $r$ strictly increasing over the domain $0$ to $\frac{\pi}{2}$.
| **B1** | Correct form at $O$. Approx tangential to initial line at $O$.
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}\int_0^{\frac{1}{2}\pi}\left(\frac{1}{\pi-\theta}-\frac{1}{\pi}\right)^2 d\theta$ | M1 | Uses $\frac{1}{2}\int r^2\, d\theta$ with correct limits |
| $\frac{1}{2}\int_0^{\frac{1}{2}\pi} \frac{1}{(\pi-\theta)^2}-\frac{2}{\pi(\pi-\theta)}+\frac{1}{\pi^2}\, d\theta$ | M1 | Expands |
| $\frac{1}{2}\left[\frac{1}{\pi-\theta}+\frac{2}{\pi}\ln(\pi-\theta)+\frac{\theta}{\pi^2}\right]_0^{\frac{1}{2}\pi}$ | M1 A1 | Integrates all terms to obtain correct form |
| $\frac{1}{2}\left(\frac{2}{\pi}+\frac{2}{\pi}\ln\frac{\pi}{2}+\frac{1}{2\pi}-\left(\frac{1}{\pi}+\frac{2}{\pi}\ln\pi\right)\right)=\frac{1}{2}\left(\frac{3}{2\pi}+\frac{2}{\pi}\ln\frac{1}{2}\right)$ | M1 | Substitute limits into correct form and simplify log terms |
| $=\dfrac{3-4\ln 2}{4\pi}$ | A1 | AG |
---5 The curve $C$ has polar equation $r = \frac { 1 } { \pi - \theta } - \frac { 1 } { \pi }$, where $0 \leqslant \theta \leqslant \frac { 1 } { 2 } \pi$.
\begin{enumerate}[label=(\alph*)]
\item Sketch $C$.
\item Show that the area of the region bounded by the half-line $\theta = \frac { 1 } { 2 } \pi$ and $C$ is $\frac { 3 - 4 \ln 2 } { 4 \pi }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2021 Q5 [9]}}