| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area of region with line boundary |
| Difficulty | Challenging +1.8 This is a challenging Further Maths polar coordinates question requiring: (a) analysis of cotangent behavior to find maximum r and determine a, (b) integration of a non-standard polar area formula with cotangent, and (c) conversion to Cartesian form involving algebraic manipulation of trigonometric identities. The cotangent function and the specific angle transformations make this harder than routine polar area questions, but the multi-part structure provides guidance through the problem. |
| 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 |
| \(\theta = 0\) [sketch of curve with initial line, correct position and shape] | B1 B1 | Initial line with correct position and shape of curve. SCB1 if correct shape with wrong starting position. Pole position must be clear |
| \(a\cot\frac{1}{6}\pi = 2\sqrt{3} \Rightarrow a = 2\) | B1 | Substitutes \(\theta = \frac{1}{6}\pi\), AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2}\int_0^{\frac{1}{6}\pi} 4\cot^2\!\left(\frac{1}{3}\pi - \theta\right)d\theta\) | M1 | Uses \(\frac{1}{2}\int r^2\,d\theta\) with correct limits |
| \(2\int_0^{\frac{1}{6}\pi}\csc^2\!\left(\frac{1}{3}\pi-\theta\right)-1\,d\theta = 2\Big[\cot\!\left(\frac{1}{3}\pi-\theta\right)-\theta\Big]_0^{\frac{1}{6}\pi}\) | M1 A1 | Uses \(\cot^2\!\left(\frac{1}{3}\pi-\theta\right) = \csc^2\!\left(\frac{1}{3}\pi-\theta\right)-1\) and integrates |
| \(2\!\left(\sqrt{3}-\frac{1}{6}\pi-\frac{1}{\sqrt{3}}\sqrt{3}\right) = \frac{4}{3}\sqrt{3}-\frac{1}{3}\pi\) | A1 | OE, must be exact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(r = 2\dfrac{\cos\frac{\pi}{3}\cos\theta + \sin\frac{\pi}{3}\sin\theta}{\sin\frac{\pi}{3}\cos\theta - \cos\frac{\pi}{3}\sin\theta}\) | M1 | Uses the identities for \(\cos(A-B)\) and \(\sin(A-B)\), or identity for \(\tan(A-B)\) |
| \(\sqrt{x^2+y^2} = 2\dfrac{r\cos\theta + \sqrt{3}\,r\sin\theta}{\sqrt{3}\,r\cos\theta - r\sin\theta}\) | M1 | Applies \(r=\sqrt{x^2+y^2}\), \(x = r\cos\theta\), \(y = r\sin\theta\) |
| \(2(x + y\sqrt{3}) = (x\sqrt{3}-y)\sqrt{x^2+y^2}\) | A1 | AG |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\theta = 0$ [sketch of curve with initial line, correct position and shape] | B1 B1 | Initial line with correct position and shape of curve. SCB1 if correct shape with wrong starting position. Pole position must be clear |
| $a\cot\frac{1}{6}\pi = 2\sqrt{3} \Rightarrow a = 2$ | B1 | Substitutes $\theta = \frac{1}{6}\pi$, AG |
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}\int_0^{\frac{1}{6}\pi} 4\cot^2\!\left(\frac{1}{3}\pi - \theta\right)d\theta$ | M1 | Uses $\frac{1}{2}\int r^2\,d\theta$ with correct limits |
| $2\int_0^{\frac{1}{6}\pi}\csc^2\!\left(\frac{1}{3}\pi-\theta\right)-1\,d\theta = 2\Big[\cot\!\left(\frac{1}{3}\pi-\theta\right)-\theta\Big]_0^{\frac{1}{6}\pi}$ | M1 A1 | Uses $\cot^2\!\left(\frac{1}{3}\pi-\theta\right) = \csc^2\!\left(\frac{1}{3}\pi-\theta\right)-1$ and integrates |
| $2\!\left(\sqrt{3}-\frac{1}{6}\pi-\frac{1}{\sqrt{3}}\sqrt{3}\right) = \frac{4}{3}\sqrt{3}-\frac{1}{3}\pi$ | A1 | OE, must be exact |
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $r = 2\dfrac{\cos\frac{\pi}{3}\cos\theta + \sin\frac{\pi}{3}\sin\theta}{\sin\frac{\pi}{3}\cos\theta - \cos\frac{\pi}{3}\sin\theta}$ | M1 | Uses the identities for $\cos(A-B)$ and $\sin(A-B)$, or identity for $\tan(A-B)$ |
| $\sqrt{x^2+y^2} = 2\dfrac{r\cos\theta + \sqrt{3}\,r\sin\theta}{\sqrt{3}\,r\cos\theta - r\sin\theta}$ | M1 | Applies $r=\sqrt{x^2+y^2}$, $x = r\cos\theta$, $y = r\sin\theta$ |
| $2(x + y\sqrt{3}) = (x\sqrt{3}-y)\sqrt{x^2+y^2}$ | A1 | AG |5 The curve $C$ has polar equation $r = \operatorname { acot } \left( \frac { 1 } { 3 } \pi - \theta \right)$, where $a$ is a positive constant and $0 \leqslant \theta \leqslant \frac { 1 } { 6 } \pi$. It is given that the greatest distance of a point on $C$ from the pole is $2 \sqrt { 3 }$.
\begin{enumerate}[label=(\alph*)]
\item Sketch $C$ and show that $a = 2$.
\item Find the exact value of the area of the region bounded by $C$, the initial line and the half-line $\theta = \frac { 1 } { 6 } \pi$.
\item Show that $C$ has Cartesian equation $2 ( x + y \sqrt { 3 } ) = ( x \sqrt { 3 } - y ) \sqrt { x ^ { 2 } + y ^ { 2 } }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2021 Q5 [10]}}