| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Polar curves with trigonometric identities |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths polar coordinates question requiring conversion between Cartesian and polar forms, sketching, and area calculation with substitution. Part (a) is routine algebraic manipulation using x=r cos θ, y=r sin θ. Part (c) involves standard polar area formula and a guided substitution with partial fractions. While it requires multiple techniques and careful algebra, the question provides significant scaffolding (the trigonometric identity, the substitution to use) and follows predictable Further Maths patterns. It's moderately above average difficulty due to length and integration complexity, but not exceptionally challenging. |
| Spec | 1.08h Integration by substitution4.09a Polar coordinates: convert to/from cartesian4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = r\cos\theta\), \(y = r\sin\theta\), \(x^2 + y^2 = r^2\) | B1 | SOI |
| \(r^2(1 + \sin\theta\cos\theta) = r^2(1 + \frac{1}{2}\sin 2\theta) = a\) | M1 | Eliminates both \(x\) and \(y\) and uses double angle formula |
| \(r^2 = \frac{2a}{2 + \sin 2\theta}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| [Polar graph with curve in correct domain] | B1 | Polar graph with curve in correct domain. Graph needs gradient \(\leq 0\) and \(r > 0\) for all \(\theta\) in domain |
| [Concave graph with \(r\) at \(\theta = \frac{\pi}{4}\) greater than half \(r\) at \(\theta = 0\)] | B1 | \(r\) strictly decreasing such that \(r\) at \(\theta = \frac{\pi}{4}\) is greater than half \(r\) at \(\theta = 0\). Concave graph. Gradient at \(\theta = 0\) and \(\frac{\pi}{4}\) must not be vertical or horizontal respectively |
| Answer | Marks | Guidance |
|---|---|---|
| \(R = \frac{1}{2}\int_0^{\frac{1}{4}\pi} r^2\, d\theta = a\int_0^{\frac{1}{4}\pi} \frac{1+\tan^2\theta}{2(1+\tan^2\theta)+2\tan\theta}\, d\theta\) | B1 | AG |
| \(t = \tan\theta\) leading to \(\frac{dt}{d\theta} = \sec^2\theta\) | M1 | Applies given substitution |
| \(\frac{dt}{d\theta} = \sec^2\theta = 1 + t^2\) | M1 | Applies \(\sec^2\theta = 1 + \tan^2\theta\) |
| \(R = \frac{1}{2}a\int_0^1 \frac{1}{t^2+t+1}\, dt\) | A1 | |
| \(t^2 + t + 1 = (t+\frac{1}{2})^2 + \frac{3}{4}\) | B1 | Completes the square |
| \(R = \frac{1}{2}a\int_0^1 \frac{1}{(t+\frac{1}{2})^2 + \frac{3}{4}}\, dt = \frac{1}{\sqrt{3}}a\left[\tan^{-1}\left(\frac{2}{\sqrt{3}}t + \frac{1}{\sqrt{3}}\right)\right]_0^1\) | M1 A1 | Applies formula |
| \(\frac{\pi a}{6\sqrt{3}}\) | A1 |
## Question 6(a):
$x = r\cos\theta$, $y = r\sin\theta$, $x^2 + y^2 = r^2$ | **B1** | SOI
$r^2(1 + \sin\theta\cos\theta) = r^2(1 + \frac{1}{2}\sin 2\theta) = a$ | **M1** | Eliminates both $x$ and $y$ and uses double angle formula
$r^2 = \frac{2a}{2 + \sin 2\theta}$ | **A1** | AG
---
## Question 6(b):
[Polar graph with curve in correct domain] | **B1** | Polar graph with curve in correct domain. Graph needs gradient $\leq 0$ and $r > 0$ for all $\theta$ in domain
[Concave graph with $r$ at $\theta = \frac{\pi}{4}$ greater than half $r$ at $\theta = 0$] | **B1** | $r$ strictly decreasing such that $r$ at $\theta = \frac{\pi}{4}$ is greater than half $r$ at $\theta = 0$. Concave graph. Gradient at $\theta = 0$ and $\frac{\pi}{4}$ must not be vertical or horizontal respectively
---
## Question 6(c):
$R = \frac{1}{2}\int_0^{\frac{1}{4}\pi} r^2\, d\theta = a\int_0^{\frac{1}{4}\pi} \frac{1+\tan^2\theta}{2(1+\tan^2\theta)+2\tan\theta}\, d\theta$ | **B1** | AG
$t = \tan\theta$ leading to $\frac{dt}{d\theta} = \sec^2\theta$ | **M1** | Applies given substitution
$\frac{dt}{d\theta} = \sec^2\theta = 1 + t^2$ | **M1** | Applies $\sec^2\theta = 1 + \tan^2\theta$
$R = \frac{1}{2}a\int_0^1 \frac{1}{t^2+t+1}\, dt$ | **A1** |
$t^2 + t + 1 = (t+\frac{1}{2})^2 + \frac{3}{4}$ | **B1** | Completes the square
$R = \frac{1}{2}a\int_0^1 \frac{1}{(t+\frac{1}{2})^2 + \frac{3}{4}}\, dt = \frac{1}{\sqrt{3}}a\left[\tan^{-1}\left(\frac{2}{\sqrt{3}}t + \frac{1}{\sqrt{3}}\right)\right]_0^1$ | **M1 A1** | Applies formula
$\frac{\pi a}{6\sqrt{3}}$ | **A1** |
---6 The curve $C$ has Cartesian equation $x ^ { 2 } + x y + y ^ { 2 } = a$, where $a$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Show that the polar equation of $C$ is $r ^ { 2 } = \frac { 2 a } { 2 + \sin 2 \theta }$.
\item Sketch the part of $C$ for $0 \leqslant \theta \leqslant \frac { 1 } { 4 } \pi$.
The region $R$ is enclosed by this part of $C$, the initial line and the half-line $\theta = \frac { 1 } { 4 } \pi$.
\item It is given that $\sin 2 \theta$ may be expressed as $\frac { 2 \tan \theta } { 1 + \tan ^ { 2 } \theta }$. Use this result to show that the area of $R$ is
$$\frac { 1 } { 2 } a \int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \frac { 1 + \tan ^ { 2 } \theta } { 1 + \tan \theta + \tan ^ { 2 } \theta } \mathrm {~d} \theta$$
and use the substitution $t = \tan \theta$ to find the exact value of this area.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2022 Q6 [13]}}