| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2009 |
| Session | November |
| Marks | 12 |
| 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 structured multi-part polar coordinates question with clear guidance at each step. Part (i) is a standard polar area integral, part (ii) requires differentiating y = r sin θ (a routine optimization setup), part (iii) provides the tan 3θ formula and just requires algebraic manipulation, and part (iv) is a straightforward sketch. While it covers several techniques, the heavy scaffolding and provided formulas make it more accessible than typical Further Maths questions, placing it slightly above average difficulty overall. |
| Spec | 4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks |
|---|---|
| (i) Area \(= (a^2/2)\int_0^{\pi/3} \sin^2 3\theta \, d\theta\) | M1 |
| \(= (a^2/4)\int_0^{\pi/3} (1 - \cos 6\theta) d\theta\) | A1 |
| \(= ... = \pi a^2/12\) (AG) | A1 |
| (ii) Considers \(y = a\sin 3\theta \sin \theta\) | B1 |
| \(dy/d\theta = 0 \Rightarrow 3\cos 3\theta \sin \theta + \sin 3\theta \cos \theta = 0\) | M1 |
| \(\Rightarrow ... \Rightarrow \tan 3\theta + 3\tan \theta = 0\) (AG) | A1 |
| (iii) Uses \(\tan 3\theta = (3\tan\theta - \tan^3\theta)/(1 - 3\tan^2\theta)\) to obtain \(\tan^2\theta = 3/5\) | M1A1 |
| Uses \(y = \sin 3\theta \sin \theta\) with \(\tan \theta = \sqrt{3/5}\) | M1 |
| Obtains \(y = 9a/16\) (Accept 0.5625a, or 0.563a) | A1 |
| (iv) Closed loop entirely in the first quadrant with lower end at the pole and clearly tangential to the initial line at the pole | B1 |
| Symmetric about line \(\theta = \pi/6\) with correct shape at \((a, \pi/6)\) | B1 |
(i) Area $= (a^2/2)\int_0^{\pi/3} \sin^2 3\theta \, d\theta$ | M1 |
$= (a^2/4)\int_0^{\pi/3} (1 - \cos 6\theta) d\theta$ | A1 |
$= ... = \pi a^2/12$ (AG) | A1 |
(ii) Considers $y = a\sin 3\theta \sin \theta$ | B1 |
$dy/d\theta = 0 \Rightarrow 3\cos 3\theta \sin \theta + \sin 3\theta \cos \theta = 0$ | M1 |
$\Rightarrow ... \Rightarrow \tan 3\theta + 3\tan \theta = 0$ (AG) | A1 |
(iii) Uses $\tan 3\theta = (3\tan\theta - \tan^3\theta)/(1 - 3\tan^2\theta)$ to obtain $\tan^2\theta = 3/5$ | M1A1 |
Uses $y = \sin 3\theta \sin \theta$ with $\tan \theta = \sqrt{3/5}$ | M1 |
Obtains $y = 9a/16$ (Accept 0.5625a, or 0.563a) | A1 |
(iv) Closed loop entirely in the first quadrant with lower end at the pole and clearly tangential to the initial line at the pole | B1 |
Symmetric about line $\theta = \pi/6$ with correct shape at $(a, \pi/6)$ | B1 |10 The curve $C$ has polar equation
$$r = a \sin 3 \theta$$
where $0 \leqslant \theta \leqslant \frac { 1 } { 3 } \pi$.\\
(i) Show that the area of the region enclosed by $C$ is $\frac { 1 } { 12 } \pi a ^ { 2 }$.\\
(ii) Show that, at the point of $C$ at maximum distance from the initial line,
$$\tan 3 \theta + 3 \tan \theta = 0 .$$
(iii) Use the formula
$$\tan 3 \theta = \frac { 3 \tan \theta - \tan ^ { 3 } \theta } { 1 - 3 \tan ^ { 2 } \theta }$$
to find this maximum distance.\\
(iv) Draw a sketch of $C$.
\hfill \mbox{\textit{CAIE FP1 2009 Q10 [12]}}