| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2008 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Polar curve intersection points |
| Difficulty | Standard +0.3 This is a straightforward Further Maths polar coordinates question requiring standard techniques: solving a quadratic equation for intersection points, sketching polar curves from given equations, and applying the standard polar area formula. While it's a Further Maths topic (inherently harder than single maths), the execution is routine with no novel insights required—just methodical application of learned procedures. |
| 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 |
|---|---|
| (i) \(\theta = 2, r = 4\) | B1B1 |
| Ignore extra values; Accept as written in MS – co-ords not required | |
| (ii) Graphs: correct location, orientation and concavity required | B1B1 |
| Separate diagrams 1/2, B1 Shapes correct; B1 Intersection correct | |
| (iii) \(A_1 = (1/2)\int_0^{\pi}(\theta + 2)^2 d\theta\) (LNR) | M1 |
| \(= ... = 28/3\) | |
| \(A_2 = (1/2)\int_0^{\pi}\theta^4 d\theta = ... = 16/5\) (LR) | A1 |
| M1 for 1 correct integral representation plus A1 if both correct | |
| Area \(= A_1 - A_2 = 92/15\) (6.13) | A1 |
| S.C. -92/15 M1 A0 A1 | |
| Alternative layout: \(\frac{1}{2}\int_0^{\pi}[(\theta + 2)^2 - \theta^4] d\theta\) | M1 (LNR) |
| \(= \frac{1}{2}\left[\frac{\theta^3}{3} + 2\theta^2 + 4\theta - \frac{\theta^5}{5}\right]_0^{\pi}\) | A1 (LR) |
| \(= 92/15\) | A1 |
**(i)** $\theta = 2, r = 4$ | B1B1 |
Ignore extra values; Accept as written in MS – co-ords not required |
**(ii)** Graphs: correct location, orientation and concavity required | B1B1 |
Separate diagrams 1/2, B1 Shapes correct; B1 Intersection correct |
**(iii)** $A_1 = (1/2)\int_0^{\pi}(\theta + 2)^2 d\theta$ (LNR) | M1 |
$= ... = 28/3$ |
$A_2 = (1/2)\int_0^{\pi}\theta^4 d\theta = ... = 16/5$ (LR) | A1 |
M1 for 1 correct integral representation plus A1 if both correct |
Area $= A_1 - A_2 = 92/15$ (6.13) | A1 |
S.C. -92/15 M1 A0 A1 |
**Alternative layout:** $\frac{1}{2}\int_0^{\pi}[(\theta + 2)^2 - \theta^4] d\theta$ | M1 (LNR) |
$= \frac{1}{2}\left[\frac{\theta^3}{3} + 2\theta^2 + 4\theta - \frac{\theta^5}{5}\right]_0^{\pi}$ | A1 (LR) |
$= 92/15$ | A1 |4 The curves $C _ { 1 }$ and $C _ { 2 }$ have polar equations
$$r = \theta + 2 \quad \text { and } \quad r = \theta ^ { 2 }$$
respectively, where $0 \leqslant \theta \leqslant \pi$.\\
(i) Find the polar coordinates of the point of intersection of $C _ { 1 }$ and $C _ { 2 }$.\\
(ii) Sketch $C _ { 1 }$ and $C _ { 2 }$ on the same diagram.\\
(iii) Find the area bounded by $C _ { 1 } , C _ { 2 }$ and the line $\theta = 0$.
\hfill \mbox{\textit{CAIE FP1 2008 Q4 [7]}}