| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Centre of mass of composite shapes |
| Difficulty | Standard +0.3 This is a straightforward composite shapes problem requiring standard formulas for sector arc length, perimeter, and area. Part (iii) involves setting up and solving a simple equation. While it requires careful bookkeeping of which sectors are major/minor, it's mostly routine application of memorized formulas with minimal problem-solving insight needed. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(r(2\pi - \alpha) + 2r\alpha + 2r\) | B1B1 | |
| \(2\pi r + r\alpha + 2r\) | B1 | ft for \(r\alpha\) instead of \(2r\alpha\) or omission 2r. SC1 for \(2r\alpha + 4r\). (Plate = shaded part) |
| (ii) \(\frac{1}{2}(2r)^2\alpha + \pi^2 - \frac{1}{2}r^2\alpha\) | B1B1 | Either B1 can be scored in (iii) |
| \(\frac{3r^2\alpha}{2} + \pi^2\) | B1 | [3] |
| (iii) \(\pi^2 - \frac{1}{2}r^2\alpha = 2r^2\alpha\) | M1 | For equating their 2 parts from (ii) |
| \(\alpha = \frac{2}{5}\pi\) | A1 | [2] |
(i) $r(2\pi - \alpha) + 2r\alpha + 2r$ | B1B1 |
$2\pi r + r\alpha + 2r$ | B1 | ft for $r\alpha$ instead of $2r\alpha$ or omission 2r. SC1 for $2r\alpha + 4r$. (Plate = shaded part) | [3]
(ii) $\frac{1}{2}(2r)^2\alpha + \pi^2 - \frac{1}{2}r^2\alpha$ | B1B1 | Either B1 can be scored in (iii)
$\frac{3r^2\alpha}{2} + \pi^2$ | B1 | [3]
(iii) $\pi^2 - \frac{1}{2}r^2\alpha = 2r^2\alpha$ | M1 | For equating their 2 parts from (ii)
$\alpha = \frac{2}{5}\pi$ | A1 | [2]6\\
\includegraphics[max width=\textwidth, alt={}, center]{02da6b6a-6db1-4bc3-ad4e-537e4f61dcac-3_412_629_258_758}
The diagram shows a metal plate made by fixing together two pieces, $O A B C D$ (shaded) and $O A E D$ (unshaded). The piece $O A B C D$ is a minor sector of a circle with centre $O$ and radius $2 r$. The piece $O A E D$ is a major sector of a circle with centre $O$ and radius $r$. Angle $A O D$ is $\alpha$ radians. Simplifying your answers where possible, find, in terms of $\alpha , \pi$ and $r$,\\
(i) the perimeter of the metal plate,\\
(ii) the area of the metal plate.
It is now given that the shaded and unshaded pieces are equal in area.\\
(iii) Find $\alpha$ in terms of $\pi$.
\hfill \mbox{\textit{CAIE P1 2013 Q6 [8]}}