| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2013 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Derive equation from area/geometry |
| Difficulty | Challenging +1.2 This is a multi-step problem requiring geometric area calculation, trigonometric manipulation to derive an equation, and then applying a given iterative formula. Part (i) involves setting up areas of sectors and segments with careful geometric reasoning (5 marks suggests moderate complexity), while part (ii) is straightforward calculator work following a provided formula. The geometric setup and algebraic manipulation elevate this above routine questions, but the iterative method is given rather than requiring students to devise it themselves. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.05l Double angle formulae: and compound angle formulae1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| (i) State or imply \(AB = 2r\cos\theta\) or \(AB^{2} = 2r^{2} - 2r^{2}\cos(\pi - 2\theta)\) | B1 | |
| Use correct formula to express the area of sector \(ABC\) in terms of \(r\) and \(\theta\) | M1 | |
| Use correct area formulae to express the area of a segment in terms of \(r\) and \(\theta\) | M1 | |
| State a correct equation in \(r\) and \(\theta\) in any form | A1 | |
| Obtain the given answer | A1 | [5] |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) Use the iterative formula correctly at least once | M1 | |
| Obtain final answer \(0.95\) | A1 | |
| Show sufficient iterations to 4 d.p. to justify \(0.95\) to 2 d.p., or show there is a sign change in the interval \((0.945, 0.955)\) | A1 | [3] |
**(i)** State or imply $AB = 2r\cos\theta$ or $AB^{2} = 2r^{2} - 2r^{2}\cos(\pi - 2\theta)$ | B1 |
Use correct formula to express the area of sector $ABC$ in terms of $r$ and $\theta$ | M1 |
Use correct area formulae to express the area of a segment in terms of $r$ and $\theta$ | M1 |
State a correct equation in $r$ and $\theta$ in any form | A1 |
Obtain the given answer | A1 | [5]
[SR: If the complete equation is approached by adding two sectors to the shaded area above $BO$ and $OC$ give the first M1 as on the scheme, and the second M1 for using correct area formulae for a triangle $AOB$ or $AOC$, and a sector $AOB$ or $AOC$.]
**(ii)** Use the iterative formula correctly at least once | M1 |
Obtain final answer $0.95$ | A1 |
Show sufficient iterations to 4 d.p. to justify $0.95$ to 2 d.p., or show there is a sign change in the interval $(0.945, 0.955)$ | A1 | [3]\includegraphics{figure_6}
In the diagram, $A$ is a point on the circumference of a circle with centre $O$ and radius $r$. A circular arc with centre $A$ meets the circumference at $B$ and $C$. The angle $OAB$ is $\theta$ radians. The shaded region is bounded by the circumference of the circle and the arc with centre $A$ joining $B$ and $C$. The area of the shaded region is equal to half the area of the circle.
\begin{enumerate}[label=(\roman*)]
\item Show that $\cos 2\theta = \frac{2\sin 2\theta - \pi}{4\theta}$. [5]
\item Use the iterative formula
$$\theta_{n+1} = \frac{1}{2}\cos^{-1}\left(\frac{2\sin 2\theta_n - \pi}{4\theta_n}\right),$$
with initial value $\theta_1 = 1$, to determine $\theta$ correct to 2 decimal places, showing the result of each iteration to 4 decimal places. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2013 Q6 [8]}}