| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2006 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Derive equation from area/geometry |
| Difficulty | Standard +0.8 This question combines circle geometry (deriving segment area formula), algebraic manipulation to reach the fixed point equation, interval verification, and iterative numerical methods. While each component is standard P2 material, the multi-step derivation requiring sector minus triangle area, followed by systematic iteration, makes this moderately challenging—above average but not requiring exceptional insight. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Obtain area of shaded segment in terms of \(r\) and \(\alpha\), e.g. \(\frac{1}{2}r^2 \alpha - \frac{1}{2}r^2 \sin \alpha\) | B1 | |
| Equate area of shaded segment to \(\frac{1}{4}\pi r^2\), or equivalent | M1 | |
| Obtain given answer correctly | A1 | 3 marks |
| (ii) Consider sign of \(x - \sin x - \frac{1}{3}\pi\) at \(x = \frac{1}{3}\pi\) and \(x = \frac{2}{3}\pi\), or equivalent | M1 | |
| Complete the argument correctly with appropriate calculations | A1 | 2 marks |
| (iii) Use the iterative formula correctly at least once | M1 | |
| Obtain final answer \(1.97\) | A1 | |
| Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval \((1.965, 1.975)\) | B1 | 3 marks |
(i) Obtain area of shaded segment in terms of $r$ and $\alpha$, e.g. $\frac{1}{2}r^2 \alpha - \frac{1}{2}r^2 \sin \alpha$ | B1 |
Equate area of shaded segment to $\frac{1}{4}\pi r^2$, or equivalent | M1 |
Obtain given answer correctly | A1 | 3 marks |
(ii) Consider sign of $x - \sin x - \frac{1}{3}\pi$ at $x = \frac{1}{3}\pi$ and $x = \frac{2}{3}\pi$, or equivalent | M1 |
Complete the argument correctly with appropriate calculations | A1 | 2 marks |
(iii) Use the iterative formula correctly at least once | M1 |
Obtain final answer $1.97$ | A1 |
Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval $(1.965, 1.975)$ | B1 | 3 marks |5\\
\includegraphics[max width=\textwidth, alt={}, center]{4029c46c-50a1-4d23-bc29-589417a6b7f5-2_396_392_1603_879}
The diagram shows a chord joining two points, $A$ and $B$, on the circumference of a circle with centre $O$ and radius $r$. The angle $A O B$ is $\alpha$ radians, where $0 < \alpha < \pi$. The area of the shaded segment is one sixth of the area of the circle.\\
(i) Show that $\alpha$ satisfies the equation
$$x = \frac { 1 } { 3 } \pi + \sin x .$$
(ii) Verify by calculation that $\alpha$ lies between $\frac { 1 } { 2 } \pi$ and $\frac { 2 } { 3 } \pi$.\\
(iii) Use the iterative formula
$$x _ { n + 1 } = \frac { 1 } { 3 } \pi + \sin x _ { n } ,$$
with initial value $x _ { 1 } = 2$, to determine $\alpha$ correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
\hfill \mbox{\textit{CAIE P2 2006 Q5 [8]}}