| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2011 |
| Session | June |
| 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 requires deriving a transcendental equation from geometric area relationships (sector minus triangle), then applying fixed point iteration. The area ratio setup demands careful reasoning about which region is larger, and the iterative method requires understanding convergence. While the individual components are A-level standard, combining geometric insight with numerical methods and managing the multi-step derivation makes this moderately challenging. |
| Spec | 1.03f Circle properties: angles, chords, tangents1.05d 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| State or imply area of segment is \(\frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta\) or \(50\theta - 50\sin\theta\) | B1 | |
| Attempt to form equation from area of segment \(= \frac{1}{5}\) of area of circle, or equivalent | M1 | |
| Confirm given result \(\theta = \frac{2}{5}\pi + \sin\theta\) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use iterative formula correctly at least once | M1 | |
| Obtain value for \(\theta\) of \(2.11\) | A1 | |
| Show sufficient iterations to justify value of \(\theta\) or show sign change in interval \((2.105, 2.115)\) | A1 | |
| Use correct trigonometry to find an expression for the length of \(AB\) | M1 | |
| e.g. \(20\sin 1.055\) or \(\sqrt{200 - 200\cos 2.11}\) | ||
| Hence \(17.4\) | A1 | [5] |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| State or imply area of segment is $\frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta$ or $50\theta - 50\sin\theta$ | B1 | |
| Attempt to form equation from area of segment $= \frac{1}{5}$ of area of circle, or equivalent | M1 | |
| Confirm given result $\theta = \frac{2}{5}\pi + \sin\theta$ | A1 | [3] |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use iterative formula correctly at least once | M1 | |
| Obtain value for $\theta$ of $2.11$ | A1 | |
| Show sufficient iterations to justify value of $\theta$ or show sign change in interval $(2.105, 2.115)$ | A1 | |
| Use correct trigonometry to find an expression for the length of $AB$ | M1 | |
| e.g. $20\sin 1.055$ or $\sqrt{200 - 200\cos 2.11}$ | | |
| Hence $17.4$ | A1 | [5] |
**Note:** $[2.1 \to 2.1198 \to 2.1097 \to 2.1149 \to 2.1122]$
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{cc85b13a-7f15-4025-a545-373cda454de8-3_456_495_255_824}
The diagram shows a circle with centre $O$ and radius 10 cm . The chord $A B$ divides the circle into two regions whose areas are in the ratio $1 : 4$ and it is required to find the length of $A B$. The angle $A O B$ is $\theta$ radians.\\
(i) Show that $\theta = \frac { 2 } { 5 } \pi + \sin \theta$.\\
(ii) Showing all your working, use an iterative formula, based on the equation in part (i), with an initial value of 2.1 , to find $\theta$ correct to 2 decimal places. Hence find the length of $A B$ in centimetres correct to 1 decimal place.
\hfill \mbox{\textit{CAIE P3 2011 Q6 [8]}}