| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2022 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Derive equation from area/geometry |
| Difficulty | Challenging +1.2 This question requires deriving a geometric relationship involving sector and triangle areas, then applying fixed point iteration. Part (a) involves standard circle geometry and area formulas (sector minus triangle), which is routine for P3. Parts (b) and (c) are mechanical calculations. The multi-step nature and combination of geometry with numerical methods elevates it slightly above average, but the techniques are all standard A-level procedures without requiring novel 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 |
|---|---|---|
| Answer | Marks | Guidance |
| State or imply angle \(AOC = \pi - 2\theta\) | B1 | Might be seen on the printed diagram |
| Use correct formulae for the area of a sector and triangle, or of a segment, and find the area of the shaded region | M1 | \(\frac{1}{2}r^2(\pi - 2\theta) - \frac{1}{2}r^2\sin(\pi - 2\theta)\) or \(\frac{1}{2}\pi r^2 - \left[\frac{1}{2}r^2(2\theta) + \frac{1}{2}r^2\sin(\pi - 2\theta)\right]\). M0 if subtraction the wrong way round |
| Equate to \(\frac{1}{6}\pi r^2\) and obtain a correct equation in any form | A1 | e.g. \(\frac{1}{6}\pi r^2 = \frac{1}{2}r^2(\pi - 2\theta) - \frac{1}{2}r^2\sin(\pi - 2\theta)\) |
| Obtain \(\theta = \frac{1}{3}(\pi - 1.5\sin 2\theta)\) correctly | A1 | AG. Condone if state/imply \(\sin(\pi - 2\theta) = \sin 2\theta\) |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Evaluate a relevant expression or pair of expressions at \(\theta = 0.5\) and \(\theta = 0.7\) | M1 | Allow work on a smaller interval. Need to evaluate for both limits, with at least one correct. When using \(x = f(x)\) embedded values are not sufficient e.g. \(f(0.5)\ldots\) is accepted but \(\frac{1}{3}(\pi - 1.5\sin 2\times 0.5) = \ldots\) is not |
| Complete the argument correctly with correct calculated values | A1 | e.g. \(0.5 < 0.626\), \(0.7 > 0.554\) or \(0.126 > 0\), \(-0.146 < 0\). If using pairs then the pairing must be clear. Need to see the inequalities or an appropriate comment. Need to see values calculated to at least 2 s.f. |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use the iterative process \(\theta_{n+1} = \frac{1}{3}(\pi - 1.5\sin 2\theta_n)\) correctly at least once | M1 | i.e. obtain one value and use that value to obtain a second value. Must be working in radians |
| Obtain final answer 0.586 | A1 | |
| Show sufficient iterations to 5 d.p. to justify 0.586 to 3 d.p., or show there is a sign change in the interval (0.5855, 0.5865) | A1 | e.g. \(0.5, 0.62646, 0.57225, 0.59195, 0.58416, 0.58715, 0.58599, 0.58644\) or \(0.6, 0.58118, 0.58833, 0.58553, 0.58661, 0.58619, 0.58636\). Allow working to more than 5 dp, but not less |
| 3 |
## Question 9(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply angle $AOC = \pi - 2\theta$ | **B1** | Might be seen on the printed diagram |
| Use correct formulae for the area of a sector and triangle, or of a segment, and find the area of the shaded region | **M1** | $\frac{1}{2}r^2(\pi - 2\theta) - \frac{1}{2}r^2\sin(\pi - 2\theta)$ or $\frac{1}{2}\pi r^2 - \left[\frac{1}{2}r^2(2\theta) + \frac{1}{2}r^2\sin(\pi - 2\theta)\right]$. M0 if subtraction the wrong way round |
| Equate to $\frac{1}{6}\pi r^2$ and obtain a correct equation in any form | **A1** | e.g. $\frac{1}{6}\pi r^2 = \frac{1}{2}r^2(\pi - 2\theta) - \frac{1}{2}r^2\sin(\pi - 2\theta)$ |
| Obtain $\theta = \frac{1}{3}(\pi - 1.5\sin 2\theta)$ correctly | **A1** | AG. Condone if state/imply $\sin(\pi - 2\theta) = \sin 2\theta$ |
| | **4** | |
## Question 9(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Evaluate a relevant expression or pair of expressions at $\theta = 0.5$ and $\theta = 0.7$ | **M1** | Allow work on a smaller interval. Need to evaluate for both limits, with at least one correct. When using $x = f(x)$ embedded values are not sufficient e.g. $f(0.5)\ldots$ is accepted but $\frac{1}{3}(\pi - 1.5\sin 2\times 0.5) = \ldots$ is not |
| Complete the argument correctly with correct calculated values | **A1** | e.g. $0.5 < 0.626$, $0.7 > 0.554$ or $0.126 > 0$, $-0.146 < 0$. If using pairs then the pairing must be clear. Need to see the inequalities or an appropriate comment. Need to see values calculated to at least 2 s.f. |
| | **2** | |
## Question 9(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use the iterative process $\theta_{n+1} = \frac{1}{3}(\pi - 1.5\sin 2\theta_n)$ correctly at least once | **M1** | i.e. obtain one value and use that value to obtain a second value. Must be working in radians |
| Obtain final answer 0.586 | **A1** | |
| Show sufficient iterations to 5 d.p. to justify 0.586 to 3 d.p., or show there is a sign change in the interval (0.5855, 0.5865) | **A1** | e.g. $0.5, 0.62646, 0.57225, 0.59195, 0.58416, 0.58715, 0.58599, 0.58644$ or $0.6, 0.58118, 0.58833, 0.58553, 0.58661, 0.58619, 0.58636$. Allow working to more than 5 dp, but not less |
| | **3** | |9\\
\includegraphics[max width=\textwidth, alt={}, center]{3c63c42a-2658-4984-93e8-b2a8d18eb37a-14_407_734_267_699}
The diagram shows a semicircle with diameter $A B$, centre $O$ and radius $r$. The shaded region is the minor segment on the chord $A C$ and its area is one third of the area of the semicircle. The angle $C A B$ is $\theta$ radians.
\begin{enumerate}[label=(\alph*)]
\item Show that $\theta = \frac { 1 } { 3 } ( \pi - 1.5 \sin 2 \theta )$.
\item Verify by calculation that $0.5 < \theta < 0.7$.
\item Use an iterative formula based on the equation in part (a) to determine $\theta$ correct to 3 decimal places. Give the result of each iteration to 5 decimal places.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2022 Q9 [9]}}