| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Show convergence to specific root |
| Difficulty | Standard +0.3 This is a standard A-level fixed point iteration question with routine steps: deriving an equation from a geometric setup, verifying a root interval by substitution, showing convergence to the correct root (algebraic rearrangement), and applying the iteration formula. While it involves multiple parts and some algebraic manipulation, each step follows predictable A-level techniques without requiring novel insight or particularly challenging problem-solving. |
| Spec | 1.05c Area of triangle: using 1/2 ab sin(C)1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.09b Sign change methods: understand failure cases1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State or imply \(CD = 2r - 2r\cos x\) | B1 | |
| Using correct formulae for area of sector and trapezium, or equivalent, form an equation in \(r\) and \(x\) | M1 | e.g. \(2\times\frac{1}{2}r^2x = \frac{0.9}{2}(2r+2r-2r\cos x)r\sin x\) |
| Obtain \(x = 0.9(2-\cos x)\sin x\) | A1 | AG, NFWW |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Calculate the values of a relevant expression or pair of expressions at \(x=0.5\) and \(x=0.7\) | M1 | Calculated for both values and correct for one value is sufficient for M1. Must be working in radians. |
| Complete the argument correctly with correct values | A1 | Must have sufficient accuracy to support the answer e.g. \(0.5>0.484\) or \(0.016>0\) or \(0.96...<1\); \(0.7<0.716\) or \(-0.016<0\) or \(1.02...>1\) |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State a suitable equation, e.g. \(\cos x = \left(2-\frac{x}{0.9\sin x}\right)\) | B1 | If working in reverse, the first B1 is for \(\frac{x}{0.9\sin x} = 2-\cos x\) |
| Rearrange this as \(x = 0.9\sin x(2-\cos x)\) | B1 | Need to see the complete sequence of changes. |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use the iterative process correctly at least once | M1 | Must be working in radians |
| Obtain answer \(0.62\) | A1 | |
| Show sufficient iterations to at least 4 d.p. to justify \(0.62\) to 2 d.p., or show there is a sign change in the interval \((0.615, 0.625)\) | A1 | Allow recovery. A candidate who starts with \(0.5\) and stops at \(0.61\) or starts at \(0.7\) and stops at \(0.63\) can score M1A0A1 if they have worked to the required accuracy. |
| Total | 3 |
## Question 10(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply $CD = 2r - 2r\cos x$ | B1 | |
| Using correct formulae for area of sector and trapezium, or equivalent, form an equation in $r$ and $x$ | M1 | e.g. $2\times\frac{1}{2}r^2x = \frac{0.9}{2}(2r+2r-2r\cos x)r\sin x$ |
| Obtain $x = 0.9(2-\cos x)\sin x$ | A1 | AG, NFWW |
| **Total** | **3** | |
---
## Question 10(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Calculate the values of a relevant expression or pair of expressions at $x=0.5$ and $x=0.7$ | M1 | Calculated for both values and correct for one value is sufficient for M1. Must be working in radians. |
| Complete the argument correctly with correct values | A1 | Must have sufficient accuracy to support the answer e.g. $0.5>0.484$ or $0.016>0$ or $0.96...<1$; $0.7<0.716$ or $-0.016<0$ or $1.02...>1$ |
| **Total** | **2** | |
---
## Question 10(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| State a suitable equation, e.g. $\cos x = \left(2-\frac{x}{0.9\sin x}\right)$ | B1 | If working in reverse, the first B1 is for $\frac{x}{0.9\sin x} = 2-\cos x$ |
| Rearrange this as $x = 0.9\sin x(2-\cos x)$ | B1 | Need to see the complete sequence of changes. |
| **Total** | **2** | |
## Question 10(d):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use the iterative process correctly at least once | M1 | Must be working in radians |
| Obtain answer $0.62$ | A1 | |
| Show sufficient iterations to at least 4 d.p. to justify $0.62$ to 2 d.p., or show there is a sign change in the interval $(0.615, 0.625)$ | A1 | Allow recovery. A candidate who starts with $0.5$ and stops at $0.61$ or starts at $0.7$ and stops at $0.63$ can score M1A0A1 if they have worked to the required accuracy. |
| **Total** | **3** | |
---\begin{enumerate}[label=(\alph*)]
\item Given that the sum of the areas of the shaded sectors is $90 \%$ of the area of the trapezium, show that $x$ satisfies the equation $x = 0.9 ( 2 - \cos x ) \sin x$.
\item Verify by calculation that $x$ lies between 0.5 and 0.7 .
\item Show that if a sequence of values in the interval $0 < x < \frac { 1 } { 2 } \pi$ given by the iterative formula
$$x _ { n + 1 } = \cos ^ { - 1 } \left( 2 - \frac { x _ { n } } { 0.9 \sin x _ { n } } \right)$$
converges, then it converges to the root of the equation in part (a).
\item Use this iterative formula to determine $x$ correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2021 Q10 [10]}}