| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| 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 fixed point iteration question requiring routine verification of a root's interval (sign change), showing convergence to the root (substitution), and applying the iteration. All steps are algorithmic with no novel insight required, making it slightly easier than average. |
| Spec | 1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Calculate the value of a relevant expression or values of a pair of expressions at \(x = 0.7\) and \(x = 0.8\) | M1 | Allow if working with a smaller interval, e.g. (0.72, 0.78). Need all relevant values but condone one error. Pairings must be clear for solutions involving four values. M0 if working in degrees e.g. \(-1.94..., 1.04...\) |
| Complete the argument correctly with correct calculated values (can use the equation in the rubric or equation in (b) or equivalent) | A1 | E.g. \(4.95... > 4.26...\) and \(4.06... < 4.49...\); \(-0.439... < 0, 0.690... > 0\); \(1.4 < 1.5029..., 1.6 > 1.449...\); \(1.1 > 1, 0.86 < 1\); \(0.0515 > 0, -0.075 < 0\). Allow values rounded or truncated to 2sf. |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State \(2x = \ln(5 + \cos 3x)\) and take exponential of both sides to obtain \(e^{2x} = 5 + \cos 3x\) | B1 | Given answer requires fully correct working or work vice versa. If working in reverse, must get to the iterative formula, including subscripts. |
| 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use the iterative process correctly at least once | M1 | M0 if working in degrees (e.g. values heading for \(0.89...\)). |
| Obtain final answer \(0.740\) | A1 | |
| Show sufficient iterations to at least 5dp to justify \(0.740\) to 3dp, or show sign change in interval \((0.7395, 0.7405)\) | A1 | E.g. \(0.7, 0.75150, 0.73719, 0.74105, 0.74000, 0.74028\); \(0.75, 0.73759, 0.74094, 0.74003, 0.74028\); \(0.8, 0.72494, 0.74443, 0.73909, 0.74053, 0.74014, 0.74025\). Allow recovery. Allow truncation or rounding and condone small differences in final decimal place. |
| 3 |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Calculate the value of a relevant expression or values of a pair of expressions at $x = 0.7$ and $x = 0.8$ | M1 | Allow if working with a smaller interval, e.g. (0.72, 0.78). Need all relevant values but condone one error. Pairings must be clear for solutions involving four values. M0 if working in degrees e.g. $-1.94..., 1.04...$ |
| Complete the argument correctly with correct calculated values (can use the equation in the rubric or equation in (b) or equivalent) | A1 | E.g. $4.95... > 4.26...$ and $4.06... < 4.49...$; $-0.439... < 0, 0.690... > 0$; $1.4 < 1.5029..., 1.6 > 1.449...$; $1.1 > 1, 0.86 < 1$; $0.0515 > 0, -0.075 < 0$. Allow values rounded or truncated to 2sf. |
| | **2** | |
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State $2x = \ln(5 + \cos 3x)$ and take exponential of both sides to obtain $e^{2x} = 5 + \cos 3x$ | B1 | Given answer requires fully correct working or work vice versa. If working in reverse, must get to the iterative formula, including subscripts. |
| | **1** | |
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use the iterative process correctly at least once | M1 | M0 if working in degrees (e.g. values heading for $0.89...$). |
| Obtain final answer $0.740$ | A1 | |
| Show sufficient iterations to at least 5dp to justify $0.740$ to 3dp, or show sign change in interval $(0.7395, 0.7405)$ | A1 | E.g. $0.7, 0.75150, 0.73719, 0.74105, 0.74000, 0.74028$; $0.75, 0.73759, 0.74094, 0.74003, 0.74028$; $0.8, 0.72494, 0.74443, 0.73909, 0.74053, 0.74014, 0.74025$. Allow recovery. Allow truncation or rounding and condone small differences in final decimal place. |
| | **3** | |5
\begin{enumerate}[label=(\alph*)]
\item It is given that the equation $\mathrm { e } ^ { 2 x } = 5 + \cos 3 x$ has only one root.\\
Show by calculation that this root lies in the interval $0.7 < x < 0.8$.
\item Show that if a sequence of values in the interval $0.7 < x < 0.8$ given by the iterative formula
$$x _ { n + 1 } = \frac { 1 } { 2 } \ln \left( 5 + \cos 3 x _ { n } \right)$$
converges then it converges to the root of the equation in part (a).
\item Use this iterative formula to determine the root correct to 3 decimal places. Give the result of each iteration to 5 decimal places.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2024 Q5 [6]}}