| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| 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 with routine steps: sketch graphs to show existence, verify interval by substitution, show convergence algebraically (rearranging the equation), perform iterations, and count steps. All techniques are textbook exercises for P3 level 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 | Mark | Guidance |
| Sketch \(y = e^x - 3\), correct shape, correct vertical intercept | B1 | |
| Sketch \(y = \cosec\dfrac{x}{2}\), correct shape, minimum above axis; mark intersection; justify given statement | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Calculate values of a relevant expression or pair of expressions at \(x=1\) and \(x=2\) | M1 | Use of degrees is M0 |
| Complete the argument correctly with correct calculated values | A1 | E.g. \(-0.282 < 2.086\), \(4.389 > 1.188\); \(1 < 1.626\), \(2 > 1.432\); \(2.36 > 0\), \(-3.2 < 0\). At least 2sf. Condone truncation. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State \(x = \ln\!\left(\cosec\dfrac{1}{2}x + 3\right)\) and rearrange to give \(\cosec\dfrac{x}{2} = e^x - 3\) | B1 | AG. Or vice versa and obtain the iterative formula. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use the iterative formula correctly at least twice | M1 | Use of degrees is M0 (might see 1.38…) |
| Obtain final answer \(1.50\) | A1 | |
| Show sufficient iterations to 4 dp to justify \(1.50\) to 4 dp or show sign change in interval \((1.495, 1.505)\) | A1 | \(1.5156, 1.4940, 1.4978, 1.4971\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(4\) | B1 |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Sketch $y = e^x - 3$, correct shape, correct vertical intercept | B1 | |
| Sketch $y = \cosec\dfrac{x}{2}$, correct shape, minimum above axis; mark intersection; justify given statement | B1 | |
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Calculate values of a relevant expression or pair of expressions at $x=1$ and $x=2$ | M1 | Use of degrees is M0 |
| Complete the argument correctly with correct calculated values | A1 | E.g. $-0.282 < 2.086$, $4.389 > 1.188$; $1 < 1.626$, $2 > 1.432$; $2.36 > 0$, $-3.2 < 0$. At least 2sf. Condone truncation. |
## Question 6(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| State $x = \ln\!\left(\cosec\dfrac{1}{2}x + 3\right)$ and rearrange to give $\cosec\dfrac{x}{2} = e^x - 3$ | B1 | AG. Or vice versa and obtain the iterative formula. |
## Question 6(d):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use the iterative formula correctly at least twice | M1 | Use of degrees is M0 (might see 1.38…) |
| Obtain final answer $1.50$ | A1 | |
| Show sufficient iterations to 4 dp to justify $1.50$ to 4 dp or show sign change in interval $(1.495, 1.505)$ | A1 | $1.5156, 1.4940, 1.4978, 1.4971$ |
## Question 6(e):
| Answer | Mark | Guidance |
|--------|------|----------|
| $4$ | B1 | |
---6
\begin{enumerate}[label=(\alph*)]
\item By sketching a suitable pair of graphs, show that the equation $\operatorname { cosec } \frac { 1 } { 2 } x = \mathrm { e } ^ { x } - 3$ has exactly one root, denoted by $\alpha$, in the interval $0 < x < \pi$.
\item Verify by calculation that $\alpha$ lies between 1 and 2 .
\item Show that if a sequence of values in the interval $0 < x < \pi$ given by the iterative formula
$$x _ { n + 1 } = \ln \left( \operatorname { cosec } \frac { 1 } { 2 } x _ { n } + 3 \right)$$
converges, then it converges to $\alpha$.
\item Use this iterative formula with an initial value of 1.4 to determine $\alpha$ correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
\item State the minimum number of calculated iterations needed with this initial value to determine $\alpha$ correct to 2 decimal places.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2024 Q6 [9]}}