| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Show convergence to specific root |
| Difficulty | Standard +0.3 This is a straightforward fixed-point iteration question requiring students to (a) algebraically verify convergence to a root by substituting the fixed point into the original equation, and (b) perform repeated iterations with a calculator. Both parts are routine applications of standard techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams1.09e Iterative method failure: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State or imply \(x = \sqrt{\dfrac{4}{5-2x}}\) and square the equation | *B1 | Could work with \(x_{n+1}\) throughout or with \(x_n\) throughout instead of \(x\) |
| Rearrange with at least one intermediate step to \(2x^3 - 5x^2 + 4 = 0\) | DB1 | |
| Alternative Method 1: Rearrange \(2x^3 - 5x^2 + 4 = 0\) to \(x^2(5-2x) = 4\), obtain \(x^2 = \dfrac{4}{5-2x}\) or \(x = \sqrt{\dfrac{4}{5-2x}}\), then obtain \(x_{n+1} = \sqrt{\dfrac{4}{5-2x_n}}\) | B2 | |
| Alternative Method 2: Rearrange to \(x^2(5-2x)=4\), to \(x^2 = \dfrac{4}{5-2x}\) and \(x_{n+1}^2 = \dfrac{4}{5-2x_n}\) | *B1 | Must have introduced \(x_{n+1}\) and \(x_n\) |
| Obtain \(x_{n+1} = \sqrt{\dfrac{4}{5-2x_n}}\) | DB1 | |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use the iterative process correctly at least once | M1 | Initial value 1.2 specified; must use formula to obtain a value then use that value in the formula |
| Obtain final answer 1.28 | A1 | Can gain this mark even if less than 4 dp shown in iteration |
| Show sufficient iterations to at least 4 dp to justify 1.28 to 2 dp, or show sign change in \((1.275, 1.285)\) | A1 | 1.2, 1.2403, 1.2601, 1.2700, 1.2752, 1.2778,... Allow small errors, truncation and recovery |
| Total | 3 |
## Question 2(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply $x = \sqrt{\dfrac{4}{5-2x}}$ and square the equation | *B1 | Could work with $x_{n+1}$ throughout or with $x_n$ throughout instead of $x$ |
| Rearrange with at least one intermediate step to $2x^3 - 5x^2 + 4 = 0$ | DB1 | |
| **Alternative Method 1:** Rearrange $2x^3 - 5x^2 + 4 = 0$ to $x^2(5-2x) = 4$, obtain $x^2 = \dfrac{4}{5-2x}$ or $x = \sqrt{\dfrac{4}{5-2x}}$, then obtain $x_{n+1} = \sqrt{\dfrac{4}{5-2x_n}}$ | B2 | |
| **Alternative Method 2:** Rearrange to $x^2(5-2x)=4$, to $x^2 = \dfrac{4}{5-2x}$ and $x_{n+1}^2 = \dfrac{4}{5-2x_n}$ | *B1 | Must have introduced $x_{n+1}$ and $x_n$ |
| Obtain $x_{n+1} = \sqrt{\dfrac{4}{5-2x_n}}$ | DB1 | |
| **Total** | **2** | |
---
## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use the iterative process correctly at least once | M1 | Initial value 1.2 specified; must use formula to obtain a value then use that value in the formula |
| Obtain final answer 1.28 | A1 | Can gain this mark even if less than 4 dp shown in iteration |
| Show sufficient iterations to at least 4 dp to justify 1.28 to 2 dp, or show sign change in $(1.275, 1.285)$ | A1 | 1.2, 1.2403, 1.2601, 1.2700, 1.2752, 1.2778,... Allow small errors, truncation and recovery |
| **Total** | **3** | |
---2 Let $\mathrm { f } ( x ) = 2 x ^ { 3 } - 5 x ^ { 2 } + 4$.
\begin{enumerate}[label=(\alph*)]
\item Show that if a sequence of values given by the iterative formula
$$x _ { n + 1 } = \sqrt { \frac { 4 } { 5 - 2 x _ { n } } }$$
converges, then it converges to a root of the equation $\mathrm { f } ( x ) = 0$.
\item The equation has a root close to 1.2 .
Use the iterative formula from part (a) and an initial value of 1.2 to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2024 Q2 [5]}}