| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2019 |
| Session | March |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Find equation satisfied by limit |
| Difficulty | Standard +0.3 This is a straightforward fixed point iteration question requiring routine application of the iterative formula (calculator work) and simple algebraic manipulation to find the equation satisfied by the limit. Part (i) is pure computation, and part (ii) requires only setting x_{n+1} = x_n = α and rearranging—a standard technique taught explicitly in P3. No novel insight or complex problem-solving needed. |
| Spec | 1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use the iterative formula correctly at least once | M1 | |
| Obtain answer \(1.3195\) | A1 | |
| Show sufficient iterations to 6 d.p. to justify \(1.3195\) to 4 d.p., or show there is a sign change in \((1.31945, 1.31955)\) | A1 | |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State \(x = \dfrac{2x^6 + 12x}{3x^5 + 8}\), or equivalent | B1 | |
| State answer \(\sqrt[3]{4}\), or exact equivalent | B1 | |
| Total | 2 |
**Question 2:**
**Part (i):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use the iterative formula correctly at least once | M1 | |
| Obtain answer $1.3195$ | A1 | |
| Show sufficient iterations to 6 d.p. to justify $1.3195$ to 4 d.p., or show there is a sign change in $(1.31945, 1.31955)$ | A1 | |
| **Total** | **3** | |
**Part (ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| State $x = \dfrac{2x^6 + 12x}{3x^5 + 8}$, or equivalent | B1 | |
| State answer $\sqrt[3]{4}$, or exact equivalent | B1 | |
| **Total** | **2** | |2 The sequence of values given by the iterative formula
$$x _ { n + 1 } = \frac { 2 x _ { n } ^ { 6 } + 12 x _ { n } } { 3 x _ { n } ^ { 5 } + 8 }$$
with initial value $x _ { 1 } = 2$, converges to $\alpha$.\\
(i) Use the formula to calculate $\alpha$ correct to 4 decimal places. Give the result of each iteration to 6 decimal places.\\
(ii) State an equation satisfied by $\alpha$ and hence find the exact value of $\alpha$.\\
\hfill \mbox{\textit{CAIE P3 2019 Q2 [5]}}