| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2005 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Find equation satisfied by limit |
| Difficulty | Moderate -0.3 This is a straightforward fixed point iteration question requiring routine application of the formula (part i) and simple algebraic manipulation to find the limit equation by setting x_{n+1} = x_n = α (part ii). The equation 4α⁴ - 3α⁴ = 8 is easily derived and solved. While it requires multiple iterations and some algebraic manipulation, it follows a standard template with no novel insight needed. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks |
|---|---|
| Use the given iterative formula correctly at least once | M1 |
| Obtain final answer \(\alpha = 1.68\) | A1 |
| Show sufficient iterations to justify the answer to 2 dp | B1 |
| Answer | Marks |
|---|---|
| State equation, e.g. \(x = \frac{3}{4} + \frac{2}{x^3}\), in any correct form | B1 |
| Derive the exact answer \(\alpha\) (or \(x\)) \(= \sqrt[4]{8}\), or equivalent | B1 |
**(i)**
| Use the given iterative formula correctly at least once | M1 |
| Obtain final answer $\alpha = 1.68$ | A1 |
| Show sufficient iterations to justify the answer to 2 dp | B1 |
**Total: 3 marks**
**(ii)**
| State equation, e.g. $x = \frac{3}{4} + \frac{2}{x^3}$, in any correct form | B1 |
| Derive the exact answer $\alpha$ (or $x$) $= \sqrt[4]{8}$, or equivalent | B1 |
**Total: 2 marks**
---3 The sequence of values given by the iterative formula
$$x _ { n + 1 } = \frac { 3 x _ { n } } { 4 } + \frac { 2 } { x _ { n } ^ { 3 } }$$
with initial value $x _ { 1 } = 2$, converges to $\alpha$.\\
(i) Use this iteration to calculate $\alpha$ correct to 2 decimal places, showing the result of each iteration to 4 decimal places.\\
(ii) State an equation which is satisfied by $\alpha$ and hence find the exact value of $\alpha$.
\hfill \mbox{\textit{CAIE P2 2005 Q3 [5]}}