| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2010 |
| Session | November |
| 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 formula (4-5 iterations with calculator work) and recognizing that at convergence x_{n+1} = x_n = α to derive the equation α = 7α/8 + 5/(2α^4), which rearranges algebraically to α^5 = 20. Both parts are standard textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Use the iterative formula correctly at least once | M1 | |
| Obtain final answer 1.82 | A1 | |
| Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (1.815, 1.825) | B1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| State equation \(x = \frac{7x}{8} + \frac{5}{2x^4}\), or equivalent | B1 | |
| Derive the exact answer \(\alpha\) (or \(x\)) = \(\sqrt[4]{20}\) | B1 | [2] |
**(i)**
| Use the iterative formula correctly at least once | M1 |
| Obtain final answer 1.82 | A1 |
| Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (1.815, 1.825) | B1 | [3] |
**(ii)**
| State equation $x = \frac{7x}{8} + \frac{5}{2x^4}$, or equivalent | B1 |
| Derive the exact answer $\alpha$ (or $x$) = $\sqrt[4]{20}$ | B1 | [2] |2 The sequence of values given by the iterative formula
$$x _ { n + 1 } = \frac { 7 x _ { n } } { 8 } + \frac { 5 } { 2 x _ { n } ^ { 4 } }$$
with initial value $x _ { 1 } = 1.7$, converges to $\alpha$.\\
(i) Use this iterative formula to determine $\alpha$ correct to 2 decimal places, giving the result of each iteration to 4 decimal places.\\
(ii) State an equation that is satisfied by $\alpha$ and hence show that $\alpha = \sqrt [ 5 ] { 20 }$.
\hfill \mbox{\textit{CAIE P2 2010 Q2 [5]}}