| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2011 |
| 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 (i) iterative calculation with a calculator to find α ≈ 1.260, and (ii) setting x = ½∛(x² + 6) and rearranging to get 8x³ - x² - 6 = 0. Both parts are routine applications of standard techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.04e Sequences: nth term and recurrence relations1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Use the iteration process correctly at least once | M1 | |
| Obtain at least two correct iterates to 5 decimal places | A1 | |
| Conclude \(a = 0.952\) \([1 \to 0.95647 \to 0.95257 \to 0.95223 \to 0.95220]\) | A1 | [3] |
| (ii) State or imply equation is \(x = \frac{1}{2}\sqrt{x^2 + 6}\) | B1 | |
| Obtain \(8x^3 - x^2 - 6 = 0\) | B1 | [2] |
**(i)** Use the iteration process correctly at least once | M1 |
Obtain at least two correct iterates to 5 decimal places | A1 |
Conclude $a = 0.952$ $[1 \to 0.95647 \to 0.95257 \to 0.95223 \to 0.95220]$ | A1 | [3]
**(ii)** State or imply equation is $x = \frac{1}{2}\sqrt{x^2 + 6}$ | B1 |
Obtain $8x^3 - x^2 - 6 = 0$ | B1 | [2]3 The sequence $x _ { 1 } , x _ { 2 } , x _ { 3 } , \ldots$ defined by
$$x _ { 1 } = 1 , \quad x _ { n + 1 } = \frac { 1 } { 2 } \sqrt [ 3 ] { } \left( x _ { n } ^ { 2 } + 6 \right)$$
converges to the value $\alpha$.\\
(i) Find the value of $\alpha$ correct to 3 decimal places. Show your working, giving each calculated value of the sequence to 5 decimal places.\\
(ii) Find, in the form $a x ^ { 3 } + b x ^ { 2 } + c = 0$, an equation of which $\alpha$ is a root.
\hfill \mbox{\textit{CAIE P2 2011 Q3 [5]}}