| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2005 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Compare iteration convergence |
| Difficulty | Standard +0.3 This is a standard fixed-point iteration question requiring: (i) simple substitution to show root location, (ii) testing two iteration formulas and recognizing divergence vs convergence through calculation. While it involves multiple steps, the techniques are routine for P3 level with no novel insight required—slightly easier than average. |
| Spec | 1.09b Sign change methods: understand failure cases1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks |
|---|---|
| Consider sign of \(x^3 - x - 3\), or equivalent | M1 |
| Justify the given statement | A1 |
| Answer | Marks |
|---|---|
| Apply an iterative formula correctly at least once, with initial value \(x_1 = 1.5\) | M1 |
| Show that \((A)\) fails to converge | A1 |
| Show that \((B)\) converges | A1 |
| Obtain final answer 1.67 | A1 |
| Show sufficient iterations to justify its accuracy to 2 d.p., or show there is a sign change in the interval (1.665, 1.675) | A1 |
**(i)**
Consider sign of $x^3 - x - 3$, or equivalent | M1 |
Justify the given statement | A1 |
**Total for (i): [2]**
**(ii)**
Apply an iterative formula correctly at least once, with initial value $x_1 = 1.5$ | M1 |
Show that $(A)$ fails to converge | A1 |
Show that $(B)$ converges | A1 |
Obtain final answer 1.67 | A1 |
Show sufficient iterations to justify its accuracy to 2 d.p., or show there is a sign change in the interval (1.665, 1.675) | A1 |
**Total for (ii): [5]**
---4 The equation $x ^ { 3 } - x - 3 = 0$ has one real root, $\alpha$.\\
(i) Show that $\alpha$ lies between 1 and 2 .
Two iterative formulae derived from this equation are as follows:
$$\begin{aligned}
& x _ { n + 1 } = x _ { n } ^ { 3 } - 3 \\
& x _ { n + 1 } = \left( x _ { n } + 3 \right) ^ { \frac { 1 } { 3 } }
\end{aligned}$$
Each formula is used with initial value $x _ { 1 } = 1.5$.\\
(ii) Show that one of these formulae produces a sequence which fails to converge, and use the other formula to calculate $\alpha$ correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
\hfill \mbox{\textit{CAIE P3 2005 Q4 [7]}}