| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2004 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Show convergence to specific root |
| Difficulty | Standard +0.3 This is a straightforward fixed point iteration question requiring: (i) simple sign change verification by substitution, (ii) algebraic manipulation to show the fixed point satisfies the original equation, and (iii) calculator-based iteration. All steps are routine applications of standard techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (i) Evaluate cubic when \(x = -1\) and \(x = 0\) | M1 | |
| Justify given statement correctly | A1 | Total: 2 |
| [If calculations not given but justification uses correct statements about signs, award B1.] | ||
| (ii) State \(x = \frac{2x^3-1}{3x^2+1}\), or equivalent | B1 | |
| Rearrange this in the form \(x^3 + x + 1 = 0\) (or vice versa) | B1 | Total: 2 |
| (iii) Use the iterative formula correctly at least once | M1 | |
| Obtain final answer \(-0.68\) | A1 | |
| Show sufficient iterations to justify accuracy to 2d.p., or show sign change in interval \((-0.685, -0.675)\) | A1 | Total: 3 |
## Question 7:
| Answer/Working | Mark | Guidance |
|---|---|---|
| **(i)** Evaluate cubic when $x = -1$ and $x = 0$ | M1 | |
| Justify given statement correctly | A1 | **Total: 2** |
| [If calculations not given but justification uses correct statements about signs, award B1.] | | |
| **(ii)** State $x = \frac{2x^3-1}{3x^2+1}$, or equivalent | B1 | |
| Rearrange this in the form $x^3 + x + 1 = 0$ (or vice versa) | B1 | **Total: 2** |
| **(iii)** Use the iterative formula correctly at least once | M1 | |
| Obtain final answer $-0.68$ | A1 | |
| Show sufficient iterations to justify accuracy to 2d.p., or show sign change in interval $(-0.685, -0.675)$ | A1 | **Total: 3** |
---7 (i) The equation $x ^ { 3 } + x + 1 = 0$ has one real root. Show by calculation that this root lies between - 1 and 0 .\\
(ii) Show that, if a sequence of values given by the iterative formula
$$x _ { n + 1 } = \frac { 2 x _ { n } ^ { 3 } - 1 } { 3 x _ { n } ^ { 2 } + 1 }$$
converges, then it converges to the root of the equation given in part (i).\\
(iii) Use this iterative formula, with initial value $x _ { 1 } = - 0.5$, to determine the root correct to 2 decimal places, showing the result of each iteration.
\hfill \mbox{\textit{CAIE P3 2004 Q7 [7]}}