| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2006 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Apply iteration to find root (pure fixed point) |
| Difficulty | Moderate -0.3 This is a straightforward application of standard C3 iteration techniques. Part (i) is routine substitution to verify sign change, and part (ii) requires mechanical application of a given formula with no rearrangement needed. The question is slightly easier than average because the iterative formula is provided and convergence is guaranteed, requiring only careful arithmetic and understanding of when to stop iterating. |
| 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 |
|---|---|---|
| (i) Attempt evaluation of cubic expression at 2 and 3 | M1 | |
| Obtain \(-11\) and \(31\) | A1 | |
| Conclude by noting change of sign | A1√ | 3 or equiv; following any calculated values provided negative then positive |
| (ii) Obtain correct first iterate | B1 | using \(x_1\) value such that \(2 \leq x_1 \leq 3\) |
| Attempt correct process to obtain at least 3 iterates | M1 | using any starting value now |
| Obtain \(2.34\) | A1 3 | answer required to 2 d.p. exactly; \(2 \to 2.3811 \to 2.3354 \to 2.3410\); \(2.5 \to 2.3208 \to 2.3428 \to 2.3401\); \(3 \to 2.2572 \to 2.3505 \to 2.3392\) |
**(i)** Attempt evaluation of cubic expression at 2 and 3 | M1 |
Obtain $-11$ and $31$ | A1 |
Conclude by noting change of sign | A1√ | 3 or equiv; following any calculated values provided negative then positive
**(ii)** Obtain correct first iterate | B1 | using $x_1$ value such that $2 \leq x_1 \leq 3$
Attempt correct process to obtain at least 3 iterates | M1 | using any starting value now
Obtain $2.34$ | A1 3 | answer required to 2 d.p. exactly; $2 \to 2.3811 \to 2.3354 \to 2.3410$; $2.5 \to 2.3208 \to 2.3428 \to 2.3401$; $3 \to 2.2572 \to 2.3505 \to 2.3392$3 The equation $2 x ^ { 3 } + 4 x - 35 = 0$ has one real root.\\
(i) Show by calculation that this real root lies between 2 and 3 .\\
(ii) Use the iterative formula
$$x _ { n + 1 } = \sqrt [ 3 ] { 17.5 - 2 x _ { n } }$$
with a suitable starting value, to find the real root of the equation $2 x ^ { 3 } + 4 x - 35 = 0$ correct to 2 decimal places. You should show the result of each iteration.
\hfill \mbox{\textit{OCR C3 2006 Q3 [6]}}