| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Year | 2010 |
| Session | June |
| Marks | 3 |
| Topic | Fixed Point Iteration |
| Type | Rearrange to iterative form |
| Difficulty | Standard +0.3 This is a straightforward iterative methods question requiring simple rearrangement (part i) and routine application of the convergence condition |g'(x)| < 1 (part ii). The algebraic manipulation is minimal and the theory is standard A-level content, making it slightly easier than average. |
| Spec | 1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams1.09e Iterative method failure: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| (i) States \(p = \frac{1}{5}\) \(q = \frac{3}{5}\) | B1 | [1] |
| (ii) Obtain \(\frac{3}{5}x^2\) | B1ft | |
| Substitute an \(x\) value between 0 and 1 and show that this gives a value < 1. Possible values of the derivative: \(x = 0\) value = 0; \(x = 0.1\) value = 0.006; \(x = 0.2\) value = 0.024; \(x = 0.3\) value = 0.054; \(x = 0.4\) value = 0.096; \(x = 0.5\) value = 0.15; \(x = 0.6\) value = 0.216; \(x = 0.7\) value = 0.294; \(x = 0.8\) value = 0.384; \(x = 0.9\) value = 0.486; \(x = 1\) value = 0.6 | B1 | [2] |
(i) States $p = \frac{1}{5}$ $q = \frac{3}{5}$ | B1 | [1]
(ii) Obtain $\frac{3}{5}x^2$ | B1ft |
Substitute an $x$ value between 0 and 1 and show that this gives a value < 1. Possible values of the derivative: $x = 0$ value = 0; $x = 0.1$ value = 0.006; $x = 0.2$ value = 0.024; $x = 0.3$ value = 0.054; $x = 0.4$ value = 0.096; $x = 0.5$ value = 0.15; $x = 0.6$ value = 0.216; $x = 0.7$ value = 0.294; $x = 0.8$ value = 0.384; $x = 0.9$ value = 0.486; $x = 1$ value = 0.6 | B1 | [2]The equation $x^3 - 5x + 3 = 0$ has a root between $x = 0$ and $x = 1$.
\begin{enumerate}[label=(\roman*)]
\item The equation can be rearranged into the form $x = g(x)$ where $g(x) = px^3 + q$. State the values of $p$ and $q$. [1]
\item By considering $|g'(x)|$, show that the iterative form $x_{n+1} = g(x_n)$ with a suitable starting value converges to the root between $x = 0$ and $x = 1$. [You are not required to find this root.] [2]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2010 Q2 [3]}}