| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2008 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Rearrange to iterative form |
| Difficulty | Standard +0.2 This is a standard textbook exercise on fixed point iteration requiring only routine application of given formulas. Part (a) is simple sign change verification, (b) is straightforward algebraic rearrangement, (c) involves calculator work with a provided iteration formula, and (d) uses basic interval checking. No problem-solving insight or novel techniques required—easier than average A-level questions. |
| 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 |
|---|---|---|
| (a) \(f(1.4) = -0.568\ldots < 0\) | M1 | |
| \(f(1.45) = 0.245\ldots > 0\) | ||
| Change of sign (and continuity) \(\Rightarrow \alpha \in (1.4, 1.45)\) | A1 | (2) |
| (b) \(3x^3 = 2x + 6\) | ||
| \(x^3 = \frac{2x}{3} + 2\) | M1 A1 | |
| \(x^2 = \frac{2}{3} + \frac{2}{x}\) | ||
| \(x = \sqrt{\left(\frac{2}{x} + \frac{2}{3}\right)}\) | cso A1 | (3) |
| (c) \(x_1 = 1.4371\) | B1 | |
| \(x_2 = 1.4347\) | B1 | |
| \(x_3 = 1.4355\) | B1 | (3) |
| (d) Choosing the interval \((1.4345, 1.4355)\) or appropriate tighter interval. | M1 | |
| \(f(1.4345) = -0.01\ldots\) | M1 | |
| \(f(1.4355) = 0.003\ldots\) | ||
| Change of sign (and continuity) \(\Rightarrow \alpha \in (1.4345, 1.4355)\) | ||
| \(\Rightarrow \alpha = 1.435\), correct to 3 decimal places | cso A1 | (3) [11] |
**(a)** $f(1.4) = -0.568\ldots < 0$ | M1 |
$f(1.45) = 0.245\ldots > 0$ | |
Change of sign (and continuity) $\Rightarrow \alpha \in (1.4, 1.45)$ | A1 | (2)
**(b)** $3x^3 = 2x + 6$ | |
$x^3 = \frac{2x}{3} + 2$ | M1 A1 |
$x^2 = \frac{2}{3} + \frac{2}{x}$ | |
$x = \sqrt{\left(\frac{2}{x} + \frac{2}{3}\right)}$ | cso A1 | (3)
**(c)** $x_1 = 1.4371$ | B1 |
$x_2 = 1.4347$ | B1 |
$x_3 = 1.4355$ | B1 | (3)
**(d)** Choosing the interval $(1.4345, 1.4355)$ or appropriate tighter interval. | M1 |
$f(1.4345) = -0.01\ldots$ | M1 |
$f(1.4355) = 0.003\ldots$ | |
Change of sign (and continuity) $\Rightarrow \alpha \in (1.4345, 1.4355)$ | |
$\Rightarrow \alpha = 1.435$, correct to 3 decimal places | cso A1 | (3) [11]
**Note:** $\alpha = 1.435304553\ldots$7.
$$f ( x ) = 3 x ^ { 3 } - 2 x - 6$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { f } ( x ) = 0$ has a root, $\alpha$, between $x = 1.4$ and $x = 1.45$
\item Show that the equation $\mathrm { f } ( x ) = 0$ can be written as
$$x = \sqrt { } \left( \frac { 2 } { x } + \frac { 2 } { 3 } \right) , \quad x \neq 0$$
\item Starting with $x _ { 0 } = 1.43$, use the iteration
$$x _ { \mathrm { n } + 1 } = \sqrt { } \left( \frac { 2 } { x _ { \mathrm { n } } } + \frac { 2 } { 3 } \right)$$
to calculate the values of $x _ { 1 } , x _ { 2 }$ and $x _ { 3 }$, giving your answers to 4 decimal places.
\item By choosing a suitable interval, show that $\alpha = 1.435$ is correct to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2008 Q7 [11]}}