| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2009 |
| 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 a given iterative formula requiring only repeated substitution and calculator work, plus a standard interval check for the root. The formula is provided, no rearrangement or derivation needed, and the convergence is well-behaved. Slightly easier than average due to minimal conceptual demand. |
| 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 |
| \(x_1 = \frac{2}{(2.5)^2} + 2\) | M1 | An attempt to substitute \(x_0 = 2.5\) into the iterative formula. Can be implied by \(x_1 = 2.32\) or \(2.320\) |
| \(x_1 = 2.32\), \(x_2 = 2.371581451...\) | A1 | Both \(x_1 = 2.32(0)\) and \(x_2\) awrt \(2.372\) |
| \(x_3 = 2.355593575...\), \(x_4 = 2.360436923...\) | A1 cso | Both \(x_3\) awrt \(2.356\) and \(x_4\) awrt \(2.360\) or \(2.36\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Let \(f(x) = -x^3 + 2x^2 + 2 = 0\); choose interval e.g. \([2.3585, 2.3595]\) | M1 | Choose suitable interval for \(x\), e.g. \([2.3585, 2.3595]\) or tighter |
| \(f(2.3585) = 0.00583577...\); any one value awrt 1sf or truncated 1sf | dM1 | Any one value awrt 1sf or truncated 1sf |
| \(f(2.3595) = -0.00142286...\); sign change (and \(f(x)\) is continuous) therefore a root; \(\alpha \in (2.3585, 2.3595) \Rightarrow \alpha = 2.359\) (3 dp) | A1 | Both values correct, sign change and conclusion. At minimum both values correct to 1sf, candidate states "change of sign, hence root" |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x_1 = \frac{2}{(2.5)^2} + 2$ | M1 | An attempt to substitute $x_0 = 2.5$ into the iterative formula. Can be implied by $x_1 = 2.32$ or $2.320$ |
| $x_1 = 2.32$, $x_2 = 2.371581451...$ | A1 | Both $x_1 = 2.32(0)$ and $x_2$ awrt $2.372$ |
| $x_3 = 2.355593575...$, $x_4 = 2.360436923...$ | A1 cso | Both $x_3$ awrt $2.356$ and $x_4$ awrt $2.360$ or $2.36$ |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $f(x) = -x^3 + 2x^2 + 2 = 0$; choose interval e.g. $[2.3585, 2.3595]$ | M1 | Choose suitable interval for $x$, e.g. $[2.3585, 2.3595]$ or tighter |
| $f(2.3585) = 0.00583577...$; any one value awrt 1sf or truncated 1sf | dM1 | Any one value awrt 1sf or truncated 1sf |
| $f(2.3595) = -0.00142286...$; sign change (and $f(x)$ is continuous) therefore a root; $\alpha \in (2.3585, 2.3595) \Rightarrow \alpha = 2.359$ (3 dp) | A1 | Both values correct, sign change and conclusion. At minimum both values correct to 1sf, candidate states "change of sign, hence root" |
---1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bcb0c693-66ae-4b97-99f8-b10fb9396886-02_579_1240_251_383}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows part of the curve with equation $y = - x ^ { 3 } + 2 x ^ { 2 } + 2$, which intersects the $x$-axis at the point $A$ where $x = \alpha$.
To find an approximation to $\alpha$, the iterative formula
$$x _ { n + 1 } = \frac { 2 } { \left( x _ { n } \right) ^ { 2 } } + 2$$
is used.
\begin{enumerate}[label=(\alph*)]
\item Taking $x _ { 0 } = 2.5$, find the values of $x _ { 1 } , x _ { 2 } , x _ { 3 }$ and $x _ { 4 }$.
Give your answers to 3 decimal places where appropriate.
\item Show that $\alpha = 2.359$ correct to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2009 Q1 [6]}}