| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2006 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Rearrange to iterative form |
| Difficulty | Standard +0.3 This is a standard C3 iteration question requiring algebraic rearrangement (routine), calculator-based iteration (mechanical), and interval verification for a root (standard technique). All parts follow textbook procedures with no novel problem-solving 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 | Marks | Guidance |
| \(2x^2 - 1 - \frac{4}{x} = 0\) | M1 | Dividing equation by \(x\) |
| \(x^2 = \frac{1}{2} + \frac{4}{2x}\) | M1 | Obtaining \(x^2 = \ldots\) |
| \(x = \sqrt{\left(\frac{2}{x} + \frac{1}{2}\right)}\) * | A1 | cso (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x_1 = 1.41,\ x_2 = 1.39,\ x_3 = 1.39\) | B1, B1, B1 | If answers given to more than 2 dp, penalise first time then accept awrt (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Choosing \((1.3915, 1.3925)\) or tighter interval | M1 | |
| \(f(1.3915) \approx -3\times10^{-3},\ f(1.3925) \approx 7\times10^{-3}\) | A1 | Both, awrt |
| Change of sign (and continuity) \(\Rightarrow \alpha \in (1.3915, 1.3925)\) \(\Rightarrow \alpha = 1.392\) to 3 d.p. * | A1 | cso (3) [9] |
# Question 5:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2x^2 - 1 - \frac{4}{x} = 0$ | M1 | Dividing equation by $x$ |
| $x^2 = \frac{1}{2} + \frac{4}{2x}$ | M1 | Obtaining $x^2 = \ldots$ |
| $x = \sqrt{\left(\frac{2}{x} + \frac{1}{2}\right)}$ * | A1 | cso **(3)** |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x_1 = 1.41,\ x_2 = 1.39,\ x_3 = 1.39$ | B1, B1, B1 | If answers given to more than 2 dp, penalise first time then accept awrt **(3)** |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Choosing $(1.3915, 1.3925)$ or tighter interval | M1 | |
| $f(1.3915) \approx -3\times10^{-3},\ f(1.3925) \approx 7\times10^{-3}$ | A1 | Both, awrt |
| Change of sign (and continuity) $\Rightarrow \alpha \in (1.3915, 1.3925)$ $\Rightarrow \alpha = 1.392$ to 3 d.p. * | A1 | cso **(3) [9]** |
---5.
$$f ( x ) = 2 x ^ { 3 } - x - 4$$
\begin{enumerate}[label=(\alph*)]
\item Show that the equation $\mathrm { f } ( x ) = 0$ can be written as
$$x = \sqrt { \left( \frac { 2 } { x } + \frac { 1 } { 2 } \right) }$$
The equation $2 x ^ { 3 } - x - 4 = 0$ has a root between 1.35 and 1.4.
\item Use the iteration formula
$$x _ { n + 1 } = \sqrt { } \left( \frac { 2 } { x _ { n } } + \frac { 1 } { 2 } \right)$$
with $x _ { 0 } = 1.35$, to find, to 2 decimal places, the values of $x _ { 1 } , x _ { 2 }$ and $x _ { 3 }$.
The only real root of $\mathrm { f } ( x ) = 0$ is $\alpha$.
\item By choosing a suitable interval, prove that $\alpha = 1.392$, to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2006 Q5 [9]}}