| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2007 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Rearrange to iterative form |
| Difficulty | Moderate -0.3 This is a standard C3 iteration question requiring algebraic rearrangement, calculator work to apply an iteration formula three times, and verification of a root to a given accuracy. All steps are routine and follow textbook procedures with no problem-solving insight 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 |
|---|---|---|
| (a) \(x^2(3-x) - 1 = 0\) o.e. (e.g. \(x^2(-x+3) = 1\)) | M1 | |
| \(x = \sqrt{\frac{1}{3-x}}\) (✱) | A1 (cso) | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| (b) \(x_2 = 0.6455\), \(x_3 = 0.6517\), \(x_4 = 0.6526\) | B1; B1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| (c) Choose values in interval \((0.6525, 0.6535)\) or tighter and evaluate both | M1 | |
| \(f(0.6525) = -0.0005\) (372..., \(f(0.6535) = 0.002\) (101... | A1 | |
| At least one correct "up to bracket", i.e. \(-0.0005\) or \(0.002\) | A1 | |
| Change of sign, \(\therefore x = 0.653\) is a root (correct) to 3 d.p. | A1 | (3 marks) |
| Requires both correct "up to bracket" and conclusion as above | ||
| Alt (i) Continued iterations at least as far as \(x_6\) | M1 | |
| \(x_5 = 0.6527\), \(x_6 = 0.6527\), \(x_7 = ...\) two correct to at least 4 s.f. | A1 | |
| Conclusion: Two values correct to 4 d.p., so 0.653 is root to 3 d.p. | A1 | |
| Alt (ii) If use \(g(0.6525) = 0.6527...>0.6525\) and \(g(0.6535) = 0.6528...<0.6535\) | M1A1 | |
| Conclusion: Both results correct, so 0.653 is root to 3 d.p. | A1 | |
| (7 marks) |
**(a)** $x^2(3-x) - 1 = 0$ o.e. (e.g. $x^2(-x+3) = 1$) | M1 | |
$x = \sqrt{\frac{1}{3-x}}$ (**✱**) | A1 (cso) | (2 marks) |
**Note(✱):** Answer is given: need to see appropriate working and A1 is cso. [Reverse process: Squaring and non-fractional equation M1, form $f(x)$ A1]
**(b)** $x_2 = 0.6455$, $x_3 = 0.6517$, $x_4 = 0.6526$ | B1; B1 | (2 marks) |
1st B1 is for one correct, 2nd B1 for other two correct. If all three are to greater accuracy, award B0 B1
**(c)** Choose values in interval $(0.6525, 0.6535)$ or tighter and evaluate both | M1 | |
$f(0.6525) = -0.0005$ (372..., $f(0.6535) = 0.002$ (101... | A1 | |
At least one correct "up to bracket", i.e. $-0.0005$ or $0.002$ | A1 | |
Change of sign, $\therefore x = 0.653$ is a root (correct) to 3 d.p. | A1 | (3 marks) |
Requires both correct "up to bracket" and conclusion as above | | |
**Alt (i)** Continued iterations at least as far as $x_6$ | M1 | |
$x_5 = 0.6527$, $x_6 = 0.6527$, $x_7 = ...$ two correct to at least 4 s.f. | A1 | |
Conclusion: Two values correct to 4 d.p., so 0.653 is root to 3 d.p. | A1 | |
**Alt (ii)** If use $g(0.6525) = 0.6527...>0.6525$ and $g(0.6535) = 0.6528...<0.6535$ | M1A1 | |
Conclusion: Both results correct, so 0.653 is root to 3 d.p. | A1 | |
| | (7 marks) |
---4.
$$f ( x ) = - x ^ { 3 } + 3 x ^ { 2 } - 1$$
\begin{enumerate}[label=(\alph*)]
\item Show that the equation $\mathrm { f } ( x ) = 0$ can be rewritten as
$$x = \sqrt { } \left( \frac { 1 } { 3 - x } \right)$$
\item Starting with $x _ { 1 } = 0.6$, use the iteration
$$\left. x _ { n + 1 } = \sqrt { ( } \frac { 1 } { 3 - x _ { n } } \right)$$
to calculate the values of $x _ { 2 } , x _ { 3 }$ and $x _ { 4 }$, giving all your answers to 4 decimal places.
\item Show that $x = 0.653$ is a root of $\mathrm { f } ( x ) = 0$ correct to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2007 Q4 [7]}}