| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2009 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Rearrange to iterative form |
| Difficulty | Moderate -0.5 This is a standard fixed-point iteration question with routine steps: showing a root exists by sign change, trivial algebraic rearrangement, mechanical iteration calculations, and drawing a cobweb diagram. All parts are textbook exercises requiring only recall and basic computation, making it slightly easier than average for A-level. |
| 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 |
| \(f(x) = x^3 + 5x - 4\); \(f(0.5) = -1.375\), \(f(1) = 2\) | M1 | Condone \(f(0.5)\) rounding to \(-1.4\) |
| Change of sign \(\therefore 0.5 < \alpha < 1\) | A1 | Both statements needed |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x^3 + 5x - 4 = 0 \Rightarrow 5x = 4 - x^3\) | Must be seen | |
| \(x = \frac{1}{5}(4 - x^3)\) | B1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x_1 = 0.5\); \(x_2 = 0.775\ \left(= 31\tfrac{}{40}\right)\) | M1 | For \(x_2\) or \(x_3\) to 2 sf |
| \(x_3 = 0.707\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Cobweb/staircase diagram from \(x=0.5\) vertical to curve then horizontal to line | M1 | From 0.5 vertical to curve then horizontal to line |
| Correct diagram | A1 | CAO |
## Question 3:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(x) = x^3 + 5x - 4$; $f(0.5) = -1.375$, $f(1) = 2$ | M1 | Condone $f(0.5)$ rounding to $-1.4$ |
| Change of sign $\therefore 0.5 < \alpha < 1$ | A1 | Both statements needed |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x^3 + 5x - 4 = 0 \Rightarrow 5x = 4 - x^3$ | | Must be seen |
| $x = \frac{1}{5}(4 - x^3)$ | B1 | AG |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x_1 = 0.5$; $x_2 = 0.775\ \left(= 31\tfrac{}{40}\right)$ | M1 | For $x_2$ or $x_3$ to 2 sf |
| $x_3 = 0.707$ | A1 | |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Cobweb/staircase diagram from $x=0.5$ vertical to curve then horizontal to line | M1 | From 0.5 vertical to curve then horizontal to line |
| Correct diagram | A1 | CAO |
**Total: 7 marks**
---3 [Figure 1, printed on the insert, is provided for use in this question.]\\
The curve with equation $y = x ^ { 3 } + 5 x - 4$ intersects the $x$-axis at the point $A$, where $x = \alpha$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\alpha$ lies between 0.5 and 1 .
\item Show that the equation $x ^ { 3 } + 5 x - 4 = 0$ can be rearranged into the form
$$x = \frac { 1 } { 5 } \left( 4 - x ^ { 3 } \right)$$
\item Use the iteration $x _ { n + 1 } = \frac { 1 } { 5 } \left( 4 - x _ { n } { } ^ { 3 } \right)$ with $x _ { 1 } = 0.5$ to find $x _ { 3 }$, giving your answer to three decimal places.
\item The sketch on Figure 1 shows parts of the graphs of $y = \frac { 1 } { 5 } \left( 4 - x ^ { 3 } \right)$ and $y = x$, and the position of $x _ { 1 }$.
On Figure 1, draw a cobweb or staircase diagram to show how convergence takes place, indicating the positions of $x _ { 2 }$ and $x _ { 3 }$ on the $x$-axis.
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2009 Q3 [7]}}