| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Rearrange to iterative form |
| Difficulty | Moderate -0.3 This is a standard textbook-style fixed point iteration question with routine steps: (a) sign change verification requiring simple substitution, (b) trivial algebraic rearrangement (1 mark), and (c) calculator-based iteration for 2 iterations. All steps are mechanical with no problem-solving or 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 |
| \includegraphics[max width=\textwidth, alt={}]{b8614dd6-2197-40c3-a673-5bef3e3653a5-2_142_116_2560_157} | \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f(2) = 8 - 12 + 1 = -3 < 0\) | M1 | Evaluate \(f(x)\) at both \(x=2\) and \(x=3\) |
| \(f(3) = 27 - 18 + 1 = 10 > 0\) | ||
| Sign change, therefore root \(\alpha\) where \(2 < \alpha < 3\) | A1 | Must state sign change and conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x^3 - 6x + 1 = 0 \Rightarrow x^3 = 6x - 1\) | B1 | Divide by \(x\): \(x^2 = 6 - \frac{1}{x}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x_2 = \sqrt{6 - \frac{1}{2.5}} = \sqrt{5.6} = 2.366...\) | M1 | Correct use of recurrence with \(x_1 = 2.5\) |
| \(x_3 = \sqrt{6 - \frac{1}{2.366...}} = 2.390\) | A1 | Answer to 4 significant figures |
# Question 1:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(2) = 8 - 12 + 1 = -3 < 0$ | M1 | Evaluate $f(x)$ at both $x=2$ and $x=3$ |
| $f(3) = 27 - 18 + 1 = 10 > 0$ | | |
| Sign change, therefore root $\alpha$ where $2 < \alpha < 3$ | A1 | Must state sign change and conclusion |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^3 - 6x + 1 = 0 \Rightarrow x^3 = 6x - 1$ | B1 | Divide by $x$: $x^2 = 6 - \frac{1}{x}$ |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x_2 = \sqrt{6 - \frac{1}{2.5}} = \sqrt{5.6} = 2.366...$ | M1 | Correct use of recurrence with $x_1 = 2.5$ |
| $x_3 = \sqrt{6 - \frac{1}{2.366...}} = 2.390$ | A1 | Answer to 4 significant figures |
---1
\begin{enumerate}[label=(\alph*)]
\item Show that the equation $x ^ { 3 } - 6 x + 1 = 0$ has a root $\alpha$, where $2 < \alpha < 3$.
\item Show that the equation $x ^ { 3 } - 6 x + 1 = 0$ can be rearranged into the form
$$x ^ { 2 } = 6 - \frac { 1 } { x }$$
(1 mark)
\item Use the recurrence relation $x _ { n + 1 } = \sqrt { 6 - \frac { 1 } { x _ { n } } }$, with $x _ { 1 } = 2.5$, to find the value of $x _ { 3 }$, giving your answer to four significant figures.\\
(2 marks)
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\includegraphics[max width=\textwidth, alt={}]{b8614dd6-2197-40c3-a673-5bef3e3653a5-2_142_116_2560_157}
& $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ \\
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\hfill \mbox{\textit{AQA C3 2013 Q1 [5]}}