| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2011 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Rearrange to iterative form |
| Difficulty | Moderate -0.3 This is a straightforward application of fixed point iteration with clear instructions. Part (a) is routine substitution, part (b) is simple algebraic rearrangement (taking square root and dividing), and part (c) requires only calculator work following the given formula. No problem-solving insight needed, just careful execution of standard procedures. |
| 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 | Mark | Guidance |
| \(x = 3.5\): \(y = (12.25 - 4)\ln(5.5) = 8.25 \times 1.704... = 14.06...\) | M1 | Evaluating at both values |
| \(x = 3.6\): \(y = (12.96 - 4)\ln(5.6) = 8.96 \times 1.722... = 15.43...\) | A1 | Both correct, change of sign, conclusion stated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((x^2 - 4)\ln(x+2) = 15 \Rightarrow \ln(x+2) = \frac{15}{x^2-4}\) | M1 | Rearranging to isolate \(\ln\) term |
| \(x^2 = 4 + \frac{15}{\ln(x+2)}\), therefore \(x = \pm\sqrt{4 + \frac{15}{\ln(x+2)}}\) | A1 | Correct completion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x_2 = \sqrt{4 + \frac{15}{\ln(5.5)}} = \sqrt{4 + \frac{15}{1.7047...}} = 3.558\) | M1 | Correct iteration applied |
| \(x_3 = \sqrt{4 + \frac{15}{\ln(5.558)}} = 3.549\) | A1 | Both correct to 3 d.p. |
# Question 2:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $x = 3.5$: $y = (12.25 - 4)\ln(5.5) = 8.25 \times 1.704... = 14.06...$ | M1 | Evaluating at both values |
| $x = 3.6$: $y = (12.96 - 4)\ln(5.6) = 8.96 \times 1.722... = 15.43...$ | A1 | Both correct, change of sign, conclusion stated |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $(x^2 - 4)\ln(x+2) = 15 \Rightarrow \ln(x+2) = \frac{15}{x^2-4}$ | M1 | Rearranging to isolate $\ln$ term |
| $x^2 = 4 + \frac{15}{\ln(x+2)}$, therefore $x = \pm\sqrt{4 + \frac{15}{\ln(x+2)}}$ | A1 | Correct completion |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $x_2 = \sqrt{4 + \frac{15}{\ln(5.5)}} = \sqrt{4 + \frac{15}{1.7047...}} = 3.558$ | M1 | Correct iteration applied |
| $x_3 = \sqrt{4 + \frac{15}{\ln(5.558)}} = 3.549$ | A1 | Both correct to 3 d.p. |
---2 A curve is defined by the equation $y = \left( x ^ { 2 } - 4 \right) \ln ( x + 2 )$ for $x \geqslant 3$.\\
The curve intersects the line $y = 15$ at a single point, where $x = \alpha$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\alpha$ lies between 3.5 and 3.6.
\item Show that the equation $\left( x ^ { 2 } - 4 \right) \ln ( x + 2 ) = 15$ can be arranged into the form
$$x = \pm \sqrt { 4 + \frac { 15 } { \ln ( x + 2 ) } }$$
(2 marks)
\item Use the iteration
$$x _ { n + 1 } = \sqrt { 4 + \frac { 15 } { \ln \left( x _ { n } + 2 \right) } }$$
with $x _ { 1 } = 3.5$ to find the values of $x _ { 2 }$ and $x _ { 3 }$, giving your answers to three decimal places.\\
(2 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2011 Q2 [6]}}