| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2016 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Solve exponential equation via iteration |
| Difficulty | Standard +0.8 This FP2 question requires multiple sophisticated techniques: verifying roots, sketching iteration functions, understanding convergence via staircase diagrams, applying fixed-point iteration, and implementing Newton-Raphson with multiple iterations. While the individual components are standard FP2 material, the combination of graphical analysis of convergence behavior and executing two different numerical methods makes this more demanding than a typical A-level question, though not exceptionally difficult for Further Maths students. |
| Spec | 1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| (i) \(x=0\) in equation satisfies as \(e^0=1\) | B1 [1] | |
| (ii)(a) Sketch with asymptote between \(x=0\) and where it crosses \(x\)-axis | B1 | Allow one branch |
| \(+ve\) roots clear | B1 | |
| LH branch going through origin; LH branch does not have to be complete | B1 | SC1 \(y=(\ln(2x-1))^2\) |
| [3] | ||
| (ii)(b) Staircase seen near middle root converging to \(\beta\) | B1 | Either starting point shown with vertical line from axis to curve or arrows on staircase lines; follow through their curve where there are two positive roots |
| [1] | ||
| (ii)(c) \(x_1=3.75\); \(x_2=3.743604\ldots\) leading to \(3.733\) | B3 | B2 for 3.734; B1 for sight of 3.7436… |
| [3] | ||
| (iii) \(f(x)=(2x-1)^2-e^x\); \(\Rightarrow f'(x)=4(2x-1)-e^x\) | B1 | \(f'(x)\) correct soi by \(x_2\) |
| \(\Rightarrow x_{r+1}=x_r-\frac{(2x-1)^2-e^x}{4(2x-1)-e^x}\) | M1 | Use of formula soi by \(x_2\); \(f(x)\) correct and their \(f'(x)\) |
| \(\Rightarrow x_2=1.629382\ldots,\ x_3=1.629053\); Root \(=1.6291\) | A1 | \(x_2\) to 2dp or better |
| A1 | Correct root stated to 5sf; at least 2 iterates shown | |
| [4] |
# Question 4:
| Answer | Mark | Guidance |
|--------|------|----------|
| **(i)** $x=0$ in equation satisfies as $e^0=1$ | B1 **[1]** | |
| **(ii)(a)** Sketch with asymptote between $x=0$ and where it crosses $x$-axis | B1 | Allow one branch |
| $+ve$ roots clear | B1 | |
| LH branch going through origin; LH branch does not have to be complete | B1 | SC1 $y=(\ln(2x-1))^2$ |
| | **[3]** | |
| **(ii)(b)** Staircase seen near middle root converging to $\beta$ | B1 | Either starting point shown with vertical line from axis to curve or arrows on staircase lines; follow through their curve where there are two positive roots |
| | **[1]** | |
| **(ii)(c)** $x_1=3.75$; $x_2=3.743604\ldots$ leading to $3.733$ | B3 | B2 for 3.734; B1 for sight of 3.7436… |
| | **[3]** | |
| **(iii)** $f(x)=(2x-1)^2-e^x$; $\Rightarrow f'(x)=4(2x-1)-e^x$ | B1 | $f'(x)$ correct **soi by** $x_2$ |
| $\Rightarrow x_{r+1}=x_r-\frac{(2x-1)^2-e^x}{4(2x-1)-e^x}$ | M1 | Use of formula **soi by** $x_2$; $f(x)$ correct and their $f'(x)$ |
| $\Rightarrow x_2=1.629382\ldots,\ x_3=1.629053$; Root $=1.6291$ | A1 | $x_2$ to 2dp or better |
| | A1 | Correct root stated to 5sf; at least 2 iterates shown |
| | **[4]** | |
---4 You are given the equation $( 2 x - 1 ) ^ { 2 } - \mathrm { e } ^ { x } = 0$.\\
(i) Verify that 0 is a root of the equation.
There are also two other roots, $\alpha$ and $\beta$, where $0 < \alpha < \beta$.\\
(ii) The iterative formula $x _ { r + 1 } = \ln \left( 2 x _ { r } - 1 \right) ^ { 2 }$ is to be used to find a root of the equation.
\begin{enumerate}[label=(\alph*)]
\item Sketch the line $y = x$ and the curve $y = \ln ( 2 x - 1 ) ^ { 2 }$ on the same axes, showing the roots $0 , \alpha$ and $\beta$.
\item By drawing a 'staircase' diagram on your sketch, starting with a value of $x$ that is between $\alpha$ and $\beta$, show that this iteration does not converge to $\alpha$.
\item Using this iterative formula with $x _ { 1 } = 3.75$, find the value of $\beta$ correct to 3 decimal places.\\
(iii) Using the Newton-Raphson method with $x _ { 1 } = 1.6$, find the root $\alpha$ of the equation $( 2 x - 1 ) ^ { 2 } - \mathrm { e } ^ { x } = 0$ correct to 5 significant figures. Show the result of each iteration.
\end{enumerate}
\hfill \mbox{\textit{OCR FP2 2016 Q4 [12]}}