| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Staircase/cobweb diagram |
| Difficulty | Standard +0.3 This is a standard FP2 question on cobweb/staircase diagrams for fixed point iteration. Part (i) requires drawing a standard convergent cobweb diagram, which is a routine graphical skill. Parts (ii)(a) and (ii)(b) require recognizing divergence patterns from the graph's gradient, which is textbook material. While this is Further Maths content, it's a straightforward application of well-practiced techniques with no novel problem-solving required. |
| Spec | 1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^{3/2} \frac{1}{(2x-1)^2+4} dx\) OR \(1/4 \int_1^{3/2} \frac{1}{(x-1/2)^2+1} dx\) | B1 | For correct denominator (in 2nd case must include \(1/4\)) |
| M1 | For integration to \(k\tan^{-1}(ax+b)\) or \(k\ln(\frac{ax+b-c}{ax+b+c})\) | |
| A1 | For correct integration | |
| M1 | For substituting limits in any \(\tan^{-1}\) expression | |
| A1 | For correct value |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x) = x^3 + 4x^2 + x - 1\) | B1 | Diffn |
| \(f'(x) = 3x^2 + 8x + 1\) | ||
| \(\Rightarrow x_{n+1} = x_n - \frac{x_n^3 + 4x_n^2 + x_n - 1}{3x_n^2 + 8x_n + 1}\) | M1 | Correct application of N-R formula |
| \(= \frac{x_n(3x_n^2 + 8x_n + 1) - (x_n^3 + 4x_n^2 + x_n - 1)}{3x_n^2 + 8x_n + 1}\) | A1 | And completed with suffices on last line |
| \(= \frac{2x_n^3 + 4x_n^2 + 1}{3x_n^2 + 8x_n + 1}\) | NB AG | |
| \(x_2 = -0.72652, x_3 = -0.72611\) | B1 | For \(x_2\) |
| \(\Rightarrow \alpha = -0.72611\) | B1 | For enough iterates to determine 3dp |
| NB \(x_4 = -0.726109\) | ||
| Sketch plus at least one tangent | B1 | At least the first tangent and vertical line to curve |
| Converges to another root | B1 | Or positive root or, for e.g. "x = 0 is the wrong side of a turning point" www |
| Use of formula to find this root numerically is not acceptable |
$\int_0^{3/2} \frac{1}{(2x-1)^2+4} dx$ OR $1/4 \int_1^{3/2} \frac{1}{(x-1/2)^2+1} dx$ | B1 | For correct denominator (in 2nd case must include $1/4$)
| M1 | For integration to $k\tan^{-1}(ax+b)$ or $k\ln(\frac{ax+b-c}{ax+b+c})$
| A1 | For correct integration
| M1 | For substituting limits in any $\tan^{-1}$ expression
| A1 | For correct value
## Question 5 (Further Pure Mathematics 2)
$f(x) = x^3 + 4x^2 + x - 1$ | B1 | Diffn
$f'(x) = 3x^2 + 8x + 1$ | |
$\Rightarrow x_{n+1} = x_n - \frac{x_n^3 + 4x_n^2 + x_n - 1}{3x_n^2 + 8x_n + 1}$ | M1 | Correct application of N-R formula
$= \frac{x_n(3x_n^2 + 8x_n + 1) - (x_n^3 + 4x_n^2 + x_n - 1)}{3x_n^2 + 8x_n + 1}$ | A1 | And completed with suffices on last line
$= \frac{2x_n^3 + 4x_n^2 + 1}{3x_n^2 + 8x_n + 1}$ | | NB AG
$x_2 = -0.72652, x_3 = -0.72611$ | B1 | For $x_2$
$\Rightarrow \alpha = -0.72611$ | B1 | For enough iterates to determine 3dp
| | NB $x_4 = -0.726109$
Sketch plus at least one tangent | B1 | At least the first tangent and vertical line to curve
Converges to another root | B1 | Or positive root or, for e.g. "x = 0 is the wrong side of a turning point" www
| | Use of formula to find this root numerically is not acceptable4 Answer the whole of this question on the insert provided.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{0ec9c4ff-8622-4dda-a000-6ffe36f38023-02_887_1273_1137_438}
\end{center}
The sketch shows the curve with equation $y = \mathrm { F } ( x )$ and the line $y = x$. The equation $x = \mathrm { F } ( x )$ has roots $x = \alpha$ and $x = \beta$ as shown.\\
(i) Use the copy of the sketch on the insert to show how an iteration of the form $x _ { n + 1 } = \mathrm { F } \left( x _ { n } \right)$, with starting value $x _ { 1 }$ such that $0 < x _ { 1 } < \alpha$ as shown, converges to the root $x = \alpha$.\\
(ii) State what happens in the iteration in the following two cases.
\begin{enumerate}[label=(\alph*)]
\item $x _ { 1 }$ is chosen such that $\alpha < x _ { 1 } < \beta$.
\item $x _ { 1 }$ is chosen such that $x _ { 1 } > \beta$.
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4 (i)\\
\includegraphics[max width=\textwidth, alt={}, center]{0ec9c4ff-8622-4dda-a000-6ffe36f38023-03_873_1259_274_484}\\
(ii) (a) $\_\_\_\_$\\
(b) $\_\_\_\_$
\section*{Jan 2006}
\end{enumerate}
\hfill \mbox{\textit{OCR FP2 Q4 [6]}}