| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Apply iteration to find root (pure fixed point) |
| Difficulty | Standard +0.3 This is a standard FP2 fixed point iteration question requiring routine application of the iterative formula, sketching a cobweb/staircase diagram, and using the convergence rate formula. While it involves multiple parts and some calculation, all techniques are textbook exercises with no novel problem-solving required. The convergence analysis in part (iv) is formulaic once the relationship is given. Slightly easier than average due to straightforward application of well-practiced methods. |
| Spec | 1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams1.09e Iterative method failure: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| (i) 3.8 | 3.868001 | |
| 3.868001 | 3.882190 | |
| 3.882190 | 3.885120 | |
| 3.885120 | 3.885723 | |
| 3.885723 | 3.885847 | |
| 3.885847 | 3.885873 | |
| 3.885873 | 3.885878 | |
| Root = 3.88588 | M1, A1, A1, A1 | N.B. Working must be seen; For \(x_2\); For \(x_3\) |
| [4] | ||
| (ii) | B1, B1 | Curve and line; Iterations showing staircase from below. At least two seen |
| [2] | ||
| (iii) 3.8 | 3.868001 | 0.068001 |
| 3.868001 | 3.88219 | 0.014189 |
| 3.88219 | 3.88512 | 0.002929 |
| 3.88512 | 3.885723 | 0.000603 |
| 3.885723 | 3.885847 | 0.000124 |
| 3.885847 | 3.885873 | |
| 3.885873 | 3.885878 | |
| \(\square_3 = 0.00293\) | M1, A1 | Working differences; Anything that rounds to 0.00293 |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| i.e. least \(n = 10\) | A1, M1, A1 | Attempt to find g' by differentiating g(x) correctly. Condone =; Take logs; If = has been used then the answer must include a justification |
| [4] |
**(i)** 3.8 | 3.868001
3.868001 | 3.882190
3.882190 | 3.885120
3.885120 | 3.885723
3.885723 | 3.885847
3.885847 | 3.885873
3.885873 | 3.885878
Root = 3.88588 | M1, A1, A1, A1 | N.B. Working must be seen; For $x_2$; For $x_3$
| [4] |
**(ii)** | B1, B1 | Curve and line; Iterations showing staircase from below. At least two seen
| [2] |
**(iii)** 3.8 | 3.868001 | 0.068001 | $\square_1$
3.868001 | 3.88219 | 0.014189 | $\square_2$
3.88219 | 3.88512 | 0.002929 | $\square_3$
3.88512 | 3.885723 | 0.000603 |
3.885723 | 3.885847 | 0.000124 |
3.885847 | 3.885873 |
3.885873 | 3.885878 |
$\square_3 = 0.00293$ | M1, A1 | Working differences; Anything that rounds to 0.00293
| [2] |
**(iv)** $g'(\alpha) = \frac{0.8}{3.88588} = 0.20587$
$g'(\alpha)^{n-1} < 10^{-6}$
$\Rightarrow n-1 \log 0.20587 < \log 10^{-6}$
$\Rightarrow n - 1 > \frac{-6}{-.68640} = 8.74\ldots$
$\Rightarrow n > 9.74\ldots$
i.e. least $n = 10$ | A1, M1, A1 | Attempt to find g' by differentiating g(x) correctly. Condone =; Take logs; If = has been used then the answer must include a justification
| [4] |
---9 The equation $10 x - 8 \ln x = 28$ has a root $\alpha$ in the interval [3,4]. The iteration $x _ { n + 1 } = \mathrm { g } \left( x _ { n } \right)$, where $\mathrm { g } ( x ) = 2.8 + 0.8 \ln x$ and $x _ { 1 } = 3.8$, is to be used to find $\alpha$.\\
(i) Find the value of $\alpha$ correct to 5 decimal places. You should show the result of each step of the iteration to 6 decimal places.\\
(ii) Illustrate this iteration by means of a sketch.\\
(iii) The difference, $\delta _ { r }$, between successive approximations is given by $\delta _ { r } = x _ { r + 1 } - x _ { r }$. Find $\delta _ { 3 }$.\\
(iv) Given that $\delta _ { n + 1 } \approx \mathrm {~g} ^ { \prime } ( \alpha ) \delta _ { n }$, for all positive integers $n$, estimate the smallest value of $n$ such that $\delta _ { n } < 10 ^ { - 6 } \delta _ { 1 }$.
\section*{OCR}
\hfill \mbox{\textit{OCR FP2 2014 Q9 [12]}}