| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton-Raphson method |
| Type | Iterative formula from rearrangement |
| Difficulty | Standard +0.3 This is a standard FP2 iterative methods question requiring routine application of given formulas, basic iteration calculations, and cobweb/staircase diagram sketching. Part (iii) applies the Newton-Raphson formula directly. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams1.09d Newton-Raphson method1.09e Iterative method failure: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x_2=1.3869\ldots\) | B1 | For correct value (4 d.p. or better) |
| \(x_3=1.3938\) | B1 | For correct value |
| Staircase diagram converging towards \(\alpha\) | B1 | For sketch showing staircase towards \(\alpha\). (Vertical lines do not need to be labelled) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Sketch of \(y=\frac{1}{2}(x^4-1)\) and \(y=x\) (curve cuts \(y\)-axis) | B1 | For sketch like \(y=\frac{1}{2}(x^4-1)\) and \(y=x\) (curve or continuation of curve cuts \(y\)-axis) |
| Staircase diagram going away from \(\alpha\) | B1 | For sketch showing staircase away from \(\alpha\). ("Away" means labelling or arrows required.) Labelling means \(x_1,x_2,\ldots\) in right place or numeric values |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x_{n+1}=x_n-\frac{x_n^4-2x_n-1}{4x_n^3-2}\) | M1, A1 | For deriving the iterative formula; For correct formula |
| \(1.35\to 1.398268\) | A1 | For 1st value |
| \(\to 1.395348\to 1.395337\) | A1 | For correct 4dp \(\alpha\) with 2 iterates equal to 4dp (i.e. last two iterates agree to 4dp) www |
| \(\Rightarrow 1.3953\) |
# Question 4:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x_2=1.3869\ldots$ | B1 | For correct value (4 d.p. or better) |
| $x_3=1.3938$ | B1 | For correct value |
| Staircase diagram converging towards $\alpha$ | B1 | For sketch showing staircase towards $\alpha$. (Vertical lines do not need to be labelled) |
**[3 marks]**
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Sketch of $y=\frac{1}{2}(x^4-1)$ and $y=x$ (curve cuts $y$-axis) | B1 | For sketch like $y=\frac{1}{2}(x^4-1)$ and $y=x$ (curve or continuation of curve cuts $y$-axis) |
| Staircase diagram going away from $\alpha$ | B1 | For sketch showing staircase away from $\alpha$. ("Away" means labelling or arrows required.) Labelling means $x_1,x_2,\ldots$ in right place or numeric values |
**[2 marks]**
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x_{n+1}=x_n-\frac{x_n^4-2x_n-1}{4x_n^3-2}$ | M1, A1 | For deriving the iterative formula; For correct formula |
| $1.35\to 1.398268$ | A1 | For 1st value |
| $\to 1.395348\to 1.395337$ | A1 | For correct 4dp $\alpha$ with 2 iterates equal to 4dp (i.e. last two iterates agree to 4dp) **www** |
| $\Rightarrow 1.3953$ | | |
**[4 marks]**
---4 It is given that the equation $x ^ { 4 } - 2 x - 1 = 0$ has only one positive root, $\alpha$, and $1.3 < \alpha < 1.5$.\\
(i)\\
\includegraphics[max width=\textwidth, alt={}, center]{72a1330a-c6dc-4f3a-9b0e-333b099f4509-2_433_424_1119_817}
The diagram shows a sketch of $y = x$ and $y = \sqrt [ 4 ] { 2 x + 1 }$ for $x \geqslant 0$. Use the iteration $x _ { n + 1 } = \sqrt [ 4 ] { 2 x _ { n } + 1 }$ with $x _ { 1 } = 1.35$ to find $x _ { 2 }$ and $x _ { 3 }$, correct to 4 decimal places. On the copy of the diagram show how the iteration converges to $\alpha$.\\
(ii) For the same equation, the iteration $x _ { n + 1 } = \frac { 1 } { 2 } \left( x _ { n } ^ { 4 } - 1 \right)$ with $x _ { 1 } = 1.35$ gives $x _ { 2 } = 1.1608$ and $x _ { 3 } = 0.4077$, correct to 4 decimal places. Draw a sketch of $y = x$ and $y = \frac { 1 } { 2 } \left( x ^ { 4 } - 1 \right)$ for $x \geqslant 0$, and show how this iteration does not converge to $\alpha$.\\
(iii) Find the positive root of the equation $x ^ { 4 } - 2 x - 1 = 0$ by using the Newton-Raphson method with $x _ { 1 } = 1.35$, giving the root correct to 4 decimal places.
\hfill \mbox{\textit{OCR FP2 2012 Q4 [9]}}