| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton-Raphson method |
| Type | Derive Newton-Raphson formula |
| Difficulty | Standard +0.3 This is a straightforward application of the Newton-Raphson formula requiring standard differentiation and algebraic manipulation to derive the given form, followed by routine iteration calculations and a basic graphical interpretation. Part (i) is mechanical algebra, part (ii) is calculator work, and part (iii) requires recognizing that x=0 gives a horizontal tangent (f'(0)=1 makes denominator small, but the key issue is the tangent doesn't intersect appropriately). This is slightly easier than average as it's a standard FP2 exercise with no novel problem-solving required. |
| Spec | 1.09d Newton-Raphson method1.09e Iterative method failure: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f(x) = x^3+4x^2+x-1\) | ||
| \(f'(x) = 3x^2+8x+1\) | B1 | Diffn |
| \(\Rightarrow x_{n+1} = x_n - \frac{x_n^3+4x_n^2+x_n-1}{3x_n^2+8x_n+1}\) | 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 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x_2 = -0.72652\) | B1 | |
| \(x_3 = -0.72611\) | B1 | NB \(x_4 = -0.726109\) |
| \(\Rightarrow \alpha = -0.72611\) | B1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 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 |
| [2] |
# Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(x) = x^3+4x^2+x-1$ | | |
| $f'(x) = 3x^2+8x+1$ | B1 | Diffn |
| $\Rightarrow x_{n+1} = x_n - \frac{x_n^3+4x_n^2+x_n-1}{3x_n^2+8x_n+1}$ | 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 |
| **[3]** | | |
---
# Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x_2 = -0.72652$ | B1 | |
| $x_3 = -0.72611$ | B1 | NB $x_4 = -0.726109$ |
| $\Rightarrow \alpha = -0.72611$ | B1 | |
| **[3]** | | |
---
# Question 5(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| 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 |
| **[2]** | | |
---5 You are given that the equation $x ^ { 3 } + 4 x ^ { 2 } + x - 1 = 0$ has a root, $\alpha$, where $- 1 < \alpha < 0$.\\
(i) Show that the Newton-Raphson iterative formula for this equation can be written in the form
$$x _ { n + 1 } = \frac { 2 x _ { n } ^ { 3 } + 4 x _ { n } ^ { 2 } + 1 } { 3 x _ { n } ^ { 2 } + 8 x _ { n } + 1 } .$$
(ii) Using the initial value $x _ { 1 } = - 0.7$, find $x _ { 2 }$ and $x _ { 3 }$ and find $\alpha$ correct to 5 decimal places.\\
(iii) The diagram shows a sketch of the curve $y = x ^ { 3 } + 4 x ^ { 2 } + x - 1$ for $- 1.5 \leqslant x \leqslant 1$.\\
\includegraphics[max width=\textwidth, alt={}, center]{a80eb21f-c273-4b65-8617-16cdee783305-3_602_926_749_566}
Using the copy of the diagram in your answer book, explain why the initial value $x _ { 1 } = 0$ will fail to find $\alpha$.
\hfill \mbox{\textit{OCR FP2 2013 Q5 [8]}}