| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | January |
| Marks | 13 |
| 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 question combining routine iteration and Newton-Raphson application. Part (i)(a) requires straightforward iteration with calculator work, (i)(b) needs a basic cobweb diagram sketch showing why iteration diverges near α, and (ii) involves algebraic manipulation to derive the N-R formula followed by standard iteration. All techniques are textbook exercises with no novel insight required, though the multi-part structure and algebraic manipulation place it slightly above average difficulty. |
| Spec | 1.06d Natural logarithm: ln(x) function and properties1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x_1=4.15,\quad x_2=4.1474\ldots\) | M1 | Using iterative formula at least once using at least 4dp; all iterates must be seen |
| \(x_3=4.1465\ldots,\quad x_4=4.1463\ldots\) | ||
| \(\beta = 4.146\) | A1 | www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Staircase diagram will always move to upper root | B1 | Sketch showing an example \(x_1 > \alpha\); ignore any statement when \(x_1>\beta\) |
| B1 | Example with \(x_1 < \alpha\) | |
| B1 | Statement dependent on first two B marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\ln(x-1) = x-3 \Rightarrow \ln(x-1)-(x-3)=0\) | M1 | Get equation in correct form |
| \(\Rightarrow f(x) = \ln(x-1)-(x-3)\) | ||
| \(\Rightarrow f'(x) = \frac{1}{x-1}-1\) | M1 | Differentiate |
| \(\Rightarrow x_{n+1} = x_n - \frac{\ln(x_n-1)-(x_n-3)}{\frac{1}{x_n-1}-1}\) | M1 | Use correct formula |
| \(= x_n - \frac{(x_n-1)(\ln(x_n-1)-(x_n-3))}{1-(x_n-1)}\) | A1 | Multiply by \((x-1)\) soi |
| \(= \frac{x_n(2-x_n)+(x_n-1)(x_n-3)-(x_n-1)\ln(x_n-1)}{2-x_n}\) | ||
| \(= \frac{2x_n - x_n^2 + x_n^2 - 4x_n + 3 - (x_n-1)\ln(x_n-1)}{2-x_n}\) | A1 | |
| \(\Rightarrow x_{n+1} = \frac{3-2x_n-(x_n-1)\ln(x_n-1)}{2-x_n}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x_2 = 1.152(359)\) | B1 | For \(x_2\); allow 3dp; \(x_2\) must be right for last B1 |
| \(1.152359 \to 1.158448 \to 1.158594 \to 1.158594\) | B1 | For enough iterates to determine 3dp; any error is likely to be self-correcting |
| Root \(= 1.159\) | B1 | www |
## Question 8:
### Part (i)(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x_1=4.15,\quad x_2=4.1474\ldots$ | M1 | Using iterative formula at least once using at least 4dp; all iterates must be seen |
| $x_3=4.1465\ldots,\quad x_4=4.1463\ldots$ | | |
| $\beta = 4.146$ | A1 | www |
### Part (i)(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Staircase diagram will always move to upper root | B1 | Sketch showing an example $x_1 > \alpha$; ignore any statement when $x_1>\beta$ |
| | B1 | Example with $x_1 < \alpha$ |
| | B1 | Statement dependent on first two B marks |
### Part (ii)(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\ln(x-1) = x-3 \Rightarrow \ln(x-1)-(x-3)=0$ | M1 | Get equation in correct form |
| $\Rightarrow f(x) = \ln(x-1)-(x-3)$ | | |
| $\Rightarrow f'(x) = \frac{1}{x-1}-1$ | M1 | Differentiate |
| $\Rightarrow x_{n+1} = x_n - \frac{\ln(x_n-1)-(x_n-3)}{\frac{1}{x_n-1}-1}$ | M1 | Use correct formula |
| $= x_n - \frac{(x_n-1)(\ln(x_n-1)-(x_n-3))}{1-(x_n-1)}$ | A1 | Multiply by $(x-1)$ soi |
| $= \frac{x_n(2-x_n)+(x_n-1)(x_n-3)-(x_n-1)\ln(x_n-1)}{2-x_n}$ | | |
| $= \frac{2x_n - x_n^2 + x_n^2 - 4x_n + 3 - (x_n-1)\ln(x_n-1)}{2-x_n}$ | A1 | |
| $\Rightarrow x_{n+1} = \frac{3-2x_n-(x_n-1)\ln(x_n-1)}{2-x_n}$ | | |
### Part (ii)(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x_2 = 1.152(359)$ | B1 | For $x_2$; allow 3dp; $x_2$ must be right for last B1 |
| $1.152359 \to 1.158448 \to 1.158594 \to 1.158594$ | B1 | For enough iterates to determine 3dp; any error is likely to be self-correcting |
| Root $= 1.159$ | B1 | www |8 It is required to solve the equation $\ln ( x - 1 ) - x + 3 = 0$.\\
You are given that there are two roots, $\alpha$ and $\beta$, where $1.1 < \alpha < 1.2$ and $4.1 < \beta < 4.2$.\\
(i) The root $\beta$ can be found using the iterative formula
$$x _ { n + 1 } = \ln \left( x _ { n } - 1 \right) + 3$$
\begin{enumerate}[label=(\alph*)]
\item Using this iterative formula with $x _ { 1 } = 4.15$, find $\beta$ correct to 3 decimal places. Show all your working.
\item Explain with the aid of a sketch why this iterative formula will not converge to $\alpha$ whatever initial value is taken.\\
(ii) (a) Show that the Newton-Raphson iterative formula for this equation can be written in the form
$$x _ { n + 1 } = \frac { 3 - 2 x _ { n } - \left( x _ { n } - 1 \right) \ln \left( x _ { n } - 1 \right) } { 2 - x _ { n } }$$
(b) Use this formula with $x _ { 1 } = 1.2$ to find $\alpha$ correct to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{OCR FP2 2013 Q8 [13]}}