| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2010 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton-Raphson method |
| Type | Newton-Raphson with complex derivative required |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on standard numerical methods (sign change, interval bisection, Newton-Raphson) with clear instructions and routine calculations. While it requires careful arithmetic with the rational term 7/x, it involves no conceptual challenges or novel problem-solving—just direct application of well-practiced FP1 techniques. Slightly easier than average due to its mechanical nature. |
| Spec | 1.09a Sign change methods: locate roots1.09b Sign change methods: understand failure cases1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(f(1.4) = -0.256\) (or \(-\frac{32}{125}\)); \(f(1.5) = 0.708...\) (or \(\frac{17}{24}\)) | M1, A1 | Change of sign \(\therefore\) root |
| Alternative method: Graphical method could earn M1 if 1.4 and 1.5 are both indicated. A1 then needs correct graph and conclusion, i.e. change of sign \(\therefore\) root | (2 marks) | |
| (b) \(f(1.45) = 0.221...\) or \(0.2\) [\(\therefore\) root is in \([1.4, 1.45]\)] | M1, M1 | |
| \(f(1.425) = -0.018...\) or \(-0.019\) or \(-0.02\) | A1 cso | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| (c) \(f'(x) = 3x^2 + 7x^{-2}\) | M1 A1, A1 ft | |
| \(f'(1.45) = 9.636...\) | Special case: if \(f'(x) = 3x^2 + 7x^{-2} + 2\) then \(f'(1.45) = 11.636...\) | |
| \(x_1 = 1.45 - \frac{f(1.45)}{f'(1.45)} = 1.45 - \frac{0.221...}{9.636...} = 1.427\) | M1 A1 cao, M1 A1 cao | M1: for attempt at Newton Raphson with their values for \(f(1.45)\) and \(f'(1.45)\). A1: is cao and needs to be correct to 3dp |
**(a)** $f(1.4) = -0.256$ (or $-\frac{32}{125}$); $f(1.5) = 0.708...$ (or $\frac{17}{24}$) | **M1, A1** | Change of sign $\therefore$ root
Alternative method: Graphical method could earn M1 if 1.4 and 1.5 are both indicated. A1 then needs correct graph and conclusion, i.e. change of sign $\therefore$ root | (2 marks)
**(b)** $f(1.45) = 0.221...$ or $0.2$ [$\therefore$ root is in $[1.4, 1.45]$] | **M1, M1** |
$f(1.425) = -0.018...$ or $-0.019$ or $-0.02$ | **A1 cso** | (3 marks)
$\therefore$ root is in $[1.425, 1.45]$
**Guidance:** A1: is cso – any slips in numerical work are penalised here even if correct region found. Answer may be written as $1.425 \le \alpha \le 1.45$ or $1.425 < \alpha < 1.45$ or $(1.425, 1.45)$ – must be correct way round. Between is sufficient.
There is no credit for linear interpolation. This is **M0 M0 A0**
Answer with no working is also **M0M0A0**
**(c)** $f'(x) = 3x^2 + 7x^{-2}$ | **M1 A1, A1 ft** |
$f'(1.45) = 9.636...$ | | Special case: if $f'(x) = 3x^2 + 7x^{-2} + 2$ then $f'(1.45) = 11.636...$
$x_1 = 1.45 - \frac{f(1.45)}{f'(1.45)} = 1.45 - \frac{0.221...}{9.636...} = 1.427$ | **M1 A1 cao, M1 A1 cao** | M1: for attempt at Newton Raphson with their values for $f(1.45)$ and $f'(1.45)$. A1: is cao and needs to be correct to 3dp
**Guidance:** Second M1: for attempt at Newton Raphson with their values. A1: is cao and needs to be correct to 3dp. Newton Raphson used more than once – isw.
Special case: $f'(x) = 3x^2 + 7x^{-2} + 2$ then $f'(1.45) = 11.636...$) is **M1 A0 A1 ft M1 A0**. This mark can also be given by implication from final answer of 1.43
**Total: 10 marks**
---3.
$$\mathrm { f } ( x ) = x ^ { 3 } - \frac { 7 } { x } + 2 , \quad x > 0$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { f } ( x ) = 0$ has a root $\alpha$ between 1.4 and 1.5\\[0pt]
\item Starting with the interval [1.4,1.5], use interval bisection twice to find an interval of width 0.025 that contains $\alpha$.
\item Taking 1.45 as a first approximation to $\alpha$, apply the Newton-Raphson procedure once to $\mathrm { f } ( x ) = x ^ { 3 } - \frac { 7 } { x } + 2$ to obtain a second approximation to $\alpha$, giving your answer to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2010 Q3 [10]}}