| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2009 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton-Raphson method |
| Type | Compare Newton-Raphson with linear interpolation |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing standard numerical methods. Part (a) requires simple substitution to verify sign change, part (b) is a single Newton-Raphson iteration with given formula and starting value, and part (c) is a single linear interpolation calculation. All techniques are routine FP1 procedures with no problem-solving insight required, making it slightly easier than average despite being Further Maths content. |
| Spec | 1.09a Sign change methods: locate roots1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(f(2.2) = 2.2^3 - 2.2^2 - 6 \quad (= -0.192)\) | M1 | M1 for attempt at \(f(2.2)\) and \(f(2.3)\) |
| \(f(2.3) = 2.3^3 - 2.3^2 - 6 \quad (= 0.877)\) | ||
| Change of sign \(\Rightarrow\) Root; need numerical values correct to 1 s.f. | A1 | Need indication of change of sign and conclusion; need numerical values |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(f'(x) = 3x^2 - 2x\) | B1 | B1 for seeing correct derivative (may be implied by later correct work) |
| \(f'(2.2) = 10.12\) | B1 | B1 for seeing 10.12 (may be implied by later work) |
| \(x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 2.2 - \frac{-0.192}{10.12}\) | M1 A1ft | M1 attempt Newton-Raphson with their values; A1ft may be implied by following answer |
| \(= 2.219\) | A1cao | Final A1 must be 2.219 exactly; answer of 2.21897 would get 4/5 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(\frac{\alpha - 2.2}{\pm\text{'0.192'}} = \frac{2.3 - \alpha}{\pm\text{'0.877'}}\) or equivalent such as \(\frac{k}{\pm\text{'0.192'}} = \frac{0.1-k}{\pm\text{'0.877'}}\) | M1 | M1 attempt at ratio with their values of \(\pm f(2.2)\) and \(\pm f(2.3)\); if \(0.192 - \alpha\) or \(0.877 - \alpha\) in fraction then M0 |
| \(\alpha(0.877 + 0.192) = 2.3 \times 0.192 + 2.2 \times 0.877\) | A1 | A1 correct linear expression and definition of variable if not \(\alpha\) |
| \(\alpha \approx 2.218\quad(2.21796\ldots)\) (allow awrt) | A1 | A1 for awrt 2.218 |
| Alternative: Line joining \((2.2, -0.192)\) to \((2.3, 0.877)\), substitute \(y=0\): \(y + 0.192 = \frac{0.192+0.877}{0.1}(x-2.2)\), \(\alpha \approx 2.218\) | M1, A1 A1 |
## Question 4:
### Part (a)
| Working/Answer | Marks | Notes |
|---|---|---|
| $f(2.2) = 2.2^3 - 2.2^2 - 6 \quad (= -0.192)$ | M1 | M1 for attempt at $f(2.2)$ and $f(2.3)$ |
| $f(2.3) = 2.3^3 - 2.3^2 - 6 \quad (= 0.877)$ | | |
| Change of sign $\Rightarrow$ Root; need numerical values correct to 1 s.f. | A1 | Need indication of change of sign and conclusion; need numerical values |
### Part (b)
| Working/Answer | Marks | Notes |
|---|---|---|
| $f'(x) = 3x^2 - 2x$ | B1 | B1 for seeing correct derivative (may be implied by later correct work) |
| $f'(2.2) = 10.12$ | B1 | B1 for seeing 10.12 (may be implied by later work) |
| $x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 2.2 - \frac{-0.192}{10.12}$ | M1 A1ft | M1 attempt Newton-Raphson with their values; A1ft may be implied by following answer |
| $= 2.219$ | A1cao | Final A1 must be 2.219 exactly; answer of 2.21897 would get 4/5 |
### Part (c)
| Working/Answer | Marks | Notes |
|---|---|---|
| $\frac{\alpha - 2.2}{\pm\text{'0.192'}} = \frac{2.3 - \alpha}{\pm\text{'0.877'}}$ or equivalent such as $\frac{k}{\pm\text{'0.192'}} = \frac{0.1-k}{\pm\text{'0.877'}}$ | M1 | M1 attempt at ratio with their values of $\pm f(2.2)$ and $\pm f(2.3)$; if $0.192 - \alpha$ or $0.877 - \alpha$ in fraction then M0 |
| $\alpha(0.877 + 0.192) = 2.3 \times 0.192 + 2.2 \times 0.877$ | A1 | A1 correct linear expression and definition of variable if not $\alpha$ |
| $\alpha \approx 2.218\quad(2.21796\ldots)$ (allow awrt) | A1 | A1 for awrt 2.218 |
| Alternative: Line joining $(2.2, -0.192)$ to $(2.3, 0.877)$, substitute $y=0$: $y + 0.192 = \frac{0.192+0.877}{0.1}(x-2.2)$, $\alpha \approx 2.218$ | M1, A1 A1 | |
---4. Given that $\alpha$ is the only real root of the equation
$$x ^ { 3 } - x ^ { 2 } - 6 = 0$$
\begin{enumerate}[label=(\alph*)]
\item show that $2.2 < \alpha < 2.3$
\item Taking 2.2 as a first approximation to $\alpha$, apply the Newton-Raphson procedure once to $\mathrm { f } ( x ) = x ^ { 3 } - x ^ { 2 } - 6$ to obtain a second approximation to $\alpha$, giving your answer to 3 decimal places.\\[0pt]
\item Use linear interpolation once on the interval [2.2, 2.3] to find another approximation to $\alpha$, giving your answer to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2009 Q4 [10]}}