| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton-Raphson method |
| Type | Compare Newton-Raphson with linear interpolation |
| Difficulty | Standard +0.3 This is a straightforward FP1 question requiring routine application of standard numerical methods. Part (a) is simple sign-checking, part (b) is one iteration of Newton-Raphson (requiring differentiation and substitution), and part (c) is one application of linear interpolation. All techniques are mechanical with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 1.09a Sign change methods: locate roots1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(-3) = 2.05555555...\) | M1 | Attempt both of \(f(-3) = \text{awrt } 2.1\) or trunc \(2\) or \(2.0\) or \(\frac{37}{18}\) and \(f(-2.5) = \text{awrt } -1.2\) or trunc \(-1.1\) or \(-\frac{139}{120}\) |
| \(f(-2.5) = -1.15833333...\) | ||
| Sign change oe (and \(f(x)\) is continuous) therefore a root \(\alpha\) {exists in the interval \([-3, -2.5].\)} | A1 | Both \(f(-3) = \text{awrt } 2.1\) and \(f(-2.5) = \text{awrt } -1.2\), sign change and 'root' or '\(\alpha\)'. Any errors award A0. |
| Answer | Marks | Guidance |
|---|---|---|
| \(f'(x) = 3x - \frac{4}{3x^2} + 2\) | M1 | \(\frac{3}{2}x^2 \to \pm Ax\) or \(\frac{4}{3x} \to \pm Bx^{-2}\) or \(2x - 5 \to 2\) Calculus must be seen for this to be awarded. At least two terms differentiated correctly |
| A1 | Correct derivative. | |
| \(\alpha = -3 - \left(\frac{"2.055..."}{"-7.148..."}\right)\) | M1 | Correct application of Newton-Raphson using their values from calculus. |
| \(= -2.71243523...\) or \(-\frac{1047}{386}\) or \(-2\frac{275}{386}\) | A1 | Exact value or awrt \(-2.712\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{-2.5 - \alpha}{"-1.158..."} = \frac{\alpha - (-3)}{"-2.055..."}" \text{ or } \frac{"-1.158..."}{"-2.055..."} = \frac{-2.5 - 3}{"-..."}\) | M1 | A correct linear interpolation statement \(\frac{-2.5 + \alpha}{\alpha - (-3)}\) or \(\frac{-3}{"-2.055..."} = \frac{-2.5 - 3}{"-..."}\) with correct signs. "1.158..." = "2.055..." provided \(\alpha\) sign changed at the end. Do not award until \(\alpha\) is seen. |
| \(\alpha = -3 + \left(\frac{"-2.055..."}{"-2.055..." + "-1.158..."}\right)(0.5)\) or \(\alpha = -3 + \left(\frac{"-2.055..."}{"-3.213..."}\right)(0.5)\) or \(\alpha = \frac{(-2.5)("-2.055..." ) - 3("-1.158...")"}{"-2.055..." + "-1.158..."}\) | dM1 | Achieves a correct linear interpolation statement with correct signs for \(\alpha = ...\) dependent on the previous method mark. |
| \(= -2.68020743...\) or \(-\frac{3101}{1157}\) or \(-2\frac{787}{1157}\) or \(= -2.680 (3 \text{ dp})\) | A1 cao | \(-2.680\): only penalise accuracy once in (b) and (c), but must be at least 3sf. |
**Part (a)**
| $f(-3) = 2.05555555...$ | M1 | Attempt both of $f(-3) = \text{awrt } 2.1$ or trunc $2$ or $2.0$ or $\frac{37}{18}$ and $f(-2.5) = \text{awrt } -1.2$ or trunc $-1.1$ or $-\frac{139}{120}$ |
| $f(-2.5) = -1.15833333...$ | | |
| Sign change oe (and $f(x)$ is continuous) therefore a root $\alpha$ {exists in the interval $[-3, -2.5].$} | A1 | Both $f(-3) = \text{awrt } 2.1$ and $f(-2.5) = \text{awrt } -1.2$, sign change and 'root' or '$\alpha$'. Any errors award A0. |
**Part (b)**
| $f'(x) = 3x - \frac{4}{3x^2} + 2$ | M1 | $\frac{3}{2}x^2 \to \pm Ax$ or $\frac{4}{3x} \to \pm Bx^{-2}$ or $2x - 5 \to 2$ Calculus must be seen for this to be awarded. At least two terms differentiated correctly |
| | A1 | Correct derivative. |
| $\alpha = -3 - \left(\frac{"2.055..."}{"-7.148..."}\right)$ | M1 | Correct application of Newton-Raphson using their values from calculus. |
| $= -2.71243523...$ or $-\frac{1047}{386}$ or $-2\frac{275}{386}$ | A1 | Exact value or awrt $-2.712$ |
**Part (c)**
| $\frac{-2.5 - \alpha}{"-1.158..."} = \frac{\alpha - (-3)}{"-2.055..."}" \text{ or } \frac{"-1.158..."}{"-2.055..."} = \frac{-2.5 - 3}{"-..."}$ | M1 | A correct linear interpolation statement $\frac{-2.5 + \alpha}{\alpha - (-3)}$ or $\frac{-3}{"-2.055..."} = \frac{-2.5 - 3}{"-..."}$ with correct signs. "1.158..." = "2.055..." provided $\alpha$ sign changed at the end. Do not award until $\alpha$ is seen. |
| $\alpha = -3 + \left(\frac{"-2.055..."}{"-2.055..." + "-1.158..."}\right)(0.5)$ or $\alpha = -3 + \left(\frac{"-2.055..."}{"-3.213..."}\right)(0.5)$ or $\alpha = \frac{(-2.5)("-2.055..." ) - 3("-1.158...")"}{"-2.055..." + "-1.158..."}$ | dM1 | Achieves a correct linear interpolation statement with correct signs for $\alpha = ...$ dependent on the previous method mark. |
| $= -2.68020743...$ or $-\frac{3101}{1157}$ or $-2\frac{787}{1157}$ or $= -2.680 (3 \text{ dp})$ | A1 cao | $-2.680$: only penalise accuracy once in (b) and (c), but must be at least 3sf. |
**Total: 10 marks**
---2.
$$f ( x ) = \frac { 3 } { 2 } x ^ { 2 } + \frac { 4 } { 3 x } + 2 x - 5 , \quad x < 0$$
The equation $\mathrm { f } ( x ) = 0$ has a single root $\alpha$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\alpha$ lies in the interval $[ - 3 , - 2.5 ]$
\item Taking - 3 as a first approximation to $\alpha$, apply the Newton-Raphson procedure once to $\mathrm { f } ( x )$ to obtain a second approximation to $\alpha$. Give your answer to 3 decimal places.
\item Use linear interpolation once on the interval $[ - 3 , - 2.5 ]$ to find another approximation to $\alpha$, giving your answer to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2018 Q2 [10]}}