| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2011 |
| Session | January |
| 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 multi-part question testing two standard numerical methods with routine differentiation. Linear interpolation and one iteration of Newton-Raphson are both textbook procedures requiring no problem-solving insight, just careful arithmetic. The differentiation in part (b) is routine application of power rule. Slightly above average only due to being Further Maths content and requiring precision in calculation. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.09a Sign change methods: locate roots1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(1.6) = -1.29543081...\) | B1 | awrt \(-1.30\) |
| \(f(1.8) = 0.5401863372...\) | B1 | awrt \(0.54\) |
| \(\frac{\alpha-1.6}{1.29543081...} = \frac{1.8-\alpha}{0.5401863372...}\) | M1 | Correct linear interpolation method with signs correct |
| \(\alpha = 1.6 + \left(\frac{1.29543081...}{0.5401863372...+1.29543081...}\right)0.2\) | ||
| \(= 1.741143899...\) | A1 | awrt \(1.741\); correct answer seen 4/4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f'(x) = 10x - 6x^{\frac{1}{2}}\) | M1 | At least one of \(\pm ax\) or \(\pm bx^{\frac{1}{2}}\) correct |
| A1 | Correct differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(1.7) = -0.4161152711...\) | B1 | \(f(1.7)=\) awrt \(-0.42\) |
| \(f'(1.7) = 9.176957114...\) | B1 | \(f'(1.7)=\) awrt \(9.18\) |
| \(\alpha_2 = 1.7 - \left(\frac{-0.4161152711...}{9.176957114...}\right)\) | M1 | Correct application of Newton-Raphson formula using their values |
| \(= 1.7453434491...\) | ||
| \(= 1.745\) (3dp) | A1 cao | Correct answer seen 4/4 |
## Question 3:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(1.6) = -1.29543081...$ | B1 | awrt $-1.30$ |
| $f(1.8) = 0.5401863372...$ | B1 | awrt $0.54$ |
| $\frac{\alpha-1.6}{1.29543081...} = \frac{1.8-\alpha}{0.5401863372...}$ | M1 | Correct linear interpolation method with signs correct |
| $\alpha = 1.6 + \left(\frac{1.29543081...}{0.5401863372...+1.29543081...}\right)0.2$ | | |
| $= 1.741143899...$ | A1 | awrt $1.741$; correct answer seen 4/4 |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f'(x) = 10x - 6x^{\frac{1}{2}}$ | M1 | At least one of $\pm ax$ or $\pm bx^{\frac{1}{2}}$ correct |
| | A1 | Correct differentiation |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(1.7) = -0.4161152711...$ | B1 | $f(1.7)=$ awrt $-0.42$ |
| $f'(1.7) = 9.176957114...$ | B1 | $f'(1.7)=$ awrt $9.18$ |
| $\alpha_2 = 1.7 - \left(\frac{-0.4161152711...}{9.176957114...}\right)$ | M1 | Correct application of Newton-Raphson formula using their values |
| $= 1.7453434491...$ | | |
| $= 1.745$ (3dp) | A1 cao | Correct answer seen 4/4 |3.
$$f ( x ) = 5 x ^ { 2 } - 4 x ^ { \frac { 3 } { 2 } } - 6 , \quad x \geqslant 0$$
The root $\alpha$ of the equation $\mathrm { f } ( x ) = 0$ lies in the interval $[ 1.6,1.8 ]$.
\begin{enumerate}[label=(\alph*)]
\item Use linear interpolation once on the interval $[ 1.6,1.8 ]$ to find an approximation to $\alpha$. Give your answer to 3 decimal places.
\item Differentiate $\mathrm { f } ( x )$ to find $\mathrm { f } ^ { \prime } ( x )$.
\item Taking 1.7 as a first approximation to $\alpha$, apply the Newton-Raphson process once to $\mathrm { f } ( x )$ to obtain a second approximation to $\alpha$. Give your answer to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2011 Q3 [10]}}