| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton-Raphson method |
| Type | Newton-Raphson with derivative given or simple |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard numerical methods. Part (a) is routine sign change verification, part (b) is mechanical interval bisection requiring no insight, and part (c) is a single Newton-Raphson iteration with a simple cubic function. All techniques are textbook exercises with no problem-solving or novel reasoning required, making it slightly easier than average despite being Further Maths content. |
| Spec | 1.09a Sign change methods: locate roots1.09b Sign change methods: understand failure cases1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(1) = -4\ (< 0)\) | B1 | |
| \(f(2) = 1\ (> 0)\) | B1 | |
| Changes sign so root in \([1,2]\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(1.5) = -2.625\); interval \([1.5, 2]\) | B1M1 | B1 for awrt \(-2.6\); M1 for attempt to find \(f(1.75)\) |
| \(f(1.75) = -1.140625\); interval \([1.75, 2]\) | A1 | \(f(1.75)=\) awrt \(-1.1\); interval \([1.75,2]\) or \((1.75,2)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f'(x) = 3x^2 - 2\) | M1A1 | At least one term differentiated correctly; correct derivative |
| \(x_1 = 1.8 - \frac{1.8^3 - 2\times1.8 - 3}{3\times1.8^2 - 2}\) | M1A1 | Correct application of Newton-Raphson; \(f(1.8) = -0.768\) |
| \(x_1 = 1.90\) to 3 s.f. | A1 | cao |
# Question 8:
## Part 8(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(1) = -4\ (< 0)$ | B1 | |
| $f(2) = 1\ (> 0)$ | B1 | |
| Changes sign so root in $[1,2]$ | B1 | |
## Part 8(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(1.5) = -2.625$; interval $[1.5, 2]$ | B1M1 | B1 for awrt $-2.6$; M1 for attempt to find $f(1.75)$ |
| $f(1.75) = -1.140625$; interval $[1.75, 2]$ | A1 | $f(1.75)=$ awrt $-1.1$; interval $[1.75,2]$ or $(1.75,2)$ |
## Part 8(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f'(x) = 3x^2 - 2$ | M1A1 | At least one term differentiated correctly; correct derivative |
| $x_1 = 1.8 - \frac{1.8^3 - 2\times1.8 - 3}{3\times1.8^2 - 2}$ | M1A1 | Correct application of Newton-Raphson; $f(1.8) = -0.768$ |
| $x_1 = 1.90$ to 3 s.f. | A1 | cao |8.
$$f ( x ) = x ^ { 3 } - 2 x - 3$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { f } ( x ) = 0$ has a root, $\alpha$, in the interval $[ 1,2 ]$.
\item Starting with the interval $[ 1,2 ]$, use interval bisection twice to find an interval of width 0.25 which contains $\alpha$.
\item Using $x _ { 0 } = 1.8$ as a first approximation to $\alpha$, apply the Newton-Raphson procedure once to $\mathrm { f } ( x )$ to find a second approximation to $\alpha$, giving your answer to 3 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2013 Q8 [11]}}