| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2010 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton-Raphson method |
| Type | Newton-Raphson with complex derivative required |
| Difficulty | Moderate -0.3 This is a straightforward computational question testing standard numerical methods (interval bisection and Newton-Raphson) with clear instructions and no conceptual challenges. While it requires careful arithmetic and multiple steps, the procedures are routine FP1 content with no problem-solving insight needed—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 |
|---|---|---|
| \(f(1.3) = -1.439\) and \(f(1.4) = 0.268\) | B1 (1) | Both answers required; accept anything rounding to these 3dp values |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(1.35) < 0\) \((-0.568...)\) \(\Rightarrow\) \(1.35 < \alpha < 1.4\) | M1 A1 | \(f(1.35)\) or awrt \(-0.6\) for M1; \(f(1.35)\) awrt \(-0.6\) AND result for first A1 |
| \(f(1.375) < 0\) \((-0.146...)\) \(\Rightarrow\) \(1.375 < \alpha < 1.4\) | A1 (3) | \(1.375 < \alpha < 1.4\) or expression using brackets/equivalent in words for second A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(f'(x) = 6x + 22x^{-3}\) | M1 A1 | One term correct for M1, both correct for A1 |
| \(x_1 = x_0 - \dfrac{f(x_0)}{f'(x_0)} = 1.4 - \dfrac{0.268}{16.417} = 1.384\) | M1 A1, A1 (5) | Correct formula seen/implied and attempt to substitute for M1; awrt 16.4 for second A1; awrt 1.384 correct answer only for final A1 |
## Question 2:
**(a)**
$f(1.3) = -1.439$ and $f(1.4) = 0.268$ | B1 (1) | Both answers required; accept anything rounding to these 3dp values
**(b)**
$f(1.35) < 0$ $(-0.568...)$ $\Rightarrow$ $1.35 < \alpha < 1.4$ | M1 A1 | $f(1.35)$ or awrt $-0.6$ for M1; $f(1.35)$ awrt $-0.6$ AND result for first A1
$f(1.375) < 0$ $(-0.146...)$ $\Rightarrow$ $1.375 < \alpha < 1.4$ | A1 (3) | $1.375 < \alpha < 1.4$ or expression using brackets/equivalent in words for second A1
**(c)**
$f'(x) = 6x + 22x^{-3}$ | M1 A1 | One term correct for M1, both correct for A1
$x_1 = x_0 - \dfrac{f(x_0)}{f'(x_0)} = 1.4 - \dfrac{0.268}{16.417} = 1.384$ | M1 A1, A1 (5) | Correct formula seen/implied and attempt to substitute for M1; awrt 16.4 for second A1; awrt 1.384 correct answer only for final A1
---2.
$$f ( x ) = 3 x ^ { 2 } - \frac { 11 } { x ^ { 2 } }$$
\begin{enumerate}[label=(\alph*)]
\item Write down, to 3 decimal places, the value of $\mathrm { f } ( 1.3 )$ and the value of $\mathrm { f } ( 1.4 )$.
The equation $\mathrm { f } ( x ) = 0$ has a root $\alpha$ between 1.3 and 1.4\\[0pt]
\item Starting with the interval [1.3, 1.4], use interval bisection to find an interval of width 0.025 which contains $\alpha$.
\item Taking 1.4 as a first approximation to $\alpha$, apply the Newton-Raphson procedure once to $\mathrm { f } ( x )$ to obtain a second approximation to $\alpha$, giving your answer to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2010 Q2 [9]}}