| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton-Raphson method |
| Type | Newton-Raphson with complex derivative required |
| Difficulty | Standard +0.3 This is a straightforward application of the Newton-Raphson method with a slightly non-trivial derivative involving fractional powers. Part (a) requires routine differentiation using power rules, and part (b) is a single iteration of the standard formula. While the fractional indices add minor complexity compared to polynomial examples, this remains a textbook exercise with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.09d Newton-Raphson method |
| VIIIV SIHI NI III IM I I I N OC | VI4V SIHI NI JAHM ION OC | VI4V SIHI NI JIIYM IONOO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (a) \(f'(x) = \dfrac{9}{2}x^{\frac{1}{2}} + \dfrac{25}{2}x^{-\frac{3}{2}}\) | M1 A1 | M1: attempting differentiation i.e. decrease a power by 1. A1: accept equivalent fractions |
| (b) \(f(12.5) = 0.5115\ldots\) (at least \(0.51\ldots\)) and \(f'(12.5) = 16.1927\ldots\) (at least \(16\ldots\)) seen | B1, B1 | Each correct value; must be explicitly seen if final answer incorrect, may be implied by correct final answer |
| \(x_1 = 12.5 - \dfrac{f(12.5)}{f'(12.5)} = 12.5 - \dfrac{0.5115}{16.1927\ldots} = 12.468\) | M1 A1 | M1: for attempting Newton–Raphson with their values. A1: cao correct to 3dp. Newton–Raphson used more than once – isw |
## Question 2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(a)** $f'(x) = \dfrac{9}{2}x^{\frac{1}{2}} + \dfrac{25}{2}x^{-\frac{3}{2}}$ | M1 A1 | M1: attempting differentiation i.e. decrease a power by 1. A1: accept equivalent fractions |
| **(b)** $f(12.5) = 0.5115\ldots$ (at least $0.51\ldots$) and $f'(12.5) = 16.1927\ldots$ (at least $16\ldots$) seen | B1, B1 | Each correct value; must be explicitly seen if final answer incorrect, may be implied by correct final answer |
| $x_1 = 12.5 - \dfrac{f(12.5)}{f'(12.5)} = 12.5 - \dfrac{0.5115}{16.1927\ldots} = 12.468$ | M1 A1 | M1: for attempting Newton–Raphson with their values. A1: cao correct to 3dp. Newton–Raphson used more than once – isw |
---2.
$$f ( x ) = 3 x ^ { \frac { 3 } { 2 } } - 25 x ^ { - \frac { 1 } { 2 } } - 125 , \quad x > 0$$
\begin{enumerate}[label=(\alph*)]
\item Find $f ^ { \prime } ( x )$.
The equation $\mathrm { f } ( x ) = 0$ has a root $\alpha$ in the interval [12, 13].
\item Using $x _ { 0 } = 12.5$ 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 decimal places.
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VIIIV SIHI NI III IM I I I N OC & VI4V SIHI NI JAHM ION OC & VI4V SIHI NI JIIYM IONOO \\
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\hfill \mbox{\textit{Edexcel FP1 2016 Q2 [6]}}