| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton-Raphson method |
| Type | Newton-Raphson with derivative given or simple |
| Difficulty | Moderate -0.3 This is a straightforward application of standard numerical methods (sign change, interval bisection, Newton-Raphson) with clear instructions and a simple polynomial. Part (a) requires basic substitution to verify a sign change, part (b) is mechanical bisection following a prescribed algorithm, and part (c) is a single iteration of Newton-Raphson with explicit starting value. While it requires careful arithmetic and knowledge of f'(x), there's no problem-solving or insight needed—just executing learned procedures on a routine polynomial. |
| Spec | 1.09a Sign change methods: locate roots1.09b Sign change methods: understand failure cases1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(f(2) = -1\) | M1 | Either any one of \(f(2) = -1\) or \(f(2.5) = \text{awrt } 3.4\) |
| \(f(2.5) = 3.40625\) | ||
| Sign change (and \(f(x)\) is continuous) therefore a root \(\alpha\) exists between \(x = 2\) and \(x = 2.5\) | A1 | both values correct, sign change and conclusion |
| (b) \(f(2.25) = 0.673828125 \left\{ = \frac{-343}{512} \right\} \Rightarrow \{2 < \alpha \lesssim 2.25\}\) | B1 | \(f(2.25) = \text{awrt } 0.7\) |
| \(f(2.125) = -0.2752685547...\) | M1 | Attempt to find \(f(2.125)\) or \(f(2.125) = \text{awrt} -0.3\) with \(2.125 < \alpha \leqslant 2.25\) or \(2.125 < \alpha < 2.25\) or \([2.125, 2.25]\) or \((2.125, 2.25)\). |
| \(\Rightarrow 2.125 < \alpha \leqslant 2.25\) | A1 | |
| (c) \(f'(x) = 2x^3 - 3x^2 + 1 \{ + 0 \}\) | M1 | At least two of the four terms differentiated correctly. Correct derivative. |
| \(f(-1.5) = 1.40625 \left(= 1\frac{13}{32}\right)\) | B1 | \(f(-1.5) = \text{awrt } 1.41\) |
| \(\{f'(-1.5) = -12.5\}\) | ||
| \(\beta_2 = -1.5 - \left( \frac{"-1.40625"}{"-12.5"} \right) = -1.3875 \left(= -1\frac{31}{80}\right)\) | M1 | Correct application of Newton-Raphson using their values. -1.3875 seen as answer to first iteration, award M1A1B1M1 |
| \(= -1.39\) (2dp) | A1 cao | -1.39 seen as answer to first iteration, award M1A1B1M1 |
**(a)** $f(2) = -1$ | M1 | Either any one of $f(2) = -1$ or $f(2.5) = \text{awrt } 3.4$
$f(2.5) = 3.40625$ | |
Sign change (and $f(x)$ is continuous) therefore a root $\alpha$ exists between $x = 2$ and $x = 2.5$ | A1 | both values correct, sign change and conclusion | [2]
**(b)** $f(2.25) = 0.673828125 \left\{ = \frac{-343}{512} \right\} \Rightarrow \{2 < \alpha \lesssim 2.25\}$ | B1 | $f(2.25) = \text{awrt } 0.7$
$f(2.125) = -0.2752685547...$ | M1 | Attempt to find $f(2.125)$ or $f(2.125) = \text{awrt} -0.3$ with $2.125 < \alpha \leqslant 2.25$ or $2.125 < \alpha < 2.25$ or $[2.125, 2.25]$ or $(2.125, 2.25)$. | [3]
$\Rightarrow 2.125 < \alpha \leqslant 2.25$ | A1 |
**(c)** $f'(x) = 2x^3 - 3x^2 + 1 \{ + 0 \}$ | M1 | At least two of the four terms differentiated correctly. Correct derivative.
$f(-1.5) = 1.40625 \left(= 1\frac{13}{32}\right)$ | B1 | $f(-1.5) = \text{awrt } 1.41$
$\{f'(-1.5) = -12.5\}$ | |
$\beta_2 = -1.5 - \left( \frac{"-1.40625"}{"-12.5"} \right) = -1.3875 \left(= -1\frac{31}{80}\right)$ | M1 | Correct application of Newton-Raphson using their values. -1.3875 seen as answer to first iteration, award M1A1B1M1
$= -1.39$ (2dp) | A1 cao | -1.39 seen as answer to first iteration, award M1A1B1M1 | [5]
**Total: [10]**
---$$f(x) = \frac{1}{2}x^4 - x^3 + x - 3$$
\begin{enumerate}[label=(\alph*)]
\item Show that the equation $f(x) = 0$ has a root $\alpha$ between $x = 2$ and $x = 2.5$ [2]
\item Starting with the interval $[2, 2.5]$ use interval bisection twice to find an interval of width $0.125$ which contains $\alpha$. [3]
\end{enumerate}
The equation $f(x) = 0$ has a root $\beta$ in the interval $[-2, -1]$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Taking $-1.5$ as a first approximation to $\beta$, apply the Newton-Raphson process once to $f(x)$ to obtain a second approximation to $\beta$.
Give your answer to 2 decimal places. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2013 Q3 [10]}}