| Exam Board | Edexcel |
|---|---|
| Module | F1 (Further Pure Mathematics 1) |
| Year | 2023 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Topic | Newton-Raphson method |
| Type | Compare Newton-Raphson with linear interpolation |
| Difficulty | Standard +0.3 This is a structured multi-part question covering standard numerical methods (sign change, bisection, Newton-Raphson, linear interpolation) with straightforward calculations. The function is simple, differentiation is routine (fractional power), and all methods are applied mechanically following textbook procedures. While it requires multiple techniques, each step is standard Further Maths content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.09a Sign change methods: locate roots1.09b Sign change methods: understand failure cases1.09d Newton-Raphson method1.09f Trapezium rule: numerical integration |
7.
$$f ( x ) = x ^ { \frac { 3 } { 2 } } + x - 3$$
\begin{enumerate}[label=(\alph*)]
\item Show that the equation $\mathrm { f } ( x ) = 0$ has a root, $\alpha$, in the interval $[ 1,2 ]$\\[0pt]
\item Starting with the interval [1, 2], use interval bisection twice to show that $\alpha$ lies in the interval [1.25, 1.5]
\item \begin{enumerate}[label=(\roman*)]
\item Determine $\mathrm { f } ^ { \prime } ( x )$
\item Using 1.375 as a first approximation for $\alpha$, apply the Newton-Raphson process once to $\mathrm { f } ( x )$ to determine a second approximation for $\alpha$, giving your answer to 3 decimal places.\\[0pt]
\end{enumerate}\item Use linear interpolation once on the interval [1.25,1.5] to obtain a different approximation for $\alpha$, giving your answer to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F1 2023 Q7 [11]}}