| Exam Board | SPS |
|---|---|
| Module | SPS FM Pure (SPS FM Pure) |
| Year | 2025 |
| Session | June |
| Marks | 11 |
| Topic | Newton-Raphson method |
| Type | Newton-Raphson convergence failure |
| Difficulty | Challenging +1.2 This is a structured Further Maths numerical methods question requiring: (i) routine sign change verification, (ii) standard Newton-Raphson formula derivation with algebraic manipulation, and (iii) analysis of iteration failure. While it involves multiple parts and some algebraic work, each component follows well-established techniques with clear guidance. The failure analysis requires understanding of Newton-Raphson behavior but is scaffolded by specific starting values to test. Slightly above average difficulty due to the algebraic manipulation and conceptual understanding required in part (iii), but remains a standard Further Maths exercise. |
| Spec | 1.09a Sign change methods: locate roots1.09d Newton-Raphson method1.09e Iterative method failure: convergence conditions |
Fig. 15 shows the graph of $f(x) = 2x + \frac{1}{x} + \ln x - 4$.
\includegraphics{figure_11}
\begin{enumerate}[label=(\roman*)]
\item Show that the equation
$$2x + \frac{1}{x} + \ln x - 4 = 0$$
has a root, $\alpha$, such that $0.1 < \alpha < 0.9$.
[2]
\item Obtain the following Newton-Raphson iteration for the equation in part (i).
$$x_{r+1} = x_r - \frac{2x_r^3 + x_r + x_r^2(\ln x_r - 4)}{2x_r^2 - 1 + x_r}$$
[3]
\item Explain why this iteration fails to find $\alpha$ using each of the following starting values.
\begin{enumerate}[label=(\Alph*)]
\item $x_0 = 0.4$
[2]
\item $x_0 = 0.5$
[2]
\item $x_0 = 0.6$
[2]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{SPS SPS FM Pure 2025 Q11 [11]}}