| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Show root in interval |
| Difficulty | Moderate -0.3 This is a straightforward numerical methods question requiring sign-change verification and iterative formula application. Part (i) involves simple substitution and arithmetic. Part (ii) is routine iteration with a given formula—no derivation or convergence analysis needed. Slightly easier than average due to its mechanical nature and clear instructions. |
| Spec | 1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
The equation $2x^3 + 4x - 35 = 0$ has one real root.
\begin{enumerate}[label=(\roman*)]
\item Show by calculation that this real root lies between 2 and 3. [3]
\item Use the iterative formula
$$x_{n+1} = \sqrt[3]{17.5 - 2x_n},$$
with a suitable starting value, to find the real root of the equation $2x^3 + 4x - 35 = 0$ correct to 2 decimal places. You should show the result of each iteration. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR C3 Q3 [9]}}