| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Topic | Newton-Raphson method |
| Type | Derive Newton-Raphson formula |
| Difficulty | Standard +0.3 This is a straightforward application of the Newton-Raphson formula requiring routine differentiation and algebraic manipulation in part (a), calculator iterations in part (b), sign-change verification in part (c), and recognition that f'(-1)=0 causes failure in part (d). All parts follow standard textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.09a Sign change methods: locate roots1.09d Newton-Raphson method1.09e Iterative method failure: convergence conditions |
9 The equation $x ^ { 3 } - x ^ { 2 } - 5 x + 10 = 0$ has exactly one real root $\alpha$.
\begin{enumerate}[label=(\alph*)]
\item Show that the Newton-Raphson iterative formula for finding this root can be written as
$$x _ { n + 1 } = \frac { 2 x _ { n } ^ { 3 } - x _ { n } ^ { 2 } - 10 } { 3 x _ { n } ^ { 2 } - 2 x _ { n } - 5 }$$
\item Apply the iterative formula in part (a) with initial value $x _ { 1 } = - 3$ to find $x _ { 2 } , x _ { 3 } , x _ { 4 }$ correct to 4 significant figures.
\item Use a change of sign method to show that $\alpha = - 2.533$ is correct to 4 significant figures.
\item Explain why the Newton-Raphson method with initial value $x _ { 1 } = - 1$ would not converge to $\alpha$.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 Q9 [9]}}