OCR H240/01 2017 Specimen — Question 9 9 marks

Exam BoardOCR
ModuleH240/01 (Pure Mathematics)
Year2017
SessionSpecimen
Marks9
TopicNewton-Raphson method
TypeDerive Newton-Raphson formula
DifficultyStandard +0.3 This is a standard Newton-Raphson question requiring routine differentiation, algebraic manipulation to derive the formula, calculator work for iterations, and sign-change verification. Part (d) tests understanding of convergence failure (likely due to stationary point or divergence), which is a common textbook scenario. All techniques are straightforward applications of A-level methods with no novel problem-solving required, making it slightly easier than average.
Spec1.09a Sign change methods: locate roots1.09d Newton-Raphson method

The equation \(x^3 - x^2 - 5x + 10 = 0\) has exactly one real root \(\alpha\).
  1. Show that the Newton-Raphson iterative formula for finding this root can be written as $$x_{n+1} = \frac{2x_n^3 - x_n^2 - 10}{3x_n^2 - 2x_n - 5}.$$ [3]
  2. 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. [1]
  3. Use a change of sign method to show that \(\alpha = -2.533\) is correct to 4 significant figures. [3]
  4. Explain why the Newton-Raphson method with initial value \(x_1 = -1\) would not converge to \(\alpha\). [2]

The equation $x^3 - x^2 - 5x + 10 = 0$ 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
$$x_{n+1} = \frac{2x_n^3 - x_n^2 - 10}{3x_n^2 - 2x_n - 5}.$$ [3]
\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. [1]
\item Use a change of sign method to show that $\alpha = -2.533$ is correct to 4 significant figures. [3]
\item Explain why the Newton-Raphson method with initial value $x_1 = -1$ would not converge to $\alpha$. [2]
\end{enumerate}

\hfill \mbox{\textit{OCR H240/01 2017 Q9 [9]}}