Edexcel C3 — Question 4 10 marks

Exam BoardEdexcel
ModuleC3 (Core Mathematics 3)
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFixed Point Iteration
TypeRearrange to iterative form
DifficultyStandard +0.2 This is a standard C3 iteration question covering routine techniques: algebraic rearrangement (straightforward manipulation), applying an iterative formula (direct substitution), and change of sign method for proving accuracy. Part (d) requires recognizing when the formula is undefined (denominator = -1 or negative under square root), but this is a common textbook exercise type. All parts follow predictable patterns with no novel problem-solving required, making it slightly easier than the average A-level question.
Spec1.09a Sign change methods: locate roots1.09b Sign change methods: understand failure cases1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams

\(\text{f}(x) = x^3 + x^2 - 4x - 1\). The equation f(x) = 0 has only one positive root, \(\alpha\).
  1. Show that f(x) = 0 can be rearranged as $$x = \sqrt{\frac{4x+1}{x+1}}, \quad x \neq -1.$$ [2]
The iterative formula \(x_{n+1} = \sqrt{\frac{4x_n+1}{x_n+1}}\) is used to find an approximation to \(\alpha\).
  1. Taking \(x_1 = 1\), find, to 2 decimal places, the values of \(x_2\), \(x_3\) and \(x_4\). [3]
  2. By choosing values of \(x\) in a suitable interval, prove that \(\alpha = 1.70\), correct to 2 decimal places. [3]
  3. Write down a value of \(x_1\) for which the iteration formula \(x_{n+1} = \sqrt{\frac{4x_n+1}{x_n+1}}\) does not produce a valid value for \(x_2\). Justify your answer. [2]

Question 4:
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Question 4:
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$\text{f}(x) = x^3 + x^2 - 4x - 1$.

The equation f(x) = 0 has only one positive root, $\alpha$.

\begin{enumerate}[label=(\alph*)]
\item Show that f(x) = 0 can be rearranged as
$$x = \sqrt{\frac{4x+1}{x+1}}, \quad x \neq -1.$$ [2]
\end{enumerate}

The iterative formula $x_{n+1} = \sqrt{\frac{4x_n+1}{x_n+1}}$ is used to find an approximation to $\alpha$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Taking $x_1 = 1$, find, to 2 decimal places, the values of $x_2$, $x_3$ and $x_4$. [3]

\item By choosing values of $x$ in a suitable interval, prove that $\alpha = 1.70$, correct to 2 decimal places. [3]

\item Write down a value of $x_1$ for which the iteration formula $x_{n+1} = \sqrt{\frac{4x_n+1}{x_n+1}}$ does not produce a valid value for $x_2$.

Justify your answer. [2]
\end{enumerate}

\hfill \mbox{\textit{Edexcel C3  Q4 [10]}}