Edexcel C3 Specimen — Question 2 8 marks

Exam BoardEdexcel
ModuleC3 (Core Mathematics 3)
SessionSpecimen
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFixed Point Iteration
TypeApply iteration to find root (pure fixed point)
DifficultyModerate -0.3 This is a standard C3 iterative methods question requiring sign change verification, calculator iteration (4 iterations), and error bound verification. All techniques are routine textbook procedures with no problem-solving insight needed, making it slightly easier than average.
Spec1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams

2. \(\quad \mathrm { f } ( x ) = x ^ { 3 } - 2 x - 5\).
  1. Show that there is a root \(\alpha\) of \(\mathrm { f } ( x ) = 0\) for \(x\) in the interval \([ 2,3 ]\). The root \(\alpha\) is to be estimated using the iterative formula $$x _ { n + 1 } = \sqrt { \left( 2 + \frac { 5 } { x _ { n } } \right) } , \quad x _ { 0 } = 2$$
  2. Calculate the values of \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to 4 significant figures.
  3. Prove that, to 5 significant figures, \(\alpha\) is 2.0946.

2. $\quad \mathrm { f } ( x ) = x ^ { 3 } - 2 x - 5$.
\begin{enumerate}[label=(\alph*)]
\item Show that there is a root $\alpha$ of $\mathrm { f } ( x ) = 0$ for $x$ in the interval $[ 2,3 ]$.

The root $\alpha$ is to be estimated using the iterative formula

$$x _ { n + 1 } = \sqrt { \left( 2 + \frac { 5 } { x _ { n } } \right) } , \quad x _ { 0 } = 2$$
\item Calculate the values of $x _ { 1 } , x _ { 2 } , x _ { 3 }$ and $x _ { 4 }$, giving your answers to 4 significant figures.
\item Prove that, to 5 significant figures, $\alpha$ is 2.0946.
\end{enumerate}

\hfill \mbox{\textit{Edexcel C3  Q2 [8]}}