AQA Paper 1 2024 June — Question 14 2 marks

Exam BoardAQA
ModulePaper 1 (Paper 1)
Year2024
SessionJune
Marks2
TopicFixed Point Iteration

14
  1. The equation $$x ^ { 3 } = \mathrm { e } ^ { 6 - 2 x }$$ has a single solution, \(x = \alpha\)
    By considering a suitable change of sign, show that \(\alpha\) lies between 0 and 4
    14
  2. Show that the equation \(x ^ { 3 } = \mathrm { e } ^ { 6 - 2 x }\) can be rearranged to give $$x = 3 - \frac { 3 } { 2 } \ln x$$ 14
    1. Use the iterative formula $$x _ { n + 1 } = 3 - \frac { 3 } { 2 } \ln x _ { n }$$ with \(x _ { 1 } = 4\), to find \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\)
      Give your answers to three decimal places.
      14
  3. (ii) Figure 1 below shows a sketch of parts of the graphs of $$y = 3 - \frac { 3 } { 2 } \ln x \text { and } y = x$$ On Figure 1, draw a staircase or cobweb diagram to show how convergence takes place.
    Label, on the \(x\)-axis, the positions of \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\)
    [0pt] [2 marks] \begin{figure}[h]
    \captionsetup{labelformat=empty} \caption{Figure 1} \includegraphics[alt={},max width=\textwidth]{0320e0a6-adc0-440a-b1da-d1a49fe06179-22_1328_1390_744_395}
    \end{figure} 14
  4. (iii) Explain why the iterative formula $$x _ { n + 1 } = 3 - \frac { 3 } { 2 } \ln x _ { n }$$ fails to converge to \(\alpha\) when the starting value is \(x _ { 1 } = 0\)