Pure Interval Bisection Only

Questions that only require applying the interval bisection method a specified number of times, without any additional methods like Newton-Raphson or linear interpolation.

5 questions

Edexcel F1 2021 January Q1
  1. (a) Show that the equation \(4 x - 2 \sin x - 1 = 0\), where \(x\) is in radians, has a root \(\alpha\) in the interval [0.2, 0.6]
    [0pt] (b) Starting with the interval [0.2, 0.6], use interval bisection twice to find an interval of width 0.1 in which \(\alpha\) lies.
    (3)
Edexcel FP1 2014 January Q1
  1. \(\mathrm { f } ( x ) = 2 x - 5 \cos x , \quad\) where \(x\) is in radians.
    1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval \([ 1,1.4 ]\).
      [0pt]
    2. Starting with the interval [1,1.4], use interval bisection twice to find an interval of width 0.1 which contains \(\alpha\).
Edexcel FP1 2011 June Q1
1. $$f ( x ) = 3 ^ { x } + 3 x - 7$$
  1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) between \(x = 1\) and \(x = 2\).
  2. Starting with the interval \([ 1,2 ]\), use interval bisection twice to find an interval of width 0.25 which contains \(\alpha\).
AQA FP1 2007 June Q2
2
  1. Show that the equation $$x ^ { 3 } + x - 7 = 0$$ has a root between 1.6 and 1.8.
  2. Use interval bisection twice, starting with the interval in part (a), to give this root to one decimal place.
Edexcel FP1 Q12
12. $$f ( x ) = 3 ^ { x } - x - 6$$
  1. Show that \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) between \(x = 1\) and \(x = 2\).
  2. Starting with the interval (1,2), use interval bisection three times to find an interval of width 0.125 which contains \(\alpha\).
    [0pt] [*P4 June 2003 Qn 4]