Show root in interval

A question is this type if and only if it requires showing that f(x) = 0 has a root in a given interval using a change of sign method, typically as a preliminary step before applying Newton-Raphson.

5 questions

Edexcel Paper 1 2024 June Q3
3. $$\mathrm { f } ( x ) = x + \tan \left( \frac { 1 } { 2 } x \right) \quad \pi < x < \frac { 3 \pi } { 2 }$$ Given that the equation \(\mathrm { f } ( x ) = 0\) has a single root \(\alpha\)
  1. show that \(\alpha\) lies in the interval [3.6, 3.7]
  2. Find \(\mathrm { f } ^ { \prime } ( x )\)
  3. Using 3.7 as a first approximation for \(\alpha\), apply the Newton-Raphson method once to obtain a second approximation for \(\alpha\). Give your answer to 3 decimal places.
Edexcel Paper 1 Specimen Q8
8. $$f ( x ) = \ln ( 2 x - 5 ) + 2 x ^ { 2 } - 30 , \quad x > 2.5$$
  1. Show that \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval [3.5,4] A student takes 4 as the first approximation to \(\alpha\).
    Given \(\mathrm { f } ( 4 ) = 3.099\) and \(\mathrm { f } ^ { \prime } ( 4 ) = 16.67\) to 4 significant figures,
  2. apply the Newton-Raphson procedure once to obtain a second approximation for \(\alpha\), giving your answer to 3 significant figures.
  3. Show that \(\alpha\) is the only root of \(\mathrm { f } ( x ) = 0\)
OCR MEI Paper 1 2021 November Q8
8 Kareem wants to solve the equation \(\sin 4 x + \mathrm { e } ^ { - x } + 0.75 = 0\). He uses his calculator to create the following table of values for \(\mathrm { f } ( x ) = \sin 4 x + \mathrm { e } ^ { - x } + 0.75\).
\(x\)0123456
\(\mathrm { f } ( x )\)1.7500.3611.8750.2630.4801.670- 0.153
He argues that because \(\mathrm { f } ( 6 )\) is the first negative value in the table, there is a root of the equation between 5 and 6 .
  1. Comment on the validity of his argument. The diagram shows the graph of \(y = \sin 4 x + e ^ { - x } + 0.75\).
    \includegraphics[max width=\textwidth, alt={}, center]{4fac72cb-85cb-48d9-8817-899ef3f80a0f-07_538_1260_920_244}
  2. Explain why Kareem failed to find other roots between 0 and 6 . Kareem decides to use the Newton-Raphson method to find the root close to 3 .
    1. Determine the iterative formula he should use for this equation.
    2. Use the Newton-Raphson method with \(x _ { 0 } = 3\) to find a root of the equation \(\mathrm { f } ( x ) = 0\). Show three iterations and give your answer to a suitable degree of accuracy. Kareem uses the Newton-Raphson method with \(x _ { 0 } = 5\) and also with \(x _ { 0 } = 6\) to try to find the root which lies between 5 and 6 . He produces the following tables.
      \(x _ { 0 }\)5
      \(x _ { 1 }\)3.97288
      \(x _ { 2 }\)4.12125
      \(x _ { 0 }\)6
      \(x _ { 1 }\)6.09036
      \(x _ { 2 }\)6.07110
    1. For the iteration beginning with \(x _ { 0 } = 5\), represent the process on the graph in the Printed Answer Booklet.
    2. Explain why the method has failed to find the root which lies between 5 and 6 .
    3. Explain how Kareem can adapt his method to find the root between 5 and 6 .
OCR MEI Paper 3 Specimen Q10
10 The function \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = x ^ { 4 } + x ^ { 3 } - 2 x ^ { 2 } - 4 x - 2\).
  1. Show that \(x = - 1\) is a root of \(\mathrm { f } ( x ) = 0\).
  2. Show that another root of \(\mathrm { f } ( x ) = 0\) lies between \(x = 1\) and \(x = 2\).
  3. Show that \(\mathrm { f } ( x ) = ( x + 1 ) \mathrm { g } ( x )\), where \(\mathrm { g } ( x ) = x ^ { 3 } + a x + b\) and \(a\) and \(b\) are integers to be determined.
  4. Without further calculation, explain why \(\mathrm { g } ( x ) = 0\) has a root between \(x = 1\) and \(x = 2\).
  5. Use the Newton-Raphson formula to show that an iteration formula for finding roots of \(\mathrm { g } ( x ) = 0\) may be written $$x _ { n + 1 } = \frac { 2 x _ { n } ^ { 3 } + 2 } { 3 x _ { n } ^ { 2 } - 2 }$$ Determine the root of \(\mathrm { g } ( x ) = 0\) which lies between \(x = 1\) and \(x = 2\) correct to 4 significant figures.
AQA FP1 2015 June Q7
4 marks
7
  1. The equation \(2 x ^ { 3 } + 5 x ^ { 2 } + 3 x - 132000 = 0\) has exactly one real root \(\alpha\).
    1. Show that \(\alpha\) lies in the interval \(39 < \alpha < 40\).
    2. Taking \(x _ { 1 } = 40\) as a first approximation to \(\alpha\), use the Newton-Raphson method to find a second approximation, \(x _ { 2 }\), to \(\alpha\). Give your answer to two decimal places.
  2. Use the formulae for \(\sum _ { r = 1 } ^ { n } r ^ { 2 }\) and \(\sum _ { r = 1 } ^ { n } r\) to show that $$\sum _ { r = 1 } ^ { n } 2 r ( 3 r + 2 ) = n ( n + p ) ( 2 n + q )$$ where \(p\) and \(q\) are integers.
    1. Express \(\log _ { 8 } 4 ^ { r }\) in the form \(\lambda r\), where \(\lambda\) is a rational number.
    2. By first finding a suitable cubic inequality for \(k\), find the greatest value of \(k\) for which \(\sum _ { r = k + 1 } ^ { 60 } ( 3 r + 2 ) \log _ { 8 } 4 ^ { r }\) is greater than 106060.
      [0pt] [4 marks]