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1.10 Numerical Methods (A-level only)

1.10 Numerical Methods (A-level only)

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Question 67

The equation x3−4x+1=0x^3 - 4x + 1 = 0x3−4x+1=0 has a single solution, x=αx = \alphax=α, in the interval [1.8,1.9][1.8, 1.9][1.8,1.9].

a.

By considering a suitable change of sign, show that α\alphaα lies between 1.8 and 1.9.

[2]
b.

Show that the equation x3−4x+1=0x^3 - 4x + 1 = 0x3−4x+1=0 can be rearranged into the form

x=4−1x x = \sqrt{4 - \frac{1}{x}} x=4−x1​​
[2]
c.

Use the iterative formula

xn+1=4−1xn x_{n+1} = \sqrt{4 - \frac{1}{x_n}} xn+1​=4−xn​1​​

with x1=1.8x_1 = 1.8x1​=1.8, to find x2,x3x_2, x_3x2​,x3​ and x4x_4x4​, giving your answers to four decimal places.

[3]
d.

Hence, deduce an interval of width 0.001 in which α\alphaα lies.

[1]
Markscheme

1.10 Numerical Methods (A-level only) Questions

  1. A Level
  2. /Maths
  3. /1.10 Numerical Methods (A-level only)

137 exam-style questions on OCR (MEI) A Level Maths 1.10 Numerical Methods (A-level only), covering 1.10.1 Locate roots by change of sign (A-level only), 1.10.2 When change of sign methods fail (A-level only), 1.10.3 Fixed point iteration (A-level only), 1.10.4 Newton-Raphson method (A-level only), 1.10.5 Convergence of iterations (A-level only), 1.10.6 Trapezium rule (A-level only), 1.10.7 Upper and lower bounds using rectangles (A-level only), and 1.10.8 Numerical methods to solve problems (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.

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