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Iterative formulae, roots and growth models

3 exam-style questions. Answers and the mark for each step are on the last page.

  1. Question 1Calculator · 3 marks

    (a) Use the iteration xₙ₊₁ = \sqrt{}(10 + xₙ), starting with x₀ = 3. Work out x₃ to 3 decimal places. (3)

  2. Question 2Calculator · 5 marks

    The iteration is xn+1=10+xnx_{n+1}=\sqrt{10+x_n} with x0=3x_0=3.

    (a) Calculate x4, keeping unrounded values until giving your answer to 3 decimal places. (3)

    (b) Write the quadratic equation satisfied by a positive fixed point. (2)

  3. Question 3Calculator · 5 marks

    The iteration formula xn+1=3+4xnx_{n+1} = 3 + \dfrac{4}{x_n} is used with x1=1x_1 = 1.

    (a) Work out x4x_4. Give your answer to 2 decimal places. (2)

    (b) The values of xnx_n get closer to a number LL. Show that L=4L = 4. (3)

Answers and marks

Question 1

(a) 3.7003.700

  • P1 Establishing 13\sqrt{13} or an equivalent valid method.
  • P1 Establishing 10+13\sqrt{10+\sqrt{13}} or an equivalent valid method.
  • A1 Correct answer: 3.7003.700

Question 2

(a) 3.7013.701

  • M1 Use the previous output as the next input; first compute x1.
  • M1 The next values are 3.688570357 and 3.699806800; use the third for the fourth update.
  • A1 Correct answer: 3.7013.701

(b) x2−x−10=0x^{2}-x-10=0

  • M1 At a fixed point the input and next output are equal.
  • A1 Correct answer: x2−x−10=0x^{2}-x-10=0

Question 3

(a) 4.124.12

  • M1 Finding x2=7x_2 = 7 and x3=3.571…x_3 = 3.571\ldots.
  • A1 The correct answer, 4.124.12.

(b) L=3+4L⇒L2−3L−4=0⇒L=4L = 3 + \frac{4}{L} \Rightarrow L^2 - 3L - 4 = 0 \Rightarrow L = 4 (positive)

  • P1 Writing L=3+4LL = 3 + \frac{4}{L}.
  • P1 Forming L2−3L−4=0L^2 - 3L - 4 = 0.
  • C1 Solving to L=4L = 4 or −1-1 and rejecting −1-1 because every iterate is positive.

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Iterative formulae, roots and growth models

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Independent practice for Pearson Edexcel GCSE Mathematics (1MA1), not endorsed by Pearson.

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