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Exam questions · Maths · Functions, Sequences and Rates of Change

Iteration

  • 6 exam questions
  • 18 marks
  • 9 quick checks
  1. 1 Work out [2 marks]

    \(x_{n+1} = 5 - \dfrac{6}{x_n}\) and \(x_0 = 6\). Work out \(x_1\) and \(x_2\). [2 marks]

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    Model answer

    \(x_1 = 5 - \dfrac{6}{6} = 4\) and \(x_2 = 5 - \dfrac{6}{4} = 3.5\).

    Mark scheme

    • \(x_1 = 4\) — B1
    • \(x_2 = 3.5\) — B1
  2. 2 Show that [2 marks]

    Show that the equation \(x^2 - 5x + 6 = 0\) can be rearranged to give \(x = 5 - \dfrac{6}{x}\). [2 marks]

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    Model answer

    Divide every term by \(x\): \(x - 5 + \dfrac{6}{x} = 0\). Then \(x = 5 - \dfrac{6}{x}\).

    Mark scheme

    • Divides by \(x\), giving \(x - 5 + \dfrac{6}{x} = 0\) — M1
    • \(x = 5 - \dfrac{6}{x}\) — Q1
  3. 3 Work out [4 marks]

    \(x_{n+1} = 1 + \dfrac{6}{x_n}\). The values \(x_n\) tend to a positive limit \(a\). Show that \(a^2 - a - 6 = 0\), and work out the value of \(a\). [4 marks]

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    Model answer

    At the limit, \(a = 1 + \dfrac{6}{a}\), so \(a^2 = a + 6\) and \(a^2 - a - 6 = 0\). Then \((a - 3)(a + 2) = 0\), and \(a\) is positive, so \(a = 3\).

    Mark scheme

    • \(a = 1 + \dfrac{6}{a}\) — M1
    • \(a^2 - a - 6 = 0\) — M1
    • \((a - 3)(a + 2) = 0\) — M1
    • \(3\) — A1
  4. 4 Show that [4 marks]

    \(f(x) = x^3 + 2x - 5\) (a) Show that the equation \(f(x) = 0\) has a root between 1 and 2. [2 marks] (b) Work out this root to 1 decimal place. Show your working. [2 marks]

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    Model answer

    (a) \(f(1) = -2\) and \(f(2) = 7\). The sign changes, so there is a root between 1 and 2. (b) \(f(1.3) = -0.203\) and \(f(1.4) = 0.544\), and \(f(1.35) = 0.160\ldots\), which is positive, so the root is between 1.3 and 1.35, which is 1.3 to 1 decimal place.

    Mark scheme

    • (a) \(f(1) = -2\) and \(f(2) = 7\) — M1
    • (a) A change of sign, so there is a root — Q1
    • (b) \(f(1.3) < 0\), \(f(1.4) > 0\) and a test of 1.35 — M1
    • (b) \(1.3\) — A1
  5. 5 Work out [4 marks]

    Use trial and improvement to find the value of \(\sqrt{13}\) correct to 1 decimal place. Show all your working. [4 marks]

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    Model answer

    \(3.6^2 = 12.96\) and \(3.7^2 = 13.69\), so \(\sqrt{13}\) is between 3.6 and 3.7. \(3.65^2 = 13.3225\), which is more than 13, so \(\sqrt{13}\) is below 3.65. So \(\sqrt{13} = 3.6\) to 1 decimal place.

    Mark scheme

    • \(3.6^2 = 12.96\) and \(3.7^2 = 13.69\) — B1
    • Tests \(3.65\) — M1
    • \(3.65^2 = 13.3225\) — A1
    • \(3.6\) — A1
  6. 6 Explain [2 marks]

    \(x_{n+1} = 5 - \dfrac{6}{x_n}\). Explain what happens to the values of \(x_n\) if \(x_0 = 3\). [2 marks]

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    Model answer

    \(x_1 = 5 - \dfrac{6}{3} = 3\), so every term is 3. The sequence stays at 3.

    Mark scheme

    • \(x_1 = 5 - 2 = 3\) — M1
    • States that every term is 3, because the value does not change — Q1

Quick check

  1. 1

    \(x_{n+1} = 3 - \dfrac{2}{x_n}\) and \(x_0 = 4\). What is \(x_1\)?

    1. A3.5
    2. B2
    3. C0.5
    4. D2.5
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    D: 2.5

    \(3 - \dfrac{2}{4} = 3 - 0.5 = 2.5\).

  2. 2

    What does \(x_0\) mean in an iteration?

    1. AThe limit
    2. BThe last value
    3. CThe starting value
    4. DThe number of steps
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    C: The starting value

    \(x_0\) is where the iteration starts.

  3. 3

    At the limit of an iteration, which statement is true?

    1. A\(x_{n+1} = 0\)
    2. B\(x_{n+1} = x_n\)
    3. C\(x_n = 1\)
    4. D\(x_{n+1} = 2x_n\)
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    B: \(x_{n+1} = x_n\)

    At the limit the values stop changing.

  4. 4

    \(x_{n+1} = 3 - \dfrac{2}{x_n}\) and \(x_1 = 2.5\). What is \(x_2\)?

    1. A2.2
    2. B2.8
    3. C2.4
    4. D1.8
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    A: 2.2

    \(3 - \dfrac{2}{2.5} = 3 - 0.8 = 2.2\).

  5. 5

    Which equation can be rearranged to \(x = 3 - \dfrac{2}{x}\)?

    1. A\(x^2 + 3x + 2 = 0\)
    2. B\(x^2 - 2x + 3 = 0\)
    3. C\(x^2 - 3x - 2 = 0\)
    4. D\(x^2 - 3x + 2 = 0\)
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    D: \(x^2 - 3x + 2 = 0\)

    Divide \(x^2 - 3x + 2 = 0\) by \(x\) to get \(x - 3 + \dfrac{2}{x} = 0\).

  6. 6

    \(f(x) = x^3 + x - 3\). What does \(f(1) = -1\) and \(f(2) = 7\) show?

    1. AThe root is 1.5
    2. BThere is no root
    3. CA root lies between 1 and 2
    4. DThe root is 0
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    C: A root lies between 1 and 2

    The sign changes, so a root lies between them.

  7. 7

    \(x_{n+1} = 1 + \dfrac{6}{x_n}\) converges to a positive number \(a\). Which equation does \(a\) satisfy?

    1. A\(a^2 + a - 6 = 0\)
    2. B\(a^2 - a - 6 = 0\)
    3. C\(a^2 - 6a - 1 = 0\)
    4. D\(a^2 - a + 6 = 0\)
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    B: \(a^2 - a - 6 = 0\)

    \(a = 1 + \dfrac{6}{a}\), so \(a^2 = a + 6\).

  8. 8

    What is the value of \(a\) in the previous question?

    1. A3
    2. B\(-2\)
    3. C6
    4. D2
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    A: 3

    \((a - 3)(a + 2) = 0\) and \(a\) is positive.

  9. 9

    \(f(x) = x^2 - 7\). \(f(2.6) = -0.24\) and \(f(2.7) = 0.29\). Why is the root 2.6 to 1 decimal place, given \(f(2.65) = 0.0225\)?

    1. AThe root is 2.7 because \(f(2.7)\) is positive
    2. BThe root is exactly 2.65
    3. CIt cannot be rounded
    4. DThe root is below 2.65, so it rounds down to 2.6
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    D: The root is below 2.65, so it rounds down to 2.6

    \(f(2.65) > 0\) means the root is between 2.6 and 2.65.