A Level Maths: Pure 3 chapter 6: Numerical solution of equations MCQs with answers

By Team Megalecture · Updated 4 Nov 2026

23 exam-style questions on Numerical solution of equations, each with the answer and a short explanation. Ten of them are below to try. The chapter test draws ten fresh questions each time and is marked the moment you finish, with your score by topic.

23 questions · 1 topics · 10 per test

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What chapter 6 covers

Ten Numerical solution of equations questions with answers

Try each one, then open the answer. These ten stay the same; the test draws new ones every time.

1. The equation ln x = 3 − x has one root, α. Which of these intervals contains α?

  1. A
    2.1 < α < 2.2
  2. B
    2.3 < α < 2.4
  3. C
    2.0 < α < 2.1
  4. D
    2.2 < α < 2.3
Show answer

Answer: D. Let f(x) = ln x + x − 3. f(2.2) = −0.012 and f(2.3) = 0.133, so f changes sign between 2.2 and 2.3. f is negative at 2.0, 2.1 and 2.2 and positive at 2.3 and 2.4, so there is no change of sign in the other intervals.

2. The equation x2 − 5x + 6 = 0 has roots 2 and 3. The iterative formula xn+1 = (xn2 + 6)/5 is used with x1 = 2.8. What happens?

  1. A
    The values converge to 2
  2. B
    The values converge to 3
  3. C
    The values converge to 2.8
  4. D
    The values diverge
Show answer

Answer: A. x2 = 2.768, x3 = 2.7324, x4 = 2.6932, … The values keep decreasing and settle at 2. Starting near a root does not guarantee convergence to that root: near 3 this formula pushes values away.

3. The curve y = x cos x, for 0 < x < ½π, has a maximum point where x = α. Setting dy/dx = 0 shows that α = tan−1(1/α). Use the iterative formula xn+1 = tan−1(1/xn), with x1 = 1, to find α correct to 2 decimal places. (x is in radians.)

  1. A
    0.79
  2. B
    0.91
  3. C
    0.87
  4. D
    0.86
Show answer

Answer: D. dy/dx = cos x − x sin x = 0 gives x tan x = 1, which is where the formula comes from. The iterates are 1, 0.7854, 0.9050, 0.8352, 0.8749, 0.8520, 0.8651, 0.8576, 0.8619, 0.8594, 0.8609, 0.8600, 0.8605, …, jumping from side to side of the root and settling at 0.8603…, so α = 0.86. The other values are early iterates: keep going until successive values agree to 2 decimal places.

4. The sequence given by xn+1 = ln(4 + xn), x1 = 2, converges to α. Which equation has α as a root?

  1. A
    ex = 4x
  2. B
    ex = x + 4
  3. C
    x = ex + 4
  4. D
    ln x = x + 4
Show answer

Answer: B. In the limit, xn+1 and xn both equal α, so α = ln(4 + α), i.e. eα = α + 4. The limit is a root of x = F(x), which here rearranges to ex = x + 4.

5. The equation ex = 3x has two real roots. Which interval contains the smaller root?

  1. A
    0.5 < x < 0.6
  2. B
    0.7 < x < 0.8
  3. C
    0.6 < x < 0.7
  4. D
    1.5 < x < 1.6
Show answer

Answer: C. Let f(x) = ex − 3x. f(0.6) = 0.022 > 0 and f(0.7) = −0.086 < 0, so the smaller root lies between 0.6 and 0.7. There is also a change of sign between 1.5 and 1.6, but that is the larger root.

6. An iteration xn+1 = F(xn) gives x1 = 1.5, x2 = 1.7333, x3 = 1.5538, x4 = 1.6871, x5 = 1.5854. How is the sequence behaving?

  1. A
    It is converging, with values increasing steadily towards the root
  2. B
    It is diverging, with values moving further away from the root
  3. C
    It is swapping between two fixed values and not converging
  4. D
    It is converging, with values alternately above and below the root
Show answer

Answer: D. The values go up, down, up, down, and the jumps shrink (0.23, 0.18, 0.13, 0.10), so they close in on the root from alternate sides: a cobweb pattern on a diagram.

7. Which iterative formula is based on a correct rearrangement of the equation 2x3 − 6x − 1 = 0?

  1. A
    xn+1 = ∛(6xn + 1)
  2. B
    xn+1 = (2xn3 + 1)/6
  3. C
    xn+1 = ∛(3xn + ½)
  4. D
    xn+1 = ∛(3xn + 1)
Show answer

Answer: C. 2x3 = 6x + 1 gives x3 = 3x + ½, so x = ∛(3x + ½). Forgetting to divide the 1 by 2 gives ∛(3x + 1); making x the subject of 6x = 2x3 − 1 gives (2x3 − 1)/6, not (2x3 + 1)/6.

8. f(x) = 1/(x − 2). A student notes that f(1) = −1 and f(3) = 1 and says that f(x) = 0 has a root between 1 and 3. Which statement is correct?

  1. A
    The claim is right: a change of sign between two values always shows a root
  2. B
    The claim is right: the root is at x = 2, which is where the sign of f changes
  3. C
    The claim is wrong: f(1) + f(3) = 0, so the two values cancel each other out
  4. D
    The claim is wrong: f is not continuous at x = 2, so the sign change proves nothing
Show answer

Answer: D. 1/(x − 2) is never zero. Its sign changes because the graph has a vertical asymptote at x = 2, not because it crosses the x-axis. The sign-change method only works for a function that is continuous on the interval.

9. The iteration xn+1 = e−xn, with x1 = 0.5, gives these values to 4 decimal places: 0.6065, 0.5452, 0.5797, 0.5601, 0.5712, 0.5649, 0.5684, 0.5664, 0.5676, 0.5669. State the root correct to 2 decimal places.

  1. A
    0.57
  2. B
    0.56
  3. C
    0.61
  4. D
    0.58
Show answer

Answer: A. The values are closing in on a number between 0.5664 and 0.5676, and every number in that range rounds to 0.57. 0.56 comes from truncating, and 0.61 or 0.58 come from stopping too early.

10. Use the iterative formula xn+1 = ∛(7 − xn), with x1 = 2, to find the root of x3 + x − 7 = 0 correct to 2 decimal places.

  1. A
    1.71
  2. B
    1.74
  3. C
    1.73
  4. D
    2.09
Show answer

Answer: B. The iterates are 2, 1.7100, 1.7424, 1.7388, 1.7392, 1.7392, …, so the root is 1.74 to 2 d.p. 1.71 is only the second iterate, 1.73 comes from truncating 1.739 instead of rounding, and 2.09 comes from the sign slip ∛(7 + xn).

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Numerical solution of equations: questions students ask

How many Numerical solution of equations MCQs are there?

23, each with the answer and a short explanation, across 1 topics: Numerical solution of equations. Every chapter test draws ten of them, so you can take it many times.

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The moment you press Finish: your score, the right answer and an explanation for every question, and the ones you got wrong saved in My mistakes to retry.

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