OxfordAQA International AS and A-level Mathematics: Numerical methods: roots, iteration, mid-ordinate rule and Simpson's rule (9660) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section P2.9 (Numerical methods) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. The specification places this content in International A-level (A2) Unit P2 (Pure Maths); it is not examined at International AS. Learn the methods from the study guide and revision notes. Course pages: hub and checklist. Unit P2 is a calculator-allowed unit, so use one, but show your working.

Questions

1. Show that the equation x⁴ + 3x − 9 = 0 has a root between x = 1.4 and x = 1.5. [3]

2. Let f(x) = 1/(x² − 4). A student finds f(1) = −1/3 and f(3) = 1/5 and concludes that f(x) = 0 has a root between 1 and 3.

(a) Calculate f(1) and f(3) to confirm the student’s values. [1] (b) Explain why the conclusion is wrong, and show that f(x) = 0 has no solution at all. [2]

3. Show that the root of x⁵ + x − 12 = 0 is 1.60 correct to 2 decimal places. [3]

4. The equation x³ + 5x − 8 = 0 has one real root, α.

(a) Show that the equation can be rearranged to x = (8 − x³)/5. [1] (b) Use xₙ₊₁ = (8 − xₙ³)/5 with x₁ = 1 to find x₂, x₃ and x₄, giving x₄ to 3 decimal places. [2] (c) A second rearrangement is xₙ₊₁ = ∛(8 − 5xₙ). For this g, g′(α) ≈ −1.10. Explain what this tells you about using it to find α. [2]

5. A sequence is defined by xₙ₊₁ = 6/(xₙ + 1) with x₁ = 1.

(a) Find x₂, x₃ and x₄. [2] (b) The sequence converges to a limit α. Show that α² + α − 6 = 0 and find α. [2] (c) Find g′(α) for g(x) = 6/(x + 1), and say whether the diagram is a staircase or a cobweb. [2]

6. Let y = √(x³ + 1).

(a) Use the mid-ordinate rule with 4 strips to estimate ∫₀² y dx, giving your answer to 3 decimal places. [4] (b) The value of the integral is 3.2413 to 4 decimal places. The curve is convex for 0 < x ≤ 2. Explain, using the geometry of the rule, why your estimate is below this value. [2]

7. (a) Use Simpson’s rule with 4 strips to estimate ∫₀² √(x³ + 1) dx, giving your answer to 3 decimal places. [4] (b) Explain why Simpson’s rule cannot be used with 3 strips. [1]

8. Water flows into a tank at a rate r litres per minute. Readings are taken every 2 minutes.

t (minutes)	0	2	4	6	8
r	3.2	4.1	5.6	5.0	3.8

Use Simpson’s rule to estimate the total volume of water that flows in during the 8 minutes. [4]

9. Consider ∫₁⁴ 1/x² dx, whose exact value is 3/4.

(a) Use the mid-ordinate rule with 3 strips to estimate the integral to 3 decimal places. [3] (b) Explain, using a tangent at a midpoint, why your answer is lower than 3/4. [1] (c) Use 6 strips to improve the estimate, giving your answer to 3 decimal places. [2]

10. (a) Use Simpson’s rule with 2 strips to estimate ∫₁³ ln x dx, giving your answer to 4 decimal places. [3] (b) Repeat with 4 strips. [3] (c) The exact value of the integral is 1.2958 to 4 decimal places. Comment on what the two estimates show. [1]

11. The curve C has equation y = 4/(1 + x²). The line y = x meets C at the point where x = α.

(a) Show that α satisfies x³ + x − 4 = 0, and that 1.3 < α < 1.4. [3] (b) Use xₙ₊₁ = ∛(4 − xₙ) with x₁ = 1 to find α correct to 3 decimal places. [4] (c) The region R is bounded by C, the line y = x and the y-axis. Use Simpson’s rule with 4 strips on ∫₀^1.4 y dx, and use 1.4 as the upper limit in place of α, to estimate the area of R to 2 decimal places. [4]

Answers

1. Let f(x) = x⁴ + 3x − 9. f(1.4) = 3.8416 + 4.2 − 9 = −0.9584 [1]. f(1.5) = 5.0625 + 4.5 − 9 = 0.5625 [1]. f is continuous (polynomial) and the sign changes, so a root lies between 1.4 and 1.5 [1]. [3] Examiner insight: Give both values, the sign change and a word on continuity; a bare “yes” with no values shows nothing.

2. (a) f(1) = 1/(1 − 4) = −1/3 and f(3) = 1/(9 − 4) = 1/5 [1]. (b) f is not continuous at x = 2, where x² − 4 = 0, and 2 lies between 1 and 3, so the sign change comes from a jump, not a crossing [1]. A fraction 1/(x² − 4) with numerator 1 is never zero, so f(x) = 0 has no solution [1]. [3] Examiner insight: A critique must name the failed condition (continuity) and the location (x = 2); “the test is unreliable” is too vague to gain credit.

3. Let f(x) = x⁵ + x − 12. f(1.595) = −0.0821 [1]. f(1.605) = 0.2556 [1]. f is continuous and changes sign, so the root is in (1.595, 1.605) and equals 1.60 to 2 d.p. [1]. [3] Examiner insight: Testing 1.6 alone shows nothing about rounding; the two bounds that round to 1.60 are what justify the decimal places.

4. (a) x³ + 5x − 8 = 0 gives 5x = 8 − x³, so x = (8 − x³)/5 [1]. (b) x₂ = (8 − 1)/5 = 1.4 [1]. x₃ = (8 − 2.744)/5 = 1.0512 and x₄ = (8 − 1.1616)/5 = 1.3677, so x₄ = 1.368 [1]. (c) |g′(α)| ≈ 1.10 > 1 [1], so the iteration does not converge: the cobweb spirals outwards, and this rearrangement cannot be used to find α [1]. [5] Examiner insight: Write the value of g′(α) beside the comparison with 1 (or −1); a statement of “diverges” with no reason is unsupported.

5. (a) x₂ = 6/2 = 3 [1]; x₃ = 6/4 = 1.5 and x₄ = 6/2.5 = 2.4 [1]. (b) At the limit α = 6/(α + 1), so α² + α = 6, giving α² + α − 6 = 0 [1]. (α + 3)(α − 2) = 0. The terms are positive, so α = 2 [1]. (c) g′(x) = −6/(x + 1)², so g′(2) = −6/9 = −2/3 [1]. It lies between −1 and 0, so the iteration converges as a cobweb spiralling inwards [1]. [6] Examiner insight: Reject the second root of the quadratic with a reason (the iterates are all positive), not silently.

6. (a) h = 2/4 = 0.5 and the midpoints are 0.25, 0.75, 1.25, 1.75 [1]. Heights: 1.0078, 1.1924, 1.7185, 2.5218 [1]. Sum = 6.4405, so estimate = 0.5 × 6.4405 [1] = 3.220 [1]. (b) On each strip the rectangle has the same area as the trapezium under the tangent at the midpoint [1]. The curve is convex, so the tangent lies below it and each strip is underestimated: 3.220 < 3.2413 [1]. [6] Examiner insight: Show the table of midpoint heights; a slip in one value is then easy to locate, and the explanation needs “tangent below the curve”, not just “convex”.

7. (a) h = 0.5; ordinates at x = 0, 0.5, 1, 1.5, 2 are 1, 1.0607, 1.4142, 2.0917, 3 [1]. Weights: ends once, odd positions ×4, even interior ×2 [1]. (0.5/3)[1 + 3 + 4(1.0607 + 2.0917) + 2(1.4142)] [1] = 3.240 [1]. (b) Simpson’s rule needs an even number of strips (an odd number of ordinates) [1]. [5] Examiner insight: Keep at least four decimal places in the ordinates; rounding them to 2 d.p. can shift the third decimal place of the answer.

8. h = 2 [1]. Volume ≈ (2/3)[3.2 + 3.8 + 4(4.1 + 5.0) + 2(5.6)] [1] = (2/3)(54.6) [1] = 36.4 litres [1]. [4] Examiner insight: The integral of a rate is an amount, so state the unit; the answer is the area under the r–t graph, not an average rate.

9. (a) h = 1 and the midpoints are 1.5, 2.5, 3.5 [1]. Heights 0.4444, 0.16, 0.0816 [1]. Sum × h = 0.686 [1]. (b) The curve is convex, so the tangent at each midpoint lies below the curve and the rule underestimates [1]. (c) h = 0.5, midpoints 1.25, 1.75, 2.25, 2.75, 3.25, 3.75 [1]. Estimate = 0.5 × (0.64 + 0.3265 + 0.1975 + 0.1322 + 0.0947 + 0.0711) = 0.731 [1]. [6] Examiner insight: A statement that more strips improve the estimate is best backed by the numbers: 0.686 and 0.731 move towards 0.75.

10. (a) h = 1; ordinates ln 1 = 0, ln 2 = 0.6931, ln 3 = 1.0986 [1]. (1/3)[0 + 1.0986 + 4(0.6931)] [1] = 1.2904 [1]. (b) h = 0.5; ordinates 0, 0.4055, 0.6931, 0.9163, 1.0986 [1]. (0.5/3)[0 + 1.0986 + 4(0.4055 + 0.9163) + 2(0.6931)] [1] = 1.2953 [1]. (c) The errors are about 0.0054 and 0.0005, so doubling the strips gave a much closer value [1]. [7] Examiner insight: A comparison needs a quantity (the size of each error); “the second is better” alone says little.

11. (a) The curves meet where 4/(1 + x²) = x, so 4 = x + x³ and x³ + x − 4 = 0 [1]. Let f(x) = x³ + x − 4: f(1.3) = −0.503 and f(1.4) = 0.144 [1]. f is continuous and changes sign, so 1.3 < α < 1.4 [1]. (b) x³ = 4 − x gives x = ∛(4 − x) [1]. x₂ = 1.4422 and x₃ = 1.3676 [1]. x₄ = 1.3808, x₅ = 1.3785, x₆ = 1.3789 [1]. x₇ = 1.3788 so α = 1.379 [1]. (c) h = 1.4/4 = 0.35; ordinates 4, 3.5635, 2.6846, 1.9025, 1.3514 [1]. (0.35/3)[4 + 1.3514 + 4(3.5635 + 1.9025) + 2(2.6846)] = 3.8015 [1]. Area under the line to 1.4 is ½ × 1.4² = 0.98 [1]. Area of R ≈ 3.8015 − 0.98 = 2.82 [1]. [11] Examiner insight: Carry unrounded values through an iteration, and when a question substitutes a convenient limit (here 1.4 for α) say so, so the answer is read as an estimate.

Where marks are usually lost

• Writing “sign change, so root” without continuity, or accepting a sign change across an asymptote.
• Testing only the rounded value when asked to justify a number of decimal places.
• Rounding iterates early, so the final values drift.
• Quoting convergence from the value of g′ at the wrong point, or without comparing with 1.
• Using the strip edges as the heights in the mid-ordinate rule.
• Applying Simpson’s rule to an odd number of strips, or mixing up the 4 and 2.
• Forgetting the factor h/3 in Simpson’s rule, or h in the mid-ordinate rule.
• Giving an answer to the wrong number of decimal places or without units.
• Calling an estimate an over- or underestimate with no geometric reason.

Next steps

• Re-read the revision notes and the study guide.
• Course pages: hub and checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2, published by OxfordAQA. This content is listed under P2.9 in International A-level Unit P2 (Pure Maths).