Cambridge International AS & A Level Mathematics 9709: Pure Mathematics 2 Numerical Solution of Equations – 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 2.6, Numerical solution of equations, of the Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027 (Version 4). Section 2.6 is part of Pure Mathematics 2 and is examined on Paper 2, the AS Level Pure Mathematics route; the same outcomes appear as section 3.6 for Paper 3. A scientific calculator is allowed on every 9709 paper, so all questions are calculator allowed, but unsupported calculator answers earn no marks. Use radians throughout. Questions 8 and 9 combine this topic with differentiation (2.4) and integration (2.5).

Related: the study guide, the revision notes, the whole-paper Pure Mathematics 2 practice set, the A Level Mathematics hub, the printable 9709 checklist and the free AS diagnostic.

Questions

1. Show by calculation that the equation e^(0.5x) = 7 − x has a root between x = 2 and x = 3. [2]

2.

(a) By sketching a suitable pair of graphs on the same diagram, show that the equation 5 − x² = e^(−x) has exactly two real roots. [2] (b) Find the pair of consecutive integers between which the negative root lies. [2]

3. It is given that f(x) = (2x + 1)/(x − 2). A student finds that f(1) = −3 and f(3) = 7 and concludes that the equation f(x) = 0 has a root between 1 and 3.

(a) Explain why the student’s reasoning is not valid. [1] (b) State the only root of f(x) = 0. [1]

4. The sequence of values given by the iterative formula xₙ₊₁ = 3xₙ/4 + 5/xₙ³, with initial value x₁ = 2, converges to α.

(a) Find x₂ and x₃, giving each correct to 4 decimal places. [2] (b) Find the exact value of α. [3]

5. The sequence of values given by the iterative formula xₙ₊₁ = ln(2xₙ + 5), with initial value x₁ = 2, converges to α.

(a) Show that α is a root of the equation eˣ − 2x − 5 = 0. [2] (b) Use the iterative formula to find α correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

6. The equation x⁴ − 2x − 7 = 0 has one positive root α. Show by calculation that α = 1.80 correct to 2 decimal places. [3]

7. The equation x ln x = 5 has one root, α.

(a) Show by calculation that α lies between 3 and 4. [2] (b) The equation can be rearranged as x = e^(5/x). Using the iterative formula xₙ₊₁ = e^(5/xₙ) with x₁ = 3.5, find x₂, x₃, x₄ and x₅ to 4 decimal places, and describe what happens. [2] (c) Show that if the sequence given by xₙ₊₁ = ½(xₙ + 5/ln xₙ) converges, then its limit is a root of x ln x = 5. [2] (d) Use the formula in part (c) with x₁ = 3.5 to find α correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

8. The curve y = x cos x, for 0 < x < ½π, has a maximum point at x = α.

(a) Find dy/dx and show that α satisfies the equation α = tan⁻¹(1/α). [3] (b) Show by calculation that α lies between 0.8 and 0.9. [2] (c) Use the iterative formula xₙ₊₁ = tan⁻¹(1/xₙ) with x₁ = 0.85 to find α correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

9. The positive constant a is such that ∫₀ᵃ (3 − 2/(x + 1)) dx = 4.

(a) Show that a = (4 + 2 ln(a + 1))/3. [3] (b) Show by calculation that a lies between 2 and 3. [2] (c) Use an iterative formula based on the equation in part (a), with a₁ = 2, to find a correct to 3 decimal places. Give the result of each iteration to 4 decimal places. [3]

10. The equation x³ − 3x² − 5 = 0 has one real root, α.

(a) Show that the equation can be written in the form x = 3 + 5/x². [1] (b) Use the iterative formula xₙ₊₁ = 3 + 5/xₙ² with x₁ = 3.5 to find α correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

Answers

1. Let f(x) = e^(0.5x) + x − 7. f(2) = e + 2 − 7 = −2.282 and f(3) = e^1.5 + 3 − 7 = 0.482 [1]. f is continuous and there is a sign change, so a root lies between 2 and 3 [1] Examiner insight: the second mark needs both the values (or clear signs) and an explicit conclusion; comparing e^(0.5x) with 7 − x at each end is equally acceptable if the change in which side is larger is stated.

2. (a) y = 5 − x²: an inverted parabola with maximum (0, 5), crossing the x-axis at ±√5; y = e^(−x): a decreasing curve through (0, 1) with asymptote y = 0 as x → ∞ [1]. The curves cross once for x > 0 and once for x < 0, so there are exactly two roots [1] (b) Let f(x) = 5 − x² − e^(−x). f(−1) = 4 − e = 1.282 and f(−2) = 1 − e² = −6.389 [1]. Sign change, so the negative root lies between −2 and −1 [1] Examiner insight: in (a) the sketch alone is not enough; the mark for “exactly two” needs a statement linking the number of intersections to the number of roots.

3. (a) f is not continuous between 1 and 3: it has a vertical asymptote at x = 2, so the sign change comes from the graph jumping across the asymptote, not crossing the axis [1] (b) f(x) = 0 when 2x + 1 = 0, so x = −½ [1] Examiner insight: “f is not defined at x = 2” or “there is an asymptote at x = 2” earns (a); saying only “the answer is wrong” does not.

4. (a) x₂ = 1.5 + 5/8 = 2.1250 [1]; x₃ = 2.1148 [1] (b) α = 3α/4 + 5/α³ [1]. α/4 = 5/α³, so α⁴ = 20 [1]. The iterates are positive, so α = ⁴√20 [1] Examiner insight: “exact value” means ⁴√20 (or 20^(1/4)); quoting 2.1147 from repeated iteration earns no marks in (b).

5. (a) At the limit, α = ln(2α + 5) [1]. So e^α = 2α + 5, which gives e^α − 2α − 5 = 0 [1] (b) Using the formula correctly at least once, x₂ = 2.1972 [1]. x₃ = 2.2401, x₄ = 2.2492, x₅ = 2.2511 [1]. x₄ and x₅ both round to 2.25, so α = 2.25 [1] Examiner insight: the final mark needs enough iterations to show two successive values agreeing to 2 d.p.; stopping at x₄ = 2.2492 is not sufficient, even though it rounds correctly.

6. Let f(x) = x⁴ − 2x − 7. Consider f(1.795) and f(1.805) [1]. f(1.795) = −0.209 and f(1.805) = 0.0047 [1]. Sign change and f is continuous, so 1.795 < α < 1.805 and α = 1.80 to 2 d.p. [1] Examiner insight: testing f(1.79) and f(1.81) does not show the root rounds to 1.80; the interval must be the rounding bounds 1.795 and 1.805.

7. (a) Let g(x) = x ln x − 5. g(3) = 3 ln 3 − 5 = −1.704 and g(4) = 4 ln 4 − 5 = 0.545 [1]. Sign change and g is continuous for x > 0, so 3 < α < 4 [1] (b) x₂ = 4.1727, x₃ = 3.3143, x₄ = 4.5204, x₅ = 3.0225 [1]. The iterates jump either side of α by increasing amounts, so the sequence does not converge [1] (c) If the limit is α, then α = ½(α + 5/ln α) [1]. So 2α = α + 5/ln α, α = 5/ln α, and α ln α = 5 [1] (d) x₂ = 3.7456 [1]. x₃ = 3.7659, x₄ = 3.7683 [1]. x₃ and x₄ both round to 3.77, so α = 3.77 [1] Examiner insight: in (b) the mark for the description needs a reason drawn from the values (moving further from α, or not settling); “it doesn’t work” with no reference to the iterates earns nothing.

8. (a) Product rule: dy/dx = (1)cos x + x(−sin x) [1] = cos x − x sin x [1]. At the maximum, cos α = α sin α, so tan α = 1/α and α = tan⁻¹(1/α) [1] (b) Let f(x) = cos x − x sin x. f(0.8) = 0.123 and f(0.9) = −0.083 [1]. Sign change and f is continuous, so 0.8 < α < 0.9 [1] (c) x₂ = 0.8663 [1]. x₃ = 0.8569, x₄ = 0.8623, x₅ = 0.8592 [1]. x₃, x₄ and x₅ all round to 0.86, so α = 0.86 [1] Examiner insight: the product-rule mark in (a) is lost by writing dy/dx = −sin x, and every value in (b) and (c) is wrong if the calculator is in degrees, so no accuracy marks can follow.

9. (a) ∫ (3 − 2/(x + 1)) dx = 3x − 2 ln(x + 1) [1]. Using limits: 3a − 2 ln(a + 1) − (0 − 2 ln 1) = 3a − 2 ln(a + 1) [1]. Setting this equal to 4: 3a = 4 + 2 ln(a + 1), so a = (4 + 2 ln(a + 1))/3 [1] (b) Let g(a) = 3a − 2 ln(a + 1) − 4. g(2) = −0.197 and g(3) = 2.227 [1]. Sign change and g is continuous for a > −1, so 2 < a < 3 [1] (c) aₙ₊₁ = (4 + 2 ln(aₙ + 1))/3; a₂ = 2.0657 [1]. a₃ = 2.0802, a₄ = 2.0833, a₅ = 2.0840, a₆ = 2.0842 [1]. a₅ and a₆ both round to 2.084, so a = 2.084 [1] Examiner insight: this is a “show that” in (a), so the lower limit must be seen to give 2 ln 1 = 0; the given answer earns nothing if it appears without the integral.

10. (a) Divide x³ − 3x² − 5 = 0 by x² (x ≠ 0): x − 3 − 5/x² = 0, so x = 3 + 5/x² [1] (b) x₂ = 3.4082 [1]. x₃ = 3.4305, x₄ = 3.4249, x₅ = 3.4263, x₆ = 3.4259 [1]. x₅ and x₆ both round to 3.43, so α = 3.43 [1] Examiner insight: x₃ rounds to 3.43 but x₄ rounds to 3.42, so stopping at x₃ is premature; the accuracy mark needs the pair x₅, x₆.

Where marks are usually lost

• Sign-change answers that give values of f but no conclusion, or no mention of the sign change.
• A sign change claimed across a vertical asymptote, as in question 3.
• Rounding bounds wrong when confirming a root to 2 d.p.: the interval is c ± 0.005, not c ± 0.01.
• Iterates retyped from rounded values instead of using the calculator’s stored answer.
• Stopping after the first iterate that rounds to the answer, before two successive iterates agree.
• Degree mode used in a question with sin, cos or tan, as in question 8.
• “Exact value” questions answered with a decimal from the calculator.
• A non-converging iteration described vaguely, with no reference to the values found.
• “Show that” rearrangements with steps missing, especially the evaluation of limits in question 9.

Next steps

• Recap the methods in the revision notes.
• Re-read the worked examples in the study guide.
• Mix this topic with the rest of the paper in the Pure Mathematics 2 practice set.
• See every Paper 2 topic on the A Level Mathematics hub and tick them off on the printable 9709 checklist.
• Take the free AS diagnostic or use the 9709 self-check bank.
• Book a free trial class.

Official syllabus

Cambridge International AS & A Level Mathematics 9709 syllabus, for exams in 2026 and 2027 (Version 4), Cambridge University Press & Assessment. Topic 2, Pure Mathematics 2 (for Paper 2): section 2.6 Numerical solution of equations.