OCR A Level Mathematics A: Numerical methods: locating roots, iteration, Newton-Raphson and numerical integration (H240) – 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 follow section 1.09, Numerical Methods, of the OCR A Level Mathematics A specification (H240, Version 3.1, August 2026, for first assessment in 2018). All of 1.09 is Stage 2, so it is A Level only. Read the study guide first and keep the revision notes beside you. The course page is OCR A Level Mathematics and the printable list is the checklist. The free diagnostics are a good warm-up. Angles are in radians.

Questions

1. The equation x + ln x = 4 has one root, α. (a) Show that α lies between 2 and 3. [2] (b) Show that α = 2.93 correct to 2 decimal places. [2]

2. f(x) = (3x − 4)/(x − 2). (a) Show that f(1) and f(3) are both positive, but that f(x) = 0 has a root between 1 and 3. [2] (b) A student evaluates f(1.9) and f(2.1) and says “there is a sign change, so there is a root between 1.9 and 2.1”. Explain the error. [2]

3. The equation x³ + 4x − 10 = 0 has a root α near 1.56. (a) Show that the equation can be written x = 10/(x² + 4). [1] (b) Use xₙ₊₁ = 10/(xₙ² + 4) with x₁ = 1.5 to find x₂, x₃ and x₄, correct to 4 d.p.. [2] (c) Find g′(x) for g(x) = 10/(x² + 4). Use g′(α) to explain what the iteration does and describe the cobweb or staircase diagram. [3]

4. An iteration is xₙ₊₁ = (xₙ² + 6)/5. (a) Show that the limit of a convergent sequence must be 2 or 3. [1] (b) Find g′(x) and decide which of the two values the iteration can approach. [3] (c) With x₁ = 2.5, find x₂ and x₃ and describe the diagram. [1]

5. f(x) = eˣ + 2x − 6. Use Newton-Raphson with x₁ = 1.5 to find x₂ and x₃, correct to 4 d.p.. [4]

6. f(x) = x³ − 3x + 3. (a) Find f′(x) and explain why Newton-Raphson cannot be used with x₁ = 1. [2] (b) Use x₁ = 0 to find x₂, and explain why the method then fails. [2] (c) Use x₁ = −2 to find x₂ and x₃ correct to 4 d.p.. [2]

7. I = ∫ (1/x) dx from 1 to 5. (a) Estimate I by applying the trapezium rule to 4 strips. [3] (b) State, with a reason, whether your answer is an overestimate or an underestimate. [2] (c) With the same 4 strips, use rectangles to give upper and lower bounds on I. [3]

8. A river is 12 m wide. Its depth, in metres, is measured every 2 m across: 0, 1.1, 1.9, 2.4, 2.2, 1.4, 0. (a) Use the trapezium rule to estimate the cross-sectional area. [3] (b) The water flows at 0.8 m per second. Estimate the volume passing per second. [1] (c) Suggest how to make the area estimate more reliable. [1]

9. I = ∫ sin x dx from 0 to π/2. (a) Find the trapezium rule estimate with 2 strips, in exact form and correct to 4 d.p.. [3] (b) Find the estimate with 4 strips, correct to 4 d.p.. [2] (c) Given that I = 1, explain why both estimates are too small, and comment on the change in the error. [2]

10. The water level h metres in a reservoir, t hours after 06:00, is modelled by h = 3 + 0.2t + 1.5 sin(0.5t). A warning is raised when h first reaches 5 metres. (a) Show that the warning time satisfies f(t) = 0, where f(t) = 0.2t + 1.5 sin(0.5t) − 2, and that it lies between t = 2 and t = 3. [2] (b) The equation is rearranged as t = 2 arcsin[(2 − 0.2t)/1.5]. The gradient of the right-hand side is −0.267/√(1 − [(2 − 0.2t)/1.5]²). Show that this iteration will not converge to the root. [3] (c) Use Newton-Raphson with t₁ = 3 to find t₂ and t₃, correct to 4 d.p.. [2] (d) Show that the root is 2.69 to 2 decimal places, and give the clock time of the warning to the nearest minute. [3]

Answers

1. (a) Let f(x) = x + ln x − 4. f(2) = 2 + 0.6931 − 4 = −1.3069 [1]. f(3) = 3 + 1.0986 − 4 = 0.0986. The function is continuous and its sign changes, which gives 2 < α < 3 [1]. (b) f(2.925) = −0.0017 [1] and f(2.935) = 0.0117. The sign change means 2.925 < α < 2.935, so α = 2.93 to 2 decimal places [1]. Examiner insight: Quote the values and say that f is continuous; a part (b) that tests only 2.93 itself does not show which way the rounding goes.

2. (a) f(1) = (−1)/(−1) = 1 and f(3) = 5/1 = 5, both positive [1]. f(4/3) = (4 − 4)/(−2/3) = 0, so 4/3 is a root in the interval [1]. (b) f has a vertical asymptote at x = 2, so it is not continuous on [1.9, 2.1] [1]. The sign change is the jump from −17 (at 1.9) to +23 (at 2.1), not a crossing of the axis, and there is no root there [1]. Examiner insight: Naming the asymptote and the failed continuity condition is the explanation; “the function is undefined” alone does not say why a sign change was misleading.

3. (a) x³ + 4x = 10, so x(x² + 4) = 10 and x = 10/(x² + 4) [1]. (b) x₂ = 10/(2.25 + 4) = 1.6000 [1]. x₃ = 1.5244, x₄ = 1.5813 [1]. (c) g′(x) = −20x/(x² + 4)² [1]. At α = 1.5568, g′(α) = −0.755 [1]. Since −1 < g′(α) < 0, the iteration converges, but slowly, with values alternating either side of α. The diagram is an inward cobweb [1]. Examiner insight: Evaluate g′ at the root, not at x₁, and give its sign: negative means a cobweb and positive a staircase.

4. (a) A limit L satisfies L = (L² + 6)/5, so L² − 5L + 6 = 0 and L = 2 or 3 [1]. (b) g′(x) = 2x/5 [1]. At 2, g′ = 0.8, so |g′| < 1 and the iteration can converge to 2 [1]. At 3, g′ = 1.2 > 1, so it cannot converge to 3 and moves away [1]. (c) x₂ = (6.25 + 6)/5 = 2.45 and x₃ = 2.4005. The values step down towards 2, so this is a staircase [1]. Examiner insight: Link each root to its own value of g′; a single statement about “the iteration” misses that one rearrangement can converge to one root and not the other.

5. f′(x) = eˣ + 2 [1]. f(1.5) = 1.4817 and f′(1.5) = 6.4817, so x₂ = 1.5 − 1.4817/6.4817 [1] = 1.2714 [1]. Then x₃ = 1.2714 − 0.1087/5.5659 = 1.2519 [1]. Examiner insight: Write the formula with the functions substituted before you use numbers; keep unrounded values for x₂ when computing x₃.

6. (a) f′(x) = 3x² − 3 [1]. f′(1) = 0, so the formula divides by zero: the tangent at x = 1 is horizontal and never meets the axis [1]. (b) x₂ = 0 − 3/(−3) = 1 [1]. Then x₃ would need f′(1) = 0, so the method fails [1]. (c) x₂ = −2 − 1/9 = −2.1111 [1]. x₃ = −2.1038 [1]. Examiner insight: A start that is not itself stationary can still lead to a stationary point on the next step, so check f′ at every iterate.

7. (a) h = 1, ordinates 1, 0.5, 0.3333, 0.25, 0.2 [1]. T = 0.5 × [1 + 0.2 + 2(0.5 + 0.3333 + 0.25)] [1] = 1.683 [1]. (b) f″(x) = 2/x³ > 0 on the interval, so the curve is convex and the chords lie above it: overestimate [2]. (c) The curve is decreasing, so left ends give the upper bound: 1 + 0.5 + 0.3333 + 0.25 = 2.083 [1]. Right ends give the lower bound: 0.5 + 0.3333 + 0.25 + 0.2 = 1.283 [1]. So 1.283 < I < 2.083, and the trapezium value is their mean [1]. Examiner insight: An over or under claim is stronger with its reason: state that the curve is convex and that the chords therefore lie above it.

8. (a) h = 2 [1]. Area ≈ (2/2) × [0 + 0 + 2(1.1 + 1.9 + 2.4 + 2.2 + 1.4)] [1] = 18.0 m² [1]. (b) 18.0 × 0.8 = 14.4 m³ per second [1]. (c) Measure the depth at closer intervals, so each strip follows the river bed more closely [1]. Examiner insight: Carry units through a context answer; giving 14.4 with no unit leaves the answer incomplete.

9. (a) h = π/4, ordinates 0, √2/2, 1 [1]. T = (π/8)[0 + 1 + 2(√2/2)] [1] = (π/8)(1 + √2) = 0.9481 [1]. (b) h = π/8; ordinates 0, 0.3827, 0.7071, 0.9239, 1 [1]. T = (π/16)[1 + 2(2.0137)] = 0.9871 [1]. (c) sin x is concave on this interval (f″ = −sin x < 0), so the chords lie below the curve [1]. The error falls from 0.0519 to 0.0129, about a quarter, when the strips are halved [1]. Examiner insight: Show the exact form first when asked; a decimal alone does not show the structure of the working.

10. (a) Setting h = 5 gives 0.2t + 1.5 sin(0.5t) = 2, so f(t) = 0 [1]. f(2) = −0.338 and f(3) = 0.096; f is continuous with a sign change, which places the root between 2 and 3 [1]. (b) At the root t ≈ 2.69, (2 − 0.2t)/1.5 = 0.975 [1], and √(1 − 0.975²) = 0.223 [1]. So g′ = −0.267/0.223 = −1.19, which has magnitude greater than 1, so the iteration diverges [1]. (c) f′(t) = 0.2 + 0.75 cos(0.5t) [1]. t₂ = 3 − 0.0962/0.2531 = 2.6197, and t₃ = 2.6879 [1]. (d) f(2.685) = −0.0019 and f(2.695) = 0.0018, so 2.685 < t < 2.695 and t = 2.69 to 2 decimal places [2]. 0.69 hours is 41 minutes, so the warning is at about 08:41 [1]. Examiner insight: Part (b) needs the numerical value of the gradient at the root and a statement comparing its magnitude with 1; saying “it looks unstable” does not support the conclusion.

Where marks are usually lost

• Showing a sign change without saying that the function is continuous.
• Testing 2.9 and 3.0 to justify a root to 2 decimal places instead of 2.925 and 2.935.
• Concluding that there is no root because the signs at the ends agree.
• Judging convergence from the first few terms, not from g′ at the root.
• Rounding intermediate values in an iteration and then reporting a last digit that is wrong.
• Forgetting that Newton-Raphson needs f′ and substituting f in both places.
• Doubling the first and last ordinates in the trapezium rule.
• Giving “overestimate” with no mention of the curve’s shape or the sign of f″.
• Swapping upper and lower rectangle bounds for a decreasing function.

Next steps

• Recap with the revision notes.
• Re-read the study guide.
• Course page: OCR A Level Mathematics and the checklist.
• Find weak spots with the free diagnostics.
• Book a free trial class.

Official syllabus

OCR, A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018. Section 1.09 Numerical Methods, 1.09a to 1.09g.