AQA A-Level Mathematics: I: Numerical methods (7357) – 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 I: Numerical methods (I1 to I4) of the AQA A-level Mathematics (7357) specification, version 1.3, for A-level exams from June 2018 onwards. Section I is Paper 1 content, and Papers 2 and 3 can also assess any Paper 1 content. A calculator is required in every 7357 paper, so every question here assumes you have one. Work in radians wherever trigonometry appears.

Learn the methods first in the Numerical methods study guide and the Numerical methods revision notes. The 7357 course hub and the printable checklist list the rest of the course, and the free 10-minute diagnostics show which topic to practise next.

Questions

1. Show that the equation x⁴ − 3x − 5 = 0 has a root between x = 1.7 and x = 1.8. [3]

2. A student says: “f(x) = 3/(x − 2) gives f(1) = −3 and f(3) = 3, so f(x) = 0 has a root between 1 and 3.” Explain why the student is wrong. [2]

3. The sequence x_(n+1) = ∛(2x_n + 7) has x₁ = 2. Find x₂, x₃ and x₄, giving each to 4 decimal places. [3]

4. The equation e^x + x² − 4 = 0 has one positive root, α.

(a) Show that the equation can be written as x = ln(4 − x²). [1] (b) Use x_(n+1) = ln(4 − x_n²) with x₁ = 1 to find x₂, x₃ and x₄ to 4 decimal places. [2] (c) Given that g(x) = ln(4 − x²) has g′(x) = −2x/(4 − x²), state whether the iteration gives a staircase or a cobweb diagram near α ≈ 1.06, and whether it converges. Give a reason. [2]

5. Use the Newton-Raphson method with x₁ = 1 to find x₂ and x₃ for the equation x e^x − 3 = 0. Give your answers to 4 decimal places. [4]

6. f(x) = x³ − 6x² + 9x + 1. The equation f(x) = 0 has one real root, near x = −0.1.

(a) Explain why x₁ = 3 cannot be used as a starting value for the Newton-Raphson method. [1] (b) Find x₂ when x₁ = 1.1, and explain why this starting value is a poor choice. [3]

7. I = ∫ from 1 to 2 of √(x³ + 1) dx.

(a) Use the trapezium rule with 4 strips to estimate I, to 3 decimal places. [4] (b) Given that the second derivative of √(x³ + 1) is 3x(x³ + 4) / [4(x³ + 1)^(3/2)], state whether your answer to (a) is an overestimate or an underestimate. Give a reason. [2]

8. The curve y = 1/(1 + x²) is decreasing for x > 0. Using four rectangles of width 0.5, find a lower bound and an upper bound for ∫ from 0 to 2 of 1/(1 + x²) dx. [5]

9. The equation x³ − 6x + 2 = 0 has three real roots, near −2.60, 0.34 and 2.26. It is rearranged as x = (x³ + 2)/6 and the iteration x_(n+1) = (x_n³ + 2)/6 is used.

(a) With x₁ = 1, find x₂, x₃ and x₄ to 4 decimal places. [2] (b) Describe the shape of the diagram for this iteration near the root 0.34, giving a reason. [2] (c) Explain why this iteration cannot be used to find the root near 2.26. [2]

10. A car moves from rest. Its velocity, v m s⁻¹, is recorded every 2 seconds.

t (s)	0	2	4	6	8
v (m s⁻¹)	0	5.41	8.39	10.02	10.91

(a) Use the trapezium rule with all the readings to estimate the distance travelled in the first 8 seconds. [3] (b) The velocity is modelled by v = 12(1 − e^(−0.3t)). Use integration to find the distance predicted by the model for 0 ≤ t ≤ 8, to 2 decimal places. [3] (c) Explain, with reference to the shape of the velocity-time graph, why your answer to (a) is smaller than your answer to (b). [2]

11. A chord of a circle with radius r subtends an angle θ radians at the centre. The minor segment it cuts off has an area equal to one third of the area of the circle.

(a) Show that θ − sin θ − 2π/3 = 0. [2] (b) Show that this equation has a root between θ = 2.5 and θ = 2.7. [2] (c) Use the Newton-Raphson method with θ₁ = 2.6 to find θ₂, to 4 decimal places. [3] (d) Show that θ = 2.61 to 3 significant figures. [2]

Answers

1. Let p(x) = x⁴ − 3x − 5. p(1.7) = −1.7479 and p(1.8) = 0.0976 [1]. One is negative and one positive: a change of sign [1]. p is a polynomial, so it is continuous, and there is a root between 1.7 and 1.8 [1]. [3] Examiner insight: the conclusion needs both the change of sign and a statement that there is a root in the interval; values with no concluding sentence usually lose the final mark.

2. f is not continuous on [1, 3]: it has a vertical asymptote at x = 2, where the sign changes [1]. 3/(x − 2) is never zero, so there is no root [1]. [2] Examiner insight: name the discontinuity at x = 2 specifically; “the graph is weird” or “f is not well-behaved” with no location is too vague for credit.

3. x₂ = ∛11 = 2.2240 [1], x₃ = 2.2538 [1], x₄ = 2.2577 [1]. [3] Examiner insight: keep full calculator values between steps (use Ans); rounding x₂ before finding x₃ can change the fourth decimal place.

4. (a) e^x = 4 − x², so taking natural logs, x = ln(4 − x²) [1]. (b) x₂ = ln 3 = 1.0986 [1]; x₃ = 1.0271, x₄ = 1.0801 [1]. (c) g′(1.06) ≈ −2.12/2.88 ≈ −0.74, which is between −1 and 0 [1]. So it is a cobweb and it converges: the iterates alternate either side of α, as in (b) [1]. [5] Examiner insight: a “show that” with one mark still needs the step e^x = 4 − x² written before taking logs; jumping straight to the printed result earns nothing.

5. f(x) = x e^x − 3, so f′(x) = (x + 1)e^x [1]. x₂ = 1 − (e − 3)/(2e) = 1 − (−0.28172)/5.43656 [1] = 1.0518 [1]. x₃ = 1.05182 − 0.01121/5.8741 = 1.0499 [1]. [4] Examiner insight: differentiating with the product rule is a method step in its own right; a wrong f′ usually loses the accuracy marks that follow, so check it before iterating.

6. (a) f′(x) = 3x² − 12x + 9, so f′(3) = 0: the tangent is horizontal and the formula would divide by zero [1]. (b) f(1.1) = 4.971 and f′(1.1) = −0.57 [1]. x₂ = 1.1 − 4.971/(−0.57) = 9.821 [1]. x₁ = 1.1 is close to the stationary point at x = 1, so f′ is small and the step is very large; x₂ is far from the root near −0.1 [1]. [4] Examiner insight: the explanation must link the failure to the stationary point (small or zero gradient), not just say “the answer is wrong”.

7. (a) h = 0.25 [1]. Ordinates: 1.4142, 1.7185, 2.0917, 2.5218, 3 [1]. I ≈ 0.125 × [1.4142 + 3 + 2(1.7185 + 2.0917 + 2.5218)] [1] = 2.135 [1]. (b) For 1 ≤ x ≤ 2 every factor is positive, so the second derivative is positive and the curve is convex [1]. The chords lie above the curve, so it is an overestimate [1]. [6] Examiner insight: “overestimate” with no reason is not enough; the mark goes to the link between the sign of the second derivative and where the chords lie.

8. y-values at x = 0, 0.5, 1, 1.5, 2: 1, 0.8, 0.5, 0.3077, 0.2 [1]. The curve is decreasing, so left-hand heights give an upper bound [1]: 0.5 × (1 + 0.8 + 0.5 + 0.3077) = 1.3038 [1]. Right-hand heights give a lower bound: 0.5 × (0.8 + 0.5 + 0.3077 + 0.2) = 0.9038 [1]. So 0.9038 < I < 1.3038 [1]. [5] Examiner insight: say which rectangles go with which bound and why; swapping them for a decreasing curve gives the right numbers attached to the wrong bounds.

9. (a) x₂ = 0.5 [1]; x₃ = 0.3542, x₄ = 0.3407 [1]. (b) g′(x) = x²/2, and g′(0.34) ≈ 0.058, which is between 0 and 1 [1]. So it is a converging staircase: the iterates approach from one side, as in (a) [1]. (c) g′(2.26) ≈ 2.55 > 1 [1], so iterates move away from that root (for example, x₁ = 3 gives 4.833, 19.15, …) and it cannot be found this way [1]. [6] Examiner insight: a numerical gradient value at the root, compared with 1, makes the reason checkable; “it diverges” alone is usually not enough.

10. (a) h = 2 [1]. Distance ≈ (2/2) × [0 + 10.91 + 2(5.41 + 8.39 + 10.02)] [1] = 58.55 m (about 58.6 m) [1]. (b) ∫ 12(1 − e^(−0.3t)) dt = 12t + 40e^(−0.3t) [1]. [12t + 40e^(−0.3t)] from 0 to 8 = (96 + 40e^(−2.4)) − 40 [1] = 59.63 m [1]. (c) The graph is concave: velocity increases at a decreasing rate [1]. Chords lie below the curve, so the trapezium rule underestimates the area, which is the distance [1]. [8] Examiner insight: units (m) are expected on a distance in context, and the explanation must connect area under the graph to distance travelled.

11. (a) Segment area = sector − triangle = ½r²θ − ½r² sin θ [1]. Setting this equal to ⅓πr² and multiplying by 2/r² gives θ − sin θ = 2π/3, so θ − sin θ − 2π/3 = 0 [1]. (b) Let S(θ) = θ − sin θ − 2π/3. S(2.5) = −0.1929 and S(2.7) = 0.1782 [1]. S is continuous, and there is a change of sign, so there is a root between 2.5 and 2.7 [1]. (c) S′(θ) = 1 − cos θ [1]. θ₂ = 2.6 − (−0.0099)/1.85689 [1] = 2.6053 [1]. (d) S(2.605) = −0.00061 and S(2.615) = 0.01801 [1]. Change of sign, so 2.605 < θ < 2.615 and θ = 2.61 to 3 s.f. [1]. [9] Examiner insight: in (d) the interval must be the rounding bounds 2.605 and 2.615; checking 2.61 and 2.62 does not show the root rounds to 2.61.

Where marks are usually lost

• Calculator left in degrees for questions 11(b) to 11(d), which gives wrong values for sin θ and cos θ.
• A change of sign stated with no values, or with no conclusion about the root.
• Rounding bounds missed when confirming a root to a given accuracy.
• Only the final iterate written when several were asked for.
• Wrong f′ in Newton-Raphson, often a missed product rule term.
• n strips confused with n + 1 ordinates, so h is wrong.
• Over- or underestimate claimed without a reason based on the curve’s shape.
• Upper and lower rectangle sums swapped for a decreasing function.
• Context answers given with no units or to an unrealistic accuracy.

Next steps

• Numerical methods revision notes
• Numerical methods study guide
• Differentiation practice questions, for the f′ work Newton-Raphson needs
• AQA A-level Mathematics course hub
• Printable 7357 checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

AQA A-level Mathematics (7357) specification, version 1.3 (31 January 2018), for A-level exams June 2018 onwards, published by AQA. Section 3.10, I: Numerical methods.