AQA A-level Mathematics: Partial Fractions and Functions in Modelling (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 B10 (partial fractions) and B11 (functions in modelling) from the AQA A-level Mathematics (7357) specification, version 1.3 (31 January 2018), for A-level exams from June 2018 onwards. Both are Section B content, assessed on Paper 1, and Papers 2 and 3 can also assess any Paper 1 content. The specification states that a calculator is required in all assessments, but each answer shows the algebra you should write down.

Learn the content first from the study guide and the revision notes. Course links: AQA A-level Mathematics hub and printable checklist. Free 10-minute tests are at /diagnostics/.

Questions

1. Express (13x + 8)/((x + 4)(2x − 3)) in partial fractions. [3]

2. Express (10x + 12)/(3x² + 8x + 4) in partial fractions. [4]

3. Express (8 − x)/(x(x + 2)²) in the form A/x + B/(x + 2) + C/(x + 2)², where A, B and C are constants to be found. [6]

4. It is given that (3x + k)/((x + 1)(x − 2)) ≡ A/(x + 1) + B/(x − 2), where A, B and k are constants and A = 2. Find k and B. [4]

5. Express (7x + 4)/(x(x + 1)(2x − 1)) in partial fractions. [5]

6. A farmer uses 72 m of fencing to make two identical rectangular pens side by side. The two end fences and the shared dividing fence each have length x metres, and the top and bottom of the whole enclosure each have length y metres.

(a) Show that the total area A m² is A = 36x − 1.5x². [2] (b) Find the largest possible area and the value of x that gives it. [3] (c) State the sensible domain of x, and one assumption made by the model. [2]

7. The height H cm of a sapling t years after it was planted is modelled by H = 150 − 130e^(−0.08t).

(a) State the height when it was planted. [1] (b) Find the time taken to reach 100 cm. Give your answer to 3 significant figures. [3] (c) State the height that the model approaches, and give one limitation of the model. [2]

8. A car park advertises £3 for each hour or part of an hour, up to a maximum of £12. This is modelled by P = 3h for 0 < h ≤ 4 and P = 12 for h > 4, where P is the charge in pounds and h is the time in hours.

(a) Find the charge for 2.5 hours and for 6 hours. [2] (b) State the range of P. [1] (c) Explain one way in which the model differs from the wording of the charge, and suggest a refinement. [1]

9. The temperature T °C in a greenhouse t hours after midnight is modelled by T = 14 + 6cos(15(t − 15))° for 0 ≤ t ≤ 24.

(a) State the maximum and minimum temperatures and the times at which they occur. [2] (b) Find the number of hours for which the temperature is above 17 °C. [3] (c) Give one limitation of the model. [1]

10. The concentration C mg/l of a drug in a patient, t hours after an injection, is modelled by C = (14t + 22)/((t + 1)(t + 3)), for t ≥ 0.

(a) Show that C = 4/(t + 1) + 10/(t + 3). [3] (b) Find the initial concentration, and state what happens to C for large t. [2] (c) Find the exact time when C = 3, and give it to 3 significant figures. [3] (d) Give one limitation of the model. [1]

11. A shop records monthly sales S (units) in month t: 45, 52, 61, 66, 75 for t = 1 to 5. A manager models sales by S = 7.5t + 38.

(a) Complete a list of the five values the model predicts. [2] (b) Find the largest error, as a percentage of the observed value, to 2 significant figures. [2] (c) Use the model to predict sales at t = 24, comment on its reliability, and suggest a refinement. [2]

Answers

1. Write (13x + 8)/((x + 4)(2x − 3)) ≡ A/(x + 4) + B/(2x − 3), so 13x + 8 ≡ A(2x − 3) + B(x + 4). [1] Put x = −4: −44 = −11A, so A = 4. [1] Put x = 3/2: 27.5 = 5.5B, so B = 5. Result: 4/(x + 4) + 5/(2x − 3). [1] Examiner insight: Writing the identity line before substituting shows your method even if a sign slips later.

2. Factorise: 3x² + 8x + 4 = (3x + 2)(x + 2). [1] Then 10x + 12 ≡ A(x + 2) + B(3x + 2). [1] Put x = −2: −8 = −4B, so B = 2. [1] Put x = −2/3: 16/3 = A(4/3), so A = 4. Result: 4/(3x + 2) + 2/(x + 2). [1] Examiner insight: The denominator is given as a quadratic, so factorising comes first.

3. Identity: 8 − x ≡ A(x + 2)² + Bx(x + 2) + Cx. [1] Put x = 0: 8 = 4A, so A = 2. [1] Put x = −2: 10 = −2C, so C = −5. [1] Compare x² terms: 0 = A + B, so B = −2. [1] Result: 2/x − 2/(x + 2) − 5/(x + 2)². [1] Check x = 1: left side 7/9; right side 2 − 2/3 − 5/9 = 7/9. [1] Examiner insight: A repeated factor needs both the (x + 2) and the (x + 2)² terms, and a final check with a spare value of x is a sensible way to protect the whole answer.

4. Identity: 3x + k ≡ A(x − 2) + B(x + 1). [1] Put x = −1 with A = 2: −3 + k = −3A = −6, so k = −3. [1] Put x = 2: 6 + k = 3B, so 3 = 3B and B = 1. [1] Check: 2(x − 2) + (x + 1) = 3x − 3, which matches 3x + k with k = −3. [1] Examiner insight: Choose substitutions that use what you are given. The final expansion confirms both values at once.

5. Form: (7x + 4)/(x(x + 1)(2x − 1)) ≡ A/x + B/(x + 1) + C/(2x − 1). [1] Identity: 7x + 4 ≡ A(x + 1)(2x − 1) + Bx(2x − 1) + Cx(x + 1). [1] Put x = 0: 4 = −A, so A = −4. [1] Put x = −1: −3 = 3B, so B = −1. [1] Put x = 1/2: 7.5 = 0.75C, so C = 10. Result: −4/x − 1/(x + 1) + 10/(2x − 1). [1] Examiner insight: With three factors, substituting the root of each factor is quicker than comparing coefficients. Keep 2x − 1 whole in the answer.

6. (a) The fence has three lengths x and two lengths y, so 3x + 2y = 72. [1] Then y = 36 − 1.5x, and A = xy = 36x − 1.5x². [1] (b) A = −1.5(x² − 24x) = −1.5[(x − 12)² − 144] = 216 − 1.5(x − 12)². [1] The largest value occurs when x = 12. [1] The largest area is 216 m² at x = 12. [1] (c) y > 0 needs x < 24, so the domain is 0 < x < 24. [1] Assumption, for example: the fencing is used with no gaps or overlaps, and the ground is level and rectangular. [1] Examiner insight: A “show that” needs each step from fencing to area. Give a reason for the domain.

7. (a) H(0) = 150 − 130 = 20 cm. [1] (b) 100 = 150 − 130e^(−0.08t), so 130e^(−0.08t) = 50. [1] Then e^(−0.08t) = 5/13, so t = ln(2.6)/0.08. [1] t = 11.9 years (3 s.f.). [1] (c) As t grows, e^(−0.08t) tends to 0, so H approaches 150 cm. [1] Limitation, for example: real growth varies with weather each year, so one smooth curve cannot capture it. [1] Examiner insight: Rearrange fully before taking logs. The model approaches 150 but never reaches it.

8. (a) 2.5 hours: P = 3 × 2.5 = £7.50. [1] 6 hours: P = £12. [1] (b) The range is 0 < P ≤ 12. [1] (c) Real charges jump in steps (2.5 hours would cost £9), but the model changes smoothly. Refinement: round the time up to the next whole hour. [1] Examiner insight: For a piecewise model, check the join at h = 4. Both rules give £12 there. A comment on the model should refer to the context, not just say “it is not realistic”.

9. (a) The cosine takes values from −1 to 1, so the maximum is 14 + 6 = 20 °C at t = 15, and the minimum is 14 − 6 = 8 °C at t = 3. [1] for both temperatures, [1] for both times. (b) Solve 14 + 6cos(15(t − 15))° > 17, so cos(15(t − 15))° > 0.5. [1] Then −60 < 15(t − 15) < 60, so 11 < t < 19. [1] The temperature is above 17 °C for 8 hours. [1] (c) Limitation, for example: the model repeats exactly every 24 hours, but real weather varies from day to day. [1] Examiner insight: A sketch of the cosine curve and its midline helps you choose the right interval. Give the duration, not just the end times.

10. (a) Write C = A/(t + 1) + B/(t + 3), so 14t + 22 ≡ A(t + 3) + B(t + 1). [1] Put t = −1: 8 = 2A, so A = 4. [1] Put t = −3: −20 = −2B, so B = 10. This gives C = 4/(t + 1) + 10/(t + 3). [1] (b) At t = 0, C = 22/3 = 7.33 mg/l (3 s.f.). [1] For large t both terms tend to 0, so C tends to 0, as the drug leaves the body. [1] (c) Set (14t + 22) = 3(t + 1)(t + 3) = 3t² + 12t + 9. [1] Then 3t² − 2t − 13 = 0. [1] The positive root is t = (1 + 2√10)/3 = 2.44 hours (3 s.f.). [1] (d) Limitation, for example: the concentration starts at its highest value, but a drug normally takes time to spread, so the model is unreliable at the start. [1] Examiner insight: In a “show that” part, derive the decomposition rather than only recombining. Reject the negative root in part (c) because time cannot be negative.

11. (a) The model gives 45.5, 53, 60.5, 68, 75.5. [2] (b) The differences between data and model are −0.5, −1, 0.5, −2, −0.5. [1] The largest error is 2 at t = 4, which is 2/66 = 3.0% of the observed value. [1] (c) At t = 24: S = 7.5 × 24 + 38 = 218 units. [1] This is far outside the data (t = 1 to 5), so it is extrapolation and may be unreliable. A refinement is to allow for seasonal changes or for sales to level off, using more data. [1] Examiner insight: Say which value you divide by when you give a percentage error. In the comment, name extrapolation and give a specific refinement.

Where marks are usually lost

• Not factorising a quadratic denominator before writing the form.
• Missing the first-power term for a repeated factor.
• Substituting a wrong value, such as x = 3 for a factor 3x − 1 instead of x = 1/3.
• Losing the coefficient inside a factor such as (2x − 3).
• Not checking an identity with a spare value of x.
• Leaving a modelling answer as a bare number with no units or context.
• Giving a limitation that is only “not accurate”.
• Missing a reason for a domain, or keeping an impossible root.
• Accepting an extrapolated value without comment.

Next steps

• Review the revision notes and the study guide.
• Use the AQA A-level Mathematics hub and printable checklist.
• Take any of the free 10-minute diagnostics.
• Book a free trial class: /trial/.

Official syllabus

AQA A-level Mathematics (7357) specification, version 1.3 (31 January 2018), A-level exams June 2018 onwards, published by AQA. Sections B10 and B11 and theme OT3 are the sources for these questions.