OxfordAQA International AS and A-level Mathematics: Parametric equations and exponential and logarithmic functions (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 follow the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2, sections P2.3 (Cartesian and parametric equations) and P2.5 (eˣ, ln x, exponential growth and decay). This content sits in Unit P2 (Pure Maths), part of the International A-level only. It applies to International AS exams from May/June 2018 and International A-level exams from May/June 2019.

Learn the content first from the study guide and the revision notes. The course page and printable checklist show the rest of the course, and the free 10-minute diagnostics help you find weak topics. Give exact answers where asked, and otherwise 3 significant figures.

Questions

1. A curve has parametric equations x = t² − 2, y = 4t. Find its Cartesian equation. [3]

2. A curve has equations x = 4cosθ, y = 3sinθ.

(a) Find the Cartesian equation of the curve. [2] (b) The point (2√3, 3/2) lies on the curve. Find the value of θ, where 0 ≤ θ < 2π. [2]

3. A curve has equations x = 1/t, y = 3t, where t ≠ 0.

(a) Find the Cartesian equation of the curve. [1] (b) The line y = 7 − 2x meets the curve at two points. Find their coordinates. [4]

4. A curve has equations x = t + 1/t, y = t − 1/t.

(a) Show that x² − y² = 4. [3] (b) For t > 0, show that x ≥ 2, and say which part of the curve is traced. [2]

5. A curve has equations x = 5cosθ, y = 2 − cos2θ. Find its Cartesian equation and state the range of values of x. [4]

6. A curve C has equations x = 2eᵗ, y = 3e⁻ᵗ, where t is real.

(a) Show that the Cartesian equation of C is y = 6/x. [2] (b) State the set of possible values of x. [1] (c) C meets the line y = x. Find the exact value of t at the point of intersection. [3]

7. The curve y = 2eˣ − 5 is drawn.

(a) Write down the equation of the asymptote and the coordinates of the y-intercept. [2] (b) The curve cuts the x-axis once. Find the exact value of x there. [2]

8. Solve e²ˣ − 7eˣ + 10 = 0, giving your answers in exact form. [4]

9. Solve ln(x + 3) + ln(x − 1) = ln 5. [4]

10. The function f is defined by f(x) = 3e²ˣ − 1 for all real x.

(a) Write down the set of values that f takes. [1] (b) Find f⁻¹(x). [3] (c) Write down the set of inputs that f⁻¹ accepts. [1]

11. The area covered by an algae bloom is modelled by P = 80eᵏᵗ square metres, where t is the time in days and k is a constant. After 5 days the area is 200 square metres.

(a) Find k. [2] (b) Find the area predicted after 12 days. [2] (c) Find the time taken to reach 1000 square metres. [2] (d) Give one reason why the model may be unsuitable for large values of t. [1]

12. A bowl of soup cools on a table. Its temperature after t minutes is T = 18 + 62e⁻ᵏᵗ degrees Celsius, where k is a positive constant.

(a) Write down the initial temperature of the soup. [1] (b) After 8 minutes the temperature is 45 degrees Celsius. Find k. [3] (c) Find how long it takes for the temperature to reach 30 degrees Celsius. [3] (d) State what happens to T as t increases, and what this represents. [1]

Answers

1. From y = 4t, t = y/4. [1] Substitute: x = (y/4)² − 2 = y²/16 − 2. [1] Hence y² = 16(x + 2). [1] [3] Examiner insight: Rearranging for t from the simpler equation first keeps the algebra short. Leave the answer as a clear equation in x and y, and check one point such as t = 1, which gives (−1, 4).

2. (a) cosθ = x/4 and sinθ = y/3. [1] Using sin²θ + cos²θ = 1: x²/16 + y²/9 = 1. [1] (b) cosθ = 2√3/4 = √3/2 and sinθ = (3/2)/3 = 1/2. [1] Both are positive and match θ = π/6. [1] [4] Examiner insight: In part (b) check both coordinates. cosθ = √3/2 alone also fits θ = 11π/6, and the sine value is what rules it out.

3. (a) xy = (1/t)(3t) = 3, so y = 3/x. [1] (b) Substitute: x(7 − 2x) = 3. [1] This gives 2x² − 7x + 3 = 0, so (2x − 1)(x − 3) = 0. [1] Then x = 1/2 or x = 3. [1] Using y = 7 − 2x gives the points (1/2, 6) and (3, 1). [1] [5] Examiner insight: Give both coordinates for each point, not just the x-values. Any equivalent route through t (t = 2 and t = 1/3) reaches the same points.

4. (a) x + y = 2t. [1] x − y = 2/t. [1] Multiply: x² − y² = (x + y)(x − y) = (2t)(2/t) = 4. [1] (b) x − 2 = t + 1/t − 2 = (t − 1)²/t. [1] For t > 0 this is at least 0, so x ≥ 2, which gives the right-hand branch only. [1] [5] Examiner insight: For a “show that” question, write the target equation at the end and make every line follow from the one before. In part (b) a few test values are not a proof.

5. Use cos2θ = 2cos²θ − 1. [1] Since cosθ = x/5, [1] y = 2 − 2(x/5)² + 1 = 3 − 2x²/25. [1] As cosθ lies between −1 and 1, −5 ≤ x ≤ 5. [1] [4] Examiner insight: The restriction on x is a separate part of the answer. Writing only y = 3 − 2x²/25 describes the whole parabola and loses this credit.

6. (a) From x = 2eᵗ, eᵗ = x/2. [1] Then y = 3/(x/2) = 6/x. [1] (b) eᵗ > 0, so x > 0. [1] (c) Put y = x: x = 6/x, so x² = 6. Since x > 0, x = √6. [1] Then 2eᵗ = √6, so eᵗ = √6/2. [1] Hence t = ln(√6/2) = (1/2)ln(3/2). [1] [6] Examiner insight: Reject x = −√6 and say why: eᵗ is always positive. An answer that gives both roots without comment shows the restriction from part (b) was not used.

7. (a) Asymptote y = −5. [1] At x = 0, y = 2 − 5, so the intercept is (0, −3). [1] (b) 2eˣ = 5, so eˣ = 5/2. [1] Hence x = ln(5/2). [1] [4] Examiner insight: The asymptote is an equation, not a point. Giving the y-intercept as (0, 2) is a common slip, since eˣ = 1 is multiplied by 2 and then 5 is subtracted.

8. Let u = eˣ, so u² − 7u + 10 = 0. [1] Factorise: (u − 2)(u − 5) = 0. [1] So eˣ = 2 or eˣ = 5. [1] Then x = ln 2 or x = ln 5. [1] [4] Examiner insight: Show the substitution or write the factors in terms of eˣ. A bare list of decimal answers is less likely to earn credit than the exact logarithms the question asks for.

9. Combine: ln[(x + 3)(x − 1)] = ln 5, [1] so (x + 3)(x − 1) = 5. [1] This gives x² + 2x − 8 = 0, so (x + 4)(x − 2) = 0. [1] Both logarithms need x > 1, so reject x = −4 and take x = 2. [1] [4] Examiner insight: After solving the quadratic, test each root in the original logarithms. Stating why x = −4 is rejected is the reasoning that earns credit on a question like this.

10. (a) e²ˣ > 0, so f takes values greater than −1. [1] (b) Let y = 3e²ˣ − 1. Then e²ˣ = (y + 1)/3. [1] So 2x = ln[(y + 1)/3]. [1] Hence f⁻¹(x) = (1/2)ln[(x + 1)/3]. [1] (c) f⁻¹ accepts exactly the values f takes: x > −1. [1] [5] Examiner insight: Do not forget the halving: ln gives 2x, not x. Link the domain of the inverse to the range of f, and use a strict inequality.

11. (a) 80e⁵ᵏ = 200, so e⁵ᵏ = 2.5. [1] Then k = (ln 2.5)/5 ≈ 0.183. [1] (b) P = 80e^(12k) = 80 × 2.5^(2.4). [1] This is ≈ 721 square metres. [1] (c) 80eᵏᵗ = 1000, so eᵏᵗ = 12.5. [1] Then t = (ln 12.5)/k ≈ 13.8 days. [1] (d) Growth is unbounded, but a real bloom is limited by space, light or nutrients, so the model cannot hold for large t. [1] [7] Examiner insight: Carry the unrounded value of k into parts (b) and (c) instead of 0.183, which could shift the third significant figure. Give units in the answers to (b) and (c).

12. (a) At t = 0, T = 18 + 62 = 80 degrees Celsius. [1] (b) 18 + 62e⁻⁸ᵏ = 45, so e⁻⁸ᵏ = 27/62. [1] Take logarithms: −8k = ln(27/62). [1] Hence k = (ln(62/27))/8 ≈ 0.104. [1] (c) 62e⁻ᵏᵗ = 12, so e⁻ᵏᵗ = 6/31. [1] Then kt = ln(31/6). [1] Hence t = ln(31/6)/k ≈ 15.8 minutes. [1] (d) e⁻ᵏᵗ → 0, so T → 18, the temperature of the table surroundings. [1] [8] Examiner insight: Subtract the background value before taking logarithms. Taking ln of 45 directly, or of 62e⁻ᵏᵗ + 18, is the usual route to a wrong k. Give the limit as a temperature in context.

Where marks are usually lost

• Leaving out the restriction on x after eliminating a trigonometric parameter.
• Writing x² + y² = 1 for an ellipse instead of x²/a² + y²/b² = 1.
• Giving the points of intersection with only the x-coordinates, or with only the parameter values.
• Not showing every step on a “show that” question, or checking only with test values.
• Splitting ln(a + b), or keeping a root that makes a logarithm undefined.
• Treating eˣ = −2 as having a solution.
• Forgetting the factor 1/2 in the inverse of e²ˣ, or giving f⁻¹ the wrong domain.
• Rounding k early and losing accuracy in later parts of a modelling question.
• Omitting units in a modelling answer, or writing nothing when asked to comment on the model.

Next steps

• Re-read the revision notes and then redo any question you lost marks on.
• Go back to the study guide for full explanations.
• Open the course page and tick the outcomes on the checklist.
• Take all the free 10-minute diagnostics to find your next topic.
• Book a free trial class if you want help from a teacher.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2, published by OxfordAQA. Sections P2.3 (Cartesian and parametric equations of curves and conversion between the two forms) and P2.5 (the function eˣ and its graph, ln x as the inverse of eˣ, exponential growth and decay including exponential functions as models).