IB DP Mathematics: Analysis and Approaches – Rational, exponential and logarithmic functions, solving equations and transformations 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 – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

This practice set covers the rational, exponential and logarithmic functions, solving equations and transformations unit of IB Diploma Programme Mathematics: Analysis and Approaches. It is aligned to the IB Mathematics: analysis and approaches guide, first assessment 2021, syllabus sections 2.8, 2.9, 2.10 and 2.11. Every question is common content for SL and HL. It follows the IB guide for first assessment 2021, which remains the examined syllabus until the new course is first assessed in May 2029, so it applies to the May and November 2026, 2027 and 2028 sessions.

Paper 1 allows no technology and Paper 2 requires it, at both SL and HL. Questions are labelled “(calculator-free)” or “(calculator allowed)” to match. Give answers exactly or to 3 significant figures unless a question says otherwise.

Learn the methods first in the study guide and the revision notes. The Functions strand overview has a separate, shorter question set; none of those questions is repeated here. The IB DP Maths AA course hub and the printable syllabus checklist show where this unit sits.

Questions

1. (calculator-free) Let f(x) = (6x + 1)/(3x − 2), x ≠ 2/3.

(a) Write down the equations of the vertical and horizontal asymptotes of the graph of f. [2] (b) Find the coordinates of the points where the graph of f meets the axes. [2]

2. (calculator-free) Let f(x) = 1/x, x ≠ 0.

(a) Show that f is self-inverse. [2] (b) Hence write down the equation of a line of symmetry of the graph of y = f(x). [1]

3. (calculator-free)

(a) Write 7ˣ in the form e^(kx), giving the exact value of k. [1] (b) Find the exact value of e^(3 ln 2). [2] (c) Simplify log₂(8ˣ). [2]

4. (calculator-free) Solve e^(2x) − 3eˣ − 10 = 0, giving your answer in exact form. [5]

5. (calculator-free) The graph of y = f(x) passes through P(−2, 6). It has a vertical asymptote x = 1 and a horizontal asymptote y = 2. For each transformed graph, write down the image of P and the equation of the stated asymptote.

(a) y = f(x + 3); the vertical asymptote. [2] (b) y = −½ f(x); the horizontal asymptote. [2] (c) y = f(2x); the vertical asymptote. [2]

6. (calculator allowed) Solve the equation eˣ = 4 − x². [3]

7. (calculator-free) Let g(x) = ln(x − 1) + 2, x > 1.

(a) Describe a sequence of two transformations that maps the graph of y = ln x onto the graph of y = g(x). [2] (b) Find the exact x-intercept of the graph of g. [2] (c) Find g⁻¹(x). [2]

8. (calculator allowed) A cup of coffee cools so that its temperature, T °C, t minutes after it is poured, is modelled by T = 22 + 68e^(−0.05t).

(a) Find the temperature when the coffee is poured. [1] (b) Find the temperature after 10 minutes. [2] (c) Find the time taken for the temperature to fall to 50 °C. [3]

9. (calculator-free) Let f(x) = (2x − 4)/(x + 1), x ≠ −1.

(a) Show that f(x) = 2 − 6/(x + 1). [2] (b) Hence describe a sequence of transformations that maps the graph of y = 1/x onto the graph of y = f(x). [3] (c) Write down the equations of the asymptotes of the graph of f. [2] (d) Find the coordinates of the axis intercepts. [2] (e) Write down the range of f. [1]

10. (calculator allowed) Two bacterial colonies are grown in a laboratory. After t hours, colony A has 200(1.5)ᵗ cells and colony B has 800e^(0.1t) cells.

(a) Write the size of colony A in the form 200e^(kt), giving k to 3 significant figures. [2] (b) Find the time at which the two colonies are the same size. [3] (c) Find the time at which the two colonies contain 10 000 cells in total. [2] (d) Find the number of cells in each colony at the time found in (b). [1]

11. (calculator-free) Let f(x) = eˣ. The graph of y = g(x) is obtained from the graph of y = f(x) by a reflection in the y-axis, then a vertical stretch with scale factor 3, then a translation of 1 unit upwards.

(a) Find g(x). [3] (b) Write down the equation of the horizontal asymptote and the y-intercept of the graph of g. [2] (c) Solve g(x) = 7, giving your answer in exact form. [2] (d) The graph of y = h(x) is obtained from y = f(x) by the same three transformations, but with the translation done first, then the reflection, then the stretch. Find h(x) and the equation of its horizontal asymptote. [2]

Answers

1. (a) 3x − 2 = 0, so x = 2/3 [1]. Horizontal asymptote y = 6/3, so y = 2 [1] (b) f(0) = 1/(−2), so (0, −1/2) [1]. 6x + 1 = 0, so (−1/6, 0) [1] Examiner insight: An asymptote must be written as an equation of a line; “2/3 and 2” on their own do not earn the marks.

2. (a) f(f(x)) = 1/(1/x) [1] = x, so f⁻¹(x) = f(x) and f is self-inverse [1] (b) A function and its inverse are reflections in y = x, so the line is y = x [1] Examiner insight: In a “show that”, stating f⁻¹(x) = 1/x without the composite (or a rearrangement of y = 1/x) earns nothing, because the result is what you are asked to prove.

3. (a) 7ˣ = e^(x ln 7), so k = ln 7 [1] (b) 3 ln 2 = ln 8 [1], so e^(ln 8) = 8 [1] (c) 8ˣ = 2^(3x) [1], so log₂ 2^(3x) = 3x [1] Examiner insight: “Exact” means k = ln 7, not 1.95; on a calculator-free paper a decimal here loses the accuracy mark.

4. Let u = eˣ, so u² − 3u − 10 = 0 [1]. (u − 5)(u + 2) = 0 [1], giving u = 5 or u = −2 [1]. eˣ > 0 for all x, so reject eˣ = −2 [1]. eˣ = 5, so x = ln 5 [1] Examiner insight: The rejection of u = −2 is a separate mark; you must give the reason (eˣ is always positive), not just cross the value out.

5. (a) Translation 3 units left: P′(−5, 6) [1], vertical asymptote x = −2 [1] (b) Reflection in the x-axis and vertical stretch factor ½: P′(−2, −3) [1], horizontal asymptote y = −1 [1] (c) Horizontal stretch factor ½: P′(−1, 6) [1], vertical asymptote x = 1/2 [1] Examiner insight: Each asymptote moves exactly as a point does; a translation of 3 left with the asymptote left at x = 1 loses the second mark even if P′ is right.

6. Graph y = eˣ and y = 4 − x² (or y = eˣ + x² − 4) and find the intersections [1]. x = −1.96 [1] and x = 1.06 [1] Examiner insight: On a technology paper the method mark comes from stating what you graphed; giving only one intersection loses an accuracy mark, so always check for a second root.

7. (a) Translation 1 unit right [1] and translation 2 units up (together, translation by vector (1, 2)) [1] (b) ln(x − 1) + 2 = 0, so ln(x − 1) = −2 [1], giving x = 1 + e⁻² [1] (c) Let y = ln(x − 1) + 2, so x − 1 = e^(y − 2) [1]. Swap x and y: g⁻¹(x) = e^(x − 2) + 1 [1] Examiner insight: Direction and size are both needed for a translation; “shift right” without “1 unit” usually earns no mark.

8. (a) T = 22 + 68e⁰ = 90 °C [1] (b) T = 22 + 68e^(−0.5) [1] = 63.2 °C [1] (c) 22 + 68e^(−0.05t) = 50, so e^(−0.05t) = 28/68 [1]. −0.05t = ln(28/68) [1], so t = 20 ln(68/28) = 17.7 minutes [1] Examiner insight: The final answer 17.7 must come from an unrounded value (17.746…); units (°C, minutes) are expected in a context question.

9. (a) 2 − 6/(x + 1) = (2(x + 1) − 6)/(x + 1) [1] = (2x − 4)/(x + 1) = f(x) [1] (b) Vertical stretch with scale factor 6 [1]; reflection in the x-axis [1]; translation 1 unit left and 2 units up, vector (−1, 2) [1] (c) x = −1 [1] and y = 2 [1] (d) f(0) = −4, so (0, −4) [1]; 2x − 4 = 0, so (2, 0) [1] (e) y ∈ ℝ, y ≠ 2 [1] Examiner insight: In (b) the vertical translation must come after the stretch and reflection; a sequence that translates 2 units up first produces a different function and loses the final mark.

10. (a) 1.5ᵗ = e^(t ln 1.5) [1], so k = ln 1.5 = 0.405 [1] (b) 200e^(t ln 1.5) = 800e^(0.1t), so e^((ln 1.5 − 0.1)t) = 4 [1]. t = ln 4 / (ln 1.5 − 0.1) [1] = 4.54 hours [1] (c) Graph y = 200(1.5)ᵗ + 800e^(0.1t) and y = 10 000 and find the intersection [1]: t = 9.10 hours [1] (d) 800e^(0.1 × 4.538…) = 1260 cells in each colony (3 s.f.) [1] Examiner insight: Using the rounded k = 0.405 in (b) gives 4.55, not 4.54; keep ln 1.5 exact or stored in your GDC to protect the accuracy mark.

11. (a) Reflection in the y-axis: e^(−x) [1]. Stretch: 3e^(−x) [1]. Translation: g(x) = 3e^(−x) + 1 [1] (b) y = 1 [1] and (0, 4) [1] (c) 3e^(−x) + 1 = 7, so e^(−x) = 2 [1], giving x = −ln 2 [1] (d) eˣ + 1, then e^(−x) + 1, then h(x) = 3e^(−x) + 3 [1], asymptote y = 3 [1] Examiner insight: Follow-through applies: an error in g(x) in (a) can still earn the method marks in (b) and (c) if you use your own g correctly.

Where marks are usually lost

• Writing an asymptote as a single number instead of the equation x = k or y = k.
• Swapping the horizontal asymptote a/c with the y-intercept b/d.
• Keeping a negative root of a quadratic in eˣ, or rejecting it without a reason.
• Moving f(x + 3) to the right, or multiplying x-coordinates by 2 for f(2x).
• Forgetting to move asymptotes along with points under a transformation.
• Translating before stretching when the target is p f(x) + b.
• Rounding ln 1.5, or any other constant, before the final step and losing the 3 s.f. answer.
• Giving a GDC decimal on a calculator-free part that asks for an exact value.
• Finding only one intersection on the GDC when there are two.
• Leaving out units (°C, minutes, hours) in a context question.

Next steps

• Revisit the revision notes for any method you lost marks on.
• Reread the matching section of the study guide.
• Check paper structure and timing in the exam preparation guide.
• See the full course on the IB DP Maths AA course hub.
• Tick off sections on the printable syllabus checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

International Baccalaureate Organization, Diploma Programme, Mathematics: analysis and approaches guide, first assessment 2021 (published February 2019, updated November 2020), sections SL 2.8, 2.9, 2.10 and 2.11.