IB DP Mathematics: Applications and Interpretation – Composite and inverse functions, transformations, further models and logarithmic scales (HL) 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.

These practice questions are for IB Diploma Programme Mathematics: Applications and Interpretation, aligned to the IB Mathematics: applications and interpretation guide, first assessment 2021. They cover syllabus sections AHL 2.7–2.10: composite and inverse functions, transformations, further models, and logarithmic scales. Every question is HL only. They follow the IB guide for first assessment 2021, which remains the examined syllabus until the new course is first assessed in May 2029, so they apply to the May and November 2026, 2027 and 2028 sessions.

All three HL papers require technology, so every question is labelled “calculator allowed”. Show written working. Give answers exact or to 3 s.f. unless told otherwise.

Revise first from the study guide or revision notes. See also the Maths AI course hub and printable checklist.

Questions

1. (HL, calculator allowed) f(x) = 3x − 1 and g(x) = x² + 2. Find (g ∘ f)(x) in the form ax² + bx + c. [2]

2. (HL, calculator allowed) A bank converts d US dollars to c(d) = 0.79d pounds, then pays out f(p) = 0.97p − 1.5 pounds. Give money to 2 d.p.

(a) Find (f ∘ c)(d). [2] (b) Find the amount paid out for $400. [1] (c) Find (f ∘ c)⁻¹(x), and hence the number of dollars needed to receive £500. [3]

3. (HL, calculator allowed) f(x) = (x + 1)² − 4.

(a) The domain of f is restricted to x ≥ p so that f⁻¹ exists. Find the smallest value of p. [1] (b) Find f⁻¹(x). [3] (c) State the domain and range of f⁻¹. [2]

4. (HL, calculator allowed) The point (3, −2) lies on the graph of y = f(x). Find the image of this point on each graph.

(a) y = f(x − 4) + 5 [1] (b) y = −2f(x) [1] (c) y = f(3x) [1] (d) y = 2f(x + 1) − 3 [2]

5. (HL, calculator allowed)

(a) Describe fully a sequence of transformations that maps the graph of y = f(x) onto the graph of y = 4 − 2f(x/3). [4] (b) The graph of y = ln x is translated by the vector (0, 3), and the result is then stretched vertically with scale factor 2. Find the equation of the final graph. [2]

6. (HL, calculator allowed) A tracer has mass M(t) = 60e^(−kt) mg after t hours. Its half-life is 6.0 hours.

(a) Find the exact value of k. [2] (b) Find the mass after 15 hours. [2] (c) Find the time taken for the mass to fall to 1 mg. [2]

7. (HL, calculator allowed; combines AHL 2.7 and 2.9) The number of fish in a lake is modelled by P(t) = 5000/(1 + Ce^(−kt)), t years after the lake is stocked. P(0) = 400 and P(3) = 1600.

(a) Find C. [2] (b) Find k. [3] (c) Show that the inverse of this model is t = (1/k) ln(CP/(5000 − P)). [3] (d) Hence find the time taken for the population to reach 4500. [1] (e) State the carrying capacity and what the model predicts in the long run. [1]

8. (HL, calculator allowed) A Ferris wheel’s lowest point is 2 m and its highest point 42 m above the ground. One rotation takes 16 minutes. A passenger boards at the lowest point at t = 0. Their height is h(t) = a sin(b(t − c)) + d metres, t in minutes, a, b > 0.

(a) Find a and d. [2] (b) Find b. [2] (c) Find the smallest positive value of c. [2] (d) Find the length of time in each rotation that the passenger is more than 35 m above the ground. [3]

9. (HL, calculator allowed) A piecewise model is defined by f(x) = 2 + 0.3x for 0 ≤ x < 5 and f(x) = ax² − 0.5x + 1 for 5 ≤ x ≤ 8. The graph of f has no break at x = 5.

(a) Find a. [2] (b) Find f(7). [1] (c) Solve f(x) = 5. [2]

10. (HL, calculator allowed) A learner’s percentage test score after n practice sessions is modelled by S(n) = a + b ln n, n ≥ 1. S(1) = 40 and S(5) = 64.

(a) Find a and b. [3] (b) Find S(12). [1] (c) Find the least number of sessions needed for a score above 90. [2] (d) Explain why the model is not suitable for large values of n. [1]

11. (HL, calculator allowed) A graph of log₁₀ y against log₁₀ x is a straight line through (0, 0.6) and (2, 1.6). Find a and b, where y = ax^b. [3]

12. (HL, calculator allowed; combines AHL 2.9 and 2.10) An experiment gives these results.

x	1	2	3	4	5	6
y	3.1	4.8	7.7	12.4	19.6	31.5

(a) Find Pearson’s correlation coefficient r for (x, ln y) and for (ln x, ln y). [2] (b) Hence state, with a reason, whether an exponential or a power model fits better. [1] (c) Find the equation of the regression line of ln y on x. [2] (d) Hence find k and a, where y = ka^x. [3] (e) Use the model to predict y when x = 9. [1] (f) Comment on the reliability of this prediction. [1]

Answers

1. (g ∘ f)(x) = g(3x − 1) = (3x − 1)² + 2 [1] = 9x² − 6x + 3 [1] [2] Examiner insight: the method mark is for putting f inside g; expanding f(g(x)) instead scores zero.

2. (a) (f ∘ c)(d) = f(0.79d) = 0.97(0.79d) − 1.5 [1] = 0.7663d − 1.5 [1] (b) 0.7663 × 400 − 1.5 = £305.02 [1] (c) y = 0.7663d − 1.5 ⇒ d = (y + 1.5)/0.7663 [1], so (f ∘ c)⁻¹(x) = (x + 1.5)/0.7663 [1]. Then (500 + 1.5)/0.7663 = $654.44 [1] Examiner insight: the question asks for 2 d.p., so $654 loses the final accuracy mark.

3. (a) The vertex is at x = −1, so p = −1 [1] (b) y + 4 = (x + 1)² [1]; x + 1 = +√(y + 4), taking the positive root because x ≥ −1 [1]; f⁻¹(x) = −1 + √(x + 4) [1] (c) Domain x ≥ −4 [1]; range f⁻¹(x) ≥ −1 [1] Examiner insight: leaving ± in the final inverse loses the last mark in (b).

4. (a) (7, 3) [1] (b) (3, 4) [1] (c) (1, −2) [1] (d) x-coordinate 3 − 1 = 2 [1]; y-coordinate 2(−2) − 3 = −7, giving (2, −7) [1] Examiner insight: the two coordinates in (d) are marked separately, so show each one’s working and you keep one mark if the other slips.

5. (a) Horizontal stretch, scale factor 3 [1]; vertical stretch, scale factor 2 [1]; reflection in the x-axis [1]; then translation by the vector (0, 4), after the stretch and reflection [1] (b) After the translation: y = ln x + 3 [1]. After the stretch: y = 2(ln x + 3), so y = 2 ln x + 6 [1] Examiner insight: “describe fully” needs the type and the scale factor or vector for each step; “shift up 4” without a vector does not earn the mark.

6. (a) 30 = 60e^(−6k) ⇒ e^(−6k) = 1/2 [1] ⇒ k = (ln 2)/6 [1] (b) M(15) = 60e^(−15 ln 2/6) [1] = 10.6 mg [1] (c) 60e^(−kt) = 1 ⇒ t = (ln 60)/k [1] = 35.4 hours [1] Examiner insight: “exact value” means ln 2/6; writing 0.116 loses the accuracy mark.

7. (a) 5000/(1 + C) = 400 [1] ⇒ C = 11.5 [1] (b) 1 + 11.5e^(−3k) = 5000/1600 = 3.125 [1]; e^(−3k) = 2.125/11.5 = 0.18478… [1]; k = 0.563 [1] (c) 1 + Ce^(−kt) = 5000/P [1]; Ce^(−kt) = (5000 − P)/P [1]; e^(kt) = CP/(5000 − P), so taking ln and dividing by k gives t = (1/k) ln(CP/(5000 − P)) [1] (d) t = (1/0.56286) ln(11.5 × 4500/500) = 8.24 years [1] (e) 5000; P approaches 5000 but never reaches it [1] Examiner insight: in a “show that” part every algebraic line must appear and the last must match the given result; jumping straight to it earns nothing.

8. (a) a = (42 − 2)/2 = 20 [1]; d = (42 + 2)/2 = 22 [1] (b) 2π/b = 16 [1] ⇒ b = π/8 [1] (c) h(0) = 2 needs sin(−πc/8) = −1, so πc/8 = π/2 [1] ⇒ c = 4 [1] (d) Solve 20 sin((π/8)(t − 4)) + 22 = 35 on a GDC [1]; t = 5.80 and t = 10.20 [1]; time above 35 m = 4.40 minutes [1] Examiner insight: in degree mode the GDC gives wrong intersections in (d); the set-up still earns its method mark, but both accuracy marks go.

9. (a) 2 + 0.3(5) = 25a − 0.5(5) + 1 [1] ⇒ 3.5 = 25a − 1.5 ⇒ a = 0.2 [1] (b) f(7) = 0.2(49) − 3.5 + 1 = 7.3 [1] (c) The first piece is at most 3.5, so use the second: 0.2x² − 0.5x − 4 = 0 [1]; x = 5.89, rejecting x = −3.39 as outside 5 ≤ x ≤ 8 [1] Examiner insight: the final mark in (c) is for the root inside the domain; giving both roots as answers can lose it.

10. (a) S(1) = a + b ln 1 = a, so a = 40 [1]; 40 + b ln 5 = 64 [1] ⇒ b = 24/ln 5 = 14.9 [1] (b) S(12) = 40 + 14.912 × ln 12 = 77.1 [1] (c) 40 + b ln n > 90 ⇒ n > e^(50/b) = 28.6 [1] ⇒ 29 sessions [1] (d) S(n) keeps increasing without limit, so the model predicts a score above 100% once n ≥ 56, which is impossible [1] Examiner insight: a “least number” answer must be a whole number that meets the condition; 28.6, or 28, loses the final mark.

11. Gradient b = (1.6 − 0.6)/2 = 0.5 [1]; intercept log₁₀ a = 0.6 [1]; a = 10^0.6 = 3.98, so y = 3.98x^0.5 [1] [3] Examiner insight: the intercept is log₁₀ a, not a; giving a = 0.6 loses only the final mark.

12. (a) For (x, ln y): r = 0.9999 [1]. For (ln x, ln y): r = 0.965 [1] (b) Exponential, because r for (x, ln y) is closer to 1 [1] (c) ln y = mx + c with m = 0.465 [1] and c = 0.652 [1] (d) ln y = ln k + x ln a, so ln k = 0.652 and ln a = 0.465 [1]; k = e^0.6517 = 1.92 [1]; a = e^0.4654 = 1.59 [1] (e) y = 1.9189 × 1.5927^9 = 127 (125 from rounded k and a) [1] (f) x = 9 is outside 1 ≤ x ≤ 6: extrapolation, so it may be unreliable [1] Examiner insight: r for (x, ln y) rounds to 1.00 at 3 s.f.; quote it to 4 d.p. so the comparison in (b) is clear.

Where marks are usually lost

• Leaving an inverse with ± or without its domain.
• Treating y = f(3x) as a stretch by 3; the scale factor is 1/3.
• Describing transformations as “moves” without a vector or scale factor.
• Rounding k early in half-life and logistic questions, then losing accuracy in the final answer.
• Answering a “least number of sessions” question with a decimal or by rounding down.
• Reading a log-graph intercept as a instead of log₁₀ a.

Next steps

• Revision notes.
• Study guide.
• Maths AI course hub.
• Printable checklist.
• All free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

International Baccalaureate Organization, Diploma Programme, Mathematics: applications and interpretation guide, first assessment 2021. Sections AHL 2.7, 2.8, 2.9 and 2.10.