IB DP Mathematics: Applications and Interpretation – Functions and modelling 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 for IB DP Mathematics: applications and interpretation is aligned to the IB Mathematics: applications and interpretation guide, first assessment 2021, syllabus sections SL 2.1–2.6, which 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.

The guide requires a GDC for every paper in this course, so every question is labelled “(calculator allowed)”. Give answers exactly or to 3 s.f. unless told otherwise.

Links: study guide · revision notes · course hub · checklist.

Questions

1. (calculator allowed) Find the equation of the line through P(−1, 4) and Q(5, 1). Give your answer in the form ax + by + d = 0, where a, b and d are integers. [3]

2. (calculator allowed) The line L₁ has equation 3x − 4y + 12 = 0.

(a) Find the gradient and the y-intercept of L₁. [2] (b) The line L₂ is perpendicular to L₁ and passes through (6, −2). Find the equation of L₂ in the form y = mx + c. [3]

3. (calculator allowed) The function f is defined by f(x) = 5 − √(x + 2).

(a) State the largest possible domain of f and the corresponding range. [2] (b) Find f⁻¹(1). [2]

4. (calculator allowed) A kiosk gives R(x) = 5.2x − 12 ringgit for x pounds, x ≥ 10, where 12 is a fixed fee.

(a) Find R(150). [1] (b) Find R⁻¹(x). [2] (c) State the domain of R⁻¹. [1] (d) Find R⁻¹(1000), to the nearest penny, and interpret it in context. [2]

5. (calculator allowed) Let f(x) = x³ − 4x² + x + 6.

(a) Find the zeros of f. [2] (b) Find the coordinates of the local maximum point. [2] (c) Find the coordinates of the local minimum point. [1]

6. (calculator allowed) Let g(x) = (4x + 3)/(x − 2).

(a) Write down the equations of the vertical and horizontal asymptotes of the graph of g. [2] (b) Find the coordinates of the points where the graph of g meets the line y = x + 1. [3]

7. (calculator allowed) The temperature of soup t minutes after serving is T(t) = 65e^(−0.08t) + 20 °C, t ≥ 0.

(a) Find the temperature when the soup is served. [1] (b) Find the temperature after 10 minutes. [2] (c) Find the time taken for the soup to cool to 30 °C. [2] (d) Write down the equation of the horizontal asymptote of the graph of T, and state what it represents. [1]

8. (calculator allowed) A seat on an observation wheel is h(t) = −16 cos(12t) + 18 metres high t minutes after boarding (12t in degrees).

(a) Write down the amplitude, the period and the equation of the principal axis. [3] (b) Write down the maximum height of the seat. [1] (c) Find the first time after boarding at which the seat is 30 m above the ground. [2]

9. (calculator allowed) Light intensity at d metres from a lamp is I = ad⁻², d > 0. When d = 2, I = 180.

(a) Find a. [2] (b) Find I when d = 5. [1] (c) Write down the equation of the vertical asymptote of the graph of I against d. [1]

10. (calculator allowed) A 25 m pool is 1.2 m deep for 0 ≤ x ≤ 8, where x m is the distance from the shallow end. The floor then slopes in a straight line to 3.0 m deep at x = 20, and stays 3.0 m deep to x = 25.

(a) Write down a piecewise linear model d(x) for the depth, stating the domain of each piece. [3] (b) Find the depth at x = 14. [1] (c) Find the value of x where the depth is 2.5 m. [2] (d) Interpret the gradient of the sloping section. [1]

11. (calculator allowed) A ball leaves a point 1.8 m above flat ground. Its height at horizontal distance x m is h(x) = ax² + bx + c metres. h = 5.0 when x = 4, and h = 3.8 when x = 10.

(a) Write down three equations in a, b and c, and solve them to find the model. [4] (b) Find the maximum height of the ball and the horizontal distance at which it occurs. [2] (c) Find the horizontal distance at which the ball hits the ground. [2] (d) State a reasonable domain for the model. [1]

12. (calculator allowed) A new car costs 24 000 dollars. Its value after t years is V(t) = k·aᵗ. After 2 years it is worth 17 340 dollars.

(a) Find k and a. [3] (b) Find the value of the car after 7 years. [2] (c) Find the time at which the value first falls below 5000 dollars. [2] (d) A linear model L(t) is fitted to the same two values. Find L(t) in the form L(t) = mt + c. [2] (e) Use both models to predict the value after 10 years, and comment on which model is more reasonable. [2]

Answers

1. Gradient m = (1 − 4)/(5 − (−1)) = −3/6 = −1/2 [1] y − 4 = −(1/2)(x + 1) [1] 2y − 8 = −x − 1, so x + 2y − 7 = 0 [1] Examiner insight: If integer general form is asked for, a correct line left as y = −0.5x + 3.5 loses the final A mark.

2. (a) 4y = 3x + 12, so y = 0.75x + 3 [1] Gradient 3/4, y-intercept 3 (the point (0, 3)) [1] (b) Perpendicular gradient = −1/(3/4) = −4/3 [1] y + 2 = −(4/3)(x − 6) [1] y = −(4/3)x + 6 [1] Examiner insight: A wrong gradient in (a) can still earn follow-through marks in (b) if you visibly apply −1/m to it.

3. (a) Need x + 2 ≥ 0: domain x ≥ −2 [1] √(x + 2) ≥ 0, so range f(x) ≤ 5 [1] (b) f⁻¹(1) is the x with f(x) = 1: 5 − √(x + 2) = 1 [1] √(x + 2) = 4, x + 2 = 16, f⁻¹(1) = 14 [1] Examiner insight: Write the range in terms of f(x) or y; “x ≤ 5” describes a different set and scores nothing.

4. (a) R(150) = 5.2 × 150 − 12 = 780 − 12 = 768 ringgit [1] (b) y = 5.2x − 12, so x = (y + 12)/5.2 [1] R⁻¹(x) = (x + 12)/5.2 [1] (c) Domain of R⁻¹ = range of R; R(10) = 40, so x ≥ 40 [1] (d) R⁻¹(1000) = 1012/5.2 = 194.615… = £194.62 [1] Exchanging £194.62 gives 1000 ringgit after the fee. [1] Examiner insight: “To the nearest penny” replaces the 3 s.f. default, so 195 loses the accuracy mark.

5. (a) Using the GDC zero function (or factorising as (x + 1)(x − 2)(x − 3)) [1] x = −1, x = 2, x = 3 [1] (b) GDC maximum: x = 0.131 [1], (0.131, 6.06) [1] (c) GDC minimum: (2.54, −0.879) [1] Examiner insight: When “coordinates” are asked for, an x-value alone loses the accuracy mark.

6. (a) x = 2 [1] and y = 4 [1] (b) (4x + 3)/(x − 2) = x + 1, solved on the GDC (or x² − 5x − 5 = 0) [1] x = −0.854 or x = 5.85 [1] (−0.854, 0.146) and (5.85, 6.85) [1] Examiner insight: Asymptotes must be equations; “2 and 4” alone does not score.

7. (a) T(0) = 65 + 20 = 85 °C [1] (b) T(10) = 65e^(−0.8) + 20 [1] = 49.2 °C [1] (c) 65e^(−0.08t) + 20 = 30 [1] t = ln(6.5)/0.08 = 23.4 minutes [1] (d) T = 20, the temperature of the room the soup cools towards. [1] Examiner insight: In (c), writing the equation T(t) = 30 earns the method mark even when the GDC does the solving.

8. (a) Amplitude 16 m [1]; period = 360/12 = 30 minutes [1]; principal axis h = 18 [1] (b) 18 + 16 = 34 m [1] (c) −16 cos(12t) + 18 = 30, solved on the GDC in degrees [1] t = 11.5 minutes [1] Examiner insight: Radian mode gives a wrong time and loses the A mark; the written equation still earns the M mark.

9. (a) 180 = a/2² [1], so a = 720 [1] (b) I = 720/25 = 28.8 [1] (c) d = 0 (the vertical axis) [1] Examiner insight: Using I = a/d instead of a/d² gives a = 360 and loses both marks in (a); (b) can then only earn follow-through.

10. (a) Gradient of slope = (3.0 − 1.2)/(20 − 8) = 0.15 [1] Middle piece: d = 1.2 + 0.15(x − 8), which simplifies to 0.15x [1] d(x) = 1.2 for 0 ≤ x ≤ 8; d(x) = 0.15x for 8 < x ≤ 20; d(x) = 3.0 for 20 < x ≤ 25 [1] (b) d(14) = 0.15 × 14 = 2.1 m [1] (c) 0.15x = 2.5 [1], x = 16.7 m [1] (d) The depth increases by 0.15 m for each metre along the pool. [1] Examiner insight: The final mark in (a) needs the domain of every piece.

11. (a) At x = 0: c = 1.8 [1] 16a + 4b + 1.8 = 5.0 and 100a + 10b + 1.8 = 3.8 [1] Solving on the GDC: a = −0.1 [1], b = 1.2 [1], so h(x) = −0.1x² + 1.2x + 1.8 (b) Vertex at x = −1.2/(2 × −0.1) = 6 [1]; maximum height 5.4 m at x = 6 m [1] (c) −0.1x² + 1.2x + 1.8 = 0, GDC gives x = −1.35 or 13.3 [1]; reject the negative root: x = 13.3 m [1] (d) 0 ≤ x ≤ 13.3 [1] Examiner insight: Listing both roots in (c) loses the final mark; reject the negative one because x ≥ 0.

12. (a) V(0) = 24 000, so k = 24 000 [1] 24 000a² = 17 340 [1], a² = 0.7225, a = 0.85 [1] (b) V(7) = 24 000 × 0.85⁷ [1] = 7690 dollars (3 s.f.) [1] (c) 24 000 × 0.85ᵗ = 5000, GDC or t = ln(5000/24 000)/ln 0.85 [1] t = 9.65, so after 9.65 years (during the 10th year) [1] (d) m = (17 340 − 24 000)/2 = −3330 [1] L(t) = −3330t + 24 000 [1] (e) L(10) = −9300 and V(10) = 4720 (3 s.f.) [1] The exponential model is more reasonable: a car cannot have a negative value, and the linear model fails when extrapolated beyond t ≈ 7.2. [1] Examiner insight: “The exponential model is better” with no reason from the numbers or context does not score.

Where marks are usually lost

• Using −m instead of −1/m for a perpendicular gradient.
• Writing a range with x instead of f(x) or y.
• Forgetting that the domain of an inverse is the range of the original function.
• Giving turning points as x-values only, or rounding to 2 s.f.
• Using radian mode for a sinusoidal model written in degrees.
• Rounding a parameter (such as a in kaᵗ) and then using the rounded value in later parts.
• Keeping a negative root, or a negative time, when the context rules it out.
• Comparing models without a reason based on the context or the numbers.

Next steps

• Review the revision notes.
• Relearn weak topics in the study guide.
• See every unit on the IB Maths AI course hub.
• Tick off each section on the printable checklist.
• Try 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 SL 2.1–2.6.