IB · MATH AI SL

Mathematics: Applications & Interpretation SL

Functions and modelling — Topic 2

Name: ____________________Date: October 2, 2026

1.
A gym charges a joining fee plus a fixed cost per visit. A member who makes 2 visits pays a total of $15, and a member who makes 6 visits pays a total of $35. Assuming the relationship is linear, find the equation for the total cost C in terms of the number of visits n, and interpret the gradient and the C-intercept in context.
[4 marks]
Marking points
- Calculates the gradient as (35 − 15)/(6 − 2) = 5.
- Finds the equation C = 5n + 5 using one of the given points.
- Interprets the gradient 5 as the cost per visit in dollars.
- Interprets the C-intercept 5 as the fixed joining fee charged regardless of visits.
Examiner tip: Treat a linear cost model exactly like any straight line: use two points to find the gradient first, then substitute to find the full equation.
2.
Marking analysis: A learner attempts the following task: “A gym charges a joining fee plus a fixed cost per visit. A member who makes 2 visits pays a total of $15, and a member who makes 6 visits pays a total of $35. Assuming the relationship is linear, find the equation for the total cost C in terms of the number of visits n, and interpret the gradient and the C-intercept in context.” Their response addresses only this point: “Calculates the gradient as (35 − 15)/(6 − 2) = 5.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Calculates the gradient as (35 − 15)/(6 − 2) = 5.
- Identifies the missing requirement: Finds the equation C = 5n + 5 using one of the given points.
- Identifies the missing requirement: Interprets the gradient 5 as the cost per visit in dollars.
- Identifies the missing requirement: Interprets the C-intercept 5 as the fixed joining fee charged regardless of visits.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
3.
The height of a ball thrown in the air is modelled by h(t) = −5t² + 20t + 1, where h is in metres and t is in seconds. (a) Find the time at which the ball reaches its maximum height. (b) Find the maximum height reached.
[3 marks]
Marking points
- Uses the vertex formula t = −b/(2a) with a = −5, b = 20 to obtain t = 2.
- Substitutes t = 2 into h(t).
- Obtains a maximum height of 21 metres.
Examiner tip: For a downward-opening parabola (negative coefficient of t²), the vertex always gives the maximum point, not a minimum.
4.
Marking analysis: A learner attempts the following task: “The height of a ball thrown in the air is modelled by h(t) = −5t² + 20t + 1, where h is in metres and t is in seconds. (a) Find the time at which the ball reaches its maximum height. (b) Find the maximum height reached.” Their response addresses only this point: “Uses the vertex formula t = −b/(2a) with a = −5, b = 20 to obtain t = 2.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]
Marking points
- Recognises credit for the stated point: Uses the vertex formula t = −b/(2a) with a = −5, b = 20 to obtain t = 2.
- Identifies the missing requirement: Substitutes t = 2 into h(t).
- Identifies the missing requirement: Obtains a maximum height of 21 metres.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
5.
A bacteria population grows exponentially according to P(t) = P0·e^(kt). The initial population is 200, and after 5 hours the population is 600. (a) Find the value of k. (b) Use your model to predict the population after 10 hours.
[4 marks]
Marking points
- Substitutes P(5) = 600 and P0 = 200 into 600 = 200e^(5k).
- Takes the natural logarithm of both sides to obtain k = ln(3)/5.
- Obtains k ≈ 0.220.
- Substitutes t = 10 into the model to predict a population of approximately 1,800.
Examiner tip: Notice that doubling the time from 5 to 10 hours exactly squares the growth factor (×3 becomes ×9), since the model is exponential.
6.
Marking analysis: A learner attempts the following task: “A bacteria population grows exponentially according to P(t) = P0·e^(kt). The initial population is 200, and after 5 hours the population is 600. (a) Find the value of k. (b) Use your model to predict the population after 10 hours.” Their response addresses only this point: “Substitutes P(5) = 600 and P0 = 200 into 600 = 200e^(5k).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Substitutes P(5) = 600 and P0 = 200 into 600 = 200e^(5k).
- Identifies the missing requirement: Takes the natural logarithm of both sides to obtain k = ln(3)/5.
- Identifies the missing requirement: Obtains k ≈ 0.220.
- Identifies the missing requirement: Substitutes t = 10 into the model to predict a population of approximately 1,800.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
7.
A radioactive substance has a half-life of 8 days. A sample initially has a mass of 50 grams. (a) Find the decay constant k in the model m(t) = 50e^(−kt). (b) Find the mass remaining after 20 days.
[4 marks]
Marking points
- Uses the half-life condition m(8) = 25 to set up 25 = 50e^(−8k).
- Solves to obtain k = ln(2)/8 ≈ 0.0866.
- Substitutes t = 20 into the model.
- Obtains a remaining mass of approximately 8.84 grams.
Examiner tip: The decay constant for a half-life model always equals ln(2) divided by the half-life — this shortcut avoids resolving the equation from scratch each time.
8.
Marking analysis: A learner attempts the following task: “A radioactive substance has a half-life of 8 days. A sample initially has a mass of 50 grams. (a) Find the decay constant k in the model m(t) = 50e^(−kt). (b) Find the mass remaining after 20 days.” Their response addresses only this point: “Uses the half-life condition m(8) = 25 to set up 25 = 50e^(−8k).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Uses the half-life condition m(8) = 25 to set up 25 = 50e^(−8k).
- Identifies the missing requirement: Solves to obtain k = ln(2)/8 ≈ 0.0866.
- Identifies the missing requirement: Substitutes t = 20 into the model.
- Identifies the missing requirement: Obtains a remaining mass of approximately 8.84 grams.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
9.
The sound intensity level L in decibels is modelled by L = 10 log10(I/I0), where I is the intensity in W/m² and I0 = 1 × 10⁻¹² W/m² is the reference intensity. Calculate the sound intensity level of a sound with intensity I = 1 × 10⁻⁶ W/m².
[3 marks]
Marking points
- Substitutes I = 1×10⁻⁶ and I0 = 1×10⁻¹² into the formula.
- Simplifies the ratio I/I0 = 10⁶.
- Obtains L = 10 log10(10⁶) = 60 decibels.
Examiner tip: When the ratio inside a log10 is a power of 10, the logarithm simplifies immediately to that power — no calculator is strictly needed.
10.
Marking analysis: A learner attempts the following task: “The sound intensity level L in decibels is modelled by L = 10 log10(I/I0), where I is the intensity in W/m² and I0 = 1 × 10⁻¹² W/m² is the reference intensity. Calculate the sound intensity level of a sound with intensity I = 1 × 10⁻⁶ W/m².” Their response addresses only this point: “Substitutes I = 1×10⁻⁶ and I0 = 1×10⁻¹² into the formula.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]
Marking points
- Recognises credit for the stated point: Substitutes I = 1×10⁻⁶ and I0 = 1×10⁻¹² into the formula.
- Identifies the missing requirement: Simplifies the ratio I/I0 = 10⁶.
- Identifies the missing requirement: Obtains L = 10 log10(10⁶) = 60 decibels.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
11.
The cost y of buying x identical notebooks varies directly with x. If 6 notebooks cost $24, find the constant of variation, and use it to find the cost of 10 notebooks.
[3 marks]
Marking points
- Sets up y = kx and substitutes y = 24, x = 6 to obtain k = 4.
- Writes the model as y = 4x.
- Substitutes x = 10 to obtain a cost of $40.
Examiner tip: In direct variation, y/x is always constant — you can check your value of k by dividing any given y by its matching x.
12.
Marking analysis: A learner attempts the following task: “The cost y of buying x identical notebooks varies directly with x. If 6 notebooks cost $24, find the constant of variation, and use it to find the cost of 10 notebooks.” Their response addresses only this point: “Sets up y = kx and substitutes y = 24, x = 6 to obtain k = 4.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]
Marking points
- Recognises credit for the stated point: Sets up y = kx and substitutes y = 24, x = 6 to obtain k = 4.
- Identifies the missing requirement: Writes the model as y = 4x.
- Identifies the missing requirement: Substitutes x = 10 to obtain a cost of $40.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
13.
The time y taken to complete a job varies inversely with the number of workers x. If 8 workers take 5 hours, find the constant of variation, and use it to find how long 20 workers would take.
[3 marks]
Marking points
- Sets up y = k/x and substitutes y = 5, x = 8 to obtain k = 40.
- Writes the model as y = 40/x.
- Substitutes x = 20 to obtain a time of 2 hours.
Examiner tip: In inverse variation, the product xy is always constant — more workers always means less time, never more.
14.
Marking analysis: A learner attempts the following task: “The time y taken to complete a job varies inversely with the number of workers x. If 8 workers take 5 hours, find the constant of variation, and use it to find how long 20 workers would take.” Their response addresses only this point: “Sets up y = k/x and substitutes y = 5, x = 8 to obtain k = 40.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]
Marking points
- Recognises credit for the stated point: Sets up y = k/x and substitutes y = 5, x = 8 to obtain k = 40.
- Identifies the missing requirement: Writes the model as y = 40/x.
- Identifies the missing requirement: Substitutes x = 20 to obtain a time of 2 hours.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
15.
A parking garage charges according to the piecewise function C(h) = 5 if 0 < h ≤ 1, C(h) = 5 + 3(h − 1) if h > 1, where h is the number of hours parked and C is the cost in dollars. Calculate the cost of parking for (a) 1 hour, and (b) 4 hours.
[3 marks]
Marking points
- Identifies that h = 1 falls in the first piece, giving a cost of $5.
- Identifies that h = 4 falls in the second piece.
- Substitutes h = 4 into C(h) = 5 + 3(h−1) to obtain a cost of $14.
Examiner tip: Always check which condition an input value satisfies before substituting it into a piecewise function — using the wrong piece is the most common error.
16.
Marking analysis: A learner attempts the following task: “A parking garage charges according to the piecewise function C(h) = 5 if 0 < h ≤ 1, C(h) = 5 + 3(h − 1) if h > 1, where h is the number of hours parked and C is the cost in dollars. Calculate the cost of parking for (a) 1 hour, and (b) 4 hours.” Their response addresses only this point: “Identifies that h = 1 falls in the first piece, giving a cost of $5.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]
Marking points
- Recognises credit for the stated point: Identifies that h = 1 falls in the first piece, giving a cost of $5.
- Identifies the missing requirement: Identifies that h = 4 falls in the second piece.
- Identifies the missing requirement: Substitutes h = 4 into C(h) = 5 + 3(h−1) to obtain a cost of $14.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
17.
Solve the equation 3e^(2x) = 45 for x, giving your answer to 3 significant figures.
[3 marks]
Marking points
- Divides both sides by 3 to obtain e^(2x) = 15.
- Takes the natural logarithm of both sides to obtain 2x = ln(15).
- Obtains x ≈ 1.35.
Examiner tip: Always isolate the exponential term completely before taking logarithms of both sides.
18.
Marking analysis: A learner attempts the following task: “Solve the equation 3e^(2x) = 45 for x, giving your answer to 3 significant figures.” Their response addresses only this point: “Divides both sides by 3 to obtain e^(2x) = 15.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]
Marking points
- Recognises credit for the stated point: Divides both sides by 3 to obtain e^(2x) = 15.
- Identifies the missing requirement: Takes the natural logarithm of both sides to obtain 2x = ln(15).
- Identifies the missing requirement: Obtains x ≈ 1.35.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
19.
Solve the equation log2(x) + log2(x − 2) = 3 for x.
[4 marks]
Marking points
- Uses the law of logarithms to combine the left side: log2[x(x − 2)] = 3.
- Converts to exponential form: x(x − 2) = 2³ = 8.
- Rearranges to x² − 2x − 8 = 0 and factorises as (x − 4)(x + 2) = 0, giving x = 4 or x = −2.
- Rejects x = −2 since log2(x) would be undefined there, confirming x = 4 as the only valid solution.
Examiner tip: Always check solutions against the original logarithmic equation's domain — a value that makes the argument of any logarithm negative or zero must be rejected.
20.
Marking analysis: A learner attempts the following task: “Solve the equation log2(x) + log2(x − 2) = 3 for x.” Their response addresses only this point: “Uses the law of logarithms to combine the left side: log2[x(x − 2)] = 3.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Uses the law of logarithms to combine the left side: log2[x(x − 2)] = 3.
- Identifies the missing requirement: Converts to exponential form: x(x − 2) = 2³ = 8.
- Identifies the missing requirement: Rearranges to x² − 2x − 8 = 0 and factorises as (x − 4)(x + 2) = 0, giving x = 4 or x = −2.
- Identifies the missing requirement: Rejects x = −2 since log2(x) would be undefined there, confirming x = 4 as the only valid solution.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
21.
Given f(x) = 2x + 1 and g(x) = x² − 3, find (a) f(g(2)) and (b) g(f(2)).
[3 marks] · no calculator
Marking points
- Calculates g(2) = 2² − 3 = 1, then f(1) = 2(1) + 1 = 3, giving f(g(2)) = 3.
- Calculates f(2) = 2(2) + 1 = 5, then g(5) = 5² − 3 = 22, giving g(f(2)) = 22.
- Notes that f(g(2)) ≠ g(f(2)), confirming that function composition is not generally commutative.
Examiner tip: Always evaluate the innermost function first in a composition — f(g(x)) means apply g first, then apply f to the result.
22.
Marking analysis: A learner attempts the following task: “Given f(x) = 2x + 1 and g(x) = x² − 3, find (a) f(g(2)) and (b) g(f(2)).” Their response addresses only this point: “Calculates g(2) = 2² − 3 = 1, then f(1) = 2(1) + 1 = 3, giving f(g(2)) = 3.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Marking points
- Recognises credit for the stated point: Calculates g(2) = 2² − 3 = 1, then f(1) = 2(1) + 1 = 3, giving f(g(2)) = 3.
- Identifies the missing requirement: Calculates f(2) = 2(2) + 1 = 5, then g(5) = 5² − 3 = 22, giving g(f(2)) = 22.
- Identifies the missing requirement: Notes that f(g(2)) ≠ g(f(2)), confirming that function composition is not generally commutative.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
23.
Find the inverse function f⁻¹(x) for f(x) = (3x − 2)/5, and verify your answer by finding f⁻¹(f(4)).
[4 marks] · no calculator
Marking points
- Sets y = (3x − 2)/5 and swaps x and y to begin solving for the inverse.
- Rearranges to obtain f⁻¹(x) = (5x + 2)/3.
- Calculates f(4) = (3(4) − 2)/5 = 2.
- Calculates f⁻¹(2) = (5(2) + 2)/3 = 4, confirming f⁻¹(f(4)) = 4 as expected.
Examiner tip: A correct inverse function must always satisfy f⁻¹(f(x)) = x — use this as a quick check on any inverse function you find.
24.
Marking analysis: A learner attempts the following task: “Find the inverse function f⁻¹(x) for f(x) = (3x − 2)/5, and verify your answer by finding f⁻¹(f(4)).” Their response addresses only this point: “Sets y = (3x − 2)/5 and swaps x and y to begin solving for the inverse.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] · no calculator
Marking points
- Recognises credit for the stated point: Sets y = (3x − 2)/5 and swaps x and y to begin solving for the inverse.
- Identifies the missing requirement: Rearranges to obtain f⁻¹(x) = (5x + 2)/3.
- Identifies the missing requirement: Calculates f(4) = (3(4) − 2)/5 = 2.
- Identifies the missing requirement: Calculates f⁻¹(2) = (5(2) + 2)/3 = 4, confirming f⁻¹(f(4)) = 4 as expected.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
25.
A model predicts the number of visitors to a museum as V(x) = 100√(20 − x), where x is the entry price in dollars. Explain why the domain of this model should be restricted to 0 ≤ x ≤ 20 in this real-world context.
[2 marks] · no calculator
Marking points
- Explains that x cannot be negative, since an entry price cannot be below zero, giving the lower bound x ≥ 0.
- Explains that x cannot exceed 20, since the expression under the square root would become negative (undefined for real numbers), giving the upper bound x ≤ 20.
Examiner tip: A model's mathematical domain (where the formula is defined) and its real-world domain (where the context makes sense) should both be considered — here they happen to agree.
26.
Marking analysis: A learner attempts the following task: “A model predicts the number of visitors to a museum as V(x) = 100√(20 − x), where x is the entry price in dollars. Explain why the domain of this model should be restricted to 0 ≤ x ≤ 20 in this real-world context.” Their response addresses only this point: “Explains that x cannot be negative, since an entry price cannot be below zero, giving the lower bound x ≥ 0.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[2 marks] · no calculator
Marking points
- Recognises credit for the stated point: Explains that x cannot be negative, since an entry price cannot be below zero, giving the lower bound x ≥ 0.
- Identifies the missing requirement: Explains that x cannot exceed 20, since the expression under the square root would become negative (undefined for real numbers), giving the upper bound x ≤ 20.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
27.
The graph of y = x² is transformed to the graph of y = (x − 3)² + 2. Describe the two transformations applied, and state the coordinates of the vertex of the transformed graph.
[2 marks] · no calculator
Marking points
- Describes a horizontal translation of 3 units to the right, and a vertical translation of 2 units up.
- States the vertex of the transformed graph as (3, 2).
Examiner tip: For y = (x − h)² + k, the vertex moves to (h, k) — note that the sign inside the brackets is opposite to the direction of the horizontal shift.
28.
Marking analysis: A learner attempts the following task: “The graph of y = x² is transformed to the graph of y = (x − 3)² + 2. Describe the two transformations applied, and state the coordinates of the vertex of the transformed graph.” Their response addresses only this point: “Describes a horizontal translation of 3 units to the right, and a vertical translation of 2 units up.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[2 marks] · no calculator
Marking points
- Recognises credit for the stated point: Describes a horizontal translation of 3 units to the right, and a vertical translation of 2 units up.
- Identifies the missing requirement: States the vertex of the transformed graph as (3, 2).
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
29.
The height of the tide in a harbour, h metres, t hours after midnight, is modelled by h(t) = 5 + 3sin(πt/6). (a) State the amplitude and the period of this model. (b) Calculate the height of the tide at t = 3 hours.
[4 marks]
Marking points
- States the amplitude as 3 metres.
- Uses period = 2π ÷ (π/6) to obtain a period of 12 hours.
- Substitutes t = 3 into h(t).
- Obtains h(3) = 8 metres.
Examiner tip: For h(t) = a + b·sin(ct), the amplitude is |b|, the vertical shift is a, and the period is 2π/c — identify each parameter separately before interpreting.
30.
Marking analysis: A learner attempts the following task: “The height of the tide in a harbour, h metres, t hours after midnight, is modelled by h(t) = 5 + 3sin(πt/6). (a) State the amplitude and the period of this model. (b) Calculate the height of the tide at t = 3 hours.” Their response addresses only this point: “States the amplitude as 3 metres.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: States the amplitude as 3 metres.
- Identifies the missing requirement: Uses period = 2π ÷ (π/6) to obtain a period of 12 hours.
- Identifies the missing requirement: Substitutes t = 3 into h(t).
- Identifies the missing requirement: Obtains h(3) = 8 metres.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
31.
Find the coordinates of the points of intersection of the line y = 2x + 1 and the curve y = x² − 4x + 5.
[5 marks]
Marking points
- Sets the two expressions for y equal: 2x + 1 = x² − 4x + 5.
- Rearranges to obtain x² − 6x + 4 = 0.
- Uses the quadratic formula (or GDC) to obtain x ≈ 5.24 or x ≈ 0.764.
- Substitutes each x-value back into y = 2x + 1.
- States the intersection points as approximately (5.24, 11.5) and (0.764, 2.53).
Examiner tip: Always substitute back into the simpler (linear) equation to find each y-value, rather than the more complex quadratic one.
32.
Marking analysis: A learner attempts the following task: “Find the coordinates of the points of intersection of the line y = 2x + 1 and the curve y = x² − 4x + 5.” Their response addresses only this point: “Sets the two expressions for y equal: 2x + 1 = x² − 4x + 5.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks]
Marking points
- Recognises credit for the stated point: Sets the two expressions for y equal: 2x + 1 = x² − 4x + 5.
- Identifies the missing requirement: Rearranges to obtain x² − 6x + 4 = 0.
- Identifies the missing requirement: Uses the quadratic formula (or GDC) to obtain x ≈ 5.24 or x ≈ 0.764.
- Identifies the missing requirement: Substitutes each x-value back into y = 2x + 1.
- Identifies the missing requirement: States the intersection points as approximately (5.24, 11.5) and (0.764, 2.53).
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.