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IB Maths AI HL · Paper 2

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2

Session: May 2023
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 7

Q1 Q2 Q3 Q4 Q5 Q6 Q7

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2023 TZ2 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–7 · 110 marks

Question 1

13 marks

Odette, Khemil and the football in a park

Geometry and trigonometry

(a)

3 marks

The two sides at A are $O A = 25.9$ and $A B = 6.36$ with included angle $12 5^{\circ}$ .

Area = \frac{1}{2} (25.9) (6.36) sin 12 5^{\circ} \approx 67.5 m^{2}

(b)

3 marks

Cosine rule in triangle AKB with $A K = 12$ , $A B = 6.36$ and angle $4 5^{\circ}$ at A.

K B^{2} = 1 2^{2} + 6.3 6^{2} - 2 (12) (6.36) cos 4 5^{\circ} \Rightarrow K B \approx 8.75 m

(c)

4 marks

Khemil is at K, so his distance to the football is $K X = 22.2$ m. In triangle OKX the angle at X is $180 - 53.8 - 51.1 = 75. 1^{\circ}$ . Use the sine rule for $O X$ , opposite the angle at K.
Odette is 21.4 m from the football and Khemil is 22.2 m, so Odette is closer.

\frac{O X}{sin 51. 1 ^{\circ}} = \frac{22.2}{sin 53. 8 ^{\circ}} \Rightarrow O X \approx 21.4 m

(d)

3 marks

The path is a semicircle with centre A and $K A Y = 4 5^{\circ}$ , so $K A X = 18 0^{\circ} - 4 5^{\circ} = 13 5^{\circ}$ .

arc K X = 12 \times \frac{135 π}{180} = 9 π \approx 28.3 m

Question 2

12 marks

Bacteria growth: exponential models and when they are equal

Number and algebra

(a)

2 marks

Put $t = 0$ .
It is the number of bacteria at the start of the experiment.

P (0) = 1200 \times k^{0} = 1200

(b)

2 marks

Use $P (3) = 18750$ .

1200 k^{3} = 18750 \Rightarrow k^{3} = 15.625 \Rightarrow k = 2.5

(c)

2 marks

1 hour 30 minutes is $t = 1.5$ .

P (1.5) = 1200 (2.5)^{1.5} \approx 4743

(d)

2 marks

Set $P (t) = S (t)$ .

1200 (2.5)^{t} = 5000 (1.65)^{t} \Rightarrow (\frac{2.5}{1.65})^{t} = \frac{25}{6} \Rightarrow t = \frac{ln ( 25/6 )}{ln ( 2.5/1.65 )} \approx 3.43

(e)

4 marks

Set $S (t) = 19000$ .
That is 2 hours and $0.6659 \times 60 = 39.95$ minutes.

1.6 5^{t} = 3.8 \Rightarrow t = \frac{ln 3.8}{ln 1.65} = 2.6659 hours m = 40

Question 3

16 marks

A garden surrounded by a path: area model and maximum

Calculus

(a)

5 marks

The park has length $x$ and width $\frac{1200}{x}$ . The garden has length $x - 3$ and width $\frac{1200}{x} - 4$ .

A = (x - 3) (\frac{1200}{x} - 4) = 1200 - 4 x - \frac{3600}{x} + 12 = 1212 - 4 x - \frac{3600}{x}

(b)

4 marks

Set $A = 800$ .
The width is $\frac{1200}{x}$ .

1212 - 4 x - \frac{3600}{x} = 800 \Rightarrow 4 x^{2} - 412 x + 3600 = 0 \Rightarrow x^{2} - 103 x + 900 = 0 x = 93.4 or x = 9.64 93.4 m by 12.9 m or 9.64 m by 124.5 m

(c)

3 marks

Differentiate $A = 1212 - 4 x - 3600 x^{- 1}$ .

\frac{d A}{d x} = - 4 + \frac{3600}{x ^{2}}

(d)

2 marks

Set the derivative to zero.
This is a maximum because $\frac{d ^{2} A}{d x ^{2}} = - \frac{7200}{x ^{3}} < 0$ .

\frac{3600}{x ^{2}} = 4 \Rightarrow x^{2} = 900 \Rightarrow x = 30

(e)

2 marks

Substitute $x = 30$ .

A = 1212 - 120 - 120 = 972 m^{2}

Question 4

19 marks

Flights between five US cities: graph terms, Kruskal and bounds

Number and algebra

(a)

2 marks

The graph is connected because it is possible to travel between any two cities, along one or more flights. It is not complete because some pairs of cities have no direct flight, for example Los Angeles and Dallas.

(b)

3 marks

Sort the flights: Chicago-Dallas 30, Dallas-New York 39, Los Angeles-Chicago 41, Los Angeles-New York 55, Seattle-Los Angeles 58, and so on. Add the cheapest edge that does not make a cycle.
The edge Los Angeles-New York (55) would make a cycle, so it is skipped. The tree has total weight $30 + 39 + 41 + 58 = 168$ .

Chicago-Dallas (30), Dallas-New York (39), Los Angeles-Chicago (41), Seattle-Los Angeles (58)

(c)

2 marks

A tour that follows the tree edges twice (once out and once back) costs twice the tree.

2 \times 168 = $336

(d)

3 marks

Start at Los Angeles and always go to the cheapest unvisited city.

L A \to D a l l a s (26) \to C hi c a g o (30) \to N e w Y or k (68) \to S e a ttl e (66) \to L A (58) 26 + 30 + 68 + 66 + 58 = $248

(e)

5 marks

Delete Chicago. The minimum spanning tree of Seattle, Los Angeles, Dallas and New York uses LA-Dallas 26, Dallas-New York 39 and Seattle-LA 58.
Delete Seattle. The minimum spanning tree of the other four uses LA-Dallas 26, Dallas-Chicago 30 and Dallas-New York 39.

26 + 39 + 58 = 123, add the two cheapest at Chicago: 30 + 41 = 71 lower bound = 123 + 71 = $194 26 + 30 + 39 = 95, add the two cheapest at Seattle: 58 + 66 = 124 lower bound = 95 + 124 = $219

(f)

2 marks

Use the smallest upper bound and the largest lower bound.

219 \leq C \leq 248

(g)

2 marks

Find a tour whose cost is between 219 and 248.

N Y C \to D a l l a s (39) \to C hi c a g o (30) \to L A (41) \to S e a ttl e (58) \to N Y C (66) 39 + 30 + 41 + 58 + 66 = $234, 219 < 234 < 248

Question 5

14 marks

Population movement between three countries: transition matrix and steady state

Number and algebra

(a)

3 marks

Use the order Belgium, Germany, Netherlands. Column 1 is people from Belgium: $94.5%$ stay, $5%$ go to Germany and $0.5%$ to the Netherlands, and similarly for the others.

T = 0.945 0.05 0.005 0.015 0.965 0.02 0.02 0.03 0.95

(b)

4 marks

Thirty years is six five-year steps.
The expected population of the German side is about 212 000.

T^{6} 2600024000050000 \approx 4213321220661661

(c)

3 marks

The steady state satisfies $T u = u$ and the proportions add to 1.
Solving the system:

- 0.055 u_{1} + 0.015 u_{2} + 0.02 u_{3} = 0, 0.05 u_{1} - 0.035 u_{2} + 0.03 u_{3} = 0, u_{1} + u_{2} + u_{3} = 1 u \approx 0.231 0.533 0.236

(d)

2 marks

The total population is $26000 + 240000 + 50000 = 316000$ .

0.5327 \times 316000 \approx 168000

(e)

2 marks

The model assumes that the movement percentages stay the same for ever, but people's choices will change over time. It also ignores births, deaths and people moving into or out of the circle, and the total population is not constant in real life.

Question 6

18 marks

Snails in a park: a Poisson model, hypothesis test and error

Statistics and probability

(a)

2 marks

A Poisson model suits counts of events that occur independently and at random, at a constant average rate. For example, the gardener might have noticed that snails are found one at a time, at random, and that the average number per square metre seems about the same all over the park.

(b)

3 marks

The number in 12 m $^{2}$ is $X \sim Po (2.4)$ .

P (X = 4) = \frac{e ^{- 2.4} 2. 4 ^{4}}{4 !} \approx 0.125

(c)

2 marks

Fewer than 3 means 0, 1 or 2 snails.

P (X < 3) = P (X \leq 2) \approx 0.570

(d)

3 marks

Each inspection must find at least one snail, and the inspections are independent.

[1 - P (X = 0)]^{3} = (1 - e^{- 2.4})^{3} \approx 0.752

(e)

2 marks

Let $λ$ be the mean number of snails in a 12 m $^{2}$ area. After heavy rain the number may have increased.

H_{0} : λ = 2.4, H_{1} : λ > 2.4

(f)

3 marks

Find the smallest $c$ with $P (X \geq c) \leq 0.01$ under $Po (2.4)$ .

P (X \geq 7) = 0.0116 > 0.01, P (X \geq 8) = 0.0033 \leq 0.01 critical region: X \geq 8

(g)

3 marks

A Type II error means $X \leq 7$ when the true mean is $0.75 \times 12 = 9$ , so $X \sim Po (9)$ .

P (X \leq 7) \approx 0.324

Question 7

18 marks

Fruit flies: eigenvalues, general solution and phase portrait

Calculus

(a)

6 marks

For $A = (- 4 9 6 - 1)$ solve $det (A - λ I) = 0$ .
For $λ = 5$ : $- 9 x + 6 y = 0$ , so $y = \frac{3}{2} x$ . For $λ = - 10$ : $6 x + 6 y = 0$ , so $y = - x$ .

(- 4 - λ) (- 1 - λ) - 54 = λ^{2} + 5 λ - 50 = 0 \Rightarrow λ = - 10 or 5 λ = 5 : (23), λ = - 10 : (1 - 1)

(b)

2 marks

Take $λ_{1} = - 10$ and $λ_{2} = 5$ (since $λ_{2} > λ_{1}$ ).

(x y) = A (1 - 1) e^{- 10 t} + B (23) e^{5 t}

(c)

2 marks

At $t = 0$ the populations are $(500, 125)$ .

A + 2 B = 500, - A + 3 B = 125 \Rightarrow A = 250, B = 125

(d)

1 mark

As $t \to \infty$ the $e^{5 t}$ term dominates, so the populations are in the ratio of $(2, 3)$ .

x : y = 2 : 3

(e)

3 marks

At $t = 0$ , with $x = 500$ and $y = 125$ :

\frac{d x}{d t} = - 4 (500) + 6 (125) = - 1250, \frac{d y}{d t} = 9 (500) - 125 = 4375 \frac{d y}{d x} = \frac{4375}{- 1250} = - 3.5

(f)

4 marks

The origin is a saddle. The line $y = \frac{3}{2} x$ (direction $(2, 3)$ ) is the unstable direction, and the line $y = - x$ is the stable direction.
Sketch: the path starts at $(500, 125)$ with gradient $- 3.5$ (up and to the left), then bends and follows the line $y = \frac{3}{2} x$ upward away from the origin.

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Number and algebra carries the most marks on this paper.

Number and algebra45 marks · 41%
Q2, Q4, Q5
Calculus34 marks · 31%
Q3, Q7
Statistics and probability18 marks · 16%
Q6
Geometry and trigonometry13 marks · 12%
Q1

Mentor insights

Traps to watch for

01
Subtract the path from both sides · Q3
The path takes $2 \times 1.5 = 3$ m from the length and $2 \times 2 = 4$ m from the width of the park.
02
Complete means every pair is joined · Q4
A complete graph has an edge between every pair of vertices. Connected means there is a path between every pair.
03
Columns add to 1 · Q5
Each column of $T$ lists where the people from one country go, so each column totals 100%.
04
Rate for 12 square metres · Q6
$0.2$ per m $^{2}$ over $12$ m $^{2}$ gives a mean of $2.4$ .
05
Larger eigenvalue dominates · Q7
In the long run the term with $λ = 5$ dominates, so the populations lie along the direction $(2, 3)$ .

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

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Our maths mentors took these exams recently. One-to-one, they'll find where your marks are going and fix it before May.

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Worked solutions

IB Maths AI HL · Paper 2

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2

Session: May 2023
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 7

Q1 Q2 Q3 Q4 Q5 Q6 Q7

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2023 TZ2 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–7 · 110 marks

Question 1

13 marks

Odette, Khemil and the football in a park

Geometry and trigonometry

(a)

3 marks

The two sides at A are $O A = 25.9$ and $A B = 6.36$ with included angle $12 5^{\circ}$ .

Area = \frac{1}{2} (25.9) (6.36) sin 12 5^{\circ} \approx 67.5 m^{2}

(b)

3 marks

Cosine rule in triangle AKB with $A K = 12$ , $A B = 6.36$ and angle $4 5^{\circ}$ at A.

K B^{2} = 1 2^{2} + 6.3 6^{2} - 2 (12) (6.36) cos 4 5^{\circ} \Rightarrow K B \approx 8.75 m

(c)

4 marks

Khemil is at K, so his distance to the football is $K X = 22.2$ m. In triangle OKX the angle at X is $180 - 53.8 - 51.1 = 75. 1^{\circ}$ . Use the sine rule for $O X$ , opposite the angle at K.
Odette is 21.4 m from the football and Khemil is 22.2 m, so Odette is closer.

\frac{O X}{sin 51. 1 ^{\circ}} = \frac{22.2}{sin 53. 8 ^{\circ}} \Rightarrow O X \approx 21.4 m

(d)

3 marks

The path is a semicircle with centre A and $K A Y = 4 5^{\circ}$ , so $K A X = 18 0^{\circ} - 4 5^{\circ} = 13 5^{\circ}$ .

arc K X = 12 \times \frac{135 π}{180} = 9 π \approx 28.3 m

Question 2

12 marks

Bacteria growth: exponential models and when they are equal

Number and algebra

(a)

2 marks

Put $t = 0$ .
It is the number of bacteria at the start of the experiment.

P (0) = 1200 \times k^{0} = 1200

(b)

2 marks

Use $P (3) = 18750$ .

1200 k^{3} = 18750 \Rightarrow k^{3} = 15.625 \Rightarrow k = 2.5

(c)

2 marks

1 hour 30 minutes is $t = 1.5$ .

P (1.5) = 1200 (2.5)^{1.5} \approx 4743

(d)

2 marks

Set $P (t) = S (t)$ .

1200 (2.5)^{t} = 5000 (1.65)^{t} \Rightarrow (\frac{2.5}{1.65})^{t} = \frac{25}{6} \Rightarrow t = \frac{ln ( 25/6 )}{ln ( 2.5/1.65 )} \approx 3.43

(e)

4 marks

Set $S (t) = 19000$ .
That is 2 hours and $0.6659 \times 60 = 39.95$ minutes.

1.6 5^{t} = 3.8 \Rightarrow t = \frac{ln 3.8}{ln 1.65} = 2.6659 hours m = 40

Question 3

16 marks

A garden surrounded by a path: area model and maximum

Calculus

(a)

5 marks

The park has length $x$ and width $\frac{1200}{x}$ . The garden has length $x - 3$ and width $\frac{1200}{x} - 4$ .

A = (x - 3) (\frac{1200}{x} - 4) = 1200 - 4 x - \frac{3600}{x} + 12 = 1212 - 4 x - \frac{3600}{x}

(b)

4 marks

Set $A = 800$ .
The width is $\frac{1200}{x}$ .

1212 - 4 x - \frac{3600}{x} = 800 \Rightarrow 4 x^{2} - 412 x + 3600 = 0 \Rightarrow x^{2} - 103 x + 900 = 0 x = 93.4 or x = 9.64 93.4 m by 12.9 m or 9.64 m by 124.5 m

(c)

3 marks

Differentiate $A = 1212 - 4 x - 3600 x^{- 1}$ .

\frac{d A}{d x} = - 4 + \frac{3600}{x ^{2}}

(d)

2 marks

Set the derivative to zero.
This is a maximum because $\frac{d ^{2} A}{d x ^{2}} = - \frac{7200}{x ^{3}} < 0$ .

\frac{3600}{x ^{2}} = 4 \Rightarrow x^{2} = 900 \Rightarrow x = 30

(e)

2 marks

Substitute $x = 30$ .

A = 1212 - 120 - 120 = 972 m^{2}

Question 4

19 marks

Flights between five US cities: graph terms, Kruskal and bounds

Number and algebra

(a)

2 marks

(b)

3 marks

Sort the flights: Chicago-Dallas 30, Dallas-New York 39, Los Angeles-Chicago 41, Los Angeles-New York 55, Seattle-Los Angeles 58, and so on. Add the cheapest edge that does not make a cycle.
The edge Los Angeles-New York (55) would make a cycle, so it is skipped. The tree has total weight $30 + 39 + 41 + 58 = 168$ .

Chicago-Dallas (30), Dallas-New York (39), Los Angeles-Chicago (41), Seattle-Los Angeles (58)

(c)

2 marks

A tour that follows the tree edges twice (once out and once back) costs twice the tree.

2 \times 168 = $336

(d)

3 marks

Start at Los Angeles and always go to the cheapest unvisited city.

L A \to D a l l a s (26) \to C hi c a g o (30) \to N e w Y or k (68) \to S e a ttl e (66) \to L A (58) 26 + 30 + 68 + 66 + 58 = $248

(e)

5 marks

Delete Chicago. The minimum spanning tree of Seattle, Los Angeles, Dallas and New York uses LA-Dallas 26, Dallas-New York 39 and Seattle-LA 58.
Delete Seattle. The minimum spanning tree of the other four uses LA-Dallas 26, Dallas-Chicago 30 and Dallas-New York 39.

26 + 39 + 58 = 123, add the two cheapest at Chicago: 30 + 41 = 71 lower bound = 123 + 71 = $194 26 + 30 + 39 = 95, add the two cheapest at Seattle: 58 + 66 = 124 lower bound = 95 + 124 = $219

(f)

2 marks

Use the smallest upper bound and the largest lower bound.

219 \leq C \leq 248

(g)

2 marks

Find a tour whose cost is between 219 and 248.

N Y C \to D a l l a s (39) \to C hi c a g o (30) \to L A (41) \to S e a ttl e (58) \to N Y C (66) 39 + 30 + 41 + 58 + 66 = $234, 219 < 234 < 248

Question 5

14 marks

Population movement between three countries: transition matrix and steady state

Number and algebra

(a)

3 marks

Use the order Belgium, Germany, Netherlands. Column 1 is people from Belgium: $94.5%$ stay, $5%$ go to Germany and $0.5%$ to the Netherlands, and similarly for the others.

T = 0.945 0.05 0.005 0.015 0.965 0.02 0.02 0.03 0.95

(b)

4 marks

Thirty years is six five-year steps.
The expected population of the German side is about 212 000.

T^{6} 2600024000050000 \approx 4213321220661661

(c)

3 marks

The steady state satisfies $T u = u$ and the proportions add to 1.
Solving the system:

- 0.055 u_{1} + 0.015 u_{2} + 0.02 u_{3} = 0, 0.05 u_{1} - 0.035 u_{2} + 0.03 u_{3} = 0, u_{1} + u_{2} + u_{3} = 1 u \approx 0.231 0.533 0.236

(d)

2 marks

The total population is $26000 + 240000 + 50000 = 316000$ .

0.5327 \times 316000 \approx 168000

(e)

2 marks

Question 6

18 marks

Snails in a park: a Poisson model, hypothesis test and error

Statistics and probability

(a)

2 marks

(b)

3 marks

The number in 12 m $^{2}$ is $X \sim Po (2.4)$ .

P (X = 4) = \frac{e ^{- 2.4} 2. 4 ^{4}}{4 !} \approx 0.125

(c)

2 marks

Fewer than 3 means 0, 1 or 2 snails.

P (X < 3) = P (X \leq 2) \approx 0.570

(d)

3 marks

Each inspection must find at least one snail, and the inspections are independent.

[1 - P (X = 0)]^{3} = (1 - e^{- 2.4})^{3} \approx 0.752

(e)

2 marks

Let $λ$ be the mean number of snails in a 12 m $^{2}$ area. After heavy rain the number may have increased.

H_{0} : λ = 2.4, H_{1} : λ > 2.4

(f)

3 marks

Find the smallest $c$ with $P (X \geq c) \leq 0.01$ under $Po (2.4)$ .

P (X \geq 7) = 0.0116 > 0.01, P (X \geq 8) = 0.0033 \leq 0.01 critical region: X \geq 8

(g)

3 marks

A Type II error means $X \leq 7$ when the true mean is $0.75 \times 12 = 9$ , so $X \sim Po (9)$ .

P (X \leq 7) \approx 0.324

Question 7

18 marks

Fruit flies: eigenvalues, general solution and phase portrait

Calculus

(a)

6 marks

For $A = (- 4 9 6 - 1)$ solve $det (A - λ I) = 0$ .
For $λ = 5$ : $- 9 x + 6 y = 0$ , so $y = \frac{3}{2} x$ . For $λ = - 10$ : $6 x + 6 y = 0$ , so $y = - x$ .

(- 4 - λ) (- 1 - λ) - 54 = λ^{2} + 5 λ - 50 = 0 \Rightarrow λ = - 10 or 5 λ = 5 : (23), λ = - 10 : (1 - 1)

(b)

2 marks

Take $λ_{1} = - 10$ and $λ_{2} = 5$ (since $λ_{2} > λ_{1}$ ).

(x y) = A (1 - 1) e^{- 10 t} + B (23) e^{5 t}

(c)

2 marks

At $t = 0$ the populations are $(500, 125)$ .

A + 2 B = 500, - A + 3 B = 125 \Rightarrow A = 250, B = 125

(d)

1 mark

As $t \to \infty$ the $e^{5 t}$ term dominates, so the populations are in the ratio of $(2, 3)$ .

x : y = 2 : 3

(e)

3 marks

At $t = 0$ , with $x = 500$ and $y = 125$ :

\frac{d x}{d t} = - 4 (500) + 6 (125) = - 1250, \frac{d y}{d t} = 9 (500) - 125 = 4375 \frac{d y}{d x} = \frac{4375}{- 1250} = - 3.5

(f)

4 marks

The origin is a saddle. The line $y = \frac{3}{2} x$ (direction $(2, 3)$ ) is the unstable direction, and the line $y = - x$ is the stable direction.
Sketch: the path starts at $(500, 125)$ with gradient $- 3.5$ (up and to the left), then bends and follows the line $y = \frac{3}{2} x$ upward away from the origin.

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Number and algebra carries the most marks on this paper.

Number and algebra45 marks · 41%
Q2, Q4, Q5
Calculus34 marks · 31%
Q3, Q7
Statistics and probability18 marks · 16%
Q6
Geometry and trigonometry13 marks · 12%
Q1

Mentor insights

Traps to watch for

01
Subtract the path from both sides · Q3
The path takes $2 \times 1.5 = 3$ m from the length and $2 \times 2 = 4$ m from the width of the park.
02
Complete means every pair is joined · Q4
A complete graph has an edge between every pair of vertices. Connected means there is a path between every pair.
03
Columns add to 1 · Q5
Each column of $T$ lists where the people from one country go, so each column totals 100%.
04
Rate for 12 square metres · Q6
$0.2$ per m $^{2}$ over $12$ m $^{2}$ gives a mean of $2.4$ .
05
Larger eigenvalue dominates · Q7
In the long run the term with $λ = 5$ dominates, so the populations lie along the direction $(2, 3)$ .

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

Want someone to go through your paper with you?

Our maths mentors took these exams recently. One-to-one, they'll find where your marks are going and fix it before May.

Find your maths mentor More past papers