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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 TZ1 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

Mean temperature of Earth: two linear models

Calculus

(a)

2 marks

Gradient through $(1708, 8.73)$ and $(1958, 9.45)$ .

m = \frac{9.45 - 8.73}{1958 - 1708} = \frac{0.72}{250} = 0.00288

(b)

2 marks

The gradient is the rate at which the mean annual temperature rises: it increases by 0.00288 degrees each year.

units:^{\circ} C per year

(c)

2 marks

Use the point $(1708, 8.73)$ .

y - 8.73 = 0.00288 (x - 1708) \Rightarrow y = 0.00288 x + 3.81

(d)

2 marks

Substitute $x = 2000$ .

y = 0.00288 (2000) + 3.811 \approx 9.57^{\circ} C

(e)

3 marks

Use linear regression of $y$ on $x$ on the seven points.

y = 0.00256 x + 4.46, r = 0.861

(f)

2 marks

Substitute $x = 2000$ .

y = 0.002557 (2000) + 4.4645 \approx 9.58^{\circ} C

Question 2

15 marks

Harbour tide: a cosine model and when boats can enter

Functions

(a)

2 marks

The amplitude is half the difference between high and low depth.

a = \frac{18 - 4}{2} = 7

(b)

2 marks

The centre line is the mean of the high and low depths.

d = \frac{18 + 4}{2} = 11

(c)

3 marks

From 06:00 to 12:15 is $6 \times 60 + 15 = 375$ minutes, half a period.

period = 2 \times 375 = 750 minutes

(d)

2 marks

The period is $\frac{2 π}{b}$ .

b = \frac{2 π}{750} = 0.00838 (rad per minute; or 0.4 8^{\circ} per minute)

(e)

4 marks

Solve $7 cos (b t) + 11 = 6$ .
The depth is at least 6 m until 282.5 minutes after 06:00, which is 10:42.5. The latest time to enter is

cos (b t) = - \frac{5}{7} \Rightarrow b t = 2.367 \Rightarrow t = 282.5 minutes 10 : 42

(f)

2 marks

The depth is below 6 m from $t = 282.5$ until the symmetrical time $750 - 282.5 = 467.5$ , which lies before 15:00 ( $t = 540$ ).

467.5 - 282.5 = 185 minutes

Question 3

17 marks

Testing athletes for banned substances: tree diagram and binomial counts

Statistics and probability

(a)

2 marks

From the user branch (0.06): positive 0.71 and negative 0.29. From the non-user branch (0.94): positive 0.02 and negative 0.98.

(b)

4 marks

Does not use and tests negative.
For two athletes the events are independent.

P = 0.94 \times 0.98 = 0.9212 0.921 2^{2} = 0.849

(c)

5 marks

Incorrect: a user who tests negative, or a non-user who tests positive.
Expected number in 1300 athletes:

P (incorrect) = 0.06 (0.29) + 0.94 (0.02) = 0.0174 + 0.0188 = 0.0362 1300 \times 0.0362 = 47.06 \approx 47

(d)

4 marks

Each Team X athlete tests negative with probability 0.98, independently.

0.9 8^{20} \approx 0.668

(e)

2 marks

The number of positive tests is $X \sim B (20, 0.02)$ .

P (X > 2) = 1 - P (X \leq 2) \approx 0.00707

Question 4

17 marks

A road network: Euler routes, nearest neighbour and bounds

Number and algebra

(a)

2 marks

Count the roads at each town: A 3, B 4, C 3, D 5, E 4, F 3, G 2.
The odd vertices are A, C, D and F, which is four, not two. So it is not possible.

(b)

2 marks

Use the roads in the diagram to find the shortest distances.

a = C F = C \to E \to F = 7 + 4 = 11, b = C G = C \to E \to G = 7 + 11 = 18 c = D G = D \to E \to G = 6 + 11 = 17, d = F G = F \to E \to G = 4 + 11 = 15

(c)

3 marks

From G take the nearest unvisited town each time.

G \to E (11) \to F (4) \to B (3) \to D (5) \to A (5) \to C (8) \to G (18) 11 + 4 + 3 + 5 + 5 + 8 + 18 = 54 km

(d)

4 marks

Kruskal's algorithm on A, B, C, D, E, F: FB 3, EF 4, AD 5, BD 5, then CD 7 (DF would make a cycle).
Sketch: the tree with edges FB, EF, AD, BD and CD.

3 + 4 + 5 + 5 + 7 = 24 km

(e)

2 marks

Add the two shortest edges from G: GE (11) and GB (13).

24 + 11 + 13 = 48 km

(f)

1 mark

Delete a different vertex, find that lower bound too, and keep the largest lower bound found.

(g)

3 marks

The optimal tour uses the roads AC 8, CE 7, EG 11, GB 13, BF 3, FD 5 and DA 5.
Each of the 7 lengths could be up to 0.5 km shorter, so the lower bound is

8 + 7 + 11 + 13 + 3 + 5 + 5 = 52 52 - 7 (0.5) = 48.5 km

Question 5

15 marks

Bird sightings: variance, Poisson expected frequencies and a goodness of fit test

Statistics and probability

(a)

3 marks

Use the frequency table of sightings with 50 weeks.

s_{n - 1}^{2} = \frac{\sum f ( x - x ˉ ) ^{2}}{49} \approx 1.70

(b)

1 mark

For a Poisson distribution the mean equals the variance. The sample mean 1.76 and variance 1.70 are close, which supports Goran's belief.

(c)

5 marks

Use $Po (1.76)$ with 50 weeks.

j = 50 P (X = 4) = 3.44 k = 50 P (X \geq 5) = 50 - (8.60 + 15.14 + 13.32 + 7.82 + 3.44) = 1.68

(d)

1 mark

The expected frequencies for 4 sightings (3.44) and for 5 or more (1.68) are less than 5, which would make the test unreliable. Combine them into "4 or more", with expected frequency 5.12 and observed frequency $3 + 2 = 5$ .

(e)

1 mark

After combining there are 5 groups, and one parameter (the mean) was estimated from the data.

ν = 5 - 1 - 1 = 3

(f)

2 marks

Run a goodness of fit test with observed $8, 16, 13, 8, 5$ and the expected frequencies above.

χ^{2} = 0.106, p \approx 0.991

(g)

2 marks

Compare with 0.05.
Do not reject $H_{0}$ : the data are consistent with a Poisson distribution.

0.991 > 0.05

Question 6

15 marks

A speedboat modelled by a linear system: eigenvalues and path

Calculus

(a)

4 marks

Solve $det (A - λ I) = 0$ .

(- 0.05 - λ)^{2} + 25 = 0 \Rightarrow λ = - 0.05 \pm 5 i

(b)

2 marks

$a \neq = 0$ means the path is not closed: the boat does not keep going around the same circle, but spirals.
The sign of $a$ is negative, so the spiral goes inwards towards the origin.

(c)

5 marks

At $(20, 0)$ :

\frac{d y}{d t} = - 5 (20) - 0.05 (0) = - 100 \frac{d x}{d t} = 5 (0) - 0.05 (20) = - 1 \Rightarrow \frac{d y}{d x} = \frac{- 100}{- 1} = 100

(d)

4 marks

At $(20, 0)$ the boat moves down very steeply (and slightly to the left), so it turns clockwise.
Sketch: a spiral starting at $(20, 0)$ , heading steeply downwards, turning clockwise around the origin and closing in on the origin with each turn.

Question 7

18 marks

A design built from a trapezoid and a family of transformation matrices

Geometry and trigonometry

(a)

2 marks

The trapezoid has two vertical parallel sides: at $x = 0$ of length $1$ (from $y = - 2$ to $- 1$ ), and at $x = sin 1 5^{\circ}$ of length 2. The distance between them is $sin 1 5^{\circ}$ .

A = \frac{1}{2} (1 + 2) sin 1 5^{\circ} = \frac{3}{2} sin 1 5^{\circ}

(b)

4 marks

With $k = 6$ : $1 - \frac{6}{12} = \frac{1}{2}$ and $6 \times 1 5^{\circ} = 9 0^{\circ}$ , so $cos 9 0^{\circ} = 0$ and $sin 9 0^{\circ} = 1$ .
The image of $(0, - 1)$ :

M_{6} = (0 \frac{1}{2} - \frac{1}{2} 0) (0 \frac{1}{2} - \frac{1}{2} 0) (0 - 1) = (\frac{1}{2} 0), (\frac{1}{2}, 0)

(c)

4 marks

Factor out $1 - \frac{k}{12}$ .

rotation: (cos (15 k^{\circ}) sin (15 k^{\circ}) - sin (15 k^{\circ}) cos (15 k^{\circ})), enlargement: (1 - \frac{k}{12} 0 0 1 - \frac{k}{12}) angle 15 k^{\circ} (anticlockwise), scale factor 1 - \frac{k}{12}

(d)

2 marks

The determinant is the area scale factor, which is the square of the length scale factor.

det M_{k} = (1 - \frac{k}{12})^{2}

(e)

4 marks

The element for $k$ has area $\frac{3}{2} sin 1 5^{\circ} (1 - \frac{k}{12})^{2}$ . Sum for $k = 0, \dots, 11$ , and write $(1 - \frac{k}{12}) = \frac{j}{12}$ with $j = 12 - k$ from 1 to 12.
The right side has this total, and the left side has the same, since reflections keep area.

k = 0 \sum 11 (1 - \frac{k}{12})^{2} = \frac{1}{144} j = 1 \sum 12 j^{2} = \frac{650}{144} Total = 2 \times \frac{3}{2} sin 1 5^{\circ} \times \frac{650}{144} = 3 sin 1 5^{\circ} \times \frac{650}{144} \approx 3.50

(f)

2 marks

A left element is the mirror image in the $y$ -axis of the right element, so apply a reflection after $M_{k}$ .

N_{k} = (- 1 0 01) M_{k}

Our analysis

Where the marks are

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

Statistics and probability32 marks · 29%
Q3, Q5
Calculus28 marks · 25%
Q1, Q6
Geometry and trigonometry18 marks · 16%
Q7
Number and algebra17 marks · 15%
Q4
Functions15 marks · 14%
Q2

Mentor insights

Traps to watch for

01
Half a period between high and low · Q2
High tide at 06:00 and the next low tide at 12:15 is 375 minutes, so the period is 750 minutes.
02
Incorrect result has two branches · Q3
A wrong result is either a user who tests negative or a non-user who tests positive.
03
Euler trail needs exactly two odd vertices · Q4
A route using every road once and ending elsewhere needs exactly two vertices of odd degree.
04
Combine small cells · Q5
Expected frequencies below 5 must be combined before the $χ^{2}$ test.
05
Matrix as rotation times enlargement · Q7
$M_{k} = (1 - \frac{k}{12}) \times$ rotation by $15 k$ degrees.

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

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 TZ1 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

Mean temperature of Earth: two linear models

Calculus

(a)

2 marks

Gradient through $(1708, 8.73)$ and $(1958, 9.45)$ .

m = \frac{9.45 - 8.73}{1958 - 1708} = \frac{0.72}{250} = 0.00288

(b)

2 marks

The gradient is the rate at which the mean annual temperature rises: it increases by 0.00288 degrees each year.

units:^{\circ} C per year

(c)

2 marks

Use the point $(1708, 8.73)$ .

y - 8.73 = 0.00288 (x - 1708) \Rightarrow y = 0.00288 x + 3.81

(d)

2 marks

Substitute $x = 2000$ .

y = 0.00288 (2000) + 3.811 \approx 9.57^{\circ} C

(e)

3 marks

Use linear regression of $y$ on $x$ on the seven points.

y = 0.00256 x + 4.46, r = 0.861

(f)

2 marks

Substitute $x = 2000$ .

y = 0.002557 (2000) + 4.4645 \approx 9.58^{\circ} C

Question 2

15 marks

Harbour tide: a cosine model and when boats can enter

Functions

(a)

2 marks

The amplitude is half the difference between high and low depth.

a = \frac{18 - 4}{2} = 7

(b)

2 marks

The centre line is the mean of the high and low depths.

d = \frac{18 + 4}{2} = 11

(c)

3 marks

From 06:00 to 12:15 is $6 \times 60 + 15 = 375$ minutes, half a period.

period = 2 \times 375 = 750 minutes

(d)

2 marks

The period is $\frac{2 π}{b}$ .

b = \frac{2 π}{750} = 0.00838 (rad per minute; or 0.4 8^{\circ} per minute)

(e)

4 marks

Solve $7 cos (b t) + 11 = 6$ .
The depth is at least 6 m until 282.5 minutes after 06:00, which is 10:42.5. The latest time to enter is

cos (b t) = - \frac{5}{7} \Rightarrow b t = 2.367 \Rightarrow t = 282.5 minutes 10 : 42

(f)

2 marks

The depth is below 6 m from $t = 282.5$ until the symmetrical time $750 - 282.5 = 467.5$ , which lies before 15:00 ( $t = 540$ ).

467.5 - 282.5 = 185 minutes

Question 3

17 marks

Testing athletes for banned substances: tree diagram and binomial counts

Statistics and probability

(a)

2 marks

From the user branch (0.06): positive 0.71 and negative 0.29. From the non-user branch (0.94): positive 0.02 and negative 0.98.

(b)

4 marks

Does not use and tests negative.
For two athletes the events are independent.

P = 0.94 \times 0.98 = 0.9212 0.921 2^{2} = 0.849

(c)

5 marks

Incorrect: a user who tests negative, or a non-user who tests positive.
Expected number in 1300 athletes:

P (incorrect) = 0.06 (0.29) + 0.94 (0.02) = 0.0174 + 0.0188 = 0.0362 1300 \times 0.0362 = 47.06 \approx 47

(d)

4 marks

Each Team X athlete tests negative with probability 0.98, independently.

0.9 8^{20} \approx 0.668

(e)

2 marks

The number of positive tests is $X \sim B (20, 0.02)$ .

P (X > 2) = 1 - P (X \leq 2) \approx 0.00707

Question 4

17 marks

A road network: Euler routes, nearest neighbour and bounds

Number and algebra

(a)

2 marks

Count the roads at each town: A 3, B 4, C 3, D 5, E 4, F 3, G 2.
The odd vertices are A, C, D and F, which is four, not two. So it is not possible.

(b)

2 marks

Use the roads in the diagram to find the shortest distances.

a = C F = C \to E \to F = 7 + 4 = 11, b = C G = C \to E \to G = 7 + 11 = 18 c = D G = D \to E \to G = 6 + 11 = 17, d = F G = F \to E \to G = 4 + 11 = 15

(c)

3 marks

From G take the nearest unvisited town each time.

G \to E (11) \to F (4) \to B (3) \to D (5) \to A (5) \to C (8) \to G (18) 11 + 4 + 3 + 5 + 5 + 8 + 18 = 54 km

(d)

4 marks

Kruskal's algorithm on A, B, C, D, E, F: FB 3, EF 4, AD 5, BD 5, then CD 7 (DF would make a cycle).
Sketch: the tree with edges FB, EF, AD, BD and CD.

3 + 4 + 5 + 5 + 7 = 24 km

(e)

2 marks

Add the two shortest edges from G: GE (11) and GB (13).

24 + 11 + 13 = 48 km

(f)

1 mark

Delete a different vertex, find that lower bound too, and keep the largest lower bound found.

(g)

3 marks

The optimal tour uses the roads AC 8, CE 7, EG 11, GB 13, BF 3, FD 5 and DA 5.
Each of the 7 lengths could be up to 0.5 km shorter, so the lower bound is

8 + 7 + 11 + 13 + 3 + 5 + 5 = 52 52 - 7 (0.5) = 48.5 km

Question 5

15 marks

Bird sightings: variance, Poisson expected frequencies and a goodness of fit test

Statistics and probability

(a)

3 marks

Use the frequency table of sightings with 50 weeks.

s_{n - 1}^{2} = \frac{\sum f ( x - x ˉ ) ^{2}}{49} \approx 1.70

(b)

1 mark

For a Poisson distribution the mean equals the variance. The sample mean 1.76 and variance 1.70 are close, which supports Goran's belief.

(c)

5 marks

Use $Po (1.76)$ with 50 weeks.

j = 50 P (X = 4) = 3.44 k = 50 P (X \geq 5) = 50 - (8.60 + 15.14 + 13.32 + 7.82 + 3.44) = 1.68

(d)

1 mark

(e)

1 mark

After combining there are 5 groups, and one parameter (the mean) was estimated from the data.

ν = 5 - 1 - 1 = 3

(f)

2 marks

Run a goodness of fit test with observed $8, 16, 13, 8, 5$ and the expected frequencies above.

χ^{2} = 0.106, p \approx 0.991

(g)

2 marks

Compare with 0.05.
Do not reject $H_{0}$ : the data are consistent with a Poisson distribution.

0.991 > 0.05

Question 6

15 marks

A speedboat modelled by a linear system: eigenvalues and path

Calculus

(a)

4 marks

Solve $det (A - λ I) = 0$ .

(- 0.05 - λ)^{2} + 25 = 0 \Rightarrow λ = - 0.05 \pm 5 i

(b)

2 marks

$a \neq = 0$ means the path is not closed: the boat does not keep going around the same circle, but spirals.
The sign of $a$ is negative, so the spiral goes inwards towards the origin.

(c)

5 marks

At $(20, 0)$ :

\frac{d y}{d t} = - 5 (20) - 0.05 (0) = - 100 \frac{d x}{d t} = 5 (0) - 0.05 (20) = - 1 \Rightarrow \frac{d y}{d x} = \frac{- 100}{- 1} = 100

(d)

4 marks

At $(20, 0)$ the boat moves down very steeply (and slightly to the left), so it turns clockwise.
Sketch: a spiral starting at $(20, 0)$ , heading steeply downwards, turning clockwise around the origin and closing in on the origin with each turn.

Question 7

18 marks

A design built from a trapezoid and a family of transformation matrices

Geometry and trigonometry

(a)

2 marks

The trapezoid has two vertical parallel sides: at $x = 0$ of length $1$ (from $y = - 2$ to $- 1$ ), and at $x = sin 1 5^{\circ}$ of length 2. The distance between them is $sin 1 5^{\circ}$ .

A = \frac{1}{2} (1 + 2) sin 1 5^{\circ} = \frac{3}{2} sin 1 5^{\circ}

(b)

4 marks

With $k = 6$ : $1 - \frac{6}{12} = \frac{1}{2}$ and $6 \times 1 5^{\circ} = 9 0^{\circ}$ , so $cos 9 0^{\circ} = 0$ and $sin 9 0^{\circ} = 1$ .
The image of $(0, - 1)$ :

M_{6} = (0 \frac{1}{2} - \frac{1}{2} 0) (0 \frac{1}{2} - \frac{1}{2} 0) (0 - 1) = (\frac{1}{2} 0), (\frac{1}{2}, 0)

(c)

4 marks

Factor out $1 - \frac{k}{12}$ .

rotation: (cos (15 k^{\circ}) sin (15 k^{\circ}) - sin (15 k^{\circ}) cos (15 k^{\circ})), enlargement: (1 - \frac{k}{12} 0 0 1 - \frac{k}{12}) angle 15 k^{\circ} (anticlockwise), scale factor 1 - \frac{k}{12}

(d)

2 marks

The determinant is the area scale factor, which is the square of the length scale factor.

det M_{k} = (1 - \frac{k}{12})^{2}

(e)

4 marks

The element for $k$ has area $\frac{3}{2} sin 1 5^{\circ} (1 - \frac{k}{12})^{2}$ . Sum for $k = 0, \dots, 11$ , and write $(1 - \frac{k}{12}) = \frac{j}{12}$ with $j = 12 - k$ from 1 to 12.
The right side has this total, and the left side has the same, since reflections keep area.

k = 0 \sum 11 (1 - \frac{k}{12})^{2} = \frac{1}{144} j = 1 \sum 12 j^{2} = \frac{650}{144} Total = 2 \times \frac{3}{2} sin 1 5^{\circ} \times \frac{650}{144} = 3 sin 1 5^{\circ} \times \frac{650}{144} \approx 3.50

(f)

2 marks

A left element is the mirror image in the $y$ -axis of the right element, so apply a reflection after $M_{k}$ .

N_{k} = (- 1 0 01) M_{k}

Our analysis

Where the marks are

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

Statistics and probability32 marks · 29%
Q3, Q5
Calculus28 marks · 25%
Q1, Q6
Geometry and trigonometry18 marks · 16%
Q7
Number and algebra17 marks · 15%
Q4
Functions15 marks · 14%
Q2

Mentor insights

Traps to watch for

01
Half a period between high and low · Q2
High tide at 06:00 and the next low tide at 12:15 is 375 minutes, so the period is 750 minutes.
02
Incorrect result has two branches · Q3
A wrong result is either a user who tests negative or a non-user who tests positive.
03
Euler trail needs exactly two odd vertices · Q4
A route using every road once and ending elsewhere needs exactly two vertices of odd degree.
04
Combine small cells · Q5
Expected frequencies below 5 must be combined before the $χ^{2}$ test.
05
Matrix as rotation times enlargement · Q7
$M_{k} = (1 - \frac{k}{12}) \times$ rotation by $15 k$ degrees.

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