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IB Maths AI HL · Paper 2

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2

Session: May 2021
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 2021 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

15 marks

A triangular field and an equal-area exchange

Geometry and trigonometry

(a)

3 marks

Use the cosine rule with the side opposite C being AB.

cos \hat{C} = \frac{100 5 ^{2} + 122 5 ^{2} - 65 0 ^{2}}{2 ( 1005 ) ( 1225 )} \approx 0.8480 \Rightarrow \hat{C} \approx 32. 0^{\circ}

(b)

3 marks

In triangle DCE use the sine rule: $\frac{D E}{sin C} = \frac{D C}{sin E}$ .

D E = \frac{210 sin 32. 0 ^{\circ}}{sin 10 0 ^{\circ}} \approx 113 m

(c)

5 marks

The third angle of triangle DCE is $18 0^{\circ} - 10 0^{\circ} - 32. 0^{\circ} = 48. 0^{\circ}$ , at D. Use $\frac{1}{2} ab sin C$ with the two known sides DE and DC.

Area = \frac{1}{2} (112.99) (210) sin 48. 0^{\circ} \approx 8820 m^{2}

(d)

4 marks

D lies on AC, so $A D = 1005 - 210 = 795$ . The angle $∠ A D F = 18 0^{\circ} - 48. 0^{\circ}$ has the same sine as $48. 0^{\circ}$ . Set the area of ADF equal to the area from part (c).

\frac{1}{2} (795) (D F) sin 48. 0^{\circ} = 8817.2 D F \approx 29.8 m

Question 2

16 marks

Weights of Persian cats: normal, binomial and conditional probability

Statistics and probability

(a)

2 marks

Sketch a bell-shaped curve centred on the mean 6.1 kg. Mark the standard deviation 0.5 kg on the axis, with $μ - 0.5 = 5.6$ and $μ + 0.5 = 6.6$ on either side.

(b)

2 marks

With $X \sim N (6.1, 0. 5^{2})$ :

P (5.5 < X < 6.5) \approx 0.673

(c)

3 marks

The proportion below 5.3 kg times 80 cats gives the expected number.

80 \times P (X < 5.3) = 80 \times 0.0548 \approx 4.38

(d)

5 marks

(i) For the females $Y \sim N (4.5, 0.4 5^{2})$ , so $P (Y > 4.62) \approx 0.3949$ . The number over 4.62 among ten is $B (10, 0.3949)$ .

(ii) For a binomial, the variance is $n p (1 - p)$ .

P (N = 1) = (1 10) (0.3949) (0.6051)^{9} \approx 0.0430 Var (N) = 10 (0.3949) (0.6051) \approx 2.39

(e)

4 marks

Half the cats are female. A female weighs over 4.7 kg with probability 0.3284, and a male with probability 0.9974.

P (F ∣ W > 4.7) = \frac{0.5 \times 0.3284}{0.5 \times 0.3284 + 0.5 \times 0.9974} \approx 0.248

Question 3

15 marks

A hexagonal chocolate box: surface area, volume and the largest volume

Geometry and trigonometry
Calculus

(a)

2 marks

A regular hexagon is made of 6 equilateral triangles with side $x$ . Each has area $\frac{1}{2} x^{2} sin 6 0^{\circ}$ .

Area = 6 \cdot \frac{1}{2} x^{2} \cdot \frac{3}{2} = \frac{3 3}{2} x^{2}

(b)

5 marks

The surface area has two hexagons (top and bottom) and six rectangles $x \times h$ .
The volume is the base area times the height.

2 \cdot \frac{3 3}{2} x^{2} + 6 x h = 1200 \Rightarrow h = \frac{1200 - 3 3 x ^{2}}{6 x} = \frac{400 - 3 x ^{2}}{2 x} V = \frac{3 3}{2} x^{2} \cdot \frac{400 - 3 x ^{2}}{2 x} = \frac{3 3}{4} x (400 - 3 x^{2}) = 3003 x - \frac{9}{4} x^{3}

(c)

2 marks

The graph of $V$ is a cubic that starts at the origin, rises to a maximum near $x \approx 8.8$ , then falls and crosses the $x$ -axis at $x = \frac{400 3}{3} \approx 15.2$ , within $0 \leq x \leq 16$ .

(d)

2 marks

Differentiate term by term.

\frac{d V}{d x} = 3003 - \frac{27}{4} x^{2}

(e)

2 marks

The maximum occurs where the derivative is zero.

x^{2} = \frac{4 \times 300 3}{27} \Rightarrow x \approx 8.77 cm

(f)

2 marks

Put $x = 8.77$ into the volume formula.

V_{m a x} \approx 3040 cm^{3}

Question 4

18 marks

Two clinics: a two-state chain, diagonalisation and the long-run ratio

Number and algebra

(a)

2 marks

Each column is "from" a clinic (Black, Green), and the entries are the proportions that stay or leave.

T = (0.965 0.035 0.05 0.95)

(b)

2 marks

Apply $T$ twice to the starting vector.
The ratio is $2294 : 3306$ , or $1147 : 1653$ , which is about 0.694.

T^{2} (21003500) \approx (22943306)

(c)

6 marks

Solve $det (T - λ I) = 0$ .
The eigenvector for $λ = 1$ : $- 0.035 x + 0.05 y = 0$ , which gives $(10, 7)$ . For $λ = 0.915$ : $x + y = 0$ , which gives $(1, - 1)$ .

(0.965 - λ) (0.95 - λ) - 0.05 \times 0.035 = λ^{2} - 1.915 λ + 0.915 = 0 \Rightarrow λ = 1, 0.915 P = (107 1 - 1)

(d)

6 marks

With $D = (10 0 0.915)$ we have $T^{n} = P D^{n} P^{- 1}$ , where $P^{- 1} = \frac{1}{17} (17 1 - 10)$ .
As $n \to \infty$ , $0.91 5^{n} \to 0$ , so $D^{n} \to (1000)$ .

T^{\infty} = (107 1 - 1) (1000) \frac{1}{17} (17 1 - 10) = (\frac{10}{17} \frac{7}{17} \frac{10}{17} \frac{7}{17})

(e)

2 marks

Apply $T^{\infty}$ to the starting numbers.

T^{\infty} (21003500) = (32942306) \Rightarrow ratio 3294 : 2306 \approx 10 : 7

Mentor tip

The ratio is the eigenvector for eigenvalue 1, which is $(10, 7)$ , whatever the starting numbers were.

Question 5

14 marks

A magpie's visits: Poisson, binomial and a sum of Poissons

Statistics and probability

(a)

1 mark

With $X \sim Po (3.1)$ :

P (X = 4) \approx 0.173

(b)

5 marks

(i) Over three months the mean is $3 \times 3.1 = 9.3$ , so $Po (9.3)$ .

(ii) Visits in months 1 and 3 and none in month 2: use $P (X = 0) = e^{- 3.1} \approx 0.04505$ and $P (X \geq 1) \approx 0.95495$ .

P (X = 12) \approx 0.0799 (0.95495) (0.04505) (0.95495) \approx 0.0411

(c)

4 marks

Each month has probability $0.04505$ of no visit, independently. Over 12 months the number of such months is $B (12, 0.04505)$ .

P (Y = 3) = (3 12) (0.04505)^{3} (0.95495)^{9} \approx 0.0133

(d)

4 marks

With Bill and $n$ baby magpies, the visits per month are $Po (3.1 + 2.1 n)$ . Find the least $n$ with $P (visits \geq 30) > 0.2$ .
So 11 baby magpies are needed, which is 12 magpies including Bill.

n = 10 : λ = 24.1, P \approx 0.137; n = 11 : λ = 26.2, P \approx 0.253

Question 6

15 marks

A particle's velocity: rest, acceleration, displacement and distance

Calculus

(a)

2 marks

The particle is at rest when $v = 0$ .

- 2 t^{2} + 16 t - 24 = 0 \Rightarrow t^{2} - 8 t + 12 = 0 \Rightarrow t = 2, 6

(b)

4 marks

Differentiate $v$ to get the acceleration, then substitute $t = 6$ .
The magnitude is $8 m s^{- 2}$ .

a = - 4 t + 16 \Rightarrow a (6) = - 8

(c)

2 marks

Check the values at the end points and at the turning point of $v$ : $v (0) = - 24$ , $v (4) = 8$ (the peak of $v$ ), and $v (6) = 0$ . The largest size is at $t = 0$ .

greatest speed = 24 m s^{- 1}

(d)

4 marks

Integrate $v$ , with $x = 0$ at $t = 0$ so no constant is added.

x = - \frac{2}{3} t^{3} + 8 t^{2} - 24 t

(e)

3 marks

$v < 0$ for $0 < t < 2$ and $v > 0$ for $2 < t < 4$ . Find the displacement at $t = 2$ and $t = 4$ .

x (2) = - \frac{16}{3} + 32 - 48 = - \frac{64}{3}, x (4) = - \frac{128}{3} + 128 - 96 = - \frac{32}{3} distance = - \frac{64}{3} + - \frac{32}{3} + \frac{64}{3} = \frac{64}{3} + \frac{32}{3} = 32 m

Question 7

17 marks

A coupled linear system: eigenvalues, stability and a phase portrait

Number and algebra

(a)

6 marks

Solve $det (- 4 - λ 3 0 - 2 - λ) = (- 4 - λ) (- 2 - λ) = 0$ .
For $λ = - 4$ : $3 x - 2 y = - 4 y$ gives $3 x + 2 y = 0$ , so an eigenvector is $(- 2 3)$ .
For $λ = - 2$ : $- 4 x = - 2 x$ forces $x = 0$ , so an eigenvector is $(01)$ .

λ = - 4, λ = - 2

(b)

2 marks

The general solution is a combination of the two eigen-solutions.

(x y) = A e^{- 4 t} (- 2 3) + B e^{- 2 t} (01)

(c)

2 marks

Both eigenvalues are real and negative, so both exponentials tend to 0 as $t \to \infty$ . Every solution tends to $(0, 0)$ , so the equilibrium is stable.

(d)

3 marks

Divide the two equations.
(i) At $(4, 0)$ : $\frac{12}{- 16} = - \frac{3}{4}$ . (ii) At $(- 4, 0)$ : $\frac{- 12}{16} = - \frac{3}{4}$ .

\frac{d y}{d x} = \frac{3 x - 2 y}{- 4 x}

(e)

4 marks

Draw axes from $- 6$ to $6$ . The origin is a stable node, so every trajectory heads in to it. Draw the two eigenvector lines, $x = 0$ (the $y$ -axis) and $y = - \frac{3}{2} x$ . Trajectories approach the origin tangent to the slower direction, the $y$ -axis ( $λ = - 2$ ), and cross the $x$ -axis with gradient $- \frac{3}{4}$ at $x = \pm 4$ . Arrows point towards $(0, 0)$ everywhere.

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 algebra35 marks · 32%
Q4, Q7
Statistics and probability30 marks · 27%
Q2, Q5
Calculus23 marks · 21%
Q3, Q6
Geometry and trigonometry22 marks · 20%
Q1, Q3

Mentor insights

Traps to watch for

01
Name the angles · Q1
The exchange triangle ADF has the same area as DCE and shares the angle $∠ A D F = 18 0^{\circ} - ∠ C D E$ . That gives an area equation with $D F$ as the only unknown.
02
Condition over the combined group · Q2
For (e) all 160 cats are equally likely to be picked, so the two groups carry the same weight, 0.5 each, and then you apply conditional probability.
03
Express height in terms of $x$ · Q3
The surface area gives a link between $h$ and $x$ . Solve it for $h$ and then put it into the volume, so the volume is a function of $x$ alone.
04
Show the limit, don't just state it · Q4
In (d) say that $0.91 5^{n} \to 0$ , so the diagonal matrix tends to $diag (1, 0)$ , and then multiply out. That justification is the heart of the answer.
05
Real negative eigenvalues mean a stable node · Q7
If every eigenvalue is real and negative, solutions decay along the eigenvector directions and the origin is a stable node. Say both parts.

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.

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Worked solutions

IB Maths AI HL · Paper 2

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2

Session: May 2021
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 2021 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

15 marks

A triangular field and an equal-area exchange

Geometry and trigonometry

(a)

3 marks

Use the cosine rule with the side opposite C being AB.

cos \hat{C} = \frac{100 5 ^{2} + 122 5 ^{2} - 65 0 ^{2}}{2 ( 1005 ) ( 1225 )} \approx 0.8480 \Rightarrow \hat{C} \approx 32. 0^{\circ}

(b)

3 marks

In triangle DCE use the sine rule: $\frac{D E}{sin C} = \frac{D C}{sin E}$ .

D E = \frac{210 sin 32. 0 ^{\circ}}{sin 10 0 ^{\circ}} \approx 113 m

(c)

5 marks

The third angle of triangle DCE is $18 0^{\circ} - 10 0^{\circ} - 32. 0^{\circ} = 48. 0^{\circ}$ , at D. Use $\frac{1}{2} ab sin C$ with the two known sides DE and DC.

Area = \frac{1}{2} (112.99) (210) sin 48. 0^{\circ} \approx 8820 m^{2}

(d)

4 marks

D lies on AC, so $A D = 1005 - 210 = 795$ . The angle $∠ A D F = 18 0^{\circ} - 48. 0^{\circ}$ has the same sine as $48. 0^{\circ}$ . Set the area of ADF equal to the area from part (c).

\frac{1}{2} (795) (D F) sin 48. 0^{\circ} = 8817.2 D F \approx 29.8 m

Question 2

16 marks

Weights of Persian cats: normal, binomial and conditional probability

Statistics and probability

(a)

2 marks

Sketch a bell-shaped curve centred on the mean 6.1 kg. Mark the standard deviation 0.5 kg on the axis, with $μ - 0.5 = 5.6$ and $μ + 0.5 = 6.6$ on either side.

(b)

2 marks

With $X \sim N (6.1, 0. 5^{2})$ :

P (5.5 < X < 6.5) \approx 0.673

(c)

3 marks

The proportion below 5.3 kg times 80 cats gives the expected number.

80 \times P (X < 5.3) = 80 \times 0.0548 \approx 4.38

(d)

5 marks

(i) For the females $Y \sim N (4.5, 0.4 5^{2})$ , so $P (Y > 4.62) \approx 0.3949$ . The number over 4.62 among ten is $B (10, 0.3949)$ .

(ii) For a binomial, the variance is $n p (1 - p)$ .

P (N = 1) = (1 10) (0.3949) (0.6051)^{9} \approx 0.0430 Var (N) = 10 (0.3949) (0.6051) \approx 2.39

(e)

4 marks

Half the cats are female. A female weighs over 4.7 kg with probability 0.3284, and a male with probability 0.9974.

P (F ∣ W > 4.7) = \frac{0.5 \times 0.3284}{0.5 \times 0.3284 + 0.5 \times 0.9974} \approx 0.248

Question 3

15 marks

A hexagonal chocolate box: surface area, volume and the largest volume

Geometry and trigonometry
Calculus

(a)

2 marks

A regular hexagon is made of 6 equilateral triangles with side $x$ . Each has area $\frac{1}{2} x^{2} sin 6 0^{\circ}$ .

Area = 6 \cdot \frac{1}{2} x^{2} \cdot \frac{3}{2} = \frac{3 3}{2} x^{2}

(b)

5 marks

The surface area has two hexagons (top and bottom) and six rectangles $x \times h$ .
The volume is the base area times the height.

2 \cdot \frac{3 3}{2} x^{2} + 6 x h = 1200 \Rightarrow h = \frac{1200 - 3 3 x ^{2}}{6 x} = \frac{400 - 3 x ^{2}}{2 x} V = \frac{3 3}{2} x^{2} \cdot \frac{400 - 3 x ^{2}}{2 x} = \frac{3 3}{4} x (400 - 3 x^{2}) = 3003 x - \frac{9}{4} x^{3}

(c)

2 marks

(d)

2 marks

Differentiate term by term.

\frac{d V}{d x} = 3003 - \frac{27}{4} x^{2}

(e)

2 marks

The maximum occurs where the derivative is zero.

x^{2} = \frac{4 \times 300 3}{27} \Rightarrow x \approx 8.77 cm

(f)

2 marks

Put $x = 8.77$ into the volume formula.

V_{m a x} \approx 3040 cm^{3}

Question 4

18 marks

Two clinics: a two-state chain, diagonalisation and the long-run ratio

Number and algebra

(a)

2 marks

Each column is "from" a clinic (Black, Green), and the entries are the proportions that stay or leave.

T = (0.965 0.035 0.05 0.95)

(b)

2 marks

Apply $T$ twice to the starting vector.
The ratio is $2294 : 3306$ , or $1147 : 1653$ , which is about 0.694.

T^{2} (21003500) \approx (22943306)

(c)

6 marks

Solve $det (T - λ I) = 0$ .
The eigenvector for $λ = 1$ : $- 0.035 x + 0.05 y = 0$ , which gives $(10, 7)$ . For $λ = 0.915$ : $x + y = 0$ , which gives $(1, - 1)$ .

(0.965 - λ) (0.95 - λ) - 0.05 \times 0.035 = λ^{2} - 1.915 λ + 0.915 = 0 \Rightarrow λ = 1, 0.915 P = (107 1 - 1)

(d)

6 marks

With $D = (10 0 0.915)$ we have $T^{n} = P D^{n} P^{- 1}$ , where $P^{- 1} = \frac{1}{17} (17 1 - 10)$ .
As $n \to \infty$ , $0.91 5^{n} \to 0$ , so $D^{n} \to (1000)$ .

T^{\infty} = (107 1 - 1) (1000) \frac{1}{17} (17 1 - 10) = (\frac{10}{17} \frac{7}{17} \frac{10}{17} \frac{7}{17})

(e)

2 marks

Apply $T^{\infty}$ to the starting numbers.

T^{\infty} (21003500) = (32942306) \Rightarrow ratio 3294 : 2306 \approx 10 : 7

Mentor tip

The ratio is the eigenvector for eigenvalue 1, which is $(10, 7)$ , whatever the starting numbers were.

Question 5

14 marks

A magpie's visits: Poisson, binomial and a sum of Poissons

Statistics and probability

(a)

1 mark

With $X \sim Po (3.1)$ :

P (X = 4) \approx 0.173

(b)

5 marks

(i) Over three months the mean is $3 \times 3.1 = 9.3$ , so $Po (9.3)$ .

(ii) Visits in months 1 and 3 and none in month 2: use $P (X = 0) = e^{- 3.1} \approx 0.04505$ and $P (X \geq 1) \approx 0.95495$ .

P (X = 12) \approx 0.0799 (0.95495) (0.04505) (0.95495) \approx 0.0411

(c)

4 marks

Each month has probability $0.04505$ of no visit, independently. Over 12 months the number of such months is $B (12, 0.04505)$ .

P (Y = 3) = (3 12) (0.04505)^{3} (0.95495)^{9} \approx 0.0133

(d)

4 marks

With Bill and $n$ baby magpies, the visits per month are $Po (3.1 + 2.1 n)$ . Find the least $n$ with $P (visits \geq 30) > 0.2$ .
So 11 baby magpies are needed, which is 12 magpies including Bill.

n = 10 : λ = 24.1, P \approx 0.137; n = 11 : λ = 26.2, P \approx 0.253

Question 6

15 marks

A particle's velocity: rest, acceleration, displacement and distance

Calculus

(a)

2 marks

The particle is at rest when $v = 0$ .

- 2 t^{2} + 16 t - 24 = 0 \Rightarrow t^{2} - 8 t + 12 = 0 \Rightarrow t = 2, 6

(b)

4 marks

Differentiate $v$ to get the acceleration, then substitute $t = 6$ .
The magnitude is $8 m s^{- 2}$ .

a = - 4 t + 16 \Rightarrow a (6) = - 8

(c)

2 marks

Check the values at the end points and at the turning point of $v$ : $v (0) = - 24$ , $v (4) = 8$ (the peak of $v$ ), and $v (6) = 0$ . The largest size is at $t = 0$ .

greatest speed = 24 m s^{- 1}

(d)

4 marks

Integrate $v$ , with $x = 0$ at $t = 0$ so no constant is added.

x = - \frac{2}{3} t^{3} + 8 t^{2} - 24 t

(e)

3 marks

$v < 0$ for $0 < t < 2$ and $v > 0$ for $2 < t < 4$ . Find the displacement at $t = 2$ and $t = 4$ .

x (2) = - \frac{16}{3} + 32 - 48 = - \frac{64}{3}, x (4) = - \frac{128}{3} + 128 - 96 = - \frac{32}{3} distance = - \frac{64}{3} + - \frac{32}{3} + \frac{64}{3} = \frac{64}{3} + \frac{32}{3} = 32 m

Question 7

17 marks

A coupled linear system: eigenvalues, stability and a phase portrait

Number and algebra

(a)

6 marks

Solve $det (- 4 - λ 3 0 - 2 - λ) = (- 4 - λ) (- 2 - λ) = 0$ .
For $λ = - 4$ : $3 x - 2 y = - 4 y$ gives $3 x + 2 y = 0$ , so an eigenvector is $(- 2 3)$ .
For $λ = - 2$ : $- 4 x = - 2 x$ forces $x = 0$ , so an eigenvector is $(01)$ .

λ = - 4, λ = - 2

(b)

2 marks

The general solution is a combination of the two eigen-solutions.

(x y) = A e^{- 4 t} (- 2 3) + B e^{- 2 t} (01)

(c)

2 marks

Both eigenvalues are real and negative, so both exponentials tend to 0 as $t \to \infty$ . Every solution tends to $(0, 0)$ , so the equilibrium is stable.

(d)

3 marks

Divide the two equations.
(i) At $(4, 0)$ : $\frac{12}{- 16} = - \frac{3}{4}$ . (ii) At $(- 4, 0)$ : $\frac{- 12}{16} = - \frac{3}{4}$ .

\frac{d y}{d x} = \frac{3 x - 2 y}{- 4 x}

(e)

4 marks

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 algebra35 marks · 32%
Q4, Q7
Statistics and probability30 marks · 27%
Q2, Q5
Calculus23 marks · 21%
Q3, Q6
Geometry and trigonometry22 marks · 20%
Q1, Q3

Mentor insights

Traps to watch for

01
Name the angles · Q1
The exchange triangle ADF has the same area as DCE and shares the angle $∠ A D F = 18 0^{\circ} - ∠ C D E$ . That gives an area equation with $D F$ as the only unknown.
02
Condition over the combined group · Q2
For (e) all 160 cats are equally likely to be picked, so the two groups carry the same weight, 0.5 each, and then you apply conditional probability.
03
Express height in terms of $x$ · Q3
The surface area gives a link between $h$ and $x$ . Solve it for $h$ and then put it into the volume, so the volume is a function of $x$ alone.
04
Show the limit, don't just state it · Q4
In (d) say that $0.91 5^{n} \to 0$ , so the diagonal matrix tends to $diag (1, 0)$ , and then multiply out. That justification is the heart of the answer.
05
Real negative eigenvalues mean a stable node · Q7
If every eigenvalue is real and negative, solutions decay along the eigenvector directions and the origin is a stable node. Say both parts.

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