Loading…

Worked solutions

IB Maths AI HL · Paper 2

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2

Session: May 2022
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 2022 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

16 marks

Eurovision scores: quartiles, outliers and rank correlation

Statistics and probability

(a)

4 marks

Order the scores: 302, 305, 331, 334, 364, 370, 472, 498. With 8 values the lower half is the first four and the upper half is the last four.

Q_{1} = \frac{305 + 331}{2} = 318, Q_{3} = \frac{370 + 472}{2} = 421 IQR = 421 - 318 = 103

(b)

3 marks

An outlier lies above $Q_{3} + 1.5 \times IQR$ .
The Netherlands scored $498 < 575.5$ , so it is not an outlier.

421 + 1.5 (103) = 575.5

(c)

2 marks

The correlation coefficient $r = 0.249$ is very close to 0, so the linear relationship is too weak.
No, a regression line would not be appropriate for predicting a country's score.

(d)

3 marks

Rank the populations from largest to smallest: Russia 145 (1), Italy 60 (2), Netherlands 17 (3), then Sweden and Azerbaijan both 10, then Switzerland 9, Norway 5 and North Macedonia 2.

a = 6, b = 4.5, c = 4.5

(e)

3 marks

Spearman's coefficient is the product-moment coefficient of the two sets of ranks.
Interpretation: a moderately strong positive association between population size and Eurovision score. Larger countries tend to rank higher.

r_{s} = 0.683

(f)

1 mark

The scores are ranked, and 478 is still the highest score. The ranks do not change, so $r_{s}$ is unchanged.

Question 2

17 marks

A goat in a square field: segment area and a rate of eating

Geometry and trigonometry
Calculus

(a)

8 marks

In the right triangle with hypotenuse 4.5 and adjacent side 4:
Sector area, with $θ = 0.9518$ rad:
Triangle area, with $A B = 2 4. 5^{2} - 4^{2} = 4.123$ :
The segment is the sector minus the triangle.

cos \frac{1}{2} A O B = \frac{4}{4.5} \Rightarrow A O B = 2 (27.2 7^{\circ}) = 54. 5^{\circ} (0.952 rad) \frac{1}{2} (4.5)^{2} (0.9518) = 9.637 m^{2} \frac{1}{2} (4.123) (4) = 8.246 m^{2} 9.637 - 8.246 = 1.39 m^{2}

(b)

4 marks

The goat reaches the circle of radius 4.5 except for the four segments outside the square.

π (4.5)^{2} - 4 (1.390) = 63.62 - 5.56 = 58.1 m^{2}

(c)

2 marks

Find the maximum of $\frac{d V}{d t} = 0.3 t e^{- t}$ . Differentiate: $0.3 e^{- t} (1 - t)$ .

0.3 e^{- t} (1 - t) = 0 \Rightarrow t = 1 hour

(d)

3 marks

The volume eaten is the total of the rate over 8 hours.

V = \int_{0}^{8} 0.3 t e^{- t} d t \approx 0.299 m^{3}

Question 3

13 marks

A school test against a national standard

Statistics and probability

(a)

1 mark

The Principal wanted fixed numbers of boys and girls and took volunteers until the groups were full. This is quota sampling.

(b)

3 marks

Use the statistics mode of the calculator on the eight marks.

\overset{x}{ˉ} = 27.125 \approx 27.1 σ = 8.298 \approx 8.30

(c)

5 marks

The population is assumed normal and its variance is unknown, so use a one-sample $t$ -test. The school mean is compared with the national mean 25.2, and the question asks whether it is higher.
Compare with 0.05.
There is insufficient evidence to reject $H_{0}$ : no significant evidence that the school's mean is above the national standard.

H_{0} : μ = 25.2, H_{1} : μ > 25.2 p \approx 0.279 0.279 > 0.05

(d)

1 mark

The students were volunteers from one class, so the sample is not random. It may not be representative of the school.

(e)

3 marks

The new scores are $2 x + 20$ .

mean = 2 (28.1) + 20 = 76.2 standard deviation = 2 (8.4) = 16.8

Question 4

13 marks

Damped motion as a system: eigenvalues and Euler's method

Calculus

(a)

5 marks

Let $y = \frac{d x}{d t}$ . Then $\frac{d y}{d t} = \frac{d ^{2} x}{d t ^{2}} = - 6 x - 5 y$ .
Eigenvalues come from $det (M - λ I) = 0$ .
Both are negative, so the particle approaches the origin and its velocity tends to 0.

(\overset{x}{˙} \overset{y}{˙}) = (0 - 6 1 - 5) (x y) λ (λ + 5) + 6 = λ^{2} + 5 λ + 6 = 0 \Rightarrow λ = - 2, - 3

(b)

8 marks

With $y = \frac{d x}{d t}$ the equation $\frac{d ^{2} x}{d t ^{2}} = 3 t + 4 - 5 \frac{d x}{d t} - 6 x$ becomes the system
Euler with $h = 0.1$ from $t_{0} = 0$ , $x_{0} = 0$ , $y_{0} = 0$ :
Ten steps give $x (1) \approx 0.644$ m.
For the long-term velocity, look for a particular solution $x = a t + b$ . Substituting gives $5 a + 6 (a t + b) = 3 t + 4$ , so $a = \frac{1}{2}$ and $b = \frac{1}{4}$ . The other terms decay, so

\frac{d x}{d t} = y, \frac{d y}{d t} = 3 t + 4 - 5 y - 6 x x_{n + 1} = x_{n} + 0.1 y_{n}, y_{n + 1} = y_{n} + 0.1 (3 t_{n} + 4 - 5 y_{n} - 6 x_{n}) \frac{d x}{d t} \to \frac{1}{2} m s^{- 1}

Question 5

15 marks

Overbooking a flight: binomial model and expected gain

Statistics and probability

(a)

3 marks

$T \sim B (74, 0.9)$ , and more than 72 passengers means 73 or 74.

P (T \geq 73) = 1 - P (T \leq 72) \approx 0.00379

(b)(i)

2 marks

The expected value of a binomial is $n p$ with $n = 72$ and $p = 0.9$ .

E (T) = 72 \times 0.9 = 64.8

(b)(ii)

2 marks

Need $0.9 n \leq 72$ .
So at most 80 tickets.

n \leq 80

(c)

8 marks

With 74 tickets sold the income is $74 \times 150 = 11100$ .
Expected compensation:
Expected income with 74 tickets sold:
Compare with 72 tickets sold, which gives $72 \times 150 = 10800$ .

P (T = 73) = 0.003380, P (T = 74) = 0.000411 300 (0.003380) + 600 (0.000411) = 1.2607 11100 - 1.2607 = 11098.74 11098.74 - 10800 \approx $299 increase

Mentor tip

A different route: compare the changes. Two extra tickets bring $+ 300$ most of the time, $0$ when 73 arrive, and $- 300$ when 74 arrive.

Question 6

18 marks

A metal plate bounded by a piecewise curve and its transformation

Calculus

(a)

4 marks

Write $y = x^{1/2}$ .
At $x = 0.16$ the gradient is $\frac{1}{2 0.16} = 1.25$ . The tangent through $(0.16, 0.4)$ is

\frac{d y}{d x} = \frac{1}{2} x^{- 1/2} y - 0.4 = 1.25 (x - 0.16) \Rightarrow y = 1.25 x + 0.2

(b)

2 marks

Apply the transformations to $A (0.5, 0.825)$ : the $x$ -coordinate is halved to 0.25, the $y$ -coordinate halved to 0.4125, then shift right 0.2.

p = 0.25 + 0.2 = 0.45, q = 0.4125

(c)

5 marks

The transformations give $g (x) = \frac{1}{2} f (2 (x - 0.2))$ , so for the square-root piece
The piece changes where $2 (x - 0.2) = 0.16$ .
For the linear piece:

h (x) = \frac{1}{2} 2 (x - 0.2) a = 0.28 g (x) = \frac{1}{2} [1.25 \cdot 2 (x - 0.2) + 0.2] = 1.25 x - 0.15 \Rightarrow b = - 0.15

(d)

7 marks

Area under $f$ : the square-root part plus the trapezoid.
The thin trapezoid under AB has parallel sides 0.4125 and 0.825 and width $0.5 - 0.45 = 0.05$ .
Shaded area:

\int_{0}^{0.16} x d x + \frac{1}{2} (0.34) (0.4 + 0.825) = 0.04267 + 0.20825 = 0.2509 m^{2} \frac{1}{2} (0.05) (0.4125 + 0.825) = 0.03094 0.250916 - 0.0627292 - 0.0309375 \approx 0.157 m^{2}

Question 7

18 marks

One transformation as a rotation and an enlargement

Geometry and trigonometry

(a)

6 marks

The origin maps to $(0, 1)$ .
Subtract $q$ from the images of $(1, 0)$ and $(0, 1)$ ; these give the columns of $P$ .

q = (01) P (10) = (\frac{3}{4} \frac{3}{4} - 1) = (\frac{3}{4} - \frac{1}{4}), P (01) = (\frac{1}{4} \frac{3}{4}) P = (\frac{3}{4} - \frac{1}{4} \frac{1}{4} \frac{3}{4})

(b)

1 mark

An enlargement with scale factor 0.5 about the origin:

S = (0.5 0 0 0.5)

(c)

7 marks

From $P = R S$ we get $R = P S^{- 1} = 2 P$ .
Compare with the rotation matrix $(cos θ sin θ - sin θ cos θ)$ : here $cos θ = \frac{3}{2}$ and $sin θ = - \frac{1}{2}$ .
The rotation is $3 0^{\circ}$ clockwise.

R = (\frac{3}{2} - \frac{1}{2} \frac{1}{2} \frac{3}{2})

(d)

4 marks

The centre $(a, b)$ is a fixed point of $T$ .
Solve.

(a b) = P (a b) + (01), so (I - P) (a b) = (01) a = \frac{5 + 2 3}{13} \approx 0.651, b = \frac{14 + 3 3}{13} \approx 1.48

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 probability44 marks · 40%
Q1, Q3, Q5
Calculus36 marks · 33%
Q2, Q4, Q6
Geometry and trigonometry30 marks · 27%
Q2, Q7

Mentor insights

Traps to watch for

01
Tied values share a rank · Q1
Populations of 10 million appear twice. They occupy ranks 4 and 5, so each gets $4.5$ , which is the mean of the two ranks.
02
Find the half-angle first · Q2
The 4 is the perpendicular distance from the centre to the chord. The half-angle $α$ satisfies $cos α = \frac{4}{4.5}$ .
03
Compensation applies to the extra passengers only · Q5
If 73 arrive, one cannot board and costs $300; if 74 arrive, two cannot board and cost $600.
04
Order of transformations matters · Q6
Stretch in $x$ by $\frac{1}{2}$ gives $f (2 x)$ . Then the $y$ stretch gives $\frac{1}{2} f (2 x)$ . Then the shift right by 0.2 replaces $x$ with $x - 0.2$ .
05
Find q first · Q7
The origin maps to $q$ , so the translation part comes directly from the image of $(0, 0)$ .

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 2022
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 2022 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

16 marks

Eurovision scores: quartiles, outliers and rank correlation

Statistics and probability

(a)

4 marks

Order the scores: 302, 305, 331, 334, 364, 370, 472, 498. With 8 values the lower half is the first four and the upper half is the last four.

Q_{1} = \frac{305 + 331}{2} = 318, Q_{3} = \frac{370 + 472}{2} = 421 IQR = 421 - 318 = 103

(b)

3 marks

An outlier lies above $Q_{3} + 1.5 \times IQR$ .
The Netherlands scored $498 < 575.5$ , so it is not an outlier.

421 + 1.5 (103) = 575.5

(c)

2 marks

The correlation coefficient $r = 0.249$ is very close to 0, so the linear relationship is too weak.
No, a regression line would not be appropriate for predicting a country's score.

(d)

3 marks

Rank the populations from largest to smallest: Russia 145 (1), Italy 60 (2), Netherlands 17 (3), then Sweden and Azerbaijan both 10, then Switzerland 9, Norway 5 and North Macedonia 2.

a = 6, b = 4.5, c = 4.5

(e)

3 marks

Spearman's coefficient is the product-moment coefficient of the two sets of ranks.
Interpretation: a moderately strong positive association between population size and Eurovision score. Larger countries tend to rank higher.

r_{s} = 0.683

(f)

1 mark

The scores are ranked, and 478 is still the highest score. The ranks do not change, so $r_{s}$ is unchanged.

Question 2

17 marks

A goat in a square field: segment area and a rate of eating

Geometry and trigonometry
Calculus

(a)

8 marks

In the right triangle with hypotenuse 4.5 and adjacent side 4:
Sector area, with $θ = 0.9518$ rad:
Triangle area, with $A B = 2 4. 5^{2} - 4^{2} = 4.123$ :
The segment is the sector minus the triangle.

cos \frac{1}{2} A O B = \frac{4}{4.5} \Rightarrow A O B = 2 (27.2 7^{\circ}) = 54. 5^{\circ} (0.952 rad) \frac{1}{2} (4.5)^{2} (0.9518) = 9.637 m^{2} \frac{1}{2} (4.123) (4) = 8.246 m^{2} 9.637 - 8.246 = 1.39 m^{2}

(b)

4 marks

The goat reaches the circle of radius 4.5 except for the four segments outside the square.

π (4.5)^{2} - 4 (1.390) = 63.62 - 5.56 = 58.1 m^{2}

(c)

2 marks

Find the maximum of $\frac{d V}{d t} = 0.3 t e^{- t}$ . Differentiate: $0.3 e^{- t} (1 - t)$ .

0.3 e^{- t} (1 - t) = 0 \Rightarrow t = 1 hour

(d)

3 marks

The volume eaten is the total of the rate over 8 hours.

V = \int_{0}^{8} 0.3 t e^{- t} d t \approx 0.299 m^{3}

Question 3

13 marks

A school test against a national standard

Statistics and probability

(a)

1 mark

The Principal wanted fixed numbers of boys and girls and took volunteers until the groups were full. This is quota sampling.

(b)

3 marks

Use the statistics mode of the calculator on the eight marks.

\overset{x}{ˉ} = 27.125 \approx 27.1 σ = 8.298 \approx 8.30

(c)

5 marks

The population is assumed normal and its variance is unknown, so use a one-sample $t$ -test. The school mean is compared with the national mean 25.2, and the question asks whether it is higher.
Compare with 0.05.
There is insufficient evidence to reject $H_{0}$ : no significant evidence that the school's mean is above the national standard.

H_{0} : μ = 25.2, H_{1} : μ > 25.2 p \approx 0.279 0.279 > 0.05

(d)

1 mark

The students were volunteers from one class, so the sample is not random. It may not be representative of the school.

(e)

3 marks

The new scores are $2 x + 20$ .

mean = 2 (28.1) + 20 = 76.2 standard deviation = 2 (8.4) = 16.8

Question 4

13 marks

Damped motion as a system: eigenvalues and Euler's method

Calculus

(a)

5 marks

Let $y = \frac{d x}{d t}$ . Then $\frac{d y}{d t} = \frac{d ^{2} x}{d t ^{2}} = - 6 x - 5 y$ .
Eigenvalues come from $det (M - λ I) = 0$ .
Both are negative, so the particle approaches the origin and its velocity tends to 0.

(\overset{x}{˙} \overset{y}{˙}) = (0 - 6 1 - 5) (x y) λ (λ + 5) + 6 = λ^{2} + 5 λ + 6 = 0 \Rightarrow λ = - 2, - 3

(b)

8 marks

With $y = \frac{d x}{d t}$ the equation $\frac{d ^{2} x}{d t ^{2}} = 3 t + 4 - 5 \frac{d x}{d t} - 6 x$ becomes the system
Euler with $h = 0.1$ from $t_{0} = 0$ , $x_{0} = 0$ , $y_{0} = 0$ :
Ten steps give $x (1) \approx 0.644$ m.
For the long-term velocity, look for a particular solution $x = a t + b$ . Substituting gives $5 a + 6 (a t + b) = 3 t + 4$ , so $a = \frac{1}{2}$ and $b = \frac{1}{4}$ . The other terms decay, so

\frac{d x}{d t} = y, \frac{d y}{d t} = 3 t + 4 - 5 y - 6 x x_{n + 1} = x_{n} + 0.1 y_{n}, y_{n + 1} = y_{n} + 0.1 (3 t_{n} + 4 - 5 y_{n} - 6 x_{n}) \frac{d x}{d t} \to \frac{1}{2} m s^{- 1}

Question 5

15 marks

Overbooking a flight: binomial model and expected gain

Statistics and probability

(a)

3 marks

$T \sim B (74, 0.9)$ , and more than 72 passengers means 73 or 74.

P (T \geq 73) = 1 - P (T \leq 72) \approx 0.00379

(b)(i)

2 marks

The expected value of a binomial is $n p$ with $n = 72$ and $p = 0.9$ .

E (T) = 72 \times 0.9 = 64.8

(b)(ii)

2 marks

Need $0.9 n \leq 72$ .
So at most 80 tickets.

n \leq 80

(c)

8 marks

With 74 tickets sold the income is $74 \times 150 = 11100$ .
Expected compensation:
Expected income with 74 tickets sold:
Compare with 72 tickets sold, which gives $72 \times 150 = 10800$ .

P (T = 73) = 0.003380, P (T = 74) = 0.000411 300 (0.003380) + 600 (0.000411) = 1.2607 11100 - 1.2607 = 11098.74 11098.74 - 10800 \approx $299 increase

Mentor tip

A different route: compare the changes. Two extra tickets bring $+ 300$ most of the time, $0$ when 73 arrive, and $- 300$ when 74 arrive.

Question 6

18 marks

A metal plate bounded by a piecewise curve and its transformation

Calculus

(a)

4 marks

Write $y = x^{1/2}$ .
At $x = 0.16$ the gradient is $\frac{1}{2 0.16} = 1.25$ . The tangent through $(0.16, 0.4)$ is

\frac{d y}{d x} = \frac{1}{2} x^{- 1/2} y - 0.4 = 1.25 (x - 0.16) \Rightarrow y = 1.25 x + 0.2

(b)

2 marks

Apply the transformations to $A (0.5, 0.825)$ : the $x$ -coordinate is halved to 0.25, the $y$ -coordinate halved to 0.4125, then shift right 0.2.

p = 0.25 + 0.2 = 0.45, q = 0.4125

(c)

5 marks

The transformations give $g (x) = \frac{1}{2} f (2 (x - 0.2))$ , so for the square-root piece
The piece changes where $2 (x - 0.2) = 0.16$ .
For the linear piece:

h (x) = \frac{1}{2} 2 (x - 0.2) a = 0.28 g (x) = \frac{1}{2} [1.25 \cdot 2 (x - 0.2) + 0.2] = 1.25 x - 0.15 \Rightarrow b = - 0.15

(d)

7 marks

Area under $f$ : the square-root part plus the trapezoid.
The thin trapezoid under AB has parallel sides 0.4125 and 0.825 and width $0.5 - 0.45 = 0.05$ .
Shaded area:

\int_{0}^{0.16} x d x + \frac{1}{2} (0.34) (0.4 + 0.825) = 0.04267 + 0.20825 = 0.2509 m^{2} \frac{1}{2} (0.05) (0.4125 + 0.825) = 0.03094 0.250916 - 0.0627292 - 0.0309375 \approx 0.157 m^{2}

Question 7

18 marks

One transformation as a rotation and an enlargement

Geometry and trigonometry

(a)

6 marks

The origin maps to $(0, 1)$ .
Subtract $q$ from the images of $(1, 0)$ and $(0, 1)$ ; these give the columns of $P$ .

q = (01) P (10) = (\frac{3}{4} \frac{3}{4} - 1) = (\frac{3}{4} - \frac{1}{4}), P (01) = (\frac{1}{4} \frac{3}{4}) P = (\frac{3}{4} - \frac{1}{4} \frac{1}{4} \frac{3}{4})

(b)

1 mark

An enlargement with scale factor 0.5 about the origin:

S = (0.5 0 0 0.5)

(c)

7 marks

From $P = R S$ we get $R = P S^{- 1} = 2 P$ .
Compare with the rotation matrix $(cos θ sin θ - sin θ cos θ)$ : here $cos θ = \frac{3}{2}$ and $sin θ = - \frac{1}{2}$ .
The rotation is $3 0^{\circ}$ clockwise.

R = (\frac{3}{2} - \frac{1}{2} \frac{1}{2} \frac{3}{2})

(d)

4 marks

The centre $(a, b)$ is a fixed point of $T$ .
Solve.

(a b) = P (a b) + (01), so (I - P) (a b) = (01) a = \frac{5 + 2 3}{13} \approx 0.651, b = \frac{14 + 3 3}{13} \approx 1.48

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 probability44 marks · 40%
Q1, Q3, Q5
Calculus36 marks · 33%
Q2, Q4, Q6
Geometry and trigonometry30 marks · 27%
Q2, Q7

Mentor insights

Traps to watch for

01
Tied values share a rank · Q1
Populations of 10 million appear twice. They occupy ranks 4 and 5, so each gets $4.5$ , which is the mean of the two ranks.
02
Find the half-angle first · Q2
The 4 is the perpendicular distance from the centre to the chord. The half-angle $α$ satisfies $cos α = \frac{4}{4.5}$ .
03
Compensation applies to the extra passengers only · Q5
If 73 arrive, one cannot board and costs $300; if 74 arrive, two cannot board and cost $600.
04
Order of transformations matters · Q6
Stretch in $x$ by $\frac{1}{2}$ gives $f (2 x)$ . Then the $y$ stretch gives $\frac{1}{2} f (2 x)$ . Then the shift right by 0.2 replaces $x$ with $x - 0.2$ .
05
Find q first · Q7
The origin maps to $q$ , so the translation part comes directly from the image of $(0, 0)$ .

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