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IB Maths AA SL · Paper 2

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2

Session: May 2022
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 9

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2022 TZ2 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Section A · Short response

Questions 1–6 · 36 marks

Question 1

6 marks

A chord and a major sector in a circle

Geometry and trigonometry

(a)

3 marks

Use the cosine rule, or split the isosceles triangle in half.

A B = 2 (5) sin \frac{1.9}{2} = 10 sin 0.95 \approx 8.13 m

(b)

3 marks

The angle of the shaded sector is the reflex angle.

θ = 2 π - 1.9 \approx 4.383, A = \frac{1}{2} (5)^{2} (4.383) \approx 54.8 m^{2}

Question 2

5 marks

Recovering a function from its derivative

Number and algebra

5 marks

Integrate term by term.
Use $g (0) = 4$ .

g (x) = \int (3 x^{2} + 5 e^{x}) d x = x^{3} + 5 e^{x} + C 4 = 0 + 5 + C \Rightarrow C = - 1 g (x) = x^{3} + 5 e^{x} - 1

Question 3

6 marks

Two salaries: a geometric sequence against a straight-line model

Number and algebra
Functions

(a)

3 marks

Each year the salary is multiplied by 1.02. In 2021, ten raises have taken place.

45000 \times 1.0 2^{10} = 54854.75 \approx $54855

(b)

3 marks

Fit $S = a x + b$ to the five data points with linear regression.
Put in $x = 2021$ .
Since $56063 > 54855$ , the model gives Kaia a higher salary than Gemma in 2021.

S = 1096.9 x - 2160754 S (2021) \approx $56063

Question 4

6 marks

Independent events with one probability three times the other

Number and algebra

6 marks

Let $b = P (B)$ , so $P (A) = 3 b$ . For independent events $P (A \cup B) = P (A) + P (B) - P (A) P (B)$ .
A probability can't exceed 1, so

3 b + b - 3 b^{2} = 0.68 \Rightarrow 3 b^{2} - 4 b + 0.68 = 0 b = \frac{4 \pm 16 - 8.16}{6} = \frac{4 \pm 2.8}{6} \Rightarrow b = 0.2 or b = 1.13 \dots P (B) = 0.2

Question 5

6 marks

Reaction times after sleeping well: a box plot, an outlier test and what it suggests

Statistics and probability

(a)

1 mark

Read the line inside the box.

median = 0.28 s

(b)

3 marks

From the box plot, $Q_{1} = 0.27$ and $Q_{3} = 0.35$ . The upper fence is $Q_{3} + 1.5 \times IQR$ .
Since $0.46 < 0.47$ , the measurement of 0.46 s is not an outlier.

IQR = 0.35 - 0.27 = 0.08, 0.35 + 1.5 (0.08) = 0.47

(c)

1 mark

The median is close to the lower quartile and the upper whisker and box are longer. The distribution is positively skewed, so a few large values pull the mean above the median.

(d)

1 mark

The box plot for not sleeping well is centred at higher times: its median equals the upper quartile of the "slept well" data. That is consistent with slower reactions after a bad night, but it doesn't prove cause, and a sample of nine is too small for strong conclusions.

Question 6

7 marks

A particle's velocity $v = \frac{( t ^{2} + 1 ) cos t}{4}$ : direction change, acceleration and greatest speed

Calculus

(a)

2 marks

The particle changes direction when $v = 0$ . Since $t^{2} + 1 > 0$ , this needs $cos t = 0$ .

t = \frac{π}{2} \approx 1.57 s

(b)

3 marks

The acceleration is $a = v^{'} (t) = \frac{2 t cos t - ( t ^{2} + 1 ) sin t}{4}$ . Solve $a (t) = - 1.9$ on the calculator for $0 \leq t \leq 3$ .

t \approx 2.26 or 2.96 s

(c)

2 marks

Compare $∣ v ∣$ at the end points and the turning points. $v (0) = 0.25$ and $v (3) = \frac{10 cos 3}{4} \approx - 2.47$ , which is the largest size. So the speed is greatest at $t = 3$ .

a (3) = \frac{6 cos 3 - 10 sin 3}{4} \approx - 1.84 m s^{- 2}

Section B · Extended response

Questions 7–9 · 44 marks

Question 7

14 marks

A triangular piece of land: cosine rule, area and splitting by area

Geometry and trigonometry

(a)

4 marks

The angle at B is $9 0^{\circ} - 6 3^{\circ} = 2 7^{\circ}$ . Use the cosine rule in triangle ABC.

A C^{2} = 17 5^{2} + 23 0^{2} - 2 (175) (230) cos 2 7^{\circ} A C \approx 109 m

(b)

2 marks

Use $\frac{1}{2} ab sin C$ with sides 175 and 230 and the angle $2 7^{\circ}$ between them.

Area = \frac{1}{2} (175) (230) sin 2 7^{\circ} \approx 9140 m^{2}

(c)

3 marks

Use the sine rule in triangle ABC.

\frac{sin C A B}{175} = \frac{sin 2 7 ^{\circ}}{108.62} \Rightarrow C A B = 47. 0^{\circ}

(d)

5 marks

Triangles ABD and CBD share the height from B, so they have equal areas when $A D = D C$ .
Use the cosine rule in triangle ABD with the angle $47. 0^{\circ}$ at A.

A D = \frac{1}{2} (108.62) = 54.31 B D^{2} = 23 0^{2} + 54.3 1^{2} - 2 (230) (54.31) cos 47. 0^{\circ} B D \approx 197 m

Question 8

12 marks

Two plants: initial heights, equal heights and growth rates

Calculus

(a)

3 marks

Put $t = 0$ into each model.

h_{B} (0) = 32 cm, h_{A} (0) = sin 6 + 27 = 26.72 \approx 26.7 cm

(b)

3 marks

Solve $h_{A} (t) = h_{B} (t)$ on a graph.

t \approx 4.01, 4.70, 5.88 weeks

(c)

6 marks

Differentiate both models.
Plant B grows faster when $h_{B}^{'} (t) > h_{A}^{'} (t)$ .
On a graph of the two rates the intervals in $0 \leq t \leq 9$ are
Each interval has length $\frac{π}{3}$ , so the total time is

h_{A}^{'} (t) = 2 cos (2 t + 6) + 9, h_{B}^{'} (t) = 8 8 > 2 cos (2 t + 6) + 9 \Rightarrow cos (2 t + 6) < - \frac{1}{2} 1.189 < t < 2.236, 4.330 < t < 5.378, 7.472 < t < 8.519 3 \times \frac{π}{3} = π \approx 3.14 weeks

Question 9

18 marks

Suzi's drive to work: normal distribution, binomial and conditional probability

Statistics and probability

(a)

4 marks

$P (T < 40) = 0.75$ , so the $z$ -score of 40 is the 75th percentile of the standard normal, $z = 0.6745$ .

\frac{40 - 35}{σ} = 0.6745 \Rightarrow σ = 7.41 minutes

(b)

2 marks

Use $T \sim N (35, 7.41 3^{2})$ .

P (T > 45) = 0.0887

(c)

3 marks

Over five days the number of late days is $L \sim B (5, 0.08867)$ .

P (L \geq 1) = 1 - P (L = 0) = 1 - 0.911 3^{5} = 0.371

(d)

5 marks

This is conditional on being late at least once.

P (L < 3 ∣ L \geq 1) = \frac{P ( L = 1 ) + P ( L = 2 )}{P ( L \geq 1 )} = \frac{0.36532}{0.37140} = 0.984

(e)

4 marks

She needs to be on time at least 20 of 22 days, so at most 2 late days. She has been late once in 16 days, so she can be late at most once in the remaining 6 days.
Let $X \sim B (6, 0.08867)$ be the number of late days in the remaining 6.

P (X \leq 1) = P (X = 0) + P (X = 1) = 0.907

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 probability24 marks · 30%
Q5, Q9
Geometry and trigonometry20 marks · 25%
Q1, Q7
Calculus19 marks · 24%
Q6, Q8
Number and algebra14 marks · 18%
Q2, Q3, Q4
Functions3 marks · 4%
Q3

Mentor insights

Traps to watch for

01
Which sector is shaded · Q1
The shaded sector is the larger one, with angle $2 π - 1.9$ at the centre.
02
Check the end of the interval · Q6
The greatest speed on $0 \leq t \leq 3$ might be at an end point. Compare $∣ v ∣$ at the turning points of $v$ and at $t = 0$ and $t = 3$ .
03
Find the angle inside the triangle · Q7
Walking west to B and then on a bearing of $06 3^{\circ}$ makes the angle at B equal to $9 0^{\circ} - 6 3^{\circ} = 2 7^{\circ}$ (the bearing is measured from north, the line AB is east-west).
04
Calculator in radians · Q8
The model contains $sin (2 t + 6)$ with $t$ in weeks, so the calculator must be in radian mode.
05
Standardise to find sigma · Q9
For $P (T > 40) = 0.25$ the z-score is $0.6745$ , so $\frac{40 - 35}{σ} = 0.6745$ .

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 AA SL · Paper 2

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2

Session: May 2022
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 9

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2022 TZ2 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Section A · Short response

Questions 1–6 · 36 marks

Question 1

6 marks

A chord and a major sector in a circle

Geometry and trigonometry

(a)

3 marks

Use the cosine rule, or split the isosceles triangle in half.

A B = 2 (5) sin \frac{1.9}{2} = 10 sin 0.95 \approx 8.13 m

(b)

3 marks

The angle of the shaded sector is the reflex angle.

θ = 2 π - 1.9 \approx 4.383, A = \frac{1}{2} (5)^{2} (4.383) \approx 54.8 m^{2}

Question 2

5 marks

Recovering a function from its derivative

Number and algebra

5 marks

Integrate term by term.
Use $g (0) = 4$ .

g (x) = \int (3 x^{2} + 5 e^{x}) d x = x^{3} + 5 e^{x} + C 4 = 0 + 5 + C \Rightarrow C = - 1 g (x) = x^{3} + 5 e^{x} - 1

Question 3

6 marks

Two salaries: a geometric sequence against a straight-line model

Number and algebra
Functions

(a)

3 marks

Each year the salary is multiplied by 1.02. In 2021, ten raises have taken place.

45000 \times 1.0 2^{10} = 54854.75 \approx $54855

(b)

3 marks

Fit $S = a x + b$ to the five data points with linear regression.
Put in $x = 2021$ .
Since $56063 > 54855$ , the model gives Kaia a higher salary than Gemma in 2021.

S = 1096.9 x - 2160754 S (2021) \approx $56063

Question 4

6 marks

Independent events with one probability three times the other

Number and algebra

6 marks

Let $b = P (B)$ , so $P (A) = 3 b$ . For independent events $P (A \cup B) = P (A) + P (B) - P (A) P (B)$ .
A probability can't exceed 1, so

3 b + b - 3 b^{2} = 0.68 \Rightarrow 3 b^{2} - 4 b + 0.68 = 0 b = \frac{4 \pm 16 - 8.16}{6} = \frac{4 \pm 2.8}{6} \Rightarrow b = 0.2 or b = 1.13 \dots P (B) = 0.2

Question 5

6 marks

Reaction times after sleeping well: a box plot, an outlier test and what it suggests

Statistics and probability

(a)

1 mark

Read the line inside the box.

median = 0.28 s

(b)

3 marks

From the box plot, $Q_{1} = 0.27$ and $Q_{3} = 0.35$ . The upper fence is $Q_{3} + 1.5 \times IQR$ .
Since $0.46 < 0.47$ , the measurement of 0.46 s is not an outlier.

IQR = 0.35 - 0.27 = 0.08, 0.35 + 1.5 (0.08) = 0.47

(c)

1 mark

The median is close to the lower quartile and the upper whisker and box are longer. The distribution is positively skewed, so a few large values pull the mean above the median.

(d)

1 mark

Question 6

7 marks

A particle's velocity $v = \frac{( t ^{2} + 1 ) cos t}{4}$ : direction change, acceleration and greatest speed

Calculus

(a)

2 marks

The particle changes direction when $v = 0$ . Since $t^{2} + 1 > 0$ , this needs $cos t = 0$ .

t = \frac{π}{2} \approx 1.57 s

(b)

3 marks

The acceleration is $a = v^{'} (t) = \frac{2 t cos t - ( t ^{2} + 1 ) sin t}{4}$ . Solve $a (t) = - 1.9$ on the calculator for $0 \leq t \leq 3$ .

t \approx 2.26 or 2.96 s

(c)

2 marks

Compare $∣ v ∣$ at the end points and the turning points. $v (0) = 0.25$ and $v (3) = \frac{10 cos 3}{4} \approx - 2.47$ , which is the largest size. So the speed is greatest at $t = 3$ .

a (3) = \frac{6 cos 3 - 10 sin 3}{4} \approx - 1.84 m s^{- 2}

Section B · Extended response

Questions 7–9 · 44 marks

Question 7

14 marks

A triangular piece of land: cosine rule, area and splitting by area

Geometry and trigonometry

(a)

4 marks

The angle at B is $9 0^{\circ} - 6 3^{\circ} = 2 7^{\circ}$ . Use the cosine rule in triangle ABC.

A C^{2} = 17 5^{2} + 23 0^{2} - 2 (175) (230) cos 2 7^{\circ} A C \approx 109 m

(b)

2 marks

Use $\frac{1}{2} ab sin C$ with sides 175 and 230 and the angle $2 7^{\circ}$ between them.

Area = \frac{1}{2} (175) (230) sin 2 7^{\circ} \approx 9140 m^{2}

(c)

3 marks

Use the sine rule in triangle ABC.

\frac{sin C A B}{175} = \frac{sin 2 7 ^{\circ}}{108.62} \Rightarrow C A B = 47. 0^{\circ}

(d)

5 marks

Triangles ABD and CBD share the height from B, so they have equal areas when $A D = D C$ .
Use the cosine rule in triangle ABD with the angle $47. 0^{\circ}$ at A.

A D = \frac{1}{2} (108.62) = 54.31 B D^{2} = 23 0^{2} + 54.3 1^{2} - 2 (230) (54.31) cos 47. 0^{\circ} B D \approx 197 m

Question 8

12 marks

Two plants: initial heights, equal heights and growth rates

Calculus

(a)

3 marks

Put $t = 0$ into each model.

h_{B} (0) = 32 cm, h_{A} (0) = sin 6 + 27 = 26.72 \approx 26.7 cm

(b)

3 marks

Solve $h_{A} (t) = h_{B} (t)$ on a graph.

t \approx 4.01, 4.70, 5.88 weeks

(c)

6 marks

Differentiate both models.
Plant B grows faster when $h_{B}^{'} (t) > h_{A}^{'} (t)$ .
On a graph of the two rates the intervals in $0 \leq t \leq 9$ are
Each interval has length $\frac{π}{3}$ , so the total time is

h_{A}^{'} (t) = 2 cos (2 t + 6) + 9, h_{B}^{'} (t) = 8 8 > 2 cos (2 t + 6) + 9 \Rightarrow cos (2 t + 6) < - \frac{1}{2} 1.189 < t < 2.236, 4.330 < t < 5.378, 7.472 < t < 8.519 3 \times \frac{π}{3} = π \approx 3.14 weeks

Question 9

18 marks

Suzi's drive to work: normal distribution, binomial and conditional probability

Statistics and probability

(a)

4 marks

$P (T < 40) = 0.75$ , so the $z$ -score of 40 is the 75th percentile of the standard normal, $z = 0.6745$ .

\frac{40 - 35}{σ} = 0.6745 \Rightarrow σ = 7.41 minutes

(b)

2 marks

Use $T \sim N (35, 7.41 3^{2})$ .

P (T > 45) = 0.0887

(c)

3 marks

Over five days the number of late days is $L \sim B (5, 0.08867)$ .

P (L \geq 1) = 1 - P (L = 0) = 1 - 0.911 3^{5} = 0.371

(d)

5 marks

This is conditional on being late at least once.

P (L < 3 ∣ L \geq 1) = \frac{P ( L = 1 ) + P ( L = 2 )}{P ( L \geq 1 )} = \frac{0.36532}{0.37140} = 0.984

(e)

4 marks

She needs to be on time at least 20 of 22 days, so at most 2 late days. She has been late once in 16 days, so she can be late at most once in the remaining 6 days.
Let $X \sim B (6, 0.08867)$ be the number of late days in the remaining 6.

P (X \leq 1) = P (X = 0) + P (X = 1) = 0.907

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 probability24 marks · 30%
Q5, Q9
Geometry and trigonometry20 marks · 25%
Q1, Q7
Calculus19 marks · 24%
Q6, Q8
Number and algebra14 marks · 18%
Q2, Q3, Q4
Functions3 marks · 4%
Q3

Mentor insights

Traps to watch for

01
Which sector is shaded · Q1
The shaded sector is the larger one, with angle $2 π - 1.9$ at the centre.
02
Check the end of the interval · Q6
The greatest speed on $0 \leq t \leq 3$ might be at an end point. Compare $∣ v ∣$ at the turning points of $v$ and at $t = 0$ and $t = 3$ .
03
Find the angle inside the triangle · Q7
Walking west to B and then on a bearing of $06 3^{\circ}$ makes the angle at B equal to $9 0^{\circ} - 6 3^{\circ} = 2 7^{\circ}$ (the bearing is measured from north, the line AB is east-west).
04
Calculator in radians · Q8
The model contains $sin (2 t + 6)$ with $t$ in weeks, so the calculator must be in radian mode.
05
Standardise to find sigma · Q9
For $P (T > 40) = 0.25$ the z-score is $0.6745$ , so $\frac{40 - 35}{σ} = 0.6745$ .

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