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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 TZ1 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 · 35 marks

Question 1

6 marks

Compound interest: nominal rate against an annual rate

Number and algebra

(a)

3 marks

Use $A = P (1 + \frac{r}{2})^{2 t}$ .

A = 1700 (1 + \frac{0.0274}{2})^{20} = $2231.71

(b)

2 marks

David's yearly rate must give the same amount after 10 years.

1700 (1 + \frac{r}{100})^{10} = 2231.71 \Rightarrow r = 2.76

(c)

1 mark

The interest is the final amount minus the amount invested.

2231.71 - 1700 = $531.71

Question 2

4 marks

Exercise hours: a missing frequency and a standard deviation

Statistics and probability

(a)

2 marks

The frequencies for 2, 3, 4 hours add to $5 + 1 + 4 = 10$ . For the median to be 4.5, half the students (10) must have exercised 4 hours or less, so the total is 20.

5 + 1 + 4 + 3 + x = 20 \Rightarrow x = 7

(b)

2 marks

The mean is $\frac{2 ( 5 ) + 3 ( 1 ) + 4 ( 4 ) + 5 ( 3 ) + 6 ( 7 )}{20} = 4.3$ . Enter the frequency table in the calculator for the standard deviation.

σ \approx 1.58

Question 3

6 marks

A logo made by removing two circular segments

Geometry and trigonometry

(a)

3 marks

Take the sector and subtract the triangle OAB.

segment = \frac{1}{2} (2)^{2} θ - \frac{1}{2} (2)^{2} sin θ = 2 θ - 2 sin θ

(b)

3 marks

The logo is the $5 \times 4 = 20$ rectangle minus two equal segments.
Solve on the calculator in radians, for $0 < θ < π$ .

20 - 2 (2 θ - 2 sin θ) = 13.4 \Rightarrow θ - sin θ = 1.65 θ \approx 2.36

Question 4

6 marks

A probability distribution with a parameter

Statistics and probability

(a)

1 mark

The probabilities add to 1.

0.41 + (k - 0.28) + 0.46 + (0.29 - 2 k^{2}) = 1 \Rightarrow 0.88 + k - 2 k^{2} = 1 \Rightarrow 2 k^{2} - k + 0.12 = 0

(b)

3 marks

Solve the quadratic.
If $k = 0.2$ then $k - 0.28 = - 0.08 < 0$ , which is not a probability. So

k = \frac{1 \pm 1 - 0.96}{4} = \frac{1 \pm 0.2}{4} \Rightarrow k = 0.3 or k = 0.2 k = 0.3

(c)

2 marks

The probabilities are $0.41$ , $0.02$ , $0.46$ , $0.11$ for $x = 0, 1, 2, 3$ .

E (X) = 0 (0.41) + 1 (0.02) + 2 (0.46) + 3 (0.11) = 1.27

Question 5

7 marks

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

Calculus

(a)

2 marks

The particle is at rest when $v = 0$ . Solve $e^{s i n t} + 4 sin t = 0$ on the calculator for $0 \leq t \leq 6$ , taking the first value.

t \approx 3.35 s

(b)

3 marks

The particle changes direction when $v = 0$ , so at $t = 3.347$ . The acceleration is $v^{'} (t) = cos t e^{s i n t} + 4 cos t$ .

a (3.347) \approx - 4.71 m s^{- 2}

(c)

2 marks

The total distance is the integral of the speed.

\int_{0}^{6} ∣ v ∣ d t = \int_{0}^{3.347} v d t - \int_{3.347}^{6} v d t \approx 20.8 m

Question 6

6 marks

Independent events and a conditional probability

Statistics and probability

(a)

4 marks

Let $P (A \cap B) = x$ . Then $P (A) = x + 0.16$ and $P (B) = x + 0.36$ .

x = (x + 0.16) (x + 0.36) \Rightarrow x^{2} - 0.48 x + 0.0576 = 0 \Rightarrow (x - 0.24)^{2} = 0 x = 0.24

(b)

2 marks

Independent events have independent complements, so $P (A^{'} ∣ B^{'}) = P (A^{'})$ . Here $P (A) = 0.24 + 0.16 = 0.4$ .

P (A^{'} ∣ B^{'}) = 1 - 0.4 = 0.6

Section B · Extended response

Questions 7–9 · 45 marks

Question 7

13 marks

A pyramid ornament and a cuboid of equal volume

Geometry and trigonometry
Calculus

(a)

2 marks

Use the distance formula with $A (- 1, 1, 6)$ and $V (1, 5, 0)$ .

A V = 2^{2} + 4^{2} + (- 6)^{2} = 56 \approx 7.48 cm

(b)

3 marks

Triangle AVB is isosceles ( $V A = V B$ ) with apex angle $4 0^{\circ}$ .

A B = 2 (7.483) sin 2 0^{\circ} \approx 5.12 cm

(c)

3 marks

Volume of a pyramid is one third of base times height.

\frac{1}{3} (5.119)^{2} h = 57.2 \Rightarrow h = 6.55 cm

(d)

5 marks

The cuboid has volume $2 x \cdot x \cdot y = 57.2$ , so $y = \frac{28.6}{x ^{2}}$ .
Surface area:
The minimum is where $\frac{d S}{d x} = 8 x - \frac{171.6}{x ^{2}} = 0$ .

S = 2 (2 x \cdot x + 2 x \cdot y + x \cdot y) = 4 x^{2} + 6 x y = 4 x^{2} + \frac{171.6}{x} x = 2.778, S = 92.6 cm^{2}

Question 8

16 marks

A rational function, its inverse and the area between them

Functions
Calculus

(a)

3 marks

The denominator is zero at $x = - 4$ . For large $∣ x ∣$ , $f (x) \to 4$ .

x = - 4, y = 4

(b)

8 marks

Swap $x$ and $y$ in $y = \frac{4 x + 1}{x + 4}$ and solve for $y$ .
Now show that the inverse is a reflection in the $y$ -axis followed by a reflection in the $x$ -axis. A reflection in the $y$ -axis replaces $x$ with $- x$ .
Reflecting in the $x$ -axis changes the sign of the whole expression.

x = \frac{4 y + 1}{y + 4} \Rightarrow x y + 4 x = 4 y + 1 \Rightarrow y (x - 4) = 1 - 4 x f^{- 1} (x) = \frac{1 - 4 x}{x - 4} = \frac{4 x - 1}{4 - x} f (- x) = \frac{- 4 x + 1}{- x + 4} = \frac{1 - 4 x}{4 - x} - f (- x) = \frac{4 x - 1}{4 - x} = f^{- 1} (x)

(c)

5 marks

Solve $f (x) = f^{- 1} (x)$ on a graph.
On $(- 1, 1)$ the curve $f$ lies above $f^{- 1}$ , so the enclosed area is

p = - 1, q = 1 A = \int_{- 1}^{1} (\frac{4 x + 1}{x + 4} - \frac{4 x - 1}{4 - x}) d x \approx 0.675

Question 9

16 marks

Chocolate and banana muffins: normal probabilities and an unknown standard deviation

Statistics and probability

(a)

2 marks

With $C \sim N (62, 2. 9^{2})$ :

P (C < 61) \approx 0.365

(b)

2 marks

The number of light muffins among 12 is $B (12, 0.365)$ .

P (X = 5) \approx 0.214

(c)

7 marks

(i) For banana muffins, $P (B < 61) = 0.0198$ with $B \sim N (68, 3. 4^{2})$ . Chocolate muffins are 60% of the total.

(ii) A conditional probability: the chocolate share of the muffins under 61 g.

P (muffin < 61) = 0.6 (0.3651) + 0.4 (0.01976) \approx 0.227 P (chocolate ∣< 61) = \frac{0.6 ( 0.3651 )}{0.227} \approx 0.965

(d)

5 marks

The overall probability is now 0.157, so $0.6 P (C < 61) + 0.4 (0.01976) = 0.157$ .
Use the inverse normal: $z = - 0.6792$ , so $\frac{61 - 62}{σ} = - 0.6792$ .

P (C < 61) = \frac{0.157 - 0.00790}{0.6} \approx 0.2485 σ = \frac{1}{0.6792} \approx 1.47 g

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 · 40%
Q2, Q4, Q6, Q9
Calculus17 marks · 21%
Q5, Q7, Q8
Geometry and trigonometry14 marks · 18%
Q3, Q7
Functions11 marks · 14%
Q8
Number and algebra6 marks · 8%
Q1

Mentor insights

Traps to watch for

01
Half-yearly compounding · Q1
A nominal 2.74% compounded twice a year gives $20$ periods of $\frac{2.74}{2} %$ over 10 years.
02
Radian mode · Q5
The velocity involves $sin t$ with $t$ in seconds, so the calculator must be in radians. The acceleration is the derivative of $v$ .
03
Pyramid base is a square · Q7
The base side AB comes from the isosceles triangle AVB. The height is not VA.
04
Reflect in both axes · Q8
Reflecting a graph in the $y$ -axis gives $f (- x)$ , and then in the $x$ -axis gives $- f (- x)$ . Show it equals $f^{- 1}$ .
05
Total probability, then Bayes · Q9
In (c)(i) combine the two muffin types using their shares of 60% and 40%. In (c)(ii) divide the chocolate part of that sum by the total.

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 TZ1 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 · 35 marks

Question 1

6 marks

Compound interest: nominal rate against an annual rate

Number and algebra

(a)

3 marks

Use $A = P (1 + \frac{r}{2})^{2 t}$ .

A = 1700 (1 + \frac{0.0274}{2})^{20} = $2231.71

(b)

2 marks

David's yearly rate must give the same amount after 10 years.

1700 (1 + \frac{r}{100})^{10} = 2231.71 \Rightarrow r = 2.76

(c)

1 mark

The interest is the final amount minus the amount invested.

2231.71 - 1700 = $531.71

Question 2

4 marks

Exercise hours: a missing frequency and a standard deviation

Statistics and probability

(a)

2 marks

The frequencies for 2, 3, 4 hours add to $5 + 1 + 4 = 10$ . For the median to be 4.5, half the students (10) must have exercised 4 hours or less, so the total is 20.

5 + 1 + 4 + 3 + x = 20 \Rightarrow x = 7

(b)

2 marks

The mean is $\frac{2 ( 5 ) + 3 ( 1 ) + 4 ( 4 ) + 5 ( 3 ) + 6 ( 7 )}{20} = 4.3$ . Enter the frequency table in the calculator for the standard deviation.

σ \approx 1.58

Question 3

6 marks

A logo made by removing two circular segments

Geometry and trigonometry

(a)

3 marks

Take the sector and subtract the triangle OAB.

segment = \frac{1}{2} (2)^{2} θ - \frac{1}{2} (2)^{2} sin θ = 2 θ - 2 sin θ

(b)

3 marks

The logo is the $5 \times 4 = 20$ rectangle minus two equal segments.
Solve on the calculator in radians, for $0 < θ < π$ .

20 - 2 (2 θ - 2 sin θ) = 13.4 \Rightarrow θ - sin θ = 1.65 θ \approx 2.36

Question 4

6 marks

A probability distribution with a parameter

Statistics and probability

(a)

1 mark

The probabilities add to 1.

0.41 + (k - 0.28) + 0.46 + (0.29 - 2 k^{2}) = 1 \Rightarrow 0.88 + k - 2 k^{2} = 1 \Rightarrow 2 k^{2} - k + 0.12 = 0

(b)

3 marks

Solve the quadratic.
If $k = 0.2$ then $k - 0.28 = - 0.08 < 0$ , which is not a probability. So

k = \frac{1 \pm 1 - 0.96}{4} = \frac{1 \pm 0.2}{4} \Rightarrow k = 0.3 or k = 0.2 k = 0.3

(c)

2 marks

The probabilities are $0.41$ , $0.02$ , $0.46$ , $0.11$ for $x = 0, 1, 2, 3$ .

E (X) = 0 (0.41) + 1 (0.02) + 2 (0.46) + 3 (0.11) = 1.27

Question 5

7 marks

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

Calculus

(a)

2 marks

The particle is at rest when $v = 0$ . Solve $e^{s i n t} + 4 sin t = 0$ on the calculator for $0 \leq t \leq 6$ , taking the first value.

t \approx 3.35 s

(b)

3 marks

The particle changes direction when $v = 0$ , so at $t = 3.347$ . The acceleration is $v^{'} (t) = cos t e^{s i n t} + 4 cos t$ .

a (3.347) \approx - 4.71 m s^{- 2}

(c)

2 marks

The total distance is the integral of the speed.

\int_{0}^{6} ∣ v ∣ d t = \int_{0}^{3.347} v d t - \int_{3.347}^{6} v d t \approx 20.8 m

Question 6

6 marks

Independent events and a conditional probability

Statistics and probability

(a)

4 marks

Let $P (A \cap B) = x$ . Then $P (A) = x + 0.16$ and $P (B) = x + 0.36$ .

x = (x + 0.16) (x + 0.36) \Rightarrow x^{2} - 0.48 x + 0.0576 = 0 \Rightarrow (x - 0.24)^{2} = 0 x = 0.24

(b)

2 marks

Independent events have independent complements, so $P (A^{'} ∣ B^{'}) = P (A^{'})$ . Here $P (A) = 0.24 + 0.16 = 0.4$ .

P (A^{'} ∣ B^{'}) = 1 - 0.4 = 0.6

Section B · Extended response

Questions 7–9 · 45 marks

Question 7

13 marks

A pyramid ornament and a cuboid of equal volume

Geometry and trigonometry
Calculus

(a)

2 marks

Use the distance formula with $A (- 1, 1, 6)$ and $V (1, 5, 0)$ .

A V = 2^{2} + 4^{2} + (- 6)^{2} = 56 \approx 7.48 cm

(b)

3 marks

Triangle AVB is isosceles ( $V A = V B$ ) with apex angle $4 0^{\circ}$ .

A B = 2 (7.483) sin 2 0^{\circ} \approx 5.12 cm

(c)

3 marks

Volume of a pyramid is one third of base times height.

\frac{1}{3} (5.119)^{2} h = 57.2 \Rightarrow h = 6.55 cm

(d)

5 marks

The cuboid has volume $2 x \cdot x \cdot y = 57.2$ , so $y = \frac{28.6}{x ^{2}}$ .
Surface area:
The minimum is where $\frac{d S}{d x} = 8 x - \frac{171.6}{x ^{2}} = 0$ .

S = 2 (2 x \cdot x + 2 x \cdot y + x \cdot y) = 4 x^{2} + 6 x y = 4 x^{2} + \frac{171.6}{x} x = 2.778, S = 92.6 cm^{2}

Question 8

16 marks

A rational function, its inverse and the area between them

Functions
Calculus

(a)

3 marks

The denominator is zero at $x = - 4$ . For large $∣ x ∣$ , $f (x) \to 4$ .

x = - 4, y = 4

(b)

8 marks

Swap $x$ and $y$ in $y = \frac{4 x + 1}{x + 4}$ and solve for $y$ .
Now show that the inverse is a reflection in the $y$ -axis followed by a reflection in the $x$ -axis. A reflection in the $y$ -axis replaces $x$ with $- x$ .
Reflecting in the $x$ -axis changes the sign of the whole expression.

x = \frac{4 y + 1}{y + 4} \Rightarrow x y + 4 x = 4 y + 1 \Rightarrow y (x - 4) = 1 - 4 x f^{- 1} (x) = \frac{1 - 4 x}{x - 4} = \frac{4 x - 1}{4 - x} f (- x) = \frac{- 4 x + 1}{- x + 4} = \frac{1 - 4 x}{4 - x} - f (- x) = \frac{4 x - 1}{4 - x} = f^{- 1} (x)

(c)

5 marks

Solve $f (x) = f^{- 1} (x)$ on a graph.
On $(- 1, 1)$ the curve $f$ lies above $f^{- 1}$ , so the enclosed area is

p = - 1, q = 1 A = \int_{- 1}^{1} (\frac{4 x + 1}{x + 4} - \frac{4 x - 1}{4 - x}) d x \approx 0.675

Question 9

16 marks

Chocolate and banana muffins: normal probabilities and an unknown standard deviation

Statistics and probability

(a)

2 marks

With $C \sim N (62, 2. 9^{2})$ :

P (C < 61) \approx 0.365

(b)

2 marks

The number of light muffins among 12 is $B (12, 0.365)$ .

P (X = 5) \approx 0.214

(c)

7 marks

(i) For banana muffins, $P (B < 61) = 0.0198$ with $B \sim N (68, 3. 4^{2})$ . Chocolate muffins are 60% of the total.

(ii) A conditional probability: the chocolate share of the muffins under 61 g.

P (muffin < 61) = 0.6 (0.3651) + 0.4 (0.01976) \approx 0.227 P (chocolate ∣< 61) = \frac{0.6 ( 0.3651 )}{0.227} \approx 0.965

(d)

5 marks

The overall probability is now 0.157, so $0.6 P (C < 61) + 0.4 (0.01976) = 0.157$ .
Use the inverse normal: $z = - 0.6792$ , so $\frac{61 - 62}{σ} = - 0.6792$ .

P (C < 61) = \frac{0.157 - 0.00790}{0.6} \approx 0.2485 σ = \frac{1}{0.6792} \approx 1.47 g

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 · 40%
Q2, Q4, Q6, Q9
Calculus17 marks · 21%
Q5, Q7, Q8
Geometry and trigonometry14 marks · 18%
Q3, Q7
Functions11 marks · 14%
Q8
Number and algebra6 marks · 8%
Q1

Mentor insights

Traps to watch for

01
Half-yearly compounding · Q1
A nominal 2.74% compounded twice a year gives $20$ periods of $\frac{2.74}{2} %$ over 10 years.
02
Radian mode · Q5
The velocity involves $sin t$ with $t$ in seconds, so the calculator must be in radians. The acceleration is the derivative of $v$ .
03
Pyramid base is a square · Q7
The base side AB comes from the isosceles triangle AVB. The height is not VA.
04
Reflect in both axes · Q8
Reflecting a graph in the $y$ -axis gives $f (- x)$ , and then in the $x$ -axis gives $- f (- x)$ . Show it equals $f^{- 1}$ .
05
Total probability, then Bayes · Q9
In (c)(i) combine the two muffin types using their shares of 60% and 40%. In (c)(ii) divide the chocolate part of that sum by the total.

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