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IB Maths AA SL · Paper 2

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2

Session: May 2023
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 2023 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 · 37 marks

Question 1

5 marks

Plant growth: regression line and an estimate

Statistics and probability

(a)

2 marks

Do linear regression of $h$ on $d$ .

h = 1.93 d + 7.22

(b)

1 mark

Read the correlation coefficient from the same regression.

r = 0.991

(c)

2 marks

Substitute $d = 20$ .

h = 1.9326 (20) + 7.2166 \approx 45.9 cm

Question 2

6 marks

Two buildings: angle at P and the taller building's height

Geometry and trigonometry

(a)

3 marks

Use the cosine rule in triangle APB.

cos A P B = \frac{15 0 ^{2} + 9 0 ^{2} - 15 4 ^{2}}{2 ( 150 ) ( 90 )} = \frac{6884}{27000} \Rightarrow A P B = 75. 2^{\circ}

(b)

3 marks

The angles on the ground at P add to $18 0^{\circ}$ , so $θ = \frac{180 - 75.2}{2} = 52. 4^{\circ}$ . In the right triangle with hypotenuse $P A = 150$ :

h = 150 sin 52. 4^{\circ} \approx 119 m

Question 3

5 marks

A drug decaying in the body: finding k and a second dose time

Calculus

(a)

1 mark

Put $t = 0$ .

A (0) = 500 mg

(b)

2 marks

At $t = 3$ the amount is 280 mg.

500 e^{- 3 k} = 280 \Rightarrow k = \frac{ln ( 500/280 )}{3} = 0.193

(c)

2 marks

Set $A (T) = 140$ .

500 e^{- k T} = 140 \Rightarrow T = \frac{ln ( 500/140 )}{0.19327} \approx 6.59 hours

Question 4

8 marks

Bags of rice: normal probabilities and a binomial count

Statistics and probability

(a)

2 marks

Use $W \sim N (204, 5^{2})$ .

P (W > 210) = 0.115

(b)

2 marks

The three parts add to 1: below $w$ , between $w$ and 210 (80%), and above 210.

P (W < w) = 1 - 0.80 - 0.1151 = 0.0849

(c)

2 marks

Use the inverse normal with the lower tail probability 0.0849.

w = 204 + 5 (- 1.375) \approx 197

(d)

2 marks

The number of bags below $w$ is $X \sim B (10, 0.0849)$ .

P (X = 1) = 10 (0.0849) (0.9151)^{9} \approx 0.382

Question 5

7 marks

A geometric sequence among binomial coefficients

Number and algebra

(a)

4 marks

Use the binomial theorem $(x + h)^{8} = \sum (r 8) x^{8 - r} h^{r}$ .

a = (1 8) h = 8 h, b = (2 8) h^{2} = 28 h^{2}, d = (4 8) h^{4} = 70 h^{4}

(b)

3 marks

In a geometric sequence $\frac{b}{a} = \frac{d}{b}$ .

\frac{28 h ^{2}}{8 h} = \frac{70 h ^{4}}{28 h ^{2}} \Rightarrow 3.5 h = 2.5 h^{2} \Rightarrow h = 1.4

Question 6

6 marks

A particle with decaying oscillating velocity

Calculus

(a)

2 marks

The acceleration is $v^{'} (t)$ .
Solve the first positive root.

v^{'} (t) = - \frac{4}{3} e^{- t /3} cos (\frac{t}{2} - \frac{π}{4}) - 2 e^{- t /3} sin (\frac{t}{2} - \frac{π}{4}) = 0 t_{1} = 0.395 s

(b)

2 marks

The particle is at rest when $v = 0$ , so $cos (\frac{t}{2} - \frac{π}{4}) = 0$ .
The second time is

t = \frac{3 π}{2} = 4.71 and t = \frac{7 π}{2} = 11.0 t_{2} = 11.0 s

(c)

2 marks

Integrate the speed from $t_{1}$ to $t_{2}$ , since $v$ changes sign at $t = \frac{3 π}{2}$ .

\int_{0.395}^{11.0} ∣ v (t) ∣ d t \approx 7.83 m

Section B · Extended response

Questions 7–9 · 43 marks

Question 7

15 marks

A square root function, its inverse and the area between them

Functions
Calculus

(a)

5 marks

Write $y = 4 x - 2$ , swap $x$ and $y$ , and solve.
The domain of $h^{- 1}$ is the range of $h$ .
The range of $h^{- 1}$ is the domain of $h$ .

x = 4 y - 2 \Rightarrow x^{2} = 4 y - 2 \Rightarrow h^{- 1} (x) = \frac{x ^{2} + 2}{4} domain: x \geq 0 h^{- 1} (x) \geq \frac{1}{2}

(b)

3 marks

A point on both graphs has $h (x) = x$ .

4 x - 2 = x \Rightarrow x^{2} - 4 x + 2 = 0 \Rightarrow x = 2 \pm 2 x = 0.586 and x = 3.41

(c)

2 marks

The enclosed area is the integral of the difference between the two functions between the intersection points.

A = \int_{0.586}^{3.414} (4 x - 2 - \frac{x ^{2} + 2}{4}) d x \approx 1.89

(d)

2 marks

Differentiate $h (x) = (4 x - 2)^{1/2}$ with the chain rule.

h^{'} (x) = \frac{1}{2} (4 x - 2)^{- 1/2} \cdot 4 = \frac{2}{4 x - 2}

(e)

3 marks

The gradient of $h^{- 1}$ is $\frac{x}{2}$ . Set it equal to $h^{'} (x)$ .
Solve on a graph.

\frac{2}{4 x - 2} = \frac{x}{2} x \approx 1.77

Question 8

13 marks

A weight on a spring: a cosine model, times and probability

Functions

(a)

2 marks

The period of $cos (7.8 t)$ is $\frac{2 π}{7.8}$ .

0.806 s

(b)

3 marks

The minimum is 1 and the maximum is 1.8.

a = - \frac{1.8 - 1}{2} = - 0.4, b = \frac{1.8 + 1}{2} = 1.4

(c)

2 marks

The maximum heights occur at $t = \frac{( 2 k - 1 ) π}{7.8} = 0.403 (2 k - 1)$ . Find how many are at most 5.
The weight reaches its maximum 6 times.

0.403 (2 k - 1) \leq 5 \Rightarrow 2 k - 1 \leq 12.4 \Rightarrow k = 1, \dots, 6

(d)

2 marks

Solve $H (t) = 1.5$ .

- 0.4 cos (7.8 t) + 1.4 = 1.5 \Rightarrow cos (7.8 t) = - 0.25 t = \frac{arccos ( - 0.25 )}{7.8} = 0.234 s

(e)

4 marks

The height is above 1.5 when $cos (7.8 t) < - 0.25$ . Find the fraction of the first 5 seconds when this holds, using a graph or by adding the intervals.

P (H > 1.5) \approx 0.406

Question 9

15 marks

Balls drawn without replacement and a game with a score

Statistics and probability

(a)

3 marks

There are $n$ balls of which 10 are green.

P (first green) = \frac{10}{n}, P (first two green) = \frac{10}{n} \times \frac{9}{n - 1} = \frac{90}{n ( n - 1 )}

(b)

2 marks

With $n = 25$ there are 15 red balls.

P (R R) = \frac{15}{25} \times \frac{14}{24} = 0.35

(c)

2 marks

Three reds in a row.

P (R R R) = \frac{15}{25} \times \frac{14}{24} \times \frac{13}{23} = \frac{91}{460} \approx 0.198

(d)

2 marks

At least one green is the complement of three reds.

1 - \frac{91}{460} = \frac{369}{460} \approx 0.802

(e)

6 marks

First green on draw 3: two reds then a green. First green on draw 4: three reds then a green.
Expected points in one game:
For $k$ games the expected score is $6.018 k$ .

P_{3} = \frac{15}{25} \times \frac{14}{24} \times \frac{10}{23} = 0.1522, P_{4} = \frac{15}{25} \times \frac{14}{24} \times \frac{13}{23} \times \frac{10}{22} = 0.0899 10 (0.1522) + 50 (0.0899) = 6.018 6.018 k > 100 \Rightarrow k > 16.6 \Rightarrow k = 17

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 probability28 marks · 35%
Q1, Q4, Q9
Functions21 marks · 26%
Q7, Q8
Calculus18 marks · 23%
Q3, Q6, Q7
Number and algebra7 marks · 9%
Q5
Geometry and trigonometry6 marks · 8%
Q2

Mentor insights

Traps to watch for

01
Three regions of the normal curve · Q4
The probability below $w$ is what is left after removing the 80% and the upper tail.
02
Coefficients of (x + h)^8 · Q5
The terms $x^{7}$ , $x^{6}$ and $x^{4}$ have coefficients $8 h$ , $28 h^{2}$ and $70 h^{4}$ .
03
Inverse swaps domain and range · Q7
The domain of $h^{- 1}$ is the range of $h$ , which is $x \geq 0$ .
04
Released at the minimum · Q8
At $t = 0$ the weight is at its lowest point, so the cosine model needs a negative $a$ .
05
Without replacement · Q9
Each draw leaves one fewer ball, and the number of green balls changes only when a green is drawn.

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 AA SL · Paper 2

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2

Session: May 2023
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 2023 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 · 37 marks

Question 1

5 marks

Plant growth: regression line and an estimate

Statistics and probability

(a)

2 marks

Do linear regression of $h$ on $d$ .

h = 1.93 d + 7.22

(b)

1 mark

Read the correlation coefficient from the same regression.

r = 0.991

(c)

2 marks

Substitute $d = 20$ .

h = 1.9326 (20) + 7.2166 \approx 45.9 cm

Question 2

6 marks

Two buildings: angle at P and the taller building's height

Geometry and trigonometry

(a)

3 marks

Use the cosine rule in triangle APB.

cos A P B = \frac{15 0 ^{2} + 9 0 ^{2} - 15 4 ^{2}}{2 ( 150 ) ( 90 )} = \frac{6884}{27000} \Rightarrow A P B = 75. 2^{\circ}

(b)

3 marks

The angles on the ground at P add to $18 0^{\circ}$ , so $θ = \frac{180 - 75.2}{2} = 52. 4^{\circ}$ . In the right triangle with hypotenuse $P A = 150$ :

h = 150 sin 52. 4^{\circ} \approx 119 m

Question 3

5 marks

A drug decaying in the body: finding k and a second dose time

Calculus

(a)

1 mark

Put $t = 0$ .

A (0) = 500 mg

(b)

2 marks

At $t = 3$ the amount is 280 mg.

500 e^{- 3 k} = 280 \Rightarrow k = \frac{ln ( 500/280 )}{3} = 0.193

(c)

2 marks

Set $A (T) = 140$ .

500 e^{- k T} = 140 \Rightarrow T = \frac{ln ( 500/140 )}{0.19327} \approx 6.59 hours

Question 4

8 marks

Bags of rice: normal probabilities and a binomial count

Statistics and probability

(a)

2 marks

Use $W \sim N (204, 5^{2})$ .

P (W > 210) = 0.115

(b)

2 marks

The three parts add to 1: below $w$ , between $w$ and 210 (80%), and above 210.

P (W < w) = 1 - 0.80 - 0.1151 = 0.0849

(c)

2 marks

Use the inverse normal with the lower tail probability 0.0849.

w = 204 + 5 (- 1.375) \approx 197

(d)

2 marks

The number of bags below $w$ is $X \sim B (10, 0.0849)$ .

P (X = 1) = 10 (0.0849) (0.9151)^{9} \approx 0.382

Question 5

7 marks

A geometric sequence among binomial coefficients

Number and algebra

(a)

4 marks

Use the binomial theorem $(x + h)^{8} = \sum (r 8) x^{8 - r} h^{r}$ .

a = (1 8) h = 8 h, b = (2 8) h^{2} = 28 h^{2}, d = (4 8) h^{4} = 70 h^{4}

(b)

3 marks

In a geometric sequence $\frac{b}{a} = \frac{d}{b}$ .

\frac{28 h ^{2}}{8 h} = \frac{70 h ^{4}}{28 h ^{2}} \Rightarrow 3.5 h = 2.5 h^{2} \Rightarrow h = 1.4

Question 6

6 marks

A particle with decaying oscillating velocity

Calculus

(a)

2 marks

The acceleration is $v^{'} (t)$ .
Solve the first positive root.

v^{'} (t) = - \frac{4}{3} e^{- t /3} cos (\frac{t}{2} - \frac{π}{4}) - 2 e^{- t /3} sin (\frac{t}{2} - \frac{π}{4}) = 0 t_{1} = 0.395 s

(b)

2 marks

The particle is at rest when $v = 0$ , so $cos (\frac{t}{2} - \frac{π}{4}) = 0$ .
The second time is

t = \frac{3 π}{2} = 4.71 and t = \frac{7 π}{2} = 11.0 t_{2} = 11.0 s

(c)

2 marks

Integrate the speed from $t_{1}$ to $t_{2}$ , since $v$ changes sign at $t = \frac{3 π}{2}$ .

\int_{0.395}^{11.0} ∣ v (t) ∣ d t \approx 7.83 m

Section B · Extended response

Questions 7–9 · 43 marks

Question 7

15 marks

A square root function, its inverse and the area between them

Functions
Calculus

(a)

5 marks

Write $y = 4 x - 2$ , swap $x$ and $y$ , and solve.
The domain of $h^{- 1}$ is the range of $h$ .
The range of $h^{- 1}$ is the domain of $h$ .

x = 4 y - 2 \Rightarrow x^{2} = 4 y - 2 \Rightarrow h^{- 1} (x) = \frac{x ^{2} + 2}{4} domain: x \geq 0 h^{- 1} (x) \geq \frac{1}{2}

(b)

3 marks

A point on both graphs has $h (x) = x$ .

4 x - 2 = x \Rightarrow x^{2} - 4 x + 2 = 0 \Rightarrow x = 2 \pm 2 x = 0.586 and x = 3.41

(c)

2 marks

The enclosed area is the integral of the difference between the two functions between the intersection points.

A = \int_{0.586}^{3.414} (4 x - 2 - \frac{x ^{2} + 2}{4}) d x \approx 1.89

(d)

2 marks

Differentiate $h (x) = (4 x - 2)^{1/2}$ with the chain rule.

h^{'} (x) = \frac{1}{2} (4 x - 2)^{- 1/2} \cdot 4 = \frac{2}{4 x - 2}

(e)

3 marks

The gradient of $h^{- 1}$ is $\frac{x}{2}$ . Set it equal to $h^{'} (x)$ .
Solve on a graph.

\frac{2}{4 x - 2} = \frac{x}{2} x \approx 1.77

Question 8

13 marks

A weight on a spring: a cosine model, times and probability

Functions

(a)

2 marks

The period of $cos (7.8 t)$ is $\frac{2 π}{7.8}$ .

0.806 s

(b)

3 marks

The minimum is 1 and the maximum is 1.8.

a = - \frac{1.8 - 1}{2} = - 0.4, b = \frac{1.8 + 1}{2} = 1.4

(c)

2 marks

The maximum heights occur at $t = \frac{( 2 k - 1 ) π}{7.8} = 0.403 (2 k - 1)$ . Find how many are at most 5.
The weight reaches its maximum 6 times.

0.403 (2 k - 1) \leq 5 \Rightarrow 2 k - 1 \leq 12.4 \Rightarrow k = 1, \dots, 6

(d)

2 marks

Solve $H (t) = 1.5$ .

- 0.4 cos (7.8 t) + 1.4 = 1.5 \Rightarrow cos (7.8 t) = - 0.25 t = \frac{arccos ( - 0.25 )}{7.8} = 0.234 s

(e)

4 marks

The height is above 1.5 when $cos (7.8 t) < - 0.25$ . Find the fraction of the first 5 seconds when this holds, using a graph or by adding the intervals.

P (H > 1.5) \approx 0.406

Question 9

15 marks

Balls drawn without replacement and a game with a score

Statistics and probability

(a)

3 marks

There are $n$ balls of which 10 are green.

P (first green) = \frac{10}{n}, P (first two green) = \frac{10}{n} \times \frac{9}{n - 1} = \frac{90}{n ( n - 1 )}

(b)

2 marks

With $n = 25$ there are 15 red balls.

P (R R) = \frac{15}{25} \times \frac{14}{24} = 0.35

(c)

2 marks

Three reds in a row.

P (R R R) = \frac{15}{25} \times \frac{14}{24} \times \frac{13}{23} = \frac{91}{460} \approx 0.198

(d)

2 marks

At least one green is the complement of three reds.

1 - \frac{91}{460} = \frac{369}{460} \approx 0.802

(e)

6 marks

First green on draw 3: two reds then a green. First green on draw 4: three reds then a green.
Expected points in one game:
For $k$ games the expected score is $6.018 k$ .

P_{3} = \frac{15}{25} \times \frac{14}{24} \times \frac{10}{23} = 0.1522, P_{4} = \frac{15}{25} \times \frac{14}{24} \times \frac{13}{23} \times \frac{10}{22} = 0.0899 10 (0.1522) + 50 (0.0899) = 6.018 6.018 k > 100 \Rightarrow k > 16.6 \Rightarrow k = 17

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 probability28 marks · 35%
Q1, Q4, Q9
Functions21 marks · 26%
Q7, Q8
Calculus18 marks · 23%
Q3, Q6, Q7
Number and algebra7 marks · 9%
Q5
Geometry and trigonometry6 marks · 8%
Q2

Mentor insights

Traps to watch for

01
Three regions of the normal curve · Q4
The probability below $w$ is what is left after removing the 80% and the upper tail.
02
Coefficients of (x + h)^8 · Q5
The terms $x^{7}$ , $x^{6}$ and $x^{4}$ have coefficients $8 h$ , $28 h^{2}$ and $70 h^{4}$ .
03
Inverse swaps domain and range · Q7
The domain of $h^{- 1}$ is the range of $h$ , which is $x \geq 0$ .
04
Released at the minimum · Q8
At $t = 0$ the weight is at its lowest point, so the cosine model needs a negative $a$ .
05
Without replacement · Q9
Each draw leaves one fewer ball, and the number of green balls changes only when a green is drawn.

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