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IB Maths AA SL · Paper 2

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2 TZ3

Session: May 2025
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 2025 TZ1 paper, which your school or the IB can provide.

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Hide our working until you've had a go.

Section A · Short response

Questions 1–6 · 35 marks

Question 1

5 marks

Splitting (x - 1)^2 over x and integrating

Calculus

(a)

2 marks

Expand the bracket and divide by $x$ .

\frac{( x - 1 ) ^{2}}{x} = \frac{x ^{2} - 2 x + 1}{x} = x - 2 + \frac{1}{x}

(b)

3 marks

Integrate the three terms (for $x > 0$ ).

\int f (x) d x = \frac{x ^{2}}{2} - 2 x + ln ∣ x ∣ + c

Question 2

6 marks

A hollow cone made from a sector of a circle

Geometry and trigonometry

(a)

3 marks

The angle $21 0^{\circ}$ is $\frac{7 π}{6}$ radians.

A = \frac{1}{2} r^{2} θ = \frac{1}{2} (19.5)^{2} (\frac{7 π}{6}) \approx 697 cm^{2}

(b)

3 marks

The arc length of the sector is the circumference of the base of the cone.

arc = 19.5 \times \frac{7 π}{6} \approx 71.47 cm 2 π r = 71.47 \Rightarrow r = \frac{91}{8} = 11.375 \approx 11.4 cm

Question 3

4 marks

A tangent function through two points

Functions

(a)

1 mark

The period is $\frac{π}{2}$ .

\frac{π}{2} \approx 1.57

(b)

3 marks

The points are $(\frac{π}{12}, 5)$ and $(\frac{π}{3}, 7)$ . Substitute into $f (x) = a tan 2 x + b$ .
Subtract the second equation from the first.
Then $b = 7 + 3 a = 7 - \frac{3}{2}$ .

a tan \frac{π}{6} + b = 5, a tan \frac{2 π}{3} + b = 7 \frac{a}{3} + b = 5, - 3 a + b = 7 a (\frac{1}{3} + 3) = - 2 \Rightarrow \frac{4 a}{3} = - 2 \Rightarrow a = - \frac{3}{2} \approx - 0.866 b = \frac{11}{2}

Question 4

6 marks

A population with a decaying rate of change

Number and algebra

6 marks

The start of 2026 is $t = 4$ . The rate is negative, so the population falls.
Add the change to the starting population.

Δ P = \int_{0}^{4} - 104000 e^{- 0.0145 t} d t = [\frac{104000}{0.0145} e^{- 0.0145 t}]_{0}^{4} \approx - 404166 P (4) = 6.78 \times 1 0^{6} - 404166 \approx 6375834 P (4) \approx 6.38 \times 1 0^{6}

Question 5

8 marks

Arm span and foot length: two regression lines, the mean point and a normal spread

Statistics and probability

(a)

2 marks

Use the line of $A$ on $F$ with $F = 19.8$ .

A = 2.89 (19.8) + 99.3 \approx 157 cm

(b)

3 marks

The mean point lies on both lines, so solve them together.

F = 0.335 (2.89 F + 99.3) - 32.6 \Rightarrow F (1 - 0.96815) = 0.66655 \overset{ˉ}{F} \approx 20.9 cm, \overset{ˉ}{A} \approx 160 cm

(c)

3 marks

The interval 153 to 173 is symmetric about the mean 163, so 6% lies below 153 and 6% above 173.

P (H < 153) = 0.06 \Rightarrow \frac{153 - 163}{σ} = - 1.5548 σ \approx 6.43 cm

Question 6

6 marks

The gradient of a tangent to a cosine curve and a given angle with the x-axis

Calculus

(a)

3 marks

Differentiate $h (x) = 15 cos (\frac{π x}{50}) + 15$ with the chain rule.

h^{'} (x) = - 15 \cdot \frac{π}{50} sin (\frac{π x}{50}) = - \frac{3 π}{10} sin (\frac{π x}{50}) gradient of T_{k} = - \frac{3 π}{10} sin (\frac{π k}{50})

(b)

3 marks

The tangent slopes downwards (the curve falls from 30 to 0), so its gradient is $- tan \frac{π}{8} = - 0.4142$ .
Solve on $0 \leq k \leq 50$ , where $\frac{π k}{50}$ is between 0 and $π$ . There are two solutions, symmetric about $k = 25$ .

- \frac{3 π}{10} sin (\frac{π k}{50}) = - 0.4142 \Rightarrow sin (\frac{π k}{50}) = 0.4395 k \approx 7.24 or k \approx 42.8

Section B · Extended response

Questions 7–9 · 45 marks

Question 7

14 marks

A dice game with scores for equal numbers and for a difference of one

Statistics and probability

(a)

2 marks

Pairs with a difference of one: $(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)$ and the five reversed ones, which is 10 outcomes.

P (5 points) = \frac{10}{36} = \frac{5}{18}

(b)

2 marks

Doubles score 10 and have probability $\frac{6}{36}$ . Use the complement.

P (0 points) = 1 - \frac{10}{36} - \frac{6}{36} = \frac{20}{36} = \frac{5}{9} \approx 0.556

(c)

4 marks

The distribution of $X$ is $P (X = 0) = \frac{20}{36}$ , $P (X = 5) = \frac{10}{36}$ and $P (X = 10) = \frac{6}{36}$ .

E (X) = 0 (\frac{20}{36}) + 5 (\frac{10}{36}) + 10 (\frac{6}{36}) = \frac{110}{36} \approx 3.06

(d)

2 marks

Multiply the expected score of one round by 90.

90 \times \frac{110}{36} = 275

(e)

4 marks

More than 40 points in five rounds needs five scores of 10 (50 points) or four scores of 10 and one of 5 (45 points). Three tens and two fives make only 40.

P (five 10s) = (\frac{1}{6})^{5} \approx 0.000129 P (four 10s and a 5) = 5 (\frac{1}{6})^{4} (\frac{5}{18}) \approx 0.00107 P (prize) \approx 0.00120

Question 8

16 marks

A pyramid of playing cards: a quadratic number sequence

Number and algebra

(a)

1 mark

From the diagram of the stack with 3 rows, count the cards.

t_{3} = 2 + 5 + 8 = 15

(b)

2 marks

A fourth row adds 11 cards (the differences between rows are 3).

t_{4} = 15 + 11 = 26

(c)

3 marks

The numbers of cards in the rows form the arithmetic sequence $2, 5, 8, 11, \dots$ with first term 2 and common difference 3. Use the sum of $n$ terms.

t_{n} = \frac{n}{2} [2 (2) + (n - 1) 3] = \frac{n}{2} (3 n + 1)

(d)

3 marks

The cards available are $14 \times 52 = 728$ . Solve $\frac{n ( 3 n + 1 )}{2} \leq 728$ .
The most rows is 21.

t_{21} = \frac{21 ( 64 )}{2} = 672 \leq 728, t_{22} = \frac{22 ( 67 )}{2} = 737 > 728

(e)

2 marks

The number of cards $t_{n}$ must be a multiple of 52. List $t_{n}$ : 2, 7, 15, 26, 40, 57, 77, 100, 126, 155, 187, 222, 260.
The first multiple of 52 is $t_{13} = 260 = 5 \times 52$ (so 5 packs).

n = 13

(f)

5 marks

Each row is an equilateral triangle with side 88 mm, so its height is
For the stack to be taller than 2000 mm we need $76.21 n > 2000$ .
The number of cards for 27 rows.

88 sin 6 0^{\circ} = 443 \approx 76.21 mm n > 26.2 \Rightarrow n = 27 t_{27} = \frac{27 ( 82 )}{2} = 1107

Question 9

15 marks

A self-inverse rational function, the closest point to the origin and a reciprocal graph

Functions

(a)

3 marks

Start with $y = \frac{2 - 2 x}{x + 2}$ and swap $x$ and $y$ .

x = \frac{2 - 2 y}{y + 2} \Rightarrow x y + 2 x = 2 - 2 y \Rightarrow y (x + 2) = 2 - 2 x f^{- 1} (x) = \frac{2 - 2 x}{x + 2} = f (x)

(b)

4 marks

(i) The graph is symmetric about $y = x$ , so the closest point to the origin is on $y = x$ . Put $f (k) = k$ .

The closest point has $k > 0$ .

(ii) On the line $y = x$ .

\frac{2 - 2 k}{k + 2} = k \Rightarrow 2 - 2 k = k^{2} + 2 k \Rightarrow k^{2} + 4 k - 2 = 0 k = - 2 \pm 6 \Rightarrow k \approx 0.449 or - 4.45 k \approx 0.449 P \approx (0.449, 0.449)

(c)

2 marks

$g (x) = \frac{2 - 3 x}{c x + d}$ .
(i) The denominator is zero.
(ii) As $x \to \infty$ , $g (x) \to - \frac{3}{c}$ .

x = - \frac{d}{c} y = - \frac{3}{c}

(d)

2 marks

For $g^{- 1} = g$ , solve for the inverse.
Equating with $g (x) = \frac{2 - 3 x}{c x + d}$ , the denominators must match.

x = \frac{2 - 3 y}{cy + d} \Rightarrow c x y + d x = 2 - 3 y \Rightarrow y (c x + 3) = 2 - d x g^{- 1} (x) = \frac{2 - d x}{c x + 3} d = 3

(e)

4 marks

With $c = 1$ , $d = 3$ , the function $g (x) = \frac{2 - 3 x}{x + 3}$ and its reciprocal is $\frac{1}{g ( x )} = \frac{x + 3}{2 - 3 x}$ .
The asymptotes of $\frac{1}{g}$ : the denominator $2 - 3 x$ is zero at $x = \frac{2}{3}$ (vertical asymptote) and as $x \to \infty$ it tends to $- \frac{1}{3}$ (horizontal asymptote).
The intercepts: when $y = 0$ , $x = - 3$ , and when $x = 0$ , $y = \frac{3}{2}$ .
Sketch: a hyperbola in two branches, one passing through $(- 3, 0)$ and $(0, \frac{3}{2})$ to the left of $x = \frac{2}{3}$ , rising towards $+ \infty$ as $x \to \frac{2}{3}$ from the left and approaching $y = - \frac{1}{3}$ from below as $x \to - \infty$ , and one branch to the right of $x = \frac{2}{3}$ coming down from $- \infty$ towards $y = - \frac{1}{3}$ .

x = \frac{2}{3}, y = - \frac{1}{3}

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 algebra22 marks · 28%
Q4, Q8
Statistics and probability22 marks · 28%
Q5, Q7
Functions19 marks · 24%
Q3, Q9
Calculus11 marks · 14%
Q1, Q6
Geometry and trigonometry6 marks · 8%
Q2

Mentor insights

Traps to watch for

01
Arc length becomes the base circumference · Q2
When the sector is rolled up, its arc becomes the circle at the base of the cone.
02
Which line to use · Q5
To estimate $A$ from $F$ use the regression line of $A$ on $F$ .
03
Sample space · Q7
There are 36 equally likely outcomes. Doubles give 6, a difference of 1 gives 10.
04
Counting · Q8
Look at how many cards are added to make each new row, and find the pattern.
05
Inverse · Q9
Swap $x$ and $y$ and solve for $y$ .

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 TZ3

Session: May 2025
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 2025 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

5 marks

Splitting (x - 1)^2 over x and integrating

Calculus

(a)

2 marks

Expand the bracket and divide by $x$ .

\frac{( x - 1 ) ^{2}}{x} = \frac{x ^{2} - 2 x + 1}{x} = x - 2 + \frac{1}{x}

(b)

3 marks

Integrate the three terms (for $x > 0$ ).

\int f (x) d x = \frac{x ^{2}}{2} - 2 x + ln ∣ x ∣ + c

Question 2

6 marks

A hollow cone made from a sector of a circle

Geometry and trigonometry

(a)

3 marks

The angle $21 0^{\circ}$ is $\frac{7 π}{6}$ radians.

A = \frac{1}{2} r^{2} θ = \frac{1}{2} (19.5)^{2} (\frac{7 π}{6}) \approx 697 cm^{2}

(b)

3 marks

The arc length of the sector is the circumference of the base of the cone.

arc = 19.5 \times \frac{7 π}{6} \approx 71.47 cm 2 π r = 71.47 \Rightarrow r = \frac{91}{8} = 11.375 \approx 11.4 cm

Question 3

4 marks

A tangent function through two points

Functions

(a)

1 mark

The period is $\frac{π}{2}$ .

\frac{π}{2} \approx 1.57

(b)

3 marks

The points are $(\frac{π}{12}, 5)$ and $(\frac{π}{3}, 7)$ . Substitute into $f (x) = a tan 2 x + b$ .
Subtract the second equation from the first.
Then $b = 7 + 3 a = 7 - \frac{3}{2}$ .

a tan \frac{π}{6} + b = 5, a tan \frac{2 π}{3} + b = 7 \frac{a}{3} + b = 5, - 3 a + b = 7 a (\frac{1}{3} + 3) = - 2 \Rightarrow \frac{4 a}{3} = - 2 \Rightarrow a = - \frac{3}{2} \approx - 0.866 b = \frac{11}{2}

Question 4

6 marks

A population with a decaying rate of change

Number and algebra

6 marks

The start of 2026 is $t = 4$ . The rate is negative, so the population falls.
Add the change to the starting population.

Δ P = \int_{0}^{4} - 104000 e^{- 0.0145 t} d t = [\frac{104000}{0.0145} e^{- 0.0145 t}]_{0}^{4} \approx - 404166 P (4) = 6.78 \times 1 0^{6} - 404166 \approx 6375834 P (4) \approx 6.38 \times 1 0^{6}

Question 5

8 marks

Arm span and foot length: two regression lines, the mean point and a normal spread

Statistics and probability

(a)

2 marks

Use the line of $A$ on $F$ with $F = 19.8$ .

A = 2.89 (19.8) + 99.3 \approx 157 cm

(b)

3 marks

The mean point lies on both lines, so solve them together.

F = 0.335 (2.89 F + 99.3) - 32.6 \Rightarrow F (1 - 0.96815) = 0.66655 \overset{ˉ}{F} \approx 20.9 cm, \overset{ˉ}{A} \approx 160 cm

(c)

3 marks

The interval 153 to 173 is symmetric about the mean 163, so 6% lies below 153 and 6% above 173.

P (H < 153) = 0.06 \Rightarrow \frac{153 - 163}{σ} = - 1.5548 σ \approx 6.43 cm

Question 6

6 marks

The gradient of a tangent to a cosine curve and a given angle with the x-axis

Calculus

(a)

3 marks

Differentiate $h (x) = 15 cos (\frac{π x}{50}) + 15$ with the chain rule.

h^{'} (x) = - 15 \cdot \frac{π}{50} sin (\frac{π x}{50}) = - \frac{3 π}{10} sin (\frac{π x}{50}) gradient of T_{k} = - \frac{3 π}{10} sin (\frac{π k}{50})

(b)

3 marks

The tangent slopes downwards (the curve falls from 30 to 0), so its gradient is $- tan \frac{π}{8} = - 0.4142$ .
Solve on $0 \leq k \leq 50$ , where $\frac{π k}{50}$ is between 0 and $π$ . There are two solutions, symmetric about $k = 25$ .

- \frac{3 π}{10} sin (\frac{π k}{50}) = - 0.4142 \Rightarrow sin (\frac{π k}{50}) = 0.4395 k \approx 7.24 or k \approx 42.8

Section B · Extended response

Questions 7–9 · 45 marks

Question 7

14 marks

A dice game with scores for equal numbers and for a difference of one

Statistics and probability

(a)

2 marks

Pairs with a difference of one: $(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)$ and the five reversed ones, which is 10 outcomes.

P (5 points) = \frac{10}{36} = \frac{5}{18}

(b)

2 marks

Doubles score 10 and have probability $\frac{6}{36}$ . Use the complement.

P (0 points) = 1 - \frac{10}{36} - \frac{6}{36} = \frac{20}{36} = \frac{5}{9} \approx 0.556

(c)

4 marks

The distribution of $X$ is $P (X = 0) = \frac{20}{36}$ , $P (X = 5) = \frac{10}{36}$ and $P (X = 10) = \frac{6}{36}$ .

E (X) = 0 (\frac{20}{36}) + 5 (\frac{10}{36}) + 10 (\frac{6}{36}) = \frac{110}{36} \approx 3.06

(d)

2 marks

Multiply the expected score of one round by 90.

90 \times \frac{110}{36} = 275

(e)

4 marks

More than 40 points in five rounds needs five scores of 10 (50 points) or four scores of 10 and one of 5 (45 points). Three tens and two fives make only 40.

P (five 10s) = (\frac{1}{6})^{5} \approx 0.000129 P (four 10s and a 5) = 5 (\frac{1}{6})^{4} (\frac{5}{18}) \approx 0.00107 P (prize) \approx 0.00120

Question 8

16 marks

A pyramid of playing cards: a quadratic number sequence

Number and algebra

(a)

1 mark

From the diagram of the stack with 3 rows, count the cards.

t_{3} = 2 + 5 + 8 = 15

(b)

2 marks

A fourth row adds 11 cards (the differences between rows are 3).

t_{4} = 15 + 11 = 26

(c)

3 marks

The numbers of cards in the rows form the arithmetic sequence $2, 5, 8, 11, \dots$ with first term 2 and common difference 3. Use the sum of $n$ terms.

t_{n} = \frac{n}{2} [2 (2) + (n - 1) 3] = \frac{n}{2} (3 n + 1)

(d)

3 marks

The cards available are $14 \times 52 = 728$ . Solve $\frac{n ( 3 n + 1 )}{2} \leq 728$ .
The most rows is 21.

t_{21} = \frac{21 ( 64 )}{2} = 672 \leq 728, t_{22} = \frac{22 ( 67 )}{2} = 737 > 728

(e)

2 marks

The number of cards $t_{n}$ must be a multiple of 52. List $t_{n}$ : 2, 7, 15, 26, 40, 57, 77, 100, 126, 155, 187, 222, 260.
The first multiple of 52 is $t_{13} = 260 = 5 \times 52$ (so 5 packs).

n = 13

(f)

5 marks

Each row is an equilateral triangle with side 88 mm, so its height is
For the stack to be taller than 2000 mm we need $76.21 n > 2000$ .
The number of cards for 27 rows.

88 sin 6 0^{\circ} = 443 \approx 76.21 mm n > 26.2 \Rightarrow n = 27 t_{27} = \frac{27 ( 82 )}{2} = 1107

Question 9

15 marks

A self-inverse rational function, the closest point to the origin and a reciprocal graph

Functions

(a)

3 marks

Start with $y = \frac{2 - 2 x}{x + 2}$ and swap $x$ and $y$ .

x = \frac{2 - 2 y}{y + 2} \Rightarrow x y + 2 x = 2 - 2 y \Rightarrow y (x + 2) = 2 - 2 x f^{- 1} (x) = \frac{2 - 2 x}{x + 2} = f (x)

(b)

4 marks

(i) The graph is symmetric about $y = x$ , so the closest point to the origin is on $y = x$ . Put $f (k) = k$ .

The closest point has $k > 0$ .

(ii) On the line $y = x$ .

\frac{2 - 2 k}{k + 2} = k \Rightarrow 2 - 2 k = k^{2} + 2 k \Rightarrow k^{2} + 4 k - 2 = 0 k = - 2 \pm 6 \Rightarrow k \approx 0.449 or - 4.45 k \approx 0.449 P \approx (0.449, 0.449)

(c)

2 marks

$g (x) = \frac{2 - 3 x}{c x + d}$ .
(i) The denominator is zero.
(ii) As $x \to \infty$ , $g (x) \to - \frac{3}{c}$ .

x = - \frac{d}{c} y = - \frac{3}{c}

(d)

2 marks

For $g^{- 1} = g$ , solve for the inverse.
Equating with $g (x) = \frac{2 - 3 x}{c x + d}$ , the denominators must match.

x = \frac{2 - 3 y}{cy + d} \Rightarrow c x y + d x = 2 - 3 y \Rightarrow y (c x + 3) = 2 - d x g^{- 1} (x) = \frac{2 - d x}{c x + 3} d = 3

(e)

4 marks

With $c = 1$ , $d = 3$ , the function $g (x) = \frac{2 - 3 x}{x + 3}$ and its reciprocal is $\frac{1}{g ( x )} = \frac{x + 3}{2 - 3 x}$ .
The asymptotes of $\frac{1}{g}$ : the denominator $2 - 3 x$ is zero at $x = \frac{2}{3}$ (vertical asymptote) and as $x \to \infty$ it tends to $- \frac{1}{3}$ (horizontal asymptote).
The intercepts: when $y = 0$ , $x = - 3$ , and when $x = 0$ , $y = \frac{3}{2}$ .
Sketch: a hyperbola in two branches, one passing through $(- 3, 0)$ and $(0, \frac{3}{2})$ to the left of $x = \frac{2}{3}$ , rising towards $+ \infty$ as $x \to \frac{2}{3}$ from the left and approaching $y = - \frac{1}{3}$ from below as $x \to - \infty$ , and one branch to the right of $x = \frac{2}{3}$ coming down from $- \infty$ towards $y = - \frac{1}{3}$ .

x = \frac{2}{3}, y = - \frac{1}{3}

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 algebra22 marks · 28%
Q4, Q8
Statistics and probability22 marks · 28%
Q5, Q7
Functions19 marks · 24%
Q3, Q9
Calculus11 marks · 14%
Q1, Q6
Geometry and trigonometry6 marks · 8%
Q2

Mentor insights

Traps to watch for

01
Arc length becomes the base circumference · Q2
When the sector is rolled up, its arc becomes the circle at the base of the cone.
02
Which line to use · Q5
To estimate $A$ from $F$ use the regression line of $A$ on $F$ .
03
Sample space · Q7
There are 36 equally likely outcomes. Doubles give 6, a difference of 1 gives 10.
04
Counting · Q8
Look at how many cards are added to make each new row, and find the pattern.
05
Inverse · Q9
Swap $x$ and $y$ and solve for $y$ .

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