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IB Maths AA SL · Paper 2

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1 TZ2TZ3

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 TZ3 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 · 34 marks

Question 1

5 marks

The surface area of a solid hemisphere from two points

Geometry and trigonometry

(a)

2 marks

The radius is the distance from $A (6, - 1, - 3)$ to $B (4, - 5, - 9)$ .

A B = (4 - 6)^{2} + (- 5 + 1)^{2} + (- 9 + 3)^{2} = 4 + 16 + 36 = 56 \approx 7.48

(b)

3 marks

The total surface area of the solid hemisphere.

S = 2 π r^{2} + π r^{2} = 3 π r^{2} = 3 π (56) = 168 π \approx 528

Question 2

6 marks

A triangle with two sides and the included angle: the cosine and sine rules

Geometry and trigonometry

(a)

3 marks

Use the cosine rule.

B C^{2} = 1 2^{2} + 7^{2} - 2 (12) (7) cos 11 6^{\circ} \approx 266.6 \Rightarrow B C \approx 16.3

(b)

3 marks

Use the sine rule for angle $C$ , the angle opposite $A B = 7$ .
The angle is acute because the angle at $A$ is already obtuse.

\frac{sin A C B}{7} = \frac{sin 11 6 ^{\circ}}{16.33} \Rightarrow sin A C B = 0.3853 A C B \approx 22. 7^{\circ}

Question 3

5 marks

The coefficient of x^7 in (x + k)^11

Number and algebra

(a)

1 mark

An expansion to the power 11 has $11 + 1$ terms.

12

(b)

4 marks

The term in $x^{7}$ .
Set the coefficient equal to 1320.
Here $k > 0$ , so only the positive root is used.

(7 11) x^{7} k^{4} = 330 k^{4} x^{7} 330 k^{4} = 1320 \Rightarrow k^{4} = 4 \Rightarrow k = 2 \approx 1.41

Question 4

6 marks

Supermarket visits: a probability distribution and a sampling method

Statistics and probability

(a)

3 marks

(i) The probabilities add to 1.

(ii) The most likely value is the one with the greatest probability. $P (X = 2) = 2 a = 0.308$ is the biggest.

1.5 a + 2 a + 0.281 + a + 0.026 + 0 = 1 \Rightarrow 4.5 a = 0.693 \Rightarrow a = 0.154 mode = 2

(b)

2 marks

Use $E (X) = \sum x P (X = x)$ with $P (X = 1) = 0.231$ , $P (X = 2) = 0.308$ , $P (X = 3) = 0.281$ , $P (X = 4) = 0.154$ , $P (X = 5) = 0.026$ .

E (X) = 0.231 + 0.616 + 0.843 + 0.616 + 0.130 = 2.436 \approx 2.44

(c)

1 mark

The manager chose the first 50 customers to arrive, which is whatever is easiest, so it is a convenience sample.

Question 5

5 marks

The arc length of a circle from the area of a triangle

Number and algebra

5 marks

The triangle $A O B$ has two sides equal to the radius $r$ , with included angle 2.51 radians.
The arc $A C B$ is the major arc with angle $2 π - 2.51 = 3.773$ radians.

\frac{1}{2} r^{2} sin 2.51 = 26 \Rightarrow r^{2} = 88.07 \Rightarrow r \approx 9.385 cm arc = r θ = 9.385 (3.773) \approx 35.4 cm

Question 6

7 marks

A rational function with a parameter: the derivative and a tangent angle

Calculus

(a)

3 marks

Differentiate with the quotient rule.
Factor out $2 (2 x + a)^{2} (x + 5)$ from the numerator and cancel one factor of $(x + 5)$ .

f^{'} (x) = \frac{6 ( 2 x + a ) ^{2} ( x + 5 ) ^{2} - 2 ( 2 x + a ) ^{3} ( x + 5 )}{( x + 5 ) ^{4}} f^{'} (x) = \frac{2 ( 2 x + a ) ^{2} [ 3 ( x + 5 ) - ( 2 x + a ) ]}{( x + 5 ) ^{3}} = \frac{2 ( 2 x + a ) ^{2} ( x + 15 - a )}{( x + 5 ) ^{3}}

(b)

4 marks

At $x = 1$ the gradient is $tan 7 0^{\circ} = 2.7475$ .
Solve for $a > 0$ with the calculator.

f^{'} (1) = \frac{2 ( 2 + a ) ^{2} ( 16 - a )}{216} = tan 7 0^{\circ} a \approx 2.73 or a \approx 15.0

Section B · Extended response

Questions 7–9 · 46 marks

Question 7

15 marks

A sine curve, normal and tangent lines and an enclosed area

Functions

(a)

3 marks

The function is $f (x) = \frac{2}{π} sin (3 π x) + 2$ .
(i) The amplitude is $\frac{2}{π} \approx 0.637$ .
(ii) The period is $\frac{2 π}{3 π} = \frac{2}{3} \approx 0.667$ .

(b)

3 marks

Evaluate the graph at $x = 1.63$ .
The point $P$ has $y = 2.16 < 2.216$ , so it lies below the graph of $f$ .

f (1.63) = \frac{2}{π} sin (3 π \times 1.63) + 2 \approx 2.216

(c)

1 mark

Rearrange $x - 6 y + 11 = 0$ as $y = \frac{x}{6} + \frac{11}{6}$ .

m = \frac{1}{6}

(d)

3 marks

(i) The tangent is perpendicular to the normal, so its gradient is $- 1 \div \frac{1}{6}$ .

(ii) It passes through $A (1, 2)$ .

m_{2} = - 6 y - 2 = - 6 (x - 1) \Rightarrow y = - 6 x + 8

(e)

2 marks

Solve $f (x) = \frac{x + 11}{6}$ for the other intersection with $x > 1.5$ .

B \approx (1.65, 2.11)

(f)

3 marks

The area between the line and the curve from $x = 1$ to $x = 1.6486$ , taking the part where the curve is above the line and the part where it is below, each with its own sign, so find the integral of the modulus.

Area = \int_{1}^{1.6486} \frac{x + 11}{6} - (\frac{2}{π} sin 3 π x + 2) d x \approx 0.253

Question 8

17 marks

A water park: binomial conditions, probabilities, a conditional probability and finding n

Statistics and probability

(a)

2 marks

There is a fixed number of trials, each trial has two possible outcomes (success and failure), and the trials are independent with the same probability of success each time.

(b)

2 marks

The expected number riding Daifong.

1900 \times 0.37 = 703

(c)

4 marks

Let $D \sim B (1900, 0.37)$ .
(i) $P (D = 712) \approx 0.0173$
(ii) $P (684 \leq D \leq 712) = P (D \leq 712) - P (D \leq 683) \approx 0.6747 - 0.1771$

P (684 \leq D \leq 712) \approx 0.498

(d)

4 marks

The condition is inside the larger range, so use conditional probability.

P (D \leq 692 ∣ 684 \leq D \leq 712) = \frac{P ( 684 \leq D \leq 692 )}{P ( 684 \leq D \leq 712 )} = \frac{0.1323}{0.4976} \approx 0.266

(e)

2 marks

Riding Daifong and riding Torbellino are independent, so multiply the probabilities.

0.37 \times 0.61 \approx 0.226

(f)

3 marks

Let $T \sim B (n, 0.61)$ with $P (T \leq 500) \approx 0.693$ . Try values of $n$ on the calculator.

n = 808 : P (T \leq 500) = 0.708, n = 809 : P (T \leq 500) = 0.693 n = 809

Question 9

14 marks

A 200 metre race: velocity models for two athletes

Calculus

(a)

3 marks

Fiona's velocity is $v (t) = \frac{8.14 t}{t ^{2} + 0.2}$ .
(i) $v (1) = \frac{8.14}{1.2} \approx 7.43 m s^{- 1}$
(ii) Solve $v (t) = 5$ .

t \approx 0.348 seconds

(b)

2 marks

Solve $v^{'} (t) = 4$ with the calculator.

t \approx 0.591 seconds

(c)

3 marks

(i) For large $t$ , $t^{2} + 0.2 \approx t$ .

(ii) The race is over once Fiona has gone 200 m, so the value of $t$ is limited. Fiona never reaches the speed 8.14 (the graph only gets close to it), so this value is not valid for the race.

t \to \infty lim v (t) = 8.14 m s^{- 1}

(d)

6 marks

Find the time $t_{f}$ when Fiona has gone 200 m, by solving the equation for the distance.
In that time Lucy, with $w (t) = \frac{8 t}{t ^{2} + 0.3}$ , has gone
So Lucy is this far from the finishing line.

\int_{0}^{t_{f}} v (t) d t = 200 \Rightarrow t_{f} \approx 25.01 seconds \int_{0}^{25.01} w (t) d t \approx 195.77 m 200 - 195.77 \approx 4.23 m

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 probability23 marks · 29%
Q4, Q8
Calculus21 marks · 26%
Q6, Q9
Functions15 marks · 19%
Q7
Geometry and trigonometry11 marks · 14%
Q1, Q2
Number and algebra10 marks · 13%
Q3, Q5

Mentor insights

Traps to watch for

01
Which rule · Q2
Two sides and the included angle give the third side by the cosine rule.
02
Quotient rule · Q6
$f^{'} (x) = \frac{u ^{'} v - u v ^{'}}{v ^{2}}$ with $u = (2 x + a)^{3}$ and $v = (x + 5)^{2}$ .
03
Perpendicular gradients · Q7
A tangent and the normal at the same point have gradients whose product is $- 1$ .
04
Binomial conditions · Q8
A fixed number of independent trials, two outcomes, and a constant probability of success.
05
Distance from velocity · Q9
Distance is the integral of velocity.

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 AA SL · Paper 2

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1 TZ2TZ3

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 TZ3 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 · 34 marks

Question 1

5 marks

The surface area of a solid hemisphere from two points

Geometry and trigonometry

(a)

2 marks

The radius is the distance from $A (6, - 1, - 3)$ to $B (4, - 5, - 9)$ .

A B = (4 - 6)^{2} + (- 5 + 1)^{2} + (- 9 + 3)^{2} = 4 + 16 + 36 = 56 \approx 7.48

(b)

3 marks

The total surface area of the solid hemisphere.

S = 2 π r^{2} + π r^{2} = 3 π r^{2} = 3 π (56) = 168 π \approx 528

Question 2

6 marks

A triangle with two sides and the included angle: the cosine and sine rules

Geometry and trigonometry

(a)

3 marks

Use the cosine rule.

B C^{2} = 1 2^{2} + 7^{2} - 2 (12) (7) cos 11 6^{\circ} \approx 266.6 \Rightarrow B C \approx 16.3

(b)

3 marks

Use the sine rule for angle $C$ , the angle opposite $A B = 7$ .
The angle is acute because the angle at $A$ is already obtuse.

\frac{sin A C B}{7} = \frac{sin 11 6 ^{\circ}}{16.33} \Rightarrow sin A C B = 0.3853 A C B \approx 22. 7^{\circ}

Question 3

5 marks

The coefficient of x^7 in (x + k)^11

Number and algebra

(a)

1 mark

An expansion to the power 11 has $11 + 1$ terms.

12

(b)

4 marks

The term in $x^{7}$ .
Set the coefficient equal to 1320.
Here $k > 0$ , so only the positive root is used.

(7 11) x^{7} k^{4} = 330 k^{4} x^{7} 330 k^{4} = 1320 \Rightarrow k^{4} = 4 \Rightarrow k = 2 \approx 1.41

Question 4

6 marks

Supermarket visits: a probability distribution and a sampling method

Statistics and probability

(a)

3 marks

(i) The probabilities add to 1.

(ii) The most likely value is the one with the greatest probability. $P (X = 2) = 2 a = 0.308$ is the biggest.

1.5 a + 2 a + 0.281 + a + 0.026 + 0 = 1 \Rightarrow 4.5 a = 0.693 \Rightarrow a = 0.154 mode = 2

(b)

2 marks

Use $E (X) = \sum x P (X = x)$ with $P (X = 1) = 0.231$ , $P (X = 2) = 0.308$ , $P (X = 3) = 0.281$ , $P (X = 4) = 0.154$ , $P (X = 5) = 0.026$ .

E (X) = 0.231 + 0.616 + 0.843 + 0.616 + 0.130 = 2.436 \approx 2.44

(c)

1 mark

The manager chose the first 50 customers to arrive, which is whatever is easiest, so it is a convenience sample.

Question 5

5 marks

The arc length of a circle from the area of a triangle

Number and algebra

5 marks

The triangle $A O B$ has two sides equal to the radius $r$ , with included angle 2.51 radians.
The arc $A C B$ is the major arc with angle $2 π - 2.51 = 3.773$ radians.

\frac{1}{2} r^{2} sin 2.51 = 26 \Rightarrow r^{2} = 88.07 \Rightarrow r \approx 9.385 cm arc = r θ = 9.385 (3.773) \approx 35.4 cm

Question 6

7 marks

A rational function with a parameter: the derivative and a tangent angle

Calculus

(a)

3 marks

Differentiate with the quotient rule.
Factor out $2 (2 x + a)^{2} (x + 5)$ from the numerator and cancel one factor of $(x + 5)$ .

f^{'} (x) = \frac{6 ( 2 x + a ) ^{2} ( x + 5 ) ^{2} - 2 ( 2 x + a ) ^{3} ( x + 5 )}{( x + 5 ) ^{4}} f^{'} (x) = \frac{2 ( 2 x + a ) ^{2} [ 3 ( x + 5 ) - ( 2 x + a ) ]}{( x + 5 ) ^{3}} = \frac{2 ( 2 x + a ) ^{2} ( x + 15 - a )}{( x + 5 ) ^{3}}

(b)

4 marks

At $x = 1$ the gradient is $tan 7 0^{\circ} = 2.7475$ .
Solve for $a > 0$ with the calculator.

f^{'} (1) = \frac{2 ( 2 + a ) ^{2} ( 16 - a )}{216} = tan 7 0^{\circ} a \approx 2.73 or a \approx 15.0

Section B · Extended response

Questions 7–9 · 46 marks

Question 7

15 marks

A sine curve, normal and tangent lines and an enclosed area

Functions

(a)

3 marks

The function is $f (x) = \frac{2}{π} sin (3 π x) + 2$ .
(i) The amplitude is $\frac{2}{π} \approx 0.637$ .
(ii) The period is $\frac{2 π}{3 π} = \frac{2}{3} \approx 0.667$ .

(b)

3 marks

Evaluate the graph at $x = 1.63$ .
The point $P$ has $y = 2.16 < 2.216$ , so it lies below the graph of $f$ .

f (1.63) = \frac{2}{π} sin (3 π \times 1.63) + 2 \approx 2.216

(c)

1 mark

Rearrange $x - 6 y + 11 = 0$ as $y = \frac{x}{6} + \frac{11}{6}$ .

m = \frac{1}{6}

(d)

3 marks

(i) The tangent is perpendicular to the normal, so its gradient is $- 1 \div \frac{1}{6}$ .

(ii) It passes through $A (1, 2)$ .

m_{2} = - 6 y - 2 = - 6 (x - 1) \Rightarrow y = - 6 x + 8

(e)

2 marks

Solve $f (x) = \frac{x + 11}{6}$ for the other intersection with $x > 1.5$ .

B \approx (1.65, 2.11)

(f)

3 marks

Area = \int_{1}^{1.6486} \frac{x + 11}{6} - (\frac{2}{π} sin 3 π x + 2) d x \approx 0.253

Question 8

17 marks

A water park: binomial conditions, probabilities, a conditional probability and finding n

Statistics and probability

(a)

2 marks

There is a fixed number of trials, each trial has two possible outcomes (success and failure), and the trials are independent with the same probability of success each time.

(b)

2 marks

The expected number riding Daifong.

1900 \times 0.37 = 703

(c)

4 marks

Let $D \sim B (1900, 0.37)$ .
(i) $P (D = 712) \approx 0.0173$
(ii) $P (684 \leq D \leq 712) = P (D \leq 712) - P (D \leq 683) \approx 0.6747 - 0.1771$

P (684 \leq D \leq 712) \approx 0.498

(d)

4 marks

The condition is inside the larger range, so use conditional probability.

P (D \leq 692 ∣ 684 \leq D \leq 712) = \frac{P ( 684 \leq D \leq 692 )}{P ( 684 \leq D \leq 712 )} = \frac{0.1323}{0.4976} \approx 0.266

(e)

2 marks

Riding Daifong and riding Torbellino are independent, so multiply the probabilities.

0.37 \times 0.61 \approx 0.226

(f)

3 marks

Let $T \sim B (n, 0.61)$ with $P (T \leq 500) \approx 0.693$ . Try values of $n$ on the calculator.

n = 808 : P (T \leq 500) = 0.708, n = 809 : P (T \leq 500) = 0.693 n = 809

Question 9

14 marks

A 200 metre race: velocity models for two athletes

Calculus

(a)

3 marks

Fiona's velocity is $v (t) = \frac{8.14 t}{t ^{2} + 0.2}$ .
(i) $v (1) = \frac{8.14}{1.2} \approx 7.43 m s^{- 1}$
(ii) Solve $v (t) = 5$ .

t \approx 0.348 seconds

(b)

2 marks

Solve $v^{'} (t) = 4$ with the calculator.

t \approx 0.591 seconds

(c)

3 marks

(i) For large $t$ , $t^{2} + 0.2 \approx t$ .

(ii) The race is over once Fiona has gone 200 m, so the value of $t$ is limited. Fiona never reaches the speed 8.14 (the graph only gets close to it), so this value is not valid for the race.

t \to \infty lim v (t) = 8.14 m s^{- 1}

(d)

6 marks

Find the time $t_{f}$ when Fiona has gone 200 m, by solving the equation for the distance.
In that time Lucy, with $w (t) = \frac{8 t}{t ^{2} + 0.3}$ , has gone
So Lucy is this far from the finishing line.

\int_{0}^{t_{f}} v (t) d t = 200 \Rightarrow t_{f} \approx 25.01 seconds \int_{0}^{25.01} w (t) d t \approx 195.77 m 200 - 195.77 \approx 4.23 m

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 probability23 marks · 29%
Q4, Q8
Calculus21 marks · 26%
Q6, Q9
Functions15 marks · 19%
Q7
Geometry and trigonometry11 marks · 14%
Q1, Q2
Number and algebra10 marks · 13%
Q3, Q5

Mentor insights

Traps to watch for

01
Which rule · Q2
Two sides and the included angle give the third side by the cosine rule.
02
Quotient rule · Q6
$f^{'} (x) = \frac{u ^{'} v - u v ^{'}}{v ^{2}}$ with $u = (2 x + a)^{3}$ and $v = (x + 5)^{2}$ .
03
Perpendicular gradients · Q7
A tangent and the normal at the same point have gradients whose product is $- 1$ .
04
Binomial conditions · Q8
A fixed number of independent trials, two outcomes, and a constant probability of success.
05
Distance from velocity · Q9
Distance is the integral of velocity.

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