Loading…

Worked solutions

IB Maths AA SL · Paper 1

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2 TZ3

Session: May 2025
Marks: 80
Duration: 1 hour 30 min
Calculator: Not 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

Writing expressions as a single logarithm

Functions

(a)

1 mark

Add the logarithms.

ln 3 + ln 4 = ln 12

(b)

2 marks

Move the 3 into the power.

3 ln 2 = ln 2^{3} = ln 8

(c)

2 marks

Use $- ln a = ln \frac{1}{a}$ .

- ln \frac{1}{2} = ln 2

Question 2

5 marks

The local minimum of a cubic

Number and algebra

5 marks

Differentiate $f (x) = \frac{4 x ^{3}}{3} - 16 x$ .
Set the derivative to zero.
The required point has $p > 0$ , so $p = 2$ .
The local minimum is $(2, - \frac{64}{3})$ .

f^{'} (x) = 4 x^{2} - 16 4 x^{2} - 16 = 0 \Rightarrow x = \pm 2 q = f (2) = \frac{4 ( 8 )}{3} - 32 = - \frac{64}{3}

Question 3

7 marks

Quarterly interest and the binomial expansion of (1 + x)^4

Number and algebra

(a)

1 mark

The quarterly rate is $\frac{4%}{4}$ .

k = 0.01

(b)

2 marks

Use Pascal's triangle for the coefficients 1, 4, 6, 4, 1.

(1 + x)^{4} = 1 + 4 x + 6 x^{2} + 4 x^{3} + x^{4}

(c)

4 marks

Put $x = 0.01$ in the expansion and multiply by 1000.
To the nearest dinar the amount is

1000 (1 + 0.04 + 0.0006 + 0.000004 + 0.00000001) = 1000 (1.04060401) = 1040.60 \dots 1041 dinar

Question 4

4 marks

The area enclosed by e^x, -e^x and two vertical lines

Number and algebra

4 marks

The vertical distance between the curves is $e^{x} - (- e^{x}) = 2 e^{x}$ .

Area = \int_{- 1}^{1} 2 e^{x} d x = 2 [e^{x}]_{- 1}^{1} 2 (e - e^{- 1}) = 2 e - \frac{2}{e}

Question 5

6 marks

Two events: an intersection and independence

Statistics and probability

(a)

3 marks

$P (A^{'}) = \frac{3}{4}$ , so $P (A) = \frac{1}{4}$ . Use $P (A \cap B) = P (B ∣ A) P (A)$ .

P (A \cap B) = \frac{2}{3} \times \frac{1}{4} = \frac{1}{6}

(b)

3 marks

Use $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ with $P (A \cup B) = \frac{3}{4}$ .
Now compare.
Since $P (A \cap B) = P (A) P (B)$ , the events are independent.

\frac{3}{4} = \frac{1}{4} + P (B) - \frac{1}{6} \Rightarrow P (B) = \frac{1}{2} + \frac{1}{6} = \frac{2}{3} P (A) P (B) = \frac{1}{4} \times \frac{2}{3} = \frac{1}{6} = P (A \cap B)

Question 6

8 marks

Picture frames growing by 50% and their mean and median area

Number and algebra

(a)

5 marks

(i) The first area is $4 \times 5 = 20$ and each area is $\frac{9}{4}$ times the one before, so the areas form a geometric sequence.

(ii) The total area is the sum of ten terms.

The mean is the sum over 10.

area of F_{n} = 20 (\frac{9}{4})^{n - 1} S_{10} = \frac{20 ( ( \frac{9}{4} ) ^{10} - 1 )}{\frac{9}{4} - 1} = 16 ((\frac{9}{4})^{10} - 1) mean = \frac{16}{10} ((\frac{9}{4})^{10} - 1) = \frac{8}{5} ((\frac{9}{4})^{10} - 1) p = \frac{8}{5}, a = 10

(b)

3 marks

The median is the mean of $F_{5}$ and $F_{6}$ .

\frac{20 ( \frac{9}{4} ) ^{4} + 20 ( \frac{9}{4} ) ^{5}}{2} = 10 (\frac{9}{4})^{4} (1 + \frac{9}{4}) = 10 (\frac{9}{4})^{4} \times \frac{13}{4} median = \frac{65}{2} (\frac{9}{4})^{4} q = \frac{65}{2}

Section B · Extended response

Questions 7–9 · 45 marks

Question 7

13 marks

An object's velocity 30 + 20t - 10t^2: maximum speed, distance and return

Calculus

(a)

4 marks

(i) The local maximum is at $t = 1$ .

(ii) Compare with the other end of the interval. At $t = 5$ :

The speed is the modulus, which is larger there.

v (1) = 30 + 20 - 10 = 40 m s^{- 1} v (5) = 30 + 100 - 250 = - 120 maximum speed = 120 m s^{- 1}

(b)

5 marks

(i) The velocity changes sign at $t = 3$ .

(ii) The object does not change direction before $t = 3$ , so the distance is the integral of $v$ .

T = 3 \int_{0}^{3} (30 + 20 t - 10 t^{2}) d t = [30 t + 10 t^{2} - \frac{10 t ^{3}}{3}]_{0}^{3} = 90 + 90 - 90 = 90 m

(c)

4 marks

The displacement from $t = 0$ to $t = 5$ is the integral of $v$ .
At $t = 3$ the object is 90 m from its start, and by $t = 5$ its displacement is negative. Displacement changes continuously, so it must have passed through zero between those times. The object does return to its initial position.

\int_{0}^{5} (30 + 20 t - 10 t^{2}) d t = 150 + 250 - \frac{1250}{3} = - \frac{50}{3}

Question 8

15 marks

A quadratic, an inequality and a composite with ln x

Functions

(a)

4 marks

The roots are $- 3$ and $- 1$ , so the vertex is at $x = - 2$ . Expand: $5 (x + 1) (x + 3) = 5 x^{2} + 20 x + 15 = 5 (x + 2)^{2} - 5$ .

f (x) = 5 (x + 2)^{2} - 5, a = 5, h = - 2, k = - 5

(b)

4 marks

The graph is a U-shaped parabola. The vertex is $(- 2, - 5)$ , the $x$ -intercepts are $(- 3, 0)$ and $(- 1, 0)$ and the $y$ -intercept is $(0, 15)$ .

(c)

4 marks

Solve $5 (x + 2)^{2} - 5 \leq 40$ .

(x + 2)^{2} \leq 9 \Rightarrow - 3 \leq x + 2 \leq 3 \Rightarrow - 5 \leq x \leq 1

(d)

3 marks

(i) Replace $x$ by $ln x$ .

(ii) Replace $x$ by $ln x$ in the answer to part (c).

(f \circ g) (x) = 5 (ln x + 1) (ln x + 3) - 5 \leq ln x \leq 1 \Rightarrow e^{- 5} \leq x \leq e

Question 9

17 marks

A cylinder inside a sphere: lengths, surface area and volume

Geometry and trigonometry

(a)

4 marks

(i) $A B = 2 r$ , $B C = 2 R$ and $A B C = θ$ and the angle at $C$ is $9 0^{\circ}$ .

(ii) In the same triangle, $A C = h$ .

cos θ = \frac{B C}{A B} = \frac{2 R}{2 r} \Rightarrow R = r cos θ sin θ = \frac{h}{2 r} \Rightarrow h = 2 r sin θ

(b)

4 marks

The two circular ends have area $2 π R^{2} = 2 π r^{2} cos^{2} θ$ and the curved surface has area $2 π R h = 2 π (r cos θ) (2 r sin θ) = 4 π r^{2} sin θ cos θ$ .
Replace $cos^{2} θ$ by $1 - sin^{2} θ$ .

S = 2 π r^{2} cos^{2} θ + 4 π r^{2} sin θ cos θ S = 2 π r^{2} (1 - sin^{2} θ + 2 sin θ cos θ) = 2 π r^{2} (1 + 2 sin θ cos θ - sin^{2} θ)

(c)

4 marks

The surface area of the sphere is $4 π r^{2} = 2 S$ .
Since $sin θ \neq = 0$ , divide by $sin θ$ .

4 π r^{2} = 4 π r^{2} (1 + 2 sin θ cos θ - sin^{2} θ) \Rightarrow 2 sin θ cos θ - sin^{2} θ = 0 2 cos θ - sin θ = 0 \Rightarrow tan θ = 2

(d)

5 marks

The volume is $V = π R^{2} h = π r^{2} cos^{2} θ \times 2 r sin θ = 2 π r^{3} cos^{2} θ sin θ$ .
With $tan θ = 2$ the right triangle has sides 1, 2 and $5$ , so $cos θ = \frac{1}{5}$ and $sin θ = \frac{2}{5}$ .

V = 2 π r^{3} \cdot \frac{1}{5} \cdot \frac{2}{5} = \frac{4 π r ^{3}}{5 5} = \frac{4}{25} π r^{3} 5 p = \frac{4}{25}

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 algebra24 marks · 30%
Q2, Q3, Q4, Q6
Functions20 marks · 25%
Q1, Q8
Geometry and trigonometry17 marks · 21%
Q9
Calculus13 marks · 16%
Q7
Statistics and probability6 marks · 8%
Q5

Mentor insights

Traps to watch for

01
Nominal rate · Q3
A nominal rate of 4% compounded quarterly is 1% per quarter, so $k = 0.01$ .
02
Geometric sequence · Q6
Each width and height is multiplied by 1.5, so each area is multiplied by $1. 5^{2} = \frac{9}{4}$ .
03
Speed against velocity · Q7
The speed is $∣ v ∣$ , so a large negative velocity counts as a large speed.
04
Complete the square · Q8
The vertex is halfway between the roots.
05
Triangle in a semicircle · Q9
$A$ and $C$ lie on the sphere with $[A B]$ a diameter, so triangle $A B C$ has a right angle at $C$ .

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 1

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2 TZ3

Session: May 2025
Marks: 80
Duration: 1 hour 30 min
Calculator: Not 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

Writing expressions as a single logarithm

Functions

(a)

1 mark

Add the logarithms.

ln 3 + ln 4 = ln 12

(b)

2 marks

Move the 3 into the power.

3 ln 2 = ln 2^{3} = ln 8

(c)

2 marks

Use $- ln a = ln \frac{1}{a}$ .

- ln \frac{1}{2} = ln 2

Question 2

5 marks

The local minimum of a cubic

Number and algebra

5 marks

Differentiate $f (x) = \frac{4 x ^{3}}{3} - 16 x$ .
Set the derivative to zero.
The required point has $p > 0$ , so $p = 2$ .
The local minimum is $(2, - \frac{64}{3})$ .

f^{'} (x) = 4 x^{2} - 16 4 x^{2} - 16 = 0 \Rightarrow x = \pm 2 q = f (2) = \frac{4 ( 8 )}{3} - 32 = - \frac{64}{3}

Question 3

7 marks

Quarterly interest and the binomial expansion of (1 + x)^4

Number and algebra

(a)

1 mark

The quarterly rate is $\frac{4%}{4}$ .

k = 0.01

(b)

2 marks

Use Pascal's triangle for the coefficients 1, 4, 6, 4, 1.

(1 + x)^{4} = 1 + 4 x + 6 x^{2} + 4 x^{3} + x^{4}

(c)

4 marks

Put $x = 0.01$ in the expansion and multiply by 1000.
To the nearest dinar the amount is

1000 (1 + 0.04 + 0.0006 + 0.000004 + 0.00000001) = 1000 (1.04060401) = 1040.60 \dots 1041 dinar

Question 4

4 marks

The area enclosed by e^x, -e^x and two vertical lines

Number and algebra

4 marks

The vertical distance between the curves is $e^{x} - (- e^{x}) = 2 e^{x}$ .

Area = \int_{- 1}^{1} 2 e^{x} d x = 2 [e^{x}]_{- 1}^{1} 2 (e - e^{- 1}) = 2 e - \frac{2}{e}

Question 5

6 marks

Two events: an intersection and independence

Statistics and probability

(a)

3 marks

$P (A^{'}) = \frac{3}{4}$ , so $P (A) = \frac{1}{4}$ . Use $P (A \cap B) = P (B ∣ A) P (A)$ .

P (A \cap B) = \frac{2}{3} \times \frac{1}{4} = \frac{1}{6}

(b)

3 marks

Use $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ with $P (A \cup B) = \frac{3}{4}$ .
Now compare.
Since $P (A \cap B) = P (A) P (B)$ , the events are independent.

\frac{3}{4} = \frac{1}{4} + P (B) - \frac{1}{6} \Rightarrow P (B) = \frac{1}{2} + \frac{1}{6} = \frac{2}{3} P (A) P (B) = \frac{1}{4} \times \frac{2}{3} = \frac{1}{6} = P (A \cap B)

Question 6

8 marks

Picture frames growing by 50% and their mean and median area

Number and algebra

(a)

5 marks

(i) The first area is $4 \times 5 = 20$ and each area is $\frac{9}{4}$ times the one before, so the areas form a geometric sequence.

(ii) The total area is the sum of ten terms.

The mean is the sum over 10.

area of F_{n} = 20 (\frac{9}{4})^{n - 1} S_{10} = \frac{20 ( ( \frac{9}{4} ) ^{10} - 1 )}{\frac{9}{4} - 1} = 16 ((\frac{9}{4})^{10} - 1) mean = \frac{16}{10} ((\frac{9}{4})^{10} - 1) = \frac{8}{5} ((\frac{9}{4})^{10} - 1) p = \frac{8}{5}, a = 10

(b)

3 marks

The median is the mean of $F_{5}$ and $F_{6}$ .

\frac{20 ( \frac{9}{4} ) ^{4} + 20 ( \frac{9}{4} ) ^{5}}{2} = 10 (\frac{9}{4})^{4} (1 + \frac{9}{4}) = 10 (\frac{9}{4})^{4} \times \frac{13}{4} median = \frac{65}{2} (\frac{9}{4})^{4} q = \frac{65}{2}

Section B · Extended response

Questions 7–9 · 45 marks

Question 7

13 marks

An object's velocity 30 + 20t - 10t^2: maximum speed, distance and return

Calculus

(a)

4 marks

(i) The local maximum is at $t = 1$ .

(ii) Compare with the other end of the interval. At $t = 5$ :

The speed is the modulus, which is larger there.

v (1) = 30 + 20 - 10 = 40 m s^{- 1} v (5) = 30 + 100 - 250 = - 120 maximum speed = 120 m s^{- 1}

(b)

5 marks

(i) The velocity changes sign at $t = 3$ .

(ii) The object does not change direction before $t = 3$ , so the distance is the integral of $v$ .

T = 3 \int_{0}^{3} (30 + 20 t - 10 t^{2}) d t = [30 t + 10 t^{2} - \frac{10 t ^{3}}{3}]_{0}^{3} = 90 + 90 - 90 = 90 m

(c)

4 marks

The displacement from $t = 0$ to $t = 5$ is the integral of $v$ .
At $t = 3$ the object is 90 m from its start, and by $t = 5$ its displacement is negative. Displacement changes continuously, so it must have passed through zero between those times. The object does return to its initial position.

\int_{0}^{5} (30 + 20 t - 10 t^{2}) d t = 150 + 250 - \frac{1250}{3} = - \frac{50}{3}

Question 8

15 marks

A quadratic, an inequality and a composite with ln x

Functions

(a)

4 marks

The roots are $- 3$ and $- 1$ , so the vertex is at $x = - 2$ . Expand: $5 (x + 1) (x + 3) = 5 x^{2} + 20 x + 15 = 5 (x + 2)^{2} - 5$ .

f (x) = 5 (x + 2)^{2} - 5, a = 5, h = - 2, k = - 5

(b)

4 marks

The graph is a U-shaped parabola. The vertex is $(- 2, - 5)$ , the $x$ -intercepts are $(- 3, 0)$ and $(- 1, 0)$ and the $y$ -intercept is $(0, 15)$ .

(c)

4 marks

Solve $5 (x + 2)^{2} - 5 \leq 40$ .

(x + 2)^{2} \leq 9 \Rightarrow - 3 \leq x + 2 \leq 3 \Rightarrow - 5 \leq x \leq 1

(d)

3 marks

(i) Replace $x$ by $ln x$ .

(ii) Replace $x$ by $ln x$ in the answer to part (c).

(f \circ g) (x) = 5 (ln x + 1) (ln x + 3) - 5 \leq ln x \leq 1 \Rightarrow e^{- 5} \leq x \leq e

Question 9

17 marks

A cylinder inside a sphere: lengths, surface area and volume

Geometry and trigonometry

(a)

4 marks

(i) $A B = 2 r$ , $B C = 2 R$ and $A B C = θ$ and the angle at $C$ is $9 0^{\circ}$ .

(ii) In the same triangle, $A C = h$ .

cos θ = \frac{B C}{A B} = \frac{2 R}{2 r} \Rightarrow R = r cos θ sin θ = \frac{h}{2 r} \Rightarrow h = 2 r sin θ

(b)

4 marks

The two circular ends have area $2 π R^{2} = 2 π r^{2} cos^{2} θ$ and the curved surface has area $2 π R h = 2 π (r cos θ) (2 r sin θ) = 4 π r^{2} sin θ cos θ$ .
Replace $cos^{2} θ$ by $1 - sin^{2} θ$ .

S = 2 π r^{2} cos^{2} θ + 4 π r^{2} sin θ cos θ S = 2 π r^{2} (1 - sin^{2} θ + 2 sin θ cos θ) = 2 π r^{2} (1 + 2 sin θ cos θ - sin^{2} θ)

(c)

4 marks

The surface area of the sphere is $4 π r^{2} = 2 S$ .
Since $sin θ \neq = 0$ , divide by $sin θ$ .

4 π r^{2} = 4 π r^{2} (1 + 2 sin θ cos θ - sin^{2} θ) \Rightarrow 2 sin θ cos θ - sin^{2} θ = 0 2 cos θ - sin θ = 0 \Rightarrow tan θ = 2

(d)

5 marks

The volume is $V = π R^{2} h = π r^{2} cos^{2} θ \times 2 r sin θ = 2 π r^{3} cos^{2} θ sin θ$ .
With $tan θ = 2$ the right triangle has sides 1, 2 and $5$ , so $cos θ = \frac{1}{5}$ and $sin θ = \frac{2}{5}$ .

V = 2 π r^{3} \cdot \frac{1}{5} \cdot \frac{2}{5} = \frac{4 π r ^{3}}{5 5} = \frac{4}{25} π r^{3} 5 p = \frac{4}{25}

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 algebra24 marks · 30%
Q2, Q3, Q4, Q6
Functions20 marks · 25%
Q1, Q8
Geometry and trigonometry17 marks · 21%
Q9
Calculus13 marks · 16%
Q7
Statistics and probability6 marks · 8%
Q5

Mentor insights

Traps to watch for

01
Nominal rate · Q3
A nominal rate of 4% compounded quarterly is 1% per quarter, so $k = 0.01$ .
02
Geometric sequence · Q6
Each width and height is multiplied by 1.5, so each area is multiplied by $1. 5^{2} = \frac{9}{4}$ .
03
Speed against velocity · Q7
The speed is $∣ v ∣$ , so a large negative velocity counts as a large speed.
04
Complete the square · Q8
The vertex is halfway between the roots.
05
Triangle in a semicircle · Q9
$A$ and $C$ lie on the sphere with $[A B]$ a diameter, so triangle $A B C$ has a right angle at $C$ .

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