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IB Maths AA SL · Paper 1

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ3

Session: November 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 November 2025 TZ3 paper, which your school or the IB can provide.

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Section A · Short response

Questions 1–6 · 35 marks

Question 1

6 marks

A probability distribution with an unknown probability

Statistics and probability

(a)

2 marks

Add the probabilities.

0.2 + 0.3 + k + 0.1 = 1 \Rightarrow k = 0.4

(b)

2 marks

The values larger than 1.5 are 2, 3 and 4.

P (X > 1.5) = 1 - P (X = 1) = 1 - 0.2 = 0.8

(c)

2 marks

Use $E (X) = \sum x P (X = x)$ .

E (X) = 1 (0.2) + 2 (0.3) + 3 (0.4) + 4 (0.1) = 0.2 + 0.6 + 1.2 + 0.4 = 2.4

Question 2

5 marks

Finding a curve from its gradient

Calculus

5 marks

Integrate $\frac{d y}{d x} = 4 cos x - sin x$ .
The curve passes through $(π, 3)$ .

y = 4 sin x + cos x + c 3 = 4 sin π + cos π + c = 0 - 1 + c \Rightarrow c = 4 y = 4 sin x + cos x + 4

Question 3

5 marks

The first term and common difference from two conditions

Number and algebra

5 marks

The 7th term is 6, and the 6th and 12th terms add to 24.
From the second equation $u_{1} + 8 d = 12$ . Subtract the first equation.

u_{1} + 6 d = 6, (u_{1} + 5 d) + (u_{1} + 11 d) = 24 \Rightarrow 2 u_{1} + 16 d = 24 2 d = 6 \Rightarrow d = 3 u_{1} = 6 - 18 = - 12

Question 4

6 marks

Two events with a given conditional probability

Statistics and probability

(a)

2 marks

Use the multiplication rule with $P (B) = \frac{1}{3}$ and $P (A ∣ B) = \frac{1}{4}$ .

P (A \cap B) = \frac{1}{4} \times \frac{1}{3} = \frac{1}{12}

(b)

4 marks

The complement of $B$ has probability $P (B^{'}) = 1 - \frac{1}{3} = \frac{2}{3}$ .
The part of $A$ outside $B$ .
Divide.

P (A \cap B^{'}) = \frac{1}{2} - \frac{1}{12} = \frac{5}{12} P (A ∣ B^{'}) = \frac{5/12}{2/3} = \frac{5}{12} \times \frac{3}{2} = \frac{5}{8}

Question 5

8 marks

A rational graph through two points and the transformations from 1/x

Functions

(a)

3 marks

Substitute $P (- 10, 1)$ .
Substitute $Q (3, - 12)$ .
Subtract the first equation from the second.

1 = \frac{- 10 A + B}{- 14} \Rightarrow - 10 A + B = - 14 - 12 = \frac{3 A + B}{- 1} \Rightarrow 3 A + B = 12 13 A = 26 \Rightarrow A = 2, B = 12 - 6 = 6

(b)

5 marks

Divide $2 x + 6$ by $x - 4$ .
Start from $y = \frac{1}{x}$ . Apply a vertical stretch with scale factor 14, to get $y = \frac{14}{x}$ .
Translate 4 units to the right, to get $y = \frac{14}{x - 4}$ .
Translate 2 units up. The vertical asymptote is then $x = 4$ and the horizontal asymptote is $y = 2$ .

y = \frac{2 x + 6}{x - 4} = 2 + \frac{14}{x - 4} y = \frac{14}{x - 4} + 2

Question 6

5 marks

A logarithmic equation with bases 8 and 4

Functions

5 marks

Change both logarithms to base 2.
Combine, using $\frac{1}{2} lo g_{2} x = lo g_{2} x$ .
Solve for $x$ .

3 lo g_{8} (10 x) - lo g_{4} x = lo g_{2} (10 x) - \frac{1}{2} lo g_{2} x = 1 lo g_{2} \frac{10 x}{x} = 1 \Rightarrow 10 x = 2 x = \frac{1}{5} \Rightarrow x = \frac{1}{25}

Section B · Extended response

Questions 7–9 · 45 marks

Question 7

14 marks

The cosine rule in a triangle with an unknown side length

Geometry and trigonometry

(a)

3 marks

Use the cosine rule in triangle $P QR$ .

P Q^{2} = 3^{2} + 2^{2} - 2 (3) (2) cos 6 0^{\circ} = 13 - 6 = 7 P Q = 7

(b)

5 marks

Use the cosine rule in triangle $A B C$ .
Expand each part, using $cos 6 0^{\circ} = \frac{1}{2}$ .
Combine.

A B^{2} = (x + 2)^{2} + (3 x - 1)^{2} - 2 (x + 2) (3 x - 1) cos 6 0^{\circ} (x + 2)^{2} = x^{2} + 4 x + 4, (3 x - 1)^{2} = 9 x^{2} - 6 x + 1 (x + 2) (3 x - 1) = 3 x^{2} + 5 x - 2 A B^{2} = x^{2} + 4 x + 4 + 9 x^{2} - 6 x + 1 - (3 x^{2} + 5 x - 2) = 7 x^{2} - 7 x + 7 A B = 7 x^{2} - 7 x + 7

(c)

6 marks

(i) Put $A B = 7$ .

A length must be positive, and $3 x - 1 > 0$ needs $x > \frac{1}{3}$ , so $x = 3$ .

(ii) The sides are $A C = 5$ and $B C = 8$ .

7 x^{2} - 7 x + 7 = 49 \Rightarrow x^{2} - x - 6 = 0 \Rightarrow (x - 3) (x + 2) = 0 x = 3 Area = \frac{1}{2} (5) (8) sin 6 0^{\circ} = 20 \times \frac{3}{2} = 103

Question 8

15 marks

A quadratic with no real roots, its minimum, a normal line and an area

Functions
Calculus

(a)

2 marks

The discriminant of $\frac{1}{2} x^{2} + k x + 13$ is $k^{2} - 4 \cdot \frac{1}{2} \cdot 13 = k^{2} - 26$ . It must be negative.
The greatest positive integer is $k = 5$ .

k^{2} < 26 \Rightarrow k < 26 \approx 5.10

(b)

3 marks

(i) With $k = 5$ , $f (x) = \frac{1}{2} x^{2} + 5 x + 13$ . The axis is at $x = - \frac{5}{2 \times \frac{1}{2}}$ .

(ii) Substitute $x = - 5$ .

x = - 5 f (- 5) = \frac{25}{2} - 25 + 13 = \frac{1}{2} \Rightarrow (- 5, \frac{1}{2})

(c)

5 marks

Evaluate at $x = - 3$ .
The gradient of the normal is $- \frac{1}{2}$ .

f (- 3) = \frac{9}{2} - 15 + 13 = \frac{5}{2}, f^{'} (x) = x + 5 \Rightarrow f^{'} (- 3) = 2 y - \frac{5}{2} = - \frac{1}{2} (x + 3) \Rightarrow y = - \frac{1}{2} x + 1

(d)

5 marks

Between $x = - 3$ and $x = 0$ the curve is above the line $L$ .
Integrate.

Area = \int_{- 3}^{0} [(\frac{1}{2} x^{2} + 5 x + 13) - (- \frac{1}{2} x + 1)] d x = \int_{- 3}^{0} (\frac{1}{2} x^{2} + \frac{11}{2} x + 12) d x [\frac{x ^{3}}{6} + \frac{11 x ^{2}}{4} + 12 x]_{- 3}^{0} = 0 - (- \frac{9}{2} + \frac{99}{4} - 36) = \frac{63}{4}

Question 9

16 marks

A rational exponential function: intercepts, range, derivative and maximum

Calculus

(a)

6 marks

(i) At $x = 0$ : $f (0) = \frac{5 - 1}{1 + 1} = 2$ .

(ii) Set $f (x) = 0$ , which needs the numerator to be zero (the denominator is always positive).

P (0, 2) 5 e^{x} - e^{2 x} = 0 \Rightarrow e^{x} (5 - e^{x}) = 0 \Rightarrow e^{x} = 5 x = ln 5, R (ln 5, 0)

(b)

1 mark

The maximum value is at $Q$ , with $y = 7 - 26 \approx 2.10$ , and the function falls towards $- \infty$ as $x$ grows.

f (x) \leq 7 - 26

(c)

3 marks

Let $u = 5 e^{x} - e^{2 x}$ and $v = e^{x} + 1$ , so $u^{'} = 5 e^{x} - 2 e^{2 x}$ and $v^{'} = e^{x}$ .
Expand the numerator.

f^{'} (x) = \frac{( 5 e ^{x} - 2 e ^{2 x} ) ( e ^{x} + 1 ) - ( 5 e ^{x} - e ^{2 x} ) e ^{x}}{( e ^{x} + 1 ) ^{2}} 5 e^{2 x} + 5 e^{x} - 2 e^{3 x} - 2 e^{2 x} - 5 e^{2 x} + e^{3 x} = - e^{3 x} - 2 e^{2 x} + 5 e^{x} f^{'} (x) = \frac{- e ^{3 x} - 2 e ^{2 x} + 5 e ^{x}}{( e ^{x} + 1 ) ^{2}}

(d)

6 marks

At the maximum point $Q$ , $f^{'} (x) = 0$ .
Divide by $- e^{x}$ (which is not zero).
Solve this quadratic in $e^{x}$ .
Since $e^{x} > 0$ , reject $- 1 - 6$ .

- e^{3 x} - 2 e^{2 x} + 5 e^{x} = 0 e^{2 x} + 2 e^{x} - 5 = 0 e^{x} = \frac{- 2 \pm 4 + 20}{2} = - 1 \pm 6 e^{x} = 6 - 1 \Rightarrow q = ln (6 - 1)

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Calculus carries the most marks on this paper.

Calculus31 marks · 39%
Q2, Q8, Q9
Functions18 marks · 23%
Q5, Q6, Q8
Geometry and trigonometry14 marks · 18%
Q7
Statistics and probability12 marks · 15%
Q1, Q4
Number and algebra5 marks · 6%
Q3

Mentor insights

Traps to watch for

01
Total · Q1
The probabilities add to 1.
02
Two equations · Q5
Substitute each point into $y = \frac{A x + B}{x - 4}$ to get two linear equations.
03
Cosine rule · Q7
$c^{2} = a^{2} + b^{2} - 2 ab cos C$ .
04
Discriminant · Q8
No real roots means $b^{2} - 4 a c < 0$ .
05
Quotient rule · Q9
$(\frac{u}{v})^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}}$ .

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 1

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ3

Session: November 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 November 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 · 35 marks

Question 1

6 marks

A probability distribution with an unknown probability

Statistics and probability

(a)

2 marks

Add the probabilities.

0.2 + 0.3 + k + 0.1 = 1 \Rightarrow k = 0.4

(b)

2 marks

The values larger than 1.5 are 2, 3 and 4.

P (X > 1.5) = 1 - P (X = 1) = 1 - 0.2 = 0.8

(c)

2 marks

Use $E (X) = \sum x P (X = x)$ .

E (X) = 1 (0.2) + 2 (0.3) + 3 (0.4) + 4 (0.1) = 0.2 + 0.6 + 1.2 + 0.4 = 2.4

Question 2

5 marks

Finding a curve from its gradient

Calculus

5 marks

Integrate $\frac{d y}{d x} = 4 cos x - sin x$ .
The curve passes through $(π, 3)$ .

y = 4 sin x + cos x + c 3 = 4 sin π + cos π + c = 0 - 1 + c \Rightarrow c = 4 y = 4 sin x + cos x + 4

Question 3

5 marks

The first term and common difference from two conditions

Number and algebra

5 marks

The 7th term is 6, and the 6th and 12th terms add to 24.
From the second equation $u_{1} + 8 d = 12$ . Subtract the first equation.

u_{1} + 6 d = 6, (u_{1} + 5 d) + (u_{1} + 11 d) = 24 \Rightarrow 2 u_{1} + 16 d = 24 2 d = 6 \Rightarrow d = 3 u_{1} = 6 - 18 = - 12

Question 4

6 marks

Two events with a given conditional probability

Statistics and probability

(a)

2 marks

Use the multiplication rule with $P (B) = \frac{1}{3}$ and $P (A ∣ B) = \frac{1}{4}$ .

P (A \cap B) = \frac{1}{4} \times \frac{1}{3} = \frac{1}{12}

(b)

4 marks

The complement of $B$ has probability $P (B^{'}) = 1 - \frac{1}{3} = \frac{2}{3}$ .
The part of $A$ outside $B$ .
Divide.

P (A \cap B^{'}) = \frac{1}{2} - \frac{1}{12} = \frac{5}{12} P (A ∣ B^{'}) = \frac{5/12}{2/3} = \frac{5}{12} \times \frac{3}{2} = \frac{5}{8}

Question 5

8 marks

A rational graph through two points and the transformations from 1/x

Functions

(a)

3 marks

Substitute $P (- 10, 1)$ .
Substitute $Q (3, - 12)$ .
Subtract the first equation from the second.

1 = \frac{- 10 A + B}{- 14} \Rightarrow - 10 A + B = - 14 - 12 = \frac{3 A + B}{- 1} \Rightarrow 3 A + B = 12 13 A = 26 \Rightarrow A = 2, B = 12 - 6 = 6

(b)

5 marks

Divide $2 x + 6$ by $x - 4$ .
Start from $y = \frac{1}{x}$ . Apply a vertical stretch with scale factor 14, to get $y = \frac{14}{x}$ .
Translate 4 units to the right, to get $y = \frac{14}{x - 4}$ .
Translate 2 units up. The vertical asymptote is then $x = 4$ and the horizontal asymptote is $y = 2$ .

y = \frac{2 x + 6}{x - 4} = 2 + \frac{14}{x - 4} y = \frac{14}{x - 4} + 2

Question 6

5 marks

A logarithmic equation with bases 8 and 4

Functions

5 marks

Change both logarithms to base 2.
Combine, using $\frac{1}{2} lo g_{2} x = lo g_{2} x$ .
Solve for $x$ .

3 lo g_{8} (10 x) - lo g_{4} x = lo g_{2} (10 x) - \frac{1}{2} lo g_{2} x = 1 lo g_{2} \frac{10 x}{x} = 1 \Rightarrow 10 x = 2 x = \frac{1}{5} \Rightarrow x = \frac{1}{25}

Section B · Extended response

Questions 7–9 · 45 marks

Question 7

14 marks

The cosine rule in a triangle with an unknown side length

Geometry and trigonometry

(a)

3 marks

Use the cosine rule in triangle $P QR$ .

P Q^{2} = 3^{2} + 2^{2} - 2 (3) (2) cos 6 0^{\circ} = 13 - 6 = 7 P Q = 7

(b)

5 marks

Use the cosine rule in triangle $A B C$ .
Expand each part, using $cos 6 0^{\circ} = \frac{1}{2}$ .
Combine.

A B^{2} = (x + 2)^{2} + (3 x - 1)^{2} - 2 (x + 2) (3 x - 1) cos 6 0^{\circ} (x + 2)^{2} = x^{2} + 4 x + 4, (3 x - 1)^{2} = 9 x^{2} - 6 x + 1 (x + 2) (3 x - 1) = 3 x^{2} + 5 x - 2 A B^{2} = x^{2} + 4 x + 4 + 9 x^{2} - 6 x + 1 - (3 x^{2} + 5 x - 2) = 7 x^{2} - 7 x + 7 A B = 7 x^{2} - 7 x + 7

(c)

6 marks

(i) Put $A B = 7$ .

A length must be positive, and $3 x - 1 > 0$ needs $x > \frac{1}{3}$ , so $x = 3$ .

(ii) The sides are $A C = 5$ and $B C = 8$ .

7 x^{2} - 7 x + 7 = 49 \Rightarrow x^{2} - x - 6 = 0 \Rightarrow (x - 3) (x + 2) = 0 x = 3 Area = \frac{1}{2} (5) (8) sin 6 0^{\circ} = 20 \times \frac{3}{2} = 103

Question 8

15 marks

A quadratic with no real roots, its minimum, a normal line and an area

Functions
Calculus

(a)

2 marks

The discriminant of $\frac{1}{2} x^{2} + k x + 13$ is $k^{2} - 4 \cdot \frac{1}{2} \cdot 13 = k^{2} - 26$ . It must be negative.
The greatest positive integer is $k = 5$ .

k^{2} < 26 \Rightarrow k < 26 \approx 5.10

(b)

3 marks

(i) With $k = 5$ , $f (x) = \frac{1}{2} x^{2} + 5 x + 13$ . The axis is at $x = - \frac{5}{2 \times \frac{1}{2}}$ .

(ii) Substitute $x = - 5$ .

x = - 5 f (- 5) = \frac{25}{2} - 25 + 13 = \frac{1}{2} \Rightarrow (- 5, \frac{1}{2})

(c)

5 marks

Evaluate at $x = - 3$ .
The gradient of the normal is $- \frac{1}{2}$ .

f (- 3) = \frac{9}{2} - 15 + 13 = \frac{5}{2}, f^{'} (x) = x + 5 \Rightarrow f^{'} (- 3) = 2 y - \frac{5}{2} = - \frac{1}{2} (x + 3) \Rightarrow y = - \frac{1}{2} x + 1

(d)

5 marks

Between $x = - 3$ and $x = 0$ the curve is above the line $L$ .
Integrate.

Area = \int_{- 3}^{0} [(\frac{1}{2} x^{2} + 5 x + 13) - (- \frac{1}{2} x + 1)] d x = \int_{- 3}^{0} (\frac{1}{2} x^{2} + \frac{11}{2} x + 12) d x [\frac{x ^{3}}{6} + \frac{11 x ^{2}}{4} + 12 x]_{- 3}^{0} = 0 - (- \frac{9}{2} + \frac{99}{4} - 36) = \frac{63}{4}

Question 9

16 marks

A rational exponential function: intercepts, range, derivative and maximum

Calculus

(a)

6 marks

(i) At $x = 0$ : $f (0) = \frac{5 - 1}{1 + 1} = 2$ .

(ii) Set $f (x) = 0$ , which needs the numerator to be zero (the denominator is always positive).

P (0, 2) 5 e^{x} - e^{2 x} = 0 \Rightarrow e^{x} (5 - e^{x}) = 0 \Rightarrow e^{x} = 5 x = ln 5, R (ln 5, 0)

(b)

1 mark

The maximum value is at $Q$ , with $y = 7 - 26 \approx 2.10$ , and the function falls towards $- \infty$ as $x$ grows.

f (x) \leq 7 - 26

(c)

3 marks

Let $u = 5 e^{x} - e^{2 x}$ and $v = e^{x} + 1$ , so $u^{'} = 5 e^{x} - 2 e^{2 x}$ and $v^{'} = e^{x}$ .
Expand the numerator.

f^{'} (x) = \frac{( 5 e ^{x} - 2 e ^{2 x} ) ( e ^{x} + 1 ) - ( 5 e ^{x} - e ^{2 x} ) e ^{x}}{( e ^{x} + 1 ) ^{2}} 5 e^{2 x} + 5 e^{x} - 2 e^{3 x} - 2 e^{2 x} - 5 e^{2 x} + e^{3 x} = - e^{3 x} - 2 e^{2 x} + 5 e^{x} f^{'} (x) = \frac{- e ^{3 x} - 2 e ^{2 x} + 5 e ^{x}}{( e ^{x} + 1 ) ^{2}}

(d)

6 marks

At the maximum point $Q$ , $f^{'} (x) = 0$ .
Divide by $- e^{x}$ (which is not zero).
Solve this quadratic in $e^{x}$ .
Since $e^{x} > 0$ , reject $- 1 - 6$ .

- e^{3 x} - 2 e^{2 x} + 5 e^{x} = 0 e^{2 x} + 2 e^{x} - 5 = 0 e^{x} = \frac{- 2 \pm 4 + 20}{2} = - 1 \pm 6 e^{x} = 6 - 1 \Rightarrow q = ln (6 - 1)

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Calculus carries the most marks on this paper.

Calculus31 marks · 39%
Q2, Q8, Q9
Functions18 marks · 23%
Q5, Q6, Q8
Geometry and trigonometry14 marks · 18%
Q7
Statistics and probability12 marks · 15%
Q1, Q4
Number and algebra5 marks · 6%
Q3

Mentor insights

Traps to watch for

01
Total · Q1
The probabilities add to 1.
02
Two equations · Q5
Substitute each point into $y = \frac{A x + B}{x - 4}$ to get two linear equations.
03
Cosine rule · Q7
$c^{2} = a^{2} + b^{2} - 2 ab cos C$ .
04
Discriminant · Q8
No real roots means $b^{2} - 4 a c < 0$ .
05
Quotient rule · Q9
$(\frac{u}{v})^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}}$ .

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