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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 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 · 37 marks

Question 1

4 marks

Matching transformations of a graph to their sketches

Number and algebra

4 marks

$y = f (x + 1)$ is a translation 1 unit to the left, so it is graph B.
$y = f (x) - 1$ is a translation 1 unit down, so it is graph A.
$y = f (- x)$ is a reflection in the $y$ -axis, so it is graph F.
$y = f (2 x)$ is a horizontal compression with scale factor $\frac{1}{2}$ , so it is graph D.

f (x + 1) : B, f (x) - 1 : A, f (- x) : F, f (2 x) : D

Question 2

6 marks

An arithmetic sequence from its first and fifth terms

Number and algebra

(a)

4 marks

Use $u_{5} = u_{1} + 4 d$ .
The 13th term.

12 = 36 + 4 d \Rightarrow d = - 6 u_{13} = 36 + 12 (- 6) = - 36

(b)

2 marks

Set $S_{n} = \frac{n}{2} [2 (36) + (n - 1) (- 6)] = 0$ and divide by $n$ (as $n \neq = 0$ ).

72 - 6 (n - 1) = 0 \Rightarrow n - 1 = 12 n = 13

Question 3

6 marks

Two events: union, independence and a conditional probability

Statistics and probability

(a)

2 marks

Use the union formula.

P (A \cup B) = 0.5 + 0.6 - 0.4 = 0.7

(b)

1 mark

Compare the product with the intersection.
So $A$ and $B$ are not independent.

P (A) P (B) = 0.5 \times 0.6 = 0.3 \neq = 0.4 = P (A \cap B)

(c)

3 marks

The part of $A$ outside $B$ is $P (A \cap B^{'}) = 0.5 - 0.4 = 0.1$ , and $P (B^{'}) = 0.4$ .

P (A ∣ B^{'}) = \frac{0.1}{0.4} = \frac{1}{4}

Question 4

7 marks

A line and a parabola through the same two points

Calculus

(a)

3 marks

(i) The parabola $y = a x^{2} + c$ passes through $P (0, 9)$ .

(ii) It also passes through $Q (3, 0)$ .

c = 9 0 = 9 a + 9 \Rightarrow a = - 1

(b)

4 marks

The parabola $y = 9 - x^{2}$ is above the line $y = 9 - 3 x$ between $x = 0$ and $x = 3$ .

Area = \int_{0}^{3} [(9 - x^{2}) - (9 - 3 x)] d x = \int_{0}^{3} (3 x - x^{2}) d x [\frac{3 x ^{2}}{2} - \frac{x ^{3}}{3}]_{0}^{3} = \frac{27}{2} - 9 = \frac{9}{2}

Question 5

6 marks

Simplifying a rational expression and solving a logarithmic equation

Number and algebra

(a)

2 marks

Factorise the quadratics, then cancel the common factors $(x - 1)$ and $(x + 1)$ , which are not zero since $x > 7$ .

\frac{x}{x ^{2} - 8 x + 7} \times \frac{x ^{2} - 1}{x + 1} = \frac{x}{( x - 1 ) ( x - 7 )} \times \frac{( x - 1 ) ( x + 1 )}{x + 1} = \frac{x}{x - 7}

(b)

4 marks

Use $lo g_{2} 2 = 1$ and combine the logarithms on the left.
The two arguments are equal.
Use part (a): the left side is $\frac{x}{x - 7}$ .

lo g_{2} [x (x^{2} - 1)] - lo g_{2} 2 = lo g_{2} [(x^{2} - 8 x + 7) (x + 1)] lo g_{2} \frac{x ( x ^{2} - 1 )}{2} = lo g_{2} [(x^{2} - 8 x + 7) (x + 1)] \frac{x ( x ^{2} - 1 )}{2} = (x^{2} - 8 x + 7) (x + 1) \Rightarrow \frac{x ( x ^{2} - 1 )}{( x ^{2} - 8 x + 7 ) ( x + 1 )} = 2 \frac{x}{x - 7} = 2 \Rightarrow x = 2 x - 14 \Rightarrow x = 14

Question 6

8 marks

Two parabolas whose vertices lie on a line

Calculus

(a)

3 marks

For $f (x) = - x^{2} + 4 x + p$ the vertex $A$ is at $x = 2$ . It lies on $y = 2 x - 1$ , so its height is $2 (2) - 1 = 3$ .

f (2) = - 4 + 8 + p = 3 \Rightarrow p = - 1

(b)

5 marks

For $g (x) = x^{2} + q x - 1$ the vertex $B$ is at $x = - \frac{q}{2}$ . On the line, its height is $y = 2 (- \frac{q}{2}) - 1 = - q - 1$ .
On the parabola, its height is
Equate the two heights.
Since $q \neq = 0$ , $q = 4$ .

g (- \frac{q}{2}) = \frac{q ^{2}}{4} - \frac{q ^{2}}{2} - 1 = - \frac{q ^{2}}{4} - 1 - \frac{q ^{2}}{4} - 1 = - q - 1 \Rightarrow q^{2} = 4 q \Rightarrow q (q - 4) = 0

Section B · Extended response

Questions 7–9 · 43 marks

Question 7

14 marks

Linear functions, inverses and compositions

Functions

(a)

3 marks

(i) Substitute.

(ii) Swap $x$ and $y$ in $y = 2 x + 3$ .

f (2) = 2 (2) + 3 = 7 x = 2 y + 3 \Rightarrow f^{- 1} (x) = \frac{x - 3}{2}

(b)

2 marks

The line passes through $(2, 0)$ and $(0, - 1)$ , so its gradient is $\frac{0 - ( - 1 )}{2 - 0} = \frac{1}{2}$ , and its $y$ -intercept is $- 1$ .

h (x) = \frac{1}{2} x - 1

(c)

2 marks

Use $h^{- 1} (x) = - 2$ , which means $h (- 2) = x$ .

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

(d)

4 marks

Write $h (x) = f^{- 1} (g (x))$ .
Solve for $g (x)$ .

\frac{g ( x ) - 3}{2} = \frac{1}{2} x - 1 g (x) - 3 = x - 2 \Rightarrow g (x) = x + 1 m = 1, c = 1

(e)

2 marks

$h (k (x)) = x$ means that $k$ is the inverse of $h$ . Swap $x$ and $y$ in $y = \frac{1}{2} x - 1$ .

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

(f)

1 mark

The graphs of $y = k (x) = h^{- 1} (x)$ and $y = h (x)$ are mirror images of each other, so the transformation is a reflection in the line $y = x$ .

Question 8

14 marks

A function from its derivative: turning points, inflexion and the function itself

Calculus

(a)

3 marks

Horizontal tangents mean $f^{'} (x) = 0$ .

3 x^{2} + 12 x - 15 = 3 (x + 5) (x - 1) = 0 a = - 5, b = 1

(b)

2 marks

At $x = a$ the sign diagram shows $f^{'}$ going from positive to negative. So there is a local maximum, since the gradient changes from positive to negative.

(c)

3 marks

Differentiate again.

f^{''} (x) = 6 x + 12 = 0 \Rightarrow c = - 2

(d)

2 marks

Yes, there is a point of inflexion at $x = c$ , because $f^{''} (x)$ changes sign there (from negative to positive, so the concavity changes).

(e)

4 marks

Integrate $f^{'} (x) = 3 x^{2} + 12 x - 15$ .
Use $f (- 2) = 36$ .

f (x) = x^{3} + 6 x^{2} - 15 x + C - 8 + 24 + 30 + C = 36 \Rightarrow C = - 10 f (x) = x^{3} + 6 x^{2} - 15 x - 10

Question 9

15 marks

A tangent identity, an equation and the exact value of tan 22.5 degrees

Functions

(a)

3 marks

Write $tan θ = \frac{sin θ}{cos θ}$ .
Use $cos^{2} θ + sin^{2} θ = 1$ and $2 sin θ cos θ = sin 2 θ$ .

\frac{2 tan θ}{1 + tan ^{2} θ} = \frac{2 sin θ / cos θ}{1 + sin ^{2} θ / cos ^{2} θ} = \frac{2 sin θ cos θ}{cos ^{2} θ + sin ^{2} θ} sin 2 θ

(b)

5 marks

By part (a) the equation is $sin 2 θ = \frac{1}{2}$ for $0^{\circ} \leq 2 θ \leq 36 0^{\circ}$ .
Halve the angles.

2 θ = 3 0^{\circ} or 15 0^{\circ} θ = 1 5^{\circ} or 7 5^{\circ}

(c)

4 marks

Put $θ = 22. 5^{\circ}$ in part (a).
Multiply out.
Divide by $2$ .

\frac{2 x}{1 + x ^{2}} = sin 4 5^{\circ} = \frac{2}{2} 4 x = 2 (1 + x^{2}) \Rightarrow 2 x^{2} - 4 x + 2 = 0 x^{2} - 22 x + 1 = 0

(d)

3 marks

Use the quadratic formula.
Since $tan 22. 5^{\circ} < 1$ , take the smaller root.

x = \frac{2 2 \pm 8 - 4}{2} = 2 \pm 1 tan 22. 5^{\circ} = 2 - 1

Our analysis

Where the marks are

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

Functions29 marks · 36%
Q7, Q9
Calculus29 marks · 36%
Q4, Q6, Q8
Number and algebra16 marks · 20%
Q1, Q2, Q5
Statistics and probability6 marks · 8%
Q3

Mentor insights

Traps to watch for

01
Area between a curve and a line · Q4
Integrate the upper graph minus the lower graph between the intersection points.
02
Vertex · Q6
The vertex of $a x^{2} + b x + c$ is at $x = - \frac{b}{2 a}$ .
03
Inverse · Q7
Swap $x$ and $y$ and solve.
04
First derivative sign · Q8
If $f^{'}$ changes from positive to negative, the point is a local maximum.
05
Identity · Q9
$\frac{2 tan θ}{1 + tan ^{2} θ} = sin 2 θ$ .

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 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 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 · 37 marks

Question 1

4 marks

Matching transformations of a graph to their sketches

Number and algebra

4 marks

$y = f (x + 1)$ is a translation 1 unit to the left, so it is graph B.
$y = f (x) - 1$ is a translation 1 unit down, so it is graph A.
$y = f (- x)$ is a reflection in the $y$ -axis, so it is graph F.
$y = f (2 x)$ is a horizontal compression with scale factor $\frac{1}{2}$ , so it is graph D.

f (x + 1) : B, f (x) - 1 : A, f (- x) : F, f (2 x) : D

Question 2

6 marks

An arithmetic sequence from its first and fifth terms

Number and algebra

(a)

4 marks

Use $u_{5} = u_{1} + 4 d$ .
The 13th term.

12 = 36 + 4 d \Rightarrow d = - 6 u_{13} = 36 + 12 (- 6) = - 36

(b)

2 marks

Set $S_{n} = \frac{n}{2} [2 (36) + (n - 1) (- 6)] = 0$ and divide by $n$ (as $n \neq = 0$ ).

72 - 6 (n - 1) = 0 \Rightarrow n - 1 = 12 n = 13

Question 3

6 marks

Two events: union, independence and a conditional probability

Statistics and probability

(a)

2 marks

Use the union formula.

P (A \cup B) = 0.5 + 0.6 - 0.4 = 0.7

(b)

1 mark

Compare the product with the intersection.
So $A$ and $B$ are not independent.

P (A) P (B) = 0.5 \times 0.6 = 0.3 \neq = 0.4 = P (A \cap B)

(c)

3 marks

The part of $A$ outside $B$ is $P (A \cap B^{'}) = 0.5 - 0.4 = 0.1$ , and $P (B^{'}) = 0.4$ .

P (A ∣ B^{'}) = \frac{0.1}{0.4} = \frac{1}{4}

Question 4

7 marks

A line and a parabola through the same two points

Calculus

(a)

3 marks

(i) The parabola $y = a x^{2} + c$ passes through $P (0, 9)$ .

(ii) It also passes through $Q (3, 0)$ .

c = 9 0 = 9 a + 9 \Rightarrow a = - 1

(b)

4 marks

The parabola $y = 9 - x^{2}$ is above the line $y = 9 - 3 x$ between $x = 0$ and $x = 3$ .

Area = \int_{0}^{3} [(9 - x^{2}) - (9 - 3 x)] d x = \int_{0}^{3} (3 x - x^{2}) d x [\frac{3 x ^{2}}{2} - \frac{x ^{3}}{3}]_{0}^{3} = \frac{27}{2} - 9 = \frac{9}{2}

Question 5

6 marks

Simplifying a rational expression and solving a logarithmic equation

Number and algebra

(a)

2 marks

Factorise the quadratics, then cancel the common factors $(x - 1)$ and $(x + 1)$ , which are not zero since $x > 7$ .

\frac{x}{x ^{2} - 8 x + 7} \times \frac{x ^{2} - 1}{x + 1} = \frac{x}{( x - 1 ) ( x - 7 )} \times \frac{( x - 1 ) ( x + 1 )}{x + 1} = \frac{x}{x - 7}

(b)

4 marks

Use $lo g_{2} 2 = 1$ and combine the logarithms on the left.
The two arguments are equal.
Use part (a): the left side is $\frac{x}{x - 7}$ .

lo g_{2} [x (x^{2} - 1)] - lo g_{2} 2 = lo g_{2} [(x^{2} - 8 x + 7) (x + 1)] lo g_{2} \frac{x ( x ^{2} - 1 )}{2} = lo g_{2} [(x^{2} - 8 x + 7) (x + 1)] \frac{x ( x ^{2} - 1 )}{2} = (x^{2} - 8 x + 7) (x + 1) \Rightarrow \frac{x ( x ^{2} - 1 )}{( x ^{2} - 8 x + 7 ) ( x + 1 )} = 2 \frac{x}{x - 7} = 2 \Rightarrow x = 2 x - 14 \Rightarrow x = 14

Question 6

8 marks

Two parabolas whose vertices lie on a line

Calculus

(a)

3 marks

For $f (x) = - x^{2} + 4 x + p$ the vertex $A$ is at $x = 2$ . It lies on $y = 2 x - 1$ , so its height is $2 (2) - 1 = 3$ .

f (2) = - 4 + 8 + p = 3 \Rightarrow p = - 1

(b)

5 marks

For $g (x) = x^{2} + q x - 1$ the vertex $B$ is at $x = - \frac{q}{2}$ . On the line, its height is $y = 2 (- \frac{q}{2}) - 1 = - q - 1$ .
On the parabola, its height is
Equate the two heights.
Since $q \neq = 0$ , $q = 4$ .

g (- \frac{q}{2}) = \frac{q ^{2}}{4} - \frac{q ^{2}}{2} - 1 = - \frac{q ^{2}}{4} - 1 - \frac{q ^{2}}{4} - 1 = - q - 1 \Rightarrow q^{2} = 4 q \Rightarrow q (q - 4) = 0

Section B · Extended response

Questions 7–9 · 43 marks

Question 7

14 marks

Linear functions, inverses and compositions

Functions

(a)

3 marks

(i) Substitute.

(ii) Swap $x$ and $y$ in $y = 2 x + 3$ .

f (2) = 2 (2) + 3 = 7 x = 2 y + 3 \Rightarrow f^{- 1} (x) = \frac{x - 3}{2}

(b)

2 marks

The line passes through $(2, 0)$ and $(0, - 1)$ , so its gradient is $\frac{0 - ( - 1 )}{2 - 0} = \frac{1}{2}$ , and its $y$ -intercept is $- 1$ .

h (x) = \frac{1}{2} x - 1

(c)

2 marks

Use $h^{- 1} (x) = - 2$ , which means $h (- 2) = x$ .

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

(d)

4 marks

Write $h (x) = f^{- 1} (g (x))$ .
Solve for $g (x)$ .

\frac{g ( x ) - 3}{2} = \frac{1}{2} x - 1 g (x) - 3 = x - 2 \Rightarrow g (x) = x + 1 m = 1, c = 1

(e)

2 marks

$h (k (x)) = x$ means that $k$ is the inverse of $h$ . Swap $x$ and $y$ in $y = \frac{1}{2} x - 1$ .

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

(f)

1 mark

The graphs of $y = k (x) = h^{- 1} (x)$ and $y = h (x)$ are mirror images of each other, so the transformation is a reflection in the line $y = x$ .

Question 8

14 marks

A function from its derivative: turning points, inflexion and the function itself

Calculus

(a)

3 marks

Horizontal tangents mean $f^{'} (x) = 0$ .

3 x^{2} + 12 x - 15 = 3 (x + 5) (x - 1) = 0 a = - 5, b = 1

(b)

2 marks

At $x = a$ the sign diagram shows $f^{'}$ going from positive to negative. So there is a local maximum, since the gradient changes from positive to negative.

(c)

3 marks

Differentiate again.

f^{''} (x) = 6 x + 12 = 0 \Rightarrow c = - 2

(d)

2 marks

Yes, there is a point of inflexion at $x = c$ , because $f^{''} (x)$ changes sign there (from negative to positive, so the concavity changes).

(e)

4 marks

Integrate $f^{'} (x) = 3 x^{2} + 12 x - 15$ .
Use $f (- 2) = 36$ .

f (x) = x^{3} + 6 x^{2} - 15 x + C - 8 + 24 + 30 + C = 36 \Rightarrow C = - 10 f (x) = x^{3} + 6 x^{2} - 15 x - 10

Question 9

15 marks

A tangent identity, an equation and the exact value of tan 22.5 degrees

Functions

(a)

3 marks

Write $tan θ = \frac{sin θ}{cos θ}$ .
Use $cos^{2} θ + sin^{2} θ = 1$ and $2 sin θ cos θ = sin 2 θ$ .

\frac{2 tan θ}{1 + tan ^{2} θ} = \frac{2 sin θ / cos θ}{1 + sin ^{2} θ / cos ^{2} θ} = \frac{2 sin θ cos θ}{cos ^{2} θ + sin ^{2} θ} sin 2 θ

(b)

5 marks

By part (a) the equation is $sin 2 θ = \frac{1}{2}$ for $0^{\circ} \leq 2 θ \leq 36 0^{\circ}$ .
Halve the angles.

2 θ = 3 0^{\circ} or 15 0^{\circ} θ = 1 5^{\circ} or 7 5^{\circ}

(c)

4 marks

Put $θ = 22. 5^{\circ}$ in part (a).
Multiply out.
Divide by $2$ .

\frac{2 x}{1 + x ^{2}} = sin 4 5^{\circ} = \frac{2}{2} 4 x = 2 (1 + x^{2}) \Rightarrow 2 x^{2} - 4 x + 2 = 0 x^{2} - 22 x + 1 = 0

(d)

3 marks

Use the quadratic formula.
Since $tan 22. 5^{\circ} < 1$ , take the smaller root.

x = \frac{2 2 \pm 8 - 4}{2} = 2 \pm 1 tan 22. 5^{\circ} = 2 - 1

Our analysis

Where the marks are

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

Functions29 marks · 36%
Q7, Q9
Calculus29 marks · 36%
Q4, Q6, Q8
Number and algebra16 marks · 20%
Q1, Q2, Q5
Statistics and probability6 marks · 8%
Q3

Mentor insights

Traps to watch for

01
Area between a curve and a line · Q4
Integrate the upper graph minus the lower graph between the intersection points.
02
Vertex · Q6
The vertex of $a x^{2} + b x + c$ is at $x = - \frac{b}{2 a}$ .
03
Inverse · Q7
Swap $x$ and $y$ and solve.
04
First derivative sign · Q8
If $f^{'}$ changes from positive to negative, the point is a local maximum.
05
Identity · Q9
$\frac{2 tan θ}{1 + tan ^{2} θ} = sin 2 θ$ .

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