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IB Maths AA SL · Paper 1

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2TZ3

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 TZ2 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

4 marks

The derivative and tangent of a cubic

Calculus

(a)

2 marks

Differentiate $f (x) = x^{3} + 5 x^{2} - 8$ .

f^{'} (x) = 3 x^{2} + 10 x \Rightarrow f^{'} (1) = 13

(b)

2 marks

The tangent passes through $(1, f (1)) = (1, - 2)$ with gradient 13.

y + 2 = 13 (x - 1) \Rightarrow y = 13 x - 15

Question 2

5 marks

Finding a function from its derivative

Number and algebra

5 marks

Integrate $g^{'} (x) = cos x + e^{2 x}$ .
Use $g (0) = 7$ .

g (x) = sin x + \frac{1}{2} e^{2 x} + C 0 + \frac{1}{2} + C = 7 \Rightarrow C = 6.5 g (x) = sin x + \frac{1}{2} e^{2 x} + 6.5

Question 3

5 marks

A rational function: value, asymptote and the range of its reflection

Functions

(a)

1 mark

Put $x = 0$ in $f (x) = \frac{3 x - 2}{2 x + 1}$ .

f (0) = - 2

(b)

1 mark

As $x \to \infty$ , $f (x) \to \frac{3}{2}$ .

y = \frac{3}{2}

(c)

3 marks

For $x \geq 0$ the function $f$ increases from $f (0) = - 2$ towards, but never reaching, the asymptote $\frac{3}{2}$ . So $- 2 \leq f (x) < \frac{3}{2}$ .
Negate to get $g = - f$ .

- \frac{3}{2} < g (x) \leq 2

Question 4

7 marks

A triangle with angles theta and 2 theta: the sine rule and an area

Geometry and trigonometry

(a)

3 marks

The side $B C = 62$ is opposite the angle $2 θ$ at $A$ , and $A B = 5$ is opposite the angle $θ$ at $C$ .
Use $sin 2 θ = 2 sin θ cos θ$ and divide by $sin θ$ (which is not zero).

\frac{sin 2 θ}{6 2} = \frac{sin θ}{5} \frac{2 cos θ}{6 2} = \frac{1}{5} \Rightarrow cos θ = \frac{6 2}{10} = \frac{3 2}{5}

(b)

2 marks

Use $sin^{2} θ + cos^{2} θ = 1$ with $θ$ acute.

sin^{2} θ = 1 - \frac{18}{25} = \frac{7}{25} \Rightarrow sin θ = \frac{7}{5}

(c)

2 marks

In triangle $B C D$ the angle at $C$ is $θ$ and $D$ is on $[A C]$ .

\frac{1}{2} (62) (D C) sin θ = 214 \Rightarrow \frac{3 14}{5} D C = 214 D C = \frac{10}{3}

Question 5

7 marks

A quadratic with two distinct roots of the same sign

Number and algebra

(a)

5 marks

Compute the discriminant of $x^{2} + k x + 15 - k$ .
Two distinct real roots need $Δ > 0$ .

Δ = k^{2} - 4 (15 - k) = k^{2} + 4 k - 60 = (k - 6) (k + 10) k < - 10 or k > 6

(b)

2 marks

The product of the roots is $15 - k$ , which must be positive.
Combine with part (a).

15 - k > 0 \Rightarrow k < 15 k < - 10 or 6 < k < 15

Question 6

7 marks

The point of a graph closest to the origin

Calculus

(a)

1 mark

The distance from the origin to the point $(x, f (x))$ is $x^{2} + f (x)^{2}$ .

l = x^{2} + (x^{2} ln x + 4 - x^{2}) = x^{2} ln x + 4

(b)

6 marks

The distance $l$ is smallest when $x^{2} ln x + 4$ is smallest. Differentiate with the product rule.
Set the derivative to zero (with $x > 0$ ).
The derivative is negative for $x < e^{- 1/2}$ and positive after it, so this is a minimum.

\frac{d}{d x} (x^{2} ln x + 4) = 2 x ln x + x^{2} \cdot \frac{1}{x} = 2 x ln x + x = x (2 ln x + 1) 2 ln x + 1 = 0 \Rightarrow ln x = - \frac{1}{2} x = e^{- 1/2}

Section B · Extended response

Questions 7–9 · 45 marks

Question 7

13 marks

A probability distribution with an unknown constant

Statistics and probability

(a)

5 marks

Add the probabilities.
A probability cannot be negative, so $k = \frac{1}{3}$ .

k + 3 k^{2} + 2 k^{2} + k^{2} = 1 \Rightarrow 6 k^{2} + k - 1 = 0 (3 k - 1) (2 k + 1) = 0 \Rightarrow k = \frac{1}{3} or k = - \frac{1}{2}

(b)

2 marks

$P (X < 3 a)$ is all values except $3 a$ .

P (X < 3 a) = 1 - k^{2} = 1 - \frac{1}{9} = \frac{8}{9}

(c)

3 marks

The outcomes with $X \geq a$ and $X < 3 a$ are $X = a$ and $X = 2 a$ .

P (X \geq a ∣ X < 3 a) = \frac{3 k ^{2} + 2 k ^{2}}{8/9} = \frac{5/9}{8/9} = \frac{5}{8}

(d)

3 marks

With $k = \frac{1}{3}$ the probabilities are $\frac{1}{3}, \frac{1}{3}, \frac{2}{9}, \frac{1}{9}$ .

E (X) = 0 (\frac{1}{3}) + a (\frac{1}{3}) + 2 a (\frac{2}{9}) + 3 a (\frac{1}{9}) = \frac{10 a}{9} = 20 a = 18

Question 8

15 marks

A Ferris wheel: a cosine model, time above a height and the fastest descent

Functions

(a)

4 marks

(i) The diameter is 72 m, so the amplitude is 36 and the seat starts at the bottom ( $h (0) = 8 = a + 44$ ).

(ii) One revolution takes 12 minutes.

a = - 36 \frac{2 π}{b} = 12 \Rightarrow b = \frac{π}{6}

(b)

2 marks

The graph of $h (t) = - 36 cos \frac{π t}{6} + 44$ starts at the end point $(0, 8)$ , rises to the maximum point $(6, 80)$ and falls back to the end point $(12, 8)$ , with a smooth wave shape.

(c)

5 marks

Solve $h (t) = 26$ .
The boundary values are $\frac{π t}{6} = \frac{π}{3}$ and $\frac{5 π}{3}$ .

- 36 cos \frac{π t}{6} + 44 \geq 26 \Rightarrow cos \frac{π t}{6} \leq \frac{1}{2} t = 2 and t = 10 T = 10 - 2 = 8 minutes

(d)

4 marks

(i) Differentiate with the chain rule.

(ii) The height decreases fastest when $h^{'} (t)$ is at its minimum value $- 6 π$ , that is when $sin \frac{π t}{6} = - 1$ .

h^{'} (t) = - 36 \cdot (- sin \frac{π t}{6}) \cdot \frac{π}{6} = 6 π sin \frac{π t}{6} \frac{π t}{6} = \frac{3 π}{2} \Rightarrow t = 9 minutes

Question 9

17 marks

Inverse functions, transformations, a composition and the area under a rational function

Functions

(a)

3 marks

Solve $f^{- 1} (8) = x$ , where $4^{x} = 8$ . Use the same base 2.

2^{2 x} = 2^{3} \Rightarrow 2 x = 3 \Rightarrow f^{- 1} (8) = \frac{3}{2}

(b)

4 marks

(i) Interchange $x$ and $y$ in $y = 1 + lo g_{2} x$ .

(ii) We want to go from $y = 2^{x - 1}$ to $y = 4^{x} = 2^{2 x}$ . First translate the graph 1 unit to the left, to get $y = 2^{x}$ . Then apply a horizontal stretch with scale factor $\frac{1}{2}$ (a horizontal compression), to get $y = 2^{2 x} = 4^{x}$ .

x = 1 + lo g_{2} y \Rightarrow y = 2^{x - 1} \Rightarrow g^{- 1} (x) = 2^{x - 1}

(c)

3 marks

Substitute $g (x)$ into $f$ .

(f \circ g) (x) = 4^{1 + l o g_{2} x} = 2^{2 + 2 l o g_{2} x} = 2^{2} \times 2^{l o g_{2} x^{2}} = 4 x^{2}

(d)

7 marks

(i) Put the left side over the common denominator $2 x + 1$ .

(ii) The region $R$ is under the graph of $h$ between $x = 1$ and $x = 3$ , so integrate using part (i).

2 x - 1 + \frac{1}{2 x + 1} = \frac{( 2 x - 1 ) ( 2 x + 1 ) + 1}{2 x + 1} = \frac{4 x ^{2} - 1 + 1}{2 x + 1} = \frac{4 x ^{2}}{2 x + 1} Area = \int_{1}^{3} (2 x - 1 + \frac{1}{2 x + 1}) d x = [x^{2} - x + \frac{1}{2} ln (2 x + 1)]_{1}^{3} (9 - 3 + \frac{1}{2} ln 7) - (1 - 1 + \frac{1}{2} ln 3) = 6 + \frac{1}{2} ln 7 - \frac{1}{2} ln 3 6 + \frac{1}{2} ln \frac{7}{3}

Our analysis

Where the marks are

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

Functions37 marks · 46%
Q3, Q8, Q9
Statistics and probability13 marks · 16%
Q7
Number and algebra12 marks · 15%
Q2, Q5
Calculus11 marks · 14%
Q1, Q6
Geometry and trigonometry7 marks · 9%
Q4

Mentor insights

Traps to watch for

01
Double angle · Q4
$sin 2 θ = 2 sin θ cos θ$ , then cancel $sin θ$ .
02
Discriminant · Q5
Two distinct real roots need $Δ > 0$ . Both roots have the same sign when the product of the roots is positive.
03
Probabilities add to 1 · Q7
This gives a quadratic in $k$ , and $k$ must be positive.
04
Cosine model · Q8
The seat starts at the lowest point, so the cosine has a negative coefficient.
05
Inverse of an exponential · Q9
$f^{- 1} (8) = lo g_{4} 8$ .

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

TZ1TZ2TZ3

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 TZ2 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

4 marks

The derivative and tangent of a cubic

Calculus

(a)

2 marks

Differentiate $f (x) = x^{3} + 5 x^{2} - 8$ .

f^{'} (x) = 3 x^{2} + 10 x \Rightarrow f^{'} (1) = 13

(b)

2 marks

The tangent passes through $(1, f (1)) = (1, - 2)$ with gradient 13.

y + 2 = 13 (x - 1) \Rightarrow y = 13 x - 15

Question 2

5 marks

Finding a function from its derivative

Number and algebra

5 marks

Integrate $g^{'} (x) = cos x + e^{2 x}$ .
Use $g (0) = 7$ .

g (x) = sin x + \frac{1}{2} e^{2 x} + C 0 + \frac{1}{2} + C = 7 \Rightarrow C = 6.5 g (x) = sin x + \frac{1}{2} e^{2 x} + 6.5

Question 3

5 marks

A rational function: value, asymptote and the range of its reflection

Functions

(a)

1 mark

Put $x = 0$ in $f (x) = \frac{3 x - 2}{2 x + 1}$ .

f (0) = - 2

(b)

1 mark

As $x \to \infty$ , $f (x) \to \frac{3}{2}$ .

y = \frac{3}{2}

(c)

3 marks

For $x \geq 0$ the function $f$ increases from $f (0) = - 2$ towards, but never reaching, the asymptote $\frac{3}{2}$ . So $- 2 \leq f (x) < \frac{3}{2}$ .
Negate to get $g = - f$ .

- \frac{3}{2} < g (x) \leq 2

Question 4

7 marks

A triangle with angles theta and 2 theta: the sine rule and an area

Geometry and trigonometry

(a)

3 marks

The side $B C = 62$ is opposite the angle $2 θ$ at $A$ , and $A B = 5$ is opposite the angle $θ$ at $C$ .
Use $sin 2 θ = 2 sin θ cos θ$ and divide by $sin θ$ (which is not zero).

\frac{sin 2 θ}{6 2} = \frac{sin θ}{5} \frac{2 cos θ}{6 2} = \frac{1}{5} \Rightarrow cos θ = \frac{6 2}{10} = \frac{3 2}{5}

(b)

2 marks

Use $sin^{2} θ + cos^{2} θ = 1$ with $θ$ acute.

sin^{2} θ = 1 - \frac{18}{25} = \frac{7}{25} \Rightarrow sin θ = \frac{7}{5}

(c)

2 marks

In triangle $B C D$ the angle at $C$ is $θ$ and $D$ is on $[A C]$ .

\frac{1}{2} (62) (D C) sin θ = 214 \Rightarrow \frac{3 14}{5} D C = 214 D C = \frac{10}{3}

Question 5

7 marks

A quadratic with two distinct roots of the same sign

Number and algebra

(a)

5 marks

Compute the discriminant of $x^{2} + k x + 15 - k$ .
Two distinct real roots need $Δ > 0$ .

Δ = k^{2} - 4 (15 - k) = k^{2} + 4 k - 60 = (k - 6) (k + 10) k < - 10 or k > 6

(b)

2 marks

The product of the roots is $15 - k$ , which must be positive.
Combine with part (a).

15 - k > 0 \Rightarrow k < 15 k < - 10 or 6 < k < 15

Question 6

7 marks

The point of a graph closest to the origin

Calculus

(a)

1 mark

The distance from the origin to the point $(x, f (x))$ is $x^{2} + f (x)^{2}$ .

l = x^{2} + (x^{2} ln x + 4 - x^{2}) = x^{2} ln x + 4

(b)

6 marks

The distance $l$ is smallest when $x^{2} ln x + 4$ is smallest. Differentiate with the product rule.
Set the derivative to zero (with $x > 0$ ).
The derivative is negative for $x < e^{- 1/2}$ and positive after it, so this is a minimum.

\frac{d}{d x} (x^{2} ln x + 4) = 2 x ln x + x^{2} \cdot \frac{1}{x} = 2 x ln x + x = x (2 ln x + 1) 2 ln x + 1 = 0 \Rightarrow ln x = - \frac{1}{2} x = e^{- 1/2}

Section B · Extended response

Questions 7–9 · 45 marks

Question 7

13 marks

A probability distribution with an unknown constant

Statistics and probability

(a)

5 marks

Add the probabilities.
A probability cannot be negative, so $k = \frac{1}{3}$ .

k + 3 k^{2} + 2 k^{2} + k^{2} = 1 \Rightarrow 6 k^{2} + k - 1 = 0 (3 k - 1) (2 k + 1) = 0 \Rightarrow k = \frac{1}{3} or k = - \frac{1}{2}

(b)

2 marks

$P (X < 3 a)$ is all values except $3 a$ .

P (X < 3 a) = 1 - k^{2} = 1 - \frac{1}{9} = \frac{8}{9}

(c)

3 marks

The outcomes with $X \geq a$ and $X < 3 a$ are $X = a$ and $X = 2 a$ .

P (X \geq a ∣ X < 3 a) = \frac{3 k ^{2} + 2 k ^{2}}{8/9} = \frac{5/9}{8/9} = \frac{5}{8}

(d)

3 marks

With $k = \frac{1}{3}$ the probabilities are $\frac{1}{3}, \frac{1}{3}, \frac{2}{9}, \frac{1}{9}$ .

E (X) = 0 (\frac{1}{3}) + a (\frac{1}{3}) + 2 a (\frac{2}{9}) + 3 a (\frac{1}{9}) = \frac{10 a}{9} = 20 a = 18

Question 8

15 marks

A Ferris wheel: a cosine model, time above a height and the fastest descent

Functions

(a)

4 marks

(i) The diameter is 72 m, so the amplitude is 36 and the seat starts at the bottom ( $h (0) = 8 = a + 44$ ).

(ii) One revolution takes 12 minutes.

a = - 36 \frac{2 π}{b} = 12 \Rightarrow b = \frac{π}{6}

(b)

2 marks

The graph of $h (t) = - 36 cos \frac{π t}{6} + 44$ starts at the end point $(0, 8)$ , rises to the maximum point $(6, 80)$ and falls back to the end point $(12, 8)$ , with a smooth wave shape.

(c)

5 marks

Solve $h (t) = 26$ .
The boundary values are $\frac{π t}{6} = \frac{π}{3}$ and $\frac{5 π}{3}$ .

- 36 cos \frac{π t}{6} + 44 \geq 26 \Rightarrow cos \frac{π t}{6} \leq \frac{1}{2} t = 2 and t = 10 T = 10 - 2 = 8 minutes

(d)

4 marks

(i) Differentiate with the chain rule.

(ii) The height decreases fastest when $h^{'} (t)$ is at its minimum value $- 6 π$ , that is when $sin \frac{π t}{6} = - 1$ .

h^{'} (t) = - 36 \cdot (- sin \frac{π t}{6}) \cdot \frac{π}{6} = 6 π sin \frac{π t}{6} \frac{π t}{6} = \frac{3 π}{2} \Rightarrow t = 9 minutes

Question 9

17 marks

Inverse functions, transformations, a composition and the area under a rational function

Functions

(a)

3 marks

Solve $f^{- 1} (8) = x$ , where $4^{x} = 8$ . Use the same base 2.

2^{2 x} = 2^{3} \Rightarrow 2 x = 3 \Rightarrow f^{- 1} (8) = \frac{3}{2}

(b)

4 marks

(i) Interchange $x$ and $y$ in $y = 1 + lo g_{2} x$ .

x = 1 + lo g_{2} y \Rightarrow y = 2^{x - 1} \Rightarrow g^{- 1} (x) = 2^{x - 1}

(c)

3 marks

Substitute $g (x)$ into $f$ .

(f \circ g) (x) = 4^{1 + l o g_{2} x} = 2^{2 + 2 l o g_{2} x} = 2^{2} \times 2^{l o g_{2} x^{2}} = 4 x^{2}

(d)

7 marks

(i) Put the left side over the common denominator $2 x + 1$ .

(ii) The region $R$ is under the graph of $h$ between $x = 1$ and $x = 3$ , so integrate using part (i).

2 x - 1 + \frac{1}{2 x + 1} = \frac{( 2 x - 1 ) ( 2 x + 1 ) + 1}{2 x + 1} = \frac{4 x ^{2} - 1 + 1}{2 x + 1} = \frac{4 x ^{2}}{2 x + 1} Area = \int_{1}^{3} (2 x - 1 + \frac{1}{2 x + 1}) d x = [x^{2} - x + \frac{1}{2} ln (2 x + 1)]_{1}^{3} (9 - 3 + \frac{1}{2} ln 7) - (1 - 1 + \frac{1}{2} ln 3) = 6 + \frac{1}{2} ln 7 - \frac{1}{2} ln 3 6 + \frac{1}{2} ln \frac{7}{3}

Our analysis

Where the marks are

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

Functions37 marks · 46%
Q3, Q8, Q9
Statistics and probability13 marks · 16%
Q7
Number and algebra12 marks · 15%
Q2, Q5
Calculus11 marks · 14%
Q1, Q6
Geometry and trigonometry7 marks · 9%
Q4

Mentor insights

Traps to watch for

01
Double angle · Q4
$sin 2 θ = 2 sin θ cos θ$ , then cancel $sin θ$ .
02
Discriminant · Q5
Two distinct real roots need $Δ > 0$ . Both roots have the same sign when the product of the roots is positive.
03
Probabilities add to 1 · Q7
This gives a quadratic in $k$ , and $k$ must be positive.
04
Cosine model · Q8
The seat starts at the lowest point, so the cosine has a negative coefficient.
05
Inverse of an exponential · Q9
$f^{- 1} (8) = lo g_{4} 8$ .

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