Loading…

Worked solutions

IB Maths AA SL · Paper 1

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1 TZ2TZ3

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 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 · 33 marks

Question 1

4 marks

Golf scores from a box plot

Statistics and probability

(a)

2 marks

The lower quartile is 68 and the upper quartile is 74.

IQR = 74 - 68 = 6

(b)

2 marks

Between 70 and 74 is the upper half of the box, which contains 25% of the golfers.

0.25 \times 80 = 20

Question 2

5 marks

Logarithms of 24 and base conversion

Functions

(a)

3 marks

Write $24 = 2^{3} \times 3$ .

lo g_{10} 24 = 3 lo g_{10} 2 + lo g_{10} 3 = 3 p + q

(b)

2 marks

Change the base to 10.

lo g_{3} 8 = \frac{lo g _{10} 8}{lo g _{10} 3} = \frac{3 lo g _{10} 2}{lo g _{10} 3} = \frac{3 p}{q}

Question 3

5 marks

Reading a function and its inverse from a graph

Functions

(a)

1 mark

Read the graph at $x = - 3$ .

f (- 3) = - 1

(b)

1 mark

The domain of $f^{- 1}$ is the range of $f$ , read from the graph of $f$ .

- 3 \leq x \leq 5

(c)

3 marks

$f^{- 1} (2 x - 7) = - 3$ means $f (- 3) = 2 x - 7$ .

- 1 = 2 x - 7 \Rightarrow x = 3

Question 4

6 marks

cos^4 x - sin^4 x and its integral

Number and algebra

(a)

3 marks

Factorise as a difference of squares.
Use $cos^{2} x + sin^{2} x = 1$ and $cos^{2} x - sin^{2} x = cos 2 x$ .

cos^{4} x - sin^{4} x = (cos^{2} x - sin^{2} x) (cos^{2} x + sin^{2} x) cos^{4} x - sin^{4} x = cos 2 x \times 1 = cos 2 x

(b)

3 marks

Use part (a).

\int (cos^{4} x - sin^{4} x) d x = \int cos 2 x d x = \frac{1}{2} sin 2 x + c

Question 5

7 marks

A line and a parabola: intersections and tangency

Calculus

(a)

2 marks

Equate the line and the curve.

x^{2} - x - 1 = m x - 3 \Rightarrow x^{2} - (m + 1) x + 2 = 0

(b)

5 marks

Tangency needs the discriminant to be zero.

Δ = (m + 1)^{2} - 4 (1) (2) = 0 \Rightarrow (m + 1)^{2} = 8 m + 1 = \pm 22 \Rightarrow m = - 1 + 22 or m = - 1 - 22

Question 6

6 marks

Two normal distributions with the same spread

Statistics and probability

(a)

2 marks

$Y$ has the same shape as $X$ with the mean 12 higher, so shift 22 down by 12.

b = 22 - (19 - 7) = 10

(b)

1 mark

About 68% of the values are within one standard deviation of the mean.

P (7 - a < X < 7 + a) \approx 0.68

(c)

3 marks

With $a = 3$ , the value 22 is $\frac{22 - 19}{3} = 1$ standard deviation above the mean of $Y$ . The area above is $\frac{1 - 0.68}{2} = 0.16$ .

P (Y < 22) = 1 - 0.16 = 0.84

Section B · Extended response

Questions 7–9 · 47 marks

Question 7

15 marks

A gardener's rope fence: expression, minimum length and second derivative

Calculus

(a)

2 marks

The eight rectangles have a total area of 60.

8 x y = 60 \Rightarrow y = \frac{15}{2 x}

(b)

2 marks

The rope is 12 lengths of $x$ and 10 lengths of $y$ in all.

T = 12 x + 10 y = 12 x + 10 \cdot \frac{15}{2 x} = 12 x + \frac{75}{x}

(c)

2 marks

Differentiate $T = 12 x + 75 x^{- 1}$ .

\frac{d T}{d x} = 12 - \frac{75}{x ^{2}}

(d)

7 marks

(i) Set the derivative to zero.

(ii) The length of rope.

(iii) Find $y$ from part (a).

12 - \frac{75}{k ^{2}} = 0 \Rightarrow k^{2} = \frac{25}{4} \Rightarrow k = \frac{5}{2} T = 12 \cdot \frac{5}{2} + \frac{75}{5/2} = 30 + 30 = 60 m y = \frac{15}{2 \cdot \frac{5}{2}} = 3 m

(e)

2 marks

(i) Differentiate again.

(ii) At $x = \frac{5}{2}$ the second derivative is $\frac{150}{( 5/2 ) ^{3}} = 9.6 > 0$ , so $T$ has a local minimum.

\frac{d ^{2} T}{d x ^{2}} = \frac{150}{x ^{3}}

Question 8

16 marks

A sequence that is arithmetic for one value of k and geometric for others

Number and algebra

(a)

5 marks

(i) Arithmetic: twice the middle term equals the sum of the other two.

(ii) Substitute $k = \frac{4}{5}$ into $u_{3}$ .

2 (3 - 2 k) = (k - 5) + (5 k + 3) \Rightarrow 6 - 4 k = 6 k - 2 \Rightarrow k = \frac{4}{5} u_{3} = 5 (\frac{4}{5}) + 3 = 7

(b)

4 marks

(i) With $k = 12$ the terms are $7$ , $- 21$ , $63$ . The ratios are equal.

So the terms form a geometric sequence with common ratio $- 3$ .

(ii) An infinite geometric series only has a sum when $∣ r ∣ < 1$ , but $∣ r ∣ = 3 \geq 1$ , so the series diverges and the sum does not exist.

\frac{- 21}{7} = - 3, \frac{63}{- 21} = - 3

(c)

7 marks

(i) For a geometric sequence, $u_{2}^{2} = u_{1} u_{3}$ .

(ii) Factorise.

The second value is $k = - 2$ , and the first three terms are

(iii) The terms alternate between $- 7$ and $7$ with ratio $- 1$ , so the sum of an even number of terms is zero.

(3 - 2 k)^{2} = (k - 5) (5 k + 3) \Rightarrow 9 - 12 k + 4 k^{2} = 5 k^{2} - 22 k - 15 k^{2} - 10 k - 24 = 0 (k - 12) (k + 2) = 0 u_{1} = - 7, u_{2} = 7, u_{3} = - 7 S_{2 m} = 0

Question 9

16 marks

A logarithm of a quadratic: domain, zeros and the base from a maximum

Number and algebra

(a)

4 marks

(i) Factorise.

(ii) The parabola is above the axis between its roots.

5 - 4 x - x^{2} = (5 + x) (1 - x) = 0 \Rightarrow x = - 5 or x = 1 - 5 < x < 1

(b)

2 marks

The vertical asymptotes are where the quadratic is zero.

a = - 5, b = 1

(c)

4 marks

$f (x) = 0$ means $5 - 4 x - x^{2} = 1$ .

x^{2} + 4 x - 4 = 0 \Rightarrow x = \frac{- 4 \pm 16 + 16}{2} = - 2 \pm 22

(d)

6 marks

The quadratic $5 - 4 x - x^{2} = 9 - (x + 2)^{2}$ is largest at $x = - 2$ , where its value is 9. So the maximum of $f$ is $lo g_{k} 9$ .
Since $k > 1$ , $k = 3$ .

lo g_{k} 9 = 2 \Rightarrow k^{2} = 9

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 algebra38 marks · 48%
Q4, Q8, Q9
Calculus22 marks · 28%
Q5, Q7
Functions10 marks · 13%
Q2, Q3
Statistics and probability10 marks · 13%
Q1, Q6

Mentor insights

Traps to watch for

01
Difference of squares · Q4
$cos^{4} x - sin^{4} x = (cos^{2} x - sin^{2} x) (cos^{2} x + sin^{2} x)$ .
02
Tangency · Q5
A line is tangent to a curve when the intersection equation has a double root, so the discriminant is zero.
03
Eight rectangles · Q7
Eight identical rectangles of area $x y$ fill the 60 m $^{2}$ .
04
Arithmetic · Q8
$u_{2} - u_{1} = u_{3} - u_{2}$ , or $2 u_{2} = u_{1} + u_{3}$ .
05
Domain · Q9
$lo g_{k}$ needs a positive argument, so $5 - 4 x - x^{2} > 0$ .

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

TZ1 TZ2TZ3

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 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 · 33 marks

Question 1

4 marks

Golf scores from a box plot

Statistics and probability

(a)

2 marks

The lower quartile is 68 and the upper quartile is 74.

IQR = 74 - 68 = 6

(b)

2 marks

Between 70 and 74 is the upper half of the box, which contains 25% of the golfers.

0.25 \times 80 = 20

Question 2

5 marks

Logarithms of 24 and base conversion

Functions

(a)

3 marks

Write $24 = 2^{3} \times 3$ .

lo g_{10} 24 = 3 lo g_{10} 2 + lo g_{10} 3 = 3 p + q

(b)

2 marks

Change the base to 10.

lo g_{3} 8 = \frac{lo g _{10} 8}{lo g _{10} 3} = \frac{3 lo g _{10} 2}{lo g _{10} 3} = \frac{3 p}{q}

Question 3

5 marks

Reading a function and its inverse from a graph

Functions

(a)

1 mark

Read the graph at $x = - 3$ .

f (- 3) = - 1

(b)

1 mark

The domain of $f^{- 1}$ is the range of $f$ , read from the graph of $f$ .

- 3 \leq x \leq 5

(c)

3 marks

$f^{- 1} (2 x - 7) = - 3$ means $f (- 3) = 2 x - 7$ .

- 1 = 2 x - 7 \Rightarrow x = 3

Question 4

6 marks

cos^4 x - sin^4 x and its integral

Number and algebra

(a)

3 marks

Factorise as a difference of squares.
Use $cos^{2} x + sin^{2} x = 1$ and $cos^{2} x - sin^{2} x = cos 2 x$ .

cos^{4} x - sin^{4} x = (cos^{2} x - sin^{2} x) (cos^{2} x + sin^{2} x) cos^{4} x - sin^{4} x = cos 2 x \times 1 = cos 2 x

(b)

3 marks

Use part (a).

\int (cos^{4} x - sin^{4} x) d x = \int cos 2 x d x = \frac{1}{2} sin 2 x + c

Question 5

7 marks

A line and a parabola: intersections and tangency

Calculus

(a)

2 marks

Equate the line and the curve.

x^{2} - x - 1 = m x - 3 \Rightarrow x^{2} - (m + 1) x + 2 = 0

(b)

5 marks

Tangency needs the discriminant to be zero.

Δ = (m + 1)^{2} - 4 (1) (2) = 0 \Rightarrow (m + 1)^{2} = 8 m + 1 = \pm 22 \Rightarrow m = - 1 + 22 or m = - 1 - 22

Question 6

6 marks

Two normal distributions with the same spread

Statistics and probability

(a)

2 marks

$Y$ has the same shape as $X$ with the mean 12 higher, so shift 22 down by 12.

b = 22 - (19 - 7) = 10

(b)

1 mark

About 68% of the values are within one standard deviation of the mean.

P (7 - a < X < 7 + a) \approx 0.68

(c)

3 marks

With $a = 3$ , the value 22 is $\frac{22 - 19}{3} = 1$ standard deviation above the mean of $Y$ . The area above is $\frac{1 - 0.68}{2} = 0.16$ .

P (Y < 22) = 1 - 0.16 = 0.84

Section B · Extended response

Questions 7–9 · 47 marks

Question 7

15 marks

A gardener's rope fence: expression, minimum length and second derivative

Calculus

(a)

2 marks

The eight rectangles have a total area of 60.

8 x y = 60 \Rightarrow y = \frac{15}{2 x}

(b)

2 marks

The rope is 12 lengths of $x$ and 10 lengths of $y$ in all.

T = 12 x + 10 y = 12 x + 10 \cdot \frac{15}{2 x} = 12 x + \frac{75}{x}

(c)

2 marks

Differentiate $T = 12 x + 75 x^{- 1}$ .

\frac{d T}{d x} = 12 - \frac{75}{x ^{2}}

(d)

7 marks

(i) Set the derivative to zero.

(ii) The length of rope.

(iii) Find $y$ from part (a).

12 - \frac{75}{k ^{2}} = 0 \Rightarrow k^{2} = \frac{25}{4} \Rightarrow k = \frac{5}{2} T = 12 \cdot \frac{5}{2} + \frac{75}{5/2} = 30 + 30 = 60 m y = \frac{15}{2 \cdot \frac{5}{2}} = 3 m

(e)

2 marks

(i) Differentiate again.

(ii) At $x = \frac{5}{2}$ the second derivative is $\frac{150}{( 5/2 ) ^{3}} = 9.6 > 0$ , so $T$ has a local minimum.

\frac{d ^{2} T}{d x ^{2}} = \frac{150}{x ^{3}}

Question 8

16 marks

A sequence that is arithmetic for one value of k and geometric for others

Number and algebra

(a)

5 marks

(i) Arithmetic: twice the middle term equals the sum of the other two.

(ii) Substitute $k = \frac{4}{5}$ into $u_{3}$ .

2 (3 - 2 k) = (k - 5) + (5 k + 3) \Rightarrow 6 - 4 k = 6 k - 2 \Rightarrow k = \frac{4}{5} u_{3} = 5 (\frac{4}{5}) + 3 = 7

(b)

4 marks

(i) With $k = 12$ the terms are $7$ , $- 21$ , $63$ . The ratios are equal.

So the terms form a geometric sequence with common ratio $- 3$ .

(ii) An infinite geometric series only has a sum when $∣ r ∣ < 1$ , but $∣ r ∣ = 3 \geq 1$ , so the series diverges and the sum does not exist.

\frac{- 21}{7} = - 3, \frac{63}{- 21} = - 3

(c)

7 marks

(i) For a geometric sequence, $u_{2}^{2} = u_{1} u_{3}$ .

(ii) Factorise.

The second value is $k = - 2$ , and the first three terms are

(iii) The terms alternate between $- 7$ and $7$ with ratio $- 1$ , so the sum of an even number of terms is zero.

(3 - 2 k)^{2} = (k - 5) (5 k + 3) \Rightarrow 9 - 12 k + 4 k^{2} = 5 k^{2} - 22 k - 15 k^{2} - 10 k - 24 = 0 (k - 12) (k + 2) = 0 u_{1} = - 7, u_{2} = 7, u_{3} = - 7 S_{2 m} = 0

Question 9

16 marks

A logarithm of a quadratic: domain, zeros and the base from a maximum

Number and algebra

(a)

4 marks

(i) Factorise.

(ii) The parabola is above the axis between its roots.

5 - 4 x - x^{2} = (5 + x) (1 - x) = 0 \Rightarrow x = - 5 or x = 1 - 5 < x < 1

(b)

2 marks

The vertical asymptotes are where the quadratic is zero.

a = - 5, b = 1

(c)

4 marks

$f (x) = 0$ means $5 - 4 x - x^{2} = 1$ .

x^{2} + 4 x - 4 = 0 \Rightarrow x = \frac{- 4 \pm 16 + 16}{2} = - 2 \pm 22

(d)

6 marks

The quadratic $5 - 4 x - x^{2} = 9 - (x + 2)^{2}$ is largest at $x = - 2$ , where its value is 9. So the maximum of $f$ is $lo g_{k} 9$ .
Since $k > 1$ , $k = 3$ .

lo g_{k} 9 = 2 \Rightarrow k^{2} = 9

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 algebra38 marks · 48%
Q4, Q8, Q9
Calculus22 marks · 28%
Q5, Q7
Functions10 marks · 13%
Q2, Q3
Statistics and probability10 marks · 13%
Q1, Q6

Mentor insights

Traps to watch for

01
Difference of squares · Q4
$cos^{4} x - sin^{4} x = (cos^{2} x - sin^{2} x) (cos^{2} x + sin^{2} x)$ .
02
Tangency · Q5
A line is tangent to a curve when the intersection equation has a double root, so the discriminant is zero.
03
Eight rectangles · Q7
Eight identical rectangles of area $x y$ fill the 60 m $^{2}$ .
04
Arithmetic · Q8
$u_{2} - u_{1} = u_{3} - u_{2}$ , or $2 u_{2} = u_{1} + u_{3}$ .
05
Domain · Q9
$lo g_{k}$ needs a positive argument, so $5 - 4 x - x^{2} > 0$ .

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