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IB Maths AA HL · Paper 3

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1 TZ2TZ3

Session: May 2025
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

Q1 Q2

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Questions

Questions 1–2 · 55 marks

Question 1

27 marks

Self-composite linear functions: a pattern, an induction proof and limits

Calculus

(a)(i)

3 marks

Apply $f$ to $f (x) = m x + c$ .

f^{2} (x) = f (m x + c) = m (m x + c) + c = m^{2} x + m c + c = m^{2} x + c (1 + m)

(a)(ii)

2 marks

Apply $f$ to $f^{2} (x)$ .

f^{3} (x) = m (m^{2} x + c (1 + m)) + c = m^{3} x + c (m + m^{2}) + c = m^{3} x + c (1 + m + m^{2})

(b)(i)

1 mark

Apply $f$ once more.

f^{4} (x) = m^{4} x + c (1 + m + m^{2} + m^{3})

(b)(ii)

2 marks

The power of $m$ on $x$ is $n$ , and the bracket is the sum of the first $n$ powers of $m$ , starting from $m^{0}$ .

f^{n} (x) = m^{n} x + c (1 + m + m^{2} + \dots + m^{n - 1})

(b)(iii)

3 marks

With $m = 1$ every term in the bracket is 1, and there are $n$ of them.

f^{n} (x) = x + n c

(c)

8 marks

Claim: $f^{n} (x) = m^{n} x + c (\frac{m ^{n} - 1}{m - 1})$ for all $n \in Z^{+}$ , with $m \neq = 1$ .
Base case $n = 1$ : the right side is $m x + c \frac{m - 1}{m - 1} = m x + c = f (x)$ , which is true.
Assume it is true for $n = k$ .
Apply $f$ once more.
Put the terms over a common denominator.
This is the formula for $n = k + 1$ . It is true for $n = 1$ and, if true for $n = k$ , it is true for $n = k + 1$ . By induction it is true for all $n \in Z^{+}$ .

f^{k} (x) = m^{k} x + c (\frac{m ^{k} - 1}{m - 1}) f^{k + 1} (x) = m (m^{k} x + c \frac{m ^{k} - 1}{m - 1}) + c = m^{k + 1} x + c (\frac{m ^{k + 1} - m}{m - 1} + 1) \frac{m ^{k + 1} - m}{m - 1} + \frac{m - 1}{m - 1} = \frac{m ^{k + 1} - 1}{m - 1} f^{k + 1} (x) = m^{k + 1} x + c (\frac{m ^{k + 1} - 1}{m - 1})

(d)

4 marks

For $- 1 < m < 1$ , $m^{n} \to 0$ as $n \to \infty$ .

f^{n} (x) = m^{n} x + c (\frac{m ^{n} - 1}{m - 1}) \to 0 \cdot x + c (\frac{- 1}{m - 1}) = \frac{c}{1 - m} L : y = \frac{c}{1 - m}

(e)(i)

2 marks

With $m = - 1$ the formula is
When $n$ is odd, $(- 1)^{n} = - 1$ .

f^{n} (x) = (- 1)^{n} x + c (\frac{( - 1 ) ^{n} - 1}{- 1 - 1}) f^{n} (x) = - x + c (\frac{- 2}{- 2}) = - x + c

(e)(ii)

2 marks

When $n$ is even, $(- 1)^{n} = 1$ , so the bracket is $\frac{1 - 1}{- 2} = 0$ .

f^{n} (x) = x

Question 2

28 marks

Open-topped boxes from a sheet: the maximum volume and integer values

Calculus

(a)(i)

1 mark

At $x = 0$ no squares are cut and the height is 0.

V = 0

(a)(ii)

1 mark

At $x = \frac{a}{2}$ the width of the base is $a - 2 x = 0$ .

V = 0

(b)

2 marks

Expand the product.

V = x (a - 2 x) (b - 2 x) = x (ab - 2 a x - 2 b x + 4 x^{2}) = 4 x^{3} - 2 (a + b) x^{2} + ab x

(c)

6 marks

Differentiate and set the derivative to zero.
Use the quadratic formula.
Simplify the discriminant: $16 (a + b)^{2} - 48 ab = 16 (a^{2} + 2 ab + b^{2} - 3 ab) = 16 (a^{2} - ab + b^{2})$ .

\frac{d V}{d x} = 12 x^{2} - 4 (a + b) x + ab = 0 x = \frac{4 ( a + b ) \pm 16 ( a + b ) ^{2} - 48 ab}{24} x = \frac{( a + b ) \pm a ^{2} - ab + b ^{2}}{6}

(d)

5 marks

Differentiate again.
At $x_{m} = \frac{( a + b ) - a ^{2} - ab + b ^{2}}{6}$ this is
Since $a^{2} - ab + b^{2} > 0$ , the second derivative is negative, so $V$ has a local maximum at $x = x_{m}$ .

\frac{d ^{2} V}{d x ^{2}} = 24 x - 4 (a + b) \frac{d ^{2} V}{d x ^{2}} = 4 [(a + b) - a^{2} - ab + b^{2}] - 4 (a + b) = - 4 a^{2} - ab + b^{2} < 0

(e)

1 mark

The volume is 0 at both end points, $x = 0$ and $x = \frac{a}{2}$ , by part (a). The local maximum at $x_{m}$ is inside the interval, where $V > 0$ . So it is higher than the values at the ends, and it is the maximum volume.

(f)

2 marks

Use $a < b$ with $a = 2 s t - s^{2}$ and $b = t^{2} - s^{2}$ .
Since $t > 0$ , divide by $t$ .

2 s t - s^{2} < t^{2} - s^{2} \Rightarrow 2 s t < t^{2} 2 s < t

(g)

3 marks

Substitute into $x_{m} = \frac{( a + b ) - a ^{2} - ab + b ^{2}}{6}$ , using $a + b = 2 s t - 2 s^{2} + t^{2}$ and $a^{2} - ab + b^{2} = s^{2} - s t + t^{2}$ .

x_{m} = \frac{2 s t - 2 s ^{2} + t ^{2} - ( s ^{2} - s t + t ^{2} )}{6} = \frac{3 s t - 3 s ^{2}}{6} x_{m} = \frac{s ( t - s )}{2}

(h)(i)

2 marks

When $s$ is even, $\frac{s}{2}$ is an integer, so $x_{m} = \frac{s}{2} (t - s)$ is an integer.
Also $a = 2 s t - s^{2}$ and $b = t^{2} - s^{2}$ are integers. So $V_{m} = x_{m} (a - 2 x_{m}) (b - 2 x_{m})$ is a product of integers, and an integer.

(h)(ii)

1 mark

If $s$ and $t$ are both odd, then $t - s$ is even, so $s (t - s)$ is even and $x_{m} = \frac{s ( t - s )}{2}$ is an integer, and so is $V_{m}$ .

s odd and t odd

(h)(iii)

4 marks

Choose $s = 1$ and $t = 3$ (both odd, with $t > 2 s$ ).
Another choice, $s = 2$ , $t = 5$ : $a = 16$ , $b = 21$ , $x_{m} = 3$ and $V_{m} = 3 (10) (15) = 450$ .

a = 2 (1) (3) - 1 = 5, b = 9 - 1 = 8, x_{m} = \frac{1 ( 3 - 1 )}{2} = 1 V_{m} = 1 (5 - 2) (8 - 2) = 18

Our analysis

Where the marks are

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

Calculus55 marks · 100%
Q1, Q2

Mentor insights

Traps to watch for

01
Composition · Q1
$f^{n + 1} (x) = f (f^{n} (x))$ , so apply $f (x) = m x + c$ to the previous result.
02
Geometric sum · Q1
$1 + m + m^{2} + \dots + m^{n - 1} = \frac{m ^{n} - 1}{m - 1}$ for $m \neq = 1$ .
03
Volume of a box · Q2
Cut squares of side $x$ from each corner of an $a$ by $b$ sheet, so the base is $(a - 2 x)$ by $(b - 2 x)$ and the height is $x$ .
04
Second derivative test · Q2
$V^{''} < 0$ at a stationary point means a local maximum.
05
Integer conditions · Q2
$x_{m} = \frac{s ( t - s )}{2}$ is an integer when $s$ is even, or when $s$ and $t$ are both odd.

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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Worked solutions

IB Maths AA HL · Paper 3

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1 TZ2TZ3

Session: May 2025
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

Q1 Q2

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.

Questions

Questions 1–2 · 55 marks

Question 1

27 marks

Self-composite linear functions: a pattern, an induction proof and limits

Calculus

(a)(i)

3 marks

Apply $f$ to $f (x) = m x + c$ .

f^{2} (x) = f (m x + c) = m (m x + c) + c = m^{2} x + m c + c = m^{2} x + c (1 + m)

(a)(ii)

2 marks

Apply $f$ to $f^{2} (x)$ .

f^{3} (x) = m (m^{2} x + c (1 + m)) + c = m^{3} x + c (m + m^{2}) + c = m^{3} x + c (1 + m + m^{2})

(b)(i)

1 mark

Apply $f$ once more.

f^{4} (x) = m^{4} x + c (1 + m + m^{2} + m^{3})

(b)(ii)

2 marks

The power of $m$ on $x$ is $n$ , and the bracket is the sum of the first $n$ powers of $m$ , starting from $m^{0}$ .

f^{n} (x) = m^{n} x + c (1 + m + m^{2} + \dots + m^{n - 1})

(b)(iii)

3 marks

With $m = 1$ every term in the bracket is 1, and there are $n$ of them.

f^{n} (x) = x + n c

(c)

8 marks

Claim: $f^{n} (x) = m^{n} x + c (\frac{m ^{n} - 1}{m - 1})$ for all $n \in Z^{+}$ , with $m \neq = 1$ .
Base case $n = 1$ : the right side is $m x + c \frac{m - 1}{m - 1} = m x + c = f (x)$ , which is true.
Assume it is true for $n = k$ .
Apply $f$ once more.
Put the terms over a common denominator.
This is the formula for $n = k + 1$ . It is true for $n = 1$ and, if true for $n = k$ , it is true for $n = k + 1$ . By induction it is true for all $n \in Z^{+}$ .

f^{k} (x) = m^{k} x + c (\frac{m ^{k} - 1}{m - 1}) f^{k + 1} (x) = m (m^{k} x + c \frac{m ^{k} - 1}{m - 1}) + c = m^{k + 1} x + c (\frac{m ^{k + 1} - m}{m - 1} + 1) \frac{m ^{k + 1} - m}{m - 1} + \frac{m - 1}{m - 1} = \frac{m ^{k + 1} - 1}{m - 1} f^{k + 1} (x) = m^{k + 1} x + c (\frac{m ^{k + 1} - 1}{m - 1})

(d)

4 marks

For $- 1 < m < 1$ , $m^{n} \to 0$ as $n \to \infty$ .

f^{n} (x) = m^{n} x + c (\frac{m ^{n} - 1}{m - 1}) \to 0 \cdot x + c (\frac{- 1}{m - 1}) = \frac{c}{1 - m} L : y = \frac{c}{1 - m}

(e)(i)

2 marks

With $m = - 1$ the formula is
When $n$ is odd, $(- 1)^{n} = - 1$ .

f^{n} (x) = (- 1)^{n} x + c (\frac{( - 1 ) ^{n} - 1}{- 1 - 1}) f^{n} (x) = - x + c (\frac{- 2}{- 2}) = - x + c

(e)(ii)

2 marks

When $n$ is even, $(- 1)^{n} = 1$ , so the bracket is $\frac{1 - 1}{- 2} = 0$ .

f^{n} (x) = x

Question 2

28 marks

Open-topped boxes from a sheet: the maximum volume and integer values

Calculus

(a)(i)

1 mark

At $x = 0$ no squares are cut and the height is 0.

V = 0

(a)(ii)

1 mark

At $x = \frac{a}{2}$ the width of the base is $a - 2 x = 0$ .

V = 0

(b)

2 marks

Expand the product.

V = x (a - 2 x) (b - 2 x) = x (ab - 2 a x - 2 b x + 4 x^{2}) = 4 x^{3} - 2 (a + b) x^{2} + ab x

(c)

6 marks

Differentiate and set the derivative to zero.
Use the quadratic formula.
Simplify the discriminant: $16 (a + b)^{2} - 48 ab = 16 (a^{2} + 2 ab + b^{2} - 3 ab) = 16 (a^{2} - ab + b^{2})$ .

\frac{d V}{d x} = 12 x^{2} - 4 (a + b) x + ab = 0 x = \frac{4 ( a + b ) \pm 16 ( a + b ) ^{2} - 48 ab}{24} x = \frac{( a + b ) \pm a ^{2} - ab + b ^{2}}{6}

(d)

5 marks

Differentiate again.
At $x_{m} = \frac{( a + b ) - a ^{2} - ab + b ^{2}}{6}$ this is
Since $a^{2} - ab + b^{2} > 0$ , the second derivative is negative, so $V$ has a local maximum at $x = x_{m}$ .

\frac{d ^{2} V}{d x ^{2}} = 24 x - 4 (a + b) \frac{d ^{2} V}{d x ^{2}} = 4 [(a + b) - a^{2} - ab + b^{2}] - 4 (a + b) = - 4 a^{2} - ab + b^{2} < 0

(e)

1 mark

(f)

2 marks

Use $a < b$ with $a = 2 s t - s^{2}$ and $b = t^{2} - s^{2}$ .
Since $t > 0$ , divide by $t$ .

2 s t - s^{2} < t^{2} - s^{2} \Rightarrow 2 s t < t^{2} 2 s < t

(g)

3 marks

Substitute into $x_{m} = \frac{( a + b ) - a ^{2} - ab + b ^{2}}{6}$ , using $a + b = 2 s t - 2 s^{2} + t^{2}$ and $a^{2} - ab + b^{2} = s^{2} - s t + t^{2}$ .

x_{m} = \frac{2 s t - 2 s ^{2} + t ^{2} - ( s ^{2} - s t + t ^{2} )}{6} = \frac{3 s t - 3 s ^{2}}{6} x_{m} = \frac{s ( t - s )}{2}

(h)(i)

2 marks

When $s$ is even, $\frac{s}{2}$ is an integer, so $x_{m} = \frac{s}{2} (t - s)$ is an integer.
Also $a = 2 s t - s^{2}$ and $b = t^{2} - s^{2}$ are integers. So $V_{m} = x_{m} (a - 2 x_{m}) (b - 2 x_{m})$ is a product of integers, and an integer.

(h)(ii)

1 mark

If $s$ and $t$ are both odd, then $t - s$ is even, so $s (t - s)$ is even and $x_{m} = \frac{s ( t - s )}{2}$ is an integer, and so is $V_{m}$ .

s odd and t odd

(h)(iii)

4 marks

Choose $s = 1$ and $t = 3$ (both odd, with $t > 2 s$ ).
Another choice, $s = 2$ , $t = 5$ : $a = 16$ , $b = 21$ , $x_{m} = 3$ and $V_{m} = 3 (10) (15) = 450$ .

a = 2 (1) (3) - 1 = 5, b = 9 - 1 = 8, x_{m} = \frac{1 ( 3 - 1 )}{2} = 1 V_{m} = 1 (5 - 2) (8 - 2) = 18

Our analysis

Where the marks are

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

Calculus55 marks · 100%
Q1, Q2

Mentor insights

Traps to watch for

01
Composition · Q1
$f^{n + 1} (x) = f (f^{n} (x))$ , so apply $f (x) = m x + c$ to the previous result.
02
Geometric sum · Q1
$1 + m + m^{2} + \dots + m^{n - 1} = \frac{m ^{n} - 1}{m - 1}$ for $m \neq = 1$ .
03
Volume of a box · Q2
Cut squares of side $x$ from each corner of an $a$ by $b$ sheet, so the base is $(a - 2 x)$ by $(b - 2 x)$ and the height is $x$ .
04
Second derivative test · Q2
$V^{''} < 0$ at a stationary point means a local maximum.
05
Integer conditions · Q2
$x_{m} = \frac{s ( t - s )}{2}$ is an integer when $s$ is even, or when $s$ and $t$ are both odd.

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