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IB Maths AA HL · Paper 3

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ3

Session: November 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 November 2025 TZ1 paper, which your school or the IB can provide.

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Questions

Questions 1–2 · 55 marks

Question 1

26 marks

Lines normal to the curve y = k squared over x

Calculus

(a)(i)

2 marks

For $H$ : $y = \frac{1}{x}$ , the gradient of the tangent is $\frac{d y}{d x} = - \frac{1}{x ^{2}}$ . At $x = t$ it is $- \frac{1}{t ^{2}}$ .
The normal has the negative reciprocal gradient.

m_{N} = - \frac{1}{- 1/ t ^{2}} = t^{2}

(a)(ii)

1 mark

The normal passes through $(t, \frac{1}{t})$ with gradient $t^{2}$ .

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

(b)(i)

4 marks

Compare with $y = m x + c$ : $m = t^{2}$ and $c = \frac{1}{t} - t^{3}$ . Since $m = t^{2} > 0$ , $t = m$ or $t = - m$ .
If $t = m$ :
If $t = - m$ :

c = \frac{1}{m} - m m = \frac{1}{m} (1 - m^{2}) c = - \frac{1}{m} + m m = \frac{1}{m} (m^{2} - 1)

(b)(ii)

1 mark

A normal exists for each $t \neq = 0$ , and its gradient $t^{2}$ can be any positive number.

m > 0

(c)(i)

1 mark

For $m > 0$ the two values of $c$ are $\pm \frac{m ^{2} - 1}{m}$ . They give the same line only when they are equal, which means $c = 0$ , so $m^{2} = 1$ . As $m > 0$ :

m = 1

(c)(ii)

2 marks

For all other positive gradients the two values of $c$ are different, so there are two different lines.

m > 0, m \neq = 1

(d)(i)

1 mark

The curve $H$ has two branches, one in the first quadrant and one in the third. It is symmetric about the origin and has the axes as asymptotes.

(d)(ii)

2 marks

Choose, for example, $m = 4$ . The two normals have $t = 2$ and $t = - 2$ .
On the sketch, draw two parallel lines with gradient 4, one with a negative $y$ -intercept crossing the first branch at $(2, \frac{1}{2})$ and one with a positive $y$ -intercept crossing the third-quadrant branch at $(- 2, - \frac{1}{2})$ , and mark those two points.

t = 2 : y = 4 x - 7.5, touching H normally at (2, \frac{1}{2}) t = - 2 : y = 4 x + 7.5, normal to H at (- 2, - \frac{1}{2})

(e)(i)

3 marks

The normal at $A$ meets $F$ : $y = \frac{k ^{2}}{x}$ where $\frac{k ^{2}}{x} = t^{2} x + \frac{k}{t} - k t^{3}$ .
Multiply by $x$ and rearrange.
Divide by $t^{2}$ .

t^{2} x^{2} + (\frac{k}{t} - k t^{3}) x - k^{2} = 0 x^{2} + k (\frac{1}{t ^{3}} - t) x - \frac{k ^{2}}{t ^{2}} = 0

(e)(ii)

3 marks

One root is the $x$ -coordinate of $A$ , which is $k t$ . The product of the roots is $- \frac{k ^{2}}{t ^{2}}$ .
Check with the sum of roots: $k t + (- \frac{k}{t ^{3}}) = - k (\frac{1}{t ^{3}} - t)$ , which matches.
The $y$ -coordinate comes from $y = \frac{k ^{2}}{x}$ .

k t \cdot x_{B} = - \frac{k ^{2}}{t ^{2}} \Rightarrow x_{B} = - \frac{k}{t ^{3}} y_{B} = \frac{k ^{2}}{- k / t ^{3}} = - k t^{3} B (- \frac{k}{t ^{3}}, - k t^{3})

(f)

6 marks

The line $O A$ has gradient $\frac{k / t}{k t} = \frac{1}{t ^{2}}$ and equation $y = \frac{x}{t ^{2}}$ . It meets $F$ where $\frac{x}{t ^{2}} = \frac{k ^{2}}{x}$ , so $x^{2} = k^{2} t^{2}$ and $x = \pm k t$ . The point $A$ has $x = k t$ , so $C$ has $x = - k t$ and
The gradient of $A C$ is the same as the gradient of $O A$ since $A$ , $O$ , $C$ are collinear.
The gradient of $B C$ .
The denominator is not zero because $t \neq = \pm 1$ . Now multiply the gradients.
The product is $- 1$ , so $B C$ is perpendicular to $C A$ , and $B C A$ is a right angle.

C (- k t, - \frac{k}{t}) m_{A C} = \frac{k / t + k / t}{k t + k t} = \frac{1}{t ^{2}} m_{B C} = \frac{- k t ^{3} + k / t}{- k / t ^{3} + k t} = \frac{k ( 1 - t ^{4} ) / t}{k ( t ^{4} - 1 ) / t ^{3}} = - t^{2} m_{A C} \times m_{B C} = \frac{1}{t ^{2}} \times (- t^{2}) = - 1

Question 2

29 marks

Modelling a wolf population: logistic, Gompertz and Euler models

Functions
Calculus

(a)(i)

1 mark

The assumption is that 200 is the largest population the area can support (the carrying capacity), because the historical stable size was 200.

(a)(ii)

2 marks

Put $t = 0$ and $x = 40$ .

40 = \frac{200}{1 + C} \Rightarrow 1 + C = 5 C = 4

(a)(iii)

2 marks

Put $t = 5$ and $x = 70$ .

70 = \frac{200}{1 + 4 e ^{- 5 k}} \Rightarrow 1 + 4 e^{- 5 k} = \frac{20}{7} \Rightarrow e^{- 5 k} = \frac{13}{28} k = \frac{1}{5} ln \frac{28}{13} \approx 0.153

(a)(iv)

2 marks

At $t = 10$ .

x = \frac{200}{1 + 4 ( \frac{13}{28} ) ^{2}} \approx 107.4 \approx 107 wolves

(b)(i)

1 mark

When $x = 200$ , $ln \frac{200}{200} = ln 1 = 0$ .

\frac{d x}{d t} = 0

(b)(ii)

1 mark

At 200 wolves the population is not changing, so 200 is the steady (equilibrium) population.

(b)(iii)

2 marks

Differentiate $f (x) = ln (ln 200 - ln x)$ with the chain rule.

f^{'} (x) = \frac{1}{ln 200 - ln x} \times (- \frac{1}{x}) = - \frac{1}{x ( ln 200 - ln x )} = - \frac{1}{x ln ( \frac{200}{x} )}

(b)(iv)

5 marks

Separate the variables in $\frac{d x}{d t} = a x ln \frac{200}{x}$ .
By part (iii) the left side is $- f (x) = - ln (ln 200 - ln x)$ .
Rearrange.

\int \frac{1}{x ln ( \frac{200}{x} )} d x = \int a d t - ln (ln 200 - ln x) = a t + c ln 200 - ln x = e^{- a t - c} = A e^{- a t}, A = e^{- c} > 0 ln x = ln 200 - A e^{- a t}

(b)(v)

2 marks

Use $x = 40$ when $t = 0$ .

ln 40 = ln 200 - A \Rightarrow A = ln 200 - ln 40 = ln 5

(b)(vi)

2 marks

Use $x = 70$ when $t = 5$ .

ln 70 = ln 200 - ln 5 \cdot e^{- 5 a} \Rightarrow e^{- 5 a} = \frac{ln ( 200/70 )}{ln 5} a = - \frac{1}{5} ln (\frac{ln ( 20/7 )}{ln 5}) \approx 0.0855

(b)(vii)

3 marks

At $t = 10$ .

ln x = ln 200 - ln 5 \cdot e^{- 10 (0.08545)} \Rightarrow x \approx 100.8 \approx 101 wolves

(c)

1 mark

After 10 years the real population is 85. The logistic model gives 107 and the Gompertz model gives 101, so both overestimate. The Gompertz prediction is closer.

(d)(i)

4 marks

Use $x_{n + 1} = x_{n} + 0.5 (0.0855 x_{n} ln \frac{200}{x _{n}} - 0.03 x_{n})$ with $x_{0} = 70$ at $t = 5$ . Ten steps are needed to reach $t = 10$ .
So the estimate is 89 wolves.

x_{1} \approx 72.09, x_{2} \approx 74.16, x_{3} \approx 76.19, x_{4} \approx 78.19, x_{5} \approx 80.16 x_{6} \approx 82.09, x_{7} \approx 83.98, x_{8} \approx 85.84, x_{9} \approx 87.65, x_{10} \approx 89.43

(d)(ii)

1 mark

The estimate of 89 is much closer to the measured 85 than either earlier model, so allowing for emigration makes the model more realistic.

Our analysis

Where the marks are

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

Calculus48 marks · 87%
Q1, Q2
Functions7 marks · 13%
Q2

Mentor insights

Traps to watch for

01
Normal · Q1
The gradient of a normal is the negative reciprocal of the gradient of the tangent.
02
Quadratic roots · Q1
For $a x^{2} + b x + c = 0$ the sum of the roots is $- \frac{b}{a}$ and the product is $\frac{c}{a}$ .
03
Logistic · Q2
$x = \frac{L}{1 + C e ^{- k t}}$ has a maximum at $L$ .
04
Separation · Q2
Separate $x$ and $t$ , integrate, and use the initial value to find the constant.
05
Euler · Q2
$x_{n + 1} = x_{n} + h \frac{d x}{d t}$ .

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

TZ1TZ3

Session: November 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 November 2025 TZ1 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

26 marks

Lines normal to the curve y = k squared over x

Calculus

(a)(i)

2 marks

For $H$ : $y = \frac{1}{x}$ , the gradient of the tangent is $\frac{d y}{d x} = - \frac{1}{x ^{2}}$ . At $x = t$ it is $- \frac{1}{t ^{2}}$ .
The normal has the negative reciprocal gradient.

m_{N} = - \frac{1}{- 1/ t ^{2}} = t^{2}

(a)(ii)

1 mark

The normal passes through $(t, \frac{1}{t})$ with gradient $t^{2}$ .

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

(b)(i)

4 marks

Compare with $y = m x + c$ : $m = t^{2}$ and $c = \frac{1}{t} - t^{3}$ . Since $m = t^{2} > 0$ , $t = m$ or $t = - m$ .
If $t = m$ :
If $t = - m$ :

c = \frac{1}{m} - m m = \frac{1}{m} (1 - m^{2}) c = - \frac{1}{m} + m m = \frac{1}{m} (m^{2} - 1)

(b)(ii)

1 mark

A normal exists for each $t \neq = 0$ , and its gradient $t^{2}$ can be any positive number.

m > 0

(c)(i)

1 mark

For $m > 0$ the two values of $c$ are $\pm \frac{m ^{2} - 1}{m}$ . They give the same line only when they are equal, which means $c = 0$ , so $m^{2} = 1$ . As $m > 0$ :

m = 1

(c)(ii)

2 marks

For all other positive gradients the two values of $c$ are different, so there are two different lines.

m > 0, m \neq = 1

(d)(i)

1 mark

The curve $H$ has two branches, one in the first quadrant and one in the third. It is symmetric about the origin and has the axes as asymptotes.

(d)(ii)

2 marks

Choose, for example, $m = 4$ . The two normals have $t = 2$ and $t = - 2$ .
On the sketch, draw two parallel lines with gradient 4, one with a negative $y$ -intercept crossing the first branch at $(2, \frac{1}{2})$ and one with a positive $y$ -intercept crossing the third-quadrant branch at $(- 2, - \frac{1}{2})$ , and mark those two points.

t = 2 : y = 4 x - 7.5, touching H normally at (2, \frac{1}{2}) t = - 2 : y = 4 x + 7.5, normal to H at (- 2, - \frac{1}{2})

(e)(i)

3 marks

The normal at $A$ meets $F$ : $y = \frac{k ^{2}}{x}$ where $\frac{k ^{2}}{x} = t^{2} x + \frac{k}{t} - k t^{3}$ .
Multiply by $x$ and rearrange.
Divide by $t^{2}$ .

t^{2} x^{2} + (\frac{k}{t} - k t^{3}) x - k^{2} = 0 x^{2} + k (\frac{1}{t ^{3}} - t) x - \frac{k ^{2}}{t ^{2}} = 0

(e)(ii)

3 marks

One root is the $x$ -coordinate of $A$ , which is $k t$ . The product of the roots is $- \frac{k ^{2}}{t ^{2}}$ .
Check with the sum of roots: $k t + (- \frac{k}{t ^{3}}) = - k (\frac{1}{t ^{3}} - t)$ , which matches.
The $y$ -coordinate comes from $y = \frac{k ^{2}}{x}$ .

k t \cdot x_{B} = - \frac{k ^{2}}{t ^{2}} \Rightarrow x_{B} = - \frac{k}{t ^{3}} y_{B} = \frac{k ^{2}}{- k / t ^{3}} = - k t^{3} B (- \frac{k}{t ^{3}}, - k t^{3})

(f)

6 marks

The line $O A$ has gradient $\frac{k / t}{k t} = \frac{1}{t ^{2}}$ and equation $y = \frac{x}{t ^{2}}$ . It meets $F$ where $\frac{x}{t ^{2}} = \frac{k ^{2}}{x}$ , so $x^{2} = k^{2} t^{2}$ and $x = \pm k t$ . The point $A$ has $x = k t$ , so $C$ has $x = - k t$ and
The gradient of $A C$ is the same as the gradient of $O A$ since $A$ , $O$ , $C$ are collinear.
The gradient of $B C$ .
The denominator is not zero because $t \neq = \pm 1$ . Now multiply the gradients.
The product is $- 1$ , so $B C$ is perpendicular to $C A$ , and $B C A$ is a right angle.

C (- k t, - \frac{k}{t}) m_{A C} = \frac{k / t + k / t}{k t + k t} = \frac{1}{t ^{2}} m_{B C} = \frac{- k t ^{3} + k / t}{- k / t ^{3} + k t} = \frac{k ( 1 - t ^{4} ) / t}{k ( t ^{4} - 1 ) / t ^{3}} = - t^{2} m_{A C} \times m_{B C} = \frac{1}{t ^{2}} \times (- t^{2}) = - 1

Question 2

29 marks

Modelling a wolf population: logistic, Gompertz and Euler models

Functions
Calculus

(a)(i)

1 mark

The assumption is that 200 is the largest population the area can support (the carrying capacity), because the historical stable size was 200.

(a)(ii)

2 marks

Put $t = 0$ and $x = 40$ .

40 = \frac{200}{1 + C} \Rightarrow 1 + C = 5 C = 4

(a)(iii)

2 marks

Put $t = 5$ and $x = 70$ .

70 = \frac{200}{1 + 4 e ^{- 5 k}} \Rightarrow 1 + 4 e^{- 5 k} = \frac{20}{7} \Rightarrow e^{- 5 k} = \frac{13}{28} k = \frac{1}{5} ln \frac{28}{13} \approx 0.153

(a)(iv)

2 marks

At $t = 10$ .

x = \frac{200}{1 + 4 ( \frac{13}{28} ) ^{2}} \approx 107.4 \approx 107 wolves

(b)(i)

1 mark

When $x = 200$ , $ln \frac{200}{200} = ln 1 = 0$ .

\frac{d x}{d t} = 0

(b)(ii)

1 mark

At 200 wolves the population is not changing, so 200 is the steady (equilibrium) population.

(b)(iii)

2 marks

Differentiate $f (x) = ln (ln 200 - ln x)$ with the chain rule.

f^{'} (x) = \frac{1}{ln 200 - ln x} \times (- \frac{1}{x}) = - \frac{1}{x ( ln 200 - ln x )} = - \frac{1}{x ln ( \frac{200}{x} )}

(b)(iv)

5 marks

Separate the variables in $\frac{d x}{d t} = a x ln \frac{200}{x}$ .
By part (iii) the left side is $- f (x) = - ln (ln 200 - ln x)$ .
Rearrange.

\int \frac{1}{x ln ( \frac{200}{x} )} d x = \int a d t - ln (ln 200 - ln x) = a t + c ln 200 - ln x = e^{- a t - c} = A e^{- a t}, A = e^{- c} > 0 ln x = ln 200 - A e^{- a t}

(b)(v)

2 marks

Use $x = 40$ when $t = 0$ .

ln 40 = ln 200 - A \Rightarrow A = ln 200 - ln 40 = ln 5

(b)(vi)

2 marks

Use $x = 70$ when $t = 5$ .

ln 70 = ln 200 - ln 5 \cdot e^{- 5 a} \Rightarrow e^{- 5 a} = \frac{ln ( 200/70 )}{ln 5} a = - \frac{1}{5} ln (\frac{ln ( 20/7 )}{ln 5}) \approx 0.0855

(b)(vii)

3 marks

At $t = 10$ .

ln x = ln 200 - ln 5 \cdot e^{- 10 (0.08545)} \Rightarrow x \approx 100.8 \approx 101 wolves

(c)

1 mark

After 10 years the real population is 85. The logistic model gives 107 and the Gompertz model gives 101, so both overestimate. The Gompertz prediction is closer.

(d)(i)

4 marks

Use $x_{n + 1} = x_{n} + 0.5 (0.0855 x_{n} ln \frac{200}{x _{n}} - 0.03 x_{n})$ with $x_{0} = 70$ at $t = 5$ . Ten steps are needed to reach $t = 10$ .
So the estimate is 89 wolves.

x_{1} \approx 72.09, x_{2} \approx 74.16, x_{3} \approx 76.19, x_{4} \approx 78.19, x_{5} \approx 80.16 x_{6} \approx 82.09, x_{7} \approx 83.98, x_{8} \approx 85.84, x_{9} \approx 87.65, x_{10} \approx 89.43

(d)(ii)

1 mark

The estimate of 89 is much closer to the measured 85 than either earlier model, so allowing for emigration makes the model more realistic.

Our analysis

Where the marks are

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

Calculus48 marks · 87%
Q1, Q2
Functions7 marks · 13%
Q2

Mentor insights

Traps to watch for

01
Normal · Q1
The gradient of a normal is the negative reciprocal of the gradient of the tangent.
02
Quadratic roots · Q1
For $a x^{2} + b x + c = 0$ the sum of the roots is $- \frac{b}{a}$ and the product is $\frac{c}{a}$ .
03
Logistic · Q2
$x = \frac{L}{1 + C e ^{- k t}}$ has a maximum at $L$ .
04
Separation · Q2
Separate $x$ and $t$ , integrate, and use the initial value to find the constant.
05
Euler · Q2
$x_{n + 1} = x_{n} + h \frac{d x}{d t}$ .

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