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IB Maths AA HL · Paper 3

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2TZ3

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

28 marks

A family of rational curves and the homogeneous differential equation it solves

Calculus

(a)(i)

1 mark

Sketch: the curve $y = \frac{x ( x ^{2} - 16 )}{x ^{2} + 16}$ is an odd function that rises from the left to a local maximum, falls through the origin to a local minimum and then rises again. It approaches the line $y = x$ at both ends.

(a)(ii)

1 mark

The curve crosses the $x$ -axis where $x (x^{2} - 16) = 0$ .

(- 4, 0), (0, 0), (4, 0)

(a)(iii)

2 marks

The turning points come from $f^{'} (x) = 0$ , that is $x^{4} + 64 x^{2} - 256 = 0$ .

x^{2} = - 32 + 1280 \approx 3.777 \Rightarrow x \approx \pm 1.943 local maximum \approx (- 1.94, 1.20), local minimum \approx (1.94, - 1.20)

(b)

2 marks

Replace $x$ by $- x$ .
So $f$ is an odd function.

f (- x) = \frac{- x ( ( - x ) ^{2} - 16 )}{( - x ) ^{2} + 16} = - \frac{x ( x ^{2} - 16 )}{x ^{2} + 16} = - f (x)

(c)

4 marks

Differentiate $f (x) = \frac{x ^{3} - A x}{x ^{2} + A}$ with the quotient rule.
Substitute $x = A$ .
This does not depend on $A$ .

f^{'} (x) = \frac{( 3 x ^{2} - A ) ( x ^{2} + A ) - ( x ^{3} - A x ) ( 2 x )}{( x ^{2} + A ) ^{2}} = \frac{x ^{4} + 4 A x ^{2} - A ^{2}}{( x ^{2} + A ) ^{2}} f^{'} (A) = \frac{A ^{2} + 4 A ^{2} - A ^{2}}{( 2 A ) ^{2}} = \frac{4 A ^{2}}{4 A ^{2}} = 1

(d)(i)

2 marks

Put the left side over the common denominator $x^{2} + A$ .

x - \frac{2 A x}{x ^{2} + A} = \frac{x ( x ^{2} + A ) - 2 A x}{x ^{2} + A} = \frac{x ^{3} - A x}{x ^{2} + A} = \frac{x ( x ^{2} - A )}{x ^{2} + A}

(d)(ii)

1 mark

From part (i), $y = x - \frac{2 A x}{x ^{2} + A}$ , and $\frac{2 A x}{x ^{2} + A} \to 0$ as $x \to \pm \infty$ .

y = x

(d)(iii)

1 mark

By part (c) the tangent at $(A, 0)$ has gradient 1, the same as the asymptote.

(A, 0)

(e)

5 marks

Write the integrand as partial fractions.
Integrate.
Write the constant as $c = \frac{1}{2} ln A$ with $A > 0$ and combine the logarithms.

\frac{1}{1 - v ^{2}} = \frac{1}{( 1 + v ) ( 1 - v )} = \frac{B}{1 + v} + \frac{C}{1 - v} \Rightarrow B = C = \frac{1}{2} \int \frac{1}{1 - v ^{2}} d v = \frac{1}{2} ln ∣1 + v ∣ - \frac{1}{2} ln ∣1 - v ∣ + c = \frac{1}{2} ln \frac{1 + v}{1 - v} + c \int \frac{1}{1 - v ^{2}} d v = \frac{1}{2} ln \frac{A ( 1 + v )}{1 - v}

(f)

5 marks

Separate the variables in $x \frac{d v}{d x} = 1 - v^{2}$ and integrate, using part (e).
Multiply by 2 and use the power rule.
Replace $v$ by $\frac{y}{x}$ , using $\frac{1 + y / x}{1 - y / x} = \frac{x + y}{x - y}$ .

\int \frac{1}{1 - v ^{2}} d v = \int \frac{1}{x} d x \Rightarrow \frac{1}{2} ln \frac{A ( 1 + v )}{1 - v} = ln ∣ x ∣ ln \frac{A ( 1 + v )}{1 - v} = ln x^{2} \Rightarrow x^{2} = \frac{A ( 1 + v )}{1 - v} x^{2} = \frac{A ( x + y )}{x - y}

(g)

4 marks

When $\frac{A ( x + y )}{x - y} > 0$ the modulus signs can be dropped.

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

Question 2

27 marks

Distances and directions between cities: a flat model and a spherical model

Geometry and trigonometry

(a)(i)

2 marks

Moscow is 6000 km north of Nairobi, and Bogota is 12 500 km west of Nairobi. Use Pythagoras.

BM = 1250 0^{2} + 600 0^{2} \approx 13865 km

(a)(ii)

3 marks

From Bogota, Moscow is 12 500 km east and 6000 km north.
The bearing is measured from north, so the bearing of Moscow from Bogota is

tan ϕ = \frac{12500}{6000} \Rightarrow ϕ \approx 64. 4^{\circ} 64. 4^{\circ}

(b)(i)

2 marks

With $p = (0, 0, 6)$ and $n = (0, 6, 0)$ , the scalar product is $p \cdot n = 0$ .

cos θ = \frac{0}{6 \times 6} = 0 \Rightarrow θ = 9 0^{\circ} = \frac{π}{2}

(b)(ii)

2 marks

The arc length is $r θ$ with radius 6000 km and angle $\frac{π}{2}$ .

6000 \times \frac{π}{2} = 3000 π km

(c)(i)

2 marks

Take the vector product of $a = (6, 0, 0)$ and $p = (0, 0, 6)$ .

a \times p = 0 \cdot 6 - 0 \cdot 0 0 \cdot 0 - 6 \cdot 6 0 = 0 - 36 0

(c)(ii)

3 marks

Similarly, with $n = (0, 6, 0)$ .
The two vectors have length 36 each. The scalar product is zero, so they are perpendicular.

a \times n = 0036 cos θ = \frac{( a \times p ) \cdot ( a \times n )}{36 \times 36} = \frac{0}{1296} = 0 \Rightarrow θ = 9 0^{\circ}

(d)

2 marks

The shortest distance is the arc $r θ$ with $r = 6$ (thousand km) and arc length 6.

6 θ = 6 \Rightarrow θ = 1 radian = 57. 3^{\circ}

(e)

5 marks

With $b = (6 sin 12 0^{\circ}, 6 cos 12 0^{\circ}, 0)$ and $m = (0, 6 cos θ, 6 sin θ)$ , use the scalar product to find the angle $B O M$ at the centre.
The arc length is $r θ$ with $r = 6$ thousand km.

cos B O M = \frac{b \cdot m}{36} = \frac{36 cos 12 0 ^{\circ} cos 57. 3 ^{\circ}}{36} = cos 12 0^{\circ} cos 57. 3^{\circ} \approx - 0.2701 B O M \approx 105.6 7^{\circ} \approx 1.844 radians 6 \times 1.8443 \approx 11.07 thousand km \approx 11100 km

(f)

6 marks

Using the method of part (c), find the vector products $b \times m$ and $b \times p$ .
The angle between them is given by the scalar product, with $∣ b \times p ∣ = 36$ and $∣ b \times m ∣ \approx 34.66$ .
The bearing of Moscow from Bogota on the sphere is about $29. 1^{\circ}$ (compared with $64. 4^{\circ}$ on the flat model).

b \times m = 36 cos 12 0^{\circ} sin θ - 36 sin 12 0^{\circ} sin θ 36 sin 12 0^{\circ} cos θ \approx - 15.15 - 26.23 16.84 b \times p = 36 cos 12 0^{\circ} - 36 sin 12 0^{\circ} 0 \approx - 18 - 31.18 0 cos ϕ = \frac{( - 15.15 ) ( - 18 ) + ( - 26.23 ) ( - 31.18 )}{36 \times 34.66} = \frac{1090.5}{1247.8} \approx 0.874 ϕ \approx 29. 1^{\circ}

Our analysis

Where the marks are

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

Calculus28 marks · 51%
Q1
Geometry and trigonometry27 marks · 49%
Q2

Mentor insights

Traps to watch for

01
Quotient rule · Q1
$f^{'} (x) = \frac{x ^{4} + 4 A x ^{2} - A ^{2}}{( x ^{2} + A ) ^{2}}$ .
02
Partial fractions · Q1
$\frac{1}{1 - v ^{2}} = \frac{1}{2} (\frac{1}{1 + v} + \frac{1}{1 - v})$ .
03
Substitution · Q1
With $y = v x$ , $\frac{d y}{d x} = v + x \frac{d v}{d x}$ .
04
Flat model · Q2
Use right-angled triangles. The bearing is measured clockwise from north.
05
Sphere · Q2
The distance along the surface is $r θ$ where $θ$ is the angle at the centre in radians.

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

TZ1TZ2TZ3

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 TZ2 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

28 marks

A family of rational curves and the homogeneous differential equation it solves

Calculus

(a)(i)

1 mark

(a)(ii)

1 mark

The curve crosses the $x$ -axis where $x (x^{2} - 16) = 0$ .

(- 4, 0), (0, 0), (4, 0)

(a)(iii)

2 marks

The turning points come from $f^{'} (x) = 0$ , that is $x^{4} + 64 x^{2} - 256 = 0$ .

x^{2} = - 32 + 1280 \approx 3.777 \Rightarrow x \approx \pm 1.943 local maximum \approx (- 1.94, 1.20), local minimum \approx (1.94, - 1.20)

(b)

2 marks

Replace $x$ by $- x$ .
So $f$ is an odd function.

f (- x) = \frac{- x ( ( - x ) ^{2} - 16 )}{( - x ) ^{2} + 16} = - \frac{x ( x ^{2} - 16 )}{x ^{2} + 16} = - f (x)

(c)

4 marks

Differentiate $f (x) = \frac{x ^{3} - A x}{x ^{2} + A}$ with the quotient rule.
Substitute $x = A$ .
This does not depend on $A$ .

f^{'} (x) = \frac{( 3 x ^{2} - A ) ( x ^{2} + A ) - ( x ^{3} - A x ) ( 2 x )}{( x ^{2} + A ) ^{2}} = \frac{x ^{4} + 4 A x ^{2} - A ^{2}}{( x ^{2} + A ) ^{2}} f^{'} (A) = \frac{A ^{2} + 4 A ^{2} - A ^{2}}{( 2 A ) ^{2}} = \frac{4 A ^{2}}{4 A ^{2}} = 1

(d)(i)

2 marks

Put the left side over the common denominator $x^{2} + A$ .

x - \frac{2 A x}{x ^{2} + A} = \frac{x ( x ^{2} + A ) - 2 A x}{x ^{2} + A} = \frac{x ^{3} - A x}{x ^{2} + A} = \frac{x ( x ^{2} - A )}{x ^{2} + A}

(d)(ii)

1 mark

From part (i), $y = x - \frac{2 A x}{x ^{2} + A}$ , and $\frac{2 A x}{x ^{2} + A} \to 0$ as $x \to \pm \infty$ .

y = x

(d)(iii)

1 mark

By part (c) the tangent at $(A, 0)$ has gradient 1, the same as the asymptote.

(A, 0)

(e)

5 marks

Write the integrand as partial fractions.
Integrate.
Write the constant as $c = \frac{1}{2} ln A$ with $A > 0$ and combine the logarithms.

\frac{1}{1 - v ^{2}} = \frac{1}{( 1 + v ) ( 1 - v )} = \frac{B}{1 + v} + \frac{C}{1 - v} \Rightarrow B = C = \frac{1}{2} \int \frac{1}{1 - v ^{2}} d v = \frac{1}{2} ln ∣1 + v ∣ - \frac{1}{2} ln ∣1 - v ∣ + c = \frac{1}{2} ln \frac{1 + v}{1 - v} + c \int \frac{1}{1 - v ^{2}} d v = \frac{1}{2} ln \frac{A ( 1 + v )}{1 - v}

(f)

5 marks

Separate the variables in $x \frac{d v}{d x} = 1 - v^{2}$ and integrate, using part (e).
Multiply by 2 and use the power rule.
Replace $v$ by $\frac{y}{x}$ , using $\frac{1 + y / x}{1 - y / x} = \frac{x + y}{x - y}$ .

\int \frac{1}{1 - v ^{2}} d v = \int \frac{1}{x} d x \Rightarrow \frac{1}{2} ln \frac{A ( 1 + v )}{1 - v} = ln ∣ x ∣ ln \frac{A ( 1 + v )}{1 - v} = ln x^{2} \Rightarrow x^{2} = \frac{A ( 1 + v )}{1 - v} x^{2} = \frac{A ( x + y )}{x - y}

(g)

4 marks

When $\frac{A ( x + y )}{x - y} > 0$ the modulus signs can be dropped.

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

Question 2

27 marks

Distances and directions between cities: a flat model and a spherical model

Geometry and trigonometry

(a)(i)

2 marks

Moscow is 6000 km north of Nairobi, and Bogota is 12 500 km west of Nairobi. Use Pythagoras.

BM = 1250 0^{2} + 600 0^{2} \approx 13865 km

(a)(ii)

3 marks

From Bogota, Moscow is 12 500 km east and 6000 km north.
The bearing is measured from north, so the bearing of Moscow from Bogota is

tan ϕ = \frac{12500}{6000} \Rightarrow ϕ \approx 64. 4^{\circ} 64. 4^{\circ}

(b)(i)

2 marks

With $p = (0, 0, 6)$ and $n = (0, 6, 0)$ , the scalar product is $p \cdot n = 0$ .

cos θ = \frac{0}{6 \times 6} = 0 \Rightarrow θ = 9 0^{\circ} = \frac{π}{2}

(b)(ii)

2 marks

The arc length is $r θ$ with radius 6000 km and angle $\frac{π}{2}$ .

6000 \times \frac{π}{2} = 3000 π km

(c)(i)

2 marks

Take the vector product of $a = (6, 0, 0)$ and $p = (0, 0, 6)$ .

a \times p = 0 \cdot 6 - 0 \cdot 0 0 \cdot 0 - 6 \cdot 6 0 = 0 - 36 0

(c)(ii)

3 marks

Similarly, with $n = (0, 6, 0)$ .
The two vectors have length 36 each. The scalar product is zero, so they are perpendicular.

a \times n = 0036 cos θ = \frac{( a \times p ) \cdot ( a \times n )}{36 \times 36} = \frac{0}{1296} = 0 \Rightarrow θ = 9 0^{\circ}

(d)

2 marks

The shortest distance is the arc $r θ$ with $r = 6$ (thousand km) and arc length 6.

6 θ = 6 \Rightarrow θ = 1 radian = 57. 3^{\circ}

(e)

5 marks

With $b = (6 sin 12 0^{\circ}, 6 cos 12 0^{\circ}, 0)$ and $m = (0, 6 cos θ, 6 sin θ)$ , use the scalar product to find the angle $B O M$ at the centre.
The arc length is $r θ$ with $r = 6$ thousand km.

cos B O M = \frac{b \cdot m}{36} = \frac{36 cos 12 0 ^{\circ} cos 57. 3 ^{\circ}}{36} = cos 12 0^{\circ} cos 57. 3^{\circ} \approx - 0.2701 B O M \approx 105.6 7^{\circ} \approx 1.844 radians 6 \times 1.8443 \approx 11.07 thousand km \approx 11100 km

(f)

6 marks

Using the method of part (c), find the vector products $b \times m$ and $b \times p$ .
The angle between them is given by the scalar product, with $∣ b \times p ∣ = 36$ and $∣ b \times m ∣ \approx 34.66$ .
The bearing of Moscow from Bogota on the sphere is about $29. 1^{\circ}$ (compared with $64. 4^{\circ}$ on the flat model).

b \times m = 36 cos 12 0^{\circ} sin θ - 36 sin 12 0^{\circ} sin θ 36 sin 12 0^{\circ} cos θ \approx - 15.15 - 26.23 16.84 b \times p = 36 cos 12 0^{\circ} - 36 sin 12 0^{\circ} 0 \approx - 18 - 31.18 0 cos ϕ = \frac{( - 15.15 ) ( - 18 ) + ( - 26.23 ) ( - 31.18 )}{36 \times 34.66} = \frac{1090.5}{1247.8} \approx 0.874 ϕ \approx 29. 1^{\circ}

Our analysis

Where the marks are

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

Calculus28 marks · 51%
Q1
Geometry and trigonometry27 marks · 49%
Q2

Mentor insights

Traps to watch for

01
Quotient rule · Q1
$f^{'} (x) = \frac{x ^{4} + 4 A x ^{2} - A ^{2}}{( x ^{2} + A ) ^{2}}$ .
02
Partial fractions · Q1
$\frac{1}{1 - v ^{2}} = \frac{1}{2} (\frac{1}{1 + v} + \frac{1}{1 - v})$ .
03
Substitution · Q1
With $y = v x$ , $\frac{d y}{d x} = v + x \frac{d v}{d x}$ .
04
Flat model · Q2
Use right-angled triangles. The bearing is measured clockwise from north.
05
Sphere · Q2
The distance along the surface is $r θ$ where $θ$ is the angle at the centre in radians.

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