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IB Maths AA HL · Paper 1

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2023
Marks: 110
Duration: 2 hours
Calculator: Not allowed
Questions: 12

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2023 TZ1 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 1–9 · 56 marks

Question 1

7 marks

A rational function: zero, asymptotes and inverse

Functions

(a)

2 marks

A zero of $f (x) = \frac{7 x + 7}{2 x - 4}$ needs the numerator to be zero.

7 x + 7 = 0 \Rightarrow x = - 1

(b)

2 marks

The vertical asymptote is where the denominator is zero. For large $∣ x ∣$ the function tends to the ratio of the leading coefficients.

x = 2, y = \frac{7}{2}

(c)

3 marks

Write $y = \frac{7 x + 7}{2 x - 4}$ and swap $x$ and $y$ .

x = \frac{7 y + 7}{2 y - 4} \Rightarrow 2 x y - 4 x = 7 y + 7 \Rightarrow y (2 x - 7) = 4 x + 7 f^{- 1} (x) = \frac{4 x + 7}{2 x - 7}, x \neq = \frac{7}{2}

Question 2

6 marks

Rides on The Dragon: probability, expected value and capacity

Statistics and probability

(a)

4 marks

Of the 40 visitors, $6$ did not ride.
The expected number of rides per visitor:

P (rides) = \frac{40 - 6}{40} = \frac{34}{40} = 0.85 E = \frac{0 ( 6 ) + 1 ( 16 ) + 2 ( 13 ) + 3 ( 2 ) + 4 ( 3 )}{40} = \frac{60}{40} = 1.5

(b)

2 marks

For 1000 visitors the expected number of rides is $1000 \times 1.5 = 1500$ . Each run carries up to 10 people.

\frac{1500}{10} = 150 runs

Question 3

6 marks

Solving cos 2x = sin x on an interval

Number and algebra

6 marks

Use $cos 2 x = 1 - 2 sin^{2} x$ .
So $sin x = \frac{1}{2}$ or $sin x = - 1$ .

1 - 2 sin^{2} x = sin x \Rightarrow 2 sin^{2} x + sin x - 1 = 0 (2 sin x - 1) (sin x + 1) = 0 x = \frac{π}{6}, \frac{5 π}{6}, - \frac{π}{2}

Question 4

6 marks

When does e^(2x) + ln k = 3e^x have a real solution?

Number and algebra

6 marks

Put $A = e^{x}$ with $A > 0$ .
Real roots need the discriminant to be non-negative.
The roots add to 3, which is positive, so at least one root is positive whenever there are real roots. Also $k > 0$ .

A^{2} - 3 A + ln k = 0 9 - 4 ln k \geq 0 \Rightarrow ln k \leq \frac{9}{4} 0 < k \leq e^{9/4}

Question 5

6 marks

f(x) = sin qx: finding m and sketching a horizontal stretch

Functions

(a)

2 marks

One full period of $sin q x$ is $\frac{2 π}{q}$ , and this is $4 m$ .

4 m = \frac{2 π}{q} \Rightarrow m = \frac{π}{2 q}

(b)

4 marks

The new period is $\frac{2 π}{2 q /3} = \frac{3 π}{q} = 6 m$ , so the graph is stretched horizontally by a factor of $\frac{3}{2}$ , and vertically by a factor of 3.
Sketch: a sine curve through $(0, 0)$ , with its maximum 3 at $x = \frac{3}{2} m$ , an intercept at $x = 3 m$ , its minimum $- 3$ at $x = \frac{9}{2} m$ and ending at $(6 m, 0)$ .

Question 6

5 marks

Rate of change of the area of an equilateral triangle

Number and algebra

5 marks

Area of an equilateral triangle: $A = \frac{1}{2} x^{2} sin 6 0^{\circ}$ .
Use the chain rule with $\frac{d x}{d t} = 4$ and $x = 53$ .

A = \frac{3}{4} x^{2} \Rightarrow \frac{d A}{d x} = \frac{3}{2} x \frac{d A}{d t} = \frac{3}{2} (53) (4) = 30 cm^{2} s^{- 1}

Question 7

6 marks

A cubic with a complex factor: finding m and all the roots

Number and algebra

6 marks

Since $z - 3 i$ is a factor and the coefficients are real, the roots are $3 i$ , $- 3 i$ and a real root $r$ .
Write the equation as $z^{3} - \frac{36}{m} z^{2} + m z - 4 m = 0$ . The sum and product of the roots are
Substitute $r$ .
The roots are $3 i$ , $- 3 i$ and $4$ .

r = \frac{36}{m}, (3 i) (- 3 i) r = 9 r = 4 m 9 \cdot \frac{36}{m} = 4 m \Rightarrow m^{2} = 81 \Rightarrow m = 9 r = \frac{36}{9} = 4

Question 8

7 marks

The graph of f(|x|) and a squared equation

Functions

(a)

4 marks

Keep the part with $x \geq 0$ and reflect it in the $y$ -axis.
Sketch: a symmetric curve with a smooth maximum at $(0, 3)$ , crossing the $x$ -axis at $(- a, 0)$ and $(a, 0)$ with sharp points there, and approaching the horizontal asymptote $y = - 2$ on both sides.

(b)

3 marks

$(g (x))^{2} = k$ means $g (x) = k$ or $g (x) = - k$ . The graph has range $- 2 < g \leq 3$ and each value strictly between $- 2$ and $3$ is reached twice, except 3 (once) and $- 2$ (never).
For $k = 0$ : $g (x) = 0$ at $x = \pm a$ , exactly two solutions.
For $0 < k < 2$ : both $k$ and $- k$ are reached twice, so four solutions.
For $k = 2$ : $g = 2$ has two solutions and $g = - 2$ has none, so exactly two. For $2 < k < 3$ : again two solutions. At $k = 3$ there is only one solution.

k = 0 or 4 \leq k < 9

Question 9

7 marks

A spherical ring: volume of a sphere with a hole through it

Number and algebra

7 marks

The ring's top and bottom are at $y = \pm \frac{h}{2}$ , where the sphere has radius $R$ .
Volume of the sphere between $y = - \frac{h}{2}$ and $y = \frac{h}{2}$ :
Subtract the cylinder $π R^{2} h = π (r^{2} - \frac{h ^{2}}{4}) h$ .
The volume is $π$ , so $r$ cancels.

R^{2} = r^{2} - \frac{h ^{2}}{4} π \int_{- h /2}^{h /2} (r^{2} - y^{2}) d y = π (r^{2} h - \frac{h ^{3}}{12}) V = π (r^{2} h - \frac{h ^{3}}{12} - r^{2} h + \frac{h ^{3}}{4}) = \frac{π h ^{3}}{6} \frac{π h ^{3}}{6} = π \Rightarrow h^{3} = 6 \Rightarrow h = 36 \approx 1.82

Section B · Extended response

Questions 10–12 · 54 marks

Question 10

14 marks

A sequence from its sum formula, then a geometric sequence

Number and algebra

(a)

4 marks

Put $n = 5$ and $n = 6$ .

S_{5} = 25 + 20 = 45 u_{6} = S_{6} - S_{5} = 60 - 45 = 15

(b)

2 marks

The first term is the first sum.

u_{1} = S_{1} = 1 + 4 = 5

(c)

3 marks

$S_{2} = 4 + 8 = 12$ , so $u_{2} = 12 - 5 = 7$ and the common difference is $d = 2$ .

u_{n} = 5 + 2 (n - 1) = 2 n + 3

(d)

3 marks

$v_{2} = 5$ and $v_{4} = 15$ , and $v_{4} = v_{2} r^{2}$ .

r^{2} = 3 \Rightarrow r = \pm 3

(e)

2 marks

$v_{99} = v_{2} r^{97}$ with $v_{2} = 5 > 0$ , so $v_{99} < 0$ needs $r < 0$ .

r = - 3, v_{5} = v_{4} r = 15 (- 3) = - 153

Question 11

19 marks

Carrying a pole round a corner: minimising a length

Calculus

(a)

2 marks

In the right triangle with the passage width $\frac{3}{4}$ : $cos α = \frac{3/4}{A C}$ . In the other triangle $sin α = \frac{6}{C B}$ .

A C = \frac{3}{4} sec α, C B = 6 csc α L = \frac{3}{4} sec α + 6 csc α

(b)

5 marks

Differentiate.
Write in terms of sine and cosine and set to zero.

\frac{d L}{d α} = \frac{3}{4} sec α tan α - 6 csc α cot α \frac{3 sin α}{4 cos ^{2} α} - \frac{6 cos α}{sin ^{2} α} = 0 \Rightarrow 3 sin^{3} α = 24 cos^{3} α \Rightarrow tan^{3} α = 8 tan α = 2 \Rightarrow α = arctan 2

(c)

7 marks

Differentiate each term with the product rule.
When $tan α = 2$ , use a right triangle with sides 1, 2 and $5$ : $sec α = 5$ , $csc α = \frac{5}{2}$ and $cot α = \frac{1}{2}$ .

\frac{d ^{2} L}{d α ^{2}} = \frac{3}{4} (sec α tan^{2} α + sec^{3} α) + 6 (csc α cot^{2} α + csc^{3} α) \frac{3}{4} (45 + 55) + 6 (\frac{5}{8} + \frac{5 5}{8}) = \frac{27 5}{4} + \frac{18 5}{4} = \frac{45 5}{4}

(d)

3 marks

The second derivative is positive at $α = arctan 2$ , so the stationary point is a minimum.
The minimum length:

L_{m i n} = \frac{3}{4} 5 + 6 \cdot \frac{5}{2} = \frac{15 5}{4} \approx 8.39 m

(e)

2 marks

The longest pole that can turn the corner has the minimum length 8.39 m. The pole is 11.25 m, which is longer.
So it is not possible to carry the pole round the corner horizontally.

Question 12

21 marks

Two lines, a plane and a cone

Geometry and trigonometry

(a)

4 marks

The acute angle between the lines satisfies $cos \frac{π}{3} = \frac{∣ P A \cdot d ∣}{∣ P A ∣∣ d ∣}$ with $d = 110$ .

\frac{1}{2} = \frac{2 t}{2 4 t ^{2} + ( 3 + t ) ^{2}} \Rightarrow 4 t = 2 5 t^{2} + 6 t + 9 4 t = 10 t^{2} + 12 t + 18

(b)

4 marks

Square both sides.
Since $t > 0$ :

16 t^{2} = 10 t^{2} + 12 t + 18 \Rightarrow t^{2} - 2 t - 3 = 0 \Rightarrow (t - 3) (t + 1) = 0 t = 3

(c)

4 marks

With $t = 3$ , $P A = 606$ and $∣ P A ∣ = 62$ . The shortest distance from A to $L_{1}$ is $∣ P A ∣ sin \frac{π}{3}$ .

62 \cdot \frac{3}{2} = 36

(d)

2 marks

The normal is perpendicular to both direction vectors, $110$ and $P A$ , which is parallel to $101$ .

110 \times 101 = 1 - 1 - 1

(e)

7 marks

The cone's base has radius $36$ (the distance from A to $L_{1}$ ). Use the volume to find the height $h$ .
The unit normal is $\frac{1}{3} 1 - 1 - 1$ , so the vertex is $h$ along the normal from $A (6, 8, 3)$ in either direction.

\frac{1}{3} π (36)^{2} h = 18 π h = 90 π 3 \Rightarrow h = 53 683 \pm 5 1 - 1 - 1 (11, 3, - 2) and (1, 13, 8)

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 algebra44 marks · 40%
Q3, Q4, Q6, Q7, Q9, Q10
Geometry and trigonometry21 marks · 19%
Q12
Functions20 marks · 18%
Q1, Q5, Q8
Calculus19 marks · 17%
Q11
Statistics and probability6 marks · 5%
Q2

Mentor insights

Traps to watch for

01
Zero is where the numerator is zero · Q1
Set $7 x + 7 = 0$ , as long as the denominator is not also zero there.
02
f(|x|) is symmetric · Q8
Keep the right-hand part of the graph of $f$ and reflect it in the $y$ -axis.
03
Term from sums · Q10
$u_{n} = S_{n} - S_{n - 1}$ , and $u_{1} = S_{1}$ .
04
Two triangles · Q11
Split AB at C into AC (in the passage) and CB (in the room) and use secant and cosecant.
05
The base is centred at A · Q12
The tangent line to the circular base means the radius of the base is the distance from A to $L_{1}$ .

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 HL · Paper 1

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2023
Marks: 110
Duration: 2 hours
Calculator: Not allowed
Questions: 12

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2023 TZ1 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 1–9 · 56 marks

Question 1

7 marks

A rational function: zero, asymptotes and inverse

Functions

(a)

2 marks

A zero of $f (x) = \frac{7 x + 7}{2 x - 4}$ needs the numerator to be zero.

7 x + 7 = 0 \Rightarrow x = - 1

(b)

2 marks

The vertical asymptote is where the denominator is zero. For large $∣ x ∣$ the function tends to the ratio of the leading coefficients.

x = 2, y = \frac{7}{2}

(c)

3 marks

Write $y = \frac{7 x + 7}{2 x - 4}$ and swap $x$ and $y$ .

x = \frac{7 y + 7}{2 y - 4} \Rightarrow 2 x y - 4 x = 7 y + 7 \Rightarrow y (2 x - 7) = 4 x + 7 f^{- 1} (x) = \frac{4 x + 7}{2 x - 7}, x \neq = \frac{7}{2}

Question 2

6 marks

Rides on The Dragon: probability, expected value and capacity

Statistics and probability

(a)

4 marks

Of the 40 visitors, $6$ did not ride.
The expected number of rides per visitor:

P (rides) = \frac{40 - 6}{40} = \frac{34}{40} = 0.85 E = \frac{0 ( 6 ) + 1 ( 16 ) + 2 ( 13 ) + 3 ( 2 ) + 4 ( 3 )}{40} = \frac{60}{40} = 1.5

(b)

2 marks

For 1000 visitors the expected number of rides is $1000 \times 1.5 = 1500$ . Each run carries up to 10 people.

\frac{1500}{10} = 150 runs

Question 3

6 marks

Solving cos 2x = sin x on an interval

Number and algebra

6 marks

Use $cos 2 x = 1 - 2 sin^{2} x$ .
So $sin x = \frac{1}{2}$ or $sin x = - 1$ .

1 - 2 sin^{2} x = sin x \Rightarrow 2 sin^{2} x + sin x - 1 = 0 (2 sin x - 1) (sin x + 1) = 0 x = \frac{π}{6}, \frac{5 π}{6}, - \frac{π}{2}

Question 4

6 marks

When does e^(2x) + ln k = 3e^x have a real solution?

Number and algebra

6 marks

Put $A = e^{x}$ with $A > 0$ .
Real roots need the discriminant to be non-negative.
The roots add to 3, which is positive, so at least one root is positive whenever there are real roots. Also $k > 0$ .

A^{2} - 3 A + ln k = 0 9 - 4 ln k \geq 0 \Rightarrow ln k \leq \frac{9}{4} 0 < k \leq e^{9/4}

Question 5

6 marks

f(x) = sin qx: finding m and sketching a horizontal stretch

Functions

(a)

2 marks

One full period of $sin q x$ is $\frac{2 π}{q}$ , and this is $4 m$ .

4 m = \frac{2 π}{q} \Rightarrow m = \frac{π}{2 q}

(b)

4 marks

The new period is $\frac{2 π}{2 q /3} = \frac{3 π}{q} = 6 m$ , so the graph is stretched horizontally by a factor of $\frac{3}{2}$ , and vertically by a factor of 3.
Sketch: a sine curve through $(0, 0)$ , with its maximum 3 at $x = \frac{3}{2} m$ , an intercept at $x = 3 m$ , its minimum $- 3$ at $x = \frac{9}{2} m$ and ending at $(6 m, 0)$ .

Question 6

5 marks

Rate of change of the area of an equilateral triangle

Number and algebra

5 marks

Area of an equilateral triangle: $A = \frac{1}{2} x^{2} sin 6 0^{\circ}$ .
Use the chain rule with $\frac{d x}{d t} = 4$ and $x = 53$ .

A = \frac{3}{4} x^{2} \Rightarrow \frac{d A}{d x} = \frac{3}{2} x \frac{d A}{d t} = \frac{3}{2} (53) (4) = 30 cm^{2} s^{- 1}

Question 7

6 marks

A cubic with a complex factor: finding m and all the roots

Number and algebra

6 marks

Since $z - 3 i$ is a factor and the coefficients are real, the roots are $3 i$ , $- 3 i$ and a real root $r$ .
Write the equation as $z^{3} - \frac{36}{m} z^{2} + m z - 4 m = 0$ . The sum and product of the roots are
Substitute $r$ .
The roots are $3 i$ , $- 3 i$ and $4$ .

r = \frac{36}{m}, (3 i) (- 3 i) r = 9 r = 4 m 9 \cdot \frac{36}{m} = 4 m \Rightarrow m^{2} = 81 \Rightarrow m = 9 r = \frac{36}{9} = 4

Question 8

7 marks

The graph of f(|x|) and a squared equation

Functions

(a)

4 marks

Keep the part with $x \geq 0$ and reflect it in the $y$ -axis.
Sketch: a symmetric curve with a smooth maximum at $(0, 3)$ , crossing the $x$ -axis at $(- a, 0)$ and $(a, 0)$ with sharp points there, and approaching the horizontal asymptote $y = - 2$ on both sides.

(b)

3 marks

$(g (x))^{2} = k$ means $g (x) = k$ or $g (x) = - k$ . The graph has range $- 2 < g \leq 3$ and each value strictly between $- 2$ and $3$ is reached twice, except 3 (once) and $- 2$ (never).
For $k = 0$ : $g (x) = 0$ at $x = \pm a$ , exactly two solutions.
For $0 < k < 2$ : both $k$ and $- k$ are reached twice, so four solutions.
For $k = 2$ : $g = 2$ has two solutions and $g = - 2$ has none, so exactly two. For $2 < k < 3$ : again two solutions. At $k = 3$ there is only one solution.

k = 0 or 4 \leq k < 9

Question 9

7 marks

A spherical ring: volume of a sphere with a hole through it

Number and algebra

7 marks

The ring's top and bottom are at $y = \pm \frac{h}{2}$ , where the sphere has radius $R$ .
Volume of the sphere between $y = - \frac{h}{2}$ and $y = \frac{h}{2}$ :
Subtract the cylinder $π R^{2} h = π (r^{2} - \frac{h ^{2}}{4}) h$ .
The volume is $π$ , so $r$ cancels.

R^{2} = r^{2} - \frac{h ^{2}}{4} π \int_{- h /2}^{h /2} (r^{2} - y^{2}) d y = π (r^{2} h - \frac{h ^{3}}{12}) V = π (r^{2} h - \frac{h ^{3}}{12} - r^{2} h + \frac{h ^{3}}{4}) = \frac{π h ^{3}}{6} \frac{π h ^{3}}{6} = π \Rightarrow h^{3} = 6 \Rightarrow h = 36 \approx 1.82

Section B · Extended response

Questions 10–12 · 54 marks

Question 10

14 marks

A sequence from its sum formula, then a geometric sequence

Number and algebra

(a)

4 marks

Put $n = 5$ and $n = 6$ .

S_{5} = 25 + 20 = 45 u_{6} = S_{6} - S_{5} = 60 - 45 = 15

(b)

2 marks

The first term is the first sum.

u_{1} = S_{1} = 1 + 4 = 5

(c)

3 marks

$S_{2} = 4 + 8 = 12$ , so $u_{2} = 12 - 5 = 7$ and the common difference is $d = 2$ .

u_{n} = 5 + 2 (n - 1) = 2 n + 3

(d)

3 marks

$v_{2} = 5$ and $v_{4} = 15$ , and $v_{4} = v_{2} r^{2}$ .

r^{2} = 3 \Rightarrow r = \pm 3

(e)

2 marks

$v_{99} = v_{2} r^{97}$ with $v_{2} = 5 > 0$ , so $v_{99} < 0$ needs $r < 0$ .

r = - 3, v_{5} = v_{4} r = 15 (- 3) = - 153

Question 11

19 marks

Carrying a pole round a corner: minimising a length

Calculus

(a)

2 marks

In the right triangle with the passage width $\frac{3}{4}$ : $cos α = \frac{3/4}{A C}$ . In the other triangle $sin α = \frac{6}{C B}$ .

A C = \frac{3}{4} sec α, C B = 6 csc α L = \frac{3}{4} sec α + 6 csc α

(b)

5 marks

Differentiate.
Write in terms of sine and cosine and set to zero.

\frac{d L}{d α} = \frac{3}{4} sec α tan α - 6 csc α cot α \frac{3 sin α}{4 cos ^{2} α} - \frac{6 cos α}{sin ^{2} α} = 0 \Rightarrow 3 sin^{3} α = 24 cos^{3} α \Rightarrow tan^{3} α = 8 tan α = 2 \Rightarrow α = arctan 2

(c)

7 marks

Differentiate each term with the product rule.
When $tan α = 2$ , use a right triangle with sides 1, 2 and $5$ : $sec α = 5$ , $csc α = \frac{5}{2}$ and $cot α = \frac{1}{2}$ .

\frac{d ^{2} L}{d α ^{2}} = \frac{3}{4} (sec α tan^{2} α + sec^{3} α) + 6 (csc α cot^{2} α + csc^{3} α) \frac{3}{4} (45 + 55) + 6 (\frac{5}{8} + \frac{5 5}{8}) = \frac{27 5}{4} + \frac{18 5}{4} = \frac{45 5}{4}

(d)

3 marks

The second derivative is positive at $α = arctan 2$ , so the stationary point is a minimum.
The minimum length:

L_{m i n} = \frac{3}{4} 5 + 6 \cdot \frac{5}{2} = \frac{15 5}{4} \approx 8.39 m

(e)

2 marks

The longest pole that can turn the corner has the minimum length 8.39 m. The pole is 11.25 m, which is longer.
So it is not possible to carry the pole round the corner horizontally.

Question 12

21 marks

Two lines, a plane and a cone

Geometry and trigonometry

(a)

4 marks

The acute angle between the lines satisfies $cos \frac{π}{3} = \frac{∣ P A \cdot d ∣}{∣ P A ∣∣ d ∣}$ with $d = 110$ .

\frac{1}{2} = \frac{2 t}{2 4 t ^{2} + ( 3 + t ) ^{2}} \Rightarrow 4 t = 2 5 t^{2} + 6 t + 9 4 t = 10 t^{2} + 12 t + 18

(b)

4 marks

Square both sides.
Since $t > 0$ :

16 t^{2} = 10 t^{2} + 12 t + 18 \Rightarrow t^{2} - 2 t - 3 = 0 \Rightarrow (t - 3) (t + 1) = 0 t = 3

(c)

4 marks

With $t = 3$ , $P A = 606$ and $∣ P A ∣ = 62$ . The shortest distance from A to $L_{1}$ is $∣ P A ∣ sin \frac{π}{3}$ .

62 \cdot \frac{3}{2} = 36

(d)

2 marks

The normal is perpendicular to both direction vectors, $110$ and $P A$ , which is parallel to $101$ .

110 \times 101 = 1 - 1 - 1

(e)

7 marks

The cone's base has radius $36$ (the distance from A to $L_{1}$ ). Use the volume to find the height $h$ .
The unit normal is $\frac{1}{3} 1 - 1 - 1$ , so the vertex is $h$ along the normal from $A (6, 8, 3)$ in either direction.

\frac{1}{3} π (36)^{2} h = 18 π h = 90 π 3 \Rightarrow h = 53 683 \pm 5 1 - 1 - 1 (11, 3, - 2) and (1, 13, 8)

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 algebra44 marks · 40%
Q3, Q4, Q6, Q7, Q9, Q10
Geometry and trigonometry21 marks · 19%
Q12
Functions20 marks · 18%
Q1, Q5, Q8
Calculus19 marks · 17%
Q11
Statistics and probability6 marks · 5%
Q2

Mentor insights

Traps to watch for

01
Zero is where the numerator is zero · Q1
Set $7 x + 7 = 0$ , as long as the denominator is not also zero there.
02
f(|x|) is symmetric · Q8
Keep the right-hand part of the graph of $f$ and reflect it in the $y$ -axis.
03
Term from sums · Q10
$u_{n} = S_{n} - S_{n - 1}$ , and $u_{1} = S_{1}$ .
04
Two triangles · Q11
Split AB at C into AC (in the passage) and CB (in the room) and use secant and cosecant.
05
The base is centred at A · Q12
The tangent line to the circular base means the radius of the base is the distance from A to $L_{1}$ .

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