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IB Maths AA SL · Paper 1

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2

Session: May 2023
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 2023 TZ2 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 · 35 marks

Question 1

6 marks

A circular sector: perimeter, angle and area

Geometry and trigonometry

(a)

2 marks

The sector is bounded by two radii and the arc PQR.

perimeter = 4 + 4 + 10 = 18 cm

(b)

2 marks

Arc length $= r θ$ .

10 = 4 θ \Rightarrow θ = 2.5

(c)

2 marks

Area of a sector.

A = \frac{1}{2} r^{2} θ = \frac{1}{2} (16) (2.5) = 20 cm^{2}

Question 2

6 marks

An arithmetic sequence in a probability distribution

Number and algebra

(a)

2 marks

Use $u_{4} = u_{1} + 3 d$ .

0.15 = 0.6 + 3 d \Rightarrow d = - 0.15

(b)

4 marks

The terms are $0.6, 0.45, 0.3, 0.15$ with sum $1.5$ .

\frac{0.6}{k} + \frac{0.45}{k} + \frac{0.3}{k} + \frac{0.15}{k} = \frac{1.5}{k} = 1 k = 1.5

Question 3

5 marks

f(x) = 1 - 1/(x - 2): asymptotes, intercepts and sketch

Functions

(a)

2 marks

The function is $f (x) = 1 - \frac{1}{x - 2}$ . The denominator is zero at $x = 2$ , and for large $∣ x ∣$ the fraction tends to 0.

x = 2, y = 1

(b)

2 marks

At $x = 0$ : $f (0) = 1 + \frac{1}{2}$ . At $f (x) = 0$ : $\frac{1}{x - 2} = 1$ .

(0, \frac{3}{2}), (3, 0)

(c)

1 mark

Sketch: two branches of a hyperbola, with the asymptotes $x = 2$ and $y = 1$ . The left branch passes through $(0, \frac{3}{2})$ and lies above $y = 1$ for $x < 2$ and rises without bound as $x \to 2^{-}$ . The right branch passes through $(3, 0)$ and goes to $- \infty$ as $x \to 2^{+}$ , approaching $y = 1$ from below for large $x$ .

Question 4

5 marks

Finding P(B) from conditional probability and the union

Number and algebra

5 marks

Addition rule: $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ .
Conditional probability: $P (A \cap B) = P (A ∣ B) P (B)$ .
Combine them.

0.55 = 0.4 + P (B) - P (A \cap B) P (A \cap B) = 0.25 P (B) 0.55 = 0.4 + P (B) - 0.25 P (B) \Rightarrow 0.75 P (B) = 0.15 \Rightarrow P (B) = 0.2

Question 5

6 marks

Area under y = x/(x^2 + 2) equal to ln 3

Number and algebra

6 marks

Set the area equal to $ln 3$ .
Combine the logs.

\int_{0}^{c} \frac{x}{x ^{2} + 2} d x = [\frac{1}{2} ln (x^{2} + 2)]_{0}^{c} = \frac{1}{2} ln (c^{2} + 2) - \frac{1}{2} ln 2 = ln 3 ln \frac{c ^{2} + 2}{2} = 2 ln 3 = ln 9 \Rightarrow \frac{c ^{2} + 2}{2} = 9 \Rightarrow c^{2} = 16 c = 4

Question 6

7 marks

Two linear functions f with (g after f) given

Number and algebra

7 marks

Substitute $f$ into $g$ .
Compare with $4 x^{2} - 14 x + 15$ .
So $a = \pm 2$ . Use the middle equation to find $b$ : for $a = 2$ , $4 b + 2 = - 14$ gives $b = - 4$ . For $a = - 2$ , $- 4 b - 2 = - 14$ gives $b = 3$ .
Check the constant: $16 - 4 + 3 = 15$ and $9 + 3 + 3 = 15$ .

(g \circ f) (x) = (a x + b)^{2} + (a x + b) + 3 = a^{2} x^{2} + (2 ab + a) x + b^{2} + b + 3 a^{2} = 4, 2 ab + a = - 14, b^{2} + b + 3 = 15 f (x) = 2 x - 4 or f (x) = - 2 x + 3

Section B · Extended response

Questions 7–9 · 45 marks

Question 7

17 marks

A quadratic with known vertex, a normal line and a chord

Functions
Calculus

(a)

1 mark

The axis of symmetry passes through the vertex.

x = - 2

(b)

2 marks

The vertex is $(- 2, - 5)$ .

h = - 2, k = - 5

(c)

2 marks

Put $x = 0$ in $f (x) = \frac{1}{4} (x + 2)^{2} - 5$ .

f (0) = \frac{1}{4} (4) - 5 = - 4

(d)

4 marks

The gradient of the curve at $P$ .
The normal has gradient $- 1$ and passes through $P (0, - 4)$ .

f^{'} (x) = \frac{1}{2} (x + 2) \Rightarrow f^{'} (0) = 1 y = - x - 4

(e)

8 marks

Find where the line meets the curve again.
At $x = - 8$ the $y$ -coordinate is $- (- 8) - 4 = 4$ , so $Q = (- 8, 4)$ .

\frac{1}{4} (x + 2)^{2} - 5 = - x - 4 \Rightarrow \frac{1}{4} x^{2} + 2 x = 0 \Rightarrow x (x + 8) = 0 x = 0 (the point P) or x = - 8 P Q = (- 8 - 0)^{2} + (4 + 4)^{2} = 128 = 82 \approx 11.3

Question 8

14 marks

Triangles with angles theta and 2 theta

Geometry and trigonometry

(a)

4 marks

Use $sin^{2} θ + cos^{2} θ = 1$ with $cos θ = \frac{2}{3}$ and $θ$ acute.

sin θ = 1 - \frac{4}{9} = \frac{5}{3} sin 2 θ = 2 sin θ cos θ = 2 \cdot \frac{5}{3} \cdot \frac{2}{3} = \frac{4 5}{9}

(b)

2 marks

Sine rule in triangle ABC with $B C = a$ opposite angle $2 θ$ and $A C = b$ opposite angle $θ$ .

\frac{b}{sin θ} = \frac{a}{sin 2 θ} \Rightarrow b = \frac{a sin θ}{2 sin θ cos θ} = \frac{a}{2 cos θ} = \frac{3 a}{4}

(c)

3 marks

BAD is a straight line, so $C A D = 18 0^{\circ} - 2 θ$ .

sin C A D = sin 2 θ = \frac{4 5}{9}

(d)

5 marks

Triangle DAC has angles $θ$ at D and $18 0^{\circ} - 2 θ$ at A, so the angle at C is also $θ$ and $A D = A C = b$ . Use area $= \frac{1}{2} ab sin C$ .

Area = \frac{1}{2} b^{2} sin C A D = \frac{1}{2} (\frac{3 a}{4})^{2} \cdot \frac{4 5}{9} = \frac{5}{8} a^{2}

Question 9

14 marks

The largest triangle inscribed in a circle

Geometry and trigonometry
Calculus

(a)

1 mark

Q lies on the circle $x^{2} + y^{2} = 9$ and in the first quadrant.

y = 9 - x^{2}

(b)

3 marks

The base QR has length $2 y$ and the height from P is $x + 3$ .

A = \frac{1}{2} (2 y) (x + 3) = (x + 3) 9 - x^{2}

(c)

4 marks

Product rule.

\frac{d A}{d x} = 9 - x^{2} - \frac{x ( x + 3 )}{9 - x ^{2}} = \frac{9 - x ^{2} - x ^{2} - 3 x}{9 - x ^{2}} \frac{d A}{d x} = \frac{9 - 3 x - 2 x ^{2}}{9 - x ^{2}}

(d)

6 marks

Set $\frac{d A}{d x} = 0$ .
Reject $x = - 3$ . For $x = \frac{3}{2}$ , $y = 9 - \frac{9}{4} = \frac{3 3}{2}$ , and $R$ is $(x, - y)$ .

2 x^{2} + 3 x - 9 = 0 \Rightarrow (2 x - 3) (x + 3) = 0 \Rightarrow x = \frac{3}{2} y_{R} = - \frac{3 3}{2} \approx - 2.60

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 algebra24 marks · 30%
Q2, Q4, Q5, Q6
Geometry and trigonometry24 marks · 30%
Q1, Q8, Q9
Calculus22 marks · 28%
Q7, Q9
Functions10 marks · 13%
Q3, Q7

Mentor insights

Traps to watch for

01
Arc length uses radians · Q1
Arc length is $r θ$ with $θ$ in radians.
02
Compare coefficients · Q6
Expand $(g \circ f) (x) = (a x + b)^{2} + (a x + b) + 3$ and match each power of $x$ .
03
Vertex form · Q7
$f (x) = \frac{1}{4} (x - h)^{2} + k$ has vertex $(h, k)$ .
04
Double angle · Q8
$sin 2 θ = 2 sin θ cos θ$ .
05
Express the area in one variable · Q9
Use $y = 9 - x^{2}$ so that $A$ depends on $x$ only.

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.

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Worked solutions

IB Maths AA SL · Paper 1

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2

Session: May 2023
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 2023 TZ2 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 · 35 marks

Question 1

6 marks

A circular sector: perimeter, angle and area

Geometry and trigonometry

(a)

2 marks

The sector is bounded by two radii and the arc PQR.

perimeter = 4 + 4 + 10 = 18 cm

(b)

2 marks

Arc length $= r θ$ .

10 = 4 θ \Rightarrow θ = 2.5

(c)

2 marks

Area of a sector.

A = \frac{1}{2} r^{2} θ = \frac{1}{2} (16) (2.5) = 20 cm^{2}

Question 2

6 marks

An arithmetic sequence in a probability distribution

Number and algebra

(a)

2 marks

Use $u_{4} = u_{1} + 3 d$ .

0.15 = 0.6 + 3 d \Rightarrow d = - 0.15

(b)

4 marks

The terms are $0.6, 0.45, 0.3, 0.15$ with sum $1.5$ .

\frac{0.6}{k} + \frac{0.45}{k} + \frac{0.3}{k} + \frac{0.15}{k} = \frac{1.5}{k} = 1 k = 1.5

Question 3

5 marks

f(x) = 1 - 1/(x - 2): asymptotes, intercepts and sketch

Functions

(a)

2 marks

The function is $f (x) = 1 - \frac{1}{x - 2}$ . The denominator is zero at $x = 2$ , and for large $∣ x ∣$ the fraction tends to 0.

x = 2, y = 1

(b)

2 marks

At $x = 0$ : $f (0) = 1 + \frac{1}{2}$ . At $f (x) = 0$ : $\frac{1}{x - 2} = 1$ .

(0, \frac{3}{2}), (3, 0)

(c)

1 mark

Question 4

5 marks

Finding P(B) from conditional probability and the union

Number and algebra

5 marks

Addition rule: $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ .
Conditional probability: $P (A \cap B) = P (A ∣ B) P (B)$ .
Combine them.

0.55 = 0.4 + P (B) - P (A \cap B) P (A \cap B) = 0.25 P (B) 0.55 = 0.4 + P (B) - 0.25 P (B) \Rightarrow 0.75 P (B) = 0.15 \Rightarrow P (B) = 0.2

Question 5

6 marks

Area under y = x/(x^2 + 2) equal to ln 3

Number and algebra

6 marks

Set the area equal to $ln 3$ .
Combine the logs.

\int_{0}^{c} \frac{x}{x ^{2} + 2} d x = [\frac{1}{2} ln (x^{2} + 2)]_{0}^{c} = \frac{1}{2} ln (c^{2} + 2) - \frac{1}{2} ln 2 = ln 3 ln \frac{c ^{2} + 2}{2} = 2 ln 3 = ln 9 \Rightarrow \frac{c ^{2} + 2}{2} = 9 \Rightarrow c^{2} = 16 c = 4

Question 6

7 marks

Two linear functions f with (g after f) given

Number and algebra

7 marks

Substitute $f$ into $g$ .
Compare with $4 x^{2} - 14 x + 15$ .
So $a = \pm 2$ . Use the middle equation to find $b$ : for $a = 2$ , $4 b + 2 = - 14$ gives $b = - 4$ . For $a = - 2$ , $- 4 b - 2 = - 14$ gives $b = 3$ .
Check the constant: $16 - 4 + 3 = 15$ and $9 + 3 + 3 = 15$ .

(g \circ f) (x) = (a x + b)^{2} + (a x + b) + 3 = a^{2} x^{2} + (2 ab + a) x + b^{2} + b + 3 a^{2} = 4, 2 ab + a = - 14, b^{2} + b + 3 = 15 f (x) = 2 x - 4 or f (x) = - 2 x + 3

Section B · Extended response

Questions 7–9 · 45 marks

Question 7

17 marks

A quadratic with known vertex, a normal line and a chord

Functions
Calculus

(a)

1 mark

The axis of symmetry passes through the vertex.

x = - 2

(b)

2 marks

The vertex is $(- 2, - 5)$ .

h = - 2, k = - 5

(c)

2 marks

Put $x = 0$ in $f (x) = \frac{1}{4} (x + 2)^{2} - 5$ .

f (0) = \frac{1}{4} (4) - 5 = - 4

(d)

4 marks

The gradient of the curve at $P$ .
The normal has gradient $- 1$ and passes through $P (0, - 4)$ .

f^{'} (x) = \frac{1}{2} (x + 2) \Rightarrow f^{'} (0) = 1 y = - x - 4

(e)

8 marks

Find where the line meets the curve again.
At $x = - 8$ the $y$ -coordinate is $- (- 8) - 4 = 4$ , so $Q = (- 8, 4)$ .

\frac{1}{4} (x + 2)^{2} - 5 = - x - 4 \Rightarrow \frac{1}{4} x^{2} + 2 x = 0 \Rightarrow x (x + 8) = 0 x = 0 (the point P) or x = - 8 P Q = (- 8 - 0)^{2} + (4 + 4)^{2} = 128 = 82 \approx 11.3

Question 8

14 marks

Triangles with angles theta and 2 theta

Geometry and trigonometry

(a)

4 marks

Use $sin^{2} θ + cos^{2} θ = 1$ with $cos θ = \frac{2}{3}$ and $θ$ acute.

sin θ = 1 - \frac{4}{9} = \frac{5}{3} sin 2 θ = 2 sin θ cos θ = 2 \cdot \frac{5}{3} \cdot \frac{2}{3} = \frac{4 5}{9}

(b)

2 marks

Sine rule in triangle ABC with $B C = a$ opposite angle $2 θ$ and $A C = b$ opposite angle $θ$ .

\frac{b}{sin θ} = \frac{a}{sin 2 θ} \Rightarrow b = \frac{a sin θ}{2 sin θ cos θ} = \frac{a}{2 cos θ} = \frac{3 a}{4}

(c)

3 marks

BAD is a straight line, so $C A D = 18 0^{\circ} - 2 θ$ .

sin C A D = sin 2 θ = \frac{4 5}{9}

(d)

5 marks

Triangle DAC has angles $θ$ at D and $18 0^{\circ} - 2 θ$ at A, so the angle at C is also $θ$ and $A D = A C = b$ . Use area $= \frac{1}{2} ab sin C$ .

Area = \frac{1}{2} b^{2} sin C A D = \frac{1}{2} (\frac{3 a}{4})^{2} \cdot \frac{4 5}{9} = \frac{5}{8} a^{2}

Question 9

14 marks

The largest triangle inscribed in a circle

Geometry and trigonometry
Calculus

(a)

1 mark

Q lies on the circle $x^{2} + y^{2} = 9$ and in the first quadrant.

y = 9 - x^{2}

(b)

3 marks

The base QR has length $2 y$ and the height from P is $x + 3$ .

A = \frac{1}{2} (2 y) (x + 3) = (x + 3) 9 - x^{2}

(c)

4 marks

Product rule.

\frac{d A}{d x} = 9 - x^{2} - \frac{x ( x + 3 )}{9 - x ^{2}} = \frac{9 - x ^{2} - x ^{2} - 3 x}{9 - x ^{2}} \frac{d A}{d x} = \frac{9 - 3 x - 2 x ^{2}}{9 - x ^{2}}

(d)

6 marks

Set $\frac{d A}{d x} = 0$ .
Reject $x = - 3$ . For $x = \frac{3}{2}$ , $y = 9 - \frac{9}{4} = \frac{3 3}{2}$ , and $R$ is $(x, - y)$ .

2 x^{2} + 3 x - 9 = 0 \Rightarrow (2 x - 3) (x + 3) = 0 \Rightarrow x = \frac{3}{2} y_{R} = - \frac{3 3}{2} \approx - 2.60

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 algebra24 marks · 30%
Q2, Q4, Q5, Q6
Geometry and trigonometry24 marks · 30%
Q1, Q8, Q9
Calculus22 marks · 28%
Q7, Q9
Functions10 marks · 13%
Q3, Q7

Mentor insights

Traps to watch for

01
Arc length uses radians · Q1
Arc length is $r θ$ with $θ$ in radians.
02
Compare coefficients · Q6
Expand $(g \circ f) (x) = (a x + b)^{2} + (a x + b) + 3$ and match each power of $x$ .
03
Vertex form · Q7
$f (x) = \frac{1}{4} (x - h)^{2} + k$ has vertex $(h, k)$ .
04
Double angle · Q8
$sin 2 θ = 2 sin θ cos θ$ .
05
Express the area in one variable · Q9
Use $y = 9 - x^{2}$ so that $A$ depends on $x$ only.

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