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IB Maths AA SL · Paper 1

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2

Session: November 2024
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 November 2024 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 1–6 · 33 marks

Question 1

6 marks

Mode, range, median and interquartile range

Statistics and probability

(a)

3 marks

(i) 27 appears twice and no value appears more often.

(ii) The largest minus the smallest.

(iii) The middle of ten values is the mean of the 5th and 6th, which are 33 and 37.

mode = 27 57 - 27 = 30 median = 35

(b)

3 marks

The lower quartile is the median of the lower five values, and the upper quartile is the median of the upper five.

Q_{1} = 31, Q_{3} = 46 IQR = 46 - 31 = 15

Question 2

5 marks

A sector with area 6 and angle 0.75

Geometry and trigonometry

(a)

3 marks

Use the area formula.

\frac{1}{2} r^{2} (0.75) = 6 \Rightarrow r^{2} = 16 r = 4 cm

(b)

2 marks

The perimeter is two radii plus the arc length.

2 r + r θ = 8 + 4 (0.75) = 11 cm

Question 3

6 marks

Two events: intersection and a conditional probability

Statistics and probability

(a)

3 marks

Rearrange the union formula.

P (A \cap B) = 0.45 + 0.65 - 0.8 = 0.3

(b)

3 marks

The probability of neither is $P (A^{'} \cap B^{'}) = 1 - 0.8 = 0.2$ , and $P (B^{'}) = 1 - 0.65 = 0.35$ .

P (A^{'} ∣ B^{'}) = \frac{0.2}{0.35} = \frac{4}{7}

Question 4

4 marks

Proving a difference of squares is a multiple of 10

Number and algebra

4 marks

Use the difference of squares with $a = 2 n + 5$ and $b = 2 n - 5$ .
Simplify each bracket.
Factor out 10.
Since $4 n$ is a positive integer, the expression is a multiple of 10 for every $n \in Z^{+}$ .

(2 n + 5)^{2} - (2 n - 5)^{2} = [(2 n + 5) - (2 n - 5)] [(2 n + 5) + (2 n - 5)] 10 \times 4 n = 40 n 40 n = 10 (4 n)

Question 5

6 marks

The exact value of cos(theta) from a triangle with angle 2 theta

Calculus

6 marks

Use the cosine rule for the angle $2 θ$ at $B$ , opposite $A C = 8$ .
Use the double angle formula.
The angle $θ$ is acute, so take the positive root.
So $p = 14$ and $q = 4$ .

cos 2 θ = \frac{3 ^{2} + 1 0 ^{2} - 8 ^{2}}{2 ( 3 ) ( 10 )} = \frac{45}{60} = \frac{3}{4} 2 cos^{2} θ - 1 = \frac{3}{4} \Rightarrow cos^{2} θ = \frac{7}{8} cos θ = \frac{7}{8} = \frac{7}{2 2} = \frac{14}{4}

Question 6

6 marks

An arithmetic sequence from u10 and S25

Number and algebra

6 marks

From $u_{10} = 14$ .
From $S_{25} = 200$ .
Subtract the first equation from the second.
Solve $u_{k} = 0$ .

u_{1} + 9 d = 14 \frac{25}{2} (2 u_{1} + 24 d) = 200 \Rightarrow u_{1} + 12 d = 8 3 d = - 6 \Rightarrow d = - 2, u_{1} = 14 + 18 = 32 32 - 2 (k - 1) = 0 \Rightarrow k - 1 = 16 k = 17

Section B · Extended response

Questions 7–9 · 47 marks

Question 7

16 marks

A line and a parabola: equation, area between them and a parallel tangent

Functions

(a)

2 marks

The line passes through $(6, 0)$ and $(0, 12)$ , so the gradient is $\frac{12 - 0}{0 - 6} = - 2$ and the $y$ -intercept is 12.

f (x) = - 2 x + 12

(b)

3 marks

The graph of $g$ passes through $(6, 0)$ .

- 36 + 6 b + 12 = 0 \Rightarrow 6 b = 24 b = 4

(c)

2 marks

Between $x = 0$ and $x = 6$ the parabola $g$ is above the line $f$ , so the area is $\int_{0}^{6} (g - f) d x$ .

g (x) - f (x) = - x^{2} + 4 x + 12 - (- 2 x + 12) = - x^{2} + 6 x Area = \int_{0}^{6} (6 x - x^{2}) d x

(d)

4 marks

Integrate.

[3 x^{2} - \frac{x ^{3}}{3}]_{0}^{6} = (108 - 72) - 0 36

(e)

5 marks

Differentiate $g$ .
The tangent is parallel to $f$ , which has gradient $- 2$ .
Find the height.

g^{'} (x) = - 2 x + 4 - 2 x + 4 = - 2 \Rightarrow x = 3 g (3) = - 9 + 12 + 12 = 15 P (3, 15)

Question 8

14 marks

A logarithmic function: inverse and transformations

Functions

(a)

3 marks

(i) Substitute.

(ii) Substitute.

f (2) = lo g_{2} (16) = 4 f (\frac{1}{8}) = lo g_{2} (1) = 0

(b)

4 marks

Swap $x$ and $y$ in $y = lo g_{2} (8 x)$ .
Solve for $y$ .

x = lo g_{2} (8 y) \Rightarrow 2^{x} = 8 y f^{- 1} (x) = \frac{2 ^{x}}{8} = 2^{x - 3}

(c)

1 mark

Substitute $x = 0$ into the inverse.

f^{- 1} (0) = 2^{- 3} = \frac{1}{8}

(d)

6 marks

Rewrite $f (4 x^{2}) = lo g_{2} (32 x^{2})$ using the log laws.
Start with $y = lo g_{2} x$ . First apply a vertical stretch with scale factor 2, which gives $y = 2 lo g_{2} x$ .
Then translate up by 5 units, which gives $y = 2 lo g_{2} x + 5$ .

lo g_{2} (32 x^{2}) = lo g_{2} 32 + lo g_{2} x^{2} = 5 + 2 lo g_{2} x

Question 9

17 marks

A hollow cylinder: surface area, volume and the maximum volume

Calculus

(a)

3 marks

The outer curved surface (radius $4 r$ ) and the inner curved surface (radius $r$ ).
The two ends are rings with area $π (4 r)^{2} - π r^{2} = 15 π r^{2}$ each.

2 π (4 r) h + 2 π r h = 8 π r h + 2 π r h = 10 π r h 2 \times 15 π r^{2} = 30 π r^{2} S = 30 π r^{2} + 10 π r h

(b)

6 marks

Set $S = 240 π$ and solve for $r h$ .
The volume is the large cylinder minus the small one.
Substitute.

30 π r^{2} + 10 π r h = 240 π \Rightarrow 3 r^{2} + r h = 24 \Rightarrow r h = 24 - 3 r^{2} V = π (4 r)^{2} h - π r^{2} h = 15 π r^{2} h = 15 π r (r h) V = 15 π r (24 - 3 r^{2}) = 360 π r - 45 π r^{3}

(c)

2 marks

Differentiate.

\frac{d V}{d r} = 360 π - 135 π r^{2}

(d)

3 marks

At the maximum $\frac{d V}{d r} = 0$ .
Compare with $r = p \frac{2}{3}$ , for which $r^{2} = \frac{2 p ^{2}}{3}$ .

135 r^{2} = 360 \Rightarrow r^{2} = \frac{8}{3} \Rightarrow r = 2 \frac{2}{3} \frac{2 p ^{2}}{3} = \frac{8}{3} \Rightarrow p^{2} = 4 \Rightarrow p = 2

(e)

3 marks

Substitute into $V = π r (360 - 45 r^{2})$ .
So $q = 480$ .

V = π \cdot 2 \frac{2}{3} (360 - 45 \cdot \frac{8}{3}) = 2 π \frac{2}{3} (360 - 120) V = 480 π \frac{2}{3}

Our analysis

Where the marks are

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

Functions30 marks · 38%
Q7, Q8
Calculus23 marks · 29%
Q5, Q9
Statistics and probability12 marks · 15%
Q1, Q3
Number and algebra10 marks · 13%
Q4, Q6
Geometry and trigonometry5 marks · 6%
Q2

Mentor insights

Traps to watch for

01
Ordered data · Q1
The data is already in order, so read off the values.
02
Union · Q3
$P (A \cup B) = P (A) + P (B) - P (A \cap B)$ .
03
Area between curves · Q7
Integrate the upper curve minus the lower curve.
04
Inverse · Q8
Swap $x$ and $y$ and make $y$ the subject.
05
Surface area · Q9
Outer curved, inner curved and two ring-shaped ends.

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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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: November 2024
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 November 2024 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 1–6 · 33 marks

Question 1

6 marks

Mode, range, median and interquartile range

Statistics and probability

(a)

3 marks

(i) 27 appears twice and no value appears more often.

(ii) The largest minus the smallest.

(iii) The middle of ten values is the mean of the 5th and 6th, which are 33 and 37.

mode = 27 57 - 27 = 30 median = 35

(b)

3 marks

The lower quartile is the median of the lower five values, and the upper quartile is the median of the upper five.

Q_{1} = 31, Q_{3} = 46 IQR = 46 - 31 = 15

Question 2

5 marks

A sector with area 6 and angle 0.75

Geometry and trigonometry

(a)

3 marks

Use the area formula.

\frac{1}{2} r^{2} (0.75) = 6 \Rightarrow r^{2} = 16 r = 4 cm

(b)

2 marks

The perimeter is two radii plus the arc length.

2 r + r θ = 8 + 4 (0.75) = 11 cm

Question 3

6 marks

Two events: intersection and a conditional probability

Statistics and probability

(a)

3 marks

Rearrange the union formula.

P (A \cap B) = 0.45 + 0.65 - 0.8 = 0.3

(b)

3 marks

The probability of neither is $P (A^{'} \cap B^{'}) = 1 - 0.8 = 0.2$ , and $P (B^{'}) = 1 - 0.65 = 0.35$ .

P (A^{'} ∣ B^{'}) = \frac{0.2}{0.35} = \frac{4}{7}

Question 4

4 marks

Proving a difference of squares is a multiple of 10

Number and algebra

4 marks

Use the difference of squares with $a = 2 n + 5$ and $b = 2 n - 5$ .
Simplify each bracket.
Factor out 10.
Since $4 n$ is a positive integer, the expression is a multiple of 10 for every $n \in Z^{+}$ .

(2 n + 5)^{2} - (2 n - 5)^{2} = [(2 n + 5) - (2 n - 5)] [(2 n + 5) + (2 n - 5)] 10 \times 4 n = 40 n 40 n = 10 (4 n)

Question 5

6 marks

The exact value of cos(theta) from a triangle with angle 2 theta

Calculus

6 marks

Use the cosine rule for the angle $2 θ$ at $B$ , opposite $A C = 8$ .
Use the double angle formula.
The angle $θ$ is acute, so take the positive root.
So $p = 14$ and $q = 4$ .

cos 2 θ = \frac{3 ^{2} + 1 0 ^{2} - 8 ^{2}}{2 ( 3 ) ( 10 )} = \frac{45}{60} = \frac{3}{4} 2 cos^{2} θ - 1 = \frac{3}{4} \Rightarrow cos^{2} θ = \frac{7}{8} cos θ = \frac{7}{8} = \frac{7}{2 2} = \frac{14}{4}

Question 6

6 marks

An arithmetic sequence from u10 and S25

Number and algebra

6 marks

From $u_{10} = 14$ .
From $S_{25} = 200$ .
Subtract the first equation from the second.
Solve $u_{k} = 0$ .

u_{1} + 9 d = 14 \frac{25}{2} (2 u_{1} + 24 d) = 200 \Rightarrow u_{1} + 12 d = 8 3 d = - 6 \Rightarrow d = - 2, u_{1} = 14 + 18 = 32 32 - 2 (k - 1) = 0 \Rightarrow k - 1 = 16 k = 17

Section B · Extended response

Questions 7–9 · 47 marks

Question 7

16 marks

A line and a parabola: equation, area between them and a parallel tangent

Functions

(a)

2 marks

The line passes through $(6, 0)$ and $(0, 12)$ , so the gradient is $\frac{12 - 0}{0 - 6} = - 2$ and the $y$ -intercept is 12.

f (x) = - 2 x + 12

(b)

3 marks

The graph of $g$ passes through $(6, 0)$ .

- 36 + 6 b + 12 = 0 \Rightarrow 6 b = 24 b = 4

(c)

2 marks

Between $x = 0$ and $x = 6$ the parabola $g$ is above the line $f$ , so the area is $\int_{0}^{6} (g - f) d x$ .

g (x) - f (x) = - x^{2} + 4 x + 12 - (- 2 x + 12) = - x^{2} + 6 x Area = \int_{0}^{6} (6 x - x^{2}) d x

(d)

4 marks

Integrate.

[3 x^{2} - \frac{x ^{3}}{3}]_{0}^{6} = (108 - 72) - 0 36

(e)

5 marks

Differentiate $g$ .
The tangent is parallel to $f$ , which has gradient $- 2$ .
Find the height.

g^{'} (x) = - 2 x + 4 - 2 x + 4 = - 2 \Rightarrow x = 3 g (3) = - 9 + 12 + 12 = 15 P (3, 15)

Question 8

14 marks

A logarithmic function: inverse and transformations

Functions

(a)

3 marks

(i) Substitute.

(ii) Substitute.

f (2) = lo g_{2} (16) = 4 f (\frac{1}{8}) = lo g_{2} (1) = 0

(b)

4 marks

Swap $x$ and $y$ in $y = lo g_{2} (8 x)$ .
Solve for $y$ .

x = lo g_{2} (8 y) \Rightarrow 2^{x} = 8 y f^{- 1} (x) = \frac{2 ^{x}}{8} = 2^{x - 3}

(c)

1 mark

Substitute $x = 0$ into the inverse.

f^{- 1} (0) = 2^{- 3} = \frac{1}{8}

(d)

6 marks

Rewrite $f (4 x^{2}) = lo g_{2} (32 x^{2})$ using the log laws.
Start with $y = lo g_{2} x$ . First apply a vertical stretch with scale factor 2, which gives $y = 2 lo g_{2} x$ .
Then translate up by 5 units, which gives $y = 2 lo g_{2} x + 5$ .

lo g_{2} (32 x^{2}) = lo g_{2} 32 + lo g_{2} x^{2} = 5 + 2 lo g_{2} x

Question 9

17 marks

A hollow cylinder: surface area, volume and the maximum volume

Calculus

(a)

3 marks

The outer curved surface (radius $4 r$ ) and the inner curved surface (radius $r$ ).
The two ends are rings with area $π (4 r)^{2} - π r^{2} = 15 π r^{2}$ each.

2 π (4 r) h + 2 π r h = 8 π r h + 2 π r h = 10 π r h 2 \times 15 π r^{2} = 30 π r^{2} S = 30 π r^{2} + 10 π r h

(b)

6 marks

Set $S = 240 π$ and solve for $r h$ .
The volume is the large cylinder minus the small one.
Substitute.

30 π r^{2} + 10 π r h = 240 π \Rightarrow 3 r^{2} + r h = 24 \Rightarrow r h = 24 - 3 r^{2} V = π (4 r)^{2} h - π r^{2} h = 15 π r^{2} h = 15 π r (r h) V = 15 π r (24 - 3 r^{2}) = 360 π r - 45 π r^{3}

(c)

2 marks

Differentiate.

\frac{d V}{d r} = 360 π - 135 π r^{2}

(d)

3 marks

At the maximum $\frac{d V}{d r} = 0$ .
Compare with $r = p \frac{2}{3}$ , for which $r^{2} = \frac{2 p ^{2}}{3}$ .

135 r^{2} = 360 \Rightarrow r^{2} = \frac{8}{3} \Rightarrow r = 2 \frac{2}{3} \frac{2 p ^{2}}{3} = \frac{8}{3} \Rightarrow p^{2} = 4 \Rightarrow p = 2

(e)

3 marks

Substitute into $V = π r (360 - 45 r^{2})$ .
So $q = 480$ .

V = π \cdot 2 \frac{2}{3} (360 - 45 \cdot \frac{8}{3}) = 2 π \frac{2}{3} (360 - 120) V = 480 π \frac{2}{3}

Our analysis

Where the marks are

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

Functions30 marks · 38%
Q7, Q8
Calculus23 marks · 29%
Q5, Q9
Statistics and probability12 marks · 15%
Q1, Q3
Number and algebra10 marks · 13%
Q4, Q6
Geometry and trigonometry5 marks · 6%
Q2

Mentor insights

Traps to watch for

01
Ordered data · Q1
The data is already in order, so read off the values.
02
Union · Q3
$P (A \cup B) = P (A) + P (B) - P (A \cap B)$ .
03
Area between curves · Q7
Integrate the upper curve minus the lower curve.
04
Inverse · Q8
Swap $x$ and $y$ and make $y$ the subject.
05
Surface area · Q9
Outer curved, inner curved and two ring-shaped ends.

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