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Worked solutions

IB Maths AA SL · Paper 2

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2

Session: May 2024
Marks: 80
Duration: 1 hour 30 min
Calculator: 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 2024 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 · 34 marks

Question 1

6 marks

Phone usage of ten students: median, IQR and an outlier

Statistics and probability

(a)

3 marks

The ten values are already in order. The median is the mean of the fifth and sixth values.
The lower quartile is the third value and the upper quartile is the eighth value.

median = \frac{4.4 + 4.5}{2} = 4.45 hours Q_{1} = 1.9, Q_{3} = 5.7 \Rightarrow IQR = 5.7 - 1.9 = 3.8 hours

(b)

3 marks

The upper fence is
Since $11.7 > 11.4$ , the value 11.7 is an outlier.

Q_{3} + 1.5 \times IQR = 5.7 + 1.5 (3.8) = 11.4

Question 2

6 marks

Area between 1 - x^2 and e^(2x) on [-1, 0]

Calculus

(a)

3 marks

Solve $1 - x^{2} = e^{2 x}$ for $- 1 \leq x \leq 0$ on a graph.

a \approx - 0.917, b = 0

(b)

3 marks

On the interval $f (x) = 1 - x^{2}$ is above $g (x) = e^{2 x}$ .

Area = \int_{- 0.9166}^{0} (1 - x^{2} - e^{2 x}) d x \approx 0.240

Question 3

5 marks

A regression line through the mean point: finding two missing values

Statistics and probability

(a)

1 mark

Substitute $\overset{x}{ˉ} = 7$ into the regression line.

\overset{y}{ˉ} = 2.1875 (7) + 0.6875 = 15.3125 + 0.6875 = 16

(b)

4 marks

The mean of the five $y$ values is 16, so they add to 80.
With $q = p + 3$ , $2 p + 3 = 37$ .

9 + 13 + p + q + 21 = 80 \Rightarrow p + q = 37 p = 17, q = 20

Question 4

6 marks

Loudness in decibels: intensity, loudness and thunder

Number and algebra

(a)

1 mark

$S_{2}$ has twice the intensity of $S_{1}$ .

I = 2 \times 1 0^{- 6}

(b)

2 marks

Substitute into the formula.

L = 10 lo g_{10} (2 \times 1 0^{- 6} \times 1 0^{12}) = 10 lo g_{10} (2 \times 1 0^{6}) \approx 63.0 decibels

(c)

3 marks

Set $L = 115$ and solve for $I$ .

115 = 10 lo g_{10} (I \times 1 0^{12}) \Rightarrow 11.5 = lo g_{10} (I \times 1 0^{12}) I \times 1 0^{12} = 1 0^{11.5} \Rightarrow I = 1 0^{- 0.5} \approx 0.316

Question 5

6 marks

A particle with velocity 1 + e^(-t) - e^(-sin 2t): velocity, maximum and acceleration

Calculus

(a)

1 mark

Substitute $t = 2$ (radians).

v (2) = 1 + e^{- 2} - e^{- s i n 4} \approx - 0.996 m s^{- 1}

(b)

2 marks

Solve $v^{'} (t) = 0$ on a graph. The maximum occurs at $t \approx 0.406$ .

v_{m a x} = v (0.4058) \approx 1.18 m s^{- 1}

(c)

3 marks

The velocity is zero at $t \approx 1.658$ and changes sign there.
Evaluate the derivative at that time.

a = v^{'} (1.658) \approx - 2.53 m s^{- 2}

Question 6

5 marks

The least n so that a binomial variable is at least 1 with probability above 0.99

Number and algebra

5 marks

Use the complement.
Take logarithms, remembering that $ln 0.75 < 0$ reverses the inequality.
Check: for $n = 16$ , $P (X \geq 1) \approx 0.98998 < 0.99$ , and for $n = 17$ , $P (X \geq 1) \approx 0.99248 > 0.99$ .

1 - 0.7 5^{n} > 0.99 \Rightarrow 0.7 5^{n} < 0.01 n > \frac{ln 0.01}{ln 0.75} \approx 16.008 n = 17

Section B · Extended response

Questions 7–9 · 46 marks

Question 7

14 marks

Carp lengths: normal probabilities, a binomial count and a recorded value

Statistics and probability

(a)

2 marks

$45.6 - 4.2 = 41.4$ is the lower end, so the upper end is

k = 45.6 + 4.2 = 49.8

(b)

2 marks

Use the normal distribution with mean 45.6 and standard deviation 4.2.

P (L > 48) \approx 0.284

(c)

2 marks

Use the inverse normal with 1% in the lower tail.

P (L < x) = 0.01 \Rightarrow x \approx 35.8 cm

(d)

5 marks

(i) The probability for one carp.

(ii) Let $X$ be the number in the interval, $X \sim B (100, 0.9021)$ .

P (40 < L < 56) \approx 0.9021 expected number = 100 \times 0.9021 \approx 90.2 P (X = 95) \approx 0.0381

(e)

3 marks

A length recorded as 45.6 means a true length between 45.55 and 45.65.

P (45.55 \leq L < 45.65) \approx 0.00950

Question 8

15 marks

A window made of a rectangle and a semicircle: area, light and optimisation

Calculus

(a)

2 marks

The rectangle has area $2 x h$ and the semicircle has area $\frac{1}{2} π x^{2}$ .

A = 2 x h + \frac{1}{2} π x^{2}

(b)

2 marks

The perimeter is the base $2 x$ , the two sides $2 h$ and the arc of the semicircle $π x$ .

P = 2 x + 2 h + π x = 10 \Rightarrow h = \frac{1}{2} (10 - 2 x - π x)

(c)

4 marks

The rectangle lets in 3 units per square metre and the semicircle 1 unit per square metre.
Substitute $h$ from part (b).

L = 3 (2 x h) + \frac{1}{2} π x^{2} = 6 x h + \frac{1}{2} π x^{2} L = 6 x \cdot \frac{1}{2} (10 - 2 x - π x) + \frac{1}{2} π x^{2} = 30 x - 6 x^{2} - 3 π x^{2} + \frac{1}{2} π x^{2} L = 30 x - 6 x^{2} - \frac{5 π}{2} x^{2}

(d)

7 marks

(i) Differentiate.

(ii) Set the derivative to zero.

$L$ is a quadratic in $x$ with a negative coefficient of $x^{2}$ , and $\frac{d ^{2} L}{d x ^{2}} = - (12 + 5 π) < 0$ , so this value of $x$ gives a maximum.

(iii) Substitute into $h = \frac{1}{2} (10 - 2 x - π x)$ .

\frac{d L}{d x} = 30 - 12 x - 5 π x = 30 - (12 + 5 π) x x = \frac{30}{12 + 5 π} \approx 1.08 m h \approx 2.22 m

Question 9

17 marks

A rotating sprinkler beside a path: chord, timing and a turtle

Geometry and trigonometry

(a)

3 marks

Let $M$ be the midpoint of $[A B]$ . Then $S M = 14$ , $S A = 20$ and $S M A = 9 0^{\circ}$ .

A M = 2 0^{2} - 1 4^{2} = 204 \approx 14.283 A B = 2 A M = 2204 \approx 28.57 m

(b)

1 mark

One revolution is $2 π$ radians in 16 seconds.

\frac{2 π}{16} = \frac{π}{8} radians per second

(c)

4 marks

The angle $A S B$ is twice $A S M$ .
The time to turn through that angle at $\frac{π}{8}$ radians per second is

A S M = cos^{- 1} \frac{14}{20} \approx 0.7954, A S B \approx 1.5908 T = \frac{1.5908}{π /8} \approx 4.05 s

(d)

1 mark

The water has turned through $\frac{π}{8}$ radians every second since $t = 0$ .

α = \frac{π t}{8}

(e)

3 marks

In triangle $A S D$ , the angle at $D$ is $π - α - β = π - 0.7754 - α \approx 2.37 - α$ .
Use the sine rule: $\frac{d}{sin α} = \frac{20}{sin A D S}$ .

d = \frac{20 sin α}{sin ( 2.37 - α )} = \frac{20 sin \frac{π t}{8}}{sin ( 2.37 - \frac{π t}{8} )}

(f)

1 mark

At $t = 0$ , $g (0) = 18$ .

18 m

(g)

4 marks

The distance between the turtle and the point $D$ is the difference between the two distances.
The water reaches the turtle when $w = 0$ . The first solution is
Then the turtle is

w = 0.05 t^{2} + 1.1 t + 18 - \frac{20 sin \frac{π t}{8}}{sin ( 2.37 - \frac{π t}{8} )} t \approx 3.35 s g (3.349) = 0.05 (3.349)^{2} + 1.1 (3.349) + 18 \approx 22.2 m south of A

Our analysis

Where the marks are

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

Calculus27 marks · 34%
Q2, Q5, Q8
Statistics and probability25 marks · 31%
Q1, Q3, Q7
Geometry and trigonometry17 marks · 21%
Q9
Number and algebra11 marks · 14%
Q4, Q6

Mentor insights

Traps to watch for

01
Outlier rule · Q1
A value above $Q_{3} + 1.5 \times IQR$ is an outlier.
02
Intersections first · Q2
The limits of the area are the intersection points. Take upper minus lower.
03
One standard deviation · Q7
Within one standard deviation means from $μ - σ$ to $μ + σ$ .
04
Perimeter and area · Q8
The rectangle has base $2 x$ , so the semicircle has diameter $2 x$ and radius $x$ . The arc of the semicircle has length $π x$ .
05
Chord from Pythagoras · Q9
The perpendicular from the sprinkler to the path meets $[A B]$ at its midpoint $M$ .

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 SL · Paper 2

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2

Session: May 2024
Marks: 80
Duration: 1 hour 30 min
Calculator: 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 2024 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 · 34 marks

Question 1

6 marks

Phone usage of ten students: median, IQR and an outlier

Statistics and probability

(a)

3 marks

The ten values are already in order. The median is the mean of the fifth and sixth values.
The lower quartile is the third value and the upper quartile is the eighth value.

median = \frac{4.4 + 4.5}{2} = 4.45 hours Q_{1} = 1.9, Q_{3} = 5.7 \Rightarrow IQR = 5.7 - 1.9 = 3.8 hours

(b)

3 marks

The upper fence is
Since $11.7 > 11.4$ , the value 11.7 is an outlier.

Q_{3} + 1.5 \times IQR = 5.7 + 1.5 (3.8) = 11.4

Question 2

6 marks

Area between 1 - x^2 and e^(2x) on [-1, 0]

Calculus

(a)

3 marks

Solve $1 - x^{2} = e^{2 x}$ for $- 1 \leq x \leq 0$ on a graph.

a \approx - 0.917, b = 0

(b)

3 marks

On the interval $f (x) = 1 - x^{2}$ is above $g (x) = e^{2 x}$ .

Area = \int_{- 0.9166}^{0} (1 - x^{2} - e^{2 x}) d x \approx 0.240

Question 3

5 marks

A regression line through the mean point: finding two missing values

Statistics and probability

(a)

1 mark

Substitute $\overset{x}{ˉ} = 7$ into the regression line.

\overset{y}{ˉ} = 2.1875 (7) + 0.6875 = 15.3125 + 0.6875 = 16

(b)

4 marks

The mean of the five $y$ values is 16, so they add to 80.
With $q = p + 3$ , $2 p + 3 = 37$ .

9 + 13 + p + q + 21 = 80 \Rightarrow p + q = 37 p = 17, q = 20

Question 4

6 marks

Loudness in decibels: intensity, loudness and thunder

Number and algebra

(a)

1 mark

$S_{2}$ has twice the intensity of $S_{1}$ .

I = 2 \times 1 0^{- 6}

(b)

2 marks

Substitute into the formula.

L = 10 lo g_{10} (2 \times 1 0^{- 6} \times 1 0^{12}) = 10 lo g_{10} (2 \times 1 0^{6}) \approx 63.0 decibels

(c)

3 marks

Set $L = 115$ and solve for $I$ .

115 = 10 lo g_{10} (I \times 1 0^{12}) \Rightarrow 11.5 = lo g_{10} (I \times 1 0^{12}) I \times 1 0^{12} = 1 0^{11.5} \Rightarrow I = 1 0^{- 0.5} \approx 0.316

Question 5

6 marks

A particle with velocity 1 + e^(-t) - e^(-sin 2t): velocity, maximum and acceleration

Calculus

(a)

1 mark

Substitute $t = 2$ (radians).

v (2) = 1 + e^{- 2} - e^{- s i n 4} \approx - 0.996 m s^{- 1}

(b)

2 marks

Solve $v^{'} (t) = 0$ on a graph. The maximum occurs at $t \approx 0.406$ .

v_{m a x} = v (0.4058) \approx 1.18 m s^{- 1}

(c)

3 marks

The velocity is zero at $t \approx 1.658$ and changes sign there.
Evaluate the derivative at that time.

a = v^{'} (1.658) \approx - 2.53 m s^{- 2}

Question 6

5 marks

The least n so that a binomial variable is at least 1 with probability above 0.99

Number and algebra

5 marks

Use the complement.
Take logarithms, remembering that $ln 0.75 < 0$ reverses the inequality.
Check: for $n = 16$ , $P (X \geq 1) \approx 0.98998 < 0.99$ , and for $n = 17$ , $P (X \geq 1) \approx 0.99248 > 0.99$ .

1 - 0.7 5^{n} > 0.99 \Rightarrow 0.7 5^{n} < 0.01 n > \frac{ln 0.01}{ln 0.75} \approx 16.008 n = 17

Section B · Extended response

Questions 7–9 · 46 marks

Question 7

14 marks

Carp lengths: normal probabilities, a binomial count and a recorded value

Statistics and probability

(a)

2 marks

$45.6 - 4.2 = 41.4$ is the lower end, so the upper end is

k = 45.6 + 4.2 = 49.8

(b)

2 marks

Use the normal distribution with mean 45.6 and standard deviation 4.2.

P (L > 48) \approx 0.284

(c)

2 marks

Use the inverse normal with 1% in the lower tail.

P (L < x) = 0.01 \Rightarrow x \approx 35.8 cm

(d)

5 marks

(i) The probability for one carp.

(ii) Let $X$ be the number in the interval, $X \sim B (100, 0.9021)$ .

P (40 < L < 56) \approx 0.9021 expected number = 100 \times 0.9021 \approx 90.2 P (X = 95) \approx 0.0381

(e)

3 marks

A length recorded as 45.6 means a true length between 45.55 and 45.65.

P (45.55 \leq L < 45.65) \approx 0.00950

Question 8

15 marks

A window made of a rectangle and a semicircle: area, light and optimisation

Calculus

(a)

2 marks

The rectangle has area $2 x h$ and the semicircle has area $\frac{1}{2} π x^{2}$ .

A = 2 x h + \frac{1}{2} π x^{2}

(b)

2 marks

The perimeter is the base $2 x$ , the two sides $2 h$ and the arc of the semicircle $π x$ .

P = 2 x + 2 h + π x = 10 \Rightarrow h = \frac{1}{2} (10 - 2 x - π x)

(c)

4 marks

The rectangle lets in 3 units per square metre and the semicircle 1 unit per square metre.
Substitute $h$ from part (b).

L = 3 (2 x h) + \frac{1}{2} π x^{2} = 6 x h + \frac{1}{2} π x^{2} L = 6 x \cdot \frac{1}{2} (10 - 2 x - π x) + \frac{1}{2} π x^{2} = 30 x - 6 x^{2} - 3 π x^{2} + \frac{1}{2} π x^{2} L = 30 x - 6 x^{2} - \frac{5 π}{2} x^{2}

(d)

7 marks

(i) Differentiate.

(ii) Set the derivative to zero.

$L$ is a quadratic in $x$ with a negative coefficient of $x^{2}$ , and $\frac{d ^{2} L}{d x ^{2}} = - (12 + 5 π) < 0$ , so this value of $x$ gives a maximum.

(iii) Substitute into $h = \frac{1}{2} (10 - 2 x - π x)$ .

\frac{d L}{d x} = 30 - 12 x - 5 π x = 30 - (12 + 5 π) x x = \frac{30}{12 + 5 π} \approx 1.08 m h \approx 2.22 m

Question 9

17 marks

A rotating sprinkler beside a path: chord, timing and a turtle

Geometry and trigonometry

(a)

3 marks

Let $M$ be the midpoint of $[A B]$ . Then $S M = 14$ , $S A = 20$ and $S M A = 9 0^{\circ}$ .

A M = 2 0^{2} - 1 4^{2} = 204 \approx 14.283 A B = 2 A M = 2204 \approx 28.57 m

(b)

1 mark

One revolution is $2 π$ radians in 16 seconds.

\frac{2 π}{16} = \frac{π}{8} radians per second

(c)

4 marks

The angle $A S B$ is twice $A S M$ .
The time to turn through that angle at $\frac{π}{8}$ radians per second is

A S M = cos^{- 1} \frac{14}{20} \approx 0.7954, A S B \approx 1.5908 T = \frac{1.5908}{π /8} \approx 4.05 s

(d)

1 mark

The water has turned through $\frac{π}{8}$ radians every second since $t = 0$ .

α = \frac{π t}{8}

(e)

3 marks

In triangle $A S D$ , the angle at $D$ is $π - α - β = π - 0.7754 - α \approx 2.37 - α$ .
Use the sine rule: $\frac{d}{sin α} = \frac{20}{sin A D S}$ .

d = \frac{20 sin α}{sin ( 2.37 - α )} = \frac{20 sin \frac{π t}{8}}{sin ( 2.37 - \frac{π t}{8} )}

(f)

1 mark

At $t = 0$ , $g (0) = 18$ .

18 m

(g)

4 marks

The distance between the turtle and the point $D$ is the difference between the two distances.
The water reaches the turtle when $w = 0$ . The first solution is
Then the turtle is

w = 0.05 t^{2} + 1.1 t + 18 - \frac{20 sin \frac{π t}{8}}{sin ( 2.37 - \frac{π t}{8} )} t \approx 3.35 s g (3.349) = 0.05 (3.349)^{2} + 1.1 (3.349) + 18 \approx 22.2 m south of A

Our analysis

Where the marks are

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

Calculus27 marks · 34%
Q2, Q5, Q8
Statistics and probability25 marks · 31%
Q1, Q3, Q7
Geometry and trigonometry17 marks · 21%
Q9
Number and algebra11 marks · 14%
Q4, Q6

Mentor insights

Traps to watch for

01
Outlier rule · Q1
A value above $Q_{3} + 1.5 \times IQR$ is an outlier.
02
Intersections first · Q2
The limits of the area are the intersection points. Take upper minus lower.
03
One standard deviation · Q7
Within one standard deviation means from $μ - σ$ to $μ + σ$ .
04
Perimeter and area · Q8
The rectangle has base $2 x$ , so the semicircle has diameter $2 x$ and radius $x$ . The arc of the semicircle has length $π x$ .
05
Chord from Pythagoras · Q9
The perpendicular from the sprinkler to the path meets $[A B]$ at its midpoint $M$ .

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