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IB Maths AA SL · Paper 2

Mathematics: Analysis & Approaches, Standard Level

Session: November 2022
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 9

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2022 Global paper, which your school or the IB can provide.

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Hide our working until you've had a go.

Section A · Short response

Questions 1–6 · 35 marks

Question 1

5 marks

Maths and Science scores: regression line and prediction

Statistics and probability

(a)

2 marks

Enter the data into the calculator and do linear regression of $y$ on $x$ .

a = 1.01, b = 2.45

(b)

1 mark

Read the correlation coefficient from the same regression output.

r = 0.981

(c)

2 marks

Substitute $x = 78$ into the regression line.

y = 1.012 (78) + 2.452 = 81.4 \approx 81

Question 2

6 marks

A triangle with two sides and a non-included angle

Geometry and trigonometry

(a)

3 marks

Use the sine rule.

\frac{sin B A C}{17} = \frac{sin 11 3 ^{\circ}}{24} \Rightarrow B A C = 40. 7^{\circ}

(b)

3 marks

The third angle is $18 0^{\circ} - 11 3^{\circ} - 40. 7^{\circ} = 26. 3^{\circ}$ . Use the sine rule again.

\frac{A B}{sin 26. 3 ^{\circ}} = \frac{24}{sin 11 3 ^{\circ}} \Rightarrow A B = 11.6

Question 3

6 marks

A function with an exponential inside a logarithm

Calculus

(a)

2 marks

Find the maximum of $f$ on $0 \leq x \leq 2$ with a graph or a derivative.

A = (0.709, 0.640)

(b)

1 mark

Solve $f (x) = 0$ for $x > 0$ .

x_{B} = 1.10

(c)

3 marks

Use absolute values so that the area below the axis is counted as positive.

Area = \int_{0}^{2} ∣ f (x) ∣ d x = \int_{0}^{1.099} f (x) d x - \int_{1.099}^{2} f (x) d x = 0.436 + 4.18 \approx 4.61

Question 4

5 marks

Geometric sequence: smallest n for a given sum

Number and algebra

5 marks

The ratio satisfies $r^{3} = \frac{86.4}{50} = 1.728$ .
The sum of the first $n$ terms:
Set $S_{n} > 33500$ .
So the smallest $n$ is 27.

r = 1.2 S_{n} = \frac{50 ( 1. 2 ^{n} - 1 )}{1.2 - 1} = 250 (1. 2^{n} - 1) 1. 2^{n} - 1 > 134 \Rightarrow 1. 2^{n} > 135 \Rightarrow n > \frac{ln 135}{ln 1.2} = 26.9

Mentor tip

Check: $S_{26} = 28369 < 33500$ and $S_{27} = 34093 > 33500$ .

Question 5

7 marks

Town population: an exponential decay model

Number and algebra

7 marks

A decrease of 11% over 8 years means $P (8) = 0.89 P (0)$ .
2041 is 27 years after 2014.

e^{8 k} = 0.89 \Rightarrow k = \frac{ln 0.89}{8} = - 0.01457 P (27) = 15000 e^{27 k} = 15000 (0.89)^{27/8} \approx 10100

Mentor tip

Keep $k$ unrounded in the calculator, or use $0.8 9^{27/8}$ directly.

Question 6

6 marks

Coefficient of x^4 in (ax + 1)^9 divided by 21x^2

Number and algebra

6 marks

The term in $x^{4}$ after dividing by $x^{2}$ comes from the $x^{6}$ term of $(a x + 1)^{9}$ .
Divide by $21 x^{2}$ .
Set this equal to the given coefficient $\frac{8}{7} a^{5}$ .

(6 9) (a x)^{6} (1)^{3} = 84 a^{6} x^{6} \frac{84 a ^{6} x ^{6}}{21 x ^{2}} = 4 a^{6} x^{4} 4 a^{6} = \frac{8}{7} a^{5} \Rightarrow a = \frac{2}{7} (a \neq = 0)

Section B · Extended response

Questions 7–9 · 45 marks

Question 7

16 marks

A particle moving in a straight line: displacement, velocity and distance

Calculus

(a)

2 marks

Put $t = 0$ .

s (0) = 3 (2) cos 0 = 6 m

(b)

2 marks

Differentiate with the product rule.

v = s^{'} (t) = 3 cos t - 3 (t + 2) sin t v (2) = 3 cos 2 - 12 sin 2 \approx - 12.2 m s^{- 1}

(c)

5 marks

Find where $s = 0$ or $v = 0$ on a graph: $v = 0$ at $t = 0.395, 3.33, 6.40$ and $s = 0$ at $t = 1.57, 4.71$ .
Check the sign of $s v$ on each interval.

0 < t < 0.395, 1.57 < t < 3.33, 4.71 < t < 6.40

(d)

4 marks

The acceleration is $a (t) = - 6 sin t - 3 (t + 2) cos t$ . Solve $a (t) = 0$ on a graph.

b = 2.03, c = 4.99

(e)

3 marks

The velocity changes sign at $t = 3.33$ , which lies between $b$ and $c$ . The distance is the integral of the speed.

\int_{2.03}^{4.99} ∣ v (t) ∣ d t \approx 31.8 m

Question 8

13 marks

A circle inscribed in a sector: kite area, a shaded corner and a special angle

Geometry and trigonometry

(a)

4 marks

The tangent AD is perpendicular to the radius OD, and AO bisects the angle $2 α$ , so $O A D = α$ .
The kite is two right triangles, each with area $\frac{1}{2} \cdot A D \cdot r$ .

tan α = \frac{r}{A D} \Rightarrow A D = \frac{r}{tan α} Area (A D O E) = 2 \cdot \frac{1}{2} \cdot \frac{r}{tan α} \cdot r = \frac{r ^{2}}{tan α}

(b)

5 marks

The angles of the kite add to $2 π$ , with two right angles at D and E and $2 α$ at A.
R is the kite minus the sector of the circle with angle $π - 2 α$ .

D O E = 2 π - \frac{π}{2} - \frac{π}{2} - 2 α = π - 2 α Area (R) = \frac{r ^{2}}{tan α} - \frac{1}{2} r^{2} (π - 2 α) = r^{2} (\frac{1}{tan α} + α - \frac{π}{2})

(c)

4 marks

Set the area of R equal to $π r^{2}$ and cancel $r^{2}$ .
Solve on a graph for $0 < α < \frac{π}{2}$ .

\frac{1}{tan α} + α - \frac{π}{2} = π \Rightarrow \frac{1}{tan α} + α = \frac{3 π}{2} α = 0.219

Question 9

16 marks

Hours worked per week: normal distribution and conditional probability

Statistics and probability

(a)

2 marks

Use $T \sim N (42, 10. 7^{2})$ .

P (T > 40) = 0.574

(b)

3 marks

The first three employees each work 40 hours or fewer, with probability $1 - 0.5741 = 0.4259$ , and the fourth works more.

0.425 9^{3} \times 0.5741 \approx 0.0443

(c)

7 marks

The probability that someone who works more than 40 hours works less than 55 is
For ten employees, let $X \sim B (10, 0.8046)$ be the number who work less than 55 hours.

P (T < 55 ∣ T > 40) = \frac{P ( 40 < T < 55 )}{P ( T > 40 )} = \frac{0.4619}{0.5741} = 0.805 P (X = 5) = (5 10) (0.8046)^{5} (0.1954)^{5} \approx 0.0242

(d)

4 marks

The two tails add to $1 - 0.904 = 0.096$ . The upper tail is twice the lower, so $P (T < a) = \frac{0.096}{3} = 0.032$ .
Use the inverse normal.
The maximum for a part-time employee is about 22.2 hours per week.

a = 42 + 10.7 (- 1.852) \approx 22.2

Our analysis

Where the marks are

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

Calculus22 marks · 28%
Q3, Q7
Statistics and probability21 marks · 26%
Q1, Q9
Geometry and trigonometry19 marks · 24%
Q2, Q8
Number and algebra18 marks · 23%
Q4, Q5, Q6

Mentor insights

Traps to watch for

01
Sine rule first, then the third angle · Q2
You know a side and its opposite angle (AC and angle B), so use the sine rule for angle A, then find angle C.
02
Find k from the 11% fall · Q5
A fall of 11% over eight years means the population is $0.89$ times as large: $e^{8 k} = 0.89$ .
03
Moving away means s and v have the same sign · Q7
The particle moves away from P when $s v > 0$ , and towards P when $s v < 0$ .
04
Split the kite into two right triangles · Q8
Radii to the tangent points meet the tangents at right angles, so ADOE is two congruent right triangles.
05
Conditional on working over 40 · Q9
In (c) the sample space is only employees who work more than 40 hours. Divide by $P (T > 40)$ .

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 SL · Paper 2

Mathematics: Analysis & Approaches, Standard Level

Session: November 2022
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 November 2022 Global 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

5 marks

Maths and Science scores: regression line and prediction

Statistics and probability

(a)

2 marks

Enter the data into the calculator and do linear regression of $y$ on $x$ .

a = 1.01, b = 2.45

(b)

1 mark

Read the correlation coefficient from the same regression output.

r = 0.981

(c)

2 marks

Substitute $x = 78$ into the regression line.

y = 1.012 (78) + 2.452 = 81.4 \approx 81

Question 2

6 marks

A triangle with two sides and a non-included angle

Geometry and trigonometry

(a)

3 marks

Use the sine rule.

\frac{sin B A C}{17} = \frac{sin 11 3 ^{\circ}}{24} \Rightarrow B A C = 40. 7^{\circ}

(b)

3 marks

The third angle is $18 0^{\circ} - 11 3^{\circ} - 40. 7^{\circ} = 26. 3^{\circ}$ . Use the sine rule again.

\frac{A B}{sin 26. 3 ^{\circ}} = \frac{24}{sin 11 3 ^{\circ}} \Rightarrow A B = 11.6

Question 3

6 marks

A function with an exponential inside a logarithm

Calculus

(a)

2 marks

Find the maximum of $f$ on $0 \leq x \leq 2$ with a graph or a derivative.

A = (0.709, 0.640)

(b)

1 mark

Solve $f (x) = 0$ for $x > 0$ .

x_{B} = 1.10

(c)

3 marks

Use absolute values so that the area below the axis is counted as positive.

Area = \int_{0}^{2} ∣ f (x) ∣ d x = \int_{0}^{1.099} f (x) d x - \int_{1.099}^{2} f (x) d x = 0.436 + 4.18 \approx 4.61

Question 4

5 marks

Geometric sequence: smallest n for a given sum

Number and algebra

5 marks

The ratio satisfies $r^{3} = \frac{86.4}{50} = 1.728$ .
The sum of the first $n$ terms:
Set $S_{n} > 33500$ .
So the smallest $n$ is 27.

r = 1.2 S_{n} = \frac{50 ( 1. 2 ^{n} - 1 )}{1.2 - 1} = 250 (1. 2^{n} - 1) 1. 2^{n} - 1 > 134 \Rightarrow 1. 2^{n} > 135 \Rightarrow n > \frac{ln 135}{ln 1.2} = 26.9

Mentor tip

Check: $S_{26} = 28369 < 33500$ and $S_{27} = 34093 > 33500$ .

Question 5

7 marks

Town population: an exponential decay model

Number and algebra

7 marks

A decrease of 11% over 8 years means $P (8) = 0.89 P (0)$ .
2041 is 27 years after 2014.

e^{8 k} = 0.89 \Rightarrow k = \frac{ln 0.89}{8} = - 0.01457 P (27) = 15000 e^{27 k} = 15000 (0.89)^{27/8} \approx 10100

Mentor tip

Keep $k$ unrounded in the calculator, or use $0.8 9^{27/8}$ directly.

Question 6

6 marks

Coefficient of x^4 in (ax + 1)^9 divided by 21x^2

Number and algebra

6 marks

The term in $x^{4}$ after dividing by $x^{2}$ comes from the $x^{6}$ term of $(a x + 1)^{9}$ .
Divide by $21 x^{2}$ .
Set this equal to the given coefficient $\frac{8}{7} a^{5}$ .

(6 9) (a x)^{6} (1)^{3} = 84 a^{6} x^{6} \frac{84 a ^{6} x ^{6}}{21 x ^{2}} = 4 a^{6} x^{4} 4 a^{6} = \frac{8}{7} a^{5} \Rightarrow a = \frac{2}{7} (a \neq = 0)

Section B · Extended response

Questions 7–9 · 45 marks

Question 7

16 marks

A particle moving in a straight line: displacement, velocity and distance

Calculus

(a)

2 marks

Put $t = 0$ .

s (0) = 3 (2) cos 0 = 6 m

(b)

2 marks

Differentiate with the product rule.

v = s^{'} (t) = 3 cos t - 3 (t + 2) sin t v (2) = 3 cos 2 - 12 sin 2 \approx - 12.2 m s^{- 1}

(c)

5 marks

Find where $s = 0$ or $v = 0$ on a graph: $v = 0$ at $t = 0.395, 3.33, 6.40$ and $s = 0$ at $t = 1.57, 4.71$ .
Check the sign of $s v$ on each interval.

0 < t < 0.395, 1.57 < t < 3.33, 4.71 < t < 6.40

(d)

4 marks

The acceleration is $a (t) = - 6 sin t - 3 (t + 2) cos t$ . Solve $a (t) = 0$ on a graph.

b = 2.03, c = 4.99

(e)

3 marks

The velocity changes sign at $t = 3.33$ , which lies between $b$ and $c$ . The distance is the integral of the speed.

\int_{2.03}^{4.99} ∣ v (t) ∣ d t \approx 31.8 m

Question 8

13 marks

A circle inscribed in a sector: kite area, a shaded corner and a special angle

Geometry and trigonometry

(a)

4 marks

The tangent AD is perpendicular to the radius OD, and AO bisects the angle $2 α$ , so $O A D = α$ .
The kite is two right triangles, each with area $\frac{1}{2} \cdot A D \cdot r$ .

tan α = \frac{r}{A D} \Rightarrow A D = \frac{r}{tan α} Area (A D O E) = 2 \cdot \frac{1}{2} \cdot \frac{r}{tan α} \cdot r = \frac{r ^{2}}{tan α}

(b)

5 marks

The angles of the kite add to $2 π$ , with two right angles at D and E and $2 α$ at A.
R is the kite minus the sector of the circle with angle $π - 2 α$ .

D O E = 2 π - \frac{π}{2} - \frac{π}{2} - 2 α = π - 2 α Area (R) = \frac{r ^{2}}{tan α} - \frac{1}{2} r^{2} (π - 2 α) = r^{2} (\frac{1}{tan α} + α - \frac{π}{2})

(c)

4 marks

Set the area of R equal to $π r^{2}$ and cancel $r^{2}$ .
Solve on a graph for $0 < α < \frac{π}{2}$ .

\frac{1}{tan α} + α - \frac{π}{2} = π \Rightarrow \frac{1}{tan α} + α = \frac{3 π}{2} α = 0.219

Question 9

16 marks

Hours worked per week: normal distribution and conditional probability

Statistics and probability

(a)

2 marks

Use $T \sim N (42, 10. 7^{2})$ .

P (T > 40) = 0.574

(b)

3 marks

The first three employees each work 40 hours or fewer, with probability $1 - 0.5741 = 0.4259$ , and the fourth works more.

0.425 9^{3} \times 0.5741 \approx 0.0443

(c)

7 marks

The probability that someone who works more than 40 hours works less than 55 is
For ten employees, let $X \sim B (10, 0.8046)$ be the number who work less than 55 hours.

P (T < 55 ∣ T > 40) = \frac{P ( 40 < T < 55 )}{P ( T > 40 )} = \frac{0.4619}{0.5741} = 0.805 P (X = 5) = (5 10) (0.8046)^{5} (0.1954)^{5} \approx 0.0242

(d)

4 marks

The two tails add to $1 - 0.904 = 0.096$ . The upper tail is twice the lower, so $P (T < a) = \frac{0.096}{3} = 0.032$ .
Use the inverse normal.
The maximum for a part-time employee is about 22.2 hours per week.

a = 42 + 10.7 (- 1.852) \approx 22.2

Our analysis

Where the marks are

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

Calculus22 marks · 28%
Q3, Q7
Statistics and probability21 marks · 26%
Q1, Q9
Geometry and trigonometry19 marks · 24%
Q2, Q8
Number and algebra18 marks · 23%
Q4, Q5, Q6

Mentor insights

Traps to watch for

01
Sine rule first, then the third angle · Q2
You know a side and its opposite angle (AC and angle B), so use the sine rule for angle A, then find angle C.
02
Find k from the 11% fall · Q5
A fall of 11% over eight years means the population is $0.89$ times as large: $e^{8 k} = 0.89$ .
03
Moving away means s and v have the same sign · Q7
The particle moves away from P when $s v > 0$ , and towards P when $s v < 0$ .
04
Split the kite into two right triangles · Q8
Radii to the tangent points meet the tangents at right angles, so ADOE is two congruent right triangles.
05
Conditional on working over 40 · Q9
In (c) the sample space is only employees who work more than 40 hours. Divide by $P (T > 40)$ .

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