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IB Maths AA SL · Paper 2

Mathematics: Analysis & Approaches, Standard Level

Session: November 2021
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 2021 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 · 37 marks

Question 1

5 marks

Piano practice and diploma scores

Statistics and probability

(a)

2 marks

Enter the eight pairs on the calculator.

r \approx 0.884

(b)

1 mark

The regression line $D = ah + b$ :

a \approx 1.37, b \approx 64.5

(c)

2 marks

Five extra hours a week changes $D$ by $5 a$ .
So her score could have been expected to rise by about 7 marks.

5 \times 1.366 \approx 6.83

Question 2

5 marks

A Gaussian-type function: its zeros and a sketch

Functions

(a)

2 marks

Solve $e^{- x^{2}} = 0.5$ .

- x^{2} = ln 0.5 \Rightarrow x^{2} = ln 2 \Rightarrow x = \pm ln 2 \approx \pm 0.833

(b)

3 marks

The graph is a bell shape, symmetric about the $y$ -axis. The maximum is at $(0, 0.5)$ , since $e^{0} - 0.5 = 0.5$ . It crosses the $x$ -axis at $x \approx \pm 0.833$ and ends at $x = \pm 2$ with $y = e^{- 4} - 0.5 \approx - 0.48$ .

Question 3

5 marks

The ambiguous case: the smallest perimeter

Number and algebra

5 marks

Apply the cosine rule with $A B = c$ unknown, using the angle $A = 2 5^{\circ}$ opposite the side $C B = 7$ .
Both roots are positive and give valid triangles.
The smaller perimeter uses the shorter side.

7^{2} = 1 2^{2} + c^{2} - 2 (12) c cos 2 5^{\circ} \Rightarrow c^{2} - 21.75 c + 95 = 0 c \approx 6.05 or c \approx 15.70 12 + 7 + 6.05 \approx 25.1

Mentor tip

Check with the sine rule: $sin B = \frac{12 s i n 2 5 ^{\circ}}{7} \approx 0.7245$ gives $B \approx 46. 4^{\circ}$ or $133. 6^{\circ}$ , the two triangles.

Question 4

7 marks

Defective lamps: a binomial and a conditional probability

Statistics and probability

(a)

3 marks

The number of defective lamps is $X \sim B (30, 0.05)$ .

P (X \geq 1) = 1 - 0.9 5^{30} \approx 0.785

(b)

4 marks

This is a conditional probability. The events "at most two" and "at least one" overlap in $1 \leq X \leq 2$ .

P (X \leq 2 ∣ X \geq 1) = \frac{P ( 1 \leq X \leq 2 )}{P ( X \geq 1 )} = \frac{0.5975}{0.7854} \approx 0.761

Question 5

6 marks

A semicircle with two equal shaded areas

Geometry and trigonometry

(a)

5 marks

One shaded region is the segment cut off by the chord QR, which has area sector minus triangle. The other is the triangle POR, whose angle at O is $π - θ$ .
Set the two areas equal and cancel $\frac{1}{2} r^{2}$ .

segment = \frac{1}{2} r^{2} θ - \frac{1}{2} r^{2} sin θ, triangle POR = \frac{1}{2} r^{2} sin (π - θ) = \frac{1}{2} r^{2} sin θ θ - sin θ = sin θ \Rightarrow θ = 2 sin θ

(b)

1 mark

Solve $θ = 2 sin θ$ on the calculator, in radians, for $0 < θ < π$ .

θ \approx 1.90

Question 6

9 marks

A geometric series given as a sum: first term, sum to infinity and when it's close

Number and algebra

(a)

2 marks

$S_{1}$ is the first term, so put $r = 1$ in the formula.

u_{1} = \frac{2}{3} \cdot \frac{7}{8} = \frac{7}{12}

(b)

3 marks

The common ratio is $\frac{7}{8}$ . Use $S_{\infty} = \frac{u _{1}}{1 - r}$ .

S_{\infty} = \frac{7/12}{1/8} = \frac{14}{3}

(c)

4 marks

The tail after $n$ terms is $S_{\infty} - S_{n} = \frac{14}{3} (\frac{7}{8})^{n}$ . Solve $\frac{14}{3} (\frac{7}{8})^{n} < 0.001$ .
At $n = 63$ the tail is 0.001036, which is too big. At $n = 64$ it is 0.000906.

n > \frac{ln ( 0.001 \times 3/14 )}{ln ( 7/8 )} \approx 63.3 n = 64

Section B · Extended response

Questions 7–9 · 43 marks

Question 7

14 marks

Crossing a river and driving to the city: minimising time and cost

Geometry and trigonometry

(a)

4 marks

(i) Boat from A to P at 42 km/h, then bus from P to B at 84 km/h.

(ii) Boat directly: $A B = 21 5^{2} + 6 5^{2} \approx 224.6$ km.

\frac{65}{42} + \frac{215}{84} \approx 4.11 hours \frac{224.6}{42} \approx 5.35 hours

(b)

6 marks

(i) $D$ is on PB with $B D = x$ , so $P D = 215 - x$ and $A D = (215 - x)^{2} + 6 5^{2}$ .

(ii) Minimise $T$ on the calculator for $0 < x < 215$ .

(iii) The minimum time is

T = \frac{( 215 - x ) ^{2} + 6 5 ^{2}}{42} + \frac{x}{84} x \approx 177 km T \approx 3.90 hours

(c)

4 marks

(i) The cost per hour is 200 for the boat and 150 for the bus.

Minimise on the calculator.

(ii) The minimum cost is

C = 200 \cdot \frac{( 215 - x ) ^{2} + 6 5 ^{2}}{42} + 150 \cdot \frac{x}{84} x \approx 189 km C \approx $671

Question 8

13 marks

Tides in Dungeness harbour: finding and using a sine model

Functions

(a)

1 mark

The time between high tides is 12 hours, so that is the period.

b = \frac{2 π}{12} = \frac{π}{6}

(b)

2 marks

The amplitude is half the range.

a = \frac{6.8 - 2.2}{2} = 2.3

(c)

2 marks

The midline is halfway between the maximum and minimum.

d = \frac{6.8 + 2.2}{2} = 4.5

(d)

3 marks

At the first high tide, $t = 4.5$ and the sine equals 1, so its argument is $\frac{π}{2}$ .

\frac{π}{6} (4.5 - c) = \frac{π}{2} \Rightarrow 4.5 - c = 3 \Rightarrow c = 1.5

(e)

2 marks

Noon is $t = 12$ .

H (12) = 2.3 sin (\frac{π}{6} \times 10.5) + 4.5 \approx 2.87 m

(f)

3 marks

Solve $2.3 sin (\frac{π}{6} (t - 1.5)) + 4.5 = 5$ on the calculator. In the first 12 hours the tide is above 5 m from about 1.92 h to 7.08 h.
That happens in each of the two high tides in 24 hours.

7.081 - 1.919 = 5.163 hours 2 \times 5.163 \approx 10.3 hours

Question 9

16 marks

Avocado weights: a normal distribution, categories and a profit target

Statistics and probability

(a)

3 marks

Standardise: $P (μ - 1.5 σ < X < μ + 1.5 σ) = P (- 1.5 < Z < 1.5)$ .

P (- 1.5 < Z < 1.5) \approx 0.866

(b)

5 marks

Small avocados are the lowest 4%, so 106.2 g is at $z = invNorm (0.04) \approx - 1.7507$ . Premium avocados are the top 9.6%, so 182.6 g is at $z \approx 1.3047$ .
Subtract to find $σ$ , then $μ$ .

106.2 = μ - 1.7507 σ, 182.6 = μ + 1.3047 σ σ \approx 25.0, μ \approx 150

(c)

4 marks

Given the avocado is not small, the sample space has probability 0.96.

P (medium) = \frac{0.576}{0.96} = 0.6, P (large) = \frac{0.288}{0.96} = 0.3, P (premium) = \frac{0.096}{0.96} = 0.1

(d)

4 marks

The expected revenue per avocado is
For $n$ avocados sold, profit is $1.243 n - 200$ . We need at least 438.

1.10 (0.6) + 1.29 (0.3) + 1.96 (0.1) = 1.243 dollars 1.243 n - 200 \geq 438 \Rightarrow n \geq 513.3 \Rightarrow n = 514

Our analysis

Where the marks are

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

Statistics and probability28 marks · 35%
Q1, Q4, Q9
Geometry and trigonometry20 marks · 25%
Q5, Q7
Functions18 marks · 23%
Q2, Q8
Number and algebra14 marks · 18%
Q3, Q6

Mentor insights

Traps to watch for

01
"At least one" is a complement · Q4
$P (X \geq 1) = 1 - P (X = 0)$ . The conditional probability in (b) divides by that value.
02
The inequality is on the remainder · Q6
$S_{\infty} - S_{n}$ is the tail of the series, which is $\frac{u _{1} r ^{n}}{1 - r}$ . Solve that against 0.001 with logs or a table, not $u_{n} < 0.001$ .
03
Set up one variable · Q7
Write the whole journey time as a function of the single variable $x$ , then minimise it on the calculator for $0 < x < 215$ . For (c) the function changes, because the costs per hour differ.
04
Read the data carefully · Q8
"The first high tide at 04:30" means the maximum occurs at $t = 4.5$ . The sine reaches its maximum when its argument is $\frac{π}{2}$ , which gives $c$ .
05
Condition on what is purchased · Q9
The supermarket buys only avocados above 106.2 g, so divide each category probability by $1 - 0.04 = 0.96$ .

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 2

Mathematics: Analysis & Approaches, Standard Level

Session: November 2021
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 2021 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 · 37 marks

Question 1

5 marks

Piano practice and diploma scores

Statistics and probability

(a)

2 marks

Enter the eight pairs on the calculator.

r \approx 0.884

(b)

1 mark

The regression line $D = ah + b$ :

a \approx 1.37, b \approx 64.5

(c)

2 marks

Five extra hours a week changes $D$ by $5 a$ .
So her score could have been expected to rise by about 7 marks.

5 \times 1.366 \approx 6.83

Question 2

5 marks

A Gaussian-type function: its zeros and a sketch

Functions

(a)

2 marks

Solve $e^{- x^{2}} = 0.5$ .

- x^{2} = ln 0.5 \Rightarrow x^{2} = ln 2 \Rightarrow x = \pm ln 2 \approx \pm 0.833

(b)

3 marks

Question 3

5 marks

The ambiguous case: the smallest perimeter

Number and algebra

5 marks

Apply the cosine rule with $A B = c$ unknown, using the angle $A = 2 5^{\circ}$ opposite the side $C B = 7$ .
Both roots are positive and give valid triangles.
The smaller perimeter uses the shorter side.

7^{2} = 1 2^{2} + c^{2} - 2 (12) c cos 2 5^{\circ} \Rightarrow c^{2} - 21.75 c + 95 = 0 c \approx 6.05 or c \approx 15.70 12 + 7 + 6.05 \approx 25.1

Mentor tip

Check with the sine rule: $sin B = \frac{12 s i n 2 5 ^{\circ}}{7} \approx 0.7245$ gives $B \approx 46. 4^{\circ}$ or $133. 6^{\circ}$ , the two triangles.

Question 4

7 marks

Defective lamps: a binomial and a conditional probability

Statistics and probability

(a)

3 marks

The number of defective lamps is $X \sim B (30, 0.05)$ .

P (X \geq 1) = 1 - 0.9 5^{30} \approx 0.785

(b)

4 marks

This is a conditional probability. The events "at most two" and "at least one" overlap in $1 \leq X \leq 2$ .

P (X \leq 2 ∣ X \geq 1) = \frac{P ( 1 \leq X \leq 2 )}{P ( X \geq 1 )} = \frac{0.5975}{0.7854} \approx 0.761

Question 5

6 marks

A semicircle with two equal shaded areas

Geometry and trigonometry

(a)

5 marks

One shaded region is the segment cut off by the chord QR, which has area sector minus triangle. The other is the triangle POR, whose angle at O is $π - θ$ .
Set the two areas equal and cancel $\frac{1}{2} r^{2}$ .

segment = \frac{1}{2} r^{2} θ - \frac{1}{2} r^{2} sin θ, triangle POR = \frac{1}{2} r^{2} sin (π - θ) = \frac{1}{2} r^{2} sin θ θ - sin θ = sin θ \Rightarrow θ = 2 sin θ

(b)

1 mark

Solve $θ = 2 sin θ$ on the calculator, in radians, for $0 < θ < π$ .

θ \approx 1.90

Question 6

9 marks

A geometric series given as a sum: first term, sum to infinity and when it's close

Number and algebra

(a)

2 marks

$S_{1}$ is the first term, so put $r = 1$ in the formula.

u_{1} = \frac{2}{3} \cdot \frac{7}{8} = \frac{7}{12}

(b)

3 marks

The common ratio is $\frac{7}{8}$ . Use $S_{\infty} = \frac{u _{1}}{1 - r}$ .

S_{\infty} = \frac{7/12}{1/8} = \frac{14}{3}

(c)

4 marks

The tail after $n$ terms is $S_{\infty} - S_{n} = \frac{14}{3} (\frac{7}{8})^{n}$ . Solve $\frac{14}{3} (\frac{7}{8})^{n} < 0.001$ .
At $n = 63$ the tail is 0.001036, which is too big. At $n = 64$ it is 0.000906.

n > \frac{ln ( 0.001 \times 3/14 )}{ln ( 7/8 )} \approx 63.3 n = 64

Section B · Extended response

Questions 7–9 · 43 marks

Question 7

14 marks

Crossing a river and driving to the city: minimising time and cost

Geometry and trigonometry

(a)

4 marks

(i) Boat from A to P at 42 km/h, then bus from P to B at 84 km/h.

(ii) Boat directly: $A B = 21 5^{2} + 6 5^{2} \approx 224.6$ km.

\frac{65}{42} + \frac{215}{84} \approx 4.11 hours \frac{224.6}{42} \approx 5.35 hours

(b)

6 marks

(i) $D$ is on PB with $B D = x$ , so $P D = 215 - x$ and $A D = (215 - x)^{2} + 6 5^{2}$ .

(ii) Minimise $T$ on the calculator for $0 < x < 215$ .

(iii) The minimum time is

T = \frac{( 215 - x ) ^{2} + 6 5 ^{2}}{42} + \frac{x}{84} x \approx 177 km T \approx 3.90 hours

(c)

4 marks

(i) The cost per hour is 200 for the boat and 150 for the bus.

Minimise on the calculator.

(ii) The minimum cost is

C = 200 \cdot \frac{( 215 - x ) ^{2} + 6 5 ^{2}}{42} + 150 \cdot \frac{x}{84} x \approx 189 km C \approx $671

Question 8

13 marks

Tides in Dungeness harbour: finding and using a sine model

Functions

(a)

1 mark

The time between high tides is 12 hours, so that is the period.

b = \frac{2 π}{12} = \frac{π}{6}

(b)

2 marks

The amplitude is half the range.

a = \frac{6.8 - 2.2}{2} = 2.3

(c)

2 marks

The midline is halfway between the maximum and minimum.

d = \frac{6.8 + 2.2}{2} = 4.5

(d)

3 marks

At the first high tide, $t = 4.5$ and the sine equals 1, so its argument is $\frac{π}{2}$ .

\frac{π}{6} (4.5 - c) = \frac{π}{2} \Rightarrow 4.5 - c = 3 \Rightarrow c = 1.5

(e)

2 marks

Noon is $t = 12$ .

H (12) = 2.3 sin (\frac{π}{6} \times 10.5) + 4.5 \approx 2.87 m

(f)

3 marks

Solve $2.3 sin (\frac{π}{6} (t - 1.5)) + 4.5 = 5$ on the calculator. In the first 12 hours the tide is above 5 m from about 1.92 h to 7.08 h.
That happens in each of the two high tides in 24 hours.

7.081 - 1.919 = 5.163 hours 2 \times 5.163 \approx 10.3 hours

Question 9

16 marks

Avocado weights: a normal distribution, categories and a profit target

Statistics and probability

(a)

3 marks

Standardise: $P (μ - 1.5 σ < X < μ + 1.5 σ) = P (- 1.5 < Z < 1.5)$ .

P (- 1.5 < Z < 1.5) \approx 0.866

(b)

5 marks

Small avocados are the lowest 4%, so 106.2 g is at $z = invNorm (0.04) \approx - 1.7507$ . Premium avocados are the top 9.6%, so 182.6 g is at $z \approx 1.3047$ .
Subtract to find $σ$ , then $μ$ .

106.2 = μ - 1.7507 σ, 182.6 = μ + 1.3047 σ σ \approx 25.0, μ \approx 150

(c)

4 marks

Given the avocado is not small, the sample space has probability 0.96.

P (medium) = \frac{0.576}{0.96} = 0.6, P (large) = \frac{0.288}{0.96} = 0.3, P (premium) = \frac{0.096}{0.96} = 0.1

(d)

4 marks

The expected revenue per avocado is
For $n$ avocados sold, profit is $1.243 n - 200$ . We need at least 438.

1.10 (0.6) + 1.29 (0.3) + 1.96 (0.1) = 1.243 dollars 1.243 n - 200 \geq 438 \Rightarrow n \geq 513.3 \Rightarrow n = 514

Our analysis

Where the marks are

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

Statistics and probability28 marks · 35%
Q1, Q4, Q9
Geometry and trigonometry20 marks · 25%
Q5, Q7
Functions18 marks · 23%
Q2, Q8
Number and algebra14 marks · 18%
Q3, Q6

Mentor insights

Traps to watch for

01
"At least one" is a complement · Q4
$P (X \geq 1) = 1 - P (X = 0)$ . The conditional probability in (b) divides by that value.
02
The inequality is on the remainder · Q6
$S_{\infty} - S_{n}$ is the tail of the series, which is $\frac{u _{1} r ^{n}}{1 - r}$ . Solve that against 0.001 with logs or a table, not $u_{n} < 0.001$ .
03
Set up one variable · Q7
Write the whole journey time as a function of the single variable $x$ , then minimise it on the calculator for $0 < x < 215$ . For (c) the function changes, because the costs per hour differ.
04
Read the data carefully · Q8
"The first high tide at 04:30" means the maximum occurs at $t = 4.5$ . The sine reaches its maximum when its argument is $\frac{π}{2}$ , which gives $c$ .
05
Condition on what is purchased · Q9
The supermarket buys only avocados above 106.2 g, so divide each category probability by $1 - 0.04 = 0.96$ .

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