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Worked solutions

IB Maths AA SL · Paper 2

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2

Session: May 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 May 2021 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 · 35 marks

Question 1

6 marks

Integrating a linear function, then fixing the constant

Calculus

(a)

3 marks

Integrate each term using $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1}$ .

\int (6 x + 7) d x = 3 x^{2} + 7 x + C

(b)

3 marks

The antiderivative has the same form, with a constant: $f (x) = 3 x^{2} + 7 x + c$ . Use $f (1.2) = 7.32$ .

3 (1.44) + 7 (1.2) + c = 7.32 \Rightarrow 4.32 + 8.4 + c = 7.32 \Rightarrow c = - 5.4 f (x) = 3 x^{2} + 7 x - 5.4

Question 2

7 marks

Regression line, a prediction and the mean point

Statistics and probability

(a)

2 marks

Enter the eight pairs in the calculator's linear regression (y on x).

a \approx 0.433, b \approx 4.50

(b)

2 marks

Substitute $x = 18$ into $y = a x + b$ .

y = 0.4331 \dots \times 18 + 4.5026 \dots \approx 12.3

Mentor tip

Using the stored unrounded values keeps the third significant figure safe.

(c)

1 mark

The regression line always passes through the point of means $(\overset{x}{ˉ}, \overset{y}{ˉ})$ .

\overset{x}{ˉ} = 15, \overset{y}{ˉ} = 11

(d)

2 marks

Draw a straight line through the mean point $(15, 11)$ that follows the trend of the points. It should meet the $y$ -axis at about 4.5 and rise by about 0.43 for each unit across.

Mentor tip

Use a ruler. Pivot the line about $(15, 11)$ so roughly equal numbers of points sit above and below it.

Question 3

6 marks

Rejected bags of sugar from a normal distribution

Statistics and probability

(a)

2 marks

Let $X \sim N (1000, 3. 5^{2})$ . The probability of a mass below 995 g comes from the normal cumulative distribution on the calculator.

P (X < 995) \approx 0.0766

(b)

1 mark

Multiply the probability by the sample size.

100 \times 0.0766 = 7.66 \approx 8 bags

(c)

3 marks

This is a conditional probability: the bag is not rejected, meaning $X \geq 995$ , and we want $X > 1005$ . The event $X > 1005$ is contained in $X \geq 995$ .

P (X > 1005 ∣ X \geq 995) = \frac{P ( X > 1005 )}{P ( X \geq 995 )} = \frac{0.07656}{1 - 0.07656} \approx 0.0829

Mentor tip

Rounding to 3 sf at the start (0.0766) changes the third figure. Keep the calculator's stored value.

Question 4

5 marks

A Ferris wheel modelled by a cosine

Number and algebra

5 marks

The amplitude is half the diameter, and starting at the lowest point means the cosine is reflected, so $a = - 55$ .
One revolution takes 20 minutes, so the period is 20 and $b = \frac{2 π}{20} = \frac{π}{10} \approx 0.314$ .
The middle height is the lowest height plus the radius, $10 + 55 = 65$ , so $c = 65$ .

h (t) = - 55 cos (\frac{π}{10} t) + 65

Mentor tip

Check: $h (0) = - 55 + 65 = 10$ , the lowest point. And $h (10) = 120$ , the top, 10 m plus the 110 m diameter.

Question 5

6 marks

Velocity, distance and acceleration from $v = t sin t - 3$

Calculus

(a)

2 marks

The particle is at rest when $v = 0$ , that is, $t sin t = 3$ . Solve on the calculator, or read the first axis crossing of the graph.

t \approx 6.74 s

(b)

2 marks

Total distance is the integral of the speed.

\int_{0}^{10} ∣ t sin t - 3∣ d t \approx 37.1 m

(c)

2 marks

Acceleration is the derivative of velocity. Differentiate with the product rule, then put in $t = 7$ .

a (t) = sin t + t cos t \Rightarrow a (7) = sin 7 + 7 cos 7 \approx 5.93 m s^{- 2}

Mentor tip

Radian mode. In degrees the answer is completely different.

Question 6

5 marks

A coefficient in a binomial expansion pins down $n$

Number and algebra

5 marks

The term in $x^{4}$ takes $x^{2}$ from two of the $n + 1$ brackets and 3 from the rest.
Set the coefficient equal to 20 412 and solve on the calculator, or by trying values.

(2 n + 1) 3^{n - 1} (x^{2})^{2} = \frac{n ( n + 1 )}{2} 3^{n - 1} x^{4} \frac{n ( n + 1 )}{2} 3^{n - 1} = 20412 \Rightarrow n = 7

Mentor tip

Check: $\frac{7 \cdot 8}{2} \cdot 3^{6} = 28 \times 729 = 20412$ .

Section B · Extended response

Questions 7–9 · 45 marks

Question 7

16 marks

Three ways to save $20 000: compound interest and a halving deposit

Number and algebra

(a)

5 marks

(i) Compound interest at 7% a year for 5 years.

(ii) Solve $9000 (1.07)^{n} = 20000$ on the calculator or with logs.

9000 (1.07)^{5} \approx 12622.97 \approx $12 600 n = \frac{ln ( 20/9 )}{ln 1.07} \approx 11.8 \Rightarrow 12 years

(b)

3 marks

Monthly compounding for 10 years is 120 periods at $\frac{r}{12} %$ each. Solve for $r$ using the finance solver.
At 8.01% the total falls just short of $20 000, so the minimum rate to two decimal places is

9000 (1 + \frac{r}{1200})^{120} = 20000 \Rightarrow r \approx 8.0117 r = 8.02%

(c)

8 marks

(i) Chris's yearly deposits are 9000, 4500, 2250, and so on: a geometric sequence with first term 9000 and ratio $\frac{1}{2}$ . Even continued forever the total is

So the safe never reaches $20 000.

(ii) With first deposit $x$ , the total after 5 deposits is

S_{\infty} = \frac{9000}{1 - 0.5} = 18000 < 20000 S_{5} = x \frac{1 - 0. 5 ^{5}}{1 - 0.5} = 1.9375 x = 20000 \Rightarrow x \approx $10 323

Question 8

14 marks

A horse tied between two fences: arc, area and bearings

Geometry and trigonometry

(a)

1 mark

The arc length of a sector is $r θ$ , and it is 28 m.

r = \frac{28}{θ}

(b)

4 marks

The horse can reach the sector $D P E$ plus the triangle $A P E$ . Here $∠ A P E = π - θ$ because $D$ , $P$ , $A$ lie on a straight line.
Replace $r$ by $\frac{28}{θ}$ , so $\frac{r ^{2}}{2} = \frac{392}{θ ^{2}}$ .

Area = \frac{1}{2} r^{2} θ + \frac{1}{2} r \cdot r sin (π - θ) = \frac{r ^{2}}{2} (θ + sin θ) Area = \frac{392}{θ ^{2}} (θ + sin θ)

(c)

2 marks

Solve $\frac{392}{θ ^{2}} (θ + sin θ) = 460$ on the calculator, with $θ$ in radians.

θ \approx 1.44

(d)

2 marks

In the isosceles triangle $A P E$ the angle at $P$ is $π - θ$ , so the base angles are equal and each is $\frac{1}{2} θ$ .

D A E = \frac{θ}{2} \approx 0.720 radians (41. 2^{\circ})

(e)

5 marks

(i) The bearing of B from A is $19 5^{\circ}$ , so the bearing of A from B is $1 5^{\circ}$ . C is due west of B, which is a bearing of $27 0^{\circ}$ . The angle between these directions is $36 0^{\circ} - (27 0^{\circ} - 1 5^{\circ})$ .

(ii) In triangle ABC the angle at A is $B A C = D A E = 0.720$ radians ( $41. 2^{\circ}$ ), since AC is along the fence AD. Use the sine rule, with $A C = 800$ opposite the angle at B.

A B C = 10 5^{\circ} \frac{B C}{sin 41. 2 ^{\circ}} = \frac{800}{sin 10 5 ^{\circ}} \Rightarrow B C \approx 546 m

Question 9

15 marks

An exponential curve, a tangent of gradient $- 1$ and enclosed areas

Calculus

(a)

2 marks

Solve $90 e^{- 0.5 x} = x$ on the calculator, or find the crossing point of the two graphs.

x \approx 5.57

(b)

4 marks

The gradient of $f$ is $f^{'} (x) = - 45 e^{- 0.5 x}$ . Set it equal to $- 1$ and solve exactly.
Then $y = 90 \cdot \frac{1}{45} = 2$ .

- 45 e^{- 0.5 x} = - 1 \Rightarrow e^{- 0.5 x} = \frac{1}{45} \Rightarrow x = 2 ln 45 Q (2 ln 45, 2)

(c)

2 marks

Use the point-gradient form with gradient $- 1$ through $Q$ .

y - 2 = - (x - 2 ln 45) \Rightarrow y = - x + 2 ln 45 + 2

(d)

5 marks

(i) The line $L$ meets $y = x$ where $x = - x + 2 ln 45 + 2$ , so $x = ln 45 + 1 \approx 4.81$ .

(ii) The region has two parts. From $x \approx 4.81$ to $x \approx 5.57$ it lies between $y = x$ (above) and $L$ (below). From $x \approx 5.57$ to $x = 2 ln 45 \approx 7.61$ it lies between $f$ (above) and $L$ (below).

A = \int_{4.81}^{5.57} (x - L (x)) d x + \int_{5.57}^{7.61} (90 e^{- 0.5 x} - L (x)) d x \approx 1.52

Mentor tip

Keep the limits unrounded in the calculator. Rounding them to 3 sf shifts the area in the third figure.

(e)

2 marks

By symmetry in $y = x$ the shaded area between $f$ , $f^{- 1}$ and $L$ is twice the area in part (d).

2 \times 1.52 \approx 3.04

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%
Q1, Q5, Q9
Number and algebra26 marks · 33%
Q4, Q6, Q7
Geometry and trigonometry14 marks · 18%
Q8
Statistics and probability13 marks · 16%
Q2, Q3

Mentor insights

Traps to watch for

01
Don't lose the constant · Q1
Part (a) needs $+ C$ . In (b) the extra condition $f (1.2) = 7.32$ is exactly what finds it.
02
Predict from the equation, not the picture · Q2
Part (b) wants a calculated value from $y = a x + b$ with unrounded or 3 sf values, not a reading off the scatter diagram.
03
Round up for the first year it is reached · Q7
In (a)(ii) the investment passes $20 000 part-way through year 12, so the answer is 12 years. Round the rate in (b) up as well, because 8.01% falls just short.
04
Find the angle at the corner from the isosceles triangle · Q8
$P A = P E = r$ , so triangle $A P E$ is isosceles. Its base angles are half of what is left over after the angle at $P$ .
05
Use the symmetry in part (e) · Q9
The line $L$ is tangent to both $f$ and $f^{- 1}$ and the graphs are reflections in $y = x$ . The area in (e) is exactly twice the area in (d).

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 2

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2

Session: May 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 May 2021 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 · 35 marks

Question 1

6 marks

Integrating a linear function, then fixing the constant

Calculus

(a)

3 marks

Integrate each term using $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1}$ .

\int (6 x + 7) d x = 3 x^{2} + 7 x + C

(b)

3 marks

The antiderivative has the same form, with a constant: $f (x) = 3 x^{2} + 7 x + c$ . Use $f (1.2) = 7.32$ .

3 (1.44) + 7 (1.2) + c = 7.32 \Rightarrow 4.32 + 8.4 + c = 7.32 \Rightarrow c = - 5.4 f (x) = 3 x^{2} + 7 x - 5.4

Question 2

7 marks

Regression line, a prediction and the mean point

Statistics and probability

(a)

2 marks

Enter the eight pairs in the calculator's linear regression (y on x).

a \approx 0.433, b \approx 4.50

(b)

2 marks

Substitute $x = 18$ into $y = a x + b$ .

y = 0.4331 \dots \times 18 + 4.5026 \dots \approx 12.3

Mentor tip

Using the stored unrounded values keeps the third significant figure safe.

(c)

1 mark

The regression line always passes through the point of means $(\overset{x}{ˉ}, \overset{y}{ˉ})$ .

\overset{x}{ˉ} = 15, \overset{y}{ˉ} = 11

(d)

2 marks

Draw a straight line through the mean point $(15, 11)$ that follows the trend of the points. It should meet the $y$ -axis at about 4.5 and rise by about 0.43 for each unit across.

Mentor tip

Use a ruler. Pivot the line about $(15, 11)$ so roughly equal numbers of points sit above and below it.

Question 3

6 marks

Rejected bags of sugar from a normal distribution

Statistics and probability

(a)

2 marks

Let $X \sim N (1000, 3. 5^{2})$ . The probability of a mass below 995 g comes from the normal cumulative distribution on the calculator.

P (X < 995) \approx 0.0766

(b)

1 mark

Multiply the probability by the sample size.

100 \times 0.0766 = 7.66 \approx 8 bags

(c)

3 marks

This is a conditional probability: the bag is not rejected, meaning $X \geq 995$ , and we want $X > 1005$ . The event $X > 1005$ is contained in $X \geq 995$ .

P (X > 1005 ∣ X \geq 995) = \frac{P ( X > 1005 )}{P ( X \geq 995 )} = \frac{0.07656}{1 - 0.07656} \approx 0.0829

Mentor tip

Rounding to 3 sf at the start (0.0766) changes the third figure. Keep the calculator's stored value.

Question 4

5 marks

A Ferris wheel modelled by a cosine

Number and algebra

5 marks

The amplitude is half the diameter, and starting at the lowest point means the cosine is reflected, so $a = - 55$ .
One revolution takes 20 minutes, so the period is 20 and $b = \frac{2 π}{20} = \frac{π}{10} \approx 0.314$ .
The middle height is the lowest height plus the radius, $10 + 55 = 65$ , so $c = 65$ .

h (t) = - 55 cos (\frac{π}{10} t) + 65

Mentor tip

Check: $h (0) = - 55 + 65 = 10$ , the lowest point. And $h (10) = 120$ , the top, 10 m plus the 110 m diameter.

Question 5

6 marks

Velocity, distance and acceleration from $v = t sin t - 3$

Calculus

(a)

2 marks

The particle is at rest when $v = 0$ , that is, $t sin t = 3$ . Solve on the calculator, or read the first axis crossing of the graph.

t \approx 6.74 s

(b)

2 marks

Total distance is the integral of the speed.

\int_{0}^{10} ∣ t sin t - 3∣ d t \approx 37.1 m

(c)

2 marks

Acceleration is the derivative of velocity. Differentiate with the product rule, then put in $t = 7$ .

a (t) = sin t + t cos t \Rightarrow a (7) = sin 7 + 7 cos 7 \approx 5.93 m s^{- 2}

Mentor tip

Radian mode. In degrees the answer is completely different.

Question 6

5 marks

A coefficient in a binomial expansion pins down $n$

Number and algebra

5 marks

The term in $x^{4}$ takes $x^{2}$ from two of the $n + 1$ brackets and 3 from the rest.
Set the coefficient equal to 20 412 and solve on the calculator, or by trying values.

(2 n + 1) 3^{n - 1} (x^{2})^{2} = \frac{n ( n + 1 )}{2} 3^{n - 1} x^{4} \frac{n ( n + 1 )}{2} 3^{n - 1} = 20412 \Rightarrow n = 7

Mentor tip

Check: $\frac{7 \cdot 8}{2} \cdot 3^{6} = 28 \times 729 = 20412$ .

Section B · Extended response

Questions 7–9 · 45 marks

Question 7

16 marks

Three ways to save $20 000: compound interest and a halving deposit

Number and algebra

(a)

5 marks

(i) Compound interest at 7% a year for 5 years.

(ii) Solve $9000 (1.07)^{n} = 20000$ on the calculator or with logs.

9000 (1.07)^{5} \approx 12622.97 \approx $12 600 n = \frac{ln ( 20/9 )}{ln 1.07} \approx 11.8 \Rightarrow 12 years

(b)

3 marks

Monthly compounding for 10 years is 120 periods at $\frac{r}{12} %$ each. Solve for $r$ using the finance solver.
At 8.01% the total falls just short of $20 000, so the minimum rate to two decimal places is

9000 (1 + \frac{r}{1200})^{120} = 20000 \Rightarrow r \approx 8.0117 r = 8.02%

(c)

8 marks

(i) Chris's yearly deposits are 9000, 4500, 2250, and so on: a geometric sequence with first term 9000 and ratio $\frac{1}{2}$ . Even continued forever the total is

So the safe never reaches $20 000.

(ii) With first deposit $x$ , the total after 5 deposits is

S_{\infty} = \frac{9000}{1 - 0.5} = 18000 < 20000 S_{5} = x \frac{1 - 0. 5 ^{5}}{1 - 0.5} = 1.9375 x = 20000 \Rightarrow x \approx $10 323

Question 8

14 marks

A horse tied between two fences: arc, area and bearings

Geometry and trigonometry

(a)

1 mark

The arc length of a sector is $r θ$ , and it is 28 m.

r = \frac{28}{θ}

(b)

4 marks

The horse can reach the sector $D P E$ plus the triangle $A P E$ . Here $∠ A P E = π - θ$ because $D$ , $P$ , $A$ lie on a straight line.
Replace $r$ by $\frac{28}{θ}$ , so $\frac{r ^{2}}{2} = \frac{392}{θ ^{2}}$ .

Area = \frac{1}{2} r^{2} θ + \frac{1}{2} r \cdot r sin (π - θ) = \frac{r ^{2}}{2} (θ + sin θ) Area = \frac{392}{θ ^{2}} (θ + sin θ)

(c)

2 marks

Solve $\frac{392}{θ ^{2}} (θ + sin θ) = 460$ on the calculator, with $θ$ in radians.

θ \approx 1.44

(d)

2 marks

In the isosceles triangle $A P E$ the angle at $P$ is $π - θ$ , so the base angles are equal and each is $\frac{1}{2} θ$ .

D A E = \frac{θ}{2} \approx 0.720 radians (41. 2^{\circ})

(e)

5 marks

(ii) In triangle ABC the angle at A is $B A C = D A E = 0.720$ radians ( $41. 2^{\circ}$ ), since AC is along the fence AD. Use the sine rule, with $A C = 800$ opposite the angle at B.

A B C = 10 5^{\circ} \frac{B C}{sin 41. 2 ^{\circ}} = \frac{800}{sin 10 5 ^{\circ}} \Rightarrow B C \approx 546 m

Question 9

15 marks

An exponential curve, a tangent of gradient $- 1$ and enclosed areas

Calculus

(a)

2 marks

Solve $90 e^{- 0.5 x} = x$ on the calculator, or find the crossing point of the two graphs.

x \approx 5.57

(b)

4 marks

The gradient of $f$ is $f^{'} (x) = - 45 e^{- 0.5 x}$ . Set it equal to $- 1$ and solve exactly.
Then $y = 90 \cdot \frac{1}{45} = 2$ .

- 45 e^{- 0.5 x} = - 1 \Rightarrow e^{- 0.5 x} = \frac{1}{45} \Rightarrow x = 2 ln 45 Q (2 ln 45, 2)

(c)

2 marks

Use the point-gradient form with gradient $- 1$ through $Q$ .

y - 2 = - (x - 2 ln 45) \Rightarrow y = - x + 2 ln 45 + 2

(d)

5 marks

(i) The line $L$ meets $y = x$ where $x = - x + 2 ln 45 + 2$ , so $x = ln 45 + 1 \approx 4.81$ .

A = \int_{4.81}^{5.57} (x - L (x)) d x + \int_{5.57}^{7.61} (90 e^{- 0.5 x} - L (x)) d x \approx 1.52

Mentor tip

Keep the limits unrounded in the calculator. Rounding them to 3 sf shifts the area in the third figure.

(e)

2 marks

By symmetry in $y = x$ the shaded area between $f$ , $f^{- 1}$ and $L$ is twice the area in part (d).

2 \times 1.52 \approx 3.04

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%
Q1, Q5, Q9
Number and algebra26 marks · 33%
Q4, Q6, Q7
Geometry and trigonometry14 marks · 18%
Q8
Statistics and probability13 marks · 16%
Q2, Q3

Mentor insights

Traps to watch for

01
Don't lose the constant · Q1
Part (a) needs $+ C$ . In (b) the extra condition $f (1.2) = 7.32$ is exactly what finds it.
02
Predict from the equation, not the picture · Q2
Part (b) wants a calculated value from $y = a x + b$ with unrounded or 3 sf values, not a reading off the scatter diagram.
03
Round up for the first year it is reached · Q7
In (a)(ii) the investment passes $20 000 part-way through year 12, so the answer is 12 years. Round the rate in (b) up as well, because 8.01% falls just short.
04
Find the angle at the corner from the isosceles triangle · Q8
$P A = P E = r$ , so triangle $A P E$ is isosceles. Its base angles are half of what is left over after the angle at $P$ .
05
Use the symmetry in part (e) · Q9
The line $L$ is tangent to both $f$ and $f^{- 1}$ and the graphs are reflections in $y = x$ . The area in (e) is exactly twice the area in (d).

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