Loading…

Worked solutions

IB Maths AA SL · Paper 2

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2

Session: November 2023
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 2023 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 · 35 marks

Question 1

7 marks

Tangent, normal and where the normal meets the parabola again (f = x^2 - 8x)

Calculus

(a)

3 marks

Differentiate and substitute $x = 3$ .
The normal is perpendicular to the tangent.

f^{'} (x) = 2 x - 8 \Rightarrow f^{'} (3) = - 2 m_{normal} = \frac{1}{2}

(b)

1 mark

Line through $A (3, - 15)$ with gradient $\frac{1}{2}$ .

y + 15 = \frac{1}{2} (x - 3) \Rightarrow y = \frac{1}{2} x - 16.5

(c)

3 marks

Set the normal equal to the curve.
At $x = 5.5$ , $y = \frac{1}{2} (5.5) - 16.5 = - 13.75$ .

x^{2} - 8 x = \frac{1}{2} x - 16.5 \Rightarrow x^{2} - 8.5 x + 16.5 = 0 \Rightarrow x = 3 or x = 5.5 B = (5.5, - 13.75)

Question 2

6 marks

A pyramid on a rectangular base: a length and an angle

Geometry and trigonometry

(a)

2 marks

$B V = (3 - 6, 4 - 8, 9 - 0) = (- 3, - 4, 9)$ .

B V = 9 + 16 + 81 = 106 \approx 10.3

(b)

4 marks

$V B = (3, 4, - 9)$ and $V C = (3, - 4, - 9)$ .

V B \cdot V C = 9 - 16 + 81 = 74, ∣ V B ∣ = ∣ V C ∣ = 106 cos B V C = \frac{74}{106} \Rightarrow B V C = 45. 7^{\circ}

Question 3

5 marks

f(x) = e^x - 3x - 4: a sketch and transformations

Functions

(a)

3 marks

Features on $- 4 \leq x \leq 3$ : $y$ -intercept $f (0) = - 3$ ; roots near $x = - 1.4$ and $x = 2.4$ ; a minimum at $x = ln 3 = 1.10$ with $f = - 4.30$ ; end points $f (- 4) = 8.02$ and $f (3) = 7.09$ .
Sketch: the curve starts at $(- 4, 8.0)$ , falls through a root in $(- 2, - 1)$ and the $y$ -intercept $(0, - 3)$ to the minimum near $(1.1, - 4.3)$ , then rises through a root in $(2, 3)$ to $(3, 7.1)$ .

(b)

2 marks

Compare $g (x) = e^{2 x} - 6 x - 7$ with $f (2 x) = e^{2 x} - 6 x - 4$ .

g (x) = f (2 x) - 3 k = \frac{1}{2}, c = - 3

Question 4

7 marks

A letter C between two circles: area and perimeter

Geometry and trigonometry

(a)

2 marks

Area of a sector is $\frac{1}{2} R^{2} θ$ with $θ = 5.2$ .

A = \frac{1}{2} (10)^{2} (5.2) - \frac{1}{2} r^{2} (5.2) = 260 - 2.6 r^{2}

(b)

5 marks

Set $260 - 2.6 r^{2} = 64$ .
The perimeter: outer arc $10 \times 5.2 = 52$ , inner arc $5.2 r = 45.1$ , and two straight ends each of length $10 - r = 1.32$ .

r^{2} = \frac{196}{2.6} = 75.38 \Rightarrow r = 8.68 P = 52 + 45.1 + 2 (1.32) \approx 99.8 cm

Question 5

5 marks

A particle with a square-root sine displacement: rest and distance

Calculus

(a)

2 marks

The displacement is $s (t) = 4.3 sin (3 t + 5)$ . The velocity is zero where $cos (3 t + 5) = 0$ .
(The value $3 (0) + 5 = 2.24$ is already past $\frac{π}{2}$ , so $\frac{3 π}{2}$ is the first one reached.)

3 t + 5 = \frac{3 π}{2} \Rightarrow 3 t + 5 = \frac{9 π ^{2}}{4} \Rightarrow q = 5.74

(b)

3 marks

Distance is the integral of speed from 0 to $q$ .
Check: $s (0) = 3.38$ and $s (q) = - 4.3$ , and the particle moves in one direction in between, so the distance is $3.38 + 4.30 = 7.68$ .

\int_{0}^{5.74} ∣ v (t) ∣ d t = 7.68 m

Question 6

5 marks

Finding a from an expected value and total probability

Number and algebra

5 marks

The probabilities are $k, k^{2}, a, k^{3}$ for $x = 1, 2, 3, 4$ .
Eliminate $a$ using $a = 1 - k - k^{2} - k^{3}$ .
Solve for $k$ with $0 < k < 1$ .

k + k^{2} + a + k^{3} = 1 E (X) = k + 2 k^{2} + 3 a + 4 k^{3} = 2.3 k + 2 k^{2} + 3 (1 - k - k^{2} - k^{3}) + 4 k^{3} = 2.3 \Rightarrow k^{3} - k^{2} - 2 k + 0.7 = 0 k = 0.316 a = 1 - 0.3159 - 0.0998 - 0.0315 \approx 0.553

Section B · Extended response

Questions 7–9 · 45 marks

Question 7

14 marks

A capsule: surface area, minimum and its justification (cylinder volume 45)

Geometry and trigonometry
Calculus

(a)

3 marks

The cylinder volume gives $h = \frac{45}{π x ^{2}}$ . The curved surface of the cylinder is $2 π x h$ and the two hemispheres together have area $4 π x^{2}$ .

S = 2 π x \cdot \frac{45}{π x ^{2}} + 4 π x^{2} = \frac{90}{x} + 4 π x^{2}

(b)

5 marks

Differentiate $S = 90 x^{- 1} + 4 π x^{2}$ .
Set it to zero.

\frac{d S}{d x} = - \frac{90}{x ^{2}} + 8 π x 8 π x^{3} = 90 \Rightarrow a = (\frac{45}{4 π})^{1/3}

(c)

6 marks

Differentiate again.
For $x > 0$ this is positive, so $S$ is a minimum at $x = a$ .
Minimum surface area:

\frac{d ^{2} S}{d x ^{2}} = \frac{180}{x ^{3}} + 8 π S_{m i n} = \frac{90}{1.530} + 4 π (1.530)^{2} \approx 88.2 cm^{2}

Question 8

15 marks

Prize money in two options and an investment (second version): series and interest

Number and algebra

(a)

5 marks

Option A is 36 payments of 5500.
Option B is a geometric series.

5500 \times 36 = $198 000 S_{36} = \frac{2000 ( 1.0 6 ^{36} - 1 )}{1.06 - 1} \approx $238 241.73

(b)

1 mark

Compound interest with 12 periods a year.

120000 (1 + \frac{0.04}{12})^{12 n}

(c)

6 marks

After six months Sorin has
Daniela's six payments of Option B total
The total value is
For the total after $m$ months to be at least 250 000, solve
After 26 months it is $249 158 and after 27 months it is $258 693.

120000 (1 + \frac{0.04}{12})^{6} = $122 420.09 \frac{2000 ( 1.0 6 ^{6} - 1 )}{0.06} = $13 950.64 122420.09 + 13950.64 = $136 370.73 120000 (1 + \frac{0.04}{12})^{m} + \frac{2000 ( 1.0 6 ^{m} - 1 )}{0.06} \geq 250000 m = 27 months

(d)

3 marks

Use a finance solver with $N = 24$ quarters, $P V = - 40000$ , $F V = 53000$ and 4 payments a year, or solve

40000 (1 + \frac{r}{400})^{24} = 53000 \Rightarrow r = 4.72 r \approx 4.72%

Question 9

16 marks

Wheat heights: finding a normal distribution and conditional probability

Statistics and probability

(a)

2 marks

The three regions add to 1.

P (94.6 < H < 98.1) = 1 - 0.288 - 0.434 = 0.278

(b)

5 marks

The z-scores are $z = - 0.5592$ for $P (H < 94.6) = 0.288$ and $z = 0.1662$ for $P (H > 98.1) = 0.434$ .
Solve the pair of equations.

\frac{94.6 - μ}{σ} = - 0.5592, \frac{98.1 - μ}{σ} = 0.1662 μ = 97.3, σ = 4.82

(c)

6 marks

A plant is ready with probability 0.434. Let $X \sim B (100, 0.434)$ .
Given fewer than 49:

P (X = 34) \approx 0.0133 P (X = 34 ∣ X < 49) = \frac{P ( X = 34 )}{P ( X \leq 48 )} = \frac{0.01332}{0.8482} \approx 0.0157

(d)

3 marks

The interquartile range of a normal distribution is $2 (0.6745) σ$ .

2 (0.6745) d = 4.82 \Rightarrow d = 3.57

Our analysis

Where the marks are

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

Calculus23 marks · 29%
Q1, Q5, Q7
Number and algebra20 marks · 25%
Q6, Q8
Geometry and trigonometry16 marks · 20%
Q2, Q4, Q7
Statistics and probability16 marks · 20%
Q9
Functions5 marks · 6%
Q3

Mentor insights

Traps to watch for

01
Normal gradient · Q1
The normal is perpendicular to the tangent: its gradient is $- \frac{1}{f ^{'} ( 3 )}$ .
02
Sector of a ring · Q4
The area is the big sector minus the small sector with the same angle.
03
Express h in terms of x · Q7
The cylinder volume $π x^{2} h = 45$ gives $h$ .
04
Option B is a geometric series · Q8
Payments rise by 6% each month, so the total of 36 payments is $a \frac{r ^{36} - 1}{r - 1}$ with $a = 2000$ and $r = 1.06$ .
05
Two inverse-normal equations · Q9
Each given probability gives a z-score and an equation in $μ$ and $σ$ .

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: November 2023
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 2023 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 · 35 marks

Question 1

7 marks

Tangent, normal and where the normal meets the parabola again (f = x^2 - 8x)

Calculus

(a)

3 marks

Differentiate and substitute $x = 3$ .
The normal is perpendicular to the tangent.

f^{'} (x) = 2 x - 8 \Rightarrow f^{'} (3) = - 2 m_{normal} = \frac{1}{2}

(b)

1 mark

Line through $A (3, - 15)$ with gradient $\frac{1}{2}$ .

y + 15 = \frac{1}{2} (x - 3) \Rightarrow y = \frac{1}{2} x - 16.5

(c)

3 marks

Set the normal equal to the curve.
At $x = 5.5$ , $y = \frac{1}{2} (5.5) - 16.5 = - 13.75$ .

x^{2} - 8 x = \frac{1}{2} x - 16.5 \Rightarrow x^{2} - 8.5 x + 16.5 = 0 \Rightarrow x = 3 or x = 5.5 B = (5.5, - 13.75)

Question 2

6 marks

A pyramid on a rectangular base: a length and an angle

Geometry and trigonometry

(a)

2 marks

$B V = (3 - 6, 4 - 8, 9 - 0) = (- 3, - 4, 9)$ .

B V = 9 + 16 + 81 = 106 \approx 10.3

(b)

4 marks

$V B = (3, 4, - 9)$ and $V C = (3, - 4, - 9)$ .

V B \cdot V C = 9 - 16 + 81 = 74, ∣ V B ∣ = ∣ V C ∣ = 106 cos B V C = \frac{74}{106} \Rightarrow B V C = 45. 7^{\circ}

Question 3

5 marks

f(x) = e^x - 3x - 4: a sketch and transformations

Functions

(a)

3 marks

Features on $- 4 \leq x \leq 3$ : $y$ -intercept $f (0) = - 3$ ; roots near $x = - 1.4$ and $x = 2.4$ ; a minimum at $x = ln 3 = 1.10$ with $f = - 4.30$ ; end points $f (- 4) = 8.02$ and $f (3) = 7.09$ .
Sketch: the curve starts at $(- 4, 8.0)$ , falls through a root in $(- 2, - 1)$ and the $y$ -intercept $(0, - 3)$ to the minimum near $(1.1, - 4.3)$ , then rises through a root in $(2, 3)$ to $(3, 7.1)$ .

(b)

2 marks

Compare $g (x) = e^{2 x} - 6 x - 7$ with $f (2 x) = e^{2 x} - 6 x - 4$ .

g (x) = f (2 x) - 3 k = \frac{1}{2}, c = - 3

Question 4

7 marks

A letter C between two circles: area and perimeter

Geometry and trigonometry

(a)

2 marks

Area of a sector is $\frac{1}{2} R^{2} θ$ with $θ = 5.2$ .

A = \frac{1}{2} (10)^{2} (5.2) - \frac{1}{2} r^{2} (5.2) = 260 - 2.6 r^{2}

(b)

5 marks

Set $260 - 2.6 r^{2} = 64$ .
The perimeter: outer arc $10 \times 5.2 = 52$ , inner arc $5.2 r = 45.1$ , and two straight ends each of length $10 - r = 1.32$ .

r^{2} = \frac{196}{2.6} = 75.38 \Rightarrow r = 8.68 P = 52 + 45.1 + 2 (1.32) \approx 99.8 cm

Question 5

5 marks

A particle with a square-root sine displacement: rest and distance

Calculus

(a)

2 marks

The displacement is $s (t) = 4.3 sin (3 t + 5)$ . The velocity is zero where $cos (3 t + 5) = 0$ .
(The value $3 (0) + 5 = 2.24$ is already past $\frac{π}{2}$ , so $\frac{3 π}{2}$ is the first one reached.)

3 t + 5 = \frac{3 π}{2} \Rightarrow 3 t + 5 = \frac{9 π ^{2}}{4} \Rightarrow q = 5.74

(b)

3 marks

Distance is the integral of speed from 0 to $q$ .
Check: $s (0) = 3.38$ and $s (q) = - 4.3$ , and the particle moves in one direction in between, so the distance is $3.38 + 4.30 = 7.68$ .

\int_{0}^{5.74} ∣ v (t) ∣ d t = 7.68 m

Question 6

5 marks

Finding a from an expected value and total probability

Number and algebra

5 marks

The probabilities are $k, k^{2}, a, k^{3}$ for $x = 1, 2, 3, 4$ .
Eliminate $a$ using $a = 1 - k - k^{2} - k^{3}$ .
Solve for $k$ with $0 < k < 1$ .

k + k^{2} + a + k^{3} = 1 E (X) = k + 2 k^{2} + 3 a + 4 k^{3} = 2.3 k + 2 k^{2} + 3 (1 - k - k^{2} - k^{3}) + 4 k^{3} = 2.3 \Rightarrow k^{3} - k^{2} - 2 k + 0.7 = 0 k = 0.316 a = 1 - 0.3159 - 0.0998 - 0.0315 \approx 0.553

Section B · Extended response

Questions 7–9 · 45 marks

Question 7

14 marks

A capsule: surface area, minimum and its justification (cylinder volume 45)

Geometry and trigonometry
Calculus

(a)

3 marks

The cylinder volume gives $h = \frac{45}{π x ^{2}}$ . The curved surface of the cylinder is $2 π x h$ and the two hemispheres together have area $4 π x^{2}$ .

S = 2 π x \cdot \frac{45}{π x ^{2}} + 4 π x^{2} = \frac{90}{x} + 4 π x^{2}

(b)

5 marks

Differentiate $S = 90 x^{- 1} + 4 π x^{2}$ .
Set it to zero.

\frac{d S}{d x} = - \frac{90}{x ^{2}} + 8 π x 8 π x^{3} = 90 \Rightarrow a = (\frac{45}{4 π})^{1/3}

(c)

6 marks

Differentiate again.
For $x > 0$ this is positive, so $S$ is a minimum at $x = a$ .
Minimum surface area:

\frac{d ^{2} S}{d x ^{2}} = \frac{180}{x ^{3}} + 8 π S_{m i n} = \frac{90}{1.530} + 4 π (1.530)^{2} \approx 88.2 cm^{2}

Question 8

15 marks

Prize money in two options and an investment (second version): series and interest

Number and algebra

(a)

5 marks

Option A is 36 payments of 5500.
Option B is a geometric series.

5500 \times 36 = $198 000 S_{36} = \frac{2000 ( 1.0 6 ^{36} - 1 )}{1.06 - 1} \approx $238 241.73

(b)

1 mark

Compound interest with 12 periods a year.

120000 (1 + \frac{0.04}{12})^{12 n}

(c)

6 marks

After six months Sorin has
Daniela's six payments of Option B total
The total value is
For the total after $m$ months to be at least 250 000, solve
After 26 months it is $249 158 and after 27 months it is $258 693.

120000 (1 + \frac{0.04}{12})^{6} = $122 420.09 \frac{2000 ( 1.0 6 ^{6} - 1 )}{0.06} = $13 950.64 122420.09 + 13950.64 = $136 370.73 120000 (1 + \frac{0.04}{12})^{m} + \frac{2000 ( 1.0 6 ^{m} - 1 )}{0.06} \geq 250000 m = 27 months

(d)

3 marks

Use a finance solver with $N = 24$ quarters, $P V = - 40000$ , $F V = 53000$ and 4 payments a year, or solve

40000 (1 + \frac{r}{400})^{24} = 53000 \Rightarrow r = 4.72 r \approx 4.72%

Question 9

16 marks

Wheat heights: finding a normal distribution and conditional probability

Statistics and probability

(a)

2 marks

The three regions add to 1.

P (94.6 < H < 98.1) = 1 - 0.288 - 0.434 = 0.278

(b)

5 marks

The z-scores are $z = - 0.5592$ for $P (H < 94.6) = 0.288$ and $z = 0.1662$ for $P (H > 98.1) = 0.434$ .
Solve the pair of equations.

\frac{94.6 - μ}{σ} = - 0.5592, \frac{98.1 - μ}{σ} = 0.1662 μ = 97.3, σ = 4.82

(c)

6 marks

A plant is ready with probability 0.434. Let $X \sim B (100, 0.434)$ .
Given fewer than 49:

P (X = 34) \approx 0.0133 P (X = 34 ∣ X < 49) = \frac{P ( X = 34 )}{P ( X \leq 48 )} = \frac{0.01332}{0.8482} \approx 0.0157

(d)

3 marks

The interquartile range of a normal distribution is $2 (0.6745) σ$ .

2 (0.6745) d = 4.82 \Rightarrow d = 3.57

Our analysis

Where the marks are

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

Calculus23 marks · 29%
Q1, Q5, Q7
Number and algebra20 marks · 25%
Q6, Q8
Geometry and trigonometry16 marks · 20%
Q2, Q4, Q7
Statistics and probability16 marks · 20%
Q9
Functions5 marks · 6%
Q3

Mentor insights

Traps to watch for

01
Normal gradient · Q1
The normal is perpendicular to the tangent: its gradient is $- \frac{1}{f ^{'} ( 3 )}$ .
02
Sector of a ring · Q4
The area is the big sector minus the small sector with the same angle.
03
Express h in terms of x · Q7
The cylinder volume $π x^{2} h = 45$ gives $h$ .
04
Option B is a geometric series · Q8
Payments rise by 6% each month, so the total of 36 payments is $a \frac{r ^{36} - 1}{r - 1}$ with $a = 2000$ and $r = 1.06$ .
05
Two inverse-normal equations · Q9
Each given probability gives a z-score and an equation in $μ$ and $σ$ .

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