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IB Maths AI HL · Paper 1

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2

Session: May 2024
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 17

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2024 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–17 · 110 marks

Question 1

6 marks

A savings account and then an annuity: finding the rate and the balance

Number and algebra

(a)

3 marks

Compound growth over 96 months.

100000 (1 + \frac{I}{1200})^{96} = 150000 \Rightarrow (1 + \frac{I}{1200})^{96} = 1.5 I = 1200 (1. 5^{1/96} - 1) \approx 5.08

(b)

3 marks

Use the finance solver with $N = 120$ , $I = 6.1%$ , $P V = 150000$ , $P M T = - 1000$ and 12 payments a year.
Check by iterating the monthly rule $B \to B (1 + 0.061/12) - 1000$ one hundred and twenty times, which gives $110 866.71.

F V \approx $110 867

Question 2

7 marks

Hours of study against marks: correlation, regression and a prediction

Statistics and probability

(a)

4 marks

Enter the five pairs $(t, m)$ in the calculator.

r \approx 0.995 m = 10.6 t + 43.9

(b)

2 marks

Each extra hour adds the gradient of the line, so 1.5 hours adds

10.6032 \times 1.5 \approx 15.9 marks

(c)

1 mark

The model comes from only five students, and it ignores other factors such as how able the students are. Also, a student on 86 marks cannot gain 15.9 more, because the test is out of 100 (the model would predict a score above the maximum).

Question 3

6 marks

Celsius and Fahrenheit: converting a formula and a mean and standard deviation

Functions
Statistics and probability

(a)

3 marks

(i) Rearrange $F = 1.8 C + 32$ .

(ii) Put $F = 77$ .

C = \frac{F - 32}{1.8} = \frac{5}{9} (F - 32) C = \frac{77 - 32}{1.8} = 25

(b)

3 marks

(i) The mean transforms like every value.

(ii) Adding 32 shifts every value and leaves the spread unchanged. The standard deviation is only multiplied by 1.8.

1.8 (17) + 32 = 62.6^{\circ} F 1.8 \times 9 = 16.2^{\circ} F

Question 4

4 marks

Euler's method for dy/dx = log10(x + y)

Number and algebra

4 marks

Start at $(x_{0}, y_{0}) = (0, 1)$ and take 20 steps of size 0.1.
Continue the same rule until $x = 2$ .

y_{1} = 1 + 0.1 lo g_{10} (0 + 1) = 1 y_{2} = 1 + 0.1 lo g_{10} (0.1 + 1) \approx 1.00414 y (2) \approx 1.61

Question 5

6 marks

Perpendicular bisectors of a triangle and its Voronoi diagram

Geometry and trigonometry

(a)

2 marks

$B (9, 6)$ and $C (11, 2)$ have midpoint $(10, 4)$ and gradient $\frac{2 - 6}{11 - 9} = - 2$ . The bisector has gradient $\frac{1}{2}$ .

y - 4 = \frac{1}{2} (x - 10) \Rightarrow y = 0.5 x - 1

(b)

2 marks

Solve $0.5 x - 1 = - 3 x + 23$ .

3.5 x = 24 \Rightarrow x = 6.857, y = 0.5 (6.857) - 1 = 2.429 V \approx (6.857, 2.429)

(c)

2 marks

The Voronoi diagram has three edges, all starting at $V$ and going off to infinity, one on each bisector.
The edge between $A$ and $B$ is on $y = - 3 x + 23$ , going up and to the left of $V$ .
The edge between $B$ and $C$ is on $y = 0.5 x - 1$ , going up and to the right of $V$ .
The edge between $A$ and $C$ is on the bisector $y = 4 x - 25$ (midpoint $(7, 3)$ and gradient 4), going down and to the left of $V$ .
These directions are where each pair of sites are the two nearest. Because angle $B$ in the triangle is obtuse, $V$ lies outside triangle $A B C$ .

Question 6

7 marks

A new car's monthly sales as a geometric sequence

Number and algebra

(a)

2 marks

The twelfth term of a geometric sequence with $u_{1} = 40$ and $r = 1.1$ .

u_{12} = 40 (1.1)^{11} \approx 114

(b)

5 marks

(i) The total for 2025 is the sum of the first 12 terms.

(ii) The total for 2026 is months 13 to 24, so $S_{24} - S_{12}$ .

S_{12} = \frac{40 ( 1. 1 ^{12} - 1 )}{1.1 - 1} \approx 855 S_{24} = \frac{40 ( 1. 1 ^{24} - 1 )}{0.1} \approx 3539.9 S_{24} - S_{12} \approx 3539.9 - 855.4 \approx 2685

Question 7

8 marks

Big cats: mass against shoulder height, and energy needs

Functions

(a)

4 marks

(i) $m = a h^{3}$ , and for the cheetah $64 = a (0.8)^{3}$ .

(ii) Substitute $h = 0.75$ .

a = \frac{64}{0.512} = 125 \Rightarrow m = 125 h^{3} m = 125 (0.75)^{3} \approx 52.7 kg

(b)

4 marks

Find the shoulder height of the lion from its mass.
Since $E = c h^{2}$ , the ratio of the lion's to the cheetah's energy needs is $(\frac{h _{L}}{h _{C}})^{2}$ .

220 = 125 h^{3} \Rightarrow h = (1.76)^{1/3} \approx 1.207 m k = (\frac{1.2074}{0.8})^{2} \approx 2.28

Question 8

5 marks

Walking all of the roads: odd vertices and the shortest closed route

Geometry and trigonometry

(a)

1 mark

Count the roads at each vertex. $B$ has roads to $A$ , $C$ and $D$ (3) and $C$ has roads to $A$ , $B$ and $F$ (3). Every other vertex has an even number.

B and C

(b)

4 marks

The route has to walk the cheapest path between $B$ and $C$ twice. There are two possible paths: the direct road of length $x$ , and $B \to A \to C$ of length $3 + 2 = 5$ .
If $x \leq 5$ , repeat the road $B C$ .
If $x \geq 5$ , repeat $B A$ and $A C$ .

33 + x + x = 33 + 2 x (x \leq 5) 33 + x + 5 = 38 + x (x \geq 5)

Question 9

7 marks

Integrating 8 over 2x + 3 and an exact area

Calculus

(a)

3 marks

The inside function is $2 x + 3$ , with derivative 2.

\int \frac{8}{2 x + 3} d x = \frac{8}{2} ln (2 x + 3) + c = 4 ln (2 x + 3) + c

(b)

4 marks

The curve is above the $x$ -axis between $x = 0$ and $x = 6$ , so the area is the definite integral.
Combine with a log law.

Area = [4 ln (2 x + 3)]_{0}^{6} = 4 ln 15 - 4 ln 3 4 ln \frac{15}{3} = 4 ln 5

Question 10

5 marks

The arc BQ of a semicircle containing a rectangle

Number and algebra

5 marks

The radius is the length $O B$ .
The angle $B O P$ is in the right triangle with sides 5 and 8.
The semicircle angle is $π$ , so the angle $B O Q$ is the rest.
Arc length with radians.

r = 5^{2} + 8^{2} = 89 \approx 9.434 B O P = tan^{- 1} \frac{5}{8} \approx 0.5586 B O Q = π - 0.5586 \approx 2.583 B Q = r θ = 89 (2.583) \approx 24.4 m

Question 11

8 marks

Rotation matrices and the double angle formulae

Number and algebra

(a)

2 marks

A rotation counter-clockwise through $θ$ has matrix $(cos θ sin θ - sin θ cos θ)$ . Put $θ = 2 α$ .

R (2 α) = (cos 2 α sin 2 α - sin 2 α cos 2 α)

(b)

2 marks

Multiply the matrix by itself.

R (α) R (α) = (cos^{2} α - sin^{2} α 2 sin α cos α - 2 sin α cos α cos^{2} α - sin^{2} α)

(c)

4 marks

(i) Doing two rotations of $α$ one after the other is a rotation through $2 α$ , so $R (α) R (α) = R (2 α)$ and the matching entries are equal. The bottom left entries give

(ii) The top left entries give $cos 2 α = cos^{2} α - sin^{2} α$ . Replace $cos^{2} α$ by $1 - sin^{2} α$ .

sin 2 α = 2 sin α cos α cos 2 α = 1 - sin^{2} α - sin^{2} α = 1 - 2 sin^{2} α

Question 12

8 marks

Charging a phone: an exponential model against a quadratic model

Number and algebra

(a)

2 marks

Solve $100 - 100 \times 2^{- t} = 40$ .

2^{- t} = 0.6 \Rightarrow t = - lo g_{2} 0.6 \approx 0.737 hours

(b)

4 marks

The model values at $t = 1, 2, 3, 4$ are $50, 75, 87.5, 93.75$ .

S S_{r es} = (48 - 50)^{2} + (74 - 75)^{2} + (86 - 87.5)^{2} + (91 - 93.75)^{2} S S_{r es} = 4 + 1 + 2.25 + 7.5625 = 14.8125 \approx 14.8

(c)

2 marks

(i) The sum of squared residuals of the quadratic (2.45) is smaller than for the exponential model (14.8), so it fits the data better.

(ii) A quadratic eventually turns downwards and would predict the charge going down, or above 100, which cannot happen. The exponential model levels off towards 100% and passes through $(0, 0)$ .

Question 13

6 marks

A triangle on the Argand diagram with vertices 0, z1 and z2

Number and algebra

(a)

2 marks

(i) The modulus.

(ii) The argument, with $z_{1}$ in the first quadrant.

∣ z_{1} ∣ = 4^{2} + 5^{2} = 41 \approx 6.40 ar g z_{1} = tan^{- 1} \frac{5}{4} \approx 0.896

(b)

4 marks

$z_{2} = 3 e^{2 i}$ has modulus 3 and argument 2. The angle at the origin between the two vertices is the difference of the arguments.
Use area $= \frac{1}{2} ab sin C$ .

2 - 0.896 = 1.104 Area = \frac{1}{2} 41 (3) sin (1.104) \approx 8.58

Question 14

4 marks

Two model cars: displacement and when the second catches the first

Functions

(a)

2 marks

Car A reaches displacement $x$ at time $t_{A} = (\frac{x}{4})^{2} = \frac{x ^{2}}{16}$ .
Car B takes half as long, $\frac{x ^{2}}{32}$ , and starts at $t = 3$ .

t = 3 + \frac{x ^{2}}{32} \Rightarrow x_{B} = 32 (t - 3) = 4 2 t - 6

(b)

2 marks

B catches A when the displacements are equal.

4 t = 4 2 t - 6 \Rightarrow t = 2 t - 6 t = 6 s

Question 15

6 marks

A phase portrait with eigenvalues -1 and 2, and when a population dies out

Calculus

(a)

3 marks

(i) The eigenvalue $- 1$ belongs to the eigenvector $(31)$ , so arrows on the line $y = \frac{1}{3} x$ point towards the origin. The eigenvalue 2 belongs to $(13)$ , so arrows on the line $y = 3 x$ point away from the origin.

(ii) The two lines split the first quadrant into three regions. A trajectory in each one comes in alongside the line $y = \frac{1}{3} x$ , bends round near the origin, and then runs off alongside the line $y = 3 x$ (or the edge of the diagram). In the region between the lines the curve bends from the stable line to the unstable one. In the region below $y = \frac{1}{3} x$ it ends by moving down to the $x$ -axis. In the region above $y = 3 x$ it ends by moving up along the $y$ -axis side.

(b)

3 marks

The population of $X$ dies out for any start on or below the line $y = \frac{1}{3} x$ (the stable eigenvector direction), and survives above it.
With $x = 252$ , the population of $Y$ must satisfy
Currently $y = 60$ , so at least 85 are needed (since $y$ is a whole number of animals).

y > \frac{1}{3} x y > \frac{252}{3} = 84 85 - 60 = 25

Question 16

8 marks

A ladder leaning on a curved wall at a given angle

Number and algebra

8 marks

The gradient of the ladder is $- tan 7 5^{\circ} \approx - 3.732$ .
Differentiate the curve.
Set the gradient equal to the ladder's and solve for $x$ (radians).
The height above the ground at the point of contact is

\frac{d y}{d x} = - 5.5 sin (1.1 x) - 5.5 sin (1.1 x) = - 3.732 \Rightarrow x \approx 0.678 y = 5 cos (1.1 \times 0.678) \approx 3.67 m

Question 17

9 marks

Two coffee shops: a transition matrix, its eigenvectors and a closed form

Number and algebra

(a)

2 marks

Solve $(M - 0.65 I) v = 0$ with $M = (0.9 0.1 0.25 0.75)$ .

(0.25 0.1 0.25 0.1) (x y) = (00) \Rightarrow y = - x v = (1 - 1)

(b)

1 mark

A power of a diagonal matrix is found by raising each diagonal entry to that power.

D^{n} = (0.6 5^{n} 0 01)

(c)

6 marks

With the eigenvectors as columns, $P = (1 - 1 52)$ and $P^{- 1} = \frac{1}{7} (21 - 5 1)$ .
The starting state is 7000 at Apollo and 0 at Phoebe's, $(70000)$ .
Apply $D^{n}$ and then $P$ .
The second entry is the number who favour Phoebe's coffee shop after $n$ months.

P^{- 1} (70000) = (20001000) (1 - 1 52) (2000 (0.65)^{n} 1000) = (2000 (0.65)^{n} + 5000 - 2000 (0.65)^{n} + 2000) 2000 - 2000 (0.65)^{n} = 2000 (1 - 0.6 5^{n})

Our analysis

Where the marks are

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

Number and algebra61 marks · 55%
Q1, Q4, Q6, Q10, Q11, Q12, Q13, Q16, Q17
Functions15 marks · 14%
Q3, Q7, Q14
Calculus13 marks · 12%
Q9, Q15
Geometry and trigonometry11 marks · 10%
Q5, Q8
Statistics and probability10 marks · 9%
Q2, Q3

Mentor insights

Traps to watch for

01
Direct proportion to a power · Q7
$m = a h^{3}$ with a constant found from the data.
02
Composition of rotations · Q11
Rotating through $α$ twice is the same as rotating through $2 α$ , so the two matrices must be equal.
03
Sum of squared residuals · Q12
$S S_{r es} = \sum (data - model)^{2}$ . A smaller value means a better fit to the data given.
04
Tangent gradient · Q16
The ladder just touches the wall, so its gradient is the gradient of the curve at the point of contact.
05
Eigenvector · Q17
Solve $(M - λ I) v = 0$ .

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 AI HL · Paper 1

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2

Session: May 2024
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 17

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2024 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–17 · 110 marks

Question 1

6 marks

A savings account and then an annuity: finding the rate and the balance

Number and algebra

(a)

3 marks

Compound growth over 96 months.

100000 (1 + \frac{I}{1200})^{96} = 150000 \Rightarrow (1 + \frac{I}{1200})^{96} = 1.5 I = 1200 (1. 5^{1/96} - 1) \approx 5.08

(b)

3 marks

Use the finance solver with $N = 120$ , $I = 6.1%$ , $P V = 150000$ , $P M T = - 1000$ and 12 payments a year.
Check by iterating the monthly rule $B \to B (1 + 0.061/12) - 1000$ one hundred and twenty times, which gives $110 866.71.

F V \approx $110 867

Question 2

7 marks

Hours of study against marks: correlation, regression and a prediction

Statistics and probability

(a)

4 marks

Enter the five pairs $(t, m)$ in the calculator.

r \approx 0.995 m = 10.6 t + 43.9

(b)

2 marks

Each extra hour adds the gradient of the line, so 1.5 hours adds

10.6032 \times 1.5 \approx 15.9 marks

(c)

1 mark

Question 3

6 marks

Celsius and Fahrenheit: converting a formula and a mean and standard deviation

Functions
Statistics and probability

(a)

3 marks

(i) Rearrange $F = 1.8 C + 32$ .

(ii) Put $F = 77$ .

C = \frac{F - 32}{1.8} = \frac{5}{9} (F - 32) C = \frac{77 - 32}{1.8} = 25

(b)

3 marks

(i) The mean transforms like every value.

(ii) Adding 32 shifts every value and leaves the spread unchanged. The standard deviation is only multiplied by 1.8.

1.8 (17) + 32 = 62.6^{\circ} F 1.8 \times 9 = 16.2^{\circ} F

Question 4

4 marks

Euler's method for dy/dx = log10(x + y)

Number and algebra

4 marks

Start at $(x_{0}, y_{0}) = (0, 1)$ and take 20 steps of size 0.1.
Continue the same rule until $x = 2$ .

y_{1} = 1 + 0.1 lo g_{10} (0 + 1) = 1 y_{2} = 1 + 0.1 lo g_{10} (0.1 + 1) \approx 1.00414 y (2) \approx 1.61

Question 5

6 marks

Perpendicular bisectors of a triangle and its Voronoi diagram

Geometry and trigonometry

(a)

2 marks

$B (9, 6)$ and $C (11, 2)$ have midpoint $(10, 4)$ and gradient $\frac{2 - 6}{11 - 9} = - 2$ . The bisector has gradient $\frac{1}{2}$ .

y - 4 = \frac{1}{2} (x - 10) \Rightarrow y = 0.5 x - 1

(b)

2 marks

Solve $0.5 x - 1 = - 3 x + 23$ .

3.5 x = 24 \Rightarrow x = 6.857, y = 0.5 (6.857) - 1 = 2.429 V \approx (6.857, 2.429)

(c)

2 marks

The Voronoi diagram has three edges, all starting at $V$ and going off to infinity, one on each bisector.
The edge between $A$ and $B$ is on $y = - 3 x + 23$ , going up and to the left of $V$ .
The edge between $B$ and $C$ is on $y = 0.5 x - 1$ , going up and to the right of $V$ .
The edge between $A$ and $C$ is on the bisector $y = 4 x - 25$ (midpoint $(7, 3)$ and gradient 4), going down and to the left of $V$ .
These directions are where each pair of sites are the two nearest. Because angle $B$ in the triangle is obtuse, $V$ lies outside triangle $A B C$ .

Question 6

7 marks

A new car's monthly sales as a geometric sequence

Number and algebra

(a)

2 marks

The twelfth term of a geometric sequence with $u_{1} = 40$ and $r = 1.1$ .

u_{12} = 40 (1.1)^{11} \approx 114

(b)

5 marks

(i) The total for 2025 is the sum of the first 12 terms.

(ii) The total for 2026 is months 13 to 24, so $S_{24} - S_{12}$ .

S_{12} = \frac{40 ( 1. 1 ^{12} - 1 )}{1.1 - 1} \approx 855 S_{24} = \frac{40 ( 1. 1 ^{24} - 1 )}{0.1} \approx 3539.9 S_{24} - S_{12} \approx 3539.9 - 855.4 \approx 2685

Question 7

8 marks

Big cats: mass against shoulder height, and energy needs

Functions

(a)

4 marks

(i) $m = a h^{3}$ , and for the cheetah $64 = a (0.8)^{3}$ .

(ii) Substitute $h = 0.75$ .

a = \frac{64}{0.512} = 125 \Rightarrow m = 125 h^{3} m = 125 (0.75)^{3} \approx 52.7 kg

(b)

4 marks

Find the shoulder height of the lion from its mass.
Since $E = c h^{2}$ , the ratio of the lion's to the cheetah's energy needs is $(\frac{h _{L}}{h _{C}})^{2}$ .

220 = 125 h^{3} \Rightarrow h = (1.76)^{1/3} \approx 1.207 m k = (\frac{1.2074}{0.8})^{2} \approx 2.28

Question 8

5 marks

Walking all of the roads: odd vertices and the shortest closed route

Geometry and trigonometry

(a)

1 mark

Count the roads at each vertex. $B$ has roads to $A$ , $C$ and $D$ (3) and $C$ has roads to $A$ , $B$ and $F$ (3). Every other vertex has an even number.

B and C

(b)

4 marks

The route has to walk the cheapest path between $B$ and $C$ twice. There are two possible paths: the direct road of length $x$ , and $B \to A \to C$ of length $3 + 2 = 5$ .
If $x \leq 5$ , repeat the road $B C$ .
If $x \geq 5$ , repeat $B A$ and $A C$ .

33 + x + x = 33 + 2 x (x \leq 5) 33 + x + 5 = 38 + x (x \geq 5)

Question 9

7 marks

Integrating 8 over 2x + 3 and an exact area

Calculus

(a)

3 marks

The inside function is $2 x + 3$ , with derivative 2.

\int \frac{8}{2 x + 3} d x = \frac{8}{2} ln (2 x + 3) + c = 4 ln (2 x + 3) + c

(b)

4 marks

The curve is above the $x$ -axis between $x = 0$ and $x = 6$ , so the area is the definite integral.
Combine with a log law.

Area = [4 ln (2 x + 3)]_{0}^{6} = 4 ln 15 - 4 ln 3 4 ln \frac{15}{3} = 4 ln 5

Question 10

5 marks

The arc BQ of a semicircle containing a rectangle

Number and algebra

5 marks

The radius is the length $O B$ .
The angle $B O P$ is in the right triangle with sides 5 and 8.
The semicircle angle is $π$ , so the angle $B O Q$ is the rest.
Arc length with radians.

r = 5^{2} + 8^{2} = 89 \approx 9.434 B O P = tan^{- 1} \frac{5}{8} \approx 0.5586 B O Q = π - 0.5586 \approx 2.583 B Q = r θ = 89 (2.583) \approx 24.4 m

Question 11

8 marks

Rotation matrices and the double angle formulae

Number and algebra

(a)

2 marks

A rotation counter-clockwise through $θ$ has matrix $(cos θ sin θ - sin θ cos θ)$ . Put $θ = 2 α$ .

R (2 α) = (cos 2 α sin 2 α - sin 2 α cos 2 α)

(b)

2 marks

Multiply the matrix by itself.

R (α) R (α) = (cos^{2} α - sin^{2} α 2 sin α cos α - 2 sin α cos α cos^{2} α - sin^{2} α)

(c)

4 marks

(i) Doing two rotations of $α$ one after the other is a rotation through $2 α$ , so $R (α) R (α) = R (2 α)$ and the matching entries are equal. The bottom left entries give

(ii) The top left entries give $cos 2 α = cos^{2} α - sin^{2} α$ . Replace $cos^{2} α$ by $1 - sin^{2} α$ .

sin 2 α = 2 sin α cos α cos 2 α = 1 - sin^{2} α - sin^{2} α = 1 - 2 sin^{2} α

Question 12

8 marks

Charging a phone: an exponential model against a quadratic model

Number and algebra

(a)

2 marks

Solve $100 - 100 \times 2^{- t} = 40$ .

2^{- t} = 0.6 \Rightarrow t = - lo g_{2} 0.6 \approx 0.737 hours

(b)

4 marks

The model values at $t = 1, 2, 3, 4$ are $50, 75, 87.5, 93.75$ .

S S_{r es} = (48 - 50)^{2} + (74 - 75)^{2} + (86 - 87.5)^{2} + (91 - 93.75)^{2} S S_{r es} = 4 + 1 + 2.25 + 7.5625 = 14.8125 \approx 14.8

(c)

2 marks

(i) The sum of squared residuals of the quadratic (2.45) is smaller than for the exponential model (14.8), so it fits the data better.

(ii) A quadratic eventually turns downwards and would predict the charge going down, or above 100, which cannot happen. The exponential model levels off towards 100% and passes through $(0, 0)$ .

Question 13

6 marks

A triangle on the Argand diagram with vertices 0, z1 and z2

Number and algebra

(a)

2 marks

(i) The modulus.

(ii) The argument, with $z_{1}$ in the first quadrant.

∣ z_{1} ∣ = 4^{2} + 5^{2} = 41 \approx 6.40 ar g z_{1} = tan^{- 1} \frac{5}{4} \approx 0.896

(b)

4 marks

$z_{2} = 3 e^{2 i}$ has modulus 3 and argument 2. The angle at the origin between the two vertices is the difference of the arguments.
Use area $= \frac{1}{2} ab sin C$ .

2 - 0.896 = 1.104 Area = \frac{1}{2} 41 (3) sin (1.104) \approx 8.58

Question 14

4 marks

Two model cars: displacement and when the second catches the first

Functions

(a)

2 marks

Car A reaches displacement $x$ at time $t_{A} = (\frac{x}{4})^{2} = \frac{x ^{2}}{16}$ .
Car B takes half as long, $\frac{x ^{2}}{32}$ , and starts at $t = 3$ .

t = 3 + \frac{x ^{2}}{32} \Rightarrow x_{B} = 32 (t - 3) = 4 2 t - 6

(b)

2 marks

B catches A when the displacements are equal.

4 t = 4 2 t - 6 \Rightarrow t = 2 t - 6 t = 6 s

Question 15

6 marks

A phase portrait with eigenvalues -1 and 2, and when a population dies out

Calculus

(a)

3 marks

(b)

3 marks

The population of $X$ dies out for any start on or below the line $y = \frac{1}{3} x$ (the stable eigenvector direction), and survives above it.
With $x = 252$ , the population of $Y$ must satisfy
Currently $y = 60$ , so at least 85 are needed (since $y$ is a whole number of animals).

y > \frac{1}{3} x y > \frac{252}{3} = 84 85 - 60 = 25

Question 16

8 marks

A ladder leaning on a curved wall at a given angle

Number and algebra

8 marks

The gradient of the ladder is $- tan 7 5^{\circ} \approx - 3.732$ .
Differentiate the curve.
Set the gradient equal to the ladder's and solve for $x$ (radians).
The height above the ground at the point of contact is

\frac{d y}{d x} = - 5.5 sin (1.1 x) - 5.5 sin (1.1 x) = - 3.732 \Rightarrow x \approx 0.678 y = 5 cos (1.1 \times 0.678) \approx 3.67 m

Question 17

9 marks

Two coffee shops: a transition matrix, its eigenvectors and a closed form

Number and algebra

(a)

2 marks

Solve $(M - 0.65 I) v = 0$ with $M = (0.9 0.1 0.25 0.75)$ .

(0.25 0.1 0.25 0.1) (x y) = (00) \Rightarrow y = - x v = (1 - 1)

(b)

1 mark

A power of a diagonal matrix is found by raising each diagonal entry to that power.

D^{n} = (0.6 5^{n} 0 01)

(c)

6 marks

With the eigenvectors as columns, $P = (1 - 1 52)$ and $P^{- 1} = \frac{1}{7} (21 - 5 1)$ .
The starting state is 7000 at Apollo and 0 at Phoebe's, $(70000)$ .
Apply $D^{n}$ and then $P$ .
The second entry is the number who favour Phoebe's coffee shop after $n$ months.

P^{- 1} (70000) = (20001000) (1 - 1 52) (2000 (0.65)^{n} 1000) = (2000 (0.65)^{n} + 5000 - 2000 (0.65)^{n} + 2000) 2000 - 2000 (0.65)^{n} = 2000 (1 - 0.6 5^{n})

Our analysis

Where the marks are

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

Number and algebra61 marks · 55%
Q1, Q4, Q6, Q10, Q11, Q12, Q13, Q16, Q17
Functions15 marks · 14%
Q3, Q7, Q14
Calculus13 marks · 12%
Q9, Q15
Geometry and trigonometry11 marks · 10%
Q5, Q8
Statistics and probability10 marks · 9%
Q2, Q3

Mentor insights

Traps to watch for

01
Direct proportion to a power · Q7
$m = a h^{3}$ with a constant found from the data.
02
Composition of rotations · Q11
Rotating through $α$ twice is the same as rotating through $2 α$ , so the two matrices must be equal.
03
Sum of squared residuals · Q12
$S S_{r es} = \sum (data - model)^{2}$ . A smaller value means a better fit to the data given.
04
Tangent gradient · Q16
The ladder just touches the wall, so its gradient is the gradient of the curve at the point of contact.
05
Eigenvector · Q17
Solve $(M - λ I) v = 0$ .

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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