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IB Maths AI HL · Paper 1

Mathematics: Applications & Interpretation, Higher Level

Session: November 2022
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 November 2022 Global paper, which your school or the IB can provide.

Practice mode

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Questions

Questions 1–17 · 110 marks

Question 1

5 marks

Is favourite berry independent of income? A chi-squared test

Statistics and probability

(a)

1 mark

The alternative hypothesis is that they are not independent.

H_{0} : a person’s favourite berry is independent of their income level

(b)

2 marks

Enter the observed table as a matrix and run a $χ^{2}$ test for independence ( $ν = 4$ ).

χ^{2} = 2.27

(c)

2 marks

Compare the statistic with the critical value.
Do not reject $H_{0}$ : there is insufficient evidence at the 10% level that favourite berry depends on income level.

2.27 < 7.78

Mentor tip

The $p$ -value, about 0.687, is also much bigger than 0.10.

Question 2

5 marks

Coffee cooling: evaluating, the asymptote and an inverse value

Functions

(a)

2 marks

Substitute $t = 16$ .

T (16) = 71 e^{- 0.0514 (16)} + 23 = 54.2^{\circ} C

(b)

1 mark

As time passes, the exponential term tends to 0.

23^{\circ} C

(c)

2 marks

$T^{- 1} (50) = k$ means $T (k) = 50$ .

71 e^{- 0.0514 k} + 23 = 50 \Rightarrow e^{- 0.0514 k} = \frac{27}{71} k = \frac{ln ( 71/27 )}{0.0514} = 18.8 minutes

Question 3

5 marks

A ski slope as a triangular prism

Geometry and trigonometry

(a)

2 marks

In the right triangle FCB, the side opposite $2 1^{\circ}$ is $C F = 17$ and the hypotenuse is $B F$ .

sin 2 1^{\circ} = \frac{17}{B F} \Rightarrow B F = 47.4 m

(b)

3 marks

The side $B C = \frac{17}{tan 2 1 ^{\circ}} = 44.29$ m. The base is a rectangle with sides 44 and 44.29, and E is 17 m above D.
The longest horizontal distance from the point below E to the base is the diagonal DB.
The least steep angle is the angle of descent from E to B.

D B = 4 4^{2} + 44.2 9^{2} = 62.4 m tan θ = \frac{17}{62.4} \Rightarrow θ = 15. 2^{\circ}

Question 4

7 marks

A maze as a graph: adjacency matrix and walks of length up to four

Number and algebra

(a)

2 marks

Read the missing entries from the graph. There are two corridors joining S and D (the straight one and the outer curve), and no corridor between A and D.

M_{S D} = M_{D S} = 2, M_{A D} = M_{D A} = 0

(b)

4 marks

The entry in row S, column T of $M^{k}$ is the number of walks of exactly $k$ edges from S to T. Compute the powers of
The S to T entries are 0 for $M$ , 4 for $M^{2}$ , 10 for $M^{3}$ and 64 for $M^{4}$ .

M = 011120101100110111111011201101001110 0 + 4 + 10 + 64 = 78 walks

(c)

1 mark

Some of the walks counted pass through T and then leave it and come back, so they would have found the treasure earlier. The number of distinct ways to reach T for the first time is smaller.

Question 5

7 marks

A bottle-throwing game: total of two games and a fair game

Statistics and probability

(a)

4 marks

List the ways to get a total of two bottles over two games: 1 and 1, 0 and 2, 2 and 0.

P (total 2) = 0.4 \times 0.4 + 0.5 \times 0.1 + 0.1 \times 0.5 = 0.16 + 0.05 + 0.05 = 0.26

(b)

3 marks

The points are $k$ with probability 0.1, $4$ with 0.4 and $- 8$ with 0.5. A fair game has an expected value of 0.

0.1 k + 0.4 (4) + 0.5 (- 8) = 0 \Rightarrow 0.1 k - 2.4 = 0 k = 24

Question 6

6 marks

A cat in an exercise wheel: a sine model for the height of a point

Functions

(a)

1 mark

At $t = 0$ , $sin (0) = 0$ , so $h (0) = c$ .

c = 78

(b)

3 marks

The amplitude is the maximum minus the central height.
The first maximum of a sine curve is a quarter of a period in, so the period is $4 \times 4 = 16$ s.

a = 143 - 78 = 65 b = \frac{2 π}{16} = \frac{π}{8} \approx 0.393

(c)

1 mark

The minimum height is the centre minus the amplitude.

78 - 65 = 13 cm

(d)

1 mark

Twice as long means a period of 32 s.

b = \frac{2 π}{32} = \frac{π}{16} \approx 0.196

Question 7

7 marks

Savings with monthly deposits: finding the number of months

Number and algebra

(a)

4 marks

The monthly rate is $\frac{7.5}{12} %$ and each period is one month. Use a finance solver with PV = $- 800$ , PMT = $- 500$ , $I = 7.5$ , $P / Y = C / Y = 12$ and FV = 10000, and solve for $N$ .
Check: after 17 months the balance is below 10000, and after 18 it is above.

N = 17.2 \Rightarrow k = 18 800 (1.00625)^{18} + 500 \cdot \frac{1.0062 5 ^{18} - 1}{0.00625} = 10389

(b)

3 marks

The total paid in is $800 + 18 \times 500 = 9800$ .

interest = 10389 - 9800 \approx 589 EUR

Question 8

7 marks

Two lines meeting at a point: coordinates and angle

Geometry and trigonometry

(a)

3 marks

Equate the coordinates of $L_{1}$ and $L_{2}$ .
From the second equation $q = 2 - 2 p$ . Put into the first.
Check the third equation: $p + 9 = 8.2$ and $2 q + 1 = 8.2$ . Substituting $p$ into $L_{1}$ :

3 p + 4 = q - 2, 2 p - 1 = 1 - q, p + 9 = 2 q + 1 3 p + 4 = - 2 p \Rightarrow p = - \frac{4}{5}, q = \frac{18}{5} M = (1.6, - 2.6, 8.2)

(b)

4 marks

The direction vectors are $321$ and $1 - 1 2$ .

cos θ = \frac{∣3 - 2 + 2∣}{14 6} = \frac{3}{84} θ = 71. 0^{\circ}

Question 9

8 marks

A matrix transformation: area scale factor and the object point

Geometry and trigonometry

(a)

2 marks

The determinant of $M$ :
The image area is the area of the pentagon times 14.

det M = 2 (1) - (- 4) (3) = 14 12 \times 14 = 168 cm^{2}

(b)

6 marks

The inverse of $(23 - 4 1)$ is $\frac{1}{14} (1 - 3 42)$ .
Multiply it by the image $(2 t - 3, 6 - 5 t)$ .

\frac{1}{14} (1 - 3 42) (2 t - 3 6 - 5 t) = \frac{1}{14} (2 t - 3 + 24 - 20 t - 6 t + 9 + 12 - 10 t) X = (\frac{21 - 18 t}{14}, \frac{21 - 16 t}{14})

Question 10

7 marks

Star magnitude and brightness on a logarithmic scale

Functions

(a)

2 marks

Put $b = 0.0525$ into the equation.

m = 1 - 2.5 lo g_{10} (0.0525) = 4.20

(b)

2 marks

Set $m = 7$ and solve for $b$ .

7 = 1 - 2.5 lo g_{10} b \Rightarrow lo g_{10} b = - 2.4 b = 1 0^{- 2.4} \approx 0.00398

(c)

3 marks

Let $m_{P}$ and $m_{N}$ be the magnitudes with $m_{P} - m_{N} = 3.2$ .
Neptune is about 19 times brighter than Proxima Centauri.

m_{P} - m_{N} = - 2.5 lo g_{10} b_{P} + 2.5 lo g_{10} b_{N} = 2.5 lo g_{10} (\frac{b _{N}}{b _{P}}) = 3.2 \frac{b _{N}}{b _{P}} = 1 0^{3.2/2.5} = 1 0^{1.28} \approx 19.1

Question 11

5 marks

Cubic regression and an enlargement of the curve

Functions

(a)

2 marks

Enter the four points into the calculator and use cubic regression.

y = - 0.00855 x^{3} - 0.234 x^{2} - 0.225 x + 3.20

(b)

3 marks

An enlargement by scale factor 2 about the origin sends a point $(x, y)$ on the life-size curve to $(2 x, 2 y)$ . So the photo curve satisfies $y = 2 g (\frac{x}{2})$ where $g$ is the life-size curve.

g (x) = \frac{1}{2} f (2 x) = \frac{1}{2} [0.0083 (2 x)^{3} - 0.075 (2 x)^{2} - 0.58 (2 x) + 2.2] g (x) = 0.0332 x^{3} - 0.15 x^{2} - 0.58 x + 1.1

Question 12

5 marks

Phase portraits from eigenvalues and eigenvectors

Calculus

(a)

3 marks

Both eigenvalues positive: all paths move away from the origin (portrait D). Eigenvalues 2 and $- 3$ : a saddle, with the outward direction along $(2, 1)$ (portrait C). Eigenvalues $- 2$ and 3: a saddle with the outward direction along $(1, - 1)$ (portrait B).

λ_{1} = 2, λ_{2} = 3 : D λ_{1} = 2, λ_{2} = - 3 : C λ_{1} = - 2, λ_{2} = 3 : B

(b)

2 marks

The real part $- 2$ is negative, so the paths spiral in towards the origin.
At $(3, 0)$ the rate $\frac{d y}{d t} = - 12 < 0$ , so the path is moving downwards at the right-hand side of the origin, which is clockwise.
Sketch: a spiral centred on the origin, starting at $(3, 0)$ and turning clockwise while closing in on the origin, crossing the axes several times.

Question 13

6 marks

Exponential and log models from straight-line plots

Functions

(a)

3 marks

The gradient of the line through $(0, - 1.3)$ and $(6, 14.9)$ :
So $lo g_{10} Q = 2.7 P - 1.3$ . Undo the logarithm.

m = \frac{14.9 - ( - 1.3 )}{6 - 0} = 2.7 Q = 1 0^{2.7 P - 1.3}

(b)

1 mark

For $Q = a ln R + b$ to be a straight line, plot $Q$ on the vertical axis and $ln R$ on the horizontal axis.

(c)

2 marks

Equate the two expressions for $Q$ .
Take $lo g_{10}$ of both sides and solve for $P$ .

1 0^{2.7 P - 1.3} = 4.3 ln R + 12.1 P = \frac{lo g _{10} ( 4.3 ln R + 12.1 ) + 1.3}{2.7}

Question 14

9 marks

A particle with velocity t sin(t^2)

Calculus

(a)

2 marks

Use the product rule with the chain rule for $sin (t^{2})$ .

a = \frac{d v}{d t} = sin (t^{2}) + t \cdot 2 t cos (t^{2}) = sin (t^{2}) + 2 t^{2} cos (t^{2})

(b)

2 marks

Find the maximum of $a (t)$ for $0 \leq t \leq 8$ on a graph.

a_{m a x} \approx 126 m s^{- 2} (at t \approx 7.93)

(c)

3 marks

Integrate $v$ (the substitution is $u = t^{2}$ ).
At $t = 0$ the displacement is 0.

s = \int t sin (t^{2}) d t = - \frac{1}{2} cos (t^{2}) + c 0 = - \frac{1}{2} + c \Rightarrow c = \frac{1}{2} s = \frac{1}{2} (1 - cos (t^{2}))

(d)

2 marks

Since $cos (t^{2}) \leq 1$ , we have $1 - cos (t^{2}) \geq 0$ .
So the displacement is never negative.

s = \frac{1}{2} (1 - cos (t^{2})) \geq 0 for all t

Question 15

5 marks

Euler's method for a second-order equation: the largest charge

Number and algebra

5 marks

Put $y = \frac{d q}{d t}$ , so $\frac{d y}{d t} = 200 - 5 y - 20 q$ . Start from $q_{0} = 1$ , $y_{0} = 8$ , $t_{0} = 0$ with $h = 0.1$ .
The first steps give $q = 1.8, 4.0, 6.74, 9.31, 11.25, 12.35, 12.66, 12.34, \dots$ at $t = 0.1, 0.2, \dots$ .
The charge rises to a maximum at $t = 0.7$ and then falls (the velocity $y$ changes sign there).

q_{n + 1} = q_{n} + 0.1 y_{n}, y_{n + 1} = y_{n} + 0.1 (200 - 5 y_{n} - 20 q_{n}) q_{m a x} \approx 12.7 coulombs

Question 16

8 marks

A hypothesis test for healthy choices: Type I and Type II errors

Statistics and probability

(a)

2 marks

The null hypothesis says nothing has changed; the alternative says the proportion has increased.

H_{0} : p = 0.3, H_{1} : p > 0.3

(b)

3 marks

A Type I error means rejecting $H_{0}$ when $p = 0.3$ . Under $H_{0}$ , $X \sim B (80, 0.3)$ and she rejects when $X \geq 31$ .

P (X \geq 31) = 1 - P (X \leq 30) \approx 0.0587

(c)

3 marks

A Type II error means not rejecting $H_{0}$ (that is, $X \leq 30$ ) when the true proportion is 0.4, so $X \sim B (80, 0.4)$ .

P (X \leq 30) \approx 0.369

Question 17

8 marks

Sunrise and sunset in Taipei: combining two cosine models

Functions

(a)

6 marks

The number of daylight hours is $D = S - R$ . The constants combine to $c = 18.9 - 4.94 = 13.96$ .
For the cosine parts, write each as the real part of a complex number times $e^{0.0165 t i}$ .

1.15 e^{- 2.97 i} - 1.08 e^{0.413 i} = 2.21 e^{- 2.85 i} a = 2.21, b = - 2.85, c = 13.96

(b)

2 marks

The maximum number of daylight hours is $a + c$ .
It occurs when $0.0165 t - 2.85 = 0$ .
The longest day is day 173 of the year (about 22 June).

D_{m a x} = 2.21 + 13.96 = 16.2 hours t = \frac{2.853}{0.0165} \approx 173

Our analysis

Where the marks are

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

Functions37 marks · 34%
Q2, Q6, Q10, Q11, Q13, Q17
Geometry and trigonometry20 marks · 18%
Q3, Q8, Q9
Statistics and probability20 marks · 18%
Q1, Q5, Q16
Number and algebra19 marks · 17%
Q4, Q7, Q15
Calculus14 marks · 13%
Q12, Q14

Mentor insights

Traps to watch for

01
Double edges count twice · Q4
Between S and D there are two corridors, so the entry is 2, not 1.
02
Area scales by the determinant · Q9
The area of the image is the original area times $∣ det M ∣$ .
03
Acceleration is the derivative of velocity · Q14
Differentiate $t sin (t^{2})$ with the product and chain rules.
04
Type II uses the true probability · Q16
A Type II error is failing to reject $H_{0}$ when the true proportion is 0.4, so use $B (80, 0.4)$ .
05
Daylight is sunset minus sunrise · Q17
$D = S - R$ . Combine the two cosines with complex numbers, since they have the same frequency.

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

Session: November 2022
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 November 2022 Global 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

5 marks

Is favourite berry independent of income? A chi-squared test

Statistics and probability

(a)

1 mark

The alternative hypothesis is that they are not independent.

H_{0} : a person’s favourite berry is independent of their income level

(b)

2 marks

Enter the observed table as a matrix and run a $χ^{2}$ test for independence ( $ν = 4$ ).

χ^{2} = 2.27

(c)

2 marks

Compare the statistic with the critical value.
Do not reject $H_{0}$ : there is insufficient evidence at the 10% level that favourite berry depends on income level.

2.27 < 7.78

Mentor tip

The $p$ -value, about 0.687, is also much bigger than 0.10.

Question 2

5 marks

Coffee cooling: evaluating, the asymptote and an inverse value

Functions

(a)

2 marks

Substitute $t = 16$ .

T (16) = 71 e^{- 0.0514 (16)} + 23 = 54.2^{\circ} C

(b)

1 mark

As time passes, the exponential term tends to 0.

23^{\circ} C

(c)

2 marks

$T^{- 1} (50) = k$ means $T (k) = 50$ .

71 e^{- 0.0514 k} + 23 = 50 \Rightarrow e^{- 0.0514 k} = \frac{27}{71} k = \frac{ln ( 71/27 )}{0.0514} = 18.8 minutes

Question 3

5 marks

A ski slope as a triangular prism

Geometry and trigonometry

(a)

2 marks

In the right triangle FCB, the side opposite $2 1^{\circ}$ is $C F = 17$ and the hypotenuse is $B F$ .

sin 2 1^{\circ} = \frac{17}{B F} \Rightarrow B F = 47.4 m

(b)

3 marks

The side $B C = \frac{17}{tan 2 1 ^{\circ}} = 44.29$ m. The base is a rectangle with sides 44 and 44.29, and E is 17 m above D.
The longest horizontal distance from the point below E to the base is the diagonal DB.
The least steep angle is the angle of descent from E to B.

D B = 4 4^{2} + 44.2 9^{2} = 62.4 m tan θ = \frac{17}{62.4} \Rightarrow θ = 15. 2^{\circ}

Question 4

7 marks

A maze as a graph: adjacency matrix and walks of length up to four

Number and algebra

(a)

2 marks

Read the missing entries from the graph. There are two corridors joining S and D (the straight one and the outer curve), and no corridor between A and D.

M_{S D} = M_{D S} = 2, M_{A D} = M_{D A} = 0

(b)

4 marks

The entry in row S, column T of $M^{k}$ is the number of walks of exactly $k$ edges from S to T. Compute the powers of
The S to T entries are 0 for $M$ , 4 for $M^{2}$ , 10 for $M^{3}$ and 64 for $M^{4}$ .

M = 011120101100110111111011201101001110 0 + 4 + 10 + 64 = 78 walks

(c)

1 mark

Some of the walks counted pass through T and then leave it and come back, so they would have found the treasure earlier. The number of distinct ways to reach T for the first time is smaller.

Question 5

7 marks

A bottle-throwing game: total of two games and a fair game

Statistics and probability

(a)

4 marks

List the ways to get a total of two bottles over two games: 1 and 1, 0 and 2, 2 and 0.

P (total 2) = 0.4 \times 0.4 + 0.5 \times 0.1 + 0.1 \times 0.5 = 0.16 + 0.05 + 0.05 = 0.26

(b)

3 marks

The points are $k$ with probability 0.1, $4$ with 0.4 and $- 8$ with 0.5. A fair game has an expected value of 0.

0.1 k + 0.4 (4) + 0.5 (- 8) = 0 \Rightarrow 0.1 k - 2.4 = 0 k = 24

Question 6

6 marks

A cat in an exercise wheel: a sine model for the height of a point

Functions

(a)

1 mark

At $t = 0$ , $sin (0) = 0$ , so $h (0) = c$ .

c = 78

(b)

3 marks

The amplitude is the maximum minus the central height.
The first maximum of a sine curve is a quarter of a period in, so the period is $4 \times 4 = 16$ s.

a = 143 - 78 = 65 b = \frac{2 π}{16} = \frac{π}{8} \approx 0.393

(c)

1 mark

The minimum height is the centre minus the amplitude.

78 - 65 = 13 cm

(d)

1 mark

Twice as long means a period of 32 s.

b = \frac{2 π}{32} = \frac{π}{16} \approx 0.196

Question 7

7 marks

Savings with monthly deposits: finding the number of months

Number and algebra

(a)

4 marks

The monthly rate is $\frac{7.5}{12} %$ and each period is one month. Use a finance solver with PV = $- 800$ , PMT = $- 500$ , $I = 7.5$ , $P / Y = C / Y = 12$ and FV = 10000, and solve for $N$ .
Check: after 17 months the balance is below 10000, and after 18 it is above.

N = 17.2 \Rightarrow k = 18 800 (1.00625)^{18} + 500 \cdot \frac{1.0062 5 ^{18} - 1}{0.00625} = 10389

(b)

3 marks

The total paid in is $800 + 18 \times 500 = 9800$ .

interest = 10389 - 9800 \approx 589 EUR

Question 8

7 marks

Two lines meeting at a point: coordinates and angle

Geometry and trigonometry

(a)

3 marks

Equate the coordinates of $L_{1}$ and $L_{2}$ .
From the second equation $q = 2 - 2 p$ . Put into the first.
Check the third equation: $p + 9 = 8.2$ and $2 q + 1 = 8.2$ . Substituting $p$ into $L_{1}$ :

3 p + 4 = q - 2, 2 p - 1 = 1 - q, p + 9 = 2 q + 1 3 p + 4 = - 2 p \Rightarrow p = - \frac{4}{5}, q = \frac{18}{5} M = (1.6, - 2.6, 8.2)

(b)

4 marks

The direction vectors are $321$ and $1 - 1 2$ .

cos θ = \frac{∣3 - 2 + 2∣}{14 6} = \frac{3}{84} θ = 71. 0^{\circ}

Question 9

8 marks

A matrix transformation: area scale factor and the object point

Geometry and trigonometry

(a)

2 marks

The determinant of $M$ :
The image area is the area of the pentagon times 14.

det M = 2 (1) - (- 4) (3) = 14 12 \times 14 = 168 cm^{2}

(b)

6 marks

The inverse of $(23 - 4 1)$ is $\frac{1}{14} (1 - 3 42)$ .
Multiply it by the image $(2 t - 3, 6 - 5 t)$ .

\frac{1}{14} (1 - 3 42) (2 t - 3 6 - 5 t) = \frac{1}{14} (2 t - 3 + 24 - 20 t - 6 t + 9 + 12 - 10 t) X = (\frac{21 - 18 t}{14}, \frac{21 - 16 t}{14})

Question 10

7 marks

Star magnitude and brightness on a logarithmic scale

Functions

(a)

2 marks

Put $b = 0.0525$ into the equation.

m = 1 - 2.5 lo g_{10} (0.0525) = 4.20

(b)

2 marks

Set $m = 7$ and solve for $b$ .

7 = 1 - 2.5 lo g_{10} b \Rightarrow lo g_{10} b = - 2.4 b = 1 0^{- 2.4} \approx 0.00398

(c)

3 marks

Let $m_{P}$ and $m_{N}$ be the magnitudes with $m_{P} - m_{N} = 3.2$ .
Neptune is about 19 times brighter than Proxima Centauri.

m_{P} - m_{N} = - 2.5 lo g_{10} b_{P} + 2.5 lo g_{10} b_{N} = 2.5 lo g_{10} (\frac{b _{N}}{b _{P}}) = 3.2 \frac{b _{N}}{b _{P}} = 1 0^{3.2/2.5} = 1 0^{1.28} \approx 19.1

Question 11

5 marks

Cubic regression and an enlargement of the curve

Functions

(a)

2 marks

Enter the four points into the calculator and use cubic regression.

y = - 0.00855 x^{3} - 0.234 x^{2} - 0.225 x + 3.20

(b)

3 marks

g (x) = \frac{1}{2} f (2 x) = \frac{1}{2} [0.0083 (2 x)^{3} - 0.075 (2 x)^{2} - 0.58 (2 x) + 2.2] g (x) = 0.0332 x^{3} - 0.15 x^{2} - 0.58 x + 1.1

Question 12

5 marks

Phase portraits from eigenvalues and eigenvectors

Calculus

(a)

3 marks

λ_{1} = 2, λ_{2} = 3 : D λ_{1} = 2, λ_{2} = - 3 : C λ_{1} = - 2, λ_{2} = 3 : B

(b)

2 marks

The real part $- 2$ is negative, so the paths spiral in towards the origin.
At $(3, 0)$ the rate $\frac{d y}{d t} = - 12 < 0$ , so the path is moving downwards at the right-hand side of the origin, which is clockwise.
Sketch: a spiral centred on the origin, starting at $(3, 0)$ and turning clockwise while closing in on the origin, crossing the axes several times.

Question 13

6 marks

Exponential and log models from straight-line plots

Functions

(a)

3 marks

The gradient of the line through $(0, - 1.3)$ and $(6, 14.9)$ :
So $lo g_{10} Q = 2.7 P - 1.3$ . Undo the logarithm.

m = \frac{14.9 - ( - 1.3 )}{6 - 0} = 2.7 Q = 1 0^{2.7 P - 1.3}

(b)

1 mark

For $Q = a ln R + b$ to be a straight line, plot $Q$ on the vertical axis and $ln R$ on the horizontal axis.

(c)

2 marks

Equate the two expressions for $Q$ .
Take $lo g_{10}$ of both sides and solve for $P$ .

1 0^{2.7 P - 1.3} = 4.3 ln R + 12.1 P = \frac{lo g _{10} ( 4.3 ln R + 12.1 ) + 1.3}{2.7}

Question 14

9 marks

A particle with velocity t sin(t^2)

Calculus

(a)

2 marks

Use the product rule with the chain rule for $sin (t^{2})$ .

a = \frac{d v}{d t} = sin (t^{2}) + t \cdot 2 t cos (t^{2}) = sin (t^{2}) + 2 t^{2} cos (t^{2})

(b)

2 marks

Find the maximum of $a (t)$ for $0 \leq t \leq 8$ on a graph.

a_{m a x} \approx 126 m s^{- 2} (at t \approx 7.93)

(c)

3 marks

Integrate $v$ (the substitution is $u = t^{2}$ ).
At $t = 0$ the displacement is 0.

s = \int t sin (t^{2}) d t = - \frac{1}{2} cos (t^{2}) + c 0 = - \frac{1}{2} + c \Rightarrow c = \frac{1}{2} s = \frac{1}{2} (1 - cos (t^{2}))

(d)

2 marks

Since $cos (t^{2}) \leq 1$ , we have $1 - cos (t^{2}) \geq 0$ .
So the displacement is never negative.

s = \frac{1}{2} (1 - cos (t^{2})) \geq 0 for all t

Question 15

5 marks

Euler's method for a second-order equation: the largest charge

Number and algebra

5 marks

Put $y = \frac{d q}{d t}$ , so $\frac{d y}{d t} = 200 - 5 y - 20 q$ . Start from $q_{0} = 1$ , $y_{0} = 8$ , $t_{0} = 0$ with $h = 0.1$ .
The first steps give $q = 1.8, 4.0, 6.74, 9.31, 11.25, 12.35, 12.66, 12.34, \dots$ at $t = 0.1, 0.2, \dots$ .
The charge rises to a maximum at $t = 0.7$ and then falls (the velocity $y$ changes sign there).

q_{n + 1} = q_{n} + 0.1 y_{n}, y_{n + 1} = y_{n} + 0.1 (200 - 5 y_{n} - 20 q_{n}) q_{m a x} \approx 12.7 coulombs

Question 16

8 marks

A hypothesis test for healthy choices: Type I and Type II errors

Statistics and probability

(a)

2 marks

The null hypothesis says nothing has changed; the alternative says the proportion has increased.

H_{0} : p = 0.3, H_{1} : p > 0.3

(b)

3 marks

A Type I error means rejecting $H_{0}$ when $p = 0.3$ . Under $H_{0}$ , $X \sim B (80, 0.3)$ and she rejects when $X \geq 31$ .

P (X \geq 31) = 1 - P (X \leq 30) \approx 0.0587

(c)

3 marks

A Type II error means not rejecting $H_{0}$ (that is, $X \leq 30$ ) when the true proportion is 0.4, so $X \sim B (80, 0.4)$ .

P (X \leq 30) \approx 0.369

Question 17

8 marks

Sunrise and sunset in Taipei: combining two cosine models

Functions

(a)

6 marks

The number of daylight hours is $D = S - R$ . The constants combine to $c = 18.9 - 4.94 = 13.96$ .
For the cosine parts, write each as the real part of a complex number times $e^{0.0165 t i}$ .

1.15 e^{- 2.97 i} - 1.08 e^{0.413 i} = 2.21 e^{- 2.85 i} a = 2.21, b = - 2.85, c = 13.96

(b)

2 marks

The maximum number of daylight hours is $a + c$ .
It occurs when $0.0165 t - 2.85 = 0$ .
The longest day is day 173 of the year (about 22 June).

D_{m a x} = 2.21 + 13.96 = 16.2 hours t = \frac{2.853}{0.0165} \approx 173

Our analysis

Where the marks are

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

Functions37 marks · 34%
Q2, Q6, Q10, Q11, Q13, Q17
Geometry and trigonometry20 marks · 18%
Q3, Q8, Q9
Statistics and probability20 marks · 18%
Q1, Q5, Q16
Number and algebra19 marks · 17%
Q4, Q7, Q15
Calculus14 marks · 13%
Q12, Q14

Mentor insights

Traps to watch for

01
Double edges count twice · Q4
Between S and D there are two corridors, so the entry is 2, not 1.
02
Area scales by the determinant · Q9
The area of the image is the original area times $∣ det M ∣$ .
03
Acceleration is the derivative of velocity · Q14
Differentiate $t sin (t^{2})$ with the product and chain rules.
04
Type II uses the true probability · Q16
A Type II error is failing to reject $H_{0}$ when the true proportion is 0.4, so use $B (80, 0.4)$ .
05
Daylight is sunset minus sunrise · Q17
$D = S - R$ . Combine the two cosines with complex numbers, since they have the same frequency.

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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