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IB Maths AI HL · Paper 1

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1 TZ2TZ3

Session: May 2025
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 16

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

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2025 TZ3 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–16 · 110 marks

Question 1

9 marks

A triangle NPV: the cosine rule, the sine rule and a shortest distance

Geometry and trigonometry

(a)

3 marks

Use the cosine rule in triangle $N P V$ .

N V^{2} = 2 0^{2} + 2 5^{2} - 2 (20) (25) cos 55. 5^{\circ} \Rightarrow N V \approx 21.4 m

(b)

3 marks

Use the sine rule.

\frac{sin P N V}{25} = \frac{sin 55. 5 ^{\circ}}{21.41} \Rightarrow P N V \approx 74. 2^{\circ}

(c)

3 marks

The shortest distance from $P$ to $[N V]$ is the perpendicular from $P$ to the line $N V$ . In the right-angled triangle with hypotenuse $N P = 20$ and angle $74. 2^{\circ}$ at $N$ ,

d = 20 sin 74.1 7^{\circ} \approx 19.2 m

Question 2

5 marks

Loudness in decibels from the intensity of a sound

Functions

(a)

2 marks

Substitute $I = 1 0^{- 5}$ and $I_{0} = 1 0^{- 12}$ .

L = 10 lo g_{10} \frac{1 0 ^{- 5}}{1 0 ^{- 12}} = 10 lo g_{10} 1 0^{7} = 70 dB

(b)

3 marks

Solve $185 = 10 lo g_{10} \frac{I}{1 0 ^{- 12}}$ .

\frac{I}{1 0 ^{- 12}} = 1 0^{18.5} \Rightarrow I = 1 0^{6.5} \approx 3.16 \times 1 0^{6} I \approx 3.16 \times 1 0^{6} W m^{- 2}

Question 3

7 marks

Customer satisfaction and type of service: a chi-squared test

Statistics and probability

(a)

1 mark

$H_{1}$ : the satisfaction level and the type of service interaction are not independent.

(b)

1 mark

The table is 3 by 3.

ν = (3 - 1) (3 - 1) = 4

(c)

2 marks

Enter the table in the calculator and run a $χ^{2}$ test.

χ_{c a l c}^{2} \approx 3.32

(d)

3 marks

(i) Since $3.32 < 9.488$ (the p-value is about 0.506, which is above 0.05), there is not enough evidence to reject $H_{0}$ . Prakash is not correct.

(ii) There is insufficient evidence at the 5% level to say that satisfaction level and type of service interaction are not independent.

Question 4

7 marks

Swimming times: quartiles, the interquartile range and an outlier

Statistics and probability

(a)

1 mark

The upper quartile is 24.70, with a quarter of the 16 swimmers above it.

\frac{1}{4} \times 16 = 4 swimmers

(b)

2 marks

From the box plot, $Q_{1} = 24.28$ and $Q_{3} = 24.70$ .

IQR = 24.70 - 24.28 = 0.42 seconds

(c)

4 marks

(i) The boundaries are

The fastest times 23.96 and 24.12 are above 23.65, and the slowest time 25.12 is below 25.33, so none of them are outliers. Only 25.40 is above 25.33.

(ii) Complete the plot: the lower whisker goes from $Q_{1}$ to 23.96, the upper whisker goes from $Q_{3}$ to 25.12, and the time 25.40 is marked separately as an outlier (with a cross), with no whisker joining it.

Q_{1} - 1.5 (0.42) = 23.65, Q_{3} + 1.5 (0.42) = 25.33

Question 5

8 marks

A 25 year loan paid off early

Number and algebra

(a)

2 marks

Use the finance solver with $N = 300$ , $I = 4.2%$ , $P V = 50000$ , $F V = 0$ , 12 payments a year.

P M T \approx 269.47 euros

(b)

6 marks

(i) After 8 years, $N = 96$ payments have been made. The balance still owed is the future value.

(ii) Without the early payment Sofia would pay $12 \times 25 \times 269.47 = 80 841.00$ . With it, she pays 96 payments plus the single payment.

N = 96, I = 4.2%, P V = 50000, P M T = - 269.47 \Rightarrow F V \approx 39 243.42 euros 12 \times 8 \times 269.47 + 39243.42 = 65 112.54 saved = 80841.00 - 65112.54 \approx 15728 euros

Question 6

6 marks

A spinner with 20 sectors whose white angles form an arithmetic sequence

Statistics and probability

(a)

3 marks

The ten white angles have first term $6^{\circ}$ and common difference $3^{\circ}$ .

S_{10} = \frac{10}{2} [2 (6) + 9 (3)] = 19 5^{\circ} P (white) = \frac{195}{360} = \frac{13}{24} \approx 0.542

(b)

3 marks

Let $X$ be the number of times the arrow stops in white, $X \sim B (4, \frac{13}{24})$ .

P (X \geq 3) \approx 0.377

Question 7

7 marks

A happiness model: sampling, correlation and a regression prediction

Statistics and probability

(a)

1 mark

The countries are divided into three wealth groups and two are chosen at random from each group.

stratified sampling

(b)

3 marks

(i) Enter the six pairs in the calculator.

(ii) Since $0.854 > 0.8$ , the model can be regarded as a valid predictor.

r \approx 0.854

(c)

1 mark

The regression line of $y$ on $x$ .

y = 0.266 x + 2.40

(d)

2 marks

Substitute $x = 17.2$ .

y = 0.2659 (17.2) + 2.3957 \approx 6.97

Question 8

5 marks

Undoing two matrix transformations

Number and algebra

5 marks

The first matrix is $(0110)$ (taking $A$ to $B$ ), and the second is $(3011)$ (taking $B$ to $C$ ). Find the combined matrix.
The point $A = (x, y)$ satisfies the equation below.

(3011) (0110) = (1130) (1130) (x y) = (47) \Rightarrow x + 3 y = 4, x = 7 y = - 1 A = (7, - 1)

Question 9

6 marks

A tangent to y = 4x sqrt(3x - 2)

Calculus

(a)

3 marks

Write $y = 4 x (3 x - 2)^{1/2}$ and use the product rule with the chain rule.

\frac{d y}{d x} = 4 (3 x - 2)^{1/2} + 4 x \cdot \frac{1}{2} (3 x - 2)^{- 1/2} \cdot 3 = 4 (3 x - 2)^{1/2} + 6 x (3 x - 2)^{- 1/2} \frac{d y}{d x} = \frac{4 ( 3 x - 2 ) + 6 x}{3 x - 2} = \frac{2 ( 9 x - 4 )}{3 x - 2}

(b)

3 marks

At $x = 6$ , $3 x - 2 = 4$ .
The tangent at $(6, 96)$ has gradient 25.

\frac{d y}{d x} = \frac{2 ( 54 - 4 )}{4} = 25 y - 96 = 25 (x - 6) \Rightarrow y = 25 x - 54

Question 10

8 marks

Gravity on a satellite: inverse square variation and straight-line graphs

Functions

(a)

4 marks

(i) $F = \frac{k}{d ^{2}}$ and $F = 40000$ when $d = 7000$ .

(ii) Substitute $d = 3500$ .

k = 40000 \times 700 0^{2} = 1.96 \times 1 0^{12} \Rightarrow F = \frac{1.96 \times 1 0 ^{12}}{d ^{2}} F = \frac{1.96 \times 1 0 ^{12}}{350 0 ^{2}} = 160000 N

(b)

4 marks

(i) $F = k d^{- 2}$ is linear against $d^{- 2}$ (graph C). Taking logs gives $lo g_{10} F = lo g_{10} k - 2 lo g_{10} d$ , which is linear against $lo g_{10} d$ (graph D).

(ii) For graph C the gradient is $k$ .

For graph D the gradient is the power of $d$ .

C and D 1.96 \times 1 0^{12} - 2

Question 11

6 marks

A coupled system, its eigenvalues and Euler's method

Calculus

(a)

1 mark

The equation $λ^{2} - 2 λ + 5 = 0$ gives $λ = 1 \pm 2 i$ . The system is $\frac{d x}{d t} = 3 x - 2 y$ , $\frac{d y}{d t} = 4 x - y$ , so the possible eigenvalue is

1 + 2 i (option C)

(b)

5 marks

Use Euler's method with $h = 0.1$ from $(x, y) = (1, 0)$ at $t = 5$ .
After five steps, at $t = 5.5$ :

x_{n + 1} = x_{n} + 0.1 (3 x_{n} - 2 y_{n}), y_{n + 1} = y_{n} + 0.1 (4 x_{n} - y_{n}) (t, x, y) : (5, 1, 0), (5.1, 1.3, 0.4), (5.2, 1.61, 0.88), (5.3, 1.917, 1.436), (5.4, 2.2049, 2.0592) x \approx 2.45, y \approx 2.74

Question 12

5 marks

A hypothesis test for the mean of a Poisson distribution

Statistics and probability

(a)

4 marks

(i) A Type I error is rejecting the null hypothesis (the mean of 12.4 cars per minute) when it is true.

(ii) Over 10 minutes the mean is 124, so under $H_{0}$ , $X \sim P o (124)$ . The test rejects when more than 130 cars arrive, that is $X \geq 131$ .

P (X \geq 131) \approx 0.276

(b)

1 mark

Cars may not arrive independently (a traffic light just before the intersection would bunch them up), or the rate of arrival may change through the day.

Question 13

9 marks

Growth of a fungus: solving a differential equation

Calculus

(a)

7 marks

Write $\frac{d P}{d t} = 0.2 (1 - P)$ and separate the variables.
Integrate (the left side by substitution $u = 1 - P$ ).
Use $P = 0$ at $t = 0$ , which gives $c = 0$ .

\int \frac{1}{1 - P} d P = \int 0.2 d t - ln (1 - P) = 0.2 t + c ln (1 - P) = - 0.2 t \Rightarrow 1 - P = e^{- 0.2 t} P = 1 - e^{- 0.2 t}

(b)

2 marks

Solve $0.9 = 1 - e^{- 0.2 t}$ .

e^{- 0.2 t} = 0.1 \Rightarrow t = \frac{ln 10}{0.2} \approx 11.5 days

Question 14

8 marks

Dropping a relief package from a plane

Calculus

(a)

2 marks

The acceleration is $(0, - 9.8)$ and the velocity at release is $(- 20, 0)$ (towards $O$ ).

(\overset{x}{˙} \overset{y}{˙}) = (- 20 - 9.8 t)

(b)

6 marks

Integrate again, with the release point at $x = d$ and $y = 70$ when $t = 0$ .
The package hits the ground when $y = 0$ .
It lands at $O$ , where $x = 0$ .

(x y) = (d - 20 t 70 - 4.9 t^{2}) 70 - 4.9 t^{2} = 0 \Rightarrow t = \frac{70}{4.9} \approx 3.780 seconds d - 20 (3.780) = 0 \Rightarrow d \approx 75.6 m

Question 15

6 marks

A two-state Markov chain: the eigenvector and the long-run proportion

Number and algebra

(a)

3 marks

With $T = (p 1 - p 1 - q q)$ , solve $T v = v$ , so $(T - I) v = 0$ .
Take $x = 1 - q$ , which makes $y = 1 - p$ .

(p - 1) x + (1 - q) y = 0 v = (1 - q 1 - p)

(b)

2 marks

Scale so that the entries add to 1.

proportion in state A = \frac{1 - q}{( 1 - q ) + ( 1 - p )} = \frac{1 - q}{2 - p - q}

(c)

1 mark

Substitute $p = 0.3$ and $q = 0.6$ .

\frac{0.4}{2 - 0.9} = \frac{0.4}{1.1} = \frac{4}{11} \approx 0.364

Question 16

8 marks

A light shade from rotating the region between two curves

Functions

(a)

2 marks

Start from $f (x) = 4 cos (\frac{π x}{8})$ .

g (x) = f (x + 0.3) - 0.2 = 4 cos (\frac{π ( x + 0.3 )}{8}) - 0.2

(b)

1 mark

Solve $g (x) = 0$ with the calculator.

a \approx 3.57

(c)

5 marks

(i) The solid is the volume from rotating $f$ over $[0, 4]$ minus the volume from rotating $g$ over $[0, a]$ , with half a turn.

(ii) Evaluate with the calculator.

V = \frac{π}{2} \int_{0}^{4} (4 cos \frac{π x}{8})^{2} d x - \frac{π}{2} \int_{0}^{3.5726} (4 cos \frac{π ( x + 0.3 )}{8} - 0.2)^{2} d x V \approx 12.9 cm^{3}

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 probability32 marks · 29%
Q3, Q4, Q6, Q7, Q12
Calculus29 marks · 26%
Q9, Q11, Q13, Q14
Functions21 marks · 19%
Q2, Q10, Q16
Number and algebra19 marks · 17%
Q5, Q8, Q15
Geometry and trigonometry9 marks · 8%
Q1

Mentor insights

Traps to watch for

01
Shortest distance to a line · Q1
It is the perpendicular distance, which is a side of a right-angled triangle.
02
Remaining balance · Q5
After 96 payments, the balance is the value found with the finance solver, with the payment entered.
03
Inverse square · Q10
$F = \frac{k}{d ^{2}}$ .
04
Separate the variables · Q13
$\frac{d P}{1 - P} = 0.2 d t$ .
05
Integrate acceleration · Q14
The velocity is the integral of the acceleration, and the position is the integral of the velocity.

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

TZ1 TZ2TZ3

Session: May 2025
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 16

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

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2025 TZ3 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–16 · 110 marks

Question 1

9 marks

A triangle NPV: the cosine rule, the sine rule and a shortest distance

Geometry and trigonometry

(a)

3 marks

Use the cosine rule in triangle $N P V$ .

N V^{2} = 2 0^{2} + 2 5^{2} - 2 (20) (25) cos 55. 5^{\circ} \Rightarrow N V \approx 21.4 m

(b)

3 marks

Use the sine rule.

\frac{sin P N V}{25} = \frac{sin 55. 5 ^{\circ}}{21.41} \Rightarrow P N V \approx 74. 2^{\circ}

(c)

3 marks

The shortest distance from $P$ to $[N V]$ is the perpendicular from $P$ to the line $N V$ . In the right-angled triangle with hypotenuse $N P = 20$ and angle $74. 2^{\circ}$ at $N$ ,

d = 20 sin 74.1 7^{\circ} \approx 19.2 m

Question 2

5 marks

Loudness in decibels from the intensity of a sound

Functions

(a)

2 marks

Substitute $I = 1 0^{- 5}$ and $I_{0} = 1 0^{- 12}$ .

L = 10 lo g_{10} \frac{1 0 ^{- 5}}{1 0 ^{- 12}} = 10 lo g_{10} 1 0^{7} = 70 dB

(b)

3 marks

Solve $185 = 10 lo g_{10} \frac{I}{1 0 ^{- 12}}$ .

\frac{I}{1 0 ^{- 12}} = 1 0^{18.5} \Rightarrow I = 1 0^{6.5} \approx 3.16 \times 1 0^{6} I \approx 3.16 \times 1 0^{6} W m^{- 2}

Question 3

7 marks

Customer satisfaction and type of service: a chi-squared test

Statistics and probability

(a)

1 mark

$H_{1}$ : the satisfaction level and the type of service interaction are not independent.

(b)

1 mark

The table is 3 by 3.

ν = (3 - 1) (3 - 1) = 4

(c)

2 marks

Enter the table in the calculator and run a $χ^{2}$ test.

χ_{c a l c}^{2} \approx 3.32

(d)

3 marks

(i) Since $3.32 < 9.488$ (the p-value is about 0.506, which is above 0.05), there is not enough evidence to reject $H_{0}$ . Prakash is not correct.

(ii) There is insufficient evidence at the 5% level to say that satisfaction level and type of service interaction are not independent.

Question 4

7 marks

Swimming times: quartiles, the interquartile range and an outlier

Statistics and probability

(a)

1 mark

The upper quartile is 24.70, with a quarter of the 16 swimmers above it.

\frac{1}{4} \times 16 = 4 swimmers

(b)

2 marks

From the box plot, $Q_{1} = 24.28$ and $Q_{3} = 24.70$ .

IQR = 24.70 - 24.28 = 0.42 seconds

(c)

4 marks

(i) The boundaries are

The fastest times 23.96 and 24.12 are above 23.65, and the slowest time 25.12 is below 25.33, so none of them are outliers. Only 25.40 is above 25.33.

Q_{1} - 1.5 (0.42) = 23.65, Q_{3} + 1.5 (0.42) = 25.33

Question 5

8 marks

A 25 year loan paid off early

Number and algebra

(a)

2 marks

Use the finance solver with $N = 300$ , $I = 4.2%$ , $P V = 50000$ , $F V = 0$ , 12 payments a year.

P M T \approx 269.47 euros

(b)

6 marks

(i) After 8 years, $N = 96$ payments have been made. The balance still owed is the future value.

(ii) Without the early payment Sofia would pay $12 \times 25 \times 269.47 = 80 841.00$ . With it, she pays 96 payments plus the single payment.

N = 96, I = 4.2%, P V = 50000, P M T = - 269.47 \Rightarrow F V \approx 39 243.42 euros 12 \times 8 \times 269.47 + 39243.42 = 65 112.54 saved = 80841.00 - 65112.54 \approx 15728 euros

Question 6

6 marks

A spinner with 20 sectors whose white angles form an arithmetic sequence

Statistics and probability

(a)

3 marks

The ten white angles have first term $6^{\circ}$ and common difference $3^{\circ}$ .

S_{10} = \frac{10}{2} [2 (6) + 9 (3)] = 19 5^{\circ} P (white) = \frac{195}{360} = \frac{13}{24} \approx 0.542

(b)

3 marks

Let $X$ be the number of times the arrow stops in white, $X \sim B (4, \frac{13}{24})$ .

P (X \geq 3) \approx 0.377

Question 7

7 marks

A happiness model: sampling, correlation and a regression prediction

Statistics and probability

(a)

1 mark

The countries are divided into three wealth groups and two are chosen at random from each group.

stratified sampling

(b)

3 marks

(i) Enter the six pairs in the calculator.

(ii) Since $0.854 > 0.8$ , the model can be regarded as a valid predictor.

r \approx 0.854

(c)

1 mark

The regression line of $y$ on $x$ .

y = 0.266 x + 2.40

(d)

2 marks

Substitute $x = 17.2$ .

y = 0.2659 (17.2) + 2.3957 \approx 6.97

Question 8

5 marks

Undoing two matrix transformations

Number and algebra

5 marks

The first matrix is $(0110)$ (taking $A$ to $B$ ), and the second is $(3011)$ (taking $B$ to $C$ ). Find the combined matrix.
The point $A = (x, y)$ satisfies the equation below.

(3011) (0110) = (1130) (1130) (x y) = (47) \Rightarrow x + 3 y = 4, x = 7 y = - 1 A = (7, - 1)

Question 9

6 marks

A tangent to y = 4x sqrt(3x - 2)

Calculus

(a)

3 marks

Write $y = 4 x (3 x - 2)^{1/2}$ and use the product rule with the chain rule.

\frac{d y}{d x} = 4 (3 x - 2)^{1/2} + 4 x \cdot \frac{1}{2} (3 x - 2)^{- 1/2} \cdot 3 = 4 (3 x - 2)^{1/2} + 6 x (3 x - 2)^{- 1/2} \frac{d y}{d x} = \frac{4 ( 3 x - 2 ) + 6 x}{3 x - 2} = \frac{2 ( 9 x - 4 )}{3 x - 2}

(b)

3 marks

At $x = 6$ , $3 x - 2 = 4$ .
The tangent at $(6, 96)$ has gradient 25.

\frac{d y}{d x} = \frac{2 ( 54 - 4 )}{4} = 25 y - 96 = 25 (x - 6) \Rightarrow y = 25 x - 54

Question 10

8 marks

Gravity on a satellite: inverse square variation and straight-line graphs

Functions

(a)

4 marks

(i) $F = \frac{k}{d ^{2}}$ and $F = 40000$ when $d = 7000$ .

(ii) Substitute $d = 3500$ .

k = 40000 \times 700 0^{2} = 1.96 \times 1 0^{12} \Rightarrow F = \frac{1.96 \times 1 0 ^{12}}{d ^{2}} F = \frac{1.96 \times 1 0 ^{12}}{350 0 ^{2}} = 160000 N

(b)

4 marks

(i) $F = k d^{- 2}$ is linear against $d^{- 2}$ (graph C). Taking logs gives $lo g_{10} F = lo g_{10} k - 2 lo g_{10} d$ , which is linear against $lo g_{10} d$ (graph D).

(ii) For graph C the gradient is $k$ .

For graph D the gradient is the power of $d$ .

C and D 1.96 \times 1 0^{12} - 2

Question 11

6 marks

A coupled system, its eigenvalues and Euler's method

Calculus

(a)

1 mark

The equation $λ^{2} - 2 λ + 5 = 0$ gives $λ = 1 \pm 2 i$ . The system is $\frac{d x}{d t} = 3 x - 2 y$ , $\frac{d y}{d t} = 4 x - y$ , so the possible eigenvalue is

1 + 2 i (option C)

(b)

5 marks

Use Euler's method with $h = 0.1$ from $(x, y) = (1, 0)$ at $t = 5$ .
After five steps, at $t = 5.5$ :

x_{n + 1} = x_{n} + 0.1 (3 x_{n} - 2 y_{n}), y_{n + 1} = y_{n} + 0.1 (4 x_{n} - y_{n}) (t, x, y) : (5, 1, 0), (5.1, 1.3, 0.4), (5.2, 1.61, 0.88), (5.3, 1.917, 1.436), (5.4, 2.2049, 2.0592) x \approx 2.45, y \approx 2.74

Question 12

5 marks

A hypothesis test for the mean of a Poisson distribution

Statistics and probability

(a)

4 marks

(i) A Type I error is rejecting the null hypothesis (the mean of 12.4 cars per minute) when it is true.

(ii) Over 10 minutes the mean is 124, so under $H_{0}$ , $X \sim P o (124)$ . The test rejects when more than 130 cars arrive, that is $X \geq 131$ .

P (X \geq 131) \approx 0.276

(b)

1 mark

Cars may not arrive independently (a traffic light just before the intersection would bunch them up), or the rate of arrival may change through the day.

Question 13

9 marks

Growth of a fungus: solving a differential equation

Calculus

(a)

7 marks

Write $\frac{d P}{d t} = 0.2 (1 - P)$ and separate the variables.
Integrate (the left side by substitution $u = 1 - P$ ).
Use $P = 0$ at $t = 0$ , which gives $c = 0$ .

\int \frac{1}{1 - P} d P = \int 0.2 d t - ln (1 - P) = 0.2 t + c ln (1 - P) = - 0.2 t \Rightarrow 1 - P = e^{- 0.2 t} P = 1 - e^{- 0.2 t}

(b)

2 marks

Solve $0.9 = 1 - e^{- 0.2 t}$ .

e^{- 0.2 t} = 0.1 \Rightarrow t = \frac{ln 10}{0.2} \approx 11.5 days

Question 14

8 marks

Dropping a relief package from a plane

Calculus

(a)

2 marks

The acceleration is $(0, - 9.8)$ and the velocity at release is $(- 20, 0)$ (towards $O$ ).

(\overset{x}{˙} \overset{y}{˙}) = (- 20 - 9.8 t)

(b)

6 marks

Integrate again, with the release point at $x = d$ and $y = 70$ when $t = 0$ .
The package hits the ground when $y = 0$ .
It lands at $O$ , where $x = 0$ .

(x y) = (d - 20 t 70 - 4.9 t^{2}) 70 - 4.9 t^{2} = 0 \Rightarrow t = \frac{70}{4.9} \approx 3.780 seconds d - 20 (3.780) = 0 \Rightarrow d \approx 75.6 m

Question 15

6 marks

A two-state Markov chain: the eigenvector and the long-run proportion

Number and algebra

(a)

3 marks

With $T = (p 1 - p 1 - q q)$ , solve $T v = v$ , so $(T - I) v = 0$ .
Take $x = 1 - q$ , which makes $y = 1 - p$ .

(p - 1) x + (1 - q) y = 0 v = (1 - q 1 - p)

(b)

2 marks

Scale so that the entries add to 1.

proportion in state A = \frac{1 - q}{( 1 - q ) + ( 1 - p )} = \frac{1 - q}{2 - p - q}

(c)

1 mark

Substitute $p = 0.3$ and $q = 0.6$ .

\frac{0.4}{2 - 0.9} = \frac{0.4}{1.1} = \frac{4}{11} \approx 0.364

Question 16

8 marks

A light shade from rotating the region between two curves

Functions

(a)

2 marks

Start from $f (x) = 4 cos (\frac{π x}{8})$ .

g (x) = f (x + 0.3) - 0.2 = 4 cos (\frac{π ( x + 0.3 )}{8}) - 0.2

(b)

1 mark

Solve $g (x) = 0$ with the calculator.

a \approx 3.57

(c)

5 marks

(i) The solid is the volume from rotating $f$ over $[0, 4]$ minus the volume from rotating $g$ over $[0, a]$ , with half a turn.

(ii) Evaluate with the calculator.

V = \frac{π}{2} \int_{0}^{4} (4 cos \frac{π x}{8})^{2} d x - \frac{π}{2} \int_{0}^{3.5726} (4 cos \frac{π ( x + 0.3 )}{8} - 0.2)^{2} d x V \approx 12.9 cm^{3}

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 probability32 marks · 29%
Q3, Q4, Q6, Q7, Q12
Calculus29 marks · 26%
Q9, Q11, Q13, Q14
Functions21 marks · 19%
Q2, Q10, Q16
Number and algebra19 marks · 17%
Q5, Q8, Q15
Geometry and trigonometry9 marks · 8%
Q1

Mentor insights

Traps to watch for

01
Shortest distance to a line · Q1
It is the perpendicular distance, which is a side of a right-angled triangle.
02
Remaining balance · Q5
After 96 payments, the balance is the value found with the finance solver, with the payment entered.
03
Inverse square · Q10
$F = \frac{k}{d ^{2}}$ .
04
Separate the variables · Q13
$\frac{d P}{1 - P} = 0.2 d t$ .
05
Integrate acceleration · Q14
The velocity is the integral of the acceleration, and the position is the integral of the velocity.

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