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IB Maths AI HL · Paper 1

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2TZ3

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 TZ2 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

7 marks

A savings plan and a second account

Number and algebra

(a)

3 marks

Use $N = 120$ , $I = 3.6%$ , $P V = 0$ , $P M T = - 1500$ , 12 payments a year.

F V \approx 216 278.58 EUR

(b)

4 marks

After the withdrawal the amount invested is
Compound quarterly for 15 years, which is 60 quarters.

216278.58 - 100000 = 116 278.58 116278.58 (1 + \frac{0.045}{4})^{60} \approx 227 515.92 EUR

Question 2

9 marks

A right-angled triangle in three dimensions

Number and algebra

(a)

2 marks

$A B = (2, 3, 1)$ .

A B = 2^{2} + 3^{2} + 1^{2} = 14 \approx 3.74

(b)

4 marks

The right angle is at $B$ , so $B A \cdot B C = 0$ . With $B A = (- 2, - 3, - 1)$ and $B C = (- 1, 3, k - 2)$ :

2 - 9 - (k - 2) = 0 \Rightarrow - 7 - k + 2 = 0 \Rightarrow k = - 5

(c)

3 marks

$A B = (2, 3, 1)$ and $A C = (1, 6, - 6)$ , with $∣ A C ∣ = 73$ .

cos B A C = \frac{2 + 18 - 6}{14 73} = \frac{14}{14 73} = \frac{14}{73} B A C \approx 64. 0^{\circ}

Question 3

6 marks

Two judges ranking eight sheepdogs: Spearman's coefficient

Statistics and probability

(a)

1 mark

Brett's ranking has $B$ in fourth place.

4

(b)

4 marks

Write the ranks of each dog.
Enter the two lists of ranks and find the correlation, or use $r_{s} = 1 - \frac{6 \sum d ^{2}}{n ( n ^{2} - 1 )}$ with $\sum d^{2} = 0 + 4 + 4 + 0 + 0 + 1 + 1 + 0 = 10$ .

A : (1, 1), B : (4, 2), C : (2, 4), D : (3, 3), E : (5, 5), F : (6, 7), G : (7, 6), H : (8, 8) r_{s} = 1 - \frac{6 ( 10 )}{8 ( 63 )} \approx 0.881

(c)

1 mark

Since $r_{s}$ is close to 1, the two judges ranked the dogs in a very similar order. Since it is not exactly 1, they did not agree on every rank.

Question 4

7 marks

A Voronoi diagram: vertices equidistant from three sites

Calculus

(a)

2 marks

(i) Read the vertex from the diagram, where the cells of $B$ , $C$ and $F$ meet.

(ii) The distance from $B (10, 5)$ .

X = (8, 4) B X = (10 - 8)^{2} + (5 - 4)^{2} = 5, n = 5

(b)

5 marks

(i) $Y$ is equidistant from $B$ and $F$ , which have the same $y$ -coordinate, so it lies on their perpendicular bisector, which is a vertical line through the midpoint.

(ii) The distances from $Y (8, b)$ to $B (10, 5)$ and to $D (5, 9)$ are equal.

a = 8 (8 - 10)^{2} + (b - 5)^{2} = (8 - 5)^{2} + (9 - b)^{2} 4 + b^{2} - 10 b + 25 = 9 + 81 - 18 b + b^{2} \Rightarrow 8 b = 61 b = \frac{61}{8}

Question 5

6 marks

A speed camera: is the car certainly speeding?

Number and algebra

6 marks

The speed is distance over time. The time is between 1.15 s and 1.25 s.
The largest possible time, 1.25 s, gives the smallest possible speed.
This is below the limit of 8.3 m s $^{- 1}$ , so the car could have been below the limit. (A time of 1.2 s gives $\frac{10}{1.2} = 8.33 > 8.3$ , so it could also have been above.)
It is not certain that the car was exceeding the speed limit.

speed = \frac{10}{1.25} = 8 m s^{- 1}

Question 6

5 marks

Coral reef temperatures: an unbiased variance and a confidence interval

Statistics and probability

(a)

2 marks

Convert the sample variance.

s_{n - 1}^{2} = \frac{5}{4} (3.2)^{2} = 12.8

(b)

2 marks

Use a 95% $t$ -interval on the calculator with mean 20.1, $s_{n - 1} = 12.8 = 3.578$ and $n = 5$ .

(15.66, 24.54)

(c)

1 mark

The value 17 is inside the interval, so it is plausible that the mean temperature is 17 degrees.

Question 7

7 marks

Integrating x e to the minus x squared and an area

Calculus

(a)

4 marks

Put $u = - x^{2}$ , so $\frac{d u}{d x} = - 2 x$ and $x d x = - \frac{1}{2} d u$ .

\int x e^{- x^{2}} d x = - \frac{1}{2} \int e^{u} d u = - \frac{1}{2} e^{u} + c - \frac{1}{2} e^{- x^{2}} + c

(b)

3 marks

The curve starts at $x = 0$ and the region is above the axis.

Area = \int_{0}^{k} x e^{- x^{2}} d x = [- \frac{1}{2} e^{- x^{2}}]_{0}^{k} = - \frac{1}{2} e^{- k^{2}} + \frac{1}{2} \frac{1}{2} - \frac{1}{2} e^{- k^{2}}

Question 8

6 marks

An adjacency matrix: a graph and closed walks of length 4

Geometry and trigonometry

(a)

2 marks

Read the matrix. Rows and columns are $A$ to $E$ , and the 1s show these edges: $A B$ , $A E$ , $B D$ , $B E$ and $C D$ . The matrix is symmetric, so each edge goes both ways.
Sketch: five labelled vertices joined by the five edges $A B$ , $A E$ , $B D$ , $B E$ , $C D$ (triangle $A B E$ , with $D$ joined to $B$ , and $C$ joined to $D$ ).

(b)

4 marks

Raise the matrix to the fourth power.
Walks that start and end in the same town are the diagonal entries.

M^{4} = 76156612426142015206566157 7 + 12 + 2 + 6 + 7 = 34

Question 9

7 marks

An ice sheet modelled as a rectangle: rates of change

Calculus

(a)

3 marks

$A = x y$ with $\frac{d x}{d t} = 10$ and $\frac{d y}{d t} = - 5$ .

\frac{d A}{d t} = 75 (- 5) + 40 (10) = - 375 + 400 = 25 km^{2} per year

(b)

1 mark

Since $\frac{d A}{d t} = 25 > 0$ at $t = 0$ , the area is increasing.

(c)

3 marks

At $t = 1$ the width is $85$ and the length is $35$ , so the area is $85 \times 35 = 2975$ . At $t = 0$ it is $75 \times 40 = 3000$ .
The area decreases by 25 km $^{2}$ .

2975 - 3000 = - 25

Question 10

5 marks

A function on five integers and its inverse

Functions

(a)

1 mark

Find the $x$ with $f (x) = 4$ .

x = 3

(b)

1 mark

Find the $x$ with $f (x) = x$ . From the table, $f (1) = 1$ .

x = 1

(c)

3 marks

Read the table backwards: $f (2) = 0$ , $f (1) = 1$ , $f (4) = 2$ , $f (0) = 3$ , $f (3) = 4$ .

f^{- 1} (0) = 2, f^{- 1} (1) = 1, f^{- 1} (2) = 4, f^{- 1} (3) = 0, f^{- 1} (4) = 3

Question 11

6 marks

A cubic relationship between bird and wasp populations

Functions

(a)

2 marks

Factorise $p^{3} - 4 p^{2} + 3 p = p (p - 1) (p - 3) = 0$ .

p = 0, 1, 3

(b)

4 marks

Sketch $w = p^{3} - 4 p^{2} + 3 p$ for $p > 0$ . It rises to a local maximum, falls to a local minimum below the axis, and then rises again.
Find the local maximum from $\frac{d w}{d p} = 3 p^{2} - 8 p + 3 = 0$ , which gives $p \approx 0.451$ .
For $0 < w < 0.631$ the line meets the curve three times with $p > 0$ . For $w > 0.631$ it meets only once, on the right.

w_{m a x} = 0.45 1^{3} - 4 (0.451)^{2} + 3 (0.451) \approx 0.631 w > 0.631

Question 12

9 marks

A differential equation for an object whose acceleration depends on v to the 1.5

Calculus

(a)

1 mark

The acceleration is proportional to $v^{1.5}$ .

\frac{d v}{d t} = k v^{1.5}

(b)

8 marks

Use $v = 4$ and $\frac{d v}{d t} = - 3$ at the start.
Separate the variables and integrate.
Use $v = 4$ at $t = 0$ , so $v^{- 0.5} = \frac{1}{2}$ .
Solve for $v$ .

- 3 = k (4)^{1.5} = 8 k \Rightarrow k = - \frac{3}{8} \int v^{- 1.5} d v = \int - \frac{3}{8} d t \Rightarrow - 2 v^{- 0.5} = - \frac{3}{8} t + c - 1 = c - 2 v^{- 0.5} = - \frac{3}{8} t - 1 \Rightarrow v^{- 0.5} = \frac{3 t + 8}{16} v = \frac{256}{( 3 t + 8 ) ^{2}}

Question 13

9 marks

Worms in a section of soil: Poisson and binomial probabilities

Statistics and probability

(a)

1 mark

Let $X \sim P o (1.2)$ be the worms in 1 m $^{2}$ .

P (X = 1) \approx 0.361

(b)

1 mark

Use the complement.

P (X \geq 1) = 1 - P (X = 0) \approx 0.699

(c)

2 marks

The total in 5 sections is $Y \sim P o (6)$ .

P (Y = 5) \approx 0.161

(d)

2 marks

Each of the five sections has exactly one worm, independently.

(P (X = 1))^{5} = 0.361 4^{5} \approx 0.00617

(e)

3 marks

Let $W$ be the number of sections with at least one worm, $W \sim B (5, 0.6988)$ .

P (W = 3) = (3 5) (0.6988)^{3} (0.3012)^{2} \approx 0.310

Question 14

6 marks

Puppy weight against age: choosing the better of two regression lines

Number and algebra

6 marks

The second graph has the larger $R^{2}$ (0.95 compared with 0.72), so choose the relationship from the line of $ln w$ on $ln d$ .
Take exponentials of both sides.

ln w = 0.302 ln d + 0.693 w = e^{0.693} d^{0.302} \approx 2.00 d^{0.302} w = 2 d^{0.302}

Question 15

8 marks

A parabolic mirror: normal, a ray of light and the angle between them

Calculus

(a)

3 marks

Differentiate $y = x^{2}$ , so the gradient of the tangent at $x = 2$ is $4$ and the normal has gradient $- \frac{1}{4}$ .

y - 4 = - \frac{1}{4} (x - 2) \Rightarrow y = - \frac{1}{4} x + \frac{9}{2}

(b)

2 marks

The ray goes from $(0, 10)$ to $(2, 4)$ .

gradient = \frac{4 - 10}{2 - 0} = - 3

(c)

3 marks

Find the angle each line makes with the $x$ -axis.
The angle between the ray and the normal is the difference.

arctan (- 3) = - 71.5 7^{\circ}, arctan (- \frac{1}{4}) = - 14.0 4^{\circ} 71.5 7^{\circ} - 14.0 4^{\circ} \approx 57. 5^{\circ} (1.00 radians)

Question 16

7 marks

A quadratic with complex roots in a circuit model

Number and algebra

(a)

2 marks

With $t = 1$ the equation is $z^{2} + 2 z + 8 = 0$ .
The required root has an argument between $\frac{π}{2}$ and $π$ , so it has a positive imaginary part.

z = \frac{- 2 \pm 4 - 32}{2} = - 1 \pm 7 i z = - 1 + 7 i \approx - 1 + 2.65 i

(b)

5 marks

Use the quadratic formula. For $0 \leq t \leq 2$ we have $t^{2} - 8 t < 0$ , so the roots are complex.
Find the modulus squared.
This increases with $t$ , so it is largest at the end of the interval.

z = - t \pm i 8 t - t^{2} ∣ z ∣^{2} = t^{2} + 8 t - t^{2} = 8 t t = 2

Our analysis

Where the marks are

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

Calculus38 marks · 35%
Q4, Q7, Q9, Q12, Q15
Number and algebra35 marks · 32%
Q1, Q2, Q5, Q14, Q16
Statistics and probability20 marks · 18%
Q3, Q6, Q13
Functions11 marks · 10%
Q10, Q11
Geometry and trigonometry6 marks · 5%
Q8

Mentor insights

Traps to watch for

01
Annuity · Q1
Use the finance solver with $P V = 0$ and a regular payment.
02
Right angle · Q2
The scalar product of $B A$ and $B C$ is zero.
03
Separate the variables · Q12
Write the equation as $v^{- 1.5} d v = k d t$ .
04
Scaling · Q13
For 5 sections the Poisson mean is $5 \times 1.2 = 6$ .
05
Normal · Q15
The normal has gradient equal to minus the reciprocal of the tangent gradient.

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

TZ1TZ2TZ3

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 TZ2 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

7 marks

A savings plan and a second account

Number and algebra

(a)

3 marks

Use $N = 120$ , $I = 3.6%$ , $P V = 0$ , $P M T = - 1500$ , 12 payments a year.

F V \approx 216 278.58 EUR

(b)

4 marks

After the withdrawal the amount invested is
Compound quarterly for 15 years, which is 60 quarters.

216278.58 - 100000 = 116 278.58 116278.58 (1 + \frac{0.045}{4})^{60} \approx 227 515.92 EUR

Question 2

9 marks

A right-angled triangle in three dimensions

Number and algebra

(a)

2 marks

$A B = (2, 3, 1)$ .

A B = 2^{2} + 3^{2} + 1^{2} = 14 \approx 3.74

(b)

4 marks

The right angle is at $B$ , so $B A \cdot B C = 0$ . With $B A = (- 2, - 3, - 1)$ and $B C = (- 1, 3, k - 2)$ :

2 - 9 - (k - 2) = 0 \Rightarrow - 7 - k + 2 = 0 \Rightarrow k = - 5

(c)

3 marks

$A B = (2, 3, 1)$ and $A C = (1, 6, - 6)$ , with $∣ A C ∣ = 73$ .

cos B A C = \frac{2 + 18 - 6}{14 73} = \frac{14}{14 73} = \frac{14}{73} B A C \approx 64. 0^{\circ}

Question 3

6 marks

Two judges ranking eight sheepdogs: Spearman's coefficient

Statistics and probability

(a)

1 mark

Brett's ranking has $B$ in fourth place.

4

(b)

4 marks

Write the ranks of each dog.
Enter the two lists of ranks and find the correlation, or use $r_{s} = 1 - \frac{6 \sum d ^{2}}{n ( n ^{2} - 1 )}$ with $\sum d^{2} = 0 + 4 + 4 + 0 + 0 + 1 + 1 + 0 = 10$ .

A : (1, 1), B : (4, 2), C : (2, 4), D : (3, 3), E : (5, 5), F : (6, 7), G : (7, 6), H : (8, 8) r_{s} = 1 - \frac{6 ( 10 )}{8 ( 63 )} \approx 0.881

(c)

1 mark

Since $r_{s}$ is close to 1, the two judges ranked the dogs in a very similar order. Since it is not exactly 1, they did not agree on every rank.

Question 4

7 marks

A Voronoi diagram: vertices equidistant from three sites

Calculus

(a)

2 marks

(i) Read the vertex from the diagram, where the cells of $B$ , $C$ and $F$ meet.

(ii) The distance from $B (10, 5)$ .

X = (8, 4) B X = (10 - 8)^{2} + (5 - 4)^{2} = 5, n = 5

(b)

5 marks

(i) $Y$ is equidistant from $B$ and $F$ , which have the same $y$ -coordinate, so it lies on their perpendicular bisector, which is a vertical line through the midpoint.

(ii) The distances from $Y (8, b)$ to $B (10, 5)$ and to $D (5, 9)$ are equal.

a = 8 (8 - 10)^{2} + (b - 5)^{2} = (8 - 5)^{2} + (9 - b)^{2} 4 + b^{2} - 10 b + 25 = 9 + 81 - 18 b + b^{2} \Rightarrow 8 b = 61 b = \frac{61}{8}

Question 5

6 marks

A speed camera: is the car certainly speeding?

Number and algebra

6 marks

The speed is distance over time. The time is between 1.15 s and 1.25 s.
The largest possible time, 1.25 s, gives the smallest possible speed.
This is below the limit of 8.3 m s $^{- 1}$ , so the car could have been below the limit. (A time of 1.2 s gives $\frac{10}{1.2} = 8.33 > 8.3$ , so it could also have been above.)
It is not certain that the car was exceeding the speed limit.

speed = \frac{10}{1.25} = 8 m s^{- 1}

Question 6

5 marks

Coral reef temperatures: an unbiased variance and a confidence interval

Statistics and probability

(a)

2 marks

Convert the sample variance.

s_{n - 1}^{2} = \frac{5}{4} (3.2)^{2} = 12.8

(b)

2 marks

Use a 95% $t$ -interval on the calculator with mean 20.1, $s_{n - 1} = 12.8 = 3.578$ and $n = 5$ .

(15.66, 24.54)

(c)

1 mark

The value 17 is inside the interval, so it is plausible that the mean temperature is 17 degrees.

Question 7

7 marks

Integrating x e to the minus x squared and an area

Calculus

(a)

4 marks

Put $u = - x^{2}$ , so $\frac{d u}{d x} = - 2 x$ and $x d x = - \frac{1}{2} d u$ .

\int x e^{- x^{2}} d x = - \frac{1}{2} \int e^{u} d u = - \frac{1}{2} e^{u} + c - \frac{1}{2} e^{- x^{2}} + c

(b)

3 marks

The curve starts at $x = 0$ and the region is above the axis.

Area = \int_{0}^{k} x e^{- x^{2}} d x = [- \frac{1}{2} e^{- x^{2}}]_{0}^{k} = - \frac{1}{2} e^{- k^{2}} + \frac{1}{2} \frac{1}{2} - \frac{1}{2} e^{- k^{2}}

Question 8

6 marks

An adjacency matrix: a graph and closed walks of length 4

Geometry and trigonometry

(a)

2 marks

Read the matrix. Rows and columns are $A$ to $E$ , and the 1s show these edges: $A B$ , $A E$ , $B D$ , $B E$ and $C D$ . The matrix is symmetric, so each edge goes both ways.
Sketch: five labelled vertices joined by the five edges $A B$ , $A E$ , $B D$ , $B E$ , $C D$ (triangle $A B E$ , with $D$ joined to $B$ , and $C$ joined to $D$ ).

(b)

4 marks

Raise the matrix to the fourth power.
Walks that start and end in the same town are the diagonal entries.

M^{4} = 76156612426142015206566157 7 + 12 + 2 + 6 + 7 = 34

Question 9

7 marks

An ice sheet modelled as a rectangle: rates of change

Calculus

(a)

3 marks

$A = x y$ with $\frac{d x}{d t} = 10$ and $\frac{d y}{d t} = - 5$ .

\frac{d A}{d t} = 75 (- 5) + 40 (10) = - 375 + 400 = 25 km^{2} per year

(b)

1 mark

Since $\frac{d A}{d t} = 25 > 0$ at $t = 0$ , the area is increasing.

(c)

3 marks

At $t = 1$ the width is $85$ and the length is $35$ , so the area is $85 \times 35 = 2975$ . At $t = 0$ it is $75 \times 40 = 3000$ .
The area decreases by 25 km $^{2}$ .

2975 - 3000 = - 25

Question 10

5 marks

A function on five integers and its inverse

Functions

(a)

1 mark

Find the $x$ with $f (x) = 4$ .

x = 3

(b)

1 mark

Find the $x$ with $f (x) = x$ . From the table, $f (1) = 1$ .

x = 1

(c)

3 marks

Read the table backwards: $f (2) = 0$ , $f (1) = 1$ , $f (4) = 2$ , $f (0) = 3$ , $f (3) = 4$ .

f^{- 1} (0) = 2, f^{- 1} (1) = 1, f^{- 1} (2) = 4, f^{- 1} (3) = 0, f^{- 1} (4) = 3

Question 11

6 marks

A cubic relationship between bird and wasp populations

Functions

(a)

2 marks

Factorise $p^{3} - 4 p^{2} + 3 p = p (p - 1) (p - 3) = 0$ .

p = 0, 1, 3

(b)

4 marks

Sketch $w = p^{3} - 4 p^{2} + 3 p$ for $p > 0$ . It rises to a local maximum, falls to a local minimum below the axis, and then rises again.
Find the local maximum from $\frac{d w}{d p} = 3 p^{2} - 8 p + 3 = 0$ , which gives $p \approx 0.451$ .
For $0 < w < 0.631$ the line meets the curve three times with $p > 0$ . For $w > 0.631$ it meets only once, on the right.

w_{m a x} = 0.45 1^{3} - 4 (0.451)^{2} + 3 (0.451) \approx 0.631 w > 0.631

Question 12

9 marks

A differential equation for an object whose acceleration depends on v to the 1.5

Calculus

(a)

1 mark

The acceleration is proportional to $v^{1.5}$ .

\frac{d v}{d t} = k v^{1.5}

(b)

8 marks

Use $v = 4$ and $\frac{d v}{d t} = - 3$ at the start.
Separate the variables and integrate.
Use $v = 4$ at $t = 0$ , so $v^{- 0.5} = \frac{1}{2}$ .
Solve for $v$ .

- 3 = k (4)^{1.5} = 8 k \Rightarrow k = - \frac{3}{8} \int v^{- 1.5} d v = \int - \frac{3}{8} d t \Rightarrow - 2 v^{- 0.5} = - \frac{3}{8} t + c - 1 = c - 2 v^{- 0.5} = - \frac{3}{8} t - 1 \Rightarrow v^{- 0.5} = \frac{3 t + 8}{16} v = \frac{256}{( 3 t + 8 ) ^{2}}

Question 13

9 marks

Worms in a section of soil: Poisson and binomial probabilities

Statistics and probability

(a)

1 mark

Let $X \sim P o (1.2)$ be the worms in 1 m $^{2}$ .

P (X = 1) \approx 0.361

(b)

1 mark

Use the complement.

P (X \geq 1) = 1 - P (X = 0) \approx 0.699

(c)

2 marks

The total in 5 sections is $Y \sim P o (6)$ .

P (Y = 5) \approx 0.161

(d)

2 marks

Each of the five sections has exactly one worm, independently.

(P (X = 1))^{5} = 0.361 4^{5} \approx 0.00617

(e)

3 marks

Let $W$ be the number of sections with at least one worm, $W \sim B (5, 0.6988)$ .

P (W = 3) = (3 5) (0.6988)^{3} (0.3012)^{2} \approx 0.310

Question 14

6 marks

Puppy weight against age: choosing the better of two regression lines

Number and algebra

6 marks

The second graph has the larger $R^{2}$ (0.95 compared with 0.72), so choose the relationship from the line of $ln w$ on $ln d$ .
Take exponentials of both sides.

ln w = 0.302 ln d + 0.693 w = e^{0.693} d^{0.302} \approx 2.00 d^{0.302} w = 2 d^{0.302}

Question 15

8 marks

A parabolic mirror: normal, a ray of light and the angle between them

Calculus

(a)

3 marks

Differentiate $y = x^{2}$ , so the gradient of the tangent at $x = 2$ is $4$ and the normal has gradient $- \frac{1}{4}$ .

y - 4 = - \frac{1}{4} (x - 2) \Rightarrow y = - \frac{1}{4} x + \frac{9}{2}

(b)

2 marks

The ray goes from $(0, 10)$ to $(2, 4)$ .

gradient = \frac{4 - 10}{2 - 0} = - 3

(c)

3 marks

Find the angle each line makes with the $x$ -axis.
The angle between the ray and the normal is the difference.

arctan (- 3) = - 71.5 7^{\circ}, arctan (- \frac{1}{4}) = - 14.0 4^{\circ} 71.5 7^{\circ} - 14.0 4^{\circ} \approx 57. 5^{\circ} (1.00 radians)

Question 16

7 marks

A quadratic with complex roots in a circuit model

Number and algebra

(a)

2 marks

With $t = 1$ the equation is $z^{2} + 2 z + 8 = 0$ .
The required root has an argument between $\frac{π}{2}$ and $π$ , so it has a positive imaginary part.

z = \frac{- 2 \pm 4 - 32}{2} = - 1 \pm 7 i z = - 1 + 7 i \approx - 1 + 2.65 i

(b)

5 marks

Use the quadratic formula. For $0 \leq t \leq 2$ we have $t^{2} - 8 t < 0$ , so the roots are complex.
Find the modulus squared.
This increases with $t$ , so it is largest at the end of the interval.

z = - t \pm i 8 t - t^{2} ∣ z ∣^{2} = t^{2} + 8 t - t^{2} = 8 t t = 2

Our analysis

Where the marks are

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

Calculus38 marks · 35%
Q4, Q7, Q9, Q12, Q15
Number and algebra35 marks · 32%
Q1, Q2, Q5, Q14, Q16
Statistics and probability20 marks · 18%
Q3, Q6, Q13
Functions11 marks · 10%
Q10, Q11
Geometry and trigonometry6 marks · 5%
Q8

Mentor insights

Traps to watch for

01
Annuity · Q1
Use the finance solver with $P V = 0$ and a regular payment.
02
Right angle · Q2
The scalar product of $B A$ and $B C$ is zero.
03
Separate the variables · Q12
Write the equation as $v^{- 1.5} d v = k d t$ .
04
Scaling · Q13
For 5 sections the Poisson mean is $5 \times 1.2 = 6$ .
05
Normal · Q15
The normal has gradient equal to minus the reciprocal of the tangent gradient.

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