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IB Maths AI HL · Paper 3

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2 TZ3

Session: May 2026
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

Q1 Q2

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2026 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–2 · 55 marks

Question 1

26 marks

Hypothesis tests on phone use and exam results

Statistics and probability

(a)(i)

1 mark

We are comparing the means of two independent groups with unknown population variances, so a $t$ -test on two samples is the right tool. It is one-tailed because we only care whether B is lower.

two-sample t -test (one-tailed)

Mentor tip

A $z$ -test needs known population standard deviations. Here we only have sample values, so it is a $t$ -test.

(a)(ii)

1 mark

The $t$ -test relies on the underlying scores being normal. (The pooled version also assumes the two schools share the same population variance; either assumption works.)

Quiz results in each school are normally distributed

Mentor tip

Give one assumption and phrase it about the populations, not the samples.

(a)(iii)

2 marks

Hypotheses are about the population means $μ_{A}$ and $μ_{B}$ , not the sample means $56$ and $51$ .
The alternative points the way we suspect: B lower than A.

H_{0} : μ_{A} = μ_{B} H_{1} : μ_{A} > μ_{B}

Mentor tip

Writing $\overset{x}{ˉ}_{A} = \overset{x}{ˉ}_{B}$ is a common slip: the sample means are known numbers, so there is nothing to test about them.

(a)(iv)

3 marks

The table gives $s_{n}$ , but the GDC wants the unbiased estimate $s_{n - 1}$ , so convert each one first.
Run the pooled two-sample $t$ -test on the GDC with $μ_{A} > μ_{B}$ .

s_{n - 1} = s_{n} \frac{n}{n - 1} s_{A} = 12.2 \frac{63}{62} = 12.298 \dots, s_{B} = 14.5 \frac{39}{38} = 14.689 \dots t = 1.851 \dots (GDC, pooled) p = 0.0336

Mentor tip

Typing $12.2$ and $14.5$ straight into the GDC gives a slightly wrong $p$ -value. Always check which standard deviation you have been handed.

(a)(v)

2 marks

Compare the $p$ -value with the $5%$ level.
It is smaller, so reject $H_{0}$ and say what that means for the two schools.

p = 0.0336 < 0.05 Reject H_{0} Evidence that the mean quiz result in school B is lower than in school A

Mentor tip

A conclusion needs both halves: the comparison with $0.05$ and a sentence in context about the schools.

(b)

1 mark

Think about what else could explain the gap. Only two schools were compared, and many things besides phones differ between schools (teachers, intake, resources).

Only two schools; other factors could explain the difference

Mentor tip

Avoid vague answers like 'not enough data'. Tie your reason to this study, for example the fact that just two schools were sampled.

(c)

3 marks

Use the midpoint of each mark class: $10, 27, 42, 57, 72, 90$ (the marks are whole numbers, so $1$ to $19$ has midpoint $10$ ).
Multiply by the unrestricted-use frequencies, add up and divide by $90$ .

\overset{m}{ˉ} = \frac{6 ( 10 ) + 10 ( 27 ) + 23 ( 42 ) + 27 ( 57 ) + 16 ( 72 ) + 8 ( 90 )}{90} = \frac{4707}{90} \overset{m}{ˉ} \approx 52.3

Mentor tip

Midpoints like $10.5$ and $27.5$ treat the data as continuous and give $52.8$ . For whole-number marks, average the smallest and largest possible values.

(d)(i)

2 marks

Expected count $= \frac{row total \times column total}{grand total}$ . The smallest one sits in the lowest mark class and the unrestricted column.
It is below $5$ , so merge that class with its neighbour: the first two rows.

E = \frac{17 \times 90}{400} = 3.825 < 5 Combine the rows 1 \leq m \leq 19 and 20 \leq m \leq 34

Mentor tip

Quote the actual expected value. Saying 'some are under 5' without showing which is not a justification.

(d)(ii)

4 marks

$H_{0}$ : achievement and mobile phone policy are independent. $H_{1}$ : they are not independent.
Enter the $5 \times 3$ table (first two rows merged into $12, 19, 16$ ) into the GDC $χ^{2}$ test.
Compare $p$ with $0.05$ and answer in terms of Ling's belief.

Merged row: 12, 19, 16 (47) χ^{2} = 13.4, ν = (5 - 1) (3 - 1) = 8 p = 0.0999 (GDC) 0.0999 > 0.05 \Rightarrow do not reject H_{0} Not enough evidence to support Ling’s belief

Mentor tip

If you forget to merge the rows the GDC gives $p = 0.201$ . Same conclusion, but the wrong test.

(e)

1 mark

A Type I error means rejecting a true $H_{0}$ . The chance of that is exactly the significance level.

P (Type I) = 0.05

(f)(i)

3 marks

Under $H_{0}$ the sample mean is normal with mean $64.8$ and standard error $\frac{15.3}{60}$ .
The test is upper one-tailed, so find the value with $5%$ above it using inverse normal.

\overset{ˉ}{X} \sim N (64.8, (\frac{15.3}{60})^{2}), \frac{15.3}{60} = 1.975 \dots P (\overset{ˉ}{X} > c) = 0.05 \Rightarrow c = 68.0489 \dots (GDC) Critical region: \overset{ˉ}{X} > 68.0

Mentor tip

Use the standard error, not $15.3$ itself: the test is about the mean of 60 students.

(f)(ii)

1 mark

A Type II error is failing to reject $H_{0}$ when the true mean is $68.3$ . Find the chance the sample mean lands outside the critical region under that true mean.

P (\overset{ˉ}{X} < 68.0489 \dots \overset{ˉ}{X} \sim N (68.3, \frac{15. 3 ^{2}}{60})) = 0.449 (GDC)

Mentor tip

Keep the unrounded critical value in the GDC. Rounding to $68.0$ first moves the answer.

(g)(i)

1 mark

A larger significance level widens the critical region, so it is easier to reject $H_{0}$ when it is false.

Increase the significance level (e.g. to 10%)

(g)(ii)

1 mark

The price of a wider critical region is that a true $H_{0}$ is also rejected more often.

The probability of a Type I error goes up

Mentor tip

For a fixed sample size, Type I and Type II errors pull against each other: lowering one raises the other.

Question 2

29 marks

Complex numbers as animation transformations

Number and algebra
Geometry and trigonometry
Calculus

(a)(i)

1 mark

Expand and use $i^{2} = - 1$ .

z i = (x + y i) i = x i + y i^{2} = - y + x i

(a)(ii)

2 marks

The point $(x, y)$ goes to $(- y, x)$ . Try $(1, 0) \to (0, 1)$ : a quarter turn anticlockwise.
Give the angle, the direction and the centre.

(x, y) \mapsto (- y, x) Rotation of 9 0^{\circ} (\frac{π}{2}) anticlockwise about the origin

Mentor tip

A rotation needs three things: angle, direction and centre. Leaving out the centre is the usual slip.

(b)

1 mark

Adding $a + bi$ adds $a$ to the $x$ -coordinate and $b$ to the $y$ -coordinate.

Translation by (a b)

(c)

1 mark

An enlargement from the origin scales both coordinates by $k$ , which is the same as multiplying the complex number by $k$ .

z \mapsto k z

(d)(i)

1 mark

$g$ adds $2 + i$ , so its inverse subtracts it.

g^{- 1} (z) = z - 2 - i

(d)(ii)

2 marks

Work from the inside out: apply $g^{- 1}$ first, then $f$ , then $g$ .

f (g^{- 1} (z)) = 2 i (z - 2 - i) = 2 i z - 4 i + 2 h (z) = g (2 i z + 2 - 4 i) = 2 i z + 2 - 4 i + 2 + i = 2 i z + 4 - 3 i

Mentor tip

In $g \circ f \circ g^{- 1}$ the right-hand function acts first. Watch the sign: $2 i \times (- i) = + 2$ .

(e)(i)

3 marks

Substitute each complex number into $h (z) = 2 i z + 4 - 3 i$ and read off the real and imaginary parts as coordinates.

h (1 + i) = 2 i - 2 + 4 - 3 i = 2 - i \Rightarrow A^{'} (2, - 1) h (2 + i) = 4 i - 2 + 4 - 3 i = 2 + i \Rightarrow B^{'} (2, 1) h (2 + 3 i) = 4 i - 6 + 4 - 3 i = - 2 + i \Rightarrow C^{'} (- 2, 1)

Mentor tip

Notice $B^{'} = B$ . A point that does not move is a strong hint about the centre in the next part.

(e)(ii)

3 marks

Compare sides: $A B$ has length $1$ and $A^{'} B^{'}$ has length $2$ , so the scale factor is $2$ .
$B A$ points left, while $B^{'} A^{'}$ points down: a $9 0^{\circ}$ anticlockwise turn.
$B$ stays fixed, so it is the centre of both.

∣ A^{'} B^{'} ∣ = 2∣ A B ∣, ∣ B^{'} C^{'} ∣ = 2∣ B C ∣ h (z) = z \Rightarrow z (1 - 2 i) = 4 - 3 i \Rightarrow z = 2 + i Rotation 9 0^{\circ} anticlockwise and enlargement, scale factor 2, both centred at (2, 1)

Another way. Write $h (z) - (2 + i) = 2 i (z - (2 + i))$ : multiplying by $2 i$ is a quarter turn with a doubling, about $2 + i$ .

Mentor tip

State the centre explicitly. Rotating and enlarging about the origin would give different points.

(f)

1 mark

Replace $30$ with $θ$ in the given rotation function.

R_{θ} (z, t) = z (cos (θ t^{\circ}) + i sin (θ t^{\circ}))

(g)

2 marks

A constant velocity means the displacement after $t$ seconds is the velocity times $t$ .
The vector $(2 - 1)$ is the complex number $2 - i$ ; add $t$ lots of it.

T (z, t) = z + (2 - i) t

Mentor tip

Check at $t = 0$ : nothing should have moved yet, and indeed $T (z, 0) = z$ .

(h)(i)

2 marks

With $k = 3$ the multiplier is $1 + 2 t$ . Every point is scaled by it, so the radius is too.

E_{3} (z, t) = (1 + 2 t) z t = 1 : r = (1 + 2) (1) = 3

(h)(ii)

1 mark

Use the same multiplier at $t = 2$ .

r = (1 + 4) (1) = 5

(i)

2 marks

The object is scaled by $p (t) = 1 + (k - 1) t$ , which is linear in $t$ .
Its rate of change is the constant $k - 1$ : growth if $k > 1$ , shrinking if $k < 1$ .

p (t) = 1 + (k - 1) t \frac{d p}{d t} = k - 1 (constant)

Mentor tip

Compare with (h): the radius went $1, 3, 5$ , rising by $k - 1 = 2$ each second.

(j)

5 marks

Shrinking: the size factor must hit $0$ at $t = 5$ , so solve $1 + 5 (k - 1) = 0$ , giving $k = 0.8$ and a factor $1 - 0.2 t$ .
Rotating: $36 0^{\circ}$ per second, so multiply by $cos (360 t^{\circ}) + i sin (360 t^{\circ})$ .
Moving: the centre goes from $0$ to $25 + 15 i$ in $5$ s, a velocity of $5 + 3 i$ per second.
Scale and rotate about the origin first, then translate, so the disc spins about its own moving centre.

1 + 5 (k - 1) = 0 \Rightarrow k - 1 = - 0.2 v = \frac{25 + 15 i}{5} = 5 + 3 i A (z, t) = (1 - 0.2 t) (cos (360 t^{\circ}) + i sin (360 t^{\circ})) z + (5 + 3 i) t

Mentor tip

Order matters. If you translate first and then rotate, the whole disc swings around the origin instead of spinning on the spot.

(k)

2 marks

At $60$ frames per second, frame $n$ is shown at $t = \frac{n}{60}$ .
Substitute into $A (z, t)$ and simplify each piece.

t = \frac{n}{60} B_{n} (z) = A (z, \frac{n}{60}) = (1 - \frac{n}{300}) (cos (6 n^{\circ}) + i sin (6 n^{\circ})) z + (\frac{1}{12} + \frac{i}{20}) n

Mentor tip

Check: frame $300$ is $t = 5$ , where the factor $1 - \frac{300}{300}$ is $0$ and the centre is at $25 + 15 i$ .

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 probability26 marks · 47%
Q1
Number and algebra21 marks · 38%
Q2
Geometry and trigonometry6 marks · 11%
Q2
Calculus2 marks · 4%
Q2

Mentor insights

Traps to watch for

01
Using s_n instead of s_{n-1} · Q1
The table gives $s_{n}$ . Convert to $s_{n - 1}$ before the $t$ -test, or the $p$ -value comes out slightly off.
02
Forgetting to merge rows · Q1
An expected count of $3.825$ is below $5$ , so the first two mark classes must be combined before running the $χ^{2}$ test.
03
Standard deviation in the critical region · Q1
The critical value uses the standard error $15.3/ 60$ , not $15.3$ , because the test is about a sample mean.
04
Missing the centre of the transformation · Q2
The rotation and enlargement in $h$ are about $(2, 1)$ , not the origin. Spot that $B$ does not move.
05
Wrong order in the animation function · Q2
Scale and rotate $z$ first, then add the translation $(5 + 3 i) t$ , so the disc spins about its own centre.

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 3

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2 TZ3

Session: May 2026
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

Q1 Q2

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2026 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–2 · 55 marks

Question 1

26 marks

Hypothesis tests on phone use and exam results

Statistics and probability

(a)(i)

1 mark

We are comparing the means of two independent groups with unknown population variances, so a $t$ -test on two samples is the right tool. It is one-tailed because we only care whether B is lower.

two-sample t -test (one-tailed)

Mentor tip

A $z$ -test needs known population standard deviations. Here we only have sample values, so it is a $t$ -test.

(a)(ii)

1 mark

The $t$ -test relies on the underlying scores being normal. (The pooled version also assumes the two schools share the same population variance; either assumption works.)

Quiz results in each school are normally distributed

Mentor tip

Give one assumption and phrase it about the populations, not the samples.

(a)(iii)

2 marks

Hypotheses are about the population means $μ_{A}$ and $μ_{B}$ , not the sample means $56$ and $51$ .
The alternative points the way we suspect: B lower than A.

H_{0} : μ_{A} = μ_{B} H_{1} : μ_{A} > μ_{B}

Mentor tip

Writing $\overset{x}{ˉ}_{A} = \overset{x}{ˉ}_{B}$ is a common slip: the sample means are known numbers, so there is nothing to test about them.

(a)(iv)

3 marks

The table gives $s_{n}$ , but the GDC wants the unbiased estimate $s_{n - 1}$ , so convert each one first.
Run the pooled two-sample $t$ -test on the GDC with $μ_{A} > μ_{B}$ .

s_{n - 1} = s_{n} \frac{n}{n - 1} s_{A} = 12.2 \frac{63}{62} = 12.298 \dots, s_{B} = 14.5 \frac{39}{38} = 14.689 \dots t = 1.851 \dots (GDC, pooled) p = 0.0336

Mentor tip

Typing $12.2$ and $14.5$ straight into the GDC gives a slightly wrong $p$ -value. Always check which standard deviation you have been handed.

(a)(v)

2 marks

Compare the $p$ -value with the $5%$ level.
It is smaller, so reject $H_{0}$ and say what that means for the two schools.

p = 0.0336 < 0.05 Reject H_{0} Evidence that the mean quiz result in school B is lower than in school A

Mentor tip

A conclusion needs both halves: the comparison with $0.05$ and a sentence in context about the schools.

(b)

1 mark

Think about what else could explain the gap. Only two schools were compared, and many things besides phones differ between schools (teachers, intake, resources).

Only two schools; other factors could explain the difference

Mentor tip

Avoid vague answers like 'not enough data'. Tie your reason to this study, for example the fact that just two schools were sampled.

(c)

3 marks

Use the midpoint of each mark class: $10, 27, 42, 57, 72, 90$ (the marks are whole numbers, so $1$ to $19$ has midpoint $10$ ).
Multiply by the unrestricted-use frequencies, add up and divide by $90$ .

\overset{m}{ˉ} = \frac{6 ( 10 ) + 10 ( 27 ) + 23 ( 42 ) + 27 ( 57 ) + 16 ( 72 ) + 8 ( 90 )}{90} = \frac{4707}{90} \overset{m}{ˉ} \approx 52.3

Mentor tip

Midpoints like $10.5$ and $27.5$ treat the data as continuous and give $52.8$ . For whole-number marks, average the smallest and largest possible values.

(d)(i)

2 marks

Expected count $= \frac{row total \times column total}{grand total}$ . The smallest one sits in the lowest mark class and the unrestricted column.
It is below $5$ , so merge that class with its neighbour: the first two rows.

E = \frac{17 \times 90}{400} = 3.825 < 5 Combine the rows 1 \leq m \leq 19 and 20 \leq m \leq 34

Mentor tip

Quote the actual expected value. Saying 'some are under 5' without showing which is not a justification.

(d)(ii)

4 marks

$H_{0}$ : achievement and mobile phone policy are independent. $H_{1}$ : they are not independent.
Enter the $5 \times 3$ table (first two rows merged into $12, 19, 16$ ) into the GDC $χ^{2}$ test.
Compare $p$ with $0.05$ and answer in terms of Ling's belief.

Merged row: 12, 19, 16 (47) χ^{2} = 13.4, ν = (5 - 1) (3 - 1) = 8 p = 0.0999 (GDC) 0.0999 > 0.05 \Rightarrow do not reject H_{0} Not enough evidence to support Ling’s belief

Mentor tip

If you forget to merge the rows the GDC gives $p = 0.201$ . Same conclusion, but the wrong test.

(e)

1 mark

A Type I error means rejecting a true $H_{0}$ . The chance of that is exactly the significance level.

P (Type I) = 0.05

(f)(i)

3 marks

Under $H_{0}$ the sample mean is normal with mean $64.8$ and standard error $\frac{15.3}{60}$ .
The test is upper one-tailed, so find the value with $5%$ above it using inverse normal.

\overset{ˉ}{X} \sim N (64.8, (\frac{15.3}{60})^{2}), \frac{15.3}{60} = 1.975 \dots P (\overset{ˉ}{X} > c) = 0.05 \Rightarrow c = 68.0489 \dots (GDC) Critical region: \overset{ˉ}{X} > 68.0

Mentor tip

Use the standard error, not $15.3$ itself: the test is about the mean of 60 students.

(f)(ii)

1 mark

A Type II error is failing to reject $H_{0}$ when the true mean is $68.3$ . Find the chance the sample mean lands outside the critical region under that true mean.

P (\overset{ˉ}{X} < 68.0489 \dots \overset{ˉ}{X} \sim N (68.3, \frac{15. 3 ^{2}}{60})) = 0.449 (GDC)

Mentor tip

Keep the unrounded critical value in the GDC. Rounding to $68.0$ first moves the answer.

(g)(i)

1 mark

A larger significance level widens the critical region, so it is easier to reject $H_{0}$ when it is false.

Increase the significance level (e.g. to 10%)

(g)(ii)

1 mark

The price of a wider critical region is that a true $H_{0}$ is also rejected more often.

The probability of a Type I error goes up

Mentor tip

For a fixed sample size, Type I and Type II errors pull against each other: lowering one raises the other.

Question 2

29 marks

Complex numbers as animation transformations

Number and algebra
Geometry and trigonometry
Calculus

(a)(i)

1 mark

Expand and use $i^{2} = - 1$ .

z i = (x + y i) i = x i + y i^{2} = - y + x i

(a)(ii)

2 marks

The point $(x, y)$ goes to $(- y, x)$ . Try $(1, 0) \to (0, 1)$ : a quarter turn anticlockwise.
Give the angle, the direction and the centre.

(x, y) \mapsto (- y, x) Rotation of 9 0^{\circ} (\frac{π}{2}) anticlockwise about the origin

Mentor tip

A rotation needs three things: angle, direction and centre. Leaving out the centre is the usual slip.

(b)

1 mark

Adding $a + bi$ adds $a$ to the $x$ -coordinate and $b$ to the $y$ -coordinate.

Translation by (a b)

(c)

1 mark

An enlargement from the origin scales both coordinates by $k$ , which is the same as multiplying the complex number by $k$ .

z \mapsto k z

(d)(i)

1 mark

$g$ adds $2 + i$ , so its inverse subtracts it.

g^{- 1} (z) = z - 2 - i

(d)(ii)

2 marks

Work from the inside out: apply $g^{- 1}$ first, then $f$ , then $g$ .

f (g^{- 1} (z)) = 2 i (z - 2 - i) = 2 i z - 4 i + 2 h (z) = g (2 i z + 2 - 4 i) = 2 i z + 2 - 4 i + 2 + i = 2 i z + 4 - 3 i

Mentor tip

In $g \circ f \circ g^{- 1}$ the right-hand function acts first. Watch the sign: $2 i \times (- i) = + 2$ .

(e)(i)

3 marks

Substitute each complex number into $h (z) = 2 i z + 4 - 3 i$ and read off the real and imaginary parts as coordinates.

h (1 + i) = 2 i - 2 + 4 - 3 i = 2 - i \Rightarrow A^{'} (2, - 1) h (2 + i) = 4 i - 2 + 4 - 3 i = 2 + i \Rightarrow B^{'} (2, 1) h (2 + 3 i) = 4 i - 6 + 4 - 3 i = - 2 + i \Rightarrow C^{'} (- 2, 1)

Mentor tip

Notice $B^{'} = B$ . A point that does not move is a strong hint about the centre in the next part.

(e)(ii)

3 marks

Compare sides: $A B$ has length $1$ and $A^{'} B^{'}$ has length $2$ , so the scale factor is $2$ .
$B A$ points left, while $B^{'} A^{'}$ points down: a $9 0^{\circ}$ anticlockwise turn.
$B$ stays fixed, so it is the centre of both.

∣ A^{'} B^{'} ∣ = 2∣ A B ∣, ∣ B^{'} C^{'} ∣ = 2∣ B C ∣ h (z) = z \Rightarrow z (1 - 2 i) = 4 - 3 i \Rightarrow z = 2 + i Rotation 9 0^{\circ} anticlockwise and enlargement, scale factor 2, both centred at (2, 1)

Another way. Write $h (z) - (2 + i) = 2 i (z - (2 + i))$ : multiplying by $2 i$ is a quarter turn with a doubling, about $2 + i$ .

Mentor tip

State the centre explicitly. Rotating and enlarging about the origin would give different points.

(f)

1 mark

Replace $30$ with $θ$ in the given rotation function.

R_{θ} (z, t) = z (cos (θ t^{\circ}) + i sin (θ t^{\circ}))

(g)

2 marks

A constant velocity means the displacement after $t$ seconds is the velocity times $t$ .
The vector $(2 - 1)$ is the complex number $2 - i$ ; add $t$ lots of it.

T (z, t) = z + (2 - i) t

Mentor tip

Check at $t = 0$ : nothing should have moved yet, and indeed $T (z, 0) = z$ .

(h)(i)

2 marks

With $k = 3$ the multiplier is $1 + 2 t$ . Every point is scaled by it, so the radius is too.

E_{3} (z, t) = (1 + 2 t) z t = 1 : r = (1 + 2) (1) = 3

(h)(ii)

1 mark

Use the same multiplier at $t = 2$ .

r = (1 + 4) (1) = 5

(i)

2 marks

The object is scaled by $p (t) = 1 + (k - 1) t$ , which is linear in $t$ .
Its rate of change is the constant $k - 1$ : growth if $k > 1$ , shrinking if $k < 1$ .

p (t) = 1 + (k - 1) t \frac{d p}{d t} = k - 1 (constant)

Mentor tip

Compare with (h): the radius went $1, 3, 5$ , rising by $k - 1 = 2$ each second.

(j)

5 marks

Shrinking: the size factor must hit $0$ at $t = 5$ , so solve $1 + 5 (k - 1) = 0$ , giving $k = 0.8$ and a factor $1 - 0.2 t$ .
Rotating: $36 0^{\circ}$ per second, so multiply by $cos (360 t^{\circ}) + i sin (360 t^{\circ})$ .
Moving: the centre goes from $0$ to $25 + 15 i$ in $5$ s, a velocity of $5 + 3 i$ per second.
Scale and rotate about the origin first, then translate, so the disc spins about its own moving centre.

1 + 5 (k - 1) = 0 \Rightarrow k - 1 = - 0.2 v = \frac{25 + 15 i}{5} = 5 + 3 i A (z, t) = (1 - 0.2 t) (cos (360 t^{\circ}) + i sin (360 t^{\circ})) z + (5 + 3 i) t

Mentor tip

Order matters. If you translate first and then rotate, the whole disc swings around the origin instead of spinning on the spot.

(k)

2 marks

At $60$ frames per second, frame $n$ is shown at $t = \frac{n}{60}$ .
Substitute into $A (z, t)$ and simplify each piece.

t = \frac{n}{60} B_{n} (z) = A (z, \frac{n}{60}) = (1 - \frac{n}{300}) (cos (6 n^{\circ}) + i sin (6 n^{\circ})) z + (\frac{1}{12} + \frac{i}{20}) n

Mentor tip

Check: frame $300$ is $t = 5$ , where the factor $1 - \frac{300}{300}$ is $0$ and the centre is at $25 + 15 i$ .

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 probability26 marks · 47%
Q1
Number and algebra21 marks · 38%
Q2
Geometry and trigonometry6 marks · 11%
Q2
Calculus2 marks · 4%
Q2

Mentor insights

Traps to watch for

01
Using s_n instead of s_{n-1} · Q1
The table gives $s_{n}$ . Convert to $s_{n - 1}$ before the $t$ -test, or the $p$ -value comes out slightly off.
02
Forgetting to merge rows · Q1
An expected count of $3.825$ is below $5$ , so the first two mark classes must be combined before running the $χ^{2}$ test.
03
Standard deviation in the critical region · Q1
The critical value uses the standard error $15.3/ 60$ , not $15.3$ , because the test is about a sample mean.
04
Missing the centre of the transformation · Q2
The rotation and enlargement in $h$ are about $(2, 1)$ , not the origin. Spot that $B$ does not move.
05
Wrong order in the animation function · Q2
Scale and rotate $z$ first, then add the translation $(5 + 3 i) t$ , so the disc spins about its own centre.

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