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IB Maths AA HL · Paper 3

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2TZ3

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 TZ2 paper, which your school or the IB can provide.

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Questions

Questions 1–2 · 55 marks

Question 1

26 marks

A sweets game: paths, recurrences and a gambler's ruin formula

Statistics and probability
Number and algebra

(a)(i)

1 mark

The only way to be back on one sweet at the start of round 3 is to win round 1 (up to two) and then lose round 2 (back to one). Multiply along that branch.

a = \frac{1}{4} \times \frac{3}{4} = \frac{3}{16}

Mentor tip

Losing round 1 ends the game, so there is no second route to count. Say so in a few words so the product is clearly complete.

(a)(ii)

2 marks

To reach three sweets at the start of round 5 you need three wins and one loss in rounds 1 to 4.
The game must not stop early, so you cannot hit $0$ or $3$ before round 5. The only valid order is win, lose, win, win: that is $a$ followed by two more wins.

b = \frac{3}{16} \times \frac{1}{4} \times \frac{1}{4} b = \frac{3}{256} \approx 0.0117

Mentor tip

Do not multiply by a binomial coefficient here. Orders such as win, win, win, lose finish the game at three sweets too early, so they do not count.

(b)

2 marks

From one sweet, the only way to be on one sweet two rounds later is to win and then lose (losing first ends the game).

\frac{1}{4} \times \frac{3}{4} = \frac{3}{16} = 0.1875

Mentor tip

This is the same branch as in (a)(i). The game has no memory, so what happened in rounds 1 to 4 does not matter.

(c)

1 mark

Put $n = 2$ into the recurrence with $p = \frac{1}{4}$ and $q = \frac{3}{4}$ .

R_{2} = \frac{3}{4} R_{1} + \frac{1}{4} R_{3}

(d)

2 marks

Substitute the given $R_{1} = 0.025$ and $R_{3} = 0.325$ into your answer to (c).

R_{2} = \frac{3}{4} (0.025) + \frac{1}{4} (0.325) R_{2} = 0.01875 + 0.08125 = 0.1

Another way. Use the recurrence at $n = 1$ with $R_{0} = 0$ : $0.025 = \frac{1}{4} R_{2}$ , which also gives $R_{2} = 0.1$ .

Mentor tip

A quick check: the three values should increase with $n$ , since more sweets means a better chance of winning. $0.025 < 0.1 < 0.325$ fits.

(e)

2 marks

Since $p + q = 1$ , replace the left-hand $R_{n}$ by $(p + q) R_{n}$ .
Collect the $p$ terms on one side and the $q$ terms on the other, then factorise and divide by $p$ .

p R_{n} + q R_{n} = q R_{n - 1} + p R_{n + 1} p R_{n + 1} - p R_{n} = q R_{n} - q R_{n - 1} p (R_{n + 1} - R_{n}) = q (R_{n} - R_{n - 1}) R_{n + 1} - R_{n} = \frac{q}{p} (R_{n} - R_{n - 1})

Mentor tip

The trick of writing $1$ as $p + q$ is worth remembering: it turns a three-term recurrence into a simple ratio between consecutive differences.

(f)(i)

1 mark

Use the result of (e) with $n = 1$ , and remember $R_{0} = 0$ .

R_{2} - R_{1} = \frac{q}{p} (R_{1} - R_{0}) = \frac{q}{p} R_{1}

(f)(ii)

2 marks

Use (e) again, this time with $n = 2$ .
Replace $R_{2} - R_{1}$ with your answer to (f)(i).

R_{3} - R_{2} = \frac{q}{p} (R_{2} - R_{1}) R_{3} - R_{2} = \frac{q}{p} \times \frac{q}{p} R_{1} = (\frac{q}{p})^{2} R_{1}

Mentor tip

Write both lines. On a show-that, the link back to (f)(i) is the whole point.

(f)(iii)

1 mark

Each step up in $n$ multiplies the difference by another $\frac{q}{p}$ , so the differences form a geometric pattern.

R_{n} - R_{n - 1} = (\frac{q}{p})^{n - 1} R_{1}

Mentor tip

Check the power with $n = 2$ : it should give $(\frac{q}{p})^{1} R_{1}$ , matching (f)(i).

(g)(i)

4 marks

Add the differences from $R_{2} - R_{1}$ up to $R_{n} - R_{n - 1}$ . Everything in the middle cancels, leaving $R_{n} - R_{1}$ .
Now add the same differences using (f)(iii): this gives $R_{1}$ times the powers of $\frac{q}{p}$ from the first up to the $(n - 1)$ th.
Set the two totals equal, move $R_{1}$ across and factorise.

(R_{2} - R_{1}) + (R_{3} - R_{2}) + \dots + (R_{n} - R_{n - 1}) = R_{n} - R_{1} (R_{2} - R_{1}) + \dots + (R_{n} - R_{n - 1}) = \frac{q}{p} R_{1} + (\frac{q}{p})^{2} R_{1} + \dots + (\frac{q}{p})^{n - 1} R_{1} R_{n} = R_{1} + \frac{q}{p} R_{1} + \dots + (\frac{q}{p})^{n - 1} R_{1} R_{n} = R_{1} (1 + \frac{q}{p} + (\frac{q}{p})^{2} + \dots + (\frac{q}{p})^{n - 1})

Mentor tip

Show the telescoping clearly by writing at least the first two and the last bracket. That makes the cancellation obvious to the reader.

(g)(ii)

4 marks

The bracket from (g)(i) is a geometric series with first term $1$ , ratio $\frac{q}{p}$ and $n$ terms. Sum it with the formula (valid since $p \neq = q$ ).
Substitute the given expression for $R_{1}$ . The factor $1 - \frac{q}{p}$ cancels.

R_{n} = R_{1} \cdot \frac{1 - ( \frac{q}{p} ) ^{n}}{1 - \frac{q}{p}} R_{n} = \frac{1 - \frac{q}{p}}{1 - ( \frac{q}{p} ) ^{w}} \cdot \frac{1 - ( \frac{q}{p} ) ^{n}}{1 - \frac{q}{p}} R_{n} = \frac{1 - ( \frac{q}{p} ) ^{n}}{1 - ( \frac{q}{p} ) ^{w}}

Mentor tip

Sanity check the result: $n = 0$ gives $0$ and $n = w$ gives $1$ , exactly as the game demands.

(h)

4 marks

She starts with $n = m$ sweets and the total in the game is $w = m + 10$ . The ratio is $\frac{q}{p} = \frac{0.505}{0.495}$ .
Use the formula from (g)(ii) and require it to be at least $0.8$ .
Solve with a GDC table or graph, then round up to the next whole number.

\frac{1 - ( \frac{0.505}{0.495} ) ^{m}}{1 - ( \frac{0.505}{0.495} ) ^{m + 10}} \geq 0.8 m \geq 108.418 \dots (GDC) m = 108 : R \approx 0.79983 < 0.8, m = 109 : R \approx 0.80024 \geq 0.8 m = 109

Mentor tip

Since $m$ counts sweets it must be a whole number, so round up, not to the nearest. Showing the table values either side of $0.8$ makes the answer convincing.

Question 2

29 marks

Pythagorean boxes generated by complex numbers

Geometry and trigonometry
Number and algebra

(a)

2 marks

The interior diagonal of a cuboid comes from Pythagoras in 3D: square the three dimensions, add and square root.

t = 9^{2} + 2^{2} + 6^{2} = 81 + 4 + 36 t = 121 = 11

(b)(i)

2 marks

The conjugate of $z_{2}$ is $c - d i$ . Expand the product and use $i^{2} = - 1$ .
Group the real and imaginary parts.

z_{1} z_{2}^{*} = (a + bi) (c - d i) = a c - a d i + b c i - b d i^{2} = (a c + b d) + (b c - a d) i

Mentor tip

Watch the sign on the $b d$ term: $- b d i^{2}$ becomes $+ b d$ .

(b)(ii)

1 mark

$v$ is twice the imaginary part of the product from (b)(i).

v = 2 (b c - a d)

Mentor tip

The imaginary part is the real number $b c - a d$ , without the $i$ .

(c)(i)

1 mark

Substitute $a = 3$ , $b = 7$ and $d = 6$ into your expression for $v$ .

v = 2 (7 c - 3 \times 6) = 2 (7 c - 18) v = 14 c - 36

(c)(ii)

4 marks

All four quantities must be positive. Use $v > 0$ to get a lower bound on $c$ .
Write $w = ∣ z_{1} ∣^{2} - ∣ z_{2} ∣^{2}$ in terms of $c$ and use $w > 0$ for an upper bound.
$c$ is a positive integer strictly between the two bounds, which leaves only $3$ and $4$ ( $u$ and $t$ are clearly positive for any $c > 0$ ).

14 c - 36 > 0 ⟹ c > \frac{18}{7} \approx 2.57 w = (3^{2} + 7^{2}) - (c^{2} + 6^{2}) = 22 - c^{2} 22 - c^{2} > 0 ⟹ c < 22 \approx 4.69 2.57 < c < 4.69, c \in Z^{+} ⟹ c = 3 or 4

Mentor tip

The second inequality is the one people miss. Using only $v > 0$ gives no upper limit, so think about which other dimension can turn negative.

(c)(iii)

5 marks

Work out $z_{1} z_{2}^{*}$ with $z_{2}^{*} = 4 - 6 i$ , then double its real and imaginary parts for $u$ and $v$ .
Find $∣ z_{1} ∣^{2}$ and $∣ z_{2} ∣^{2}$ , then subtract for $w$ and add for $t$ .

z_{1} z_{2}^{*} = (3 + 7 i) (4 - 6 i) = 12 - 18 i + 28 i + 42 = 54 + 10 i u = 2 \times 54 = 108, v = 2 \times 10 = 20 ∣ z_{1} ∣^{2} = 9 + 49 = 58, ∣ z_{2} ∣^{2} = 16 + 36 = 52 w = 58 - 52 = 6, t = 58 + 52 = 110

Mentor tip

Check with Pythagoras: $10 8^{2} + 2 0^{2} + 6^{2} = 12100 = 11 0^{2}$ . You can also check $v$ against (c)(i): $14 (4) - 36 = 20$ .

(d)

6 marks

Since $t = ∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2} = a^{2} + b^{2} + c^{2} + d^{2}$ , the values $a, b, c, d$ must be $1, 2, 3, 3$ in some order.
For $w > 0$ the larger modulus must belong to $z_{1}$ . Try $z_{1} = 3 + 3 i$ (so $∣ z_{1} ∣^{2} = 18$ ) and $z_{2}$ built from $1$ and $2$ (so $∣ z_{2} ∣^{2} = 5$ ).
For $v = 2 (b c - a d) > 0$ take $c = 2$ , $d = 1$ , so $z_{2} = 2 + i$ . Then compute $u$ , $v$ and $w$ .

a = 3, b = 3, c = 2, d = 1 u = 2 (a c + b d) = 2 (6 + 3) = 18 v = 2 (b c - a d) = 2 (6 - 3) = 6 w = 18 - 5 = 13, t = 18 + 5 = 23 18 \times 6 \times 13 box

Another way. Other valid orders give the other two boxes: $z_{1} = 2 + 3 i$ , $z_{2} = 3 + i$ gives $18 \times 14 \times 3$ , and $z_{1} = 3 + 2 i$ , $z_{2} = 3 + i$ gives $22 \times 6 \times 3$ .

Mentor tip

Always check the answer: $1 8^{2} + 6^{2} + 1 3^{2} = 324 + 36 + 169 = 529 = 2 3^{2}$ . If $v$ or $w$ comes out negative, swap values rather than starting again.

(e)

1 mark

Substitute $u = 26$ , $v = 38$ , $t = 63$ and $w = 43$ , then simplify.

z = \frac{26 + 38 i}{63 - 43} = \frac{26 + 38 i}{20} = \frac{13 + 19 i}{10}

(f)(i)

2 marks

Subtract the definitions of $w$ and $t$ : the $∣ z_{1} ∣^{2}$ terms cancel, leaving twice $∣ z_{2} ∣^{2}$ .
Use $t - w = 63 - 43 = 20$ and $∣ z_{2} ∣^{2} = c^{2} + d^{2}$ .

t - w = (∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2}) - (∣ z_{1} ∣^{2} - ∣ z_{2} ∣^{2}) = 2∣ z_{2} ∣^{2} 2 (c^{2} + d^{2}) = 63 - 43 = 20 c^{2} + d^{2} = 10

Mentor tip

Do not just read $∣ z_{2} ∣^{2} = 10$ off the denominator in (e). The link comes from $t - w$ , so show it.

(f)(ii)

5 marks

The only positive integers with $c^{2} + d^{2} = 10$ and $c > d$ are $c = 3$ , $d = 1$ , so $z_{2} = 3 + i$ .
Multiply the given form by $∣ z_{2} ∣^{2} = 10$ : $z_{1} z_{2}^{*} = 10 z = 13 + 19 i$ .
With $z_{2}^{*} = 3 - i$ , expand $(a + bi) (3 - i)$ and compare real and imaginary parts to get two equations in $a$ and $b$ .
Solve them simultaneously.

z_{2} = 3 + i, z_{1} z_{2}^{*} = 10 \times \frac{13 + 19 i}{10} = 13 + 19 i (a + bi) (3 - i) = (3 a + b) + (3 b - a) i 3 a + b = 13, 3 b - a = 19 a = 2, b = 7 z_{1} = 2 + 7 i, z_{2} = 3 + i

Another way. Divide directly: $z_{1} = \frac{13 + 19 i}{3 - i} = \frac{( 13 + 19 i ) ( 3 + i )}{10} = \frac{20 + 70 i}{10} = 2 + 7 i$ .

Mentor tip

Check: $∣ z_{1} ∣^{2} = 53$ , so $t = 53 + 10 = 63$ and $w = 53 - 10 = 43$ , both matching the box.

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 algebra44 marks · 80%
Q1, Q2
Statistics and probability9 marks · 16%
Q1
Geometry and trigonometry2 marks · 4%
Q2

Mentor insights

Traps to watch for

01
Counting paths that end the game early · Q1
In (a)(ii), only one order of wins and losses works, because reaching three sweets or none stops the game. Multiplying by a binomial coefficient overcounts.
02
Forgetting that R0 is zero · Q1
The first difference $R_{1} - R_{0}$ is just $R_{1}$ . Leaving $R_{0}$ in makes (f) and (g) look messier than they are.
03
Rounding m to the nearest integer · Q1
In (h) the inequality gives $m \geq 108.4$ , so the smallest whole number is 109, not 108. Check the table either side of $0.8$ .
04
Missing the upper bound on c · Q2
In (c)(ii), $v > 0$ only gives a lower bound. The upper bound comes from needing $w > 0$ .
05
Negative v or w in the 23 box · Q2
In (d), a careless order of $1, 2, 3, 3$ gives a negative dimension. Put the larger modulus in $z_{1}$ and swap $c$ and $d$ if $v$ is negative.

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 AA HL · Paper 3

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2TZ3

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 TZ2 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

A sweets game: paths, recurrences and a gambler's ruin formula

Statistics and probability
Number and algebra

(a)(i)

1 mark

The only way to be back on one sweet at the start of round 3 is to win round 1 (up to two) and then lose round 2 (back to one). Multiply along that branch.

a = \frac{1}{4} \times \frac{3}{4} = \frac{3}{16}

Mentor tip

Losing round 1 ends the game, so there is no second route to count. Say so in a few words so the product is clearly complete.

(a)(ii)

2 marks

To reach three sweets at the start of round 5 you need three wins and one loss in rounds 1 to 4.
The game must not stop early, so you cannot hit $0$ or $3$ before round 5. The only valid order is win, lose, win, win: that is $a$ followed by two more wins.

b = \frac{3}{16} \times \frac{1}{4} \times \frac{1}{4} b = \frac{3}{256} \approx 0.0117

Mentor tip

Do not multiply by a binomial coefficient here. Orders such as win, win, win, lose finish the game at three sweets too early, so they do not count.

(b)

2 marks

From one sweet, the only way to be on one sweet two rounds later is to win and then lose (losing first ends the game).

\frac{1}{4} \times \frac{3}{4} = \frac{3}{16} = 0.1875

Mentor tip

This is the same branch as in (a)(i). The game has no memory, so what happened in rounds 1 to 4 does not matter.

(c)

1 mark

Put $n = 2$ into the recurrence with $p = \frac{1}{4}$ and $q = \frac{3}{4}$ .

R_{2} = \frac{3}{4} R_{1} + \frac{1}{4} R_{3}

(d)

2 marks

Substitute the given $R_{1} = 0.025$ and $R_{3} = 0.325$ into your answer to (c).

R_{2} = \frac{3}{4} (0.025) + \frac{1}{4} (0.325) R_{2} = 0.01875 + 0.08125 = 0.1

Another way. Use the recurrence at $n = 1$ with $R_{0} = 0$ : $0.025 = \frac{1}{4} R_{2}$ , which also gives $R_{2} = 0.1$ .

Mentor tip

A quick check: the three values should increase with $n$ , since more sweets means a better chance of winning. $0.025 < 0.1 < 0.325$ fits.

(e)

2 marks

Since $p + q = 1$ , replace the left-hand $R_{n}$ by $(p + q) R_{n}$ .
Collect the $p$ terms on one side and the $q$ terms on the other, then factorise and divide by $p$ .

p R_{n} + q R_{n} = q R_{n - 1} + p R_{n + 1} p R_{n + 1} - p R_{n} = q R_{n} - q R_{n - 1} p (R_{n + 1} - R_{n}) = q (R_{n} - R_{n - 1}) R_{n + 1} - R_{n} = \frac{q}{p} (R_{n} - R_{n - 1})

Mentor tip

The trick of writing $1$ as $p + q$ is worth remembering: it turns a three-term recurrence into a simple ratio between consecutive differences.

(f)(i)

1 mark

Use the result of (e) with $n = 1$ , and remember $R_{0} = 0$ .

R_{2} - R_{1} = \frac{q}{p} (R_{1} - R_{0}) = \frac{q}{p} R_{1}

(f)(ii)

2 marks

Use (e) again, this time with $n = 2$ .
Replace $R_{2} - R_{1}$ with your answer to (f)(i).

R_{3} - R_{2} = \frac{q}{p} (R_{2} - R_{1}) R_{3} - R_{2} = \frac{q}{p} \times \frac{q}{p} R_{1} = (\frac{q}{p})^{2} R_{1}

Mentor tip

Write both lines. On a show-that, the link back to (f)(i) is the whole point.

(f)(iii)

1 mark

Each step up in $n$ multiplies the difference by another $\frac{q}{p}$ , so the differences form a geometric pattern.

R_{n} - R_{n - 1} = (\frac{q}{p})^{n - 1} R_{1}

Mentor tip

Check the power with $n = 2$ : it should give $(\frac{q}{p})^{1} R_{1}$ , matching (f)(i).

(g)(i)

4 marks

Add the differences from $R_{2} - R_{1}$ up to $R_{n} - R_{n - 1}$ . Everything in the middle cancels, leaving $R_{n} - R_{1}$ .
Now add the same differences using (f)(iii): this gives $R_{1}$ times the powers of $\frac{q}{p}$ from the first up to the $(n - 1)$ th.
Set the two totals equal, move $R_{1}$ across and factorise.

(R_{2} - R_{1}) + (R_{3} - R_{2}) + \dots + (R_{n} - R_{n - 1}) = R_{n} - R_{1} (R_{2} - R_{1}) + \dots + (R_{n} - R_{n - 1}) = \frac{q}{p} R_{1} + (\frac{q}{p})^{2} R_{1} + \dots + (\frac{q}{p})^{n - 1} R_{1} R_{n} = R_{1} + \frac{q}{p} R_{1} + \dots + (\frac{q}{p})^{n - 1} R_{1} R_{n} = R_{1} (1 + \frac{q}{p} + (\frac{q}{p})^{2} + \dots + (\frac{q}{p})^{n - 1})

Mentor tip

Show the telescoping clearly by writing at least the first two and the last bracket. That makes the cancellation obvious to the reader.

(g)(ii)

4 marks

The bracket from (g)(i) is a geometric series with first term $1$ , ratio $\frac{q}{p}$ and $n$ terms. Sum it with the formula (valid since $p \neq = q$ ).
Substitute the given expression for $R_{1}$ . The factor $1 - \frac{q}{p}$ cancels.

R_{n} = R_{1} \cdot \frac{1 - ( \frac{q}{p} ) ^{n}}{1 - \frac{q}{p}} R_{n} = \frac{1 - \frac{q}{p}}{1 - ( \frac{q}{p} ) ^{w}} \cdot \frac{1 - ( \frac{q}{p} ) ^{n}}{1 - \frac{q}{p}} R_{n} = \frac{1 - ( \frac{q}{p} ) ^{n}}{1 - ( \frac{q}{p} ) ^{w}}

Mentor tip

Sanity check the result: $n = 0$ gives $0$ and $n = w$ gives $1$ , exactly as the game demands.

(h)

4 marks

She starts with $n = m$ sweets and the total in the game is $w = m + 10$ . The ratio is $\frac{q}{p} = \frac{0.505}{0.495}$ .
Use the formula from (g)(ii) and require it to be at least $0.8$ .
Solve with a GDC table or graph, then round up to the next whole number.

\frac{1 - ( \frac{0.505}{0.495} ) ^{m}}{1 - ( \frac{0.505}{0.495} ) ^{m + 10}} \geq 0.8 m \geq 108.418 \dots (GDC) m = 108 : R \approx 0.79983 < 0.8, m = 109 : R \approx 0.80024 \geq 0.8 m = 109

Mentor tip

Since $m$ counts sweets it must be a whole number, so round up, not to the nearest. Showing the table values either side of $0.8$ makes the answer convincing.

Question 2

29 marks

Pythagorean boxes generated by complex numbers

Geometry and trigonometry
Number and algebra

(a)

2 marks

The interior diagonal of a cuboid comes from Pythagoras in 3D: square the three dimensions, add and square root.

t = 9^{2} + 2^{2} + 6^{2} = 81 + 4 + 36 t = 121 = 11

(b)(i)

2 marks

The conjugate of $z_{2}$ is $c - d i$ . Expand the product and use $i^{2} = - 1$ .
Group the real and imaginary parts.

z_{1} z_{2}^{*} = (a + bi) (c - d i) = a c - a d i + b c i - b d i^{2} = (a c + b d) + (b c - a d) i

Mentor tip

Watch the sign on the $b d$ term: $- b d i^{2}$ becomes $+ b d$ .

(b)(ii)

1 mark

$v$ is twice the imaginary part of the product from (b)(i).

v = 2 (b c - a d)

Mentor tip

The imaginary part is the real number $b c - a d$ , without the $i$ .

(c)(i)

1 mark

Substitute $a = 3$ , $b = 7$ and $d = 6$ into your expression for $v$ .

v = 2 (7 c - 3 \times 6) = 2 (7 c - 18) v = 14 c - 36

(c)(ii)

4 marks

All four quantities must be positive. Use $v > 0$ to get a lower bound on $c$ .
Write $w = ∣ z_{1} ∣^{2} - ∣ z_{2} ∣^{2}$ in terms of $c$ and use $w > 0$ for an upper bound.
$c$ is a positive integer strictly between the two bounds, which leaves only $3$ and $4$ ( $u$ and $t$ are clearly positive for any $c > 0$ ).

14 c - 36 > 0 ⟹ c > \frac{18}{7} \approx 2.57 w = (3^{2} + 7^{2}) - (c^{2} + 6^{2}) = 22 - c^{2} 22 - c^{2} > 0 ⟹ c < 22 \approx 4.69 2.57 < c < 4.69, c \in Z^{+} ⟹ c = 3 or 4

Mentor tip

The second inequality is the one people miss. Using only $v > 0$ gives no upper limit, so think about which other dimension can turn negative.

(c)(iii)

5 marks

Work out $z_{1} z_{2}^{*}$ with $z_{2}^{*} = 4 - 6 i$ , then double its real and imaginary parts for $u$ and $v$ .
Find $∣ z_{1} ∣^{2}$ and $∣ z_{2} ∣^{2}$ , then subtract for $w$ and add for $t$ .

z_{1} z_{2}^{*} = (3 + 7 i) (4 - 6 i) = 12 - 18 i + 28 i + 42 = 54 + 10 i u = 2 \times 54 = 108, v = 2 \times 10 = 20 ∣ z_{1} ∣^{2} = 9 + 49 = 58, ∣ z_{2} ∣^{2} = 16 + 36 = 52 w = 58 - 52 = 6, t = 58 + 52 = 110

Mentor tip

Check with Pythagoras: $10 8^{2} + 2 0^{2} + 6^{2} = 12100 = 11 0^{2}$ . You can also check $v$ against (c)(i): $14 (4) - 36 = 20$ .

(d)

6 marks

Since $t = ∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2} = a^{2} + b^{2} + c^{2} + d^{2}$ , the values $a, b, c, d$ must be $1, 2, 3, 3$ in some order.
For $w > 0$ the larger modulus must belong to $z_{1}$ . Try $z_{1} = 3 + 3 i$ (so $∣ z_{1} ∣^{2} = 18$ ) and $z_{2}$ built from $1$ and $2$ (so $∣ z_{2} ∣^{2} = 5$ ).
For $v = 2 (b c - a d) > 0$ take $c = 2$ , $d = 1$ , so $z_{2} = 2 + i$ . Then compute $u$ , $v$ and $w$ .

a = 3, b = 3, c = 2, d = 1 u = 2 (a c + b d) = 2 (6 + 3) = 18 v = 2 (b c - a d) = 2 (6 - 3) = 6 w = 18 - 5 = 13, t = 18 + 5 = 23 18 \times 6 \times 13 box

Another way. Other valid orders give the other two boxes: $z_{1} = 2 + 3 i$ , $z_{2} = 3 + i$ gives $18 \times 14 \times 3$ , and $z_{1} = 3 + 2 i$ , $z_{2} = 3 + i$ gives $22 \times 6 \times 3$ .

Mentor tip

Always check the answer: $1 8^{2} + 6^{2} + 1 3^{2} = 324 + 36 + 169 = 529 = 2 3^{2}$ . If $v$ or $w$ comes out negative, swap values rather than starting again.

(e)

1 mark

Substitute $u = 26$ , $v = 38$ , $t = 63$ and $w = 43$ , then simplify.

z = \frac{26 + 38 i}{63 - 43} = \frac{26 + 38 i}{20} = \frac{13 + 19 i}{10}

(f)(i)

2 marks

Subtract the definitions of $w$ and $t$ : the $∣ z_{1} ∣^{2}$ terms cancel, leaving twice $∣ z_{2} ∣^{2}$ .
Use $t - w = 63 - 43 = 20$ and $∣ z_{2} ∣^{2} = c^{2} + d^{2}$ .

t - w = (∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2}) - (∣ z_{1} ∣^{2} - ∣ z_{2} ∣^{2}) = 2∣ z_{2} ∣^{2} 2 (c^{2} + d^{2}) = 63 - 43 = 20 c^{2} + d^{2} = 10

Mentor tip

Do not just read $∣ z_{2} ∣^{2} = 10$ off the denominator in (e). The link comes from $t - w$ , so show it.

(f)(ii)

5 marks

The only positive integers with $c^{2} + d^{2} = 10$ and $c > d$ are $c = 3$ , $d = 1$ , so $z_{2} = 3 + i$ .
Multiply the given form by $∣ z_{2} ∣^{2} = 10$ : $z_{1} z_{2}^{*} = 10 z = 13 + 19 i$ .
With $z_{2}^{*} = 3 - i$ , expand $(a + bi) (3 - i)$ and compare real and imaginary parts to get two equations in $a$ and $b$ .
Solve them simultaneously.

z_{2} = 3 + i, z_{1} z_{2}^{*} = 10 \times \frac{13 + 19 i}{10} = 13 + 19 i (a + bi) (3 - i) = (3 a + b) + (3 b - a) i 3 a + b = 13, 3 b - a = 19 a = 2, b = 7 z_{1} = 2 + 7 i, z_{2} = 3 + i

Another way. Divide directly: $z_{1} = \frac{13 + 19 i}{3 - i} = \frac{( 13 + 19 i ) ( 3 + i )}{10} = \frac{20 + 70 i}{10} = 2 + 7 i$ .

Mentor tip

Check: $∣ z_{1} ∣^{2} = 53$ , so $t = 53 + 10 = 63$ and $w = 53 - 10 = 43$ , both matching the box.

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 algebra44 marks · 80%
Q1, Q2
Statistics and probability9 marks · 16%
Q1
Geometry and trigonometry2 marks · 4%
Q2

Mentor insights

Traps to watch for

01
Counting paths that end the game early · Q1
In (a)(ii), only one order of wins and losses works, because reaching three sweets or none stops the game. Multiplying by a binomial coefficient overcounts.
02
Forgetting that R0 is zero · Q1
The first difference $R_{1} - R_{0}$ is just $R_{1}$ . Leaving $R_{0}$ in makes (f) and (g) look messier than they are.
03
Rounding m to the nearest integer · Q1
In (h) the inequality gives $m \geq 108.4$ , so the smallest whole number is 109, not 108. Check the table either side of $0.8$ .
04
Missing the upper bound on c · Q2
In (c)(ii), $v > 0$ only gives a lower bound. The upper bound comes from needing $w > 0$ .
05
Negative v or w in the 23 box · Q2
In (d), a careless order of $1, 2, 3, 3$ gives a negative dimension. Put the larger modulus in $z_{1}$ and swap $c$ and $d$ if $v$ is negative.

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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