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Worked solutions

IB Maths AI HL · Paper 2

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1 TZ2TZ3

Session: May 2026
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 8

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2026 TZ3 paper, which your school or the IB can provide.

Practice mode

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Questions

Questions 1–8 · 110 marks

Question 1

14 marks

Compound interest, savings and an arithmetic sequence

Number and algebra

(a)

3 marks

Monthly compounding for three years means $36$ periods at $\frac{2.3}{12} %$ each.
Use the finance solver or the compound interest formula, and give the answer to 2 d.p. as the paper asks.

N = 36, I % = 2.3, P V = - 14500, P M T = 0, P / Y = C / Y = 12 F V = 14500 (1 + \frac{2.3}{100 \times 12})^{36} F V = 15 534.80 GBP (GDC)

Mentor tip

Money answers to 2 d.p. here: $15 534.8$ on its own is not in the form asked for.

(b)

3 marks

The balance from (a) becomes the new present value.
Add the monthly payment of $250$ for $24$ months, with the same rate and monthly compounding.

N = 24, I % = 2.3, P V = - 15 534.80, P M T = - 250, P / Y = C / Y = 12 F V = 22 399.50 GBP (GDC)

Mentor tip

Give $P V$ and $P M T$ the same sign: both are money going into the account. Mixed signs make the solver treat the payments as withdrawals.

(c)

5 marks

(i) The costs drop by $1876$ each year, so $d = - 1876$ . Write $u_{n} = u_{1} + (n - 1) d$ and tidy it into the form $a + bn$ .

(ii) Substitute $n = 7$ .

(iii) Give one real weakness of a straight-line model for car prices.

u_{n} = 28492 + (n - 1) (- 1876) (i) u_{n} = 30368 - 1876 n (ii) u_{7} = 30368 - 1876 \times 7 = 17236 GBP (iii) e.g. u_{n} < 0 once n \geq 17, but a car never has a negative price

Mentor tip

For (iii), say why the formula fails, not just that cars differ. Other good points: it only works for whole-number ages, and real prices usually fall faster at first rather than linearly.

(d)

3 marks

Younger cars cost more, so look for the smallest age whose price fits within $22 399.50$ .
Solve $u_{n} \leq 22 399.50$ and round $n$ up to the next whole number.
Subtract that car's price from the balance.

30368 - 1876 n \leq 22 399.50 \Rightarrow n \geq 4.25 u_{4} = 22864 (too dear), u_{5} = 20988 22 399.50 - 20988 = 1411.50 GBP

Mentor tip

Rounding $4.25$ down to $4$ gives a car she cannot afford. Always check the two neighbouring whole numbers.

Question 2

12 marks

Sine model for body temperature

Functions

(a)

2 marks

The period of $sin (b t)$ is $\frac{36 0 ^{\circ}}{b}$ (degrees) or $\frac{2 π}{b}$ (radians). Set it equal to $24$ .

b = \frac{360}{24} = 15 (degrees) or b = \frac{2 π}{24} = \frac{π}{12} \approx 0.262 (radians)

Mentor tip

Pick degrees or radians now and keep your GDC in that mode for the rest of the question.

(b)

6 marks

(i) At 18:00, $t = 18$ and $15 \times 18 = 27 0^{\circ}$ , where $sin 27 0^{\circ} = - 1$ .

(ii) At 13:00, $t = 13$ and $15 \times 13 = 19 5^{\circ}$ .

(iii) Solve the two linear equations together, by subtracting or with the GDC's simultaneous equation solver.

(i) 36.9 = a sin 27 0^{\circ} + c = - a + c (ii) 36.6 = a sin 19 5^{\circ} + c = - 0.258819 \dots a + c (iii) 0.3 = - 0.741180 \dots a a = - 0.405 (- 0.404759 \dots) c = 36.9 + a = 36.5 (36.4952 \dots)

Mentor tip

A negative $a$ is fine: it just flips the sine curve, which matches the graph dipping first and peaking late in the day.

(c)

2 marks

Put $t = 8$ into the model and compare the result with $36.2$ .

T (8) = - 0.404759 \dots sin 12 0^{\circ} + 36.4952 \dots = 36.1 (36.1447 \dots) 36.2 > 36.1 \Rightarrow the point is above the curve

Mentor tip

Use the unrounded $a$ and $c$ here. Say clearly which value is bigger before you give the conclusion.

(d)

2 marks

With $a < 0$ , the curve falls wherever $sin (15 t)$ is rising.
$sin (15 t)$ rises from $t = 0$ to its peak at $t = 6$ , and again from its trough at $t = 18$ up to $t = 24$ .

0 < t < 6 and 18 < t < 24 i.e. 00:00 to 06:00 and 18:00 to 24:00

Mentor tip

There are two intervals. The model is decreasing on both sides of midnight, so do not stop at the first one.

Question 3

17 marks

Wind turbine power, bounds and a quartic model

Number and algebra
Calculus
Functions

(a)

3 marks

Substitute $r = 40$ and $v = 10$ into the formula.
Write the result in standard form to 3 s.f.

P = 0.245 π (40)^{2} (10)^{3} = 1231 504.32 \dots \approx 1.23 \times 1 0^{6}

Mentor tip

Standard form means $1.23 \times 1 0^{6}$ , not $1.23 E 6$ copied off the calculator screen.

(b)

5 marks

(i) Category 6 starts at $10.8$ to 1 d.p., so the smallest possible true speed is the lower bound $10.75$ .

(ii) The top of the category is $13.8$ , so the largest true speed is the upper bound $13.85$ .

Work out the power at both bounds and subtract.

(i) P_{m i n} = 0.245 π (40)^{2} (10.75)^{3} = 1529 893.96 \dots \approx 1.53 \times 1 0^{6} (ii) P_{m a x} = 0.245 π (40)^{2} (13.85)^{3} = 3271 788.78 \dots P_{m a x} - P_{m i n} = 1741 894.8 \dots \approx 1.74 \times 1 0^{6}

Mentor tip

Using $10.8$ and $13.8$ as they stand ignores the rounding in the table. The bounds sit half a unit of the last digit either side.

(c)

6 marks

(i) Differentiate term by term with the power rule; the constant disappears.

(ii) Solve $\frac{d P}{d v} = 0$ on the GDC and keep only the root inside $3.35 \leq v < 17.15$ .

(iii) Check the sign or the graph of $P$ : it changes from increasing to decreasing there.

(i) \frac{d P}{d v} = 17.92 v^{3} - 708 v^{2} + 7140 v - 8740 (ii) roots 1.42, 14.8, 23.3 (GDC) \Rightarrow v = 14.8 (14.7728 \dots) m s^{- 1} (iii) the wind speed at which this turbine gives its greatest power

Mentor tip

The other two roots lie outside the operating range, so they have no meaning here. Say so rather than listing all three.

(d)

3 marks

Graph $P$ and the line $P = 90000$ on the operating domain and find where they cross.
Power is at least $90000$ from that crossing up to the end of the domain, since $P (17.15) \approx 96500$ is still above $90000$ .

4.48 v^{4} - 236 v^{3} + 3570 v^{2} - 8740 v - 735 = 90000 v = 11.4550 \dots (GDC; the other roots lie outside the domain) 11.46 \leq v < 17.15

Mentor tip

The upper end comes from the domain, not from a second crossing: the crossing at $18.5$ is outside the range the turbine can work in.

Question 4

13 marks

Normal distributions and a difference of variables

Statistics and probability

(a)

4 marks

(i) Use normal cdf with mean $14.2$ and standard deviation $2.1$ , upper limit $14$ .

(ii) Three separate races, each with the same probability, so cube it.

(i) P (A < 14) = 0.462 (0.462062 \dots) (GDC) (ii) (0.462062 \dots)^{3} = 0.0987 (0.0986513 \dots)

Another way. In (ii) you could use the binomial distribution $B (3, 0.462 \dots)$ and find $P (X = 3)$ .

Mentor tip

Cube the full stored value. Cubing $0.462$ gives $0.0986$ , which is slightly off.

(b)

1 mark

Multiplying the probabilities needs the three race times to be independent of each other.

Anne’s time in one race does not affect her time in any other race

Mentor tip

Name the assumption in context: it is about the races, not just the word independent on its own.

(c)

2 marks

Sofia wins if Anne takes longer than $13.2$ minutes.

P (A > 13.2) = 0.683 (0.683030 \dots) (GDC)

Mentor tip

In a race, the smaller time wins. Finding $P (A < 13.2)$ answers the opposite question.

(d)

1 mark

$D > 0$ means $A > S$ : Anne's time is longer.

D > 0 ⟺ Sofia finishes the race faster than Anne (Sofia wins)

(e)

3 marks

(i) Means subtract.

(ii) For independent variables the variances add, even for a difference. Square the standard deviations first.

(i) E (D) = 14.2 - 13.4 = 0.8 minutes (ii) Var (D) = 2. 1^{2} + 1. 7^{2} = 4.41 + 2.89 = 7.3 minutes^{2}

Mentor tip

Subtracting variances ( $4.41 - 2.89$ ) is the classic slip. A difference is more spread out than either time alone.

(f)

2 marks

Anne wins when her time is shorter, so $D < 0$ .
$D$ is normal with mean $0.8$ and standard deviation $7.3$ .

D \sim N (0.8, 7.3) P (D < 0) = 0.384 (0.383579 \dots) (GDC)

Mentor tip

The GDC wants the standard deviation, $7.3 \approx 2.70$ , not the variance $7.3$ .

Question 5

12 marks

Route inspection on a road network

Geometry and trigonometry

(a)

3 marks

Count degrees: only $A$ and $B$ are odd (both degree $3$ ).
The shortest path from $A$ to $B$ is the direct edge of $8$ , so walk it twice. The second walk costs half, $4$ .
(ii) Any closed walk from $A$ that uses every edge once, plus $A B$ a second time, will do.

(i) 122 + \frac{1}{2} (8) = 126 minutes (ii) e.g. A B D E F G E H C B A H I A

Mentor tip

Remember the half-time rule. Adding the full $8$ gives $130$ , which ignores the faster second walk.

(b)

6 marks

The new road makes $C$ and $D$ odd as well, so the odd vertices are $A, B, C, D$ .
List all three pairings, using shortest paths where there is no direct road: $A C$ via $B$ is $15$ , $A D$ via $B$ is $26$ .
The pairing with the smallest total is the one to repeat.
(ii) Add the new road's $14$ to $122$ , plus half of the repeated total.

A B + C D = 8 + 14 = 22 A C + B D = 15 + 18 = 33 A D + B C = 26 + 7 = 33 (i) repeat A B and C D (ii) 122 + 14 + \frac{1}{2} (22) = 147 minutes

Mentor tip

Showing all three pairings is the justification. Picking the smallest-looking one without the others is not enough.

(c)

3 marks

With free start and finish, two of the four odd vertices can be the ends, so only one pair needs repeating.
Repeat the cheapest single connection, $B C = 7$ , and start and finish at the other two, $A$ and $D$ .
Compare with the $11$ minutes of repeats in (b).

(i) start at A and finish at D (or the other way round) repeat cost = \frac{1}{2} (7) = 3.5 instead of \frac{1}{2} (22) = 11 (ii) time saved = 11 - 3.5 = 7.5 minutes

Mentor tip

The ends are the pair you do not repeat. Students often choose $A$ and $B$ because $A B$ was repeated before, but that is the wrong way round.

Question 6

13 marks

Velocity vectors, bearings and journey times

Geometry and trigonometry

(a)

5 marks

(i) Speed is the magnitude of the velocity, so set $a^{2} + (- 10)^{2} = 26$ .

(ii) The boat moves $24$ east and $10$ south each hour. Find the angle below east, then add $9 0^{\circ}$ to measure clockwise from north.

(i) a^{2} + 1 0^{2} = 2 6^{2} \Rightarrow a = 24 (ii) arctan (\frac{10}{24}) = 22.6198 \dots^{\circ} bearing = 9 0^{\circ} + 22.6198 \dots^{\circ} = 11 3^{\circ} (112.619 \dots^{\circ})

Mentor tip

Bearings are in degrees, clockwise from north, and usually three figures. A quick sketch of east and south stops you giving $6 7^{\circ}$ .

(b)

2 marks

From 08:00 to 11:00 is $3$ hours. Start at $H$ and add three lots of the velocity.

r = (520) + 3 (24 - 10) (77, - 10)

(c)

3 marks

Due east for two hours at $26$ adds $52$ to the $x$ -coordinate only.
Use the distance formula between $H (5, 20)$ and $P$ .

P = (77 + 52, - 10) = (129, - 10) H P = (129 - 5)^{2} + (- 10 - 20)^{2} = 16276 \approx 128 km (127.577 \dots)

Another way. Use the cosine rule in triangle $H$ , 11:00 position, $P$ with sides $78$ and $52$ and the included angle $18 0^{\circ} - 22. 6^{\circ}$ .

(d)

3 marks

Boat $B$ reaches $P$ at 13:00.
Time for the second boat is distance over speed; subtract it from 13:00 and convert the decimal hours into minutes.

time = \frac{127.577 \dots}{40} = 3.18943 \dots h 13 - 3.18943 \dots = 9.81056 \dots h 0.81056 \dots \times 60 = 48.6 \dots min \Rightarrow 09 : 49

Mentor tip

Keep the unrounded distance: using $128$ shifts the answer by about a minute. And $0.81$ hours is not $81$ minutes.

Question 7

12 marks

Matrix powers, determinants and a geometric series

Number and algebra

(a)

3 marks

(i) Multiply $A$ by itself, row by column.

(ii) Multiply your $A^{2}$ by $A$ .

(i) A^{2} = (a 0 21) (a 0 21) = (a^{2} 0 2 a + 2 1) (ii) A^{3} = (a^{2} 0 2 a + 2 1) (a 0 21) = (a^{3} 0 2 a^{2} + 2 a + 2 1)

Mentor tip

Leave the top-right entry expanded as a sum of powers of $a$ : that pattern is exactly what (c) needs.

(b)

3 marks

A transformation scales area by the absolute value of its determinant.
$det A^{3} = a^{3} \times 1 - 0 = a^{3}$ , so the area goes from $16$ to $16 a^{3}$ .

16 a^{3} = 2 a^{3} = \frac{1}{8} a = 0.5

Mentor tip

It is $det (A^{3}) = (det A)^{3}$ that matters, not $det A$ once.

(c)

3 marks

(i) The top-left entries go $a, a^{2}, a^{3}, \dots$

(ii) The top-right entry is $2 (1 + a + a^{2} + \dots + a^{n - 1})$ , a geometric series with $n$ terms, first term $2$ and ratio $a = 0.5$ .

(i) p_{n} = 0. 5^{n} (ii) q_{n} = \frac{2 ( 1 - 0. 5 ^{n} )}{1 - 0.5} = 4 (1 - 0. 5^{n})

Mentor tip

Check with $n = 1$ : $4 (1 - 0.5) = 2$ , which matches the entry in $A$ .

(d)

3 marks

Multiply $A^{n}$ by the column vector of the point; the $y$ -coordinate stays $96$ .
Set the new $x$ -coordinate equal to $383.75$ and solve for $n$ .

128 (0. 5^{n}) + 96 \times 4 (1 - 0. 5^{n}) = 383.75 384 - 256 (0. 5^{n}) = 383.75 0. 5^{n} = \frac{1}{1024} \Rightarrow n = 10

Another way. Try powers of $A$ on the GDC with $a = 0.5$ until $A^{n}$ sends $(128, 96)$ to $(383.75, 96)$ .

Question 8

17 marks

Rates of change and a separable differential equation

Calculus

(a)

3 marks

Proportional means $\frac{d V}{d t} = k h$ .
The volume is going down, so the rate is $- 0.4$ when $h = 0.5$ . Solve for $k$ .

- 0.4 = k (0.5) \Rightarrow k = - 0.8 \frac{d V}{d t} = - 0.8 h

Mentor tip

Water leaving the tank is a negative rate of change of volume. Leaving out the minus sign causes trouble in (b).

(b)

4 marks

From $t = 5$ , water comes in at $2.0$ and goes out at $0.8 h$ .
For a cylinder with base $16$ , $V = 16 h$ , so $\frac{d V}{d h} = 16$ .
Link the rates with the chain rule.

\frac{d V}{d t} = 2.0 - 0.8 h \frac{d h}{d t} = \frac{d V}{d t} \div \frac{d V}{d h} = \frac{2.0 - 0.8 h}{16} \frac{d h}{d t} = 0.125 - 0.05 h

(c)

8 marks

Put everything with $h$ on one side and integrate both sides.
The left side gives a log of a linear function; watch the factor $- \frac{1}{0.05} = - 20$ .
Rearrange, taking exponentials, to make $h$ the subject.
Use $h = 0.5$ at $t = 5$ to find the constant.

\int \frac{1}{0.125 - 0.05 h} d h = \int 1 d t - 20 ln (0.125 - 0.05 h) = t + c 0.125 - 0.05 h = A e^{- 0.05 t} \Rightarrow h = 2.5 - B e^{- 0.05 t} 0.5 = 2.5 - B e^{- 0.25} \Rightarrow B = 2 e^{0.25} = 2.57 (2.56805 \dots) h = 2.5 - 2.57 e^{- 0.05 t}

Mentor tip

No modulus is needed inside the log because $h$ stays below $2.5$ , so $0.125 - 0.05 h$ stays positive. Check your answer gives $h = 0.5$ at $t = 5$ .

(d)

2 marks

As $t$ grows, $e^{- 0.05 t} \to 0$ , and $\frac{d h}{d t} > 0$ for all $h < 2.5$ .

h increases, levelling off towards 2.5 m t \to \infty lim h = 2.5

Mentor tip

Give both the direction and the limit. The height never actually reaches $2.5$ m, it only approaches it.

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 algebra34 marks · 31%
Q1, Q3, Q7
Geometry and trigonometry25 marks · 23%
Q5, Q6
Calculus23 marks · 21%
Q3, Q8
Functions15 marks · 14%
Q2, Q3
Statistics and probability13 marks · 12%
Q4

Mentor insights

Traps to watch for

01
Using table values instead of bounds · Q3
The speeds are rounded to 1 d.p., so category 6 really runs from 10.75 to 13.85. Using 10.8 and 13.8 gives the wrong difference.
02
Subtracting variances for a difference · Q4
For independent times, the variance of A minus S is 2.1 squared plus 1.7 squared, which is 7.3. Also enter the standard deviation, not 7.3, in the GDC.
03
Forgetting the half-time repeat rule · Q5
A road walked a second time costs only half its weight. Adding the full weight of repeated roads inflates every total.
04
Rounding the car age the wrong way · Q1
Solving gives n = 4.25; a 4-year-old car costs more than she has, so the answer uses the 5-year-old car.
05
Bearing measured from the wrong line · Q6
The angle from arctan is measured below east. Add 90 degrees to get a bearing clockwise from north.

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 2

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1 TZ2TZ3

Session: May 2026
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 8

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2026 TZ3 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–8 · 110 marks

Question 1

14 marks

Compound interest, savings and an arithmetic sequence

Number and algebra

(a)

3 marks

Monthly compounding for three years means $36$ periods at $\frac{2.3}{12} %$ each.
Use the finance solver or the compound interest formula, and give the answer to 2 d.p. as the paper asks.

N = 36, I % = 2.3, P V = - 14500, P M T = 0, P / Y = C / Y = 12 F V = 14500 (1 + \frac{2.3}{100 \times 12})^{36} F V = 15 534.80 GBP (GDC)

Mentor tip

Money answers to 2 d.p. here: $15 534.8$ on its own is not in the form asked for.

(b)

3 marks

The balance from (a) becomes the new present value.
Add the monthly payment of $250$ for $24$ months, with the same rate and monthly compounding.

N = 24, I % = 2.3, P V = - 15 534.80, P M T = - 250, P / Y = C / Y = 12 F V = 22 399.50 GBP (GDC)

Mentor tip

Give $P V$ and $P M T$ the same sign: both are money going into the account. Mixed signs make the solver treat the payments as withdrawals.

(c)

5 marks

(i) The costs drop by $1876$ each year, so $d = - 1876$ . Write $u_{n} = u_{1} + (n - 1) d$ and tidy it into the form $a + bn$ .

(ii) Substitute $n = 7$ .

(iii) Give one real weakness of a straight-line model for car prices.

u_{n} = 28492 + (n - 1) (- 1876) (i) u_{n} = 30368 - 1876 n (ii) u_{7} = 30368 - 1876 \times 7 = 17236 GBP (iii) e.g. u_{n} < 0 once n \geq 17, but a car never has a negative price

Mentor tip

For (iii), say why the formula fails, not just that cars differ. Other good points: it only works for whole-number ages, and real prices usually fall faster at first rather than linearly.

(d)

3 marks

Younger cars cost more, so look for the smallest age whose price fits within $22 399.50$ .
Solve $u_{n} \leq 22 399.50$ and round $n$ up to the next whole number.
Subtract that car's price from the balance.

30368 - 1876 n \leq 22 399.50 \Rightarrow n \geq 4.25 u_{4} = 22864 (too dear), u_{5} = 20988 22 399.50 - 20988 = 1411.50 GBP

Mentor tip

Rounding $4.25$ down to $4$ gives a car she cannot afford. Always check the two neighbouring whole numbers.

Question 2

12 marks

Sine model for body temperature

Functions

(a)

2 marks

The period of $sin (b t)$ is $\frac{36 0 ^{\circ}}{b}$ (degrees) or $\frac{2 π}{b}$ (radians). Set it equal to $24$ .

b = \frac{360}{24} = 15 (degrees) or b = \frac{2 π}{24} = \frac{π}{12} \approx 0.262 (radians)

Mentor tip

Pick degrees or radians now and keep your GDC in that mode for the rest of the question.

(b)

6 marks

(i) At 18:00, $t = 18$ and $15 \times 18 = 27 0^{\circ}$ , where $sin 27 0^{\circ} = - 1$ .

(ii) At 13:00, $t = 13$ and $15 \times 13 = 19 5^{\circ}$ .

(iii) Solve the two linear equations together, by subtracting or with the GDC's simultaneous equation solver.

(i) 36.9 = a sin 27 0^{\circ} + c = - a + c (ii) 36.6 = a sin 19 5^{\circ} + c = - 0.258819 \dots a + c (iii) 0.3 = - 0.741180 \dots a a = - 0.405 (- 0.404759 \dots) c = 36.9 + a = 36.5 (36.4952 \dots)

Mentor tip

A negative $a$ is fine: it just flips the sine curve, which matches the graph dipping first and peaking late in the day.

(c)

2 marks

Put $t = 8$ into the model and compare the result with $36.2$ .

T (8) = - 0.404759 \dots sin 12 0^{\circ} + 36.4952 \dots = 36.1 (36.1447 \dots) 36.2 > 36.1 \Rightarrow the point is above the curve

Mentor tip

Use the unrounded $a$ and $c$ here. Say clearly which value is bigger before you give the conclusion.

(d)

2 marks

With $a < 0$ , the curve falls wherever $sin (15 t)$ is rising.
$sin (15 t)$ rises from $t = 0$ to its peak at $t = 6$ , and again from its trough at $t = 18$ up to $t = 24$ .

0 < t < 6 and 18 < t < 24 i.e. 00:00 to 06:00 and 18:00 to 24:00

Mentor tip

There are two intervals. The model is decreasing on both sides of midnight, so do not stop at the first one.

Question 3

17 marks

Wind turbine power, bounds and a quartic model

Number and algebra
Calculus
Functions

(a)

3 marks

Substitute $r = 40$ and $v = 10$ into the formula.
Write the result in standard form to 3 s.f.

P = 0.245 π (40)^{2} (10)^{3} = 1231 504.32 \dots \approx 1.23 \times 1 0^{6}

Mentor tip

Standard form means $1.23 \times 1 0^{6}$ , not $1.23 E 6$ copied off the calculator screen.

(b)

5 marks

(i) Category 6 starts at $10.8$ to 1 d.p., so the smallest possible true speed is the lower bound $10.75$ .

(ii) The top of the category is $13.8$ , so the largest true speed is the upper bound $13.85$ .

Work out the power at both bounds and subtract.

(i) P_{m i n} = 0.245 π (40)^{2} (10.75)^{3} = 1529 893.96 \dots \approx 1.53 \times 1 0^{6} (ii) P_{m a x} = 0.245 π (40)^{2} (13.85)^{3} = 3271 788.78 \dots P_{m a x} - P_{m i n} = 1741 894.8 \dots \approx 1.74 \times 1 0^{6}

Mentor tip

Using $10.8$ and $13.8$ as they stand ignores the rounding in the table. The bounds sit half a unit of the last digit either side.

(c)

6 marks

(i) Differentiate term by term with the power rule; the constant disappears.

(ii) Solve $\frac{d P}{d v} = 0$ on the GDC and keep only the root inside $3.35 \leq v < 17.15$ .

(iii) Check the sign or the graph of $P$ : it changes from increasing to decreasing there.

(i) \frac{d P}{d v} = 17.92 v^{3} - 708 v^{2} + 7140 v - 8740 (ii) roots 1.42, 14.8, 23.3 (GDC) \Rightarrow v = 14.8 (14.7728 \dots) m s^{- 1} (iii) the wind speed at which this turbine gives its greatest power

Mentor tip

The other two roots lie outside the operating range, so they have no meaning here. Say so rather than listing all three.

(d)

3 marks

Graph $P$ and the line $P = 90000$ on the operating domain and find where they cross.
Power is at least $90000$ from that crossing up to the end of the domain, since $P (17.15) \approx 96500$ is still above $90000$ .

4.48 v^{4} - 236 v^{3} + 3570 v^{2} - 8740 v - 735 = 90000 v = 11.4550 \dots (GDC; the other roots lie outside the domain) 11.46 \leq v < 17.15

Mentor tip

The upper end comes from the domain, not from a second crossing: the crossing at $18.5$ is outside the range the turbine can work in.

Question 4

13 marks

Normal distributions and a difference of variables

Statistics and probability

(a)

4 marks

(i) Use normal cdf with mean $14.2$ and standard deviation $2.1$ , upper limit $14$ .

(ii) Three separate races, each with the same probability, so cube it.

(i) P (A < 14) = 0.462 (0.462062 \dots) (GDC) (ii) (0.462062 \dots)^{3} = 0.0987 (0.0986513 \dots)

Another way. In (ii) you could use the binomial distribution $B (3, 0.462 \dots)$ and find $P (X = 3)$ .

Mentor tip

Cube the full stored value. Cubing $0.462$ gives $0.0986$ , which is slightly off.

(b)

1 mark

Multiplying the probabilities needs the three race times to be independent of each other.

Anne’s time in one race does not affect her time in any other race

Mentor tip

Name the assumption in context: it is about the races, not just the word independent on its own.

(c)

2 marks

Sofia wins if Anne takes longer than $13.2$ minutes.

P (A > 13.2) = 0.683 (0.683030 \dots) (GDC)

Mentor tip

In a race, the smaller time wins. Finding $P (A < 13.2)$ answers the opposite question.

(d)

1 mark

$D > 0$ means $A > S$ : Anne's time is longer.

D > 0 ⟺ Sofia finishes the race faster than Anne (Sofia wins)

(e)

3 marks

(i) Means subtract.

(ii) For independent variables the variances add, even for a difference. Square the standard deviations first.

(i) E (D) = 14.2 - 13.4 = 0.8 minutes (ii) Var (D) = 2. 1^{2} + 1. 7^{2} = 4.41 + 2.89 = 7.3 minutes^{2}

Mentor tip

Subtracting variances ( $4.41 - 2.89$ ) is the classic slip. A difference is more spread out than either time alone.

(f)

2 marks

Anne wins when her time is shorter, so $D < 0$ .
$D$ is normal with mean $0.8$ and standard deviation $7.3$ .

D \sim N (0.8, 7.3) P (D < 0) = 0.384 (0.383579 \dots) (GDC)

Mentor tip

The GDC wants the standard deviation, $7.3 \approx 2.70$ , not the variance $7.3$ .

Question 5

12 marks

Route inspection on a road network

Geometry and trigonometry

(a)

3 marks

Count degrees: only $A$ and $B$ are odd (both degree $3$ ).
The shortest path from $A$ to $B$ is the direct edge of $8$ , so walk it twice. The second walk costs half, $4$ .
(ii) Any closed walk from $A$ that uses every edge once, plus $A B$ a second time, will do.

(i) 122 + \frac{1}{2} (8) = 126 minutes (ii) e.g. A B D E F G E H C B A H I A

Mentor tip

Remember the half-time rule. Adding the full $8$ gives $130$ , which ignores the faster second walk.

(b)

6 marks

The new road makes $C$ and $D$ odd as well, so the odd vertices are $A, B, C, D$ .
List all three pairings, using shortest paths where there is no direct road: $A C$ via $B$ is $15$ , $A D$ via $B$ is $26$ .
The pairing with the smallest total is the one to repeat.
(ii) Add the new road's $14$ to $122$ , plus half of the repeated total.

A B + C D = 8 + 14 = 22 A C + B D = 15 + 18 = 33 A D + B C = 26 + 7 = 33 (i) repeat A B and C D (ii) 122 + 14 + \frac{1}{2} (22) = 147 minutes

Mentor tip

Showing all three pairings is the justification. Picking the smallest-looking one without the others is not enough.

(c)

3 marks

With free start and finish, two of the four odd vertices can be the ends, so only one pair needs repeating.
Repeat the cheapest single connection, $B C = 7$ , and start and finish at the other two, $A$ and $D$ .
Compare with the $11$ minutes of repeats in (b).

(i) start at A and finish at D (or the other way round) repeat cost = \frac{1}{2} (7) = 3.5 instead of \frac{1}{2} (22) = 11 (ii) time saved = 11 - 3.5 = 7.5 minutes

Mentor tip

The ends are the pair you do not repeat. Students often choose $A$ and $B$ because $A B$ was repeated before, but that is the wrong way round.

Question 6

13 marks

Velocity vectors, bearings and journey times

Geometry and trigonometry

(a)

5 marks

(i) Speed is the magnitude of the velocity, so set $a^{2} + (- 10)^{2} = 26$ .

(ii) The boat moves $24$ east and $10$ south each hour. Find the angle below east, then add $9 0^{\circ}$ to measure clockwise from north.

(i) a^{2} + 1 0^{2} = 2 6^{2} \Rightarrow a = 24 (ii) arctan (\frac{10}{24}) = 22.6198 \dots^{\circ} bearing = 9 0^{\circ} + 22.6198 \dots^{\circ} = 11 3^{\circ} (112.619 \dots^{\circ})

Mentor tip

Bearings are in degrees, clockwise from north, and usually three figures. A quick sketch of east and south stops you giving $6 7^{\circ}$ .

(b)

2 marks

From 08:00 to 11:00 is $3$ hours. Start at $H$ and add three lots of the velocity.

r = (520) + 3 (24 - 10) (77, - 10)

(c)

3 marks

Due east for two hours at $26$ adds $52$ to the $x$ -coordinate only.
Use the distance formula between $H (5, 20)$ and $P$ .

P = (77 + 52, - 10) = (129, - 10) H P = (129 - 5)^{2} + (- 10 - 20)^{2} = 16276 \approx 128 km (127.577 \dots)

Another way. Use the cosine rule in triangle $H$ , 11:00 position, $P$ with sides $78$ and $52$ and the included angle $18 0^{\circ} - 22. 6^{\circ}$ .

(d)

3 marks

Boat $B$ reaches $P$ at 13:00.
Time for the second boat is distance over speed; subtract it from 13:00 and convert the decimal hours into minutes.

time = \frac{127.577 \dots}{40} = 3.18943 \dots h 13 - 3.18943 \dots = 9.81056 \dots h 0.81056 \dots \times 60 = 48.6 \dots min \Rightarrow 09 : 49

Mentor tip

Keep the unrounded distance: using $128$ shifts the answer by about a minute. And $0.81$ hours is not $81$ minutes.

Question 7

12 marks

Matrix powers, determinants and a geometric series

Number and algebra

(a)

3 marks

(i) Multiply $A$ by itself, row by column.

(ii) Multiply your $A^{2}$ by $A$ .

(i) A^{2} = (a 0 21) (a 0 21) = (a^{2} 0 2 a + 2 1) (ii) A^{3} = (a^{2} 0 2 a + 2 1) (a 0 21) = (a^{3} 0 2 a^{2} + 2 a + 2 1)

Mentor tip

Leave the top-right entry expanded as a sum of powers of $a$ : that pattern is exactly what (c) needs.

(b)

3 marks

A transformation scales area by the absolute value of its determinant.
$det A^{3} = a^{3} \times 1 - 0 = a^{3}$ , so the area goes from $16$ to $16 a^{3}$ .

16 a^{3} = 2 a^{3} = \frac{1}{8} a = 0.5

Mentor tip

It is $det (A^{3}) = (det A)^{3}$ that matters, not $det A$ once.

(c)

3 marks

(i) The top-left entries go $a, a^{2}, a^{3}, \dots$

(ii) The top-right entry is $2 (1 + a + a^{2} + \dots + a^{n - 1})$ , a geometric series with $n$ terms, first term $2$ and ratio $a = 0.5$ .

(i) p_{n} = 0. 5^{n} (ii) q_{n} = \frac{2 ( 1 - 0. 5 ^{n} )}{1 - 0.5} = 4 (1 - 0. 5^{n})

Mentor tip

Check with $n = 1$ : $4 (1 - 0.5) = 2$ , which matches the entry in $A$ .

(d)

3 marks

Multiply $A^{n}$ by the column vector of the point; the $y$ -coordinate stays $96$ .
Set the new $x$ -coordinate equal to $383.75$ and solve for $n$ .

128 (0. 5^{n}) + 96 \times 4 (1 - 0. 5^{n}) = 383.75 384 - 256 (0. 5^{n}) = 383.75 0. 5^{n} = \frac{1}{1024} \Rightarrow n = 10

Another way. Try powers of $A$ on the GDC with $a = 0.5$ until $A^{n}$ sends $(128, 96)$ to $(383.75, 96)$ .

Question 8

17 marks

Rates of change and a separable differential equation

Calculus

(a)

3 marks

Proportional means $\frac{d V}{d t} = k h$ .
The volume is going down, so the rate is $- 0.4$ when $h = 0.5$ . Solve for $k$ .

- 0.4 = k (0.5) \Rightarrow k = - 0.8 \frac{d V}{d t} = - 0.8 h

Mentor tip

Water leaving the tank is a negative rate of change of volume. Leaving out the minus sign causes trouble in (b).

(b)

4 marks

From $t = 5$ , water comes in at $2.0$ and goes out at $0.8 h$ .
For a cylinder with base $16$ , $V = 16 h$ , so $\frac{d V}{d h} = 16$ .
Link the rates with the chain rule.

\frac{d V}{d t} = 2.0 - 0.8 h \frac{d h}{d t} = \frac{d V}{d t} \div \frac{d V}{d h} = \frac{2.0 - 0.8 h}{16} \frac{d h}{d t} = 0.125 - 0.05 h

(c)

8 marks

Put everything with $h$ on one side and integrate both sides.
The left side gives a log of a linear function; watch the factor $- \frac{1}{0.05} = - 20$ .
Rearrange, taking exponentials, to make $h$ the subject.
Use $h = 0.5$ at $t = 5$ to find the constant.

\int \frac{1}{0.125 - 0.05 h} d h = \int 1 d t - 20 ln (0.125 - 0.05 h) = t + c 0.125 - 0.05 h = A e^{- 0.05 t} \Rightarrow h = 2.5 - B e^{- 0.05 t} 0.5 = 2.5 - B e^{- 0.25} \Rightarrow B = 2 e^{0.25} = 2.57 (2.56805 \dots) h = 2.5 - 2.57 e^{- 0.05 t}

Mentor tip

No modulus is needed inside the log because $h$ stays below $2.5$ , so $0.125 - 0.05 h$ stays positive. Check your answer gives $h = 0.5$ at $t = 5$ .

(d)

2 marks

As $t$ grows, $e^{- 0.05 t} \to 0$ , and $\frac{d h}{d t} > 0$ for all $h < 2.5$ .

h increases, levelling off towards 2.5 m t \to \infty lim h = 2.5

Mentor tip

Give both the direction and the limit. The height never actually reaches $2.5$ m, it only approaches it.

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 algebra34 marks · 31%
Q1, Q3, Q7
Geometry and trigonometry25 marks · 23%
Q5, Q6
Calculus23 marks · 21%
Q3, Q8
Functions15 marks · 14%
Q2, Q3
Statistics and probability13 marks · 12%
Q4

Mentor insights

Traps to watch for

01
Using table values instead of bounds · Q3
The speeds are rounded to 1 d.p., so category 6 really runs from 10.75 to 13.85. Using 10.8 and 13.8 gives the wrong difference.
02
Subtracting variances for a difference · Q4
For independent times, the variance of A minus S is 2.1 squared plus 1.7 squared, which is 7.3. Also enter the standard deviation, not 7.3, in the GDC.
03
Forgetting the half-time repeat rule · Q5
A road walked a second time costs only half its weight. Adding the full weight of repeated roads inflates every total.
04
Rounding the car age the wrong way · Q1
Solving gives n = 4.25; a 4-year-old car costs more than she has, so the answer uses the 5-year-old car.
05
Bearing measured from the wrong line · Q6
The angle from arctan is measured below east. Add 90 degrees to get a bearing clockwise from north.

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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