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Worked solutions

IB Maths AA SL · Paper 2

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1 TZ2TZ3

Session: May 2026
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 9

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9

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.

Section A · Short response

Questions 1–6 · 36 marks

Question 1

5 marks

Linear regression and a prediction

Statistics and probability

(a)

2 marks

Enter leg length as the $x$ -list and race time as the $y$ -list, then run linear regression on the GDC.

a = - 0.767415 \dots \approx - 0.767 b = 130.345 \dots \approx 130 (GDC)

Mentor tip

Keep the full values stored in the calculator. Rounded coefficients can shift the prediction in (c).

(b)

1 mark

The same regression screen shows $r$ .

r = - 0.996248 \dots \approx - 0.996

Mentor tip

Longer legs go with shorter times, so $r$ must be negative. A positive value means the lists were swapped or mistyped.

(c)

2 marks

Substitute $L = 79$ into the regression equation.

T = - 0.767415 \dots (79) + 130.345 \dots T \approx 69.7 seconds

Mentor tip

$79$ cm sits inside the data range ( $76$ to $91$ ), so this is interpolation and the prediction is reasonable.

Question 2

6 marks

Angles of elevation and a cliff height

Geometry and trigonometry

(a)

2 marks

(i) In the right-angled triangle at $B$ , $h$ is opposite and $x$ is adjacent.

(ii) From $A$ the adjacent side is the whole distance $x + 70$ .

(i) tan 4 0^{\circ} = \frac{h}{x} (ii) tan 1 5^{\circ} = \frac{h}{x + 70}

Mentor tip

Label the horizontal distance from each point carefully. The distance from $A$ includes the $70$ m already sailed.

(b)

2 marks

Make $h$ the subject of both equations and set them equal.
Collect the $x$ terms and solve.

x tan 4 0^{\circ} = (x + 70) tan 1 5^{\circ} x (tan 4 0^{\circ} - tan 1 5^{\circ}) = 70 tan 1 5^{\circ} x = \frac{70 tan 1 5 ^{\circ}}{tan 4 0 ^{\circ} - tan 1 5 ^{\circ}} = 32.8397 \dots \approx 32.8 m

Another way. You could instead use the sine rule in the triangle formed by $A$ , $B$ and the cliff top (its angle at the top is $2 5^{\circ}$ ) to find the slant distance first.

Mentor tip

Check the calculator is in degrees before you press anything.

(c)

2 marks

Use your unrounded $x$ in the equation from (a)(i).

h = 32.8397 \dots \times tan 4 0^{\circ} h = 27.5558 \dots \approx 27.6 m

Mentor tip

Quick check: $tan 1 5^{\circ} = \frac{27.6}{102.8} \approx 0.268$ , which matches.

Question 3

5 marks

Exponential decay of a volume

Functions

(a)

1 mark

The initial volume is at $t = 0$ , where $e^{0} = 1$ .

V (0) = 5 + 20 = 25 m^{3}

(b)

2 marks

Half of $25$ is $12.5$ . Set $V = 12.5$ with $t = 4$ and solve for $k$ .

12.5 = 5 + 20 e^{- 4 k} e^{- 4 k} = 0.375 k = - \frac{1}{4} ln 0.375 = 0.245207 \dots \approx 0.245

Mentor tip

Halving applies to the whole volume, not to the $20$ . A common slip is setting $20 e^{- 4 k} = 10$ .

(c)

2 marks

Set $V = 8$ using the stored value of $k$ , and solve (logs or GDC intersection).

8 = 5 + 20 e^{- k t} ⟹ e^{- k t} = 0.15 t = \frac{- ln 0.15}{0.245207 \dots} = 7.73680 \dots \approx 7.74 minutes

Mentor tip

Sense check: the volume passes half at $4$ minutes and keeps falling towards $5$ , so a time a little under $8$ minutes for $8$ m $^{3}$ is believable.

Question 4

6 marks

A probability distribution with halving terms

Statistics and probability

(a)

2 marks

Each probability is half the one before, so step along the table from $k$ by factors of $\frac{1}{2}$ .

(i) a = \frac{k}{2} (ii) b = \frac{k}{16}

Mentor tip

Check against the entry you are given: $\frac{k}{8}$ in the fourth column fits the pattern $k, \frac{k}{2}, \frac{k}{4}, \frac{k}{8}, \dots$

(b)

2 marks

The probabilities add to $1$ . Write every entry in terms of $k$ and sum.

k + \frac{k}{2} + \frac{k}{4} + \frac{k}{8} + \frac{k}{16} + \frac{k}{32} = 1 \frac{63 k}{32} = 1 ⟹ k = \frac{32}{63} \approx 0.508

Another way. Faster still: the last entry is $\frac{k}{32}$ , so matching it to the given value finds $k$ in one line.

Mentor tip

Confirm with the other given entry: $\frac{k}{4} = \frac{8}{63}$ also gives $k = \frac{32}{63}$ .

(c)

2 marks

Multiply each value of $x$ by its probability and add. With $k = \frac{32}{63}$ the probabilities are $\frac{32}{63}, \frac{16}{63}, \dots, \frac{1}{63}$ .

E (X) = \frac{1 ( 32 ) + 2 ( 16 ) + 3 ( 8 ) + 4 ( 4 ) + 5 ( 2 ) + 6 ( 1 )}{63} = \frac{120}{63} = \frac{40}{21} \approx 1.90

Mentor tip

Most of the probability sits at $x = 1$ and $x = 2$ , so an expected value just under $2$ makes sense.

Question 5

7 marks

Sectors: area, perimeter and radius

Geometry and trigonometry

(a)

5 marks

Write the area of sector $O A B$ as $\frac{1}{2} R^{2} θ = 50$ .
The perimeter is two radii plus the arc: $2 R + R θ = 40$ , so $R = \frac{40}{2 + θ}$ .
Substitute into the area equation and solve for $θ$ .
Two roots appear; only the acute angle is allowed.

R^{2} θ = 100 (\frac{40}{2 + θ})^{2} θ = 100 1600 θ = 100 (2 + θ)^{2} ⟹ θ^{2} - 12 θ + 4 = 0 θ = 6 \pm 42 = 0.343145 \dots or 11.6568 \dots θ \approx 0.343 radians

Another way. Eliminating $θ$ instead gives $R^{2} - 20 R + 50 = 0$ , so $R = 10 + 52$ , which leads to the same angle.

Mentor tip

Do not forget the two radii in the perimeter. Arc length alone is only $R θ$ .

(b)

2 marks

Sector $O D C$ has the same angle. Use $\frac{1}{2} r^{2} θ = 18$ with your stored $θ$ .

r^{2} = \frac{36}{0.343145 \dots} r = 10.2426 \dots \approx 10.2 cm

Mentor tip

Check that $r < R \approx 17.1$ cm, as the smaller circle must be.

Question 6

7 marks

Kinematics: displacement and acceleration

Calculus

(a)

4 marks

Displacement has a turning point where $v (t) = 0$ . Graph $v$ on $0 \leq t \leq 6$ to find the zeros after the start ( $v (0) = 0$ ): $t = 1.96227 \dots$ and $t = 5.39302 \dots$
Velocity is negative until the first zero, positive between the zeros, then negative again, so the largest displacement comes at the second zero.
Starting at $O$ , the displacement is the integral of $v$ from $0$ .

v (t) = 0 ⟹ t = 1.96227 \dots, 5.39302 \dots (GDC) s (5.39302 \dots) = \int_{0}^{5.39302 \dots} v (t) d t = 4.80592 \dots \approx 4.81 m (GDC)

Another way. Graph $s (t) = \int_{0}^{t} v (u) d u$ directly on the GDC and read off its maximum.

Mentor tip

Compare every possibility. At $t = 1.96$ the displacement is about $- 5.30$ m (the lowest point) and at $t = 6$ it is about $4.37$ m, so the maximum is $4.81$ m.

(b)

3 marks

Acceleration is the gradient of the velocity graph, so look for where $v$ is decreasing.
The turning points of $v$ (the zeros of $a = v^{'}$ ) are at $t = 0.791904 \dots$ and $t = 3.50201 \dots$
The graph of $v$ falls before the first of these and after the second.

v^{'} (t) = 0 ⟹ t = 0.791904 \dots, 3.50201 \dots (GDC) a (t) < 0 for 0 \leq t < 0.792 and 3.50 < t \leq 6

Mentor tip

Give both intervals and include the ends of the domain. Negative acceleration is about the slope of $v$ , not the sign of $v$ .

Section B · Extended response

Questions 7–9 · 44 marks

Question 7

15 marks

A sphere, then minimising a cuboid surface

Geometry and trigonometry
Functions
Calculus

(a)

4 marks

(i) The centre is the midpoint of the diameter: average each coordinate of $R$ and $S$ .

(ii) The radius is the distance from the centre to either end.

centre = (\frac{5 - 3}{2}, \frac{- 3 + 3}{2}, \frac{2 - 4}{2}) = (1, 0, - 1) r = (5 - 1)^{2} + (- 3 - 0)^{2} + (2 + 1)^{2} = 34 r \approx 5.83 m

Mentor tip

If you find the full length $R S = 136$ , remember to halve it.

(b)

2 marks

Use the sphere volume formula with $r = 34$ .

V = \frac{4}{3} π (34)^{3} = 830.437 \dots \approx 830 m^{3}

Mentor tip

Use the exact radius; cubing $5.83$ gives a slightly different figure.

(c)

4 marks

(i) Volume is length times width times height, with width $x^{2}$ .

(ii) The surface is three pairs of faces: $x \times x^{2}$ , $x \times y$ and $x^{2} \times y$ .

Make $y$ the subject of the volume equation and substitute, then take out the factor $2$ .

(i) x \cdot x^{2} \cdot y = x^{3} y = 1000 A = 2 (x^{3} + x y + x^{2} y) y = \frac{1000}{x ^{3}} A = 2 (x^{3} + \frac{1000}{x ^{2}} + \frac{1000}{x})

Mentor tip

In a show that, write the substitution out in full before simplifying, so every step can be followed.

(d)

5 marks

(i) Graph $A$ for $x > 0$ and find the minimum, or solve $\frac{d A}{d x} = 0$ .

(ii) Differentiate again and evaluate at $x = k$ .

(iii) A positive second derivative means the curve is concave up there, so the stationary point is a minimum.

\frac{d A}{d x} = 6 x^{2} - \frac{4000}{x ^{3}} - \frac{2000}{x ^{2}} = 0 ⟹ k = 4.67105 \dots (GDC) A_{m i n} = 723.666 \dots \approx 724 m^{2} \frac{d ^{2} A}{d x ^{2}} = 12 x + \frac{12000}{x ^{4}} + \frac{4000}{x ^{3}} \frac{d ^{2} A}{d x ^{2}}_{x = k} = 120.507 \dots \approx 121 121 > 0 ⟹ A is a minimum at x = k

Mentor tip

For (iii), state the sign and the conclusion together: the value is positive, therefore a minimum. The number alone is not a justification.

Question 8

14 marks

Normal waiting times and a binomial count

Statistics and probability

(a)

7 marks

(i) Convert each percentage to a $z$ -score with inverse normal: the lower tail $0.038$ and the upper tail $0.073$ (a left area of $0.927$ ).

Write $z = \frac{x - μ}{σ}$ for each and solve the pair of linear equations.

(ii) Use the normal CDF with your stored $μ$ and $σ$ .

z_{1} = - 1.77438 \dots, z_{2} = 1.45380 \dots 20 - μ = - 1.77438 \dots σ, 30 - μ = 1.45380 \dots σ μ = 25.4965 \dots \approx 25.5, σ = 3.09771 \dots \approx 3.10 P (15 < T < 22) = 0.129152 \dots \approx 0.129 (GDC)

Another way. Subtracting the two equations directly gives $σ = \frac{10}{1.77438 \dots + 1.45380 \dots}$ .

Mentor tip

Solve with $z$ -scores, not the percentages themselves. A quick sketch of the curve with both tails shaded keeps the signs right.

(b)

7 marks

Let $Y$ be the number who attend, so $Y \sim B (20, 0.826)$ .
(i) At least $13$ means $1 - P (Y \leq 12)$ .
(ii) More than $2$ absent means fewer than $18$ attend, so $Y \leq 17$ .
Conditional probability: the overlap of the two events divided by the given event.

P (Y \geq 13) = 1 - P (Y \leq 12) = 0.985473 \dots \approx 0.985 P (Y \leq 17 ∣ Y \geq 13) = \frac{P ( 13 \leq Y \leq 17 )}{P ( Y \geq 13 )} = \frac{0.687241 \dots}{0.985473 \dots} = 0.697372 \dots \approx 0.697

Mentor tip

Translate the non-attenders into attenders before you type anything: more than $2$ absent out of $20$ is at most $17$ present.

Question 9

15 marks

Geometric series, squares and a log sequence

Number and algebra

(a)

1 mark

Rearrange the sum to infinity formula with $A_{\infty} = 5$ .

\frac{a _{1}}{1 - r} = 5 ⟹ a_{1} = 5 (1 - r)

(b)

2 marks

Divide the second term of the new sequence by the first, writing $a_{2} = r a_{1}$ .

q = \frac{a _{2}^{2}}{a _{1}^{2}} = \frac{( r a _{1} ) ^{2}}{a _{1}^{2}} q = r^{2}

Mentor tip

Show the substitution $a_{2} = r a_{1}$ explicitly; that line carries the whole argument.

(c)

5 marks

(i) Use $B_{\infty} = \frac{a _{1}^{2}}{1 - q} = 16$ with $a_{1} = 5 (1 - r)$ and $q = r^{2}$ .

Expand and collect everything on one side.

(ii) Solve the quadratic. $r = 1$ makes the series diverge, so reject it.

(iii) Substitute into $a_{1} = 5 (1 - r)$ .

\frac{25 ( 1 - r ) ^{2}}{1 - r ^{2}} = 16 25 (1 - 2 r + r^{2}) = 16 - 16 r^{2} 41 r^{2} - 50 r + 9 = 0 (41 r - 9) (r - 1) = 0 ⟹ r = \frac{9}{41} \approx 0.220 a_{1} = 5 (1 - \frac{9}{41}) = \frac{160}{41} \approx 3.90

Another way. Cancel a factor: $1 - r^{2} = (1 - r) (1 + r)$ , so $\frac{25 ( 1 - r )}{1 + r} = 16$ gives $r = \frac{9}{41}$ straight away.

Mentor tip

Say why $r = 1$ is rejected: a sum to infinity needs $∣ r ∣ < 1$ .

(d)

4 marks

Find $c_{n + 1} - c_{n}$ in general.
The constant $k$ cancels, and the log laws turn the difference of logs into the log of a quotient.
That quotient is the common ratio $r$ , a constant, so the difference is constant.

c_{n + 1} - c_{n} = (lo g_{k} a_{n + 1} + k) - (lo g_{k} a_{n} + k) = lo g_{k} \frac{a _{n + 1}}{a _{n}} = lo g_{k} r lo g_{k} r is constant ⟹ arithmetic

Mentor tip

Finish with a sentence: the difference does not depend on $n$ , so the sequence is arithmetic. Checking only the first few terms is not a proof.

(e)

3 marks

The first term is $lo g_{k} a_{1} + k$ and the common difference is $lo g_{k} r$ .
Use $S_{20} = \frac{20}{2} (2 c_{1} + 19 d)$ with $a_{1} = \frac{160}{41}$ and $r = \frac{9}{41}$ .
Graph $C_{20}$ against $k$ and find where it crosses zero, then take the largest integer below it.

C_{20} = 10 (2 lo g_{k} \frac{160}{41} + 2 k + 19 lo g_{k} \frac{9}{41}) C_{20} = 0 ⟹ k = 6.80296 \dots (GDC) k \in Z^{+} ⟹ k = 6

Another way. Use $lo g_{k} x = \frac{l n x}{l n k}$ to enter the expression on a GDC that has no base- $k$ log.

Mentor tip

Remember $k$ is a positive integer. The crossing at $6.80$ is not the answer; check $C_{20} (6) < 0$ and $C_{20} (7) > 0$ .

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 probability25 marks · 31%
Q1, Q4, Q8
Geometry and trigonometry19 marks · 24%
Q2, Q5, Q7
Number and algebra15 marks · 19%
Q9
Calculus12 marks · 15%
Q6, Q7
Functions9 marks · 11%
Q3, Q7

Mentor insights

Traps to watch for

01
Perimeter of a sector · Q5
The perimeter is the arc plus two radii. Using the arc alone gives the wrong equation, and you must then keep only the acute root.
02
Which zero of v gives the maximum · Q6
Velocity is zero twice. Integrate to both and compare with the end point; the first zero is actually the lowest displacement.
03
Justifying a minimum · Q7
Give the sign of the second derivative and the conclusion it leads to. A number on its own is not a justification.
04
Translating non-attenders · Q8
More than 2 absent means at most 17 present. Getting this inequality wrong spoils the whole conditional probability.
05
k must be an integer · Q9
The GDC gives a crossing at about 6.80, but k is a positive integer, so the largest value that works is 6.

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Worked solutions

IB Maths AA SL · Paper 2

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1 TZ2TZ3

Session: May 2026
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 9

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9

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.

Section A · Short response

Questions 1–6 · 36 marks

Question 1

5 marks

Linear regression and a prediction

Statistics and probability

(a)

2 marks

Enter leg length as the $x$ -list and race time as the $y$ -list, then run linear regression on the GDC.

a = - 0.767415 \dots \approx - 0.767 b = 130.345 \dots \approx 130 (GDC)

Mentor tip

Keep the full values stored in the calculator. Rounded coefficients can shift the prediction in (c).

(b)

1 mark

The same regression screen shows $r$ .

r = - 0.996248 \dots \approx - 0.996

Mentor tip

Longer legs go with shorter times, so $r$ must be negative. A positive value means the lists were swapped or mistyped.

(c)

2 marks

Substitute $L = 79$ into the regression equation.

T = - 0.767415 \dots (79) + 130.345 \dots T \approx 69.7 seconds

Mentor tip

$79$ cm sits inside the data range ( $76$ to $91$ ), so this is interpolation and the prediction is reasonable.

Question 2

6 marks

Angles of elevation and a cliff height

Geometry and trigonometry

(a)

2 marks

(i) In the right-angled triangle at $B$ , $h$ is opposite and $x$ is adjacent.

(ii) From $A$ the adjacent side is the whole distance $x + 70$ .

(i) tan 4 0^{\circ} = \frac{h}{x} (ii) tan 1 5^{\circ} = \frac{h}{x + 70}

Mentor tip

Label the horizontal distance from each point carefully. The distance from $A$ includes the $70$ m already sailed.

(b)

2 marks

Make $h$ the subject of both equations and set them equal.
Collect the $x$ terms and solve.

x tan 4 0^{\circ} = (x + 70) tan 1 5^{\circ} x (tan 4 0^{\circ} - tan 1 5^{\circ}) = 70 tan 1 5^{\circ} x = \frac{70 tan 1 5 ^{\circ}}{tan 4 0 ^{\circ} - tan 1 5 ^{\circ}} = 32.8397 \dots \approx 32.8 m

Another way. You could instead use the sine rule in the triangle formed by $A$ , $B$ and the cliff top (its angle at the top is $2 5^{\circ}$ ) to find the slant distance first.

Mentor tip

Check the calculator is in degrees before you press anything.

(c)

2 marks

Use your unrounded $x$ in the equation from (a)(i).

h = 32.8397 \dots \times tan 4 0^{\circ} h = 27.5558 \dots \approx 27.6 m

Mentor tip

Quick check: $tan 1 5^{\circ} = \frac{27.6}{102.8} \approx 0.268$ , which matches.

Question 3

5 marks

Exponential decay of a volume

Functions

(a)

1 mark

The initial volume is at $t = 0$ , where $e^{0} = 1$ .

V (0) = 5 + 20 = 25 m^{3}

(b)

2 marks

Half of $25$ is $12.5$ . Set $V = 12.5$ with $t = 4$ and solve for $k$ .

12.5 = 5 + 20 e^{- 4 k} e^{- 4 k} = 0.375 k = - \frac{1}{4} ln 0.375 = 0.245207 \dots \approx 0.245

Mentor tip

Halving applies to the whole volume, not to the $20$ . A common slip is setting $20 e^{- 4 k} = 10$ .

(c)

2 marks

Set $V = 8$ using the stored value of $k$ , and solve (logs or GDC intersection).

8 = 5 + 20 e^{- k t} ⟹ e^{- k t} = 0.15 t = \frac{- ln 0.15}{0.245207 \dots} = 7.73680 \dots \approx 7.74 minutes

Mentor tip

Sense check: the volume passes half at $4$ minutes and keeps falling towards $5$ , so a time a little under $8$ minutes for $8$ m $^{3}$ is believable.

Question 4

6 marks

A probability distribution with halving terms

Statistics and probability

(a)

2 marks

Each probability is half the one before, so step along the table from $k$ by factors of $\frac{1}{2}$ .

(i) a = \frac{k}{2} (ii) b = \frac{k}{16}

Mentor tip

Check against the entry you are given: $\frac{k}{8}$ in the fourth column fits the pattern $k, \frac{k}{2}, \frac{k}{4}, \frac{k}{8}, \dots$

(b)

2 marks

The probabilities add to $1$ . Write every entry in terms of $k$ and sum.

k + \frac{k}{2} + \frac{k}{4} + \frac{k}{8} + \frac{k}{16} + \frac{k}{32} = 1 \frac{63 k}{32} = 1 ⟹ k = \frac{32}{63} \approx 0.508

Another way. Faster still: the last entry is $\frac{k}{32}$ , so matching it to the given value finds $k$ in one line.

Mentor tip

Confirm with the other given entry: $\frac{k}{4} = \frac{8}{63}$ also gives $k = \frac{32}{63}$ .

(c)

2 marks

Multiply each value of $x$ by its probability and add. With $k = \frac{32}{63}$ the probabilities are $\frac{32}{63}, \frac{16}{63}, \dots, \frac{1}{63}$ .

E (X) = \frac{1 ( 32 ) + 2 ( 16 ) + 3 ( 8 ) + 4 ( 4 ) + 5 ( 2 ) + 6 ( 1 )}{63} = \frac{120}{63} = \frac{40}{21} \approx 1.90

Mentor tip

Most of the probability sits at $x = 1$ and $x = 2$ , so an expected value just under $2$ makes sense.

Question 5

7 marks

Sectors: area, perimeter and radius

Geometry and trigonometry

(a)

5 marks

Write the area of sector $O A B$ as $\frac{1}{2} R^{2} θ = 50$ .
The perimeter is two radii plus the arc: $2 R + R θ = 40$ , so $R = \frac{40}{2 + θ}$ .
Substitute into the area equation and solve for $θ$ .
Two roots appear; only the acute angle is allowed.

R^{2} θ = 100 (\frac{40}{2 + θ})^{2} θ = 100 1600 θ = 100 (2 + θ)^{2} ⟹ θ^{2} - 12 θ + 4 = 0 θ = 6 \pm 42 = 0.343145 \dots or 11.6568 \dots θ \approx 0.343 radians

Another way. Eliminating $θ$ instead gives $R^{2} - 20 R + 50 = 0$ , so $R = 10 + 52$ , which leads to the same angle.

Mentor tip

Do not forget the two radii in the perimeter. Arc length alone is only $R θ$ .

(b)

2 marks

Sector $O D C$ has the same angle. Use $\frac{1}{2} r^{2} θ = 18$ with your stored $θ$ .

r^{2} = \frac{36}{0.343145 \dots} r = 10.2426 \dots \approx 10.2 cm

Mentor tip

Check that $r < R \approx 17.1$ cm, as the smaller circle must be.

Question 6

7 marks

Kinematics: displacement and acceleration

Calculus

(a)

4 marks

Displacement has a turning point where $v (t) = 0$ . Graph $v$ on $0 \leq t \leq 6$ to find the zeros after the start ( $v (0) = 0$ ): $t = 1.96227 \dots$ and $t = 5.39302 \dots$
Velocity is negative until the first zero, positive between the zeros, then negative again, so the largest displacement comes at the second zero.
Starting at $O$ , the displacement is the integral of $v$ from $0$ .

v (t) = 0 ⟹ t = 1.96227 \dots, 5.39302 \dots (GDC) s (5.39302 \dots) = \int_{0}^{5.39302 \dots} v (t) d t = 4.80592 \dots \approx 4.81 m (GDC)

Another way. Graph $s (t) = \int_{0}^{t} v (u) d u$ directly on the GDC and read off its maximum.

Mentor tip

Compare every possibility. At $t = 1.96$ the displacement is about $- 5.30$ m (the lowest point) and at $t = 6$ it is about $4.37$ m, so the maximum is $4.81$ m.

(b)

3 marks

Acceleration is the gradient of the velocity graph, so look for where $v$ is decreasing.
The turning points of $v$ (the zeros of $a = v^{'}$ ) are at $t = 0.791904 \dots$ and $t = 3.50201 \dots$
The graph of $v$ falls before the first of these and after the second.

v^{'} (t) = 0 ⟹ t = 0.791904 \dots, 3.50201 \dots (GDC) a (t) < 0 for 0 \leq t < 0.792 and 3.50 < t \leq 6

Mentor tip

Give both intervals and include the ends of the domain. Negative acceleration is about the slope of $v$ , not the sign of $v$ .

Section B · Extended response

Questions 7–9 · 44 marks

Question 7

15 marks

A sphere, then minimising a cuboid surface

Geometry and trigonometry
Functions
Calculus

(a)

4 marks

(i) The centre is the midpoint of the diameter: average each coordinate of $R$ and $S$ .

(ii) The radius is the distance from the centre to either end.

centre = (\frac{5 - 3}{2}, \frac{- 3 + 3}{2}, \frac{2 - 4}{2}) = (1, 0, - 1) r = (5 - 1)^{2} + (- 3 - 0)^{2} + (2 + 1)^{2} = 34 r \approx 5.83 m

Mentor tip

If you find the full length $R S = 136$ , remember to halve it.

(b)

2 marks

Use the sphere volume formula with $r = 34$ .

V = \frac{4}{3} π (34)^{3} = 830.437 \dots \approx 830 m^{3}

Mentor tip

Use the exact radius; cubing $5.83$ gives a slightly different figure.

(c)

4 marks

(i) Volume is length times width times height, with width $x^{2}$ .

(ii) The surface is three pairs of faces: $x \times x^{2}$ , $x \times y$ and $x^{2} \times y$ .

Make $y$ the subject of the volume equation and substitute, then take out the factor $2$ .

(i) x \cdot x^{2} \cdot y = x^{3} y = 1000 A = 2 (x^{3} + x y + x^{2} y) y = \frac{1000}{x ^{3}} A = 2 (x^{3} + \frac{1000}{x ^{2}} + \frac{1000}{x})

Mentor tip

In a show that, write the substitution out in full before simplifying, so every step can be followed.

(d)

5 marks

(i) Graph $A$ for $x > 0$ and find the minimum, or solve $\frac{d A}{d x} = 0$ .

(ii) Differentiate again and evaluate at $x = k$ .

(iii) A positive second derivative means the curve is concave up there, so the stationary point is a minimum.

\frac{d A}{d x} = 6 x^{2} - \frac{4000}{x ^{3}} - \frac{2000}{x ^{2}} = 0 ⟹ k = 4.67105 \dots (GDC) A_{m i n} = 723.666 \dots \approx 724 m^{2} \frac{d ^{2} A}{d x ^{2}} = 12 x + \frac{12000}{x ^{4}} + \frac{4000}{x ^{3}} \frac{d ^{2} A}{d x ^{2}}_{x = k} = 120.507 \dots \approx 121 121 > 0 ⟹ A is a minimum at x = k

Mentor tip

For (iii), state the sign and the conclusion together: the value is positive, therefore a minimum. The number alone is not a justification.

Question 8

14 marks

Normal waiting times and a binomial count

Statistics and probability

(a)

7 marks

(i) Convert each percentage to a $z$ -score with inverse normal: the lower tail $0.038$ and the upper tail $0.073$ (a left area of $0.927$ ).

Write $z = \frac{x - μ}{σ}$ for each and solve the pair of linear equations.

(ii) Use the normal CDF with your stored $μ$ and $σ$ .

z_{1} = - 1.77438 \dots, z_{2} = 1.45380 \dots 20 - μ = - 1.77438 \dots σ, 30 - μ = 1.45380 \dots σ μ = 25.4965 \dots \approx 25.5, σ = 3.09771 \dots \approx 3.10 P (15 < T < 22) = 0.129152 \dots \approx 0.129 (GDC)

Another way. Subtracting the two equations directly gives $σ = \frac{10}{1.77438 \dots + 1.45380 \dots}$ .

Mentor tip

Solve with $z$ -scores, not the percentages themselves. A quick sketch of the curve with both tails shaded keeps the signs right.

(b)

7 marks

Let $Y$ be the number who attend, so $Y \sim B (20, 0.826)$ .
(i) At least $13$ means $1 - P (Y \leq 12)$ .
(ii) More than $2$ absent means fewer than $18$ attend, so $Y \leq 17$ .
Conditional probability: the overlap of the two events divided by the given event.

P (Y \geq 13) = 1 - P (Y \leq 12) = 0.985473 \dots \approx 0.985 P (Y \leq 17 ∣ Y \geq 13) = \frac{P ( 13 \leq Y \leq 17 )}{P ( Y \geq 13 )} = \frac{0.687241 \dots}{0.985473 \dots} = 0.697372 \dots \approx 0.697

Mentor tip

Translate the non-attenders into attenders before you type anything: more than $2$ absent out of $20$ is at most $17$ present.

Question 9

15 marks

Geometric series, squares and a log sequence

Number and algebra

(a)

1 mark

Rearrange the sum to infinity formula with $A_{\infty} = 5$ .

\frac{a _{1}}{1 - r} = 5 ⟹ a_{1} = 5 (1 - r)

(b)

2 marks

Divide the second term of the new sequence by the first, writing $a_{2} = r a_{1}$ .

q = \frac{a _{2}^{2}}{a _{1}^{2}} = \frac{( r a _{1} ) ^{2}}{a _{1}^{2}} q = r^{2}

Mentor tip

Show the substitution $a_{2} = r a_{1}$ explicitly; that line carries the whole argument.

(c)

5 marks

(i) Use $B_{\infty} = \frac{a _{1}^{2}}{1 - q} = 16$ with $a_{1} = 5 (1 - r)$ and $q = r^{2}$ .

Expand and collect everything on one side.

(ii) Solve the quadratic. $r = 1$ makes the series diverge, so reject it.

(iii) Substitute into $a_{1} = 5 (1 - r)$ .

\frac{25 ( 1 - r ) ^{2}}{1 - r ^{2}} = 16 25 (1 - 2 r + r^{2}) = 16 - 16 r^{2} 41 r^{2} - 50 r + 9 = 0 (41 r - 9) (r - 1) = 0 ⟹ r = \frac{9}{41} \approx 0.220 a_{1} = 5 (1 - \frac{9}{41}) = \frac{160}{41} \approx 3.90

Another way. Cancel a factor: $1 - r^{2} = (1 - r) (1 + r)$ , so $\frac{25 ( 1 - r )}{1 + r} = 16$ gives $r = \frac{9}{41}$ straight away.

Mentor tip

Say why $r = 1$ is rejected: a sum to infinity needs $∣ r ∣ < 1$ .

(d)

4 marks

Find $c_{n + 1} - c_{n}$ in general.
The constant $k$ cancels, and the log laws turn the difference of logs into the log of a quotient.
That quotient is the common ratio $r$ , a constant, so the difference is constant.

c_{n + 1} - c_{n} = (lo g_{k} a_{n + 1} + k) - (lo g_{k} a_{n} + k) = lo g_{k} \frac{a _{n + 1}}{a _{n}} = lo g_{k} r lo g_{k} r is constant ⟹ arithmetic

Mentor tip

Finish with a sentence: the difference does not depend on $n$ , so the sequence is arithmetic. Checking only the first few terms is not a proof.

(e)

3 marks

The first term is $lo g_{k} a_{1} + k$ and the common difference is $lo g_{k} r$ .
Use $S_{20} = \frac{20}{2} (2 c_{1} + 19 d)$ with $a_{1} = \frac{160}{41}$ and $r = \frac{9}{41}$ .
Graph $C_{20}$ against $k$ and find where it crosses zero, then take the largest integer below it.

C_{20} = 10 (2 lo g_{k} \frac{160}{41} + 2 k + 19 lo g_{k} \frac{9}{41}) C_{20} = 0 ⟹ k = 6.80296 \dots (GDC) k \in Z^{+} ⟹ k = 6

Another way. Use $lo g_{k} x = \frac{l n x}{l n k}$ to enter the expression on a GDC that has no base- $k$ log.

Mentor tip

Remember $k$ is a positive integer. The crossing at $6.80$ is not the answer; check $C_{20} (6) < 0$ and $C_{20} (7) > 0$ .

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 probability25 marks · 31%
Q1, Q4, Q8
Geometry and trigonometry19 marks · 24%
Q2, Q5, Q7
Number and algebra15 marks · 19%
Q9
Calculus12 marks · 15%
Q6, Q7
Functions9 marks · 11%
Q3, Q7

Mentor insights

Traps to watch for

01
Perimeter of a sector · Q5
The perimeter is the arc plus two radii. Using the arc alone gives the wrong equation, and you must then keep only the acute root.
02
Which zero of v gives the maximum · Q6
Velocity is zero twice. Integrate to both and compare with the end point; the first zero is actually the lowest displacement.
03
Justifying a minimum · Q7
Give the sign of the second derivative and the conclusion it leads to. A number on its own is not a justification.
04
Translating non-attenders · Q8
More than 2 absent means at most 17 present. Getting this inequality wrong spoils the whole conditional probability.
05
k must be an integer · Q9
The GDC gives a crossing at about 6.80, but k is a positive integer, so the largest value that works is 6.

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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