Worked solutions

IB Maths AA HL · Paper 2

Mathematics: Analysis & Approaches, Higher Level

Timezone

Session: May 2026
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 12

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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Section A · Short response

Questions 1–9 · 55 marks

Question 1

6 marks

Summary statistics from a frequency table

Statistics and probability

(a)

1 mark

The mode is the value with the biggest frequency. Read down the frequency column and pick the largest, $15$ .

mode = 3

Mentor tip

The mode is a data value, not a frequency. Writing $15$ is the classic slip.

(b)

3 marks

Put the values in one GDC list and the frequencies in another, then run one-variable statistics with the frequency list.
(i) The total is $n = 60$ and $\sum f x = 183$ .
(ii) With $60$ values, the median sits between the $30$ th and $31$ st. Both are $3$ .
(iii) Use the population standard deviation, $σ$ , from the GDC.

\overset{x}{ˉ} = \frac{183}{60} = 3.05 median = 3 σ = \frac{719}{60} - 3.0 5^{2} = 1.63732 \dots \approx 1.64 (GDC)

Mentor tip

Forgetting to attach the frequency list is the usual GDC error: you get statistics for the eight values $0$ to $7$ instead of all $60$ students. Check that $n = 60$ on screen.

(c)

2 marks

You need five numbers: minimum, lower quartile, median, upper quartile and maximum.
The lower half is the first $30$ values, so $Q_{1}$ is the average of the $15$ th and $16$ th values, both $2$ . Likewise $Q_{3}$ is the average of the $45$ th and $46$ th values, both $4$ .
Draw the box from $2$ to $4$ with a vertical line at $3$ , and whiskers out to $0$ and $7$ .

min = 0, Q_{1} = 2, median = 3, Q_{3} = 4, max = 7

Mentor tip

Use a ruler and line the ends up with the printed scale. Mark the median clearly inside the box; it is easy to leave out.

Question 2

6 marks

Minimising the surface area of a cylinder

Geometry and trigonometry
Calculus

(a)

1 mark

Write the cylinder's volume formula, set it equal to $500$ and make $h$ the subject.

π r^{2} h = 500 h = \frac{500}{π r ^{2}}

Mentor tip

On a one-mark show-that, the key is writing $π r^{2} h = 500$ before the rearranged form, not just the final line.

(b)

2 marks

A closed cylinder has two circular ends and a curved side, so start from $A = 2 π r^{2} + 2 π r h$ .
Replace $h$ with your result from part (a) and simplify.

A = 2 π r^{2} + 2 π r h A = 2 π r^{2} + 2 π r \times \frac{500}{π r ^{2}} A = 2 π r^{2} + \frac{1000}{r}

Mentor tip

Show the substitution line with $\frac{500}{π r ^{2}}$ written in. Jumping straight to the given answer skips the step that matters.

(c)

3 marks

Graph $A = 2 π r^{2} + \frac{1000}{r}$ for $r > 0$ on the GDC and find the minimum point.
(i) The $y$ -coordinate of the minimum is the smallest area.
(ii) The $x$ -coordinate is the radius that gives it.

minimum at (4.30127 \dots, 348.734 \dots) (GDC) A_{m i n} \approx 349 cm^{2} r \approx 4.30 cm

Another way. Differentiate: $A^{'} (r) = 4 π r - \frac{1000}{r ^{2}} = 0$ gives $r^{3} = \frac{250}{π}$ , so $r = 4.30$ .

Mentor tip

Include units, and make a quick sketch of the graph with the minimum marked when you rely on the GDC.

Question 3

5 marks

Kinematics with an exponential velocity

Calculus

(a)

1 mark

Initial means $t = 0$ . Substitute and remember $e^{0} = 1$ .

v (0) = 10 (0 - e^{0}) = - 10 m s^{- 1}

Mentor tip

The answer is negative: the particle starts off moving in the negative direction. Keep the sign.

(b)

2 marks

Distance is the integral of the speed, $∣ v ∣$ , not of $v$ , because the velocity changes sign during the first five seconds.
Let the GDC evaluate the integral of the absolute value from $0$ to $5$ .

distance = \int_{0}^{5} 10 (\frac{t ^{2}}{5} - e^{- 0.2 t}) d t = 74.2121 \dots \approx 74.2 m (GDC)

Mentor tip

Without the modulus the negative stretch at the start cancels part of the positive stretch and you get displacement instead of distance.

(c)

2 marks

Acceleration is $\frac{d v}{d t}$ . Differentiate, or use the GDC's derivative function.
Set it equal to $19.3$ and solve with the GDC (graph the derivative and the line $y = 19.3$ ).

a = \frac{d v}{d t} = 4 t + 2 e^{- 0.2 t} 4 t + 2 e^{- 0.2 t} = 19.3 t = 4.62680 \dots \approx 4.63 s (GDC)

Mentor tip

There is only one solution with $t \geq 0$ . Give just that one; listing extra roots makes the answer ambiguous.

Question 4

6 marks

Binomial distribution: mean, spread, conditioning

Statistics and probability

(a)

3 marks

For $B (n, p)$ the mean is $n p$ and the standard deviation is $n p (1 - p)$ . Put in $n = 15$ and set mean $= 3 \times$ standard deviation.
Square both sides, then divide by $15 p$ (allowed because $p \neq = 0$ ).

15 p = 3 15 p (1 - p) 225 p^{2} = 135 p (1 - p) 225 p = 135 - 135 p ⟹ 360 p = 135 p = \frac{3}{8} = 0.375

Mentor tip

Squaring can bring in the root $p = 0$ . The paper rules that out, so say why you discard it.

(b)

3 marks

Use $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ . Since $9 > 6$ , the event $X = 9$ sits inside $X > 6$ , so the top is just $P (X = 9)$ .
For the bottom, $P (X > 6) = P (X \geq 7) = 1 - P (X \leq 6)$ .
Get both values from the GDC's binomial functions with $n = 15$ , $p = 0.375$ .

P (X = 9 ∣ X > 6) = \frac{P ( X = 9 )}{P ( X \geq 7 )} = \frac{0.0437487 \dots}{0.314775 \dots} (GDC) \approx 0.139

Mentor tip

For a discrete variable, $X > 6$ means $X \geq 7$ . Using $1 - P (X \leq 7)$ by mistake drops the value $7$ .

Question 5

5 marks

Volume of revolution about the y-axis

Calculus

5 marks

Rotating about the $y$ -axis needs $V = π \int x^{2} d y$ , so rearrange to get $x$ in terms of $y$ .
The $y$ -limits come from the two points: $y = 4$ to $y = 10$ .
Expand $x^{2}$ , integrate term by term (or use the GDC) and evaluate.

y = \frac{20}{x + 1} ⟹ x = \frac{20}{y} - 1 V = π \int_{4}^{10} (\frac{20}{y} - 1)^{2} d y = π \int_{4}^{10} (\frac{400}{y ^{2}} - \frac{40}{y} + 1) d y = π [- \frac{400}{y} - 40 ln y + y]_{4}^{10} = π (66 - 40 ln \frac{5}{2}) V = 92.2006 \dots \approx 92.2

Mentor tip

Using the $x$ -values $1$ and $4$ as limits is the common error. The integral is in $y$ , so the limits must be $y$ -values too. Give the final answer as a decimal to 3 s.f.

Question 6

5 marks

A trigonometric probability density function

Statistics and probability

(a)

2 marks

The total probability must be $1$ , so integrate the density over $0 \leq x \leq π$ and set the result to $1$ .

\frac{1}{a} \int_{0}^{π} (1 - cos 2 x) d x = 1 \frac{1}{a} [x - \frac{1}{2} sin 2 x]_{0}^{π} = \frac{π}{a} = 1 a = π

Mentor tip

The question wants an exact value, so do the integral by hand rather than relying on a decimal from the GDC.

(b)

3 marks

The graph of $1 - cos 2 x$ is symmetric about $x = \frac{π}{2}$ , so $E (X) = \frac{π}{2}$ .
Use $Var (X) = E (X^{2}) - [E (X)]^{2}$ , finding $E (X^{2})$ with the GDC.

E (X) = \frac{π}{2} E (X^{2}) = \int_{0}^{π} x^{2} \cdot \frac{1 - cos 2 x}{π} d x = 2.78986 \dots (GDC) Var (X) = 2.78986 \dots - (\frac{π}{2})^{2} = 0.322467 \dots \approx 0.322

Another way. Integrate $(x - \frac{π}{2})^{2} f (x)$ over $[0, π]$ directly; the exact value is $\frac{π ^{2}}{12} - \frac{1}{2}$ .

Mentor tip

Do not forget to subtract $[E (X)]^{2}$ . Stopping at $E (X^{2}) \approx 2.79$ is a frequent slip.

Question 7

6 marks

Three planes meeting in a line

Number and algebra

(a)

3 marks

Eliminate one variable in two different ways. Adding the first two equations removes $z$ .
Adding the second and third also removes $z$ . For infinitely many solutions the two results must be the same equation.

(1) + (2) : 4 x - 4 y = 4 ⟹ x - y = 1 (2) + (3) : 6 x - 6 y = 3 + k ⟹ x - y = \frac{3 + k}{6} \frac{3 + k}{6} = 1 ⟹ k = 3

Another way. Row-reduce the augmented matrix: the last row becomes all zeros apart from $k - 3$ on the right, which must be $0$ .

Mentor tip

Check it: with $k = 3$ , equation (3) equals $\frac{3}{2}$ of (1) plus $\frac{1}{2}$ of (2), so it adds no new information.

(b)

3 marks

(i) Infinitely many solutions to three planes, none of them parallel, means they share a common line (a sheaf).

(ii) Let $z = λ$ . From $x - y = 1$ and equation (1), solve for $x$ and $y$ in terms of $λ$ .

(i) the three planes meet in a single line 2 x - y = 1 - λ, x - y = 1 x = - λ, y = - 1 - λ, z = λ

Mentor tip

Substitute your parametric answer back into all three equations. It takes seconds and catches sign errors.

Question 8

8 marks

An odd rational function from its turning points

Functions
Calculus

(a)

2 marks

Replace $x$ with $- x$ and simplify. Since $(- x)^{2} = x^{2}$ , the denominator does not change.

f (- x) = \frac{a ( - x )}{b ( - x ) ^{2} + c} = \frac{- a x}{b x ^{2} + c} = - f (x) so f is odd

Mentor tip

End with the statement $f (- x) = - f (x)$ written in full. That line is the whole point of the show-that.

(b)

6 marks

(i) Differentiate with the quotient rule and simplify the numerator.

The turning point at $x = 1$ gives $f^{'} (1) = 0$ . Since $a \neq = 0$ , this forces $c = b$ .

The point $(1, 2)$ lies on the curve: substitute with $c = b$ to get $a = 4 b$ .

(ii) Replace $a$ with $4 b$ and $c$ with $b$ , then cancel the $b$ .

f^{'} (x) = \frac{a ( b x ^{2} + c ) - a x ( 2 b x )}{( b x ^{2} + c ) ^{2}} = \frac{a ( c - b x ^{2} )}{( b x ^{2} + c ) ^{2}} f^{'} (1) = 0 ⟹ a (c - b) = 0 ⟹ c = b f (1) = \frac{a}{b + b} = 2 ⟹ a = 4 b f (x) = \frac{4 b x}{b x ^{2} + b} = \frac{4 x}{x ^{2} + 1}

Mentor tip

Say why you can divide by $a$ : it is given as positive, so it cannot be zero. Show-that answers need that reasoning written down.

Question 9

8 marks

Integrating sec x with partial fractions

Geometry and trigonometry
Calculus

(a)

2 marks

Multiply out the denominator as a difference of two squares, then use $1 - sin^{2} x = cos^{2} x$ .

\frac{cos x}{( 1 - sin x ) ( 1 + sin x )} = \frac{cos x}{1 - sin ^{2} x} = \frac{cos x}{cos ^{2} x} = \frac{1}{cos x} = sec x

Mentor tip

Work on one side only and finish with $sec x$ . Moving terms across both sides is harder to follow.

(b)

6 marks

Write $sec x$ using part (a). With $u = sin x$ you have $d u = cos x d x$ , which soaks up the $cos x$ on top.
Split $\frac{1}{( 1 - u ) ( 1 + u )}$ into partial fractions.
Integrate each piece into a logarithm and combine them.
Put $u = sin x$ back and add the constant.

\int sec x d x = \int \frac{cos x}{( 1 - sin x ) ( 1 + sin x )} d x = \int \frac{d u}{( 1 - u ) ( 1 + u )} \frac{1}{( 1 - u ) ( 1 + u )} = \frac{1}{2 ( 1 + u )} + \frac{1}{2 ( 1 - u )} \int (\frac{1}{2 ( 1 + u )} + \frac{1}{2 ( 1 - u )}) d u = \frac{1}{2} ln ∣1 + u ∣ - \frac{1}{2} ln ∣1 - u ∣ + c = \frac{1}{2} ln \frac{1 + sin x}{1 - sin x} + c

Mentor tip

The integral of $\frac{1}{1 - u}$ is $- ln ∣1 - u ∣$ : the minus sign from the inside is easy to lose. Finish with both the modulus and $+ c$ .

Section B · Extended response

Questions 10–12 · 55 marks

Question 10

15 marks

Normal distributions and an unknown mean and SD

Statistics and probability

(a)

6 marks

Let $X \sim N (173, 2 1^{2})$ and use the GDC's normal cdf for (i) and (ii).
(iii) This is conditional: the event 'between $180$ and $200$ ' is already inside 'more than $180$ ', so divide (ii) by $P (X > 180)$ .

(i) P (X < 180) = 0.630558 \dots \approx 0.631 (ii) P (180 < X < 200) = 0.270169 \dots \approx 0.270 (iii) \frac{P ( 180 < X < 200 )}{P ( X > 180 )} = \frac{0.270169 \dots}{1 - 0.630558 \dots} = 0.731293 \dots \approx 0.731

Mentor tip

Keep the full calculator values for the conditional step. Dividing rounded answers can move the third figure.

(b)

6 marks

Let $Y \sim N (μ, σ^{2})$ . Add the two given probabilities to get $P (Y < 180) = 0.85901$ .
Use inverse normal on the standard normal to turn each cumulative probability into a $z$ -score.
Write $z = \frac{y - μ}{σ}$ for $y = 170$ and $y = 180$ , and solve the two equations together.

P (Y < 180) = 0.62102 + 0.23799 = 0.85901 \frac{170 - μ}{σ} = 0.308160 \dots, \frac{180 - μ}{σ} = 1.07588 \dots μ = 165.986 \dots \approx 166 g, σ = 13.0255 \dots \approx 13.0 g

Mentor tip

The inverse normal needs a cumulative (left-tail) probability. Feeding it $0.23799$ on its own, the middle strip, gives nonsense.

(c)

3 marks

'At least one' is quickest through the complement: one minus the chance that neither potato is over $180$ g.
The two choices are independent, so multiply the two 'under $180$ ' probabilities.

P (neither over 180) = 0.630558 \dots \times 0.85901 P (at least one) = 1 - 0.630558 \dots \times 0.85901 = 0.458343 \dots \approx 0.458

Another way. Add the three cases directly: red only over, purple only over, and both over.

Mentor tip

For continuous weights, 'more than $180$ ' and 'at least $180$ ' give the same probability, so you can reuse the values from (a) and (b).

Question 11

21 marks

Vectors: a pyramid, a plane and skew lines

Geometry and trigonometry

(a)

5 marks

(i) Find $A B = B - A$ and $A C = C - A$ , then take their cross product.

(ii) The magnitude of the cross product is the area of the parallelogram, so halve it for the triangle.

A B = 1 - 3 2, A C = 3 - 1 0 A B \times A C = (- 3) (0) - (2) (- 1) (2) (3) - (1) (0) (1) (- 1) - (- 3) (3) = 268 area = \frac{1}{2} 2^{2} + 6^{2} + 8^{2} = \frac{1}{2} 104 = 26 \approx 5.10

Mentor tip

Forgetting the half is the common slip. The cross product's length is a parallelogram, not a triangle.

(b)

3 marks

Find M as the average of B and C.
Work out $A M$ , then go from A two-thirds of the way along it.

M = (\frac{2 + 4}{2}, \frac{- 1 + 1}{2}, \frac{5 + 3}{2}) = (3, 0, 4) A M = 2 - 2 1 O K = 123 + \frac{2}{3} 2 - 2 1 = 7/3 2/3 11/3

Mentor tip

$A K$ is a displacement, not a position. Add it to A's position vector to reach K.

(c)

1 mark

A line perpendicular to the plane runs along the normal $(1, 3, 4)$ , read straight from $x + 3 y + 4 z = 19$ . Start it at K.

r = 7/3 2/3 11/3 + s 134

Mentor tip

Write the full equation starting with $r =$ . A bare pair of vectors is not an equation.

(d)

3 marks

The height of the pyramid is the distance KH, because KH lies along the normal to the base.
Use volume $= \frac{1}{3} \times$ base area $\times$ height, with the base area from (a).

K H = 2 - 7/3 - 1/3 - 2/3 7/3 - 11/3 = - 1/3 - 1 - 4/3 ∣ K H ∣ = \frac{1}{3} 1 + 9 + 16 = \frac{26}{3} V = \frac{1}{3} \times 26 \times \frac{26}{3} = \frac{26}{9} \approx 2.89

Mentor tip

The volume formula for a pyramid has a $\frac{1}{3}$ in it. Leaving it out triples the answer.

(e)

3 marks

Find the angle between the direction of $L_{2}$ and the plane's normal using the dot product.
The angle with the plane is the complement: using $sin$ instead of $cos$ gives it directly.

1 - 3 - 2 \cdot 134 = 1 - 9 - 8 = - 16 sin α = \frac{∣ - 16∣}{14 26} = 0.83862 \dots α = 56.995 \dots^{\circ} \approx 5 7^{\circ}

Mentor tip

If you use $cos$ you get the angle with the normal ( $3 3^{\circ}$ or $14 7^{\circ}$ ). Subtract the acute one from $9 0^{\circ}$ , and round to the nearest degree as asked.

(f)

6 marks

Show the lines are not parallel: the direction vectors $(1, 3, 4)$ and $(1, - 3, - 2)$ are not multiples of each other.
Set the general points of $L_{1}$ and $L_{2}$ equal to get three equations in $s$ and $t$ .
Solve two of them together, then test the third. A contradiction means the lines never meet.
Not parallel and not meeting means skew.

\frac{7}{3} + s = 1 + t, \frac{2}{3} + 3 s = 2 - 3 t, \frac{11}{3} + 4 s = 3 - 2 t first two: s = - \frac{4}{9}, t = \frac{8}{9} third: \frac{11}{3} + 4 (- \frac{4}{9}) = \frac{17}{9}, 3 - 2 (\frac{8}{9}) = \frac{11}{9} \frac{17}{9} \neq = \frac{11}{9} ⟹ no intersection not parallel and not intersecting ⟹ L_{1}, L_{2} are skew

Mentor tip

Skew needs both conditions. Showing only that the lines do not meet leaves open the possibility that they are parallel.

Question 12

19 marks

Transformed arctan: inverse, tangent and area

Functions
Calculus

(a)

3 marks

Arctan runs between $- \frac{π}{2}$ and $\frac{π}{2}$ , so $2 arctan (x - 1)$ runs between $- π$ and $π$ . Adding $\frac{π}{2}$ gives the horizontal asymptotes.
Find a point to pin it down: $f (0) = 2 arctan (- 1) + \frac{π}{2} = 0$ , so it goes through the origin.
Describe the sketch: an increasing S-shaped curve, flattening towards $y = - \frac{π}{2}$ on the left and $y = \frac{3 π}{2}$ on the right, passing through $(0, 0)$ and centred on $(1, \frac{π}{2})$ .

y = - \frac{π}{2} and y = \frac{3 π}{2} (horizontal asymptotes) f (0) = - \frac{π}{2} + \frac{π}{2} = 0

Mentor tip

Label both asymptotes with their equations, and draw the curve getting close to them without touching or crossing.

(b)

4 marks

(i) Work from the inside out: $x - 1$ is a horizontal translation, the $2$ is a vertical stretch, and $+ \frac{π}{2}$ is a vertical translation.

The order matters for the vertical shift: if you translate before stretching, the shift must be $\frac{π}{4}$ instead.

(ii) The point of inflexion of $arctan x$ is the origin. Follow it through the transformations.

translate 1 unit right, then stretch vertically by factor 2, then translate up \frac{π}{2} (0, 0) \mapsto (1, 0) \mapsto (1, 0) \mapsto (1, \frac{π}{2}) point of inflexion (1, \frac{π}{2})

Mentor tip

Name each transformation properly, with its size and direction: 'translation', 'stretch' and 'parallel to the $y$ -axis'. Vague words like 'move' or 'shrink' do not describe the change precisely.

(c)

6 marks

(i) Swap $x$ and $y$ , then make $y$ the subject step by step: subtract $\frac{π}{2}$ , divide by $2$ , take tan, add $1$ .

The domain of $f^{- 1}$ is the range of $f$ , which comes from the asymptotes in (a).

(ii) The graphs of $f$ and $f^{- 1}$ meet on the line $y = x$ here, so solve $f (x) = x$ on the GDC and take the solution with $x > 0$ .

x = 2 arctan (y - 1) + \frac{π}{2} arctan (y - 1) = \frac{2 x - π}{4} f^{- 1} (x) = tan (\frac{2 x - π}{4}) + 1, - \frac{π}{2} < x < \frac{3 π}{2} f (x) = x ⟹ x = 4.08557 \dots (GDC) intersection (4.09, 4.09)

Mentor tip

Use strict inequalities in the domain: the asymptote values are never reached. And give only the positive intersection, as the part asks.

(d)

2 marks

Differentiate: $f^{'} (x) = \frac{2}{1 + ( x - 1 ) ^{2}}$ , so the gradient at $x = 3$ is $\frac{2}{5}$ .
Find $f (3)$ on the GDC and use $y - y_{1} = m (x - x_{1})$ .

f^{'} (3) = \frac{2}{1 + 4} = 0.4 f (3) = 2 arctan 2 + \frac{π}{2} = 3.78509 \dots y - 3.78509 \dots = 0.4 (x - 3) ⟹ y = 0.4 x + 2.59

Mentor tip

Carry the unrounded intercept into part (e). Rounding it to $2.59$ too early shifts the intersection point slightly.

(e)

4 marks

(i) Solve $f (x) = 0.4 x + 2.58509 \dots$ on the GDC. One solution is $x = 3$ (where the tangent touches); the other is P.

(ii) Integrate the gap between the two graphs from P to $x = 3$ , using the absolute value so the order of subtraction does not matter.

P = (- 9.93369 \dots, - 1.38838 \dots) \approx (- 9.93, - 1.39) (GDC) area = \int_{- 9.93369 \dots}^{3} ∣ f (x) - (0.4 x + 2.58509 \dots) ∣ d x = 19.9265 \dots \approx 19.9 (GDC)

Mentor tip

Store P's $x$ -value in the calculator and use it as the lower limit. Typing $- 9.93$ is fine for the answer but loses accuracy in the integral.

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 probability32 marks · 29%
Q1, Q4, Q6, Q10
Calculus31 marks · 28%
Q2, Q3, Q5, Q8, Q9, Q12
Geometry and trigonometry26 marks · 24%
Q2, Q9, Q11
Functions15 marks · 14%
Q8, Q12
Number and algebra6 marks · 5%
Q7

Mentor insights

Traps to watch for

01
Distance is not displacement · Q3
The velocity is negative at the start, so integrate |v|. Integrating v alone lets the backwards stretch cancel part of the journey.
02
Limits in the wrong variable · Q5
Rotating about the y-axis means integrating in y, so the limits are the y-values 4 and 10, not the x-values 1 and 4.
03
Variance needs the mean squared off · Q6
E(X²) is not the variance. Subtract (π/2)² to finish.
04
Inverse normal wants a left tail · Q10
Turn 'between 170 and 180' into P(Y < 180) = 0.85901 before using inverse normal; the middle strip on its own gives a meaningless z-score.
05
Skew needs two checks · Q11
Show the lines do not meet and that their directions are not parallel. One without the other is not enough.

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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