HL only · Paper 3

IB Maths AA HL Paper 3: guide and practice questions

Q: How long is IB Maths AA HL Paper 3?

1 hour 15 minutes, with a graphic display calculator (GDC) and the formula booklet. It is two compulsory extended-response problem-solving questions and counts for 20% of your HL grade.

Q: Can I use a calculator in Paper 3?

Yes. Paper 3 is a technology-allowed paper, so you need your GDC. Use it to explore cases, check answers and do heavy calculations, but write down what you did.

Q: How do I revise for Paper 3?

Learn the four techniques on this page, then work through investigation-style questions slowly with the solutions. Once that feels natural, sit full Paper 3s timed and mark them against the markscheme.

Q: Is Paper 3 changing?

For the new courses (first exams in May 2029), the IB has announced that HL Paper 3 will be 1 hour instead of 1 hour 15 minutes. Exams before then keep the current format.

Q: Are these official IB questions?

No. Every question on this page is our own, written in the style of Paper 3. This site is independent and is not connected to the IB.

Paper 3 is the HL paper that tests problem-solving rather than recall: two long questions that each take one new idea and lead you through it. This page covers the format, how it is marked and the techniques that score, then gives you worked investigations and full-length practice.

By Pete Bromfield, former Head of Maths with IB examining experience (independent) · Updated 25 October 2026

Time1 hour 15 minutes
QuestionsTwo, both compulsory
CalculatorGDC needed
Weight20% of your HL grade

Try a worked investigation Full Paper 3s

What Paper 3 is

Paper 3 is sat by HL students only, after Papers 1 and 2. It has two compulsory extended-response questions, and you have the formula booklet and your GDC throughout. Each question starts with something short and concrete, then asks you to spot a pattern, make a conjecture, prove or justify it, and extend it. Any topic in the course can come up, including the SL core, and one question can join several: calculus with sequences, say, or proof with complex numbers.

AA HL paper	Time	Calculator	Weight
Paper 1	2 hours	No calculator	30%
Paper 2	2 hours	Calculator (GDC)	30%
Paper 3	1 hour 15 minutes	Calculator (GDC)	20%
Internal assessment	Coursework	–	20%

Our full Paper 3s are 55 marks over the two questions. From the first exams of the new courses in May 2029, the IB has announced HL Paper 3 will be 1 hour (what is changing). More on every paper: AA HL past papers · AA HL formula booklet guide.

How Paper 3 is marked

Paper 3 uses the same kinds of marks as Papers 1 and 2, but more of them reward reasoning, because the questions ask you to explain and prove. Knowing what each mark is for tells you what to write.

M: method
For a valid method, shown in your working. You can earn it even if your answer goes wrong later, so always write the method down.
A: accuracy
For a correct answer or step, usually only after the method mark before it. A bare answer with no working can score nothing.
R: reasoning
For explaining why: a justification in words, a reason a step is valid, a concluding sentence. Paper 3 has many of these.
AG: answer given
On "show that" parts the answer is printed, so the marks are for the working that reaches it. If you cannot show it, use the given result in the next part anyway.

An early slip is usually penalised once if you then use your wrong value correctly (follow through). More: how IB Maths mark schemes work · where AA HL students lose marks.

The techniques that score

Spot the pattern

Work out the first three or four cases exactly and put them in a table. Leave fractions and powers unsimplified, factorise, and look at differences or ratios: the pattern is easier to see before you tidy it.

Conjecture, then prove

Write the conjecture as a formula with its domain (for example $n \in Z^{+}$ ). Then prove it: by induction, by direct algebra, or with an earlier part. An induction proof needs the base case, the assumption, the step and the concluding sentence.

Use your GDC

Graph to count roots and see behaviour, use the table or sequence mode to generate many cases, and check an answer numerically. The GDC finds the pattern; the proof still needs algebra.

Communicate

Define any letter you introduce, say which earlier result you use ("from part (b)") and finish with a sentence that answers the question.

When you are stuck: read to the end of the question to see where it is heading, use any given result and move on to the next part; later parts are often easier than the one before. Timing: about half the time for each question.

Two worked mini-investigations

Short Paper 3-style questions, with the mark codes on each solution so you can see where the marks are. Try each part before you open its solution. Every answer was checked.

Investigation 1: Repeated derivatives of $x e^{2 x}$

Spot the pattern
Conjecture, then prove
Use your result

Let $f (x) = x e^{2 x}$ , and write $f^{(n)} (x)$ for its $n$ th derivative.

(a) Find $f^{'} (x)$ , $f^{″} (x)$ and $f^{‴} (x)$ , writing each in the form $(a x + b) e^{2 x}$ . [3]

Worked solution (a)

Use the product rule each time.

$f^{'} (x) = e^{2 x} + 2 x e^{2 x} = (2 x + 1) e^{2 x}$ A1

$f^{″} (x) = 2 e^{2 x} + 2 (2 x + 1) e^{2 x} = (4 x + 4) e^{2 x}$ A1

$f^{‴} (x) = 4 e^{2 x} + 2 (4 x + 4) e^{2 x} = (8 x + 12) e^{2 x}$ A1

(b) Conjecture an expression for $f^{(n)} (x)$ , where $n \in Z^{+}$ . [2]

Worked solution (b)

Look at the two numbers in each bracket separately. The coefficients of $x$ are 2, 4, 8: they double, so they are $2^{n}$ . The constants are 1, 4, 12, which are $1 \times 1$ , $2 \times 2$ , $4 \times 3$ : that is $2^{n - 1} \times n$ . M1

So $f^{(n)} (x) = (2^{n} x + n 2^{n - 1}) e^{2 x} = 2^{n - 1} (2 x + n) e^{2 x}$ . A1

Put the results in a table before you guess. Taking out the factor $2^{n - 1}$ makes the pattern easy to see.

(c) Prove your conjecture by mathematical induction. [5]

Worked solution (c)

Let $P (n)$ be the statement $f^{(n)} (x) = 2^{n - 1} (2 x + n) e^{2 x}$ .

$n = 1$ : $2^{0} (2 x + 1) e^{2 x} = f^{'} (x)$ , from (a), so $P (1)$ is true. A1

Assume $P (k)$ is true for some $k \in Z^{+}$ : $f^{(k)} (x) = 2^{k - 1} (2 x + k) e^{2 x}$ . M1

Then $f^{(k + 1)} (x) = \frac{d}{d x} [2^{k - 1} (2 x + k) e^{2 x}] = 2^{k - 1} [2 e^{2 x} + 2 (2 x + k) e^{2 x}]$ M1

$= 2^{k} (2 x + k + 1) e^{2 x} = 2^{(k + 1) - 1} (2 x + (k + 1)) e^{2 x}$ , which is $P (k + 1)$ . A1

$P (1)$ is true, and $P (k)$ true implies $P (k + 1)$ true, so $P (n)$ is true for all $n \in Z^{+}$ by mathematical induction. R1

The last sentence earns a mark of its own. Write it out in full.

(d) Hence show that the graph of $y = f^{(n)} (x)$ has exactly one stationary point, and find its coordinates in terms of $n$ . [4]

Worked solution (d)

The gradient of $y = f^{(n)} (x)$ is $f^{(n + 1)} (x) = 2^{n} (2 x + n + 1) e^{2 x}$ , from (c) with $n + 1$ in place of $n$ . M1

Since $2^{n} e^{2 x} > 0$ , the gradient is zero only when $2 x + n + 1 = 0$ . So there is exactly one stationary point, at $x = - \frac{n + 1}{2}$ . R1 A1

$y = 2^{n - 1} (- (n + 1) + n) e^{- (n + 1)} = - 2^{n - 1} e^{- (n + 1)}$ A1

The stationary point is $(- \frac{n + 1}{2}, - 2^{n - 1} e^{- (n + 1)})$ .

Check it on your GDC. For $n = 2$ , graph $y = 4 (x + 1) e^{2 x}$ : the minimum is at $(- 1.5, - 0.0996)$ , and $- 2 e^{- 3} \approx - 0.0996$ .

What this shows: Three small cases, a pattern, a proof, then a result that uses the proof: the shape of many Paper 3 questions. "Hence" in (d) means the formula from (c) does the work. 14 marks

Investigation 2: How many real roots does $1 + x + \frac{x^{2}}{2!} + \dots + \frac{x^{n}}{n!}$ have?

Use your GDC
Spot the pattern
Conjecture, then prove

For $n \in N$ let $p_{n} (x) = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots + \frac{x^{n}}{n!}$ , so $p_{0} (x) = 1$ , $p_{1} (x) = 1 + x$ , and so on.

(a) Use your GDC to find the number of real solutions of $p_{n} (x) = 0$ for $n = 1, 2, 3, 4, 5, 6$ . [3]

Worked solution (a)

Graph each $p_{n}$ and count where it crosses the $x$ -axis:

$n$	1	2	3	4	5	6
Real roots	1	0	1	0	1	0
Root	$- 1$	–	$- 1.60$	–	$- 2.18$	–

A1 A1 A1 For $n = 2$ you can confirm it without the GDC: $p_{2} (x) = \frac{1}{2} x^{2} + x + 1$ has discriminant $1 - 2 = - 1 < 0$ .

(b) Make a conjecture about the number of real roots of $p_{n} (x) = 0$ . [1]

Worked solution (b)

$p_{n} (x) = 0$ has exactly one real root when $n$ is odd, and no real roots when $n$ is even. A1

(c) Show that $p_{n}^{'} (x) = p_{n - 1} (x)$ for $n \geq 1$ . [2]

Worked solution (c)

Differentiate term by term. For $r \geq 1$ , $\frac{d}{d x} (\frac{x^{r}}{r!}) = \frac{r x^{r - 1}}{r!} = \frac{x^{r - 1}}{(r - 1)!}$ , and the constant 1 differentiates to 0. M1

So $p_{n}^{'} (x) = 1 + x + \dots + \frac{x^{n - 1}}{(n - 1)!} = p_{n - 1} (x)$ . A1 AG

(d) Prove your conjecture. [6]

Worked solution (d)

Each case needs a fact about the one before, so prove a slightly stronger statement by induction. Let $P (n)$ be: if $n$ is even, $p_{n} (x) > 0$ for all $x$ ; if $n$ is odd, $p_{n}$ is strictly increasing and $p_{n} (x) = 0$ has exactly one real root. M1

$n = 0$ : $p_{0} (x) = 1 > 0$ , so $P (0)$ is true. Now assume $P (n - 1)$ for some $n \geq 1$ .

$n$ odd. Then $n - 1$ is even, so $p_{n}^{'} (x) = p_{n - 1} (x) > 0$ for all $x$ : $p_{n}$ is strictly increasing. M1 It is a polynomial of odd degree with a positive leading coefficient, so $p_{n} (x) \to - \infty$ as $x \to - \infty$ and $p_{n} (x) \to \infty$ as $x \to \infty$ . A continuous, strictly increasing function with these limits crosses zero exactly once. R1

$n$ even. Now $n - 1$ is odd, so $p_{n}^{'} = p_{n - 1}$ is strictly increasing with exactly one root, $c$ . So $p_{n}^{'} (x) < 0$ for $x < c$ and $p_{n}^{'} (x) > 0$ for $x > c$ : the least value of $p_{n}$ is $p_{n} (c)$ . M1

$p_{n} (c) = p_{n - 1} (c) + \frac{c^{n}}{n!} = 0 + \frac{c^{n}}{n!}$ . Since $p_{n - 1} (0) = 1 \neq 0$ , $c \neq 0$ ; and $n$ is even, so $\frac{c^{n}}{n!} > 0$ . A1 Hence $p_{n} (x) \geq p_{n} (c) > 0$ for all $x$ . R1

So $P (n - 1)$ implies $P (n)$ . With $P (0)$ true, $P (n)$ is true for all $n \in N$ by induction, which proves the conjecture.

(e) Explain why every real root of $p_{n} (x) = 0$ is negative. [2]

Worked solution (e)

For $x \geq 0$ every term of $p_{n} (x)$ is at least 0 and the first term is 1, so $p_{n} (x) \geq 1 > 0$ . R1 So no root is $\geq 0$ : every real root is negative. A1

The GDC agrees: for $n = 1, 3, 5, 7$ the roots are $- 1$ , $- 1.60$ , $- 2.18$ and $- 2.76$ , moving further left as $n$ grows.

What this shows: The GDC finds the pattern; it cannot prove it. Part (c) is the key that unlocks (d), which is typical: an early "show that" is usually the tool for the hardest part. 14 marks

Full-length practice questions

Two full-length questions in our Paper 3 style: iteration, binary expansions and an induction proof; then a random walk solved with matrices, recurrences and limits. Together they make a full paper: sit them in 1 hour 15 minutes, then check each part.

Question 1 · The doubling map and chaotic dynamics

Topics: functions, iteration, sequences, logarithms · 28 marks

Define the doubling map $T : [0, 1) \to [0, 1)$ by

$T (x) = 2 x mod 1 = {\begin{cases} 2 x & if 0 \leq x < \frac{1}{2} \\ 2 x - 1 & if \frac{1}{2} \leq x < 1 \end{cases}$

The orbit of a starting point $x_{0}$ is the sequence $x_{0}, x_{1}, x_{2}, \dots$ where $x_{n + 1} = T (x_{n})$ . Throughout this question you may express numbers in binary using the notation $0. b_{1} b_{2} b_{3} \dots_{2}$ where each $b_{i} \in {0, 1}$ .

(a) [3 marks]

Compute the first five terms of the orbit starting at $x_{0} = \frac{1}{3}$ .

Reveal worked solution

Step 1. Apply $T$ repeatedly.

x_{1} = T (\frac{1}{3}) = \frac{2}{3}

(since

\frac{1}{3} < \frac{1}{2}

)

x_{2} = T (\frac{2}{3}) = 2 \cdot \frac{2}{3} - 1 = \frac{1}{3}

x_{3} = T (\frac{1}{3}) = \frac{2}{3}, x_{4} = \frac{1}{3}

Observation: the orbit is a period-2 cycle $\frac{1}{3} \to \frac{2}{3} \to \frac{1}{3} \to \dots$ (A1 · A1 · A1)

(b) [4 marks]

Let $x_{0} = 0. b_{1} b_{2} b_{3} \dots_{2}$ be a binary expansion. Prove that $x_{1} = T (x_{0}) = 0. b_{2} b_{3} b_{4} \dots_{2}$ .

Reveal worked solution

Setup. By definition, $x_{0} = \sum_{i = 1}^{\infty} b_{i} 2^{- i}$ .

2 x_{0} = \sum_{i = 1}^{\infty} b_{i} 2^{- (i - 1)} = b_{1} + \sum_{i = 2}^{\infty} b_{i} 2^{- (i - 1)} = b_{1} + 0. b_{2} b_{3} \dots_{2}

Case split.

If $b_{1} = 0$ : $x_{0} < \frac{1}{2}$ , so $x_{1} = 2 x_{0} = 0. b_{2} b_{3} \dots_{2}$ . (M1)
If $b_{1} = 1$ : $x_{0} \geq \frac{1}{2}$ , so $x_{1} = 2 x_{0} - 1 = b_{1} + 0. b_{2} b_{3} \dots_{2} - 1 = 0. b_{2} b_{3} \dots_{2}$ . (M1)

In both cases $x_{1} = 0. b_{2} b_{3} b_{4} \dots_{2}$ — the doubling map is a binary shift. (R1 · AG1)

Deduce that every rational $x_{0} \in (0, 1)$ has an eventually periodic orbit.

Reveal worked solution

Argument. A rational has a binary expansion that is either terminating or eventually periodic. (R1)

By part (b), $T$ shifts this expansion left by one bit each step, so after finitely many applications the tail is purely periodic; the orbit then repeats every $p$ steps where $p$ is the binary period. (R1 · A1)

(d) [5 marks]

Two starting points $x_{0} = 0.0101 {\overset{―}{01}}_{2}$ and $y_{0} = 0.0101 {\overset{―}{10}}_{2}$ differ from the 5th binary digit onwards. Prove that after $n$ applications of $T$ , $| x_{n} - y_{n} | \leq 2^{- (4 - n)}$ for $n \leq 4$ , and that the orbits diverge to be at least $\frac{1}{4}$ apart by step 4.

Reveal worked solution

Setup. Because the doubling map is a binary shift (part b), the difference in the first surviving bit is what matters after $n$ shifts.

Initial difference: agreeing on bits 1–4, differing at bit 5. So $| x_{0} - y_{0} | \leq 2^{- 4}$ . (M1)

After $n$ applications of $T$ , the leading disagreeing bit is at position $5 - n$ . In general, $| x_{n} - y_{n} |$ can double each step (approximately). (M1)

$n = 1$ : bits 4 disagree ⇒ $| x_{1} - y_{1} | \leq 2^{- 3} = \frac{1}{8}$
$n = 2$ : bits 3 disagree ⇒ $| x_{2} - y_{2} | \leq 2^{- 2} = \frac{1}{4}$
$n = 3$ : bits 2 disagree ⇒ $| x_{3} - y_{3} | \leq \frac{1}{2}$
$n = 4$ : bits 1 disagree; the numbers are macroscopically apart. (A1 · A1)

Sensitivity to initial conditions: tiny changes in the starting point ( $\leq 2^{- 4}$ ) grow to differences of $\geq \frac{1}{4}$ within 4 steps — a signature of chaos. (R1)

(e) [5 marks]

Prove by mathematical induction that for any $x_{0} \in [0, 1)$ whose binary expansion starts with $k$ zeros ( $b_{1} = b_{2} = \dots = b_{k} = 0$ ), $x_{k} = 2^{k} x_{0}$ .

Reveal worked solution

Base case ( $k = 0$ ): $x_{0} = 2^{0} x_{0} = x_{0}$ . Trivially true. (M1)

Inductive step. Assume $x_{j} = 2^{j} x_{0}$ for some $0 \leq j < k$ . Because bits $b_{1}, \dots, b_{k}$ are all $0$ , $x_{j}$ has its bit- $1$ position equal to $b_{j + 1} = 0$ , so $x_{j} < \frac{1}{2}$ . (M1)

Hence

x_{j + 1} = T (x_{j}) = 2 x_{j} = 2 \cdot 2^{j} x_{0} = 2^{j + 1} x_{0} .

Conclusion. By induction, $x_{k} = 2^{k} x_{0}$ for all $0 \leq k \leq$ (first non-zero bit position). (R1 · AG1 · A1)

(f) [4 marks]

The Lyapunov exponent of the doubling map at $x_{0}$ is defined as $λ (x_{0}) = lim_{n \to \infty} \frac{1}{n} \sum_{i = 0}^{n - 1} \ln | T^{'} (x_{i}) |$ wherever $T^{'}$ exists. Show that $λ (x_{0}) = \ln 2$ for every $x_{0}$ where the derivative is defined.

Reveal worked solution

Compute $T^{'}$ . On each of $[0, \frac{1}{2})$ and $[\frac{1}{2}, 1)$ , $T (x)$ is linear with slope $2$ . So $T^{'} (x) = 2$ wherever it exists. (M1 · A1)

Apply the definition.

λ (x_{0}) = lim_{n \to \infty} \frac{1}{n} \sum_{i = 0}^{n - 1} \ln 2 = lim_{n \to \infty} \frac{n \ln 2}{n} = \ln 2.

Interpretation. A positive Lyapunov exponent ( $\ln 2 \approx 0.693$ ) confirms the chaos observed in (d): nearby orbits diverge exponentially fast, doubling their separation each step. (R1 · AG1)

(g) [4 marks]

Explain, using parts (b) and (f), why despite being deterministic the doubling map is unpredictable in practice.

Reveal worked solution

Argument. By part (b), each application of $T$ discards one bit of the initial condition. After $n$ steps, the current state depends only on bits $b_{n + 1}, b_{n + 2}, \dots$ of $x_{0}$ . (R1)

In practice we know $x_{0}$ only to finite precision — say $B$ bits. After $B$ iterations, all known information about $x_{0}$ has been "shifted out"; the future orbit is determined by bits we never measured. (R1)

Equivalently, from part (f), errors grow by a factor of $2$ per step ( $e^{λ} = 2$ ). Any starting-value uncertainty of $2^{- B}$ becomes $O (1)$ after $B$ steps. (R1 · A1)

Question 2 · Random walk on a triangle

Topics: probability, matrices, sequences, limits · 27 marks

A robot sits at one of three vertices of an equilateral triangle labelled $A$ , $B$ , $C$ . At each step the robot moves to one of the other two vertices with probability $\frac{1}{2}$ each, independently of past moves. Let $p_{n}$ , $q_{n}$ , $r_{n}$ denote the probability that the robot is at $A$ , $B$ , $C$ respectively after $n$ steps.

Let $v_{n} = (\begin{matrix} p_{n} \\ q_{n} \\ r_{n} \end{matrix})$ be the state vector at step $n$ .

(a) [3 marks]

Write down the transition matrix $M$ such that $v_{n + 1} = M v_{n}$ .

Reveal worked solution

From each vertex the robot goes to the other two with probability $\frac{1}{2}$ , and to itself with probability $0$ .

M = (\begin{matrix} 0 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & 0 \end{matrix})

(M1 · A1 · A1)

(b) [3 marks]

The robot starts at $A$ : $v_{0} = (1, 0, 0)^{T}$ . Compute $v_{1}$ , $v_{2}$ and $v_{3}$ .

Reveal worked solution

v_{1} = M v_{0} = (0, \frac{1}{2}, \frac{1}{2})^{T}

v_{2} = M v_{1} = (\frac{1}{2}, \frac{1}{4}, \frac{1}{4})^{T}

v_{3} = M v_{2} = (\frac{1}{4}, \frac{3}{8}, \frac{3}{8})^{T}

(A1 · A1 · A1)

By symmetry, $q_{n} = r_{n}$ for all $n \geq 1$ . Show that $p_{n + 1} = 1 - p_{n} - \frac{1}{2} (1 - p_{n})$ and deduce the recurrence $p_{n + 1} = \frac{1}{2} (1 - p_{n})$ .

Reveal worked solution

Use symmetry. Since $q_{n} = r_{n}$ and $p_{n} + q_{n} + r_{n} = 1$ , we have $q_{n} = r_{n} = \frac{1 - p_{n}}{2}$ . (M1)

Compute $p_{n + 1}$ . The robot arrives at $A$ from $B$ or $C$ with probability $\frac{1}{2}$ each:

p_{n + 1} = \frac{1}{2} q_{n} + \frac{1}{2} r_{n} = q_{n} = \frac{1 - p_{n}}{2} .

Hence $p_{n + 1} = \frac{1}{2} (1 - p_{n})$ . (M1 · A1 · AG1)

(d) [6 marks]

Solve the recurrence $p_{n + 1} = \frac{1}{2} (1 - p_{n})$ with $p_{0} = 1$ to obtain a closed form for $p_{n}$ .

Reveal worked solution

Find the fixed point. Setting $p^{*} = \frac{1}{2} (1 - p^{*})$ gives $p^{*} = \frac{1}{3}$ . (M1)

Substitute $u_{n} = p_{n} - \frac{1}{3}$ .

u_{n + 1} = p_{n + 1} - \frac{1}{3} = \frac{1}{2} (1 - p_{n}) - \frac{1}{3} = \frac{1}{2} - \frac{1}{2} p_{n} - \frac{1}{3} = \frac{1}{6} - \frac{1}{2} p_{n}

= - \frac{1}{2} (p_{n} - \frac{1}{3}) = - \frac{1}{2} u_{n} .

So $u_{n}$ is geometric with ratio $- \frac{1}{2}$ and $u_{0} = 1 - \frac{1}{3} = \frac{2}{3}$ . (M1 · A1)

Closed form.

u_{n} = \frac{2}{3} {(- \frac{1}{2})}^{n} ⟹ p_{n} = \frac{1}{3} + \frac{2}{3} {(- \frac{1}{2})}^{n} .

(M1 · A1 · AG1)

(e) [3 marks]

Deduce $lim_{n \to \infty} v_{n}$ and interpret it.

Reveal worked solution

${(- \frac{1}{2})}^{n} \to 0$ as $n \to \infty$ , so $p_{n} \to \frac{1}{3}$ and by symmetry $q_{n}, r_{n} \to \frac{1}{3}$ . (M1)

lim_{n \to \infty} v_{n} = {(\frac{1}{3}, \frac{1}{3}, \frac{1}{3})}^{T}

— the stationary distribution. (A1)

Interpretation: in the long run the robot is equally likely to be at any vertex, regardless of where it started. (R1)

(f) [4 marks]

The matrix $M$ has eigenvalues $λ_{1} = 1$ and $λ_{2} = λ_{3} = - \frac{1}{2}$ . Explain — without diagonalising $M$ — how the eigenvalues predict both the long-run limit in (e) and the geometric decay rate in (d).

Reveal worked solution

Eigenvalue interpretation. Any state vector can be decomposed as a linear combination of eigenvectors: one along the $λ = 1$ eigenspace (the stationary distribution $(\frac{1}{3}, \frac{1}{3}, \frac{1}{3})^{T}$ ), the rest along the $λ = - \frac{1}{2}$ eigenspace. (R1)

After $n$ applications of $M$ , the $λ = 1$ component is unchanged, and the transient components are multiplied by $(- \frac{1}{2})^{n}$ . (R1)

Long-run limit: as $n \to \infty$ , $(- \frac{1}{2})^{n} \to 0$ , leaving only the stationary distribution — matches (e). (A1)
Decay rate: convergence is geometric with ratio $| λ_{2} | = \frac{1}{2}$ — matches the closed form in (d), where the transient term $\frac{2}{3} (- \frac{1}{2})^{n}$ decays by exactly a factor of $\frac{1}{2}$ each step. (A1)

(g) [4 marks]

A biased robot now stays put with probability $b \in [0, 1)$ and moves to each other vertex with probability $\frac{1 - b}{2}$ . Predict the new dominant transient eigenvalue as a function of $b$ , and hence how many steps roughly are needed for the state to be within $10^{- 3}$ of the stationary distribution when $b = 0.9$ .

Reveal worked solution

New transition matrix.

M_{b} = b I + (1 - b) M

Since $M$ has eigenvalues $1, - \frac{1}{2}, - \frac{1}{2}$ , $M_{b}$ has eigenvalues $b + (1 - b) λ$ — so $1$ (unchanged, since $b + (1 - b) \cdot 1 = 1$ ) and $b - \frac{1 - b}{2} = \frac{3 b - 1}{2}$ (transient). (M1 · A1)

Decay to $10^{- 3}$ . Require $| \frac{3 b - 1}{2} |^{n} < 10^{- 3}$ . With $b = 0.9$ , transient eigenvalue is $\frac{1.7}{2} = 0.85$ :

{0.85}^{n} < 10^{- 3} ⟺ n > \frac{- 3 \ln 10}{\ln 0.85} \approx \frac{6.908}{0.1625} \approx 42.5

So about $43$ steps. Compare with the unbiased robot ( $b = 0$ , transient eigenvalue $\frac{1}{2}$ ): ${0.5}^{n} < 10^{- 3} \Rightarrow n \geq 10$ . The bias dramatically slows convergence. (M1 · A1)

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AA HL Prediction Paper 3s

Full Paper 3s we wrote in the format of the current IB guide, each with a full markscheme (M, A and R marks) and worked solutions. Every answer is checked twice, the second time by an independent solver.

November 2026 Paper 3 1 hour 15 minutes · 55 marks · 2 questions · out 14 October 2026
May 2027 Paper 3 1 hour 15 minutes · 55 marks · 2 questions · out 18 April 2027
New-syllabus practice Paper 3 (first exams May 2029) 1 hour · 50 marks · 2 questions · out 4 October 2026

They open with IB Pro (€99 a year) or the 7-day free trial. Paper 1 of each set is the free sample: a free account opens the whole paper.

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Full mock, timed and marked

AA HL Mock A Paper 3

Our own full AA HL mock, separate from the Prediction Papers: 1 hour 15 minutes, 55 marks, with a full markscheme. Sit it timed and your score is saved. It opens with an IB plan or the free 7-day trial.

AA HL Mock A Paper 3

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More Paper 3 practice

AA HL Paper 3 practice questions · The two full-length questions on this page, with worked solutions.
Paper 3 toolkit · The problem-solving moves for unfamiliar contexts, with tips for when you are stuck.
How to prepare for Paper 3 · A deeper guide for both HL courses, with a walkthrough and a section for teachers.
Paper 3 to IA · Two step-by-step investigations, with the IA criterion each step earns.

Practise Paper 3 properly: free account

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Common questions

How long is IB Maths AA HL Paper 3?

1 hour 15 minutes, with a graphic display calculator (GDC) and the formula booklet. It is two compulsory extended-response problem-solving questions and counts for 20% of your HL grade.

Can I use a calculator in Paper 3?

Yes. Paper 3 is a technology-allowed paper, so you need your GDC. Use it to explore cases, check answers and do heavy calculations, but write down what you did.

What topics come up on AA HL Paper 3?

Any topic in the course can come up, including the SL core, and one question can join several: calculus with sequences, say, or proof with complex numbers. Each question introduces an idea and leads you through it, so you are tested on using what you know in a new setting.

How do I revise for Paper 3?

Learn the four techniques on this page, then work through investigation-style questions slowly with the solutions. Once that feels natural, sit full Paper 3s timed and mark them against the markscheme.

Is Paper 3 changing?

For the new courses (first exams in May 2029), the IB has announced that HL Paper 3 will be 1 hour instead of 1 hour 15 minutes. Exams before then keep the current format.

Are these official IB questions?

No. Every question on this page is our own, written in the style of Paper 3. This site is independent and is not connected to the IB.

More for AA HL

IB Maths AA HL · AA HL mock exam guide · AA HL questions by topic · Prediction Papers · AI HL Paper 3

IB Math Revision is independent and is not affiliated with, endorsed by or connected to the International Baccalaureate Organization. This work/product/service has been developed independently from and is not endorsed by the International Baccalaureate Organization. International Baccalaureate, Baccalauréat International, Bachillerato Internacional and IB are registered trademarks owned by the International Baccalaureate Organization.

What Paper 3 is

How Paper 3 is marked

M: method

A: accuracy

R: reasoning

AG: answer given

The techniques that score

Spot the pattern

Conjecture, then prove

Use your GDC

Communicate

Two worked mini-investigations

Investigation 1: Repeated derivatives of xe2x

Investigation 2: How many real roots does 1+x+x22!+⋯+xnn! have?

Full-length practice questions

Question 1 · The doubling map and chaotic dynamics

Question 2 · Random walk on a triangle

Full Paper 3s to sit

AA HL Prediction Paper 3s

AA HL Mock A Paper 3

More Paper 3 practice

Practise Paper 3 properly: free account

Common questions

More for AA HL

Investigation 1: Repeated derivatives of $x e^{2 x}$

Investigation 2: How many real roots does $1 + x + \frac{x^{2}}{2!} + \dots + \frac{x^{n}}{n!}$ have?