IB Maths AA HL Paper 3: how the problem-solving paper works and how to score on it

IB Maths AA HL Paper 3 is a 1 hour 15 minute, 55-mark paper with two compulsory problem-solving questions, and you need a graphing calculator for it. Each question takes one idea and walks you from easy special cases to a general result, often through a conjecture you then prove. You score well by reading the structure of each question before you start, using every printed result, and continuing to collect marks after you get stuck rather than abandoning the question.

What is AA HL Paper 3?

Papers 1 and 2 are each 30% at HL and the exploration is 20%. If you read elsewhere that Paper 3 lasts one hour, that comes from the original 2021 course documents; the official May 2026 timetable gives 1 hour 15 minutes. For how Paper 3 sits alongside the other papers, see IB Maths AA vs AI exams.

What does a Paper 3 question look like?

Paper 1 and Paper 2 questions test topics you can name in advance. A Paper 3 question tests whether you can think through an unfamiliar situation with the maths you already know. The topics are still from the course (sequences, calculus, complex numbers, probability, functions), but the set-up is new. Many questions follow a similar shape:

1. Simple cases. Compute something for n = 1, 2, 3, or for a specific value of a parameter. Routine marks.
2. Pattern. Organise the results, often in a table, and spot what is changing.
3. Conjecture. Write a general formula or statement.
4. Proof. Prove it, often by induction or by direct algebra.
5. Extension. Use the result, or generalise it further (a second parameter, a different starting function, a limit as n → ∞).

The early parts are accessible by design. The later parts are harder, but printed results often give you a way back in.

How should you read the question before starting?

Spend the first two or three minutes of each question reading all of it, not just part (a). You are looking for:

• Printed results. Anything in a "show that" is a fact you can use later, whether or not you showed it.
• "Hence" chains. Mark which parts depend on which. A part that starts with "Hence" is telling you the method.
• The destination. If the last part asks for a limit or a general formula, the earlier parts are steps towards it. Knowing that tells you what form your conjecture should take (a power of 2? a factorial? a polynomial in n?).
• The marks per part. A 1-mark "write down" needs a line; a 6-mark proof needs structure.

Command terms matter more than usual on Paper 3: "write down" (no working needed), "show that" (route required), "conjecture" (a clear general statement), "prove" (a complete argument). Our command terms guide covers how each one changes what the examiner expects.

Conjecture, test, prove: what does each step need?

Conjecture

A conjecture is a precise statement in terms of the variable the question uses, not a description. "The derivatives keep doubling" is not a conjecture; "f⁽ⁿ⁾(x) = 2ⁿ⁻¹(2x + n)e^2x" is. Put your simple cases in a table with a column for n and write each result in a form that shows the pattern: 2, 6, 24, 120 hides less as 2!, 3!, 4!, 5!.

Test

Before trying to prove anything, check your conjecture on a case you did not use to build it. Your calculator is useful here (see below). A two-minute test can save fifteen minutes of trying to prove a false statement.

Prove

For statements about every positive integer n, induction is the usual tool. The marks are spread across the parts of the argument, so each needs to be on the page:

• the base case, checked explicitly (both sides calculated);
• the assumption for n = k, written out;
• the step to n = k + 1, using the assumption, ending in the statement for k + 1;
• a concluding sentence linking the base case and the inductive step.

The last point is the one most often left out: the algebra is done and the answer stops. The conclusion carries a mark, and it usually needs the earlier steps to be in place.

How should you use the calculator on Paper 3?

The GDC is required, but it is a tool for exploring, not a replacement for proof.

• Generate cases fast. Use the table feature, a sequence mode or a list to get the first several terms without arithmetic slips.
• Test conjectures. Graph your conjectured formula on top of the original, or evaluate both at a value you haven't used. Matching values at n = 7 is strong evidence; a mismatch tells you to rethink.
• Find features numerically (a minimum, an intersection, a definite integral) to check an exact answer you found by hand.
• Watch your settings. Radians for any calculus with trig, and keep full precision. Our guide to GDC mistakes in IB maths covers the settings that cost marks.

A calculator result never proves a statement for all n. If a part says "prove" or "show", the marks are for algebra on the page.

How do you pick up marks when you're stuck?

• Use the printed result. If you can't prove part (c) but part (d) says "hence", use the result from (c) anyway. Examiners mark (d) on its own.
• Write your conjecture even if you can't prove it. The conjecture mark is separate from the proof marks.
• Do the parts of a proof you can. Base case and assumption are often worth marks on their own.
• Skip to a later part. Extension parts sometimes restart from a new special case that is easier than the proof before it.
• Write down partial progress. Setting up the right derivative, integral or recurrence can earn a method mark even if you can't finish.

How should you split the time?

75 minutes for 55 marks is about 1.4 minutes per mark, more generous than Papers 1 and 2 because the thinking is slower. Split the time between the two questions in proportion to their marks; a 27-mark question gets about 37 minutes. Within that:

• 2–3 minutes reading the whole question;
• move briskly through the simple cases (they are routine, don't polish them);
• if a proof stalls for more than about twice its marks in minutes, write what you have and move to the next part;
• switch questions on time. The opening parts of the second question are usually the cheapest marks left on the paper.

A worked mini Paper 3 task

This is an original task in Paper 3 style, shorter than a real question. Try it first.

Let f(x) = x e^2x, and write f⁽ⁿ⁾(x) for the nth derivative.
(a) Find f′(x), f″(x) and f‴(x), writing each in the form (px + q)e^2x. [4]
(b) Conjecture an expression for f⁽ⁿ⁾(x), n ∈ ℤ⁺. [2]
(c) Prove your conjecture by mathematical induction. [6]
(d) Hence find, in terms of n, the coordinates of the minimum point on the graph of y = f⁽ⁿ⁾(x). [5]
(e) Suggest an expression for the nth derivative of g(x) = x e^ax, where a ≠ 0. [2]

(a) Simple cases

Product rule each time:

• f′(x) = e^2x + 2x e^2x = (2x + 1)e^2x
• f″(x) = 2e^2x + 2(2x + 1)e^2x = (4x + 4)e^2x
• f‴(x) = 4e^2x + 2(4x + 4)e^2x = (8x + 12)e^2x

(b) Pattern and conjecture

Taking out the power of 2 makes the pattern visible. Conjecture: f⁽ⁿ⁾(x) = 2ⁿ⁻¹(2x + n)e^2x.

Test on a case not used: the conjecture gives f⁽⁴⁾(x) = 8(2x + 4)e^2x = (16x + 32)e^2x. Differentiating f‴ directly: 8e^2x + 2(8x + 12)e^2x = (16x + 32)e^2x. It agrees.

(c) Proof by induction

Let P(n) be the statement f⁽ⁿ⁾(x) = 2ⁿ⁻¹(2x + n)e^2x.

n = 1: the formula gives 2⁰(2x + 1)e^2x = (2x + 1)e^2x, which is f′(x) from (a). So P(1) is true.

Assume P(k) is true for some k ∈ ℤ⁺: f^(k)(x) = 2^k−1(2x + k)e^2x.

Then f^(k+1)(x) = d/dx[2^k−1(2x + k)e^2x]
= 2^k−1[2e^2x + 2(2x + k)e^2x]
= 2^k−1 × 2 × (1 + 2x + k)e^2x
= 2^k(2x + (k + 1))e^2x,
which is P(k + 1).

Conclusion: P(1) is true, and P(k) true implies P(k + 1) true, so P(n) is true for all n ∈ ℤ⁺ by mathematical induction.

(d) Using the result

This part works even if you couldn't do (c): the "hence" means use the formula. Stationary points of y = f⁽ⁿ⁾(x) are where its derivative, f⁽ⁿ⁺¹⁾(x), is zero. By the formula with n + 1 in place of n:

f⁽ⁿ⁺¹⁾(x) = 2ⁿ(2x + n + 1)e^2x = 0. Since 2ⁿe^2x > 0, we need 2x + n + 1 = 0, so x = −(n + 1)/2.

It is a minimum: 2x + n + 1 changes from negative to positive as x increases through −(n + 1)/2, so f⁽ⁿ⁺¹⁾ changes sign from negative to positive.

y-coordinate: f⁽ⁿ⁾(−(n + 1)/2) = 2ⁿ⁻¹(−(n + 1) + n)e⁻⁽ⁿ⁺¹⁾ = −2ⁿ⁻¹e⁻⁽ⁿ⁺¹⁾.

Minimum at (−(n + 1)/2, −2ⁿ⁻¹e⁻⁽ⁿ⁺¹⁾).

GDC check for n = 3: graph y = (8x + 12)e^2x and find the minimum. The calculator gives x ≈ −2, y ≈ −0.0733. The formula gives x = −4/2 = −2 and y = −4e⁻⁴ ≈ −0.0733.

(e) Generalising

Repeat (a) quickly with a in place of 2: g′(x) = (ax + 1)e^ax, g″(x) = a(ax + 2)e^ax. The same pattern suggests g⁽ⁿ⁾(x) = aⁿ⁻¹(ax + n)e^ax, which gives the part (b) formula when a = 2. A "suggest" part like this usually only needs the statement and a sentence of support, not a second induction.

Notice where the marks were: 4 routine marks in (a), 2 for a clear conjecture, and 5 in (d) that needed only the conjectured formula. A student who could not prove (c) could still reach 13 of 19 marks. For how mark schemes split method and accuracy marks across a proof, see our guide to IB mark schemes.