IB Maths AA Paper 1 tips: how to score on the non-calculator paper

To do well on IB Maths AA Paper 1, you need two things: hand skills (fractions, surds, logs, exact trig values, factorising) that are automatic, and working that is written so clearly that an examiner can give you method marks even when the final answer slips. AA Paper 1 is the only IB maths paper without a calculator. That changes how you should revise for it and how you should spend the time in the exam.

Three original Paper 1-style questions are worked in full at the end. For how the AA and AI papers compare overall, see IB Maths AA vs AI exams.

What is on AA Paper 1?

Both levels have a Section A of short-response questions and a Section B of extended-response questions, and every question is compulsory. You get a clean copy of the AA formula booklet. HL Paper 1 can draw on HL-only content (complex numbers, proof by induction, partial fractions, further calculus), so the list of hand skills below is longer at HL.

How should you approach Section A and Section B?

Section A questions are short and usually test one or two skills: solve a log equation, find an exact value, differentiate and find a tangent. They reward speed and accuracy. Most students should work through them in order, but if one has eaten twice its time, leave a gap and come back.

Section B questions are long, multi-part and built like a staircase: part (a) is often a routine step, and later parts use earlier answers. Two habits matter most here:

• Attempt every part. A later part often does not depend on you having solved the earlier one, especially when the earlier result is printed for you ("show that…"). Abandoning a Section B question at the first block throws away marks you could have had.
• Read the whole question before starting. Part (d) sometimes tells you what part (a) was for, and seeing the direction stops you choosing a clumsy method.

Which skills must be automatic without a calculator?

Paper 1 questions are designed so the arithmetic comes out cleanly, but only if you can do it fast and without slips. Drill these until they feel like times tables.

Fractions, surds and indices

• Fractions: 2/3 + 5/12 = 8/12 + 5/12 = 13/12. Keep improper fractions; don't convert to mixed numbers.
• Surds: √72 = √36 × √2 = 6√2. Rationalise: 6/(3 − √3) = 6(3 + √3)/(9 − 3) = 3 + √3.
• Fractional and negative indices: 8^2/3 = (∛8)² = 4, and 27^−1/3 = 1/3.

Logs and exponentials

• log₃ 81 = 4, ln e⁵ = 5, e^{ln 7} = 7.
• Laws: log₂ 12 − log₂ 3 = log₂ 4 = 2. The log of a sum has no law: ln(a + b) is not ln a + ln b.
• Domain: the argument of a log must be positive. This is where most "reject one solution" marks come from.

Exact trig values

Know these cold, and know how to get the other quadrants from the unit circle (sign and reference angle).

Example: cos(5π/3). The angle 5π/3 is in the fourth quadrant, reference angle π/3, cosine positive there, so cos(5π/3) = 1/2.

Algebra

• Factorising non-monic quadratics: 6x² − x − 2 = (3x − 2)(2x + 1). Expand to check: 6x² + 3x − 4x − 2.
• Completing the square, the discriminant (b² − 4ac) for "how many roots" questions, and spotting hidden quadratics such as e^2x − 5e^x + 6 = 0.

Calculus by hand

• Chain, product and quotient rules without hesitation, and the standard derivatives and integrals of e^x, ln x, sin x and cos x.
• Check an integral by differentiating it: if you claim ∫ x e^x2 dx = ½e^x2 + C, differentiate to get ½ × 2x e^x2 = x e^x2. It matches, so it's right.

HL extras

At HL, add complex numbers in polar and exponential form, which turn big powers into exact arithmetic. For example (1 + i)⁸: 1 + i = √2 e^iπ/4, so (1 + i)⁸ = (√2)⁸ e^2πi = 16. (Check: (1 + i)² = 2i, and (2i)⁴ = 16.) Partial fractions, the binomial expansion with negative and fractional powers, and proof by induction can also appear on HL Paper 1.

Why do exact answers matter so much?

Without a calculator, an exact answer is both easier to produce and what the question expects. Unless a question tells you otherwise, leave answers as fractions, surds, multiples of π, or in terms of e and ln.

The same goes for simplifying. 4/√2 is correct but unfinished; 2√2 is what a mark scheme will show. If a question says "give your answer in the form a + b√3", anything else is not in the requested form.

How do "show that" and "hence" questions work?

"Show that" gives you the answer. The marks are for the route, so every step must be visible and each line must follow from the one before. Two rules:

• Start from what you are given and end at the printed result. Don't start from the result and work backwards, and don't write the result down and "verify" it unless the question allows that.
• Writing the final printed line earns nothing by itself. The marks sit on the lines before it.

"Hence" means use the previous result. If you solve the part some other way, you may get no marks even with the right answer. "Hence or otherwise" lets you use any valid method, but the "hence" route is usually the short one. More on how command terms set the expected answer in our guide to IB command terms.

Where do method marks come from?

IB mark schemes split marks into method marks (M), accuracy marks (A) that usually depend on the method before them, and reasoning marks (R). Results you were asked to show are labelled AG, answer given, and carry no mark of their own. Follow-through often lets you keep later marks after an early slip, provided you use your wrong value correctly. Our guide to IB mark schemes goes through the codes in detail.

What this means on Paper 1:

• Write the method before the arithmetic. "f′(x) = 0 at a maximum" followed by the equation you set up is a method mark, even if your algebra then goes wrong.
• Substitute visibly. Writing "area = ∫₁^e (ln x)/x dx" before integrating shows the examiner your limits and integrand.
• Don't cross out working you haven't replaced. Crossed-out work is normally not marked. If you have two attempts, cross out the one you don't want marked.

How long should each question take?

SL Paper 1 is 90 minutes for 80 marks; HL Paper 1 is 120 minutes for 110 marks. Both work out at a little over one minute per mark. A 6-mark Section A question gets about 6–7 minutes; a 15-mark Section B question about 17.

• Note your start time and the time at which you want to reach Section B, based on the marks in each section of your paper.
• If a question has taken twice its share, write down what you have (it may earn method marks), leave a gap and move on.
• Keep 5–10 minutes at the end for checking and for gaps you left. That buffer usually earns more than squeezing out the last part of a question you are stuck on. For a fuller plan, see time management in IB exams.

How do you check answers without a calculator?

• Substitute back. Put your solutions into the original equation, not your rearranged one.
• Check the domain. Log arguments positive, square roots of non-negatives, denominators not zero, solutions inside the stated interval.
• Differentiate your integrals and expand your factorisations.
• Estimate. Carry a few anchors in your head: √2 ≈ 1.41, √3 ≈ 1.73, π ≈ 3.14, e ≈ 2.72, ln 2 ≈ 0.69. An area of −3 or a probability of 1.2 is wrong whatever the algebra says.
• Count solutions against a sketch. For trig equations on an interval, a quick sketch of the graph tells you how many solutions to expect.

Worked examples in Paper 1 style

These are original questions written to match the style of the paper. Try each one before reading the solution.

Example 1 (Section A style, 5 marks): a log equation with a trap

Solve log₂(x + 3) + log₂(x − 3) = 4.

1. Combine the logs: log₂[(x + 3)(x − 3)] = 4, so log₂(x² − 9) = 4.
2. Remove the log: x² − 9 = 2⁴ = 16.
3. x² = 25, so x = 5 or x = −5.
4. Check the domain: log₂(x − 3) needs x > 3, so x = −5 is rejected.
5. x = 5. Check: log₂ 8 + log₂ 2 = 3 + 1 = 4.

Leaving x = −5 in costs the final mark. Writing "x = −5 rejected because x − 3 < 0" earns it.

Example 2 (7 marks): show that, then hence

(a) Show that the equation cos 2x + 3 sin x = 2 can be written as 2 sin² x − 3 sin x + 1 = 0. [2]
(b) Hence solve cos 2x + 3 sin x = 2 for 0 ≤ x ≤ 2π. [5]

(a) Use cos 2x = 1 − 2 sin² x (chosen because everything else is in sin x):

1 − 2 sin² x + 3 sin x = 2
−2 sin² x + 3 sin x − 1 = 0
2 sin² x − 3 sin x + 1 = 0, as required.

(b) Factorise the quadratic in sin x: (2 sin x − 1)(sin x − 1) = 0.

• sin x = 1/2 gives x = π/6 or x = 5π/6 (first and second quadrants, where sine is positive).
• sin x = 1 gives x = π/2.

x = π/6, π/2, 5π/6. Check x = 5π/6 in the original: cos(5π/3) + 3 × 1/2 = 1/2 + 3/2 = 2.

The most common loss here is missing 5π/6. The sketch check catches it: y = sin x crosses y = 1/2 twice between 0 and π.

Example 3 (Section B style, 9 marks): calculus, exactly

Let f(x) = (ln x)/x for x > 0.
(a) Show that f′(x) = (1 − ln x)/x². [2]
(b) Find the exact coordinates of the maximum point of the graph of f. [3]
(c) Find the exact area of the region enclosed by the graph of f, the x-axis and the line x = e. [4]

(a) Quotient rule with u = ln x, v = x:

f′(x) = (x × 1/x − ln x × 1)/x² = (1 − ln x)/x², as required.

(b) Set f′(x) = 0: 1 − ln x = 0, so ln x = 1 and x = e. Then f(e) = (ln e)/e = 1/e. For x < e, ln x < 1 so f′(x) > 0; for x > e, f′(x) < 0, so this is a maximum. Maximum at (e, 1/e).

(c) The graph meets the x-axis where ln x = 0, at x = 1, and f(x) ≥ 0 for 1 ≤ x ≤ e. So the region runs from x = 1 to x = e:

Area = ∫₁^e (ln x)/x dx. Since the derivative of ln x is 1/x, the integrand is (ln x) × (ln x)′, so the integral is ½(ln x)².

Area = [½(ln x)²]₁^e = ½(1)² − ½(0)² = ½.

Check by differentiating: d/dx[½(ln x)²] = ln x × 1/x. Matches the integrand.

Notice how the parts connect: (b) needs (a), and (c) needs you to find the lower limit yourself, which the question never states. That is typical of Section B. For the slips that cost AA marks across all the papers, see Maths AA common mistakes.