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100+ Free IB Mathematics: Analysis and Approaches HL Practice Questions

Pass your International Baccalaureate Mathematics: Analysis and Approaches Higher Level exam on the first try — instant access, no signup required.

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Key Facts: IB Mathematics: Analysis and Approaches HL Exam

1-7

IB grading scale

IBO Diploma Programme

240 hours

Recommended teaching time

IB AA HL subject guide

5 hours

Total written exam time (Papers 1+2+3)

IB AA HL subject guide

20%

Internal Assessment weighting

IB AA HL subject guide

May 2021

First exams under current syllabus

IB AA subject brief

100

Free practice questions here

OpenExamPrep

IB Math AA HL is the most rigorous of the four DP maths courses, with 240 teaching hours, three exam papers (5 hours total) and a 20% Internal Assessment. The current syllabus (first exams 2021) covers number, algebra, functions, trigonometry, statistics and calculus with substantial HL extensions including complex numbers, vectors in 3D, proof by induction and differential equations.

Sample IB Mathematics: Analysis and Approaches HL Practice Questions

Try these sample questions to test your IB Mathematics: Analysis and Approaches HL exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1Prove by induction that 1+2+3+...+n = n(n+1)/2. The inductive step assumes P(k): 1+2+...+k = k(k+1)/2. What is the next step to show P(k+1)?
A.Add (k+1) to both sides
B.Multiply both sides by (k+1)
C.Differentiate both sides with respect to k
D.Substitute k = 1 again
Explanation: To go from P(k) to P(k+1), add the next term (k+1) to the left side: k(k+1)/2 + (k+1) = (k+1)(k/2 + 1) = (k+1)(k+2)/2, which matches P(k+1). This is the standard inductive step structure.
2Prove by contradiction that sqrt(2) is irrational. Assume sqrt(2) = p/q in lowest terms; what contradiction emerges?
A.Both p and q must be even, contradicting lowest terms
B.p must equal q
C.q must be zero
D.p must be prime
Explanation: Squaring gives p^2 = 2q^2 so p is even, p = 2m, then 4m^2 = 2q^2, so q^2 = 2m^2 and q is also even. But p and q being both even contradicts the assumption that p/q is in lowest terms.
3Disprove by counter-example: 'n^2 + n + 41 is prime for every positive integer n.' Smallest n that fails is:
A.40
B.41
C.10
D.5
Explanation: When n = 40, the expression equals 1600 + 40 + 41 = 1681 = 41^2, which is not prime. The classic Euler quadratic produces primes for n = 0..39 and fails at n = 40.
4Express (3x + 5) / ((x + 1)(x + 2)) in partial fractions.
A.2/(x+1) + 1/(x+2)
B.1/(x+1) + 2/(x+2)
C.3/(x+1) + 5/(x+2)
D.2/(x+1) - 1/(x+2)
Explanation: Write 3x+5 = A(x+2) + B(x+1). Set x = -1: 2 = A, so A = 2. Set x = -2: -1 = -B, so B = 1. Therefore (3x+5)/((x+1)(x+2)) = 2/(x+1) + 1/(x+2).
5Find the modulus |z| of the complex number z = 3 + 4i.
A.5
B.7
C.sqrt(7)
D.25
Explanation: The modulus of z = a + bi is |z| = sqrt(a^2 + b^2). Here |z| = sqrt(9 + 16) = sqrt(25) = 5.
6Find the argument arg(z) for z = -1 + sqrt(3) i, with -pi < arg(z) <= pi.
A.2pi/3
B.pi/3
C.-pi/3
D.5pi/6
Explanation: z lies in the second quadrant (real negative, imaginary positive). |z| = 2 and reference angle arctan(sqrt(3)/1) = pi/3, so arg(z) = pi - pi/3 = 2pi/3.
7Convert z = 1 - i to polar (modulus-argument) form r(cos theta + i sin theta).
A.sqrt(2)(cos(-pi/4) + i sin(-pi/4))
B.sqrt(2)(cos(pi/4) + i sin(pi/4))
C.2(cos(-pi/4) + i sin(-pi/4))
D.1(cos(-pi/2) + i sin(-pi/2))
Explanation: |z| = sqrt(1 + 1) = sqrt(2). z is in the fourth quadrant so arg(z) = -pi/4. Polar form is sqrt(2)(cos(-pi/4) + i sin(-pi/4)).
8Write z = -2 + 2i in exponential form r e^(i theta).
A.2 sqrt(2) e^(i 3pi/4)
B.2 sqrt(2) e^(i pi/4)
C.2 e^(i 3pi/4)
D.4 e^(i pi/2)
Explanation: |z| = sqrt(4+4) = 2 sqrt(2). z is in the second quadrant; reference angle is pi/4, so arg = pi - pi/4 = 3pi/4. Hence z = 2 sqrt(2) e^(i 3pi/4).
9Use De Moivre's theorem to evaluate (cos(pi/6) + i sin(pi/6))^6.
A.-1
B.1
C.i
D.-i
Explanation: De Moivre gives (cos theta + i sin theta)^n = cos(n theta) + i sin(n theta). Here n theta = 6 (pi/6) = pi, so the value is cos(pi) + i sin(pi) = -1 + 0i = -1.
10Find all cube roots of unity (solutions to z^3 = 1).
A.1, e^(i 2pi/3), e^(i 4pi/3)
B.1, -1, i
C.1, i, -i
D.1, e^(i pi/3), e^(i 2pi/3)
Explanation: Write 1 = e^(i 2pi k) for integer k. Then z = e^(i 2pi k / 3). For k = 0, 1, 2 we get 1, e^(i 2pi/3), e^(i 4pi/3). These are the three cube roots of unity.

About the IB Mathematics: Analysis and Approaches HL Exam

IB Mathematics: Analysis and Approaches Higher Level is one of the four DP mathematics options, designed for students with strong algebraic skills who enjoy abstract reasoning, proof, and theoretical mathematics. The course covers five core topic areas at greater depth and breadth than the SL course, adds 80 extra HL hours on proof, complex numbers, vectors, series and differential equations, and is assessed via three external papers (Paper 1 no-calculator, Paper 2 with a graphic display calculator, Paper 3 extended response) plus an Internal Assessment exploration worth 20% of the final grade.

Questions

100 scored questions

Time Limit

5 hours total (Paper 1: 2h, Paper 2: 2h, Paper 3: 1h)

Passing Score

Grade 4 standard pass on 1-7 scale; final grade combines three papers and the Internal Assessment

Exam Fee

Set by school; IB subject registration fees typically USD 119 per subject (International Baccalaureate Organization (IBO))

IB Mathematics: Analysis and Approaches HL Exam Content Outline

55 hours (SL+HL)

Number and Algebra

Sequences and series, exponents and logarithms, binomial theorem, proof by induction and contradiction, complex numbers (Cartesian, polar, exponential), De Moivre's theorem, partial fractions

43 hours (SL+HL)

Functions

Linear, quadratic, exponential, logarithmic, rational, polynomial and modulus functions; transformations; factor and remainder theorems; sum and product of roots; sketching reciprocals, moduli and squared graphs

69 hours (SL+HL)

Geometry and Trigonometry

Radian measure, unit circle, identities, compound and double angle formulae, sine and cosine rules, 3D vectors, dot and cross products, equations of lines and planes, angles and intersections

37 hours (SL+HL)

Statistics and Probability

Data presentation and statistics, probability rules, Bayes' theorem, binomial, normal and geometric distributions, expectation and variance, linear combinations of normal variables, central limit theorem

60 hours (SL+HL)

Calculus

Limits and L'Hopital's rule, differentiation including implicit and related rates, integration by parts and substitution, partial fractions, Maclaurin series, separable and integrating-factor ODEs, Euler's method, volumes of revolution

30 hours (IA)

Internal Assessment

Mathematical exploration: an individual written investigation of 12-20 pages assessed against five criteria (presentation, mathematical communication, personal engagement, reflection, use of mathematics). Worth 20% of the final grade.

How to Pass the IB Mathematics: Analysis and Approaches HL Exam

What You Need to Know

  • Passing score: Grade 4 standard pass on 1-7 scale; final grade combines three papers and the Internal Assessment
  • Exam length: 100 questions
  • Time limit: 5 hours total (Paper 1: 2h, Paper 2: 2h, Paper 3: 1h)
  • Exam fee: Set by school; IB subject registration fees typically USD 119 per subject

Keys to Passing

  • Complete 500+ practice questions
  • Score 80%+ consistently before scheduling
  • Focus on highest-weighted sections
  • Use our AI tutor for tough concepts

IB Mathematics: Analysis and Approaches HL Study Tips from Top Performers

1Practice every Paper 1 topic without a calculator — the no-calculator paper rewards strong algebraic manipulation and memorised exact values
2Use the IB formula booklet during practice so you know exactly what is and is not provided in the exam
3Drill Paper 3 style extended-response problems — they reward sustained reasoning across several connected sub-parts
4Master complex numbers in all three forms (Cartesian, polar, exponential) and switch between them fluently
5Build a personal proof library: induction, contradiction and counter-example questions appear nearly every session

Frequently Asked Questions

How is IB Math AA HL different from AA SL?

AA HL has 240 teaching hours versus 150 for SL, adds an extra Paper 3 extended-response section, and covers significant additional content including complex numbers (Cartesian, polar and exponential forms), De Moivre's theorem, 3D vectors with dot and cross products, proof by induction, Maclaurin series, L'Hopital's rule and differential equations.

What are the three exam papers in IB Math AA HL?

Paper 1 is 2 hours, 110 marks, no calculator. Paper 2 is 2 hours, 110 marks, requires a graphic display calculator (GDC). Paper 3 is 1 hour, 55 marks, consists of two extended-response problem-solving questions and uses a GDC. The three papers together count for 80% of the final grade; the Internal Assessment exploration is the other 20%.

When are IB Math AA HL exams sat?

IB Diploma exams are held twice a year: the May session (most schools in the Northern Hemisphere) and the November session (most Southern Hemisphere schools). The current AA syllabus was first examined in May 2021 and remains in force for 2026 exams.

How is IB Math AA HL graded?

Each subject is graded on a 1-7 scale, with 7 the highest. A 4 is generally considered a pass. Grades are determined by combining the marks from Paper 1, Paper 2, Paper 3 and the Internal Assessment exploration against grade boundaries set after each session. Most top universities expect a 6 or 7 in AA HL for mathematics-heavy degrees.

Which calculator is allowed in IB Math AA HL?

A graphic display calculator (GDC) from the IBO-approved list is required for Paper 2 and Paper 3, with statistical and graphing functionality. Paper 1 is taken without any calculator. Approved models include TI-84 Plus CE, TI-Nspire CX (non-CAS), Casio fx-CG50 and similar; CAS-enabled calculators are not allowed.