IB DP Mathematics: Analysis and Approaches – Counting principles, partial fractions, proof and linear systems (HL) Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

This practice set covers counting principles, the extended binomial theorem, partial fractions, proof and systems of linear equations for IB Diploma Programme Mathematics: Analysis and Approaches. It is aligned to the IB Mathematics: analysis and approaches guide, first assessment 2021, syllabus sections 1.10, 1.11, 1.15 and 1.16, and every question is HL only (AHL). It follows the IB guide for first assessment 2021, which remains the examined syllabus until the new course is first assessed in May 2029, so it applies to the May and November 2026, 2027 and 2028 HL sessions.

HL Paper 1 allows no technology; Papers 2 and 3 require it. Questions are labelled “(calculator-free)” or “(calculator allowed)” to match.

Learn the methods first in the study guide for this unit and the revision notes. The IB DP Maths AA course hub and the printable syllabus checklist show where this unit sits.

Questions

1. (calculator allowed) Use technology to solve the system

3x + 2y − z = 1
x − y + 4z = 15
2x + 5y + 3z = 8

[2]

2. (calculator-free) A team of 4 is chosen from 7 girls and 5 boys.

(a) Find the number of possible teams. [1] (b) Find the number of teams with exactly 2 boys. [2] (c) Find the number of teams with at least one boy. [2]

3. (calculator allowed) Five different maths books and three different physics books are placed in a row on a shelf.

(a) Find the number of possible arrangements. [1] (b) Find the number of arrangements in which the three physics books are next to each other. [2] (c) Find the number of arrangements in which no two physics books are next to each other. [3]

4. (calculator-free) Expand (1 + 3x)^(−1/3) in ascending powers of x up to and including the term in x³, and state the values of x for which the expansion is valid. [5]

5. (calculator-free)

(a) Find the first three terms of the expansion of ∛(8 − x) in ascending powers of x, and state the values of x for which it is valid. [4] (b) Use your answer to find an approximation to ∛7, giving your answer as a fraction. [2]

6. (calculator-free) Express (x + 13)/(x² + x − 6) in partial fractions. [4]

7. (calculator-free) Prove by contradiction that log₂ 5 is irrational. [4]

8. (calculator-free) Show that the statement “n² + n + 17 is prime for every n ∈ ℤ⁺” is not always true. [2]

9. (calculator-free) Prove by mathematical induction that 5ⁿ + 2 × 11ⁿ is divisible by 3 for all n ∈ ℤ⁺. [6]

10. (calculator-free) Let f(x) = xe^(3x). Prove by mathematical induction that the n-th derivative is f⁽ⁿ⁾(x) = 3^(n−1)(3x + n)e^(3x) for all n ∈ ℤ⁺. [7]

11. (calculator-free) Consider the system of equations

x + 2y − z = 3
2x − y + z = 1
x − 8y + az = b

where a, b ∈ ℝ.

(a) Find the value of a for which the system does not have a unique solution. [4] (b) For this value of a, find the value of b for which the system has infinitely many solutions. [2] (c) For these values of a and b, find the general solution. [3] (d) State the number of solutions when a takes the value found in (a) and b = 0. [1]

12. (calculator-free) Let g(x) = (x + 8)/((1 + 2x)(2 − x)).

(a) Express g(x) in partial fractions. [3] (b) Hence find the expansion of g(x) in ascending powers of x up to and including the term in x². [5] (c) State the set of values of x for which this expansion is valid. [2]

Answers

1. Enter the system in the GDC’s simultaneous equation solver, or row-reduce the augmented matrix [1]. x = 2, y = −1, z = 3 [1] Examiner insight: Write the system or matrix you entered; if you mistype one coefficient and give only the final values, there is no working to earn the method mark.

2. (a) ¹²C₄ = 495 [1] (b) ⁵C₂ × ⁷C₂ [1] = 10 × 21 = 210 [1] (c) No boys: ⁷C₄ = 35 [1]. At least one boy: 495 − 35 = 460 [1] Examiner insight: In (c), choosing one boy (5 ways) and then any 3 of the other 11 counts some teams more than once, and that method earns no marks.

3. (a) 8! = 40320 [1] (b) Treat the physics books as one block, giving 6 items: 6! [1]. Arrange within the block: 3!. Total 720 × 6 = 4320 [1] (c) Arrange the maths books: 5! = 120 [1]. There are 6 gaps (including the ends), and the physics books go in 3 of them in order: ⁶P₃ = 120 [1]. Total 120 × 120 = 14400 [1] Examiner insight: 8! − 4320 counts arrangements where not all three are together, which still allows two to touch; that subtraction scores nothing in (c).

4. n = −1/3 with X = 3x [1]. Terms: 1 + (−1/3)(3x) = 1 − x [1]; ((−1/3)(−4/3)/2)(9x²) = 2x² [1]; ((−1/3)(−4/3)(−7/3)/6)(27x³) = −(14/3)x³ [1]. So (1 + 3x)^(−1/3) = 1 − x + 2x² − (14/3)x³, valid for |3x| < 1, that is |x| < 1/3 [1] Examiner insight: The accuracy marks need (3x)² = 9x² and (3x)³ = 27x³; writing 3x² is a common slip that loses both later terms.

5. (a) ∛(8 − x) = 8^(1/3)(1 − x/8)^(1/3) = 2(1 − x/8)^(1/3) [1]. Then (1 − x/8)^(1/3) = 1 + (1/3)(−x/8) + ((1/3)(−2/3)/2)(x²/64) [1] = 1 − x/24 − x²/576, so ∛(8 − x) = 2 − x/12 − x²/288 [1], valid for |x/8| < 1, that is |x| < 8 [1] (b) 8 − x = 7 gives x = 1 [1]. ∛7 ≈ 2 − 1/12 − 1/288 = (576 − 24 − 1)/288 = 551/288 [1] Examiner insight: Forgetting the factor 8^(1/3) = 2 is a method error, so the later accuracy marks are lost even if the bracket is expanded correctly.

6. x² + x − 6 = (x − 2)(x + 3) [1]. x + 13 ≡ A(x + 3) + B(x − 2) [1]. x = 2: 15 = 5A, so A = 3 [1]. x = −3: 10 = −5B, so B = −2 [1]. Answer: 3/(x − 2) − 2/(x + 3) Examiner insight: A final answer written as 3/(x − 2) + (−2)/(x + 3) is acceptable, but a sign error in B loses the last accuracy mark, so check with x = 0: 13/(−6) = −3/2 − 2/3.

7. Assume log₂ 5 = p/q where p, q ∈ ℤ⁺ (both positive, since 5 > 1 means log₂ 5 > 0) [1]. Then 2^(p/q) = 5, so 2ᵖ = 5^q [1]. Since p ≥ 1, 2ᵖ is even, but 5^q is odd [1]. An even number cannot equal an odd number, a contradiction, so log₂ 5 is irrational [1] Examiner insight: The first mark needs the assumption stated in full, with p and q named as integers; “suppose it is rational” alone does not set up the contradiction.

8. When n = 16: 16² + 16 + 17 = 289 [1]. 289 = 17², so it is not prime, and the statement is not always true [1] Examiner insight: The guide states that giving the counterexample alone is not sufficient; you must show the calculation and say why it breaks the statement.

9. n = 1: 5 + 22 = 27 = 3 × 9, so true for n = 1 [1]. Assume 5ᵏ + 2 × 11ᵏ = 3m for some k ∈ ℤ⁺ and integer m [1]. Then 5^(k+1) + 2 × 11^(k+1) = 5 × 5ᵏ + 22 × 11ᵏ [1] = 5(5ᵏ + 2 × 11ᵏ) + 12 × 11ᵏ [1] = 15m + 12 × 11ᵏ = 3(5m + 4 × 11ᵏ), which is divisible by 3 [1]. It is true for n = 1, and true for n = k implies true for n = k + 1, so it is true for all n ∈ ℤ⁺ by mathematical induction [1] Examiner insight: The final reasoning mark needs all parts of the conclusion; “hence proved” on its own is not enough.

10. n = 1: f′(x) = e^(3x) + 3xe^(3x) [1] = (3x + 1)e^(3x) = 3⁰(3x + 1)e^(3x), so true for n = 1 [1]. Assume f⁽ᵏ⁾(x) = 3^(k−1)(3x + k)e^(3x) for some k ∈ ℤ⁺ [1]. Differentiate using the product rule: f⁽ᵏ⁺¹⁾(x) = 3^(k−1)[3e^(3x) [1] + 3(3x + k)e^(3x)] [1] = 3^(k−1) × 3(3x + k + 1)e^(3x) = 3ᵏ(3x + (k + 1))e^(3x) [1]. This is the formula with n = k + 1. True for n = 1, and true for n = k implies true for n = k + 1, so true for all n ∈ ℤ⁺ by mathematical induction [1] Examiner insight: Write the final form as 3ᵏ(3x + (k + 1))e^(3x) so the match with the n = k + 1 statement is visible; the accuracy mark is for reaching that form.

11. (a) Eliminate x: (2) − 2(1) gives −5y + 3z = −5 [1]; (3) − (1) gives −10y + (a + 1)z = b − 3 [1]. Subtract twice the first of these: (a − 5)z = b + 7 [1]. There is no unique solution when a = 5 [1] (b) With a = 5 the last equation is 0 = b + 7 [1]. Infinitely many solutions need 0 = 0, so b = −7 [1] (c) Let z = 5t [1]. From −5y + 3z = −5: y = 1 + 3t [1]. From (1): x = 3 − 2(1 + 3t) + 5t = 1 − t [1]. General solution x = 1 − t, y = 1 + 3t, z = 5t, t ∈ ℝ (any equivalent parameter form is fine) (d) 0 = 7 is impossible, so the system is inconsistent: no solutions [1] Examiner insight: In (c), one particular solution such as (1, 1, 0) is not a general solution and earns only the first mark at most.

12. (a) x + 8 ≡ A(2 − x) + B(1 + 2x) [1]. x = 2: 10 = 5B, so B = 2 [1]. x = −1/2: 15/2 = (5/2)A, so A = 3 [1]. g(x) = 3/(1 + 2x) + 2/(2 − x) (b) 3(1 + 2x)^(−1) [1] = 3 − 6x + 12x² + … [1]. 2/(2 − x) = (1 − x/2)^(−1) [1] = 1 + x/2 + x²/4 + … [1]. Adding: g(x) = 4 − (11/2)x + (49/4)x² + … [1] (c) Need |2x| < 1 and |x/2| < 1 [1]. Both hold when |x| < 1/2 [1] Examiner insight: The combined validity is the smaller interval; giving |x| < 2 (or listing both conditions without combining them) loses the final mark.

Where marks are usually lost

• Using combinations where roles or positions make order matter, or permutations for plain groups.
• Double-counting in “at least one” questions by fixing one person first.
• Subtracting “all together” from the total when the question says “no two together”.
• Dropping the aⁿ factor after rewriting (a + b)ⁿ as aⁿ(1 + b/a)ⁿ.
• Raising only x, not the whole term (such as 3x or −x/8), to the power in each binomial term.
• Validity intervals left out, or not combined into the smaller interval for a sum of two series.
• Induction proofs where the P(k) assumption is never used, or the conclusion is missing.
• Counterexamples stated with no calculation showing why they fail.
• General solutions replaced by a single particular solution.

Next steps

• Recap the key facts in the revision notes for this unit.
• Re-learn any weak topic in the study guide for this unit.
• See the rest of the course on the IB DP Maths AA course hub and tick off the printable syllabus checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

International Baccalaureate Organization, Diploma Programme, Mathematics: analysis and approaches guide, first assessment 2021 (published February 2019, updated November 2020).