IB Mathematics AA SL Paper 1 Practice Paper (90 minutes, 80 marks, no calculator) with mark scheme

2026-10-29

Practice paper | IB DP | Mathematics: Analysis and Approaches SL | Paper 1

IB Maths AA SL Paper 1 paper: 90 minutes, 80 marks, no calculator. Ten original questions on sequences, logs, trig, calculus, vectors and probability with a full mark scheme.

Download　Question paper PDF (for students)　|　Answer PDF (M/A/R mark scheme)

Paper details　Level: IB DP (SL, years 1–2) | Subject: Mathematics: Analysis and Approaches SL | Time: 90 minutes | Marks: 80 (no calculator)

Paper structure and marks

Section	Questions	No. of questions	Marks	Content
Section A	Questions 1–8 (short response, 5–8 marks each)	8	52	Sequences, quadratics, logarithms, trigonometry, differentiation, binomial, vectors, probability
Section B	Questions 9–10 (extended response, 14 marks each)	2	28	Logistic model with differentiation; cosine rule and triangle problem solving
Total	10 questions in total	10	80	All topics above

Section A (short response, no calculator)

Question 1 | Arithmetic sequence (5 marks)

The fourth term of an arithmetic sequence is 11 and the sum of the first 10 terms is 155.

(a)　Find the first term a and the common difference d.　(3 marks)

(b)　Find S₂₀.　(2 marks)

Question 2 | Completing the square (6 marks)

Let f(x) = 2x² − 8x + 3.

(a)　Express f(x) in the form a(x − h)² + k.　(3 marks)

(b)　Write down the coordinates of the vertex of y = f(x).　(1 marks)

(c)　Solve f(x) = 0, giving exact answers in surd form.　(2 marks)

Question 3 | Logarithmic equations (6 marks)

Give exact answers throughout.

(a)　Solve log₃(2x + 1) − log₃(x − 1) = 2.　(4 marks)

(b)　Solve log₂x + log₂(x − 6) = 4.　(2 marks)

Question 4 | Trigonometric equation (6 marks)

Consider the equation 2 sin²θ − sinθ − 1 = 0 for 0 ≤ θ ≤ 2π.

(a)　Factorise the equation and hence find all possible values of sinθ.　(3 marks)

(b)　Hence solve the equation for 0 ≤ θ ≤ 2π, giving answers in terms of π.　(3 marks)

Question 5 | Differentiation and stationary points (7 marks)

Let f(x) = x³ − 6x² + 9x + 2.

(a)　Find f′(x) and write it as f′(x) = 3(x − p)(x − q).　(2 marks)

(b)　Find the coordinates of all stationary points of y = f(x).　(3 marks)

(c)　Determine the nature of each stationary point, justifying your answer.　(2 marks)

Question 6 | The binomial theorem (8 marks)

In the expansion of (2 + 3x)⁵.

(a)　Find the coefficient of x³.　(3 marks)

(b)　Find the coefficient of x⁵.　(1 marks)

(c)　Find the coefficient of x².　(2 marks)

(d)　Hence find the sum of all the coefficients in the expansion.　(2 marks)

Question 7 | Vectors in three dimensions (6 marks)

The vectors are a = 3i − j + 2k and b = −i + 4j + k.

(a)　Find a · b.　(2 marks)

(b)　Find the exact values of |a| and |b| (simplify any surd).　(2 marks)

(c)　Find the constant k such that a + kb is perpendicular to a.　(2 marks)

Question 8 | Probability without replacement (8 marks)

A bag holds 5 red balls and 3 blue balls, 8 balls in total. Two balls are drawn at random without replacement.

(a)　Find the probability that both balls are red.　(2 marks)

(b)　Find the probability that the two balls are of different colours.　(3 marks)

(c)　Find the probability that at least one ball is red.　(3 marks)

Section B (extended response)

Question 9 (14 marks) | A logistic growth model

The number of fish, N, in a pond t months after stocking is modelled by N(t) = 600/(1 + 5e^−0.4t), t ≥ 0.

(a)　Find the number of fish initially placed in the pond (t = 0).　(2 marks)

(b)　Find the limiting value of N as t → ∞ and interpret it.　(2 marks)

(c)　Find the time taken for the number of fish to reach 300, in terms of ln, and correct to three significant figures.　(4 marks)

(d)　Show that N′(t) = 1200e^−0.4t/(1 + 5e^−0.4t)² and find the growth rate at t = 0.　(4 marks)

(e)　Show that N′ = N(600 − N)/1500 and hence explain why the growth rate is greatest when there are 300 fish.　(2 marks)

Question 10 (14 marks) | The cosine rule and a triangle

In triangle ABC, AB = 8 cm, AC = 5 cm and angle BAC = 60°. (No calculator; give exact answers.)

(a)　Find the length of BC.　(3 marks)

(b)　Find the area of triangle ABC, giving an exact surd answer.　(3 marks)

(c)　Find the exact value of sin(angle ABC) and hence find angle ABC to one decimal place.　(3 marks)

(d)　Let M be the midpoint of BC. Find the exact length of the median AM.　(3 marks)

(e)　Find the shortest distance from A to BC.　(2 marks)

Answers and mark scheme

Question 1 (5 marks)

(a)　3 marks

•　M1 writes u₄ = a + 3d = 11 and S₁₀ = 5(2a + 9d) = 155

•　M1 substitutes a = 11 − 3d to get 22 − 6d + 9d = 31, so d = 3

•　A1 d = 3 and a = 2 (both required)

(b)　2 marks

•　M1 substitutes S₂₀ = 10(2 × 2 + 19 × 3)

•　A1 610

Question 2 (6 marks)

(a)　3 marks

•　M1 factors out 2: 2(x² − 4x) + 3

•　M1 completes the square: 2[(x − 2)² − 4] + 3

•　A1 f(x) = 2(x − 2)² − 5

(b)　1 marks

•　A1 (2, −5)

•　M1 from 2(x − 2)² = 5, x − 2 = ±√(5/2) = ±√10/2

•　A1 x = 2 + √10/2 or x = 2 − √10/2 (≈ 3.581, 0.419)

Question 3 (6 marks)

(a)　4 marks

•　M1 combines into log₃[(2x + 1)/(x − 1)] = 2

•　M1 converts to exponential form (2x + 1)/(x − 1) = 9

•　M1 cross-multiplies: 2x + 1 = 9x − 9, so 7x = 10

•　A1 x = 10/7 (≈ 1.43); the domain check x − 1 > 0 holds

(b)　2 marks

•　M1 x(x − 6) = 16, so x² − 6x − 16 = 0 = (x − 8)(x + 2)

•　A1 x = 8 (x = −2 is rejected as it is outside the domain)

Question 4 (6 marks)

(a)　3 marks

•　M1 treats sinθ as the variable: (2 sinθ + 1)(sinθ − 1) = 0

•　A1 sinθ = −1/2

•　A1 sinθ = 1

(b)　3 marks

•　M1 sinθ = 1 gives θ = π/2

•　M1 sinθ = −1/2 gives reference angle π/6 in the third and fourth quadrants: θ = π + π/6 and 2π − π/6

•　A1 θ = π/2, 7π/6, 11π/6 (all three required)

Question 5 (7 marks)

(a)　2 marks

•　M1 differentiates term by term: 3x² − 12x + 9

•　A1 f′(x) = 3(x − 1)(x − 3)

(b)　3 marks

•　M1 sets f′(x) = 0, giving x = 1 or x = 3

•　A1 f(1) = 1 − 6 + 9 + 2 = 6, so (1, 6)

•　A1 f(3) = 27 − 54 + 27 + 2 = 2, so (3, 2)

•　M1 uses f″(x) = 6x − 12 (or a sign table for f′)

•　A1 f″(1) = −6 < 0 so (1, 6) is a maximum; f″(3) = 6 > 0 so (3, 2) is a minimum

Question 6 (8 marks)

(a)　3 marks

•　M1 uses the general term C(5, r) 2^5−r (3x)^r with r = 3

•　M1 evaluates 10 × 2² × 3³ = 10 × 4 × 27

•　A1 1080

(b)　1 marks

•　A1 3⁵ = 243

•　M1 takes r = 2: C(5, 2) 2³ 3² = 10 × 8 × 9

•　A1 720

(d)　2 marks

•　M1 sets x = 1, so the sum of coefficients is (2 + 3)⁵

•　A1 5⁵ = 3125

Question 7 (6 marks)

(a)　2 marks

•　M1 3(−1) + (−1)(4) + 2(1)

•　A1 −5

(b)　2 marks

•　|a| = √(9 + 1 + 4) = √14

•　|b| = √(1 + 16 + 1) = √18 = 3√2

•　M1 perpendicularity gives (a + kb) · a = |a|² + k(a · b) = 14 − 5k = 0

•　A1 k = 14/5 = 2.8

Question 8 (8 marks)

(a)　2 marks

•　M1 C(5, 2)/C(8, 2) = 10/28

•　A1 5/14

(b)　3 marks

•　M1 counts both orders (red then blue, blue then red): 5 × 3 × 2 = 30 ordered outcomes

•　M1 the sample space has 8 × 7 = 56 ordered outcomes, so the probability is 30/56

•　A1 15/28 (≈ 0.536)

•　M1 uses the complement: P(at least one red) = 1 − P(both blue)

•　M1 P(both blue) = C(3, 2)/C(8, 2) = 3/28

•　A1 25/28 (≈ 0.893)

Question 9 (14 marks)

(a)　2 marks

•　M1 substitutes t = 0: e⁰ = 1, so the denominator is 1 + 5 = 6

•　A1 N(0) = 600/6 = 100 fish

(b)　2 marks

•　M1 as t → ∞, e^−0.4t → 0, so the denominator → 1

•　A1 the limit is 600 fish — the long-run carrying capacity of the pond

•　M1 sets 600/(1 + 5e^−0.4t) = 300, giving 1 + 5e^−0.4t = 2

•　M1 simplifies to e^−0.4t = 1/5

•　M1 takes natural logs: −0.4t = −ln 5, so t = ln 5/0.4 = 2.5 ln 5

•　A1 t = 2.5 ln 5 ≈ 4.02 months (3 s.f.)

(d)　4 marks

•　M1 writes N = 600(1 + 5e^−0.4t)⁻¹ and uses the chain rule

•　M1 the inner derivative is 5 × (−0.4)e^−0.4t = −2e^−0.4t, so N′ = −600(1 + 5e^−0.4t)⁻² × (−2e^−0.4t)

•　R1 this simplifies to N′(t) = 1200e^−0.4t/(1 + 5e^−0.4t)², as required

•　A1 N′(0) = 1200/36 = 100/3 ≈ 33.3 fish per month

(e)　2 marks

•　R1 from N = 600/(1 + 5e^−0.4t) we get e^−0.4t = (600 − N)/(5N); substituting into N′ gives N′ = N(600 − N)/1500

•　R1 this is a downward parabola in N with its vertex at N = 300 (i.e. t = 2.5 ln 5 ≈ 4.02), where N′ = 300 × 300/1500 = 60 fish per month, the maximum

Question 10 (14 marks)

(a)　3 marks

•　M1 applies the cosine rule: BC² = 8² + 5² − 2(8)(5)cos 60°

•　M1 cos 60° = 1/2, so BC² = 64 + 25 − 40 = 49

•　A1 BC = 7 cm

(b)　3 marks

•　M1 area = (1/2)(8)(5)sin 60°

•　M1 sin 60° = √3/2, so the area is 20 × √3/2

•　A1 10√3 cm² (≈ 17.3 cm²)

•　M1 applies the sine rule: sin B/5 = sin 60°/7

•　A1 sin B = 5√3/14 (≈ 0.6186)

•　A1 angle ABC ≈ 38.2° (B is acute, so take 38.2° and not 141.8°)

(d)　3 marks

•　M1 uses the median formula AM² = (2AB² + 2AC² − BC²)/4

•　M1 substitutes: (2 × 64 + 2 × 25 − 49)/4 = 129/4

•　A1 AM = √129/2 cm (≈ 5.68 cm)

(e)　2 marks

•　M1 the shortest distance h satisfies (1/2)(BC)h = area, so h = 2 × 10√3/7

•　A1 20√3/7 cm (≈ 4.95 cm)

Grade guide (out of 80, this paper only)

Grade	Marks	Description
7	68–80	Secure across all topics; exact values and reasoning both complete.
6	59–67	Mostly correct; loses marks only on surd simplification or units.
5	50–58	Short questions sound; gaps in the modelling or proof parts of Section B.
4	39–49	Basic manipulation works; clear errors in completing the square, logs or trig equations.
3	28–38	Only some short questions completed; no method established in Section B.
1–2	0–27	Revise the standard question types and practise working without a calculator.

Suggested timing (90 minutes)

Stage	Minutes
Section A questions 1–3 (sequences, quadratics, logs)	10
Section A questions 4–5 (trig equations, differentiation)	15
Section A questions 6–8 (binomial, vectors, probability)	15
Section B question 9 (logistic model)	12
Section B question 10 (cosine rule and triangles)	12
Review: exact values and units, especially surd simplification	26

Notices

•　Time allowed: 1 hour 30 minutes. Total: 80 marks. Section A contains short-response questions; Section B contains extended-response questions.

•　No calculator is permitted (this is a no-calculator paper). Give exact answers: fractions, simplified surds, multiples of π or natural logarithms.

•　Unless a question says otherwise, justify every answer. An answer without working earns no method (M) mark.

•　In the mark scheme M = method, A = answer, R = reasoning. An A mark is only awarded when the preceding M mark is earned.

•　Simplify all surds (write √252 as 6√7) and leave angles in radians in terms of π where required.

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All questions are original and are not IB past-paper questions. The grade bands below are this paper’s own guide; IB grades are awarded on Paper 1, Paper 2 and the Internal Assessment combined.