IB Mathematics AA SL Paper 1 Practice Paper | 10 Questions Without Calculator, 100 Marks

2026-11-01

IB Mathematics AA SL

IB Mathematics AA SL Paper 1 practice paper: indicial and logarithmic equations, composite and inverse functions, completing the square, geometric sequences and sums to infinity, binomial coefficients, the cosine rule and triangle area, stationary points and tangents, definite integrals and the area between curves, vector dot products and angles, and probability with standard deviation; 10 questions for 100 marks in 90 minutes with no calculator, plus full answers and marking notes.

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Section A: Indices, logarithms and functions (Questions 1-3, 30 marks)

Answer all questions. No calculator allowed. Show all working; answers without working score no marks.

#	Question	Answer and marking notes
1	(a) Solve 3^(2x-1) = 27^(x-2). (4 marks)(b) Without a calculator, find the value of log2 48 – log2 3. (3 marks)(c) Solve log5 x + log5 (x-4) = 1. (3 marks)	(a) 27 = 3^3, so 3^(2x-1) = 3^(3x-6); equating indices gives 2x-1 = 3x-6, hence x = 5.(b) log2 48 – log2 3 = log2 (48 / 3) = log2 16 = 4.(c) log5 [x(x-4)] = 1, so x^2 – 4x = 5, x^2 – 4x – 5 = 0, (x-5)(x+1) = 0, x = 5 or x = -1; x = -1 is rejected because the logarithm requires a positive argument, so x = 5.
2	Let f(x) = 2x + 3 and g(x) = x^2 – 1.(a) Find g(f(x)) in simplified form. (3 marks)(b) Find f inverse of x. (3 marks)(c) Solve g(f(x)) = 24. (4 marks)	(a) g(f(x)) = (2x+3)^2 – 1 = 4x^2 + 12x + 9 – 1 = 4x^2 + 12x + 8.(b) Let y = 2x + 3, then x = (y – 3) / 2, so the inverse function is (x – 3) / 2.(c) 4x^2 + 12x + 8 = 24 gives 4x^2 + 12x – 16 = 0, x^2 + 3x – 4 = 0, (x-1)(x+4) = 0, so x = 1 or x = -4; both satisfy the original equation.
3	The curve y = 2x^2 – 12x + 7 is given.(a) Express it in the form y = a(x – h)^2 + k. (3 marks)(b) State the coordinates of the vertex and the minimum value. (3 marks)(c) Solve 2x^2 – 12x + 7 = 0, giving surd form and values to 3 decimal places. (4 marks)	(a) y = 2(x^2 – 6x) + 7 = 2[(x-3)^2 – 9] + 7 = 2(x-3)^2 – 11.(b) Vertex (3, -11); since a = 2 is positive the minimum value is -11.(c) 2(x-3)^2 = 11, (x-3)^2 = 5.5, x = 3 plus or minus sqrt(22)/2, that is x = 5.345 or x = 0.655 (discriminant 144 – 56 = 88, two real roots).

Section B: Sequences, binomial and trigonometry (Questions 4-6, 30 marks)

Answer all questions. No calculator allowed. Show all working; answers without working score no marks.

#	Question	Answer and marking notes
4	A geometric sequence has first term 3 and common ratio 2.(a) Find the 7th term. (3 marks)(b) Find the sum of the first 8 terms. (4 marks)(c) Another geometric sequence has first term 8 and common ratio 1/3; find its sum to infinity. (3 marks)	(a) T(7) = 3 x 2^6 = 3 x 64 = 192.(b) S(8) = 3 x (2^8 – 1) / (2 – 1) = 3 x 255 = 765.(c) Since the ratio is between -1 and 1, S infinity = 8 / (1 – 1/3) = 8 / (2/3) = 12.
5	The expansion of (1 – 2x)^6 in ascending powers of x is required.(a) Find the constant term. (2 marks)(b) Find the coefficient of x^2. (4 marks)(c) Find the coefficient of x^3. (4 marks)	(a) Constant term = C(6,0) x 1^6 = 1.(b) The x^2 term is C(6,2) x (-2x)^2 = 15 x 4x^2, so the coefficient is 60.(c) The x^3 term is C(6,3) x (-2x)^3 = 20 x (-8x^3), so the coefficient is -160.
6	In triangle ABC, a = 7 cm, b = 9 cm and angle C = 60 degrees.(a) Find c exactly and to 3 decimal places. (4 marks)(b) Find the area of the triangle to 3 significant figures. (3 marks)(c) Find angle A to 1 decimal place. (3 marks)	(a) c^2 = 49 + 81 – 2 x 7 x 9 x cos 60 = 49 + 81 – 63 = 67, so c = sqrt(67) = 8.185 cm.(b) Area = 0.5 x 7 x 9 x sin 60 = 31.5 x 0.8660 = 27.3 cm^2.(c) By the sine rule sinA / 7 = sin60 / c, so sinA = 7 x 0.8660 / 8.185 = 0.7406, giving A = 47.8 degrees and B = 72.2 degrees.

Section C: Differentiation and integration (Questions 7-8, 20 marks)

Answer all questions. No calculator allowed. Show all working; answers without working score no marks.

#	Question	Answer and marking notes
7	Let f(x) = x^3 – 6x^2 + 9x + 2.(a) Find f dash of x, and the x values of all stationary points with their f(x) values. (4 marks)(b) Find the equation of the tangent to the curve at x = 2. (3 marks)(c) Use the second derivative to classify each stationary point. (3 marks)	(a) f dash of x = 3x^2 – 12x + 9 = 3(x-1)(x-3); setting it to zero gives x = 1 or x = 3, with f(1) = 6 and f(3) = 2.(b) f dash of 2 = 12 – 24 + 9 = -3 and f(2) = 4, so y – 4 = -3(x – 2), that is y = -3x + 10.(c) f double dash of x = 6x – 12; at x = 1 it is -6 (maximum at (1, 6)), at x = 3 it is 6 (minimum at (3, 2)).
8	(a) Evaluate the integral from 1 to 4 of (3x^2 – 2x + 1) dx. (4 marks)(b) Find the area enclosed by y = x^2 and y = 2x. (3 marks)(c) Evaluate the integral from 0 to 2 of (4 – x^2) dx, giving a fraction. (3 marks)	(a) An antiderivative is x^3 – x^2 + x; at 4 it is 52 and at 1 it is 1, so the value is 51.(b) The curves meet at x = 0 and x = 2; area = integral of (2x – x^2) = [x^2 – x^3/3] from 0 to 2 = 4 – 8/3 = 4/3 = 1.333.(c) Integral of (4 – x^2) = [4x – x^3/3] from 0 to 2 = 8 – 8/3 = 16/3 = 5.333.

Section D: Vectors, probability and statistics (Questions 9-10, 20 marks)

Answer all questions. No calculator allowed. Show all working; answers without working score no marks.

#	Question	Answer and marking notes
9	Given the vectors a = (2, -1) and b = (-1, 3).(a) Find a dot b, the magnitude of a and the magnitude of b. (4 marks)(b) Find the angle between a and b to 1 decimal place. (3 marks)(c) Find 2a – 3b and its magnitude to 3 significant figures. (3 marks)	(a) a dot b = 2 x (-1) + (-1) x 3 = -5; magnitude of a = sqrt(5) = 2.236; magnitude of b = sqrt(10) = 3.162.(b) cos of the angle = -5 / (2.236 x 3.162) = -0.7071, so the angle is 135.0 degrees.(c) 2a – 3b = (7, -11), magnitude = sqrt(170) = 13.0.
10	A bag contains 5 red, 3 blue and 2 green balls, 10 balls in total.(a) Two balls are drawn at random without replacement; find the probability both are red as a fraction. (4 marks)(b) Find the probability that at least one ball is red as a fraction. (3 marks)(c) For the data 4, 7, 7, 9, 12, 15 find the mean, median and standard deviation to 3 significant figures. (3 marks)	(a) The probability is (5/10) x (4/9) = 20/90 = 2/9 = 0.222.(b) The complement is drawing two non-red balls, also 2/9, so the answer is 1 – 2/9 = 7/9 = 0.778.(c) Mean = 54/6 = 9; median = (7 + 9)/2 = 8; variance = 78/6 = 13, so the standard deviation is sqrt(13) = 3.61.

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