S4 Mathematics Practice Paper: Quadratic Equations, Factor Theorem, Indices And Logarithms, Sequences, Variation, Coordinate Geometry, Trigonometry, Statistics And Probability (12 Questions, 100 Marks, 75 Minutes, Answers Included)

S4 Mathematics Practice Paper: Quadratic Equations, Factor Theorem, Indices and Logarithms, Sequences, Variation, Coordinate Geometry, Trigonometry, Statistics and Probability (12 Questions, 100 Marks, 75 Minutes, Answers Included)

2026-11-06

S4 Maths: Quadratics, Logarithms, Sequences and Coordinate Geometry

S4 Maths paper: 12 questions, 100 marks, 75 minutes on quadratics, factor theorem, logarithms, sequences, variation, coordinate geometry, trigonometry, statistics and probability.

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Section A: Quadratic Equations and Polynomials (24 marks)

Each question carries 6 marks; show your working or reasoning.

#	Question	Answer and marking notes
1	Let α and β be the roots of 2x² − 7x + 3 = 0. (a) Solve the equation by factorisation. (b) Find α + β and αβ, and hence find the value of the discriminant. (c) Form the quadratic equation whose roots are α + 1 and β + 1, giving it in the form ax² + bx + c = 0 with integer coefficients.	(a) 2x² − 7x + 3 = (2x − 1)(x − 3) = 0, so x = 1/2 or x = 3. (b) α + β = 7/2 and αβ = 3/2; the discriminant is 49 − 24 = 25, which is positive, so the roots are real and distinct. (c) New sum = 7/2 + 2 = 11/2 and new product = (1/2 + 1)(3 + 1) = 6, so the equation is x² − (11/2)x + 6 = 0, that is 2x² − 11x + 12 = 0. Marking points: correct factorisation, sum and product of roots, forming an equation from its roots.
2	Let p(x) = 2x³ + ax² − 5x + 6, where a is a constant. It is given that p(1) = 0. (a) Find a. (b) Factorise p(x) into a product of linear factors. (c) Hence solve p(x) = 0 and list all the roots.	(a) p(1) = 2 + a − 5 + 6 = a + 3 = 0, so a = −3. (b) p(x) = 2x³ − 3x² − 5x + 6 = (x − 1)(2x² − x − 6) = (x − 1)(2x + 3)(x − 2). (c) Hence x = 1, x = 2 or x = −3/2. Marking points: substitution into the factor theorem, long division or comparing coefficients, complete factorisation.
3	(a) Without using a calculator, simplify (27^(2/3) × 9^(−1/2)) ÷ 81^(1/4), showing each step. (b) Solve 4^(x+1) = 8^(2x−3). (c) Given 5^a = 2 and 5^b = 3, find the value of 5^(2a+b).	(a) 27^(2/3) = 9, 9^(−1/2) = 1/3 and 81^(1/4) = 3, so the expression equals 9 × (1/3) ÷ 3 = 1. (b) 4^(x+1) = 2^(2x+2) and 8^(2x−3) = 2^(6x−9); comparing indices gives 2x + 2 = 6x − 9, so x = 11/4 = 2.75. (c) 5^(2a+b) = (5^a)² × 5^b = 2² × 3 = 12. Marking points: handling fractional and negative indices, comparing indices with the same base, combining index laws.
4	A rectangle has perimeter 26 cm and area 40 cm². Let its length be x cm. (a) Show that x satisfies x² − 13x + 40 = 0. (b) Find the length and the width. (c) Find the length of a diagonal, correct to 2 decimal places.	(a) The width is (13 − x) cm, so x(13 − x) = 40, which expands to x² − 13x + 40 = 0. (b) (x − 5)(x − 8) = 0, so the sides are 8 cm and 5 cm. (c) The diagonal is √(8² + 5²) = √89 ≈ 9.43 cm. Marking points: defining the unknown and forming the equation, factorisation, taking the positive root.

Section B: Indices, Logarithms and Sequences (36 marks)

Each question carries 9 marks; note the marks for each part.

#	Question	Answer and marking notes
5	(a) Solve log₂(x + 3) + log₂(x − 3) = 4, explaining why one of the solutions must be rejected. (b) Solve log₄(x + 12) = 2. (c) Without using a calculator, evaluate log 8 ÷ log 2 + log 25 + log 4, where the logarithms are to base 10.	(a) Combining the logarithms gives log₂(x² − 9) = 4, so x² − 9 = 16, x² = 25, x = 5 or x = −5. Since the argument of a logarithm must be positive (x greater than 3), x = −5 is rejected and the answer is x = 5. (b) log₄(x + 12) = 2 gives x + 12 = 16, so x = 4. (c) log 8 ÷ log 2 = 3 and log 25 + log 4 = log 100 = 2, so the total is 5. Marking points: log laws, converting to index form, checking the domain.
6	The value V (in dollars) of a car n years after purchase is V = 240000 × 0.85ⁿ. (a) Find its value 3 years after purchase, correct to the nearest dollar. (b) Find the least number of years after which the value first falls below $100,000, showing your use of logarithms. (c) Another car depreciates at a constant rate r per year and is worth exactly half its original value after 4 years. Find r, correct to 0.1%.	(a) V = 240000 × 0.85³ = $147,390. (b) We need 0.85ⁿ less than 5/12; taking logarithms gives n greater than 5.387, so the least number of years is 6 (after 5 years the value is $106,489 and after 6 years it is $90,516). (c) (1 − r)⁴ = 0.5, so 1 − r = 0.5^(1/4) = 0.8409 and r = 0.1591, that is about 15.9% per year. Marking points: correct substitution, logarithms applied to an inequality, fourth root expressed as a percentage.
7	(a) An arithmetic sequence has first term 7 and common difference 4. Find its 20th term. (b) Find the sum of the first 20 terms. (c) A geometric sequence has first term 3 and common ratio 2; find the sum of its first 8 terms. (d) Using (a) and (b), find the least n such that the sum of the first n terms of the arithmetic sequence exceeds 1000, supporting your answer with calculation.	(a) 20th term = 7 + 19 × 4 = 83. (b) Sum of the first 20 terms = 20/2 × (7 + 83) = 900.0. (c) Sum of the first 8 terms = 3(2⁸ − 1)/(2 − 1) = 765. (d) The sum of the first n terms is n(2n + 5); it is 987 when n = 21 and 1078 when n = 22, so the least n is 22. Marking points: general term, arithmetic and geometric sum formulae, checking the boundary value.
8	It is given that y varies directly as x² and inversely as z. When x = 4 and z = 2, y = 24. (a) Find the constant of variation and write down the relation between y, x and z. (b) Find y when x = 6 and z = 9. (c) Find z when y = 9 and x = 3.	(a) Let y = kx²/z; substituting gives 24 = k × 16 ÷ 2, so k = 3.0 and y = 3x²/z. (b) y = 3 × 36 ÷ 9 = 12.0. (c) From 9 = 3 × 9 ÷ z we get z = 3.0. Marking points: writing the variation equation, finding the constant, substituting back.

Section C: Coordinate Geometry, Trigonometry, Statistics and Probability (40 marks)

Each question carries 10 marks; give full working or explanations.

#	Question	Answer and marking notes
9	A(−2, 1) and B(4, 5) are two points on a coordinate plane. (a) Find the length of AB, in surd form and correct to 2 decimal places. (b) Find the coordinates of the midpoint of AB. (c) Find the equation of the perpendicular bisector of AB in the form Ax + By + C = 0. (d) Find where this perpendicular bisector meets the y-axis, and describe its distance from A and from B.	(a) AB = √[(4 + 2)² + (5 − 1)²] = √52 = 2√13 ≈ 7.21. (b) Midpoint = ((−2 + 4)/2, (1 + 5)/2) = (1, 3). (c) The slope of AB is (5 − 1)/(4 + 2) = 2/3, so the perpendicular bisector has slope −3/2; through (1, 3) it is y − 3 = −(3/2)(x − 1), that is 3x + 2y − 9 = 0. (d) Putting x = 0 gives y = 4.5, so the y-intercept is (0, 4.5); substituting shows it is equidistant from A and B, as every point on a perpendicular bisector must be. Marking points: distance formula, midpoint formula, product of perpendicular slopes, standard form.
10	In triangle ABC, AB = 10 cm, AC = 14 cm and angle A = 60°. (a) Find BC, correct to 2 decimal places. (b) Find the area of triangle ABC, in surd form and correct to 2 decimal places. (c) Find angle B, correct to 0.1°, stating the formula you use.	(a) BC² = 10² + 14² − 2 × 10 × 14 × cos60° = 100 + 196 − 140 = 156, so BC = 2√39 ≈ 12.49 cm. (b) Area = ½ × 10 × 14 × sin60° = 35√3 ≈ 60.62 cm². (c) By the sine rule sinB ÷ 14 = sin60° ÷ BC, so sinB ≈ 0.9707 and angle B ≈ 76.1°. Marking points: cosine rule, area formula, sine rule with a sensible angle.
11	The data set is 4, 7, 7, 9, 11, 13, 15, 18. (a) Find the mean. (b) Find the median. (c) Find the mode. (d) Find the standard deviation, correct to 2 decimal places. (e) If 3 is added to every value, describe how the mean, median, standard deviation and range change, and explain why.	(a) Mean = (4 + 7 + 7 + 9 + 11 + 13 + 15 + 18) ÷ 8 = 10.5. (b) Median = (9 + 11) ÷ 2 = 10. (c) Mode = 7. (d) The sum of squared deviations is 152, so the variance is 19 and the standard deviation is about 4.36. (e) Adding 3 to every value makes the mean 13.5 and the median 13; the standard deviation and the range are unchanged (still 4.36 and 14) because a translation does not change the spread of the data. Marking points: definitions of mean and median, sum of squared deviations, understanding of translation.
12	A bag contains 5 red, 3 blue and 2 green balls, identical except in colour. (a) One ball is drawn at random; find the probability that it is red. (b) Two balls are drawn without replacement; find the probability that both are red. (c) Two balls are drawn without replacement; find the probability that at least one is green. (d) Two balls are drawn with replacement; find the probability of exactly one red and one blue.	(a) P(red) = 5/10 = 0.5. (b) P(both red) = (5/10) × (4/9) = 2/9 ≈ 0.222. (c) P(at least one green) = 1 − (8/10)(7/9) = 1 − 28/45 = 17/45 ≈ 0.378. (d) P(exactly one red and one blue) = 2 × (5/10) × (3/10) = 0.3. Marking points: decreasing denominators without replacement, complement for at least one, allowing for both orders.

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