S6 Mathematics Extended Part Module 2 Practice Paper: Binomial Theorem, Sequences, Trigonometry, Calculus, Vectors And Matrices (100 Marks)

S6 Mathematics Extended Part Module 2 Practice Paper: Binomial Theorem, Sequences, Trigonometry, Calculus, Vectors and Matrices (100 marks)

2026-11-13

S6 M2 – algebra and calculus

S6 M2 practice paper, 6 questions and 24 parts, 100 marks in 60 minutes, on binomial theorem, sequences, trigonometry, calculus, vectors and matrices, with full solutions.

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Paper details　S6 M2 practice paper with 6 questions and 24 parts on the binomial theorem, sequences, induction, trigonometry, limits, differentiation, integration, vectors and matrices, with full solutions.

Section A: Binomial Theorem (20 marks)

Show your working. Give exact values unless a decimal is requested.

#	Question	Answer and marking notes
1	In the expansion of (1 + 2x)^8, find the coefficient of x^3.	Answer: the x^3 term is C(8,3)(2x)^3 = 56 x 8 x^3 = 448x^3, so the coefficient is 448. (Marking: binomial coefficient 2, power 2)
2	Find the coefficient of x^4 in the same expansion and write down that term.	Answer: C(8,4)(2x)^4 = 70 x 16 x^4 = 1120x^4, coefficient 1120 and the term is 1120x^4. (Marking: coefficient 2, term 2)
3	Find the term with the largest coefficient in the expansion and explain how you decide it.	Answer: the ratio of consecutive coefficients is 2(8 – k)/(k + 1), which exceeds 1 for k less than 5, equals 1 at k = 5 and falls below 1 afterwards; so the ratio equals 1 at k = 5, so the largest coefficient appears at both k = 5 and k = 6, C(8,5)(2)^5 = C(8,6)(2)^6 = 56 x 32 = 1792, giving 1792x^5 and 1792x^6. (Marking: ratio method 3, k = 5 and 6 1, answer 2)
4	Find the sum of all coefficients in the expansion and state the method used.	Answer: put x = 1 to get (1 + 2)^8 = 3^8 = 6561, so the sum of the coefficients is 6561. (Marking: idea of x = 1 3, computation 2, explanation 1)

Section B: Sequences and Mathematical Induction (16 marks)

Each part carries 4 marks. Quote the formula you use.

#	Question	Answer and marking notes
5	An arithmetic sequence has first term 7 and common difference 4. Find the 20th term.	Answer: T(20) = 7 + 19 x 4 = 83. (Marking: formula 2, answer 2)
6	For the same sequence, find the sum of the first 20 terms.	Answer: S(20) = 20(7 + 83)/2 = 20 x 90/2 = 900. (Marking: pairing or formula 2, answer 2)
7	A geometric sequence has first term 3 and common ratio 2. Find the sum of the first 10 terms.	Answer: S(10) = 3(2^10 – 1)/(2 – 1) = 3 x 1023 = 3069. (Marking: formula 2, 2^10 1, answer 1)
8	Prove by mathematical induction that the sum of the first n odd numbers is n^2, and check with n = 10.	Answer: for n = 1, 1 = 1^2 holds; assume it holds for n = k, then 1 + 3 + … + (2k – 1) + (2k + 1) = k^2 + 2k + 1 = (k + 1)^2, so it holds for all positive integers. Check n = 10: the sum is 100 = 10^2. (Marking: base case 1, induction step 2, conclusion and check 1)

Section C: Trigonometry (16 marks)

Unless stated otherwise, give angles in degrees.

#	Question	Answer and marking notes
9	Solve sin θ = 0.5 and write the general solution with all values between 0 and 360 degrees.	Answer: the basic angle is 30 degrees; between 0 and 360 degrees, θ = 30 or 150 degrees; the general solution is θ = 30 + 360k or 150 + 360k for integer k. (Marking: basic angle 1, two solutions 2, general form 1)
10	Given tan θ = 3/4 and θ is in the third quadrant, find the exact values of sin θ and cos θ.	Answer: from the 3-4-5 triangle the hypotenuse is 5; in the third quadrant both sine and cosine are negative, so sin θ = -3/5 and cos θ = -4/5. (Marking: hypotenuse 1, signs 2, values 1)
11	Prove the identity (1 – cos 2θ) / sin 2θ = tan θ, where the denominator is non-zero.	Answer: 1 – cos 2θ = 2sin^2 θ and sin 2θ = 2 sin θ cos θ, so the quotient is (2sin^2 θ)/(2 sin θ cos θ) = sin θ / cos θ = tan θ. (Marking: double angle formulas 2, simplification 1, conclusion 1)
12	Find the maximum value of 3 sin θ + 4 cos θ and the value of tan θ when it is attained.	Answer: combine as R sin(θ + α): the maximum is 5 and it occurs when tan θ = 3/4. (Marking: R = 5 2, angle 2)

Section D: Limits and Differentiation (16 marks)

Show the rule or limiting step you use.

#	Question	Answer and marking notes
13	Evaluate the limit as x tends to 3 of (x^2 – 9)/(x – 3).	Answer: factorise the numerator as (x – 3)(x + 3), cancel x – 3 to get x + 3, then substitute x = 3 to get 6. (Marking: factorisation 2, substitution 2)
14	For the curve y = x^3 – 2x, find the equation of the tangent at x = 1.	Answer: y(1) = 1 – 2 = -1 and dy/dx = 3x^2 – 2, so the slope at x = 1 is 1; the tangent is y + 1 = 1(x – 1), that is y = x – 2. (Marking: point 1, slope 2, equation 1)
15	Find d/dx [x^3 ln x] for x greater than 0.	Answer: by the product rule, 3x^2 ln x + x^3 x (1/x) = 3x^2 ln x + x^2. (Marking: product rule 2, simplification 2)
16	Find the stationary points of y = x^3 – 3x and determine their nature.	Answer: dy/dx = 3x^2 – 3 = 0 gives x = plus or minus 1; the second derivative 6x is -6 at x = -1 (maximum, y = 2) and 6 at x = 1 (minimum, y = -2). (Marking: stationary points 2, nature 2)

Section E: Integration and Area (16 marks)

For definite integrals, show the primitive and the limits.

#	Question	Answer and marking notes
17	Evaluate the integral of 3x^2 + 2x from 0 to 2.	Answer: the primitive is x^3 + x^2; at the upper limit 2 it is 8 + 4 = 12 and at 0 it is 0, so the integral is 12. (Marking: primitive 2, evaluation 2)
18	Find the area enclosed by y = x and y = x^2 between 0 and 1.	Answer: area = integral of (x – x^2) from 0 to 1 = 1/2 – 1/3 = 1/6, about 0.167 square units. (Marking: integrand 2, computation 1, answer 1)
19	Use integration by parts to evaluate the integral of x e^x from 0 to 1.	Answer: take u = x and dv = e^x dx to get the primitive e^x(x – 1); at 1 it is 0 and at 0 it is -1, so the integral is 1. (Marking: by parts 2, limits 2)
20	Evaluate the integral of 1/x from 1 to 2, giving the answer in terms of ln and to 3 decimal places.	Answer: the primitive is ln x, so the integral is ln 2 – ln 1 = ln 2, about 0.693. (Marking: primitive 2, exact value 1, decimal 1)

Section F: Vectors and Matrices (16 marks)

Give non-integer answers to 3 decimal places.

#	Question	Answer and marking notes
21	Given a = (2, -1, 3) and b = (1, 4, -2), find a dot b.	Answer: a dot b = 2(1) + (-1)(4) + 3(-2) = 2 – 4 – 6 = -8. (Marking: term by term 2, total 2)
22	Find the magnitudes of a and b, and the angle between them to 1 decimal place.	Answer: \|a\| = sqrt 14 is about 3.742 and \|b\| = sqrt 21 is about 4.583; cos θ = -8/(sqrt 14 x sqrt 21) is about -0.467, so θ is about 117.8 degrees. (Marking: magnitudes 2, cosine 1, angle 1)
23	Find a cross b and state how the result relates to a and b.	Answer: a cross b = ((-1)(-2) – 3(4), 3(1) – 2(-2), 2(4) – (-1)(1)) = (-10, 7, 9); this vector is perpendicular to both a and b. (Marking: cross product 3, perpendicularity 1)
24	The matrix A has rows (3, 1) and (2, 4). Find det A and the inverse of A.	Answer: det A = 3(4) – 1(2) = 10; the inverse is (1/10) times the matrix with rows (4, -1) and (-2, 3), that is rows (0.4, -0.1) and (-0.2, 0.3). (Marking: determinant 2, inverse 2)

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