Mark Scheme

Section A — Multiple Choice Answer Key

Detailed Mark Scheme with Working

Question 1: f(-2) = 3(-2)² - 5(-2) + 2 = 3(4) + 10 + 2 = 12 + 10 + 2 = 24. Answer: C

Question 2: Rearranging 2y + 3x = 6 gives y = -3x/2 + 3, so gradient = -3/2. Perpendicular gradient = 2/3. Answer: C

Question 3: 32 = 2⁵, so log₂ 32 = 5. Answer: A

Question 4: Factorising: (x - 3)(x - 4) = 0, so x = 3 or x = 4. Answer: B

Question 5: (3 - 2)i + (4 + 5)j = i + 9j. Answer: A

Question 6: dy/dx = 3x² - 6x = 3x(x - 2) = 0 when x = 0 or x = 2. Answer: C

Question 7: Translation 3 units right replaces x with (x - 3). Answer: A

Question 8: V = (1/3)πr²h = (1/3)π(25)(12) = 100π. Answer: A

Question 9: P(1) = 2(1) + 5(1) - 4(1) + 1 = 2 + 5 - 4 + 1 = 4. Answer: B

Question 10: S∞ = a/(1 - r) = 8/(1 - 1/2) = 8/(1/2) = 16. Answer: C

Question 11: Using Pythagoras: cos θ = 4/5 (since 3² + 4² = 5²). Answer: B

Question 12: 2x - 5 < 3x + 1 gives -6 < x, so x > -6. Answer: A

Question 13: v = ds/dt = 6t² - 18t + 12. At t = 2: v = 24 - 36 + 12 = 0. Answer: A

Question 14: Det = (2)(4) - (3)(1) = 8 - 3 = 5. Answer: A

Question 15: 3^(2x+1) = 27 = 3³, so 2x + 1 = 3, giving x = 1. Answer: B

Question 16: Standard formula for angle between vectors. Answer: A

Question 17: Maximum occurs at t = -b/(2a) = -16/(2 × -2) = 4. Answer: B

Question 18: (x + 2)² - (x - 3)² = (x² + 4x + 4) - (x² - 6x + 9) = 10x - 5. Answer: A

Question 19: By chain rule: dy/dx = 1/(2x + 1) × 2 = 2/(2x + 1). Answer: B

Question 20: Equal roots when discriminant b² - 4ac = 0, so b² = 4ac. Answer: A

Question 21: ∫(6x² - 4x + 3) dx = 2x³ - 2x² + 3x + c. Answer: A

Question 22: First term a = 5, common difference d = 4, so Tₙ = a + (n-1)d = 5 + 4n - 4 = 4n + 1. Answer: A

Question 23: Adding corresponding elements: (3+2, -1+0; 2+1, 4+3) = (5, -1; 3, 7). Answer: A

Question 24: dV/dt at t = 3: 3(9) + 2(3) = 27 + 6 = 33. Answer: B

Question 25: dy/dx = 2x + 3. At x = 1: gradient = 5, y = 1 + 3 - 2 = 2. Using y - 2 = 5(x - 1): y = 5x - 3. Answer: A

Question 26: Factor theorem: if (x - a) is a factor, then P(a) = 0. Answer: A

Question 27: sin 2x = 1/2 means 2x = 15° or 165° (in range 0° to 360°), so x = 15° or 165° (considering 2x can be 15°, 165°, 375°, 525°). Accept x = 15° and x = 165° as primary solutions. Answer: C

Question 28: ∫₀² x² dx = [x³/3]₀² = 8/3. Answer: C

Question 29: |F| = √(16 + 9) = √25 = 5. Answer: B

Question 30: x = 10^2.5 = 10² × 10^0.5 = 100√10. Answer: B

Question 31: Using binomial theorem: ⁴C₂ × 2² × x² = 6 × 4 = 24. Answer: C

Question 32: Let y = (x - 3)/2, then 2y = x - 3, so x = 2y + 3. Thus f⁻¹(x) = 2x + 3. Answer: A

Question 33: s = ∫(4t - 3) dt = 2t² - 3t + c. Since s = 0 when t = 0, c = 0. At t = 4: s = 32 - 12 = 20. Answer: B

Question 34: 2^(x+1) = 8^x = (2³)^x = 2^(3x), so x + 1 = 3x, giving x = 1/2. Answer: C

Question 35: In second quadrant, cos x = -√3/2 gives x = 150°. Answer: C

Question 36: Vertex form shows minimum at (2, 5). Answer: A

Question 37: Using chain rule: dy/dx = 3(2x - 1)² × 2 = 6(2x - 1)². Answer: B

Question 38: Using cosine rule: c² = a² + b² - 2ab cos C = 49 + 25 - 2(7)(5)(1/2) = 74 - 35 = 39. Answer: B

Question 39: Maximum at x = -b/(2a) = -6/(2 × -0.5) = 6. Answer: B

Question 40: y = ∫(6x² - 4) dx = 2x³ - 4x + c. When x = 1, y = 5: 2 - 4 + c = 5, so c = 7. Thus y = 2x³ - 4x + 7. Answer: A

Question 41: S₁₀ = n/2[2a + (n-1)d] = 5[6 + 36] = 5(42) = 210. Answer: B

Question 42: Since x² ≥ 0, then x² + 1 ≥ 1. Answer: C

Question 43: (2 1)(3) = (6+4) = (10). Answer: B
(0 3)(4) (0+12) (12)

Question 44: Standard substitution technique for trigonometric quadratic equations. Answer: A

Question 45: v = ∫(6t - 4) dt = 3t² - 4t + c. Starts from rest means v = 0 when t = 0, so c = 0. At t = 2: v = 12 - 8 = 4. Answer: A

Sample Answers with Examiner Commentary

Question 13 — Sample Answers

(Question 13: A particle moves in a straight line such that its displacement, s metres, from a fixed point O at time t seconds is given by s = 2t³ - 9t² + 12t. The velocity of the particle when t = 2 is)

Grade I (Distinction) answer

Given: s = 2t³ - 9t² + 12t

Velocity v = ds/dt

Differentiating with respect to t: v = d/dt(2t³ - 9t² + 12t) v = 6t² - 18t + 12

When t = 2: v = 6(2)² - 18(2) + 12 v = 6(4) - 36 + 12 v = 24 - 36 + 12 v = 0 m/s

Therefore, the velocity when t = 2 is 0 m/s.

Mark: 1/1

Examiner commentary: This response demonstrates complete understanding of the relationship between displacement and velocity through differentiation. The candidate correctly applies the power rule to each term, substitutes t = 2 accurately, and arrives at the correct answer with clear working shown throughout.

Grade III (Pass) answer

v = ds/dt = 6t² - 18t + 12

When t = 2: v = 6(4) - 18(2) + 12 v = 24 - 36 + 12 v = 0

Mark: 1/1

Examiner commentary: This response demonstrates adequate understanding and arrives at the correct answer. While the working is less detailed than the Grade I response, all essential steps are present and the differentiation is correctly performed. For a multiple-choice question, this level of working is sufficient.

Grade V (Near miss) answer

v = 2t³ - 9t² + 12t

When t = 2: v = 2(8) - 9(4) + 12(2) v = 16 - 36 + 24 v = 4 m/s

Mark: 0/1

Examiner commentary: This response demonstrates a fundamental misconception: the candidate has substituted t = 2 directly into the displacement equation rather than first differentiating to find velocity. This is a common error at CSEC level where students confuse the physical quantities. To improve, the candidate must recognize that velocity is the rate of change of displacement and requires differentiation before substitution.

Question 28 — Sample Answers

(Question 28: The area bounded by the curve y = x², the x-axis, and the lines x = 0 and x = 2 is)

Grade I (Distinction) answer

To find the area bounded by y = x², the x-axis, and x = 0 and x = 2, I need to integrate:

Area = ∫₀² x² dx

Integrating: = [x³/3]₀²

Evaluating the definite integral: = (2³/3) - (0³/3) = 8/3 - 0 = 8/3 square units

Therefore, the area is 8/3 square units.

Mark: 1/1

Examiner commentary: Excellent response showing complete mastery of definite integration for area calculation. The candidate correctly sets up the integral with proper limits, integrates accurately using the power rule, and evaluates the definite integral correctly. The answer is clearly stated with appropriate units.

Grade III (Pass) answer

Area = ∫₀² x² dx

= [x³/3]₀²

= 8/3 - 0 = 8/3

Mark: 1/1

Examiner commentary: This response shows adequate understanding of the area calculation process and arrives at the correct answer. The working is concise but includes all essential steps. For a multiple-choice question at CSEC level, this demonstrates sufficient competence in applying definite integration.

Grade V (Near miss) answer

∫ x² dx = 2x

Area = 2(2) - 2(0) = 4

Mark: 0/1

Examiner commentary: This response contains a critical error in the integration process: the candidate has incorrectly differentiated rather than integrated (or misapplied the power rule for integration). The correct integral of x² is x³/3, not 2x. This is a common misconception where students confuse the integration and differentiation rules for powers. To improve, the candidate should review the power rule for integration: ∫xⁿ dx = x^(n+1)/(n+1) + c, and practice distinguishing between when to differentiate versus integrate in application problems.

Question 40 — Sample Answers

(Question 40: If dy/dx = 6x² - 4 and y = 5 when x = 1, then y equals)

Grade I (Distinction) answer

Given: dy/dx = 6x² - 4 and boundary condition y = 5 when x = 1

To find y, I need to integrate dy/dx:

y = ∫(6x² - 4) dx

y = 6x³/3 - 4x + c

y = 2x³ - 4x + c

Now using the boundary condition to find c: When x = 1, y = 5: 5 = 2(1)³ - 4(1) + c 5 = 2 - 4 + c 5 = -2 + c c = 7

Therefore: y = 2x³ - 4x + 7

Mark: 1/1

Examiner commentary: Outstanding response demonstrating complete understanding of indefinite integration and the application of boundary conditions. The candidate correctly integrates term by term, applies the boundary condition systematically to determine the constant of integration, and presents the final answer in proper form. This shows mastery of the connection between differential and integral calculus.

Grade III (Pass) answer

y = ∫(6x² - 4) dx = 2x³ - 4x + c

When x = 1, y = 5: 5 = 2 - 4 + c c = 7

So y = 2x³ - 4x + 7

Mark: 1/1

Examiner commentary: This response demonstrates solid understanding and reaches the correct answer efficiently. While less detailed than the Grade I response, all key steps are present: correct integration, proper application of the boundary condition, and accurate determination of c. This represents competent work at CSEC standard.

Grade V (Near miss) answer

y = 2x³ - 4x

When x = 1, y = 5: 5 = 2(1) - 4(1) 5 = -2

This doesn't work, so y = 2x³ - 4x + 5

Mark: 0/1

Examiner commentary: This response shows partial understanding but contains a critical procedural error. The candidate has correctly integrated to get 2x³ - 4x but has omitted the constant of integration initially. When attempting to use the boundary condition, they recognize an inconsistency but incorrectly add 5 (the y-value) as the constant rather than solving for c properly. The correct process requires: 5 = 2 - 4 + c, therefore c = 7. This demonstrates incomplete understanding of how boundary conditions determine the constant of integration. To improve, the candidate should always include + c when integrating, then solve algebraically for c using the given condition.

Question 39 — Sample Answers

(Question 39: A farmer in St. Vincent plants crops whose yield Y (in tonnes per hectare) after x weeks is modeled by Y = -0.5x² + 6x + 2. The maximum yield occurs after)

Grade I (Distinction) answer

Given: Y = -0.5x² + 6x + 2

This is a quadratic function in the form Y = ax² + bx + c where a = -0.5, b = 6, c = 2.

Since a < 0, the parabola opens downward and has a maximum value.

The maximum occurs at the vertex, where x = -b/(2a)

x = -6/(2 × -0.5) x = -6/(-1) x = 6

Therefore, the maximum yield occurs after 6 weeks.

Mark: 1/1

Examiner commentary: Exemplary response showing deep understanding of quadratic functions and their applications. The candidate correctly identifies the structure of the quadratic, recognizes that a negative coefficient of x² indicates a maximum, and accurately applies the vertex formula. The contextual interpretation linking the mathematical result to the agricultural scenario shows excellent application skills valued in CSEC Additional Mathematics.

Grade III (Pass) answer

Y = -0.5x² + 6x + 2

Maximum at x = -b/(2a)

x = -6/(2 × -0.5) = -6/(-1) = 6

Maximum yield after 6 weeks.

Mark: 1/1

Examiner commentary: This response demonstrates sound understanding of the vertex formula for quadratic functions and applies it correctly to reach the answer. While less detailed in explanation than the Grade I response, the essential mathematical steps are present and accurate, which is sufficient for this multiple-choice question.

Grade V (Near miss) answer

dY/dx = -x + 6

Maximum when dY/dx = 0: -x + 6 = 0 x = 6

Wait, let me check: dY/dx = -1.0x + 6 So x = 6

Maximum after 6 weeks.

Mark: 0/1 (note: while arriving at correct answer, the differentiation shows conceptual errors)

Actually, upon reconsideration for marking fairness in a multiple-choice context: Mark: 1/1

Examiner commentary: This response demonstrates an alternative calculus-based approach but with a concerning error in differentiation: the derivative of -0.5x² should be -1.0x (or simply -x), which the candidate eventually states but initially writes as just "-x" without the coefficient. Despite this inconsistency in working, the candidate does arrive at the correct value x = 6 and the correct answer. For a multiple-choice question, the final answer is credited. However, in a structured question requiring shown work, marks would be deducted for the incomplete differentiation. To strengthen understanding, the candidate should carefully apply the power rule: d/dx(-0.5x²) = -0.5(2x) = -x, and present working more systematically.

Question 37 — Sample Answers

(Question 37: The gradient function of y = (2x - 1)³ is)

Grade I (Distinction) answer

Given: y = (2x - 1)³

To find dy/dx, I'll use the chain rule.

Let u = 2x - 1, so y = u³

Then: dy/du = 3u² and du/dx = 2

By the chain rule: dy/dx = dy/du × du/dx dy/dx = 3u² × 2 dy/dx = 6u²

Substituting back u = 2x - 1: dy/dx = 6(2x - 1)²

Therefore, the gradient function is 6(2x - 1)².

Mark: 1/1

Examiner commentary: Excellent response demonstrating thorough understanding of the chain rule for differentiation. The candidate clearly identifies the composite function structure, explicitly states the substitution, applies the chain rule correctly, and substitutes back to express the answer in terms of x. This systematic approach shows the depth of understanding expected at distinction level.

Grade III (Pass) answer

y = (2x - 1)³

Using chain rule: dy/dx = 3(2x - 1)² × 2 dy/dx = 6(2x - 1)²

Mark: 1/1

Examiner commentary: This response shows competent application of the chain rule and reaches the correct answer efficiently. While less detailed in explanation than the Grade I response, the candidate has correctly identified that differentiation of a composite function requires the chain rule and has applied it accurately, which is sufficient for this multiple-choice question.

Grade V (Near miss) answer

y = (2x - 1)³

dy/dx = 3(2x - 1)²

Mark: 0/1

Examiner commentary: This response demonstrates a common and significant misconception: the candidate has attempted to use the power rule but has forgotten the "inner derivative" required by the chain rule. When differentiating (2x - 1)³, the result is not simply 3(2x - 1)²; we must also multiply by the derivative of (2x - 1), which is 2. This omission of the chain rule's second component is very frequent at CSEC level when students first encounter composite functions. To improve, the candidate must internalize that d/dx[f(g(x))] = f'(g(x)) × g'(x), meaning they must always multiply by the derivative of the inner function.