S6 Mathematics Practice Paper: Differentiation And Integration Basics (12 Questions, 100 Marks, 75 Minutes)

S6 Mathematics Practice Paper: Differentiation and Integration Basics (12 Questions, 100 Marks, 75 Minutes)

2026-11-10

S6 Mathematics: Differentiation, Integration and Applications

S6 Mathematics paper (12 questions, 100 marks, 75 minutes) on differentiation and integration, including tangents, stationary points, areas and related rates, with full answers.

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Paper details　Subject: Mathematics | Time: 75 minutes | Marks: 100 | Questions: 12

Instructions: This paper has three sections, A, B and C. Answer all questions. Total marks: 100. Time allowed: 75 minutes.

Show the formula used and the values substituted. Give numerical answers to three significant figures with correct units.

State the antiderivative in integration questions, and name the method used in applied questions, for example the second derivative test, the chain rule or the trapezoidal rule.

Spend about 20 minutes on Section A, 30 minutes on Section B and the last 25 minutes on Section C.

Section A Derivatives, Rules of Differentiation, Tangents and Normals (4 questions, 28 marks)

Answer all questions and show your working.

#	Question	Answer and marking notes
1	Let f(x) = 3x^2 – 4x + 1. (a) Find f'(x) using the rules of differentiation. (b) Hence find f'(2).	f'(x) = 6x – 4, 3 marks for both terms. f'(2) = 6(2) – 4 = 8, 3 marks.
2	Use first principles to find the derivative of g(x) = x^2 at x = 3, showing all steps clearly.	g'(3) = limit as h tends to 0 of [((3+h)^2 – 9)]/h = limit of (6h + h^2)/h = limit of (6 + h) = 6. Expansion 2 marks, simplification 2 marks, taking the limit 2 marks.
3	Find the stationary points of the curve y = x^3 – 6x^2 + 9x, and use the second derivative to determine the nature of each point.	y’ = 3x^2 – 12x + 9 = 3(x – 1)(x – 3), 2 marks; stationary points at x = 1 and x = 3, 2 marks; y” = 6x – 12, at x = 1, y” = -6 which is less than 0 so (1, 4) is a maximum point, 2 marks; at x = 3, y” = 6 which is greater than 0 so (3, 0) is a minimum point, 2 marks.
4	Find the maximum and minimum values of y = 2x^3 – 3x^2 – 12x + 5, stating the corresponding values of x.	y’ = 6x^2 – 6x – 12 = 6(x – 2)(x + 1), 2 marks; x = 2 or x = -1, 2 marks; y” = 12x – 6, at x = -1, y” = -18 which is less than 0 giving a maximum value of 12, 2 marks; at x = 2, y” = 18 which is greater than 0 giving a minimum value of -15, 2 marks.

Section B Integration, Areas and Kinematics (4 questions, 32 marks)

For integration, state the antiderivative and show the limits clearly.

#	Question	Answer and marking notes
5	Find the following indefinite integrals: (a) the integral of (6x^2 – 4x + 3) dx; (b) the integral of (8x^(-3) + 5) dx.	(a) 2x^3 – 2x^2 + 3x + C, 4 marks (1 mark per term and 1 for C). (b) 8x^(-2)/(-2) + 5x + C = -4/x^2 + 5x + C, 4 marks (1.5 marks per term and 1 for C).
6	Evaluate the definite integral of (x^2 + 2x) dx from x = 1 to x = 3, showing the antiderivative and the substitution.	Antiderivative x^3/3 + x^2, 3 marks; at the upper limit 9 + 9 = 18; at the lower limit 1/3 + 1 = 4/3; 18 – 4/3 = 50/3, about 16.7, 5 marks.
7	Find the area of the region bounded by the curve y = 4x – x^2 and the x-axis.	The curve meets the x-axis at x = 0 and x = 4, 2 marks; area = the integral of (4x – x^2) from 0 to 4, 2 marks; = [2x^2 – x^3/3] from 0 to 4 = 32 – 64/3 = 32/3, about 10.7 square units, 4 marks.
8	A particle moves along a straight line with velocity v(t) = 3t^2 – 12t + 9 in metres per second, where t is at least 0. (a) Find the acceleration at t = 2. (b) Find the displacement from t = 0 to t = 4.	(a) a(t) = v'(t) = 6t – 12, so the acceleration at t = 2 is 0 metres per second squared, 3 marks. (b) displacement = the integral of v from 0 to 4 = [t^3 – 6t^2 + 9t] from 0 to 4 = 64 – 96 + 36 = 4 m, 5 marks.

Section C Optimisation, Related Rates and Estimation (4 questions, 40 marks)

For applied questions, define the variable, write the function or relation, and state the units.

#	Question	Answer and marking notes
9	A rectangular field has one side along a wall and the other three sides fenced with 60 m of fencing. Let each side perpendicular to the wall be x m. (a) Express the area A in terms of x. (b) Find the maximum area and the corresponding value of x.	(a) The other side is (60 – 2x) m, so A = x(60 – 2x) = 60x – 2x^2, 3 marks. (b) dA/dx = 60 – 4x = 0 gives x = 15, 3 marks; d2A/dx2 = -4 which is less than 0 so this is a maximum, 2 marks; A = 15 times 30 = 450 square metres, 2 marks.
10	The curve is y = x^2 – 3x + 4. Find the equations of the tangent and the normal at x = 2.	The point of contact is (2, 2), 1 mark; y’ = 2x – 3 so the gradient of the tangent is 1, 3 marks; tangent y – 2 = 1(x – 2), that is y = x, 3 marks; gradient of the normal is -1, y – 2 = -(x – 2), that is y = -x + 4, 3 marks.
11	The volume of a spherical balloon increases at 30 cubic centimetres per second, with V = (4/3) pi r^3. Find the rate at which the radius increases when r = 5 cm.	dV/dr = 4 pi r^2, 3 marks; by the chain rule dV/dt = (dV/dr)(dr/dt), 2 marks; 30 = 4 pi (25)(dr/dt), 2 marks; dr/dt = 30/(100 pi) = 0.0955, about 0.096 cm per second, 3 marks.
12	The inflow rate R(t) of a tank is measured every 2 minutes: at t = 0, 2, 4, 6 and 8 minutes the readings are 4, 7, 9, 8 and 5 litres per minute. (a) Use the trapezoidal rule to estimate the total inflow in the first 8 minutes. (b) State whether this estimate is likely too high or too low, and explain why.	(a) Trapezoidal rule with h = 2: area is about (2/2)[4 + 2(7 + 9 + 8) + 5] = 57 litres, 6 marks (3 for the formula and 3 for the answer). (b) The rate peaks near t = 4 and the curve is concave down there, so joining the points with straight lines overestimates the integral; the estimate is therefore likely too high, 4 marks.

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