IB DP Mathematics: Analysis and Approaches -- Calculus Strand Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

Related: Calculus revision notes and the IB DP Mathematics: Analysis and Approaches syllabus guide.

Section A

1. Find dy/dx for y = 3x^4 - 2x^2 + 5. [2]

2. State what a definite integral represents geometrically. [1]

3. Find the indefinite integral of f(x) = 6x^2 + 4x. [2]

Section B

4. A particle moves so that its displacement s (in metres) at time t (in seconds) is given by s = t^3 - 6t^2 + 9t.

(a) Find an expression for the particle’s velocity at time t. [2] (b) Find the times at which the particle is momentarily at rest. [3]

5. Find the coordinates of the stationary point(s) of y = x^2 - 4x + 7, and determine, using the second derivative, whether each is a maximum or minimum. [4]

6. Use the product rule to differentiate y = x^2(2x - 1)^3. Leave your answer unexpanded but fully simplified in factorised form where possible. [4]

Section C

7. A farmer wants to fence a rectangular field with one side against an existing straight wall (so no fencing is needed on that side), using 80 m of fencing for the remaining three sides.

(a) Show that the area A of the field can be expressed as A = 80x - 2x^2, where x is the length (in metres) of each of the two sides perpendicular to the wall. [3] (b) Find the value of x that maximises the area, and justify that this value gives a maximum. [3] (c) Calculate the maximum possible area. [2]

Worked answers

1. dy/dx = 12x^3 - 4x. [2] (1 mark for each correctly differentiated term)

2. A definite integral represents the (signed) area between the curve and the x-axis, between the two given bounds. [1]

3. ∫(6x^2 + 4x) dx = 2x^3 + 2x^2 + C. [2] (1 mark for correct terms, 1 mark for including +C)

4. (a) v = ds/dt = 3t^2 - 12t + 9. [2] (b) At rest, v = 0: 3t^2 - 12t + 9 = 0, so t^2 - 4t + 3 = 0, (t-1)(t-3) = 0, giving t = 1 or t = 3 seconds. [3] (1 mark for correct method/factorising, 2 for both correct values)

5. dy/dx = 2x - 4; setting to zero: 2x - 4 = 0, so x = 2. When x = 2, y = 4 - 8 + 7 = 3, giving the stationary point (2, 3). [2] Second derivative: d²y/dx² = 2, which is positive, confirming (2, 3) is a minimum. [2]

6. Let u = x^2, v = (2x-1)^3. Then du/dx = 2x, dv/dx = 3(2x-1)^2 x 2 = 6(2x-1)^2 (by the chain rule). By the product rule: dy/dx = u(dv/dx) + v(du/dx) = x^2 x 6(2x-1)^2 + (2x-1)^3 x 2x = 6x^2(2x-1)^2 + 2x(2x-1)^3. Factorising out 2x(2x-1)^2: dy/dx = 2x(2x-1)^2[3x + (2x-1)] = 2x(2x-1)^2(5x-1). [4] (1 mark for correct u, v derivatives via chain rule, 1 for correctly applying the product rule, 2 for correct factorised simplification)

7. (a) The three fenced sides consist of two sides of length x (perpendicular to the wall) and one side of length (80 - 2x) parallel to the wall, since the total fencing used is 80 m. The area is therefore A = x(80 - 2x) = 80x - 2x^2. [3] (b) dA/dx = 80 - 4x. Setting dA/dx = 0: 80 - 4x = 0, so x = 20. Since d²A/dx² = -4, which is negative, this confirms x = 20 gives a maximum. [3] (c) A = 80(20) - 2(20)^2 = 1600 - 800 = 800 m². [2]

A note on the second-derivative test

Questions 5 and 7(b) both use the second-derivative test to classify a stationary point: if d²y/dx² > 0 at the stationary point, it is a local minimum (the gradient is increasing through zero, curving upwards); if d²y/dx² < 0, it is a local maximum (the gradient is decreasing through zero, curving downwards). If d²y/dx² = 0, the test is inconclusive and a sign check of the first derivative either side of the point is needed instead – this edge case does not arise in questions 5 or 7 here, but it is worth revising separately since exam questions occasionally set up a stationary point exactly at this boundary to test whether students default to the second-derivative test without checking it actually applies.

Why this set builds up to question 7

Questions 1-3 isolate the mechanical differentiation and integration rules the revision notes call “completely automatic” prerequisites; questions 4-6 test each application named in the notes (kinematics, stationary points, and the product/chain rule combination) in one clean context each; question 7 then requires the full optimisation method the notes describe end-to-end – setting up a function from a described real-world scenario, differentiating, solving, and justifying the result – exactly the kind of multi-step problem the notes identify as the most direct preparation for Paper 3’s extended-response questions.

Official syllabus

International Baccalaureate Organization, Diploma Programme Subject Brief – Mathematics: Analysis and Approaches, first assessment 2021, (c) 2019 – the same source cited by the Calculus revision notes and the IB DP Mathematics: Analysis and Approaches syllabus guide.