AP Calculus AB Section II Free-Response Practice Paper | 6 questions, 100 marks, 1 hour 30 minutes

2026-10-31

AP Calculus AB Section II

AP Calculus AB Section II practice paper: 6 free-response questions, 100 marks, 90 minutes; Part A calculator kinematics and rates (36), Part B volumes, differential equations and Riemann sums (64), full solutions included.

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Paper details　Level: AP | Subject: Calculus AB | Time: 1 hour 30 minutes | Marks: 100 | Questions: 6

Part A: calculator permitted (36 marks)

Free-response questions: show all working; answers without supporting work earn no credit. Give correct units and appropriate significant figures. Total 100 marks, 90 minutes.

#	Question	Answer and marking notes
1	1. A particle moves along a straight line. Its displacement s(t) metres from a fixed point O at time t seconds (t ≥ 0) is s(t) = t³ − 6t² + 9t + 2. (a) Find the velocity v(t) and the times at which the particle is instantaneously at rest, giving a reason. (4 marks) (b) Find the acceleration a(t) and the time at which it is zero. (3 marks) (c) Find the total distance travelled by the particle from t = 0 to t = 4. (5 marks) (d) Find the average velocity on the interval 0 ≤ t ≤ 4 and explain why it differs from the total distance. (3 marks) (e) State the intervals on which the speed is increasing, justifying each with the signs of v and a. (3 marks)	Answer: (a) v(t) = s′(t) = 3t² − 12t + 9 = 3(t − 1)(t − 3), so the particle is at rest at t = 1 s and t = 3 s (velocity zero). (b) a(t) = v′(t) = 6t − 12, so a = 0 at t = 2 s. (c) s(0) = 2, s(1) = 6, s(3) = 2, s(4) = 6; the particle moves 4 m forward in the first second, 4 m backward from t = 1 to 3, then 4 m forward from t = 3 to 4, giving a total distance of 12 m. (d) average velocity = (s(4) − s(0)) ÷ 4 = (6 − 2) ÷ 4 = 1 m s⁻¹; the total distance also counts the backward 4 m as positive, so it exceeds the displacement of 4 m. (e) speed = \|v\|: for 1 ≤ t ≤ 2 both v and a are negative, so speed increases; for 2 ≤ t ≤ 3 v is negative and a positive, so speed decreases; for 3 ≤ t ≤ 4 both are positive, so speed increases. Speed is increasing on 1 ≤ t ≤ 2 and 3 ≤ t ≤ 4.
2	2. An inverted conical tank has top radius 4 m and depth 8 m. Water is pumped in at 2 m³ per minute. (a) Using similar triangles, show that the surface radius r and the depth h satisfy r = h ÷ 2, and find the rate at which the water level rises when h = 5 m. (5 marks) (b) Find the rate when h = 4 m and explain why it is greater than the answer in (a). (4 marks) (c) A 13 ft ladder leans against a vertical wall; its foot slides away from the wall at 2 ft per second. Find the rate at which the top descends when the foot is 5 ft from the wall. (4 marks) (d) A cylinder is inscribed in a sphere of radius 6. Find the maximum possible volume of the cylinder. (5 marks)	Answer: (a) similar triangles give r ÷ h = 4 ÷ 8, so r = h ÷ 2. Then V = (1/3)πr²h = πh³ ÷ 12, so dV/dt = (πh² ÷ 4)(dh/dt). With dV/dt = 2 and h = 5, dh/dt = 8 ÷ (25π) ≈ 0.1019 m min⁻¹. (b) at h = 4, dh/dt = 8 ÷ (16π) = 1 ÷ (2π) ≈ 0.1592 m min⁻¹; the surface area of the water grows with the square of the depth, so the same inflow lifts a shallow surface faster. (c) let x be the distance of the foot from the wall and y the height of the top; x² + y² = 169, so y = 12 when x = 5. Differentiating, 2x(dx/dt) + 2y(dy/dt) = 0, hence dy/dt = −(x ÷ y)(dx/dt) = −(5 ÷ 12)(2) = −5/6 ≈ −0.833 ft s⁻¹, i.e. the top descends at about 0.833 ft per second. (d) for height h and radius r, r² + (h ÷ 2)² = 36, so V = πr²h = π(36 − h² ÷ 4)h = 36πh − (π ÷ 4)h³. Setting V′ = 36π − (3π ÷ 4)h² = 0 gives h² = 48, h = 4√3 ≈ 6.928 and r² = 24. Then V = 24π(4√3) = 96π√3 ≈ 522.37, and V″ is negative there, so this is the maximum.

Part B: no calculator (64 marks)

Free-response questions: show all working; answers without supporting work earn no credit. Give correct units and appropriate significant figures. Total 100 marks, 90 minutes.

#	Question	Answer and marking notes
3	3. (a) For the curve x² + xy + y² = 7, use implicit differentiation to find dy/dx and hence the equation of the tangent at the point (1, 2). (6 marks) (b) Find the coordinates of all points on the curve where the tangent is horizontal (exact values). (5 marks) (c) Let f(x) = x²e^(−x). Find the maximum point and the maximum value (exact, and to three decimal places). (4 marks) (d) Find the value of d/dx [ln(x² + 1)] at x = 1. (3 marks)	Answer: (a) differentiating with respect to x, 2x + y + x(dy/dx) + 2y(dy/dx) = 0, so dy/dx = −(2x + y) ÷ (x + 2y). At (1, 2), dy/dx = −4 ÷ 5 = −0.8, so the tangent is y − 2 = −0.8(x − 1), i.e. y = 2.8 − 0.8x. (b) a horizontal tangent needs dy/dx = 0, i.e. 2x + y = 0, so y = −2x. Substituting: x² − 2x² + 4x² = 3x² = 7, giving x = ±√(7/3) ≈ ±1.5275 with y = ∓3.0551; the points are (1.5275, −3.0551) and (−1.5275, 3.0551). (c) f′(x) = 2xe^(−x) − x²e^(−x) = xe^(−x)(2 − x), so x = 0 and x = 2 are critical points. f(0) = 0 and f(2) = 4e^(−2) ≈ 0.541; f′ is positive between them and negative after x = 2, so x = 2 gives the maximum value 4 ÷ e² ≈ 0.541. (d) d/dx ln(x² + 1) = 2x ÷ (x² + 1), which equals 1 at x = 1.
4	4. Let f(x) = 4 − x² and g(x) = x + 2. (a) Find the points of intersection and the area of the region enclosed by the two curves (exact value). (6 marks) (b) The region is rotated about the x-axis; find the volume of the solid generated (exact value, and to three decimal places). (6 marks) (c) Find the average value of f on 0 ≤ x ≤ 2. (4 marks)	Answer: (a) 4 − x² = x + 2 gives x² + x − 2 = 0, so x = 1 or x = −2. Area = ∫(from −2 to 1)(2 − x − x²)dx = [2x − x²/2 − x³/3] from −2 to 1 = 7/6 − (−10/3) = 9/2 = 4.5. (b) Volume = π∫(from −2 to 1)[(4 − x²)² − (x + 2)²]dx = π∫(x⁴ − 9x² − 4x + 12)dx = π[x⁵/5 − 3x³ − 2x² + 12x] from −2 to 1 = π(36/5 + 72/5) = 108π/5 ≈ 67.858. (c) average value = (1/2)∫(from 0 to 2)(4 − x²)dx = (1/2)(8 − 8/3) = 8/3 ≈ 2.667.
5	5. Consider the differential equation dy/dx = 2xy with initial condition y(0) = 3. (a) Solve it by separation of variables, find the particular solution and the value of y(1) (four decimal places). (6 marks) (b) Use Euler’s method from x = 0 with step 0.5 to estimate y(1), showing each step. (5 marks) (c) Let g(x) = ∫(from 0 to x)(t² − 2t)dt. Find the minimum point and minimum value of g for x ≥ 0. (5 marks)	Answer: (a) dy ÷ y = 2x dx, so ln\|y\| = x² + C and y = Ae^(x²); y(0) = 3 gives A = 3, hence y = 3e^(x²) and y(1) = 3e ≈ 8.1548. (b) at x = 0, y = 3, slope = 2(0)(3) = 0, so y(0.5) ≈ 3 + 0.5(0) = 3; at x = 0.5, slope = 2(0.5)(3) = 3, so y(1) ≈ 3 + 0.5(3) = 4.5. The Euler estimate 4.5 is far below the true value 8.155 because the solution curve is concave up and each tangent step undershoots. (c) g′(x) = x² − 2x = x(x − 2), so x = 0 and x = 2 are critical points; g″(2) = 2 is positive, so x = 2 is the minimum point with g(2) = 8/3 − 4 = −4/3 ≈ −1.333.
6	6. (a) Find lim(x → 0) sin(4x) ÷ (5x). (3 marks) (b) Use a right Riemann sum with n = 4 to estimate ∫(from 0 to 2)x²dx; give the estimate, the exact value and the error (exact values). (5 marks) (c) Let f(x) = ax² + b for x ≤ 1 and f(x) = 2x + 3 for x greater than 1. Find a and b so that f is both continuous and differentiable at x = 1. (3 marks) (d) Find lim(x → ∞) (3x² + 2x) ÷ (x² − 5). (3 marks)	Answer: (a) by l’Hôpital’s rule, the limit is 4cos(4x) ÷ 5, which at x = 0 equals 4/5 = 0.8 (equivalently use sin u ÷ u → 1). (b) Δx = 0.5 with right endpoints 0.5, 1, 1.5 and 2, so the sum is 0.5(0.25 + 1 + 2.25 + 4) = 3.75; the exact value is 8/3 ≈ 2.667 and the error is 15/4 − 8/3 = 13/12 ≈ 1.083, an overestimate because x² is increasing. (c) continuity requires a + b = 2(1) + 3 = 5; differentiability requires the left derivative 2a(1) to equal the right derivative 2, so a = 1 and b = 4. Check: both one-sided limits are 5 and both one-sided derivatives are 2. (d) dividing numerator and denominator by x² gives (3 + 2 ÷ x) ÷ (1 − 5 ÷ x²) → 3.

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