Format: Section I — 20 multiple-choice questions · Section II — 2 free-response questions
Suggested time: 40 min (MCQ) + 30 min (FRQ) · 29 points in total
Coverage: Limits and Continuity · Differentiation (product, quotient, chain and implicit) · Applications of Derivatives (linearization, Mean Value Theorem, related rates, optimization) · Integration and Accumulation (substitution, Riemann sums, the Fundamental Theorem) · Differential Equations · Applications of Integration (area, volume, average value).
On the real AP exam, part of each section is done without a calculator. Every question here can be done by hand, so work without one for the full effect.
Section I — Multiple Choice (20 points, 1 point each)
1. The limit of (sin 3x)/x as x → 0 is
A) 0 B) 1 C) 3 D) 1/3
2. d/dx [x/(x + 1)] =
A) 1/(x + 1)² B) −1/(x + 1)² C) 1/(x + 1) D) (2x + 1)/(x + 1)²
3. d/dx [(3x² + 1)⁴] =
A) 4(3x² + 1)³ B) 24x(3x² + 1)³ C) 6x(3x² + 1)³ D) 12x(3x² + 1)⁴
4. The function f is defined by f(x) = x² + k for x < 2 and f(x) = 3x − 1 for x ≥ 2. For what value of k is f continuous at x = 2?
A) 5 B) −1 C) 3 D) 1
5. The curve x² + xy + y² = 7 passes through (1, 2). What is dy/dx at this point?
A) −4/5 B) 4/5 C) −5/4 D) −2
6. d/dx [ln(x² + 1)] =
A) 1/(x² + 1) B) 2x ln(x² + 1) C) 2x/(x² + 1) D) 2/(x² + 1)
7. Using the tangent line to y = √x at x = 9, the approximate value of √9.2 is
A) 3.067 B) 3.033 C) 3.100 D) 3.020
8. For f(x) = x² on [1, 3], the value of c guaranteed by the Mean Value Theorem is
A) 1.5 B) 2.5 C) 4 D) 2
9. The radius of a circle is increasing at 3 cm per second. How fast is the area increasing when the radius is 5 cm?
A) 15π cm² per second B) 10π cm² per second C) 30π cm² per second D) 75π cm² per second
10. The limit of (eˣ − 1)/x as x → 0 is
A) 1 B) 0 C) e D) ∞
11. A particle moves along a line with velocity v(t) = t² − 4t + 3 for t ≥ 0. On which interval is the particle moving to the left?
A) 0 < t < 1 B) t > 3 C) 2 < t < 4 D) 1 < t < 3
12. The function f(x) = x³ − 3x has a local minimum at
A) x = −1 B) x = 1 C) x = 0 D) x = 3
13. ∫ x(x² + 1)³ dx =
A) (x² + 1)⁴/8 + C B) (x² + 1)⁴/4 + C C) x²(x² + 1)⁴/8 + C D) (x² + 1)⁴/2 + C
14. ∫₀^(π/2) cos x dx =
A) 0 B) −1 C) 1 D) π/2
15. A left Riemann sum with 2 subintervals of equal width is used to approximate ∫₀² x² dx. The approximation is
A) 5 B) 8/3 C) 4 D) 1
16. d/dx ∫₀^(x²) sin t dt =
A) sin(x²) B) 2x sin(x²) C) cos(x²) D) 2x cos(x²)
17. The area of the region between the graph of y = 4 − x² and the x-axis is
A) 16/3 B) 8 C) 32/3 D) 16
18. Which function is the solution of dy/dx = xy with y(0) = 2?
A) y = 2e^(x²/2) B) y = e^(x²/2) + 1 C) y = 2e^(x²) D) y = 2 + x²/2
19. The region under y = √x from x = 0 to x = 4 is rotated about the x-axis. The volume of the solid is
A) 4π B) 16π C) 32π/3 D) 8π
20. The average value of f(x) = sin x on the interval [0, π] is
A) 1/π B) 2/π C) 2 D) π/2
Section II — Free Response (9 points)
FRQ 1 (Analysis of a Function — 5 points).
Let f(x) = x³ − 6x² + 9x.
(a) Find f′(x) and the x-coordinates of all critical points of f. (2 pts)
(b) Classify each critical point as a local maximum or a local minimum. Justify your answer. (1 pt)
(c) Find the x-coordinate of the point of inflection of the graph of f. Justify your answer. (1 pt)
(d) Evaluate ∫₀³ f(x) dx. (1 pt)
FRQ 2 (Optimization — 4 points).
An open-topped box is made from a 12 cm by 12 cm square of card by cutting a square of side x cm from each corner and folding up the sides.
(a) Show that the volume of the box is V(x) = x(12 − 2x)², and state the domain of x for which a box can be made. (1 pt)
(b) Find the value of x that gives the maximum volume, and justify that it is a maximum. (2 pts)
(c) Find the maximum volume. (1 pt)
Answer key (Section I)
| Q | Ans | Q | Ans | Q | Ans | Q | Ans |
|---|---|---|---|---|---|---|---|
| 1 | C | 6 | C | 11 | D | 16 | B |
| 2 | A | 7 | B | 12 | B | 17 | C |
| 3 | B | 8 | D | 13 | A | 18 | A |
| 4 | D | 9 | C | 14 | C | 19 | D |
| 5 | A | 10 | A | 15 | D | 20 | B |
Key distribution: A×5, B×5, C×5, D×5.
Worked solutions (Section I)
1. (C) Write (sin 3x)/x = 3 × (sin 3x)/(3x). As x → 0, (sin 3x)/(3x) → 1, so the limit is 3. (1/3 inverts the factor; 1 ignores it.)
2. (A) Quotient rule: [(x + 1)(1) − x(1)]/(x + 1)² = 1/(x + 1)².
3. (B) Chain rule: 4(3x² + 1)³ × 6x = 24x(3x² + 1)³. (4(3x² + 1)³ forgets the inner derivative.)
4. (D) For continuity the two pieces must meet at x = 2: 2² + k = 3(2) − 1, so 4 + k = 5 and k = 1.
5. (A) Differentiate implicitly: 2x + y + x(dy/dx) + 2y(dy/dx) = 0, so dy/dx = −(2x + y)/(x + 2y). At (1, 2): −(2 + 2)/(1 + 4) = −4/5.
6. (C) Chain rule: (1/(x² + 1)) × 2x = 2x/(x² + 1).
7. (B) At x = 9, y = 3 and dy/dx = 1/(2√x) = 1/6. So √9.2 ≈ 3 + (1/6)(0.2) = 3.033.
8. (D) Average rate of change = (9 − 1)/(3 − 1) = 4. Set f′(c) = 2c = 4, so c = 2. (4 is the slope, not c.)
9. (C) A = πr², so dA/dt = 2πr(dr/dt) = 2π(5)(3) = 30π cm² per second.
10. (A) The limit has the form 0/0. By L'Hôpital's rule it equals the limit of eˣ/1, which is e⁰ = 1.
11. (D) v(t) = (t − 1)(t − 3), which is negative between its zeros, so the particle moves left for 1 < t < 3.
12. (B) f′(x) = 3x² − 3 = 0 gives x = ±1. f″(x) = 6x, and f″(1) = 6 > 0, so x = 1 is a local minimum. (x = −1 is the local maximum.)
13. (A) Let u = x² + 1, so du = 2x dx and x dx = du/2. Then ∫ u³ (du/2) = u⁴/8 + C = (x² + 1)⁴/8 + C.
14. (C) [sin x] from 0 to π/2 = sin(π/2) − sin 0 = 1.
15. (D) Width 1; left endpoints 0 and 1: 1 × (0² + 1²) = 1. (5 is the right Riemann sum; 8/3 is the exact value.)
16. (B) FTC with the chain rule: sin(x²) × d/dx(x²) = 2x sin(x²).
17. (C) The curve meets the x-axis at x = ±2. Area = ∫₋₂² (4 − x²) dx = [4x − x³/3] from −2 to 2 = 16/3 − (−16/3) = 32/3.
18. (A) Separate: dy/y = x dx, so ln|y| = x²/2 + c and y = Ae^(x²/2). y(0) = 2 gives A = 2.
19. (D) Disc method: V = π∫₀⁴ (√x)² dx = π∫₀⁴ x dx = π × 16/2 = 8π.
20. (B) Average value = (1/π)∫₀^π sin x dx = (1/π) × 2 = 2/π. (2 is the integral before dividing by the interval length.)
FRQ solutions
FRQ 1 (5 pts).
(a) (2) f′(x) = 3x² − 12x + 9 (1). Setting f′(x) = 0: 3(x² − 4x + 3) = 3(x − 1)(x − 3) = 0, so x = 1 and x = 3 (1).
(b) (1) f′ changes from positive to negative at x = 1, so f has a local maximum there; f′ changes from negative to positive at x = 3, so f has a local minimum there. (Using f″: f″(1) = −6 < 0 and f″(3) = 6 > 0 gives the same result.)
(c) (1) f″(x) = 6x − 12, which is 0 at x = 2. f″ changes sign from negative to positive at x = 2, so there is a point of inflection at x = 2. (f″ = 0 alone does not earn the point; the sign change is the justification.)
(d) (1) ∫₀³ (x³ − 6x² + 9x) dx = [x⁴/4 − 2x³ + 9x²/2] from 0 to 3 = 81/4 − 54 + 81/2 = 27/4 = 6.75.
FRQ 2 (4 pts).
(a) (1) After folding, the base is (12 − 2x) by (12 − 2x) and the height is x, so V(x) = x(12 − 2x)². A box needs 0 < x < 6.
(b) (2) V′(x) = (12 − 2x)² − 4x(12 − 2x) = (12 − 2x)(12 − 6x). On 0 < x < 6, V′(x) = 0 only at x = 2 (1). V′ > 0 for 0 < x < 2 and V′ < 0 for 2 < x < 6, so V has its maximum at x = 2 (1).
(c) (1) V(2) = 2 × 8² = 128 cm³.
Scoring guidance
- Section I: 20 points (1 each).
- Section II: FRQ 1 = 5 points, FRQ 2 = 4 points.
Pedagogy: AP free-response points reward justification. "x = 2 is an inflection point because f″ changes sign there" earns the point; "f″(2) = 0" alone does not, because f″ can be zero without changing sign.