Format: Section I — 20 multiple-choice questions · Section II — 2 free-response questions
Suggested time: 40 min (MCQ) + 30 min (FRQ) · all arithmetic verified
Coverage: Limits · Differentiation · Applications of Derivatives · Integration · Differential Equations · Applications of Integration. (Correct answers spread across A–D.)
Section I — Multiple Choice
1. The limit of (x² + 3x) as x → 2 is
A) 8 B) 10 C) 6 D) 12
2. d/dx(4x³ − x²) =
A) 12x² − x B) 12x³ − 2x C) 4x² − 2x D) 12x² − 2x
3. The limit of (3x + 2)/(6x − 1) as x → ∞ is
A) 1/2 B) 3 C) ∞ D) 0
4. d/dx(cos x) =
A) −sin x B) cos x C) −cos x D) sin x
5. ∫ 4x³ dx =
A) 12x² + C B) 4x⁴ + C C) x⁴ + C D) x⁴
6. If f(x) = e²ˣ, then f'(x) =
A) e² B) 2x e²ˣ C) 2e²ˣ D) e²ˣ
7. The function f(x) = 1/(x − 5) is discontinuous at x =
A) 5 B) 1 C) 0 D) −5
8. Critical points of g(x) = x² − 8x occur at x =
A) −4 B) 4 C) 8 D) 0
9. ∫₀³ 2x dx =
A) 6 B) 12 C) 9 D) 3
10. Using the chain rule, d/dx sin(4x) =
A) cos x B) 4 cos(4x) C) cos(4x) D) 4 sin(4x)
11. The solution to dy/dt = ky with y(0) = 50 is y =
A) 50 + kt B) kt² C) 50e^(kt) D) 50kᵗ
12. The slope of the tangent to y = x² at x = 4 is:
A) 2 B) 8 C) 4 D) 16
13. For implicit differentiation of x² + y² = 9, dy/dx =
A) −x/y B) y/x C) x/y D) −y/x
14. The average value of f(x) = 3x² on [0, 2] is:
A) 12 B) 4 C) 8 D) 2
15. If f'(x) > 0 and f''(x) > 0, the graph is:
A) increasing, concave up B) increasing, concave down C) decreasing, concave down D) decreasing, concave up
16. ∫ 1/x dx =
A) −1/x² + C B) x⁻² + C C) ln|x| + C D) 1 + C
17. A particle's velocity is v(t) = 4t − 8. It is at rest at t =
A) 0 B) 4 C) 8 D) 2
18. The area between y = x and y = x² from x = 0 to x = 1 is:
A) 1/3 B) 1 C) 1/2 D) 1/6
19. By the Fundamental Theorem of Calculus, d/dx ∫₁ˣ cos t dt =
A) sin x B) cos x − cos 1 C) −sin x D) cos x
20. ∫₀² (3x² + 1) dx =
A) 12 B) 6 C) 8 D) 10
Section II — Free Response
FRQ 1 (Applications of Derivatives — 5 points). A particle moves with position s(t) = t³ − 9t² + 24t for t ≥ 0.
(a) Find v(t) and a(t). (2 pts)
(b) Find all times the particle is at rest. (1 pt)
(c) State the intervals where the particle moves left (negative direction). (1 pt)
(d) Find the acceleration at t = 1 and state whether the particle is speeding up or slowing down at that time. Give a reason. (1 pt)
FRQ 2 (Integration — 4 points). The rate of sales of a product is R(t) = 6t + 2 units per day for 0 ≤ t ≤ 4.
(a) Write and evaluate an integral for total units sold in the first 4 days. (2 pts)
(b) Find the average rate of sales over [0, 4]. (1 pt)
(c) Interpret the meaning of ∫₀⁴ R(t) dt in context. (1 pt)
Answer key (Section I)
| Q | Ans | Q | Ans | Q | Ans | Q | Ans |
|---|---|---|---|---|---|---|---|
| 1 | B | 6 | C | 11 | C | 16 | C |
| 2 | D | 7 | A | 12 | B | 17 | D |
| 3 | A | 8 | B | 13 | A | 18 | D |
| 4 | A | 9 | C | 14 | B | 19 | D |
| 5 | C | 10 | B | 15 | A | 20 | D |
Key distribution: A×5, B×5, C×5, D×5.
Worked solutions (Section I)
1. (B) 4 + 6 = 10.
2. (D) 12x² − 2x.
3. (A) Ratio of leading coefficients 3/6 = 1/2.
4. (A) −sin x.
5. (C) ∫ 4x³ dx = 4 × x⁴/4 + C = x⁴ + C.
6. (C) Chain rule: e²ˣ × 2 = 2e²ˣ.
7. (A) Denominator zero at x = 5.
8. (B) g'(x) = 2x − 8 = 0 ⇒ x = 4.
9. (C) [x²]₀³ = 9.
10. (B) cos(4x) × 4 = 4 cos(4x).
11. (C) Separating variables gives y = Ce^(kt), and y(0) = 50 gives C = 50, so y = 50e^(kt).
12. (B) y' = 2x = 8.
13. (A) 2x + 2y y' = 0 ⇒ y' = −x/y.
14. (B) Average value = (1/2) ∫₀² 3x² dx = (1/2)[x³]₀² = 8/2 = 4.
15. (A) Both positive → increasing, concave up.
16. (C) ∫ (1/x) dx = ln|x| + C.
17. (D) 4t − 8 = 0 ⇒ t = 2.
18. (D) On 0 < x < 1, x > x², so the area is ∫₀¹ (x − x²) dx = 1/2 − 1/3 = 1/6.
19. (D) FTC Part 1: the derivative is the integrand evaluated at x, which is cos x.
20. (D) [x³ + x]₀² = 8 + 2 = 10.
FRQ solutions
FRQ 1 (5 pts).
(a) (2) v(t) = 3t² − 18t + 24; a(t) = 6t − 18.
(b) (1) v = 3(t² − 6t + 8) = 3(t − 2)(t − 4) = 0 ⇒ t = 2, 4.
(c) (1) v < 0 where (t − 2)(t − 4) < 0: on (2, 4).
(d) (1) a(1) = 6(1) − 18 = −12 and v(1) = 3 − 18 + 24 = 9. Velocity and acceleration have opposite signs, so the particle is slowing down at t = 1. (The point needs both the value of a(1) and the sign-based reason.)
FRQ 2 (4 pts).
(a) (2) ∫₀⁴ (6t + 2) dt = [3t² + 2t]₀⁴ = 48 + 8 = 56 units.
(b) (1) Average rate = (1/4) ∫₀⁴ (6t + 2) dt = 56/4 = 14 units per day.
(c) (1) It represents the total number of units sold over the first 4 days (56 units).
Pedagogy: FRQ 1(d) rewards reasoning about the signs of velocity and acceleration at the same instant — speeding up when they agree in sign, slowing down when they differ.