AP Calculus AB — Practice Exam 1
Total: 60 points · ~1 hour 45 min. Section I: 15 multiple-choice (1 pt each). Section II: 1 free-response (45 pts, condensed). Representative practice set.
Instructions
- Section I, Part A is non-calculator on the real exam; here, work analytically.
- Show all work in Section II — points are awarded for reasoning.
Section I — Multiple Choice
- lim(x→2) (x² − 4)/(x − 2) = (A) 0 (B) 4 (C) 2 (D) does not exist
- d/dx [x³] = (A) 3x² (B) x⁴/4 (C) 3x (D) x²/2
- d/dx [sin x] = (A) cos x (B) −cos x (C) sec²x (D) −sin x
- If f(x) = 5x² − 3x, then f′(x) = (A) 5x − 3 (B) 10x − 3 (C) 10x − 3x (D) 10x
- ∫ 2x dx = (A) 2 + C (B) 2x² + C (C) x² + C (D) x²/2 + C
- The slope of the tangent to y = x² at x = 3 is (A) 12 (B) 9 (C) 6 (D) 3
- lim(x→∞) (3x² + 1)/(x² + 5) = (A) ∞ (B) 0 (C) 3 (D) 1
- d/dx [eˣ] = (A) x eˣ⁻¹ (B) eˣ⁺¹ (C) eˣ (D) 1
- If f is continuous on [a,b] and f(a) < 0 < f(b), the IVT guarantees (A) a vertical asymptote (B) a maximum (C) differentiability (D) a root in (a,b)
- d/dx [ln x] = (A) e^x (B) 1/x (C) ln x (D) x
- The function f(x) = 1/(x−4) has a vertical asymptote at (A) x = −4 (B) x = 0 (C) y = 4 (D) x = 4
- By the power rule, d/dx [x⁻²] = (A) −2x⁻³ (B) x⁻³ (C) −2x⁻¹ (D) 2x⁻³
- ∫₀¹ 3x² dx = (A) 2 (B) 6 (C) 3 (D) 1
- If position s(t) = t² , velocity at t = 3 is (A) 12 (B) 6 (C) 9 (D) 3
- A function has f′(x) > 0 on an interval. There f is (A) decreasing (B) concave down (C) constant (D) increasing
Section II — Free Response (condensed)
Q1 (~45 pts). Let f(x) = x³ − 6x² + 9x.
- (a) Find f′(x). (6)
- (b) Find the x-coordinates of all critical points. (9)
- (c) Classify each critical point as a local max or min using the first derivative test. (12)
- (d) Find f″(x) and the x-coordinate of the inflection point. (9)
- (e) Evaluate ∫₀³ f(x) dx. (9)
Answer Key (Section I)
| Q | Ans | Q | Ans | Q | Ans |
|---|---|---|---|---|---|
| 1 | B | 6 | C | 11 | D |
| 2 | A | 7 | C | 12 | A |
| 3 | A | 8 | C | 13 | D |
| 4 | B | 9 | D | 14 | B |
| 5 | C | 10 | B | 15 | D |
FRQ worked solution (Q1)
- (a) f′(x) = 3x² − 12x + 9.
- (b) 3x² − 12x + 9 = 0 → x² − 4x + 3 = 0 → (x−1)(x−3)=0 → x = 1, 3.
- (c) f′ changes + → − at x = 1 (local max), − → + at x = 3 (local min).
- (d) f″(x) = 6x − 12 = 0 → x = 2 (inflection point).
- (e) ∫₀³ (x³ − 6x² + 9x) dx = [x⁴/4 − 2x³ + 9x²/2]₀³ = (81/4 − 54 + 81/2) = 81/4 + 162/4 − 216/4 = 27/4 = 6.75.
AP score guide (approx.)
Section I (15) + Section II (45) = 60 points. Map: 5 ≈ 68%+, 4 ≈ 56–67%, 3 ≈ 42–55%, 2 ≈ 30–41%, 1 ≈ below. Official cut scores vary — use as a guide.