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Paper A
AP Calculus AB — Practice Paper A
Original unofficial practice questions · paper A · answer key on the last page
Total time: see section headers · No guessing penalty
| Section | Questions | Format |
|---|---|---|
| Section I: Multiple Choice | ||
| Section II: Free Response |
Section I — Multiple Choice
The derivative of f(x) = x³ - 4x is
A. 3x² - 4B. x³ - 4C. 3x²D. 3x² - 4xIf g(x) = eˣ, find g'(x).
A. eˣB. xeˣC. eˣ ln xD. x e^(x-1)∫ 2x dx =
A. x² + CB. 2x² + CC. x + CD. 2 + CThe limit as x→2 of (x² - 4)/(x - 2) is
A. 4B. 0C. 2D. Does not existA function has a relative minimum where f ' changes
A. negative to positiveB. positive to negativeC. zero to nonzeroD. undefined to negativeIf f(x) = sin x, then f'(x) =
A. cos xB. -cos xC. sin xD. -sin x∫₀¹ 3x² dx =
A. 1B. 3C. 9D. 0A particle moves with velocity v(t) = 3t². Its acceleration is
A. 6tB. 3tC. 6D. t³The average value of f(x) = x on [0, 4] is
A. 2B. 4C. 8D. 1If h(x) = ln(x), h'(x) =
A. 1/xB. xC. ln xD. 1/ln xd/dx (x²·eˣ) by product rule =
A. 2x eˣ + x² eˣB. x² eˣC. 2x eˣD. 2x + eˣThe line tangent to f(x) = x² at x = 1 has slope
A. 2B. 1C. 0D. 4The integral of sec²x is
A. tan x + CB. sec x + CC. sin x + CD. -cot x + CIf f is continuous on [a,b] and differentiable on (a,b), the Mean Value Theorem guarantees
A. a c where f'(c) = (f(b)-f(a))/(b-a)B. f has an extremumC. f'' existsD. a zero where f(c)=0e^(ln 5) =
A. 5B. ln 5C. 10D. 1The limit of f(x) = (x² − 4)/(x − 2) as x → 2 is
A. 0B. 4C. 2D. Does not existA function with a hole where the limit exists is said to have a
A. jump discontinuityB. removable discontinuityC. infinite discontinuityD. no discontinuityThe derivative of f(x) = x⁵ is
A. 5x⁴B. x⁴C. 5x⁵D. x⁶/6Where a function has a local maximum, its derivative
A. is zero or undefinedB. is always positiveC. is always negativeD. has an inflection pointThe derivative of e^(2x) is
A. 2e^(2x)B. e^(2x)C. 2eˣD. e^(2x)/2The derivative of arcsin(x) is
A. 1/√(1−x²)B. 1/(1+x²)C. √(1−x²)D. cos⁻¹(x)If s(t) = t³ − 6t², the velocity at t = 2 is
A. −12B. 0C. 12D. −6Acceleration is the derivative of
A. positionB. velocityC. speedD. distanceThe Mean Value Theorem requires that f is
A. continuous on [a,b] and differentiable on (a,b)B. differentiable everywhereC. increasingD. concave upAn inflection point occurs where
A. f'' changes signB. f' = 0C. f = 0D. f'' = 0 onlyThe antiderivative of x³ is
A. x⁴/4 + CB. 3x² + CC. x²/3 + CD. x⁴ + C∫cos x dx =
A. sin x + CB. −sin x + CC. tan x + CD. cos x + CThe solution of dy/dt = 3y is
A. y = Ce^(3t)B. y = Ce^(t/3)C. y = 3CtD. y = Ce³A slope field shows
A. tiny tangent segments with slope dy/dxB. integral values onlyC. the second derivativeD. critical points onlyThe area between y = x and y = 0 on [0, 3] is
A. 9/2B. 3C. 9D. 6Section II — Free Response
Let f(x) = x³ - 3x. (a) Find the critical points. (b) Classify each as a local max or min. (c) On what intervals is f increasing?
9 points · rubric: Each part 3 pts: (a) f'=0 solvable, (b) sign analysis correct, (c) correct intervals.
Using the graph data: a car's velocity is v(t) = 4t - t² meters/sec. (a) When is the car at rest? (b) Find the total distance traveled on [0, 4].
6 points · rubric: (a) set v=0, (b) integrate |v| correctly over each interval.
Evaluate ∫ x·eˣ dx using integration by parts with u = x, dv = eˣ dx.
4 points · rubric: Correct u/dv choice 1 pt, correct integration 2 pts, +C 1 pt.
A ladder 10 ft long slides down a wall. When the base is 6 ft from the wall and sliding out at 2 ft/s, how fast is the top falling?
6 points · rubric: Set up Pythagorean relation, differentiate implicitly, plug in.
Let f(x) = (x − 3)/(x² − 9). (a) Simplify f. (b) Find lim(x→3) f(x). (c) Describe the discontinuity at x = 3.
6 points · rubric: Simplify 2 pts, limit 2 pts, classification 2 pts.
Find and classify all local extrema of f(x) = 2x³ − 3x² − 12x using the first-derivative test.
6 points · rubric: Critical points 2 pts, sign analysis 2 pts, classification 2 pts.
Answer Key
1. 3x² - 4 — Power rule: 3x² - 4.
2. eˣ — The derivative of eˣ is itself.
3. x² + C — Reverse power rule.
4. 4 — Factor to x + 2; the hole is removable, limit = 4.
5. negative to positive — First-derivative test at a local minimum.
6. cos x — Standard derivative.
7. 1 — Integral = x³ evaluated 0 to 1 = 1.
8. 6t — Acceleration = v'(t) = 6t.
9. 2 — (1/4)∫₀⁴ x dx = (1/4)(8) = 2.
10. 1/x — Derivative of ln x.
11. 2x eˣ + x² eˣ — Product rule: 2x·eˣ + x²·eˣ.
12. 2 — f'(x)=2x; at x=1 slope = 2.
13. tan x + C — Known antiderivative.
14. a c where f'(c) = (f(b)-f(a))/(b-a) — The MVT slope condition.
15. 5 — e and ln are inverse.
16. 4 — Factor to x + 2; the limit is 4 even though f is undefined at 2.
17. removable discontinuity — A hole with a defined limit is a removable discontinuity.
18. 5x⁴ — Power rule: bring down 5, reduce exponent to 4.
19. is zero or undefined — Flat spots have derivative zero (or undefined at corners).
20. 2e^(2x) — Chain rule: e^(2x)·2.
21. 1/√(1−x²) — The standard inverse trig derivative is 1/√(1−x²).
22. −12 — v(t) = 3t² − 12t; at t = 2, v = 12 − 24 = −12.
23. velocity — Acceleration is the derivative of velocity (second derivative of position).
24. continuous on [a,b] and differentiable on (a,b) — Continuity on the closed interval and differentiability on the open interval are required.
25. f'' changes sign — Concavity changes at inflection points.
26. x⁴/4 + C — Power rule for integration: add 1 to the exponent, divide by the new exponent.
27. sin x + C — The antiderivative of cos x is sin x.
28. y = Ce^(3t) — Separate and integrate to get y = Ce^(3t).
29. tiny tangent segments with slope dy/dx — Slope fields plot tangent segments from dy/dx.
30. 9/2 — ∫₀³ x dx = x²/2 = 9/2.
Free response — rubric notes
1. Each part 3 pts: (a) f'=0 solvable, (b) sign analysis correct, (c) correct intervals. · model: f' = 3x² - 3 = 0 at x = ±1. f'' = 6x so x=1 is a min, x=-1 is a max. Increasing on (-inf,-1) and (1,inf).
2. (a) set v=0, (b) integrate |v| correctly over each interval. · model: v=0 at t=0 and t=4. Distance = ∫₀⁴ |4t-t²| dt = 32/3 m.
3. Correct u/dv choice 1 pt, correct integration 2 pts, +C 1 pt. · model: = x eˣ - eˣ + C.
4. Set up Pythagorean relation, differentiate implicitly, plug in. · model: x²+y²=100; 2x dx/dt + 2y dy/dt = 0; y=8, dx/dt=2 → dy/dt = -1.5 ft/s.
5. Simplify 2 pts, limit 2 pts, classification 2 pts. · model: (a) f = 1/(x + 3). (b) The limit as x→3 is 1/6. (c) The function is undefined at 3 but the limit exists, so x = 3 is a removable discontinuity.
6. Critical points 2 pts, sign analysis 2 pts, classification 2 pts. · model: f'(x) = 6x² − 6x − 12 = 6(x−2)(x+1). Critical at x = −1, 2. Sign of f': + to − at −1 (local max), − to + at 2 (local min).
Paper B
AP Calculus AB — Practice Paper B
Original unofficial practice questions · paper B · answer key on the last page
Total time: see section headers · No guessing penalty
| Section | Questions | Format |
|---|---|---|
| Section I: Multiple Choice | ||
| Section II: Free Response |
Section I — Multiple Choice
The derivative of f(x) = x³ - 4x is
A. 3x² - 4xB. x³ - 4C. 3x² - 4D. 3x²If g(x) = eˣ, find g'(x).
A. xeˣB. eˣ ln xC. x e^(x-1)D. eˣ∫ 2x dx =
A. 2x² + CB. x² + CC. 2 + CD. x + CThe limit as x→2 of (x² - 4)/(x - 2) is
A. Does not existB. 2C. 4D. 0A function has a relative minimum where f ' changes
A. zero to nonzeroB. negative to positiveC. positive to negativeD. undefined to negativeIf f(x) = sin x, then f'(x) =
A. sin xB. -sin xC. cos xD. -cos x∫₀¹ 3x² dx =
A. 3B. 0C. 9D. 1A particle moves with velocity v(t) = 3t². Its acceleration is
A. t³B. 6tC. 6D. 3tThe average value of f(x) = x on [0, 4] is
A. 4B. 8C. 2D. 1If h(x) = ln(x), h'(x) =
A. xB. 1/ln xC. ln xD. 1/xd/dx (x²·eˣ) by product rule =
A. 2x + eˣB. 2x eˣ + x² eˣC. x² eˣD. 2x eˣThe line tangent to f(x) = x² at x = 1 has slope
A. 4B. 1C. 0D. 2The integral of sec²x is
A. -cot x + CB. sin x + CC. tan x + CD. sec x + CIf f is continuous on [a,b] and differentiable on (a,b), the Mean Value Theorem guarantees
A. f has an extremumB. a zero where f(c)=0C. a c where f'(c) = (f(b)-f(a))/(b-a)D. f'' existse^(ln 5) =
A. ln 5B. 1C. 10D. 5The limit of f(x) = (x² − 4)/(x − 2) as x → 2 is
A. 0B. 4C. 2D. Does not existA function with a hole where the limit exists is said to have a
A. jump discontinuityB. no discontinuityC. removable discontinuityD. infinite discontinuityThe derivative of f(x) = x⁵ is
A. x⁶/6B. 5x⁴C. x⁴D. 5x⁵Where a function has a local maximum, its derivative
A. has an inflection pointB. is zero or undefinedC. is always negativeD. is always positiveThe derivative of e^(2x) is
A. e^(2x)/2B. e^(2x)C. 2eˣD. 2e^(2x)The derivative of arcsin(x) is
A. 1/√(1−x²)B. √(1−x²)C. 1/(1+x²)D. cos⁻¹(x)If s(t) = t³ − 6t², the velocity at t = 2 is
A. 0B. −12C. −6D. 12Acceleration is the derivative of
A. distanceB. velocityC. speedD. positionThe Mean Value Theorem requires that f is
A. increasingB. continuous on [a,b] and differentiable on (a,b)C. concave upD. differentiable everywhereAn inflection point occurs where
A. f = 0B. f'' changes signC. f' = 0D. f'' = 0 onlyThe antiderivative of x³ is
A. x⁴ + CB. 3x² + CC. x²/3 + CD. x⁴/4 + C∫cos x dx =
A. sin x + CB. −sin x + CC. cos x + CD. tan x + CThe solution of dy/dt = 3y is
A. y = Ce^(3t)B. y = Ce^(t/3)C. y = 3CtD. y = Ce³A slope field shows
A. the second derivativeB. integral values onlyC. critical points onlyD. tiny tangent segments with slope dy/dxThe area between y = x and y = 0 on [0, 3] is
A. 3B. 9/2C. 9D. 6Section II — Free Response
Let f(x) = x³ - 3x. (a) Find the critical points. (b) Classify each as a local max or min. (c) On what intervals is f increasing?
9 points · rubric: Each part 3 pts: (a) f'=0 solvable, (b) sign analysis correct, (c) correct intervals.
Using the graph data: a car's velocity is v(t) = 4t - t² meters/sec. (a) When is the car at rest? (b) Find the total distance traveled on [0, 4].
6 points · rubric: (a) set v=0, (b) integrate |v| correctly over each interval.
Evaluate ∫ x·eˣ dx using integration by parts with u = x, dv = eˣ dx.
4 points · rubric: Correct u/dv choice 1 pt, correct integration 2 pts, +C 1 pt.
A ladder 10 ft long slides down a wall. When the base is 6 ft from the wall and sliding out at 2 ft/s, how fast is the top falling?
6 points · rubric: Set up Pythagorean relation, differentiate implicitly, plug in.
Let f(x) = (x − 3)/(x² − 9). (a) Simplify f. (b) Find lim(x→3) f(x). (c) Describe the discontinuity at x = 3.
6 points · rubric: Simplify 2 pts, limit 2 pts, classification 2 pts.
Find and classify all local extrema of f(x) = 2x³ − 3x² − 12x using the first-derivative test.
6 points · rubric: Critical points 2 pts, sign analysis 2 pts, classification 2 pts.
Answer Key
1. 3x² - 4 — Power rule: 3x² - 4.
2. eˣ — The derivative of eˣ is itself.
3. x² + C — Reverse power rule.
4. 4 — Factor to x + 2; the hole is removable, limit = 4.
5. negative to positive — First-derivative test at a local minimum.
6. cos x — Standard derivative.
7. 1 — Integral = x³ evaluated 0 to 1 = 1.
8. 6t — Acceleration = v'(t) = 6t.
9. 2 — (1/4)∫₀⁴ x dx = (1/4)(8) = 2.
10. 1/x — Derivative of ln x.
11. 2x eˣ + x² eˣ — Product rule: 2x·eˣ + x²·eˣ.
12. 2 — f'(x)=2x; at x=1 slope = 2.
13. tan x + C — Known antiderivative.
14. a c where f'(c) = (f(b)-f(a))/(b-a) — The MVT slope condition.
15. 5 — e and ln are inverse.
16. 4 — Factor to x + 2; the limit is 4 even though f is undefined at 2.
17. removable discontinuity — A hole with a defined limit is a removable discontinuity.
18. 5x⁴ — Power rule: bring down 5, reduce exponent to 4.
19. is zero or undefined — Flat spots have derivative zero (or undefined at corners).
20. 2e^(2x) — Chain rule: e^(2x)·2.
21. 1/√(1−x²) — The standard inverse trig derivative is 1/√(1−x²).
22. −12 — v(t) = 3t² − 12t; at t = 2, v = 12 − 24 = −12.
23. velocity — Acceleration is the derivative of velocity (second derivative of position).
24. continuous on [a,b] and differentiable on (a,b) — Continuity on the closed interval and differentiability on the open interval are required.
25. f'' changes sign — Concavity changes at inflection points.
26. x⁴/4 + C — Power rule for integration: add 1 to the exponent, divide by the new exponent.
27. sin x + C — The antiderivative of cos x is sin x.
28. y = Ce^(3t) — Separate and integrate to get y = Ce^(3t).
29. tiny tangent segments with slope dy/dx — Slope fields plot tangent segments from dy/dx.
30. 9/2 — ∫₀³ x dx = x²/2 = 9/2.
Free response — rubric notes
1. Each part 3 pts: (a) f'=0 solvable, (b) sign analysis correct, (c) correct intervals. · model: f' = 3x² - 3 = 0 at x = ±1. f'' = 6x so x=1 is a min, x=-1 is a max. Increasing on (-inf,-1) and (1,inf).
2. (a) set v=0, (b) integrate |v| correctly over each interval. · model: v=0 at t=0 and t=4. Distance = ∫₀⁴ |4t-t²| dt = 32/3 m.
3. Correct u/dv choice 1 pt, correct integration 2 pts, +C 1 pt. · model: = x eˣ - eˣ + C.
4. Set up Pythagorean relation, differentiate implicitly, plug in. · model: x²+y²=100; 2x dx/dt + 2y dy/dt = 0; y=8, dx/dt=2 → dy/dt = -1.5 ft/s.
5. Simplify 2 pts, limit 2 pts, classification 2 pts. · model: (a) f = 1/(x + 3). (b) The limit as x→3 is 1/6. (c) The function is undefined at 3 but the limit exists, so x = 3 is a removable discontinuity.
6. Critical points 2 pts, sign analysis 2 pts, classification 2 pts. · model: f'(x) = 6x² − 6x − 12 = 6(x−2)(x+1). Critical at x = −1, 2. Sign of f': + to − at −1 (local max), − to + at 2 (local min).
Full-length study package exam
AP Calculus AB — Full-Length Practice Exam
Timing Breakdown
SECTION I, PART A: MULTIPLE CHOICE (NO CALCULATOR)
- 30 questions | 60 minutes
- Suggested pacing: 2 minutes per question
SECTION I, PART B: MULTIPLE CHOICE (CALCULATOR ALLOWED)
- 15 questions | 45 minutes
- Suggested pacing: 3 minutes per question
SECTION II, PART A: FREE RESPONSE (CALCULATOR ALLOWED)
- 2 questions | 30 minutes
- Suggested pacing: 15 minutes per question
SECTION II, PART B: FREE RESPONSE (NO CALCULATOR)
- 4 questions | 60 minutes
- Suggested pacing: 15 minutes per questionSECTION I: MULTIPLE CHOICE
PART A — NO CALCULATOR
Questions 1–30. Do not use a calculator. Select the best answer.
Unit 1: Limits and Continuity
1. What is lim(x→2) (x² + x − 6) / (x − 2) ?
(A) 0   (B) 3   (C) 5   (D) The limit does not exist.
2. What is lim(x→∞) (5x³ + 2) / (3x³ − x) ?
(A) 0   (B) 5/3   (C) 5   (D) ∞
3. Let f be the function defined by
f(x) = { x² − 1, x < 3
{ ax + 5, x ≥ 3For what value of a is f continuous at x = 3 ?
(A) 1   (B) 3   (C) 8   (D) −1
4. lim(x→0) sin(5x) / (3x) =
(A) 0   (B) 3/5   (C) 5/3   (D) 1
Unit 2: Definition of the Derivative
5. Let f(x) = 2x² + 3x. Using the limit definition of the derivative, what is f′(−1) ?
(A) −7   (B) −1   (C) 1   (D) 7
6. If f(x) = 4x³ − x, then f′(x) =
(A) 4x² − 1   (B) 12x² − 1   (C) 12x − 1   (D) 12x² − x
7. If f(x) = (3x − 1)(x + 2), then f′(x) =
(A) 3x + 2   (B) 6x − 5   (C) 6x + 5   (D) 3x + 5
8. If f(x) = x² cos x, then f′(π) =
(A) −2π   (B) 2π   (C) π²   (D) −π
Unit 3: Chain Rule, Implicit Differentiation, and Inverse Functions
9. If f(x) = (2x + 1)⁵, then f′(x) =
(A) 5(2x + 1)⁴   (B) 10(2x + 1)⁴   (C) (2x + 1)⁴   (D) 10x(2x + 1)⁴
10. If f(x) = sin(3x²), then f′(x) =
(A) cos(3x²)   (B) 6 cos(3x²)   (C) 6x cos(3x²)   (D) 3x² cos(3x²)
11. If x² + y² = 25, then dy/dx at the point (3, 4) is
(A) −3/4   (B) 3/4   (C) −4/3   (D) 4/3
12. If f(x) = e^(2x−1), then f″(0) =
(A) 2/e   (B) 4/e   (C) 2e   (D) 4e
13. If f(2) = 5, f′(2) = −3, and g(x) = f(√x), then g′(4) =
(A) −3/4   (B) −3   (C) 3/4   (D) −12
Unit 4: Contextual Applications of Differentiation
14. A 10-foot ladder leans against a wall. The bottom of the ladder slides away from the wall at 2 ft/s. At what rate is the top of the ladder sliding down the wall when the bottom is 6 ft from the wall?
(A) −3/4 ft/s   (B) −4/3 ft/s   (C) 3/4 ft/s   (D) −2 ft/s
15. lim(x→0) (e^x − 1) / x =
(A) 0   (B) 1   (C) e   (D) The limit does not exist.
16. The tangent line to f(x) = √x at x = 9 is used to approximate √10. What is the approximation?
(A) 3.100   (B) 19/6   (C) 3.200   (D) 10/3
Unit 5: Analytical Applications of Differentiation
17. If f(x) = x³ − 3x + 1 on [−2, 2], then by the Mean Value Theorem there exists c in (−2, 2) such that f′(c) =
(A) 0   (B) 1   (C) 2   (D) 3
18. f(x) = x⁴ − 4x³. The x-coordinates of all critical points of f are
(A) 0 only   (B) 3 only   (C) 0 and 3   (D) 0, 3, and 4
19. f(x) = x³ − 6x² + 9x − 1. On which of the following intervals is f increasing?
(A) (−∞, 3) only   (B) (1, 3)   (C) (−∞, 1) ∪ (3, ∞)   (D) (0, 1) ∪ (3, ∞)
20. Which of the following is an inflection point of f(x) = x³ − 3x ?
(A) (1, −2)   (B) (0, 0)   (C) (−1, 2)   (D) (3, 18)
21. A rectangle has its base on the x-axis and its upper vertices on the parabola y = 12 − x². What is the maximum possible area of the rectangle?
(A) 24   (B) 32   (C) 36   (D) 48
Unit 6: Integration and Accumulation of Change
22. ∫ (4x³ − 6x + 2) dx =
(A) x⁴ − 3x² + 2x + C   (B) 4x² − 6x + C   (C) x⁴ − 3x² + C   (D) 12x² − 6 + C
23. d/dx [ ∫₀ˣ sin(t²) dt ] =
(A) cos(x²)   (B) sin(x²)   (C) 2x cos(x²)   (D) −sin(x²)
24. ∫₀² (3x² − 1) dx =
(A) 4   (B) 6   (C) 8   (D) 10
25. ∫ 2x√(x² + 1) dx =
(A) (2/3)(x² + 1)^(3/2) + C   (B) (x² + 1)^(3/2) + C   (C) (2/3)(x² + 1)^(1/2) + C   (D) 2√(x² + 1) + C
26. A Riemann sum is given by Σᵢ₌₁ⁿ (1 + 2i/n) · (2/n) for the function f on [0, 2] using right endpoints and n subintervals. Which of the following is f(x) ?
(A) 1 + x   (B) 1 + 2x   (C) x   (D) 2x + 1
Unit 7: Differential Equations
27. The general solution to dy/dx = x/y is
(A) y = x² + C   (B) y² = x² + C   (C) y = x + C   (D) y² = 2x + C
28. A population P(t) grows at a rate proportional to the population. If P(0) = 500 and P(2) = 2000, what is P(5) ?
(A) 8,000   (B) 16,000   (C) 32,000   (D) 64,000
Unit 8: Applications of Integration
29. The area of the region bounded by y = x² and y = x on [0, 1] is
(A) 1/6   (B) 1/3   (C) 1/2   (D) 1
30. The volume of the solid formed by revolving the region bounded by y = √x, x = 4, and y = 0 about the x-axis is
(A) 4π   (B) 8π   (C) 16π   (D) 32π
PART B — CALCULATOR ALLOWED
Questions 31–45. A graphing calculator is required. Select the best answer.
Unit 1: Limits and Continuity
31. The table below gives selected values of a function f.
| x | 2.99 | 2.999 | 3.001 | 3.01 |
|---|---|---|---|---|
| f(x) | 5.9850 | 5.9985 | 6.0015 | 6.0150 |
What is the best estimate for lim(x→3) f(x) ?
(A) 5.99   (B) 6.0   (C) 6.15   (D) The limit does not exist.
Unit 2: Definition of the Derivative
32. Selected values of a differentiable function f are given below.
| x | 2 | 3 | 4 | 5 |
|---|---|---|---|---|
| f(x) | 8 | 15 | 24 | 35 |
Using this data, what is the best approximation for f′(3.5) ?
(A) 7.5   (B) 9   (C) 10.5   (D) 11
Unit 3: Chain Rule, Implicit Differentiation, and Inverse Functions
33. Let f(x) = x³ + x. The function g is the inverse of f. What is g′(2) ?
(A) 1/4   (B) 1/2   (C) 2   (D) 4
34. Let f(x) = xe^(−x²). At which of the following x-values does f have a relative maximum?
(A) x = −1/√2   (B) x = 0   (C) x = 1/√2   (D) x = √2
35. A curve is defined implicitly by x³ + y³ − 6xy = 0. Given that dy/dx = (2y − x²)/(y² − 2x), what is the slope of the tangent line at (3, 3) ?
(A) −1   (B) −1/3   (C) 1/3   (D) 1
Unit 4: Contextual Applications of Differentiation
36. The radius of a circle is increasing at a rate of 2 cm/s. At what rate is the area of the circle increasing when the radius is 5 cm?
(A) 10π cm²/s   (B) 20π cm²/s   (C) 25π cm²/s   (D) 50π cm²/s
37. The velocity of a particle moving along the x-axis is given by v(t) = t² − 4t + 3 for 0 ≤ t ≤ 5. At which of the following times does the particle change direction?
(A) t = 1 only   (B) t = 3 only   (C) t = 1 and t = 3   (D) t = 0, 1, and 3
Unit 5: Analytical Applications of Differentiation
38. Let f(x) = x⁴ − 4x² + 3. What is the absolute minimum value of f on the interval [−1, 2] ?
(A) −1   (B) 0   (C) 3   (D) −3
39. The derivative of a function f is given by f′(x) = x³ − 6x² + 8x. On which of the following intervals is f decreasing?
(A) (0, 2) ∪ (4, ∞)   (B) (−∞, 0) ∪ (2, 4)   (C) (0, 4)   (D) (−∞, 2) only
Unit 6: Integration and Accumulation of Change
40. ∫₀¹ e^(x²) · 2x dx =
(A) e − 1   (B) 2(e − 1)   (C) e² − 1   (D) 1
41. If g(x) = ∫₀^(x²) √(t³ + 1) dt, then g′(1) =
(A) 2√2   (B) √2   (C) 4√2   (D) 2
42. The table gives values of a function f at selected points.
| x | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| f(x) | 3 | 5 | 4 | 2 | 4 |
Using a right Riemann sum with 4 subintervals of equal width, which of the following best approximates ∫₀⁴ f(x) dx ?
(A) 12   (B) 15   (C) 16   (D) 18
Unit 7: Differential Equations
43. A slope field for the differential equation dy/dx = xy/2 is shown (conceptually). Which of the following could be a particular solution with f(0) = 3 ?
(A) y = 3e^(x²/4)   (B) y = 3e^(x²/2)   (C) y = e^(x²/4) + 2   (D) y = 3e^(−x²/4)
Unit 8: Applications of Integration
44. Let R be the region bounded by y = x + 2 sin x, y = x, x = 0, and x = π. Which of the following gives the area of R ?
(A) ∫₀π (x + 2 sin x − x) dx   (B) ∫₀π (2 sin x) dx   (C) ∫₀π (x²/2 + 2 cos x) dx   (D) Both A and B give the correct area.
45. The table gives the velocity of a particle moving along a line.
| t (s) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| v(t) (m/s) | 3 | 5 | 4 | 1 | −2 |
Using a left Riemann sum with 4 subintervals, the total distance traveled by the particle on [0, 4] is approximately
(A) 11 m   (B) 12 m   (C) 13 m   (D) 15 m
SECTION II: FREE RESPONSE
PART A — CALCULATOR ALLOWED
Questions 1–2. A graphing calculator is required. Show all your work.
FRQ 1 (Integration and Particle Motion)
A particle moves along the x-axis so that its velocity at time t, for 0 ≤ t ≤ 2, is given by
v(t) = e^(−t²) + t − 2The position of the particle at time t = 0 is s(0) = 1.
(a) Find the acceleration of the particle at time t = 1.
(b) Find all times t in [0, 2] at which the particle changes direction. Justify your answer.
(c) Find the total distance traveled by the particle on the interval [0, 2].
(d) Find the position of the particle at time t = 2.
FRQ 2 (Area and Volume)
Let R be the region bounded by the graphs of y = 2 + sin(x²), y = e^(−x), x = 0, and x = 2.
(a) Find the area of region R.
(b) Find the volume of the solid generated when R is revolved about the x-axis.
(c) The region R is the base of a solid. For each x in [0, 2], the cross section perpendicular to the x-axis is a square. Find the volume of this solid.
(d) Write, but do not evaluate, an integral expression for the volume of the solid generated when R is revolved about the line y = −1.
PART B — NO CALCULATOR
Questions 3–6. No calculator is allowed. Show all your work.
FRQ 3 (Differential Equations)
Consider the differential equation dy/dx = (x + 1) / y.
(a) On the axes provided, sketch a slope field for the given differential equation at the nine points where x and y are each −1, 0, or 1.
(b) Let f be the function that satisfies the differential equation and f(0) = 1. Write an equation for the tangent line to the graph of f at x = 0.
(c) Find the particular solution y = f(x) to the differential equation with the initial condition f(0) = 1.
(d) Determine the domain of the particular solution found in part (c).
FRQ 4 (Motion and Rates)
A particle moves along the x-axis so that its velocity at time t ≥ 0 is given by
v(t) = t³ − 6t² + 11t − 6(a) Find the acceleration of the particle at time t = 2.
(b) Find all values of t ≥ 0 at which the particle changes direction. Classify each as a change from moving left to right or from right to left.
(c) Find the total distance traveled by the particle on the interval 0 ≤ t ≤ 4.
(d) Is the speed of the particle increasing or decreasing at time t = 3 ? Give a reason for your answer.
FRQ 5 (Table and Graph Analysis)
The function f is twice differentiable. Selected values of f and f′ are given in the table below.
| x | −2 | −1 | 0 | 1 | 2 |
|---|---|---|---|---|---|
| f(x) | 3 | 1 | 2 | 4 | 3 |
| f′(x) | −3 | −1 | 2 | 1 | −2 |
(a) Find the x-coordinate of each critical point of f in the interval (−2, 2). Classify each as a relative minimum, relative maximum, or neither. Justify your answer.
(b) On what open intervals, if any, is the graph of f concave up? Justify your answer.
(c) Let g(x) = f(x²). Find g′(−1).
(d) Let h(x) = f(f(x)). Find h′(0).
FRQ 6 (Area and Volume — No Calculator)
Let R be the region enclosed by the graphs of y = x² and y = 2x.
(a) Find the area of region R.
(b) Find the volume of the solid generated when R is revolved about the x-axis.
(c) Find the volume of the solid generated when R is revolved about the y-axis.
(d) A horizontal line y = k divides R into two regions of equal area. Set up, but do not evaluate, an equation involving an integral whose solution gives the value of k.
END OF EXAM
Answer Key & Rubric
AP Calculus AB — Full Practice Exam: Answer Key
SECTION I: MULTIPLE CHOICE ANSWERS
PART A — NO CALCULATOR (Questions 1–30)
Unit 1: Limits and Continuity
1. Answer: C (5)
Factor the numerator: x² + x − 6 = (x + 3)(x − 2). Cancel (x − 2) and substitute x = 2 to get 2 + 3 = 5. The function has a removable discontinuity at x = 2, not an infinite one.
- (A) 0 — This would result from incorrectly setting the numerator to zero or canceling incorrectly.
- (B) 3 — This comes from evaluating only the (x + 3) factor at x = 2 and dropping the constant, or from misreading the factored form.
- (D) The limit does not exist — The student likely saw the 0/0 form and concluded the limit is undefined without attempting to simplify.
2. Answer: B (5/3)
Divide every term by the highest power of x in the denominator (x³). The leading coefficients dominate: the limit equals 5/3. The lower-degree terms become negligible as x → ∞.
- (A) 0 — The student divided by the highest power in the numerator instead of the denominator.
- (C) 5 — The student only looked at the leading coefficient of the numerator and ignored the denominator.
- (D) ∞ — The student may have incorrectly concluded that exponential or polynomial growth always yields infinity, ignoring the matching degrees.
3. Answer: A (1)
For continuity at x = 3, the left-hand limit must equal the right-hand limit. The left-hand limit is 3² − 1 = 8. Setting 3a + 5 = 8 gives a = 1.
- (B) 3 — The student confused the x-value (3) with the required y-value or set the function equal to x.
- (C) 8 — This is the y-value at x = 3, not the value of a. The student solved for f(3) instead of a.
- (D) −1 — The student set the two expressions equal but made an arithmetic sign error when solving.
4. Answer: C (5/3)
Rewrite as (5/3) · sin(5x)/(5x). Since lim(u→0) sin(u)/u = 1, the limit equals 5/3. This is a direct application of the special trigonometric limit.
- (A) 0 — The student likely assumed sin(5x) approaches 0 and ignored the denominator also approaching 0.
- (B) 3/5 — The student inverted the fraction; the correct ratio is numerator coefficient over denominator coefficient.
- (D) 1 — The student correctly recalled that lim sin(u)/u = 1 but forgot to account for the coefficient ratio 5/3.
Unit 2: Definition of the Derivative
5. Answer: B (−1)
f′(−1) = lim(h→0) [2(−1 + h)² + 3(−1 + h) − (−1)] / h = lim(h→0) [2(1 − 2h + h²) − 3 + 3h + 1] / h = lim(h→0) [−h + 2h²] / h = −1. The h² term vanishes.
- (A) −7 — The student computed f′(x) first and then evaluated f′(−1) incorrectly, possibly mixing up the derivative formula.
- (C) 1 — The student lost the negative sign when simplifying −h/h.
- (D) 7 — The student may have plugged −1 into f(x) instead of using the limit definition, or made an algebraic error in expanding.
6. Answer: B (12x² − 1)
Apply the power rule to each term: d/dx[4x³] = 12x² and d/dx[−x] = −1. The derivative of a constant is zero.
- (A) 4x² − 1 — The student applied the power rule incorrectly, multiplying by the original exponent minus two instead of one.
- (C) 12x − 1 — The student reduced the exponent by one but forgot to multiply by the original exponent 3.
- (D) 12x² − x — The student treated −x as if it were a constant and forgot to take its derivative.
7. Answer: C (6x + 5)
By the product rule: f′(x) = 3(x + 2) + (3x − 1)(1) = 3x + 6 + 3x − 1 = 6x + 5. Each factor is differentiated in turn.
- (A) 3x + 2 — The student only differentiated the first factor and did not apply the full product rule.
- (B) 6x − 5 — The student made a sign error when combining the constant terms (6 − 1 = 5, not −5).
- (D) 3x + 5 — The student differentiated the first factor correctly but set the derivative of the second factor to zero.
8. Answer: A (−2π)
By the product rule: f′(x) = 2x cos x − x² sin x. Evaluating at x = π: f′(π) = 2π cos π − π² sin π = 2π(−1) − π²(0) = −2π. The sin π term vanishes.
- (B) 2π — The student forgot that cos π = −1 and used cos π = 1.
- (C) π² — The student only computed the −x² sin x term and set 2x cos x to zero.
- (D) −π — The student used the power rule on cos x incorrectly, treating it like xⁿ.
Unit 3: Chain Rule, Implicit Differentiation, and Inverse Functions
9. Answer: B (10(2x + 1)⁴)
Apply the chain rule: the derivative of (2x + 1)⁵ is 5(2x + 1)⁴ times the derivative of the inside function (2x + 1), which is 2. This gives 10(2x + 1)⁴.
- (A) 5(2x + 1)⁴ — The student applied the outer power rule but forgot to multiply by the derivative of the inner function (the chain rule step).
- (C) (2x + 1)⁴ — The student dropped the coefficient 5 entirely and also missed the chain rule.
- (D) 10x(2x + 1)⁴ — The student incorrectly distributed the 2 only to x instead of applying it to the entire inner derivative.
10. Answer: C (6x cos(3x²))
By the chain rule: the derivative of sin(3x²) is cos(3x²) times the derivative of 3x², which is 6x. So f′(x) = 6x cos(3x²).
- (A) cos(3x²) — The student forgot the chain rule entirely and only differentiated the outer sine function.
- (B) 6 cos(3x²) — The student differentiated the inside as if it were 3x (getting 3), then multiplied by 2, but lost the x factor.
- (D) 3x² cos(3x²) — The student used the inside function 3x² instead of its derivative 6x.
11. Answer: A (−3/4)
Differentiate implicitly: 2x + 2y(dy/dx) = 0, so dy/dx = −x/y. At (3, 4): dy/dx = −3/4. This gives the slope of the tangent line to the circle at that point.
- (B) 3/4 — The student dropped the negative sign when solving for dy/dx.
- (C) −4/3 — The student inverted the ratio, computing −y/x instead of −x/y.
- (D) 4/3 — The student both dropped the negative sign and inverted the ratio.
12. Answer: B (4/e)
f′(x) = 2e^(2x−1) and f″(x) = 4e^(2x−1). Evaluating at x = 0: f″(0) = 4e^(−1) = 4/e. Each differentiation of e^(2x−1) brings down a factor of 2.
- (A) 2/e — The student only differentiated once and evaluated f′(0) instead of f″(0).
- (C) 2e — The student used the wrong exponent sign, computing 4e^(1) / 2 instead of 4e^(−1).
- (D) 4e — The student forgot that e^(−1) = 1/e and used e^(+1) = e.
13. Answer: A (−3/4)
By the chain rule: g′(x) = f′(√x) · (1/(2√x)). At x = 4: g′(4) = f′(2) · (1/(2·2)) = (−3)(1/4) = −3/4.
- (B) −3 — The student computed f′(2) = −3 but forgot the chain rule factor 1/(2√x).
- (C) 3/4 — The student made a sign error, perhaps using f′(2) = 3.
- (D) −12 — The student inverted the chain rule factor, using 2√x = 4 instead of 1/(2√x) = 1/4.
Unit 4: Contextual Applications of Differentiation
14. Answer: A (−3/4 ft/s)
By the Pythagorean theorem: x² + y² = 100. Differentiating with respect to t: 2x(dx/dt) + 2y(dy/dt) = 0. When x = 6, y = 8. So 12(2) + 16(dy/dt) = 0, giving dy/dt = −3/4 ft/s.
- (B) −4/3 ft/s — The student inverted the ratio, computing dy/dt = −x·(dx/dt)/y incorrectly.
- (C) 3/4 ft/s — The student dropped the negative sign; the top slides down, so dy/dt must be negative.
- (D) −2 ft/s — The student confused dy/dt with dx/dt, using the given rate directly.
15. Answer: B (1)
This is the limit definition of the derivative of e^x at x = 0. Since (e^x)′ = e^x and e⁰ = 1, the limit equals 1. This can also be verified by L'Hôpital's rule.
- (A) 0 — The student assumed that since e⁰ − 1 = 0, the limit is 0, ignoring the 0/0 indeterminate form.
- (C) e — The student confused this with the limit of (e^x − 1)/x as x → 1, or with the value of e^x at x = 0.
- (D) The limit does not exist — The student may have incorrectly identified this as an indeterminate form with no resolution.
16. Answer: B (19/6)
f(9) = 3 and f′(9) = 1/(2√9) = 1/6. The tangent line is L(x) = 3 + (1/6)(x − 9). So L(10) = 3 + 1/6 = 19/6.
- (A) 3.100 — This is close to the true value of √10 ≈ 3.162, but the linearization should give 19/6 ≈ 3.167.
- (C) 3.200 — The student used a slope of 1/5 instead of 1/6, perhaps using f′(9) = 1/(2·9) incorrectly.
- (D) 10/3 — The student used L(x) = √x + (x − 9)/x or some other incorrect linearization formula.
Unit 5: Analytical Applications of Differentiation
17. Answer: B (1)
By the Mean Value Theorem: f′(c) = [f(2) − f(−2)] / [2 − (−2)] = [3 − (−1)] / 4 = 1. Verifying: f′(x) = 3x² − 3, and 3x² − 3 = 1 gives x² = 4/3, which has solutions in (−2, 2).
- (A) 0 — The student may have confused MVT with the fact that f′(1) = 0 for this function.
- (C) 2 — The student computed [f(2) − f(−2)] / 2 instead of dividing by 4.
- (D) 3 — The student may have used f(2) directly without computing the difference quotient.
18. Answer: C (0 and 3)
f′(x) = 4x³ − 12x² = 4x²(x − 3). Setting equal to zero: x² = 0 gives x = 0, and x − 3 = 0 gives x = 3. Both are in the domain of f.
- (A) 0 only — The student missed the factor (x − 3) when factoring f′(x).
- (B) 3 only — The student missed the factor x² when factoring f′(x).
- (D) 0, 3, and 4 — The student incorrectly included x = 4, which makes f′(x) = 64 ≠ 0.
19. Answer: C ((−∞, 1) ∪ (3, ∞))
f′(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3). The sign of f′ is positive (f is increasing) when both factors have the same sign: x < 1 or x > 3.
- (A) (−∞, 3) only — The student only found one critical point and misidentified the increasing interval.
- (B) (1, 3) — This is where f′(x) < 0, so f is decreasing, not increasing.
- (D) (0, 1) ∪ (3, ∞) — The student incorrectly restricted the left interval to x > 0.
20. Answer: B ((0, 0))
f″(x) = 6x. Setting f″(x) = 0 gives x = 0. Since f″ changes from negative to positive at x = 0, the point (0, f(0)) = (0, 0) is an inflection point.
- (A) (1, −2) — This is a relative minimum (where f′(x) = 0), not an inflection point.
- (C) (−1, 2) — This is a relative maximum (where f′(x) = 0), not an inflection point.
- (D) (3, 18) — This point is not on the graph of f(x) = x³ − 3x, since f(3) = 18.
21. Answer: B (32)
The upper vertices are at (x, 12 − x²) and (−x, 12 − x²). Area A = 2x(12 − x²) = 24x − 2x³. A′(x) = 24 − 6x² = 0 gives x = 2 (rejecting the negative root). A(2) = 2(2)(12 − 4) = 32.
- (A) 24 — The student found the maximum height of the rectangle (at x = 0) rather than the maximum area.
- (C) 36 — The student may have used x(12 − x²) instead of 2x(12 − x²), then found the wrong optimum.
- (D) 48 — The student computed 2x · 12 = 24x and maximized this without the −2x³ correction term.
Unit 6: Integration and Accumulation of Change
22. Answer: A (x⁴ − 3x² + 2x + C)
Integrate term by term using the power rule: ∫4x³ dx = x⁴, ∫(−6x) dx = −3x², ∫2 dx = 2x. Add the constant of integration C.
- (B) 4x² − 6x + C — The student added 1 to each exponent instead of the correct procedure (dividing by the new exponent).
- (C) x⁴ − 3x² + C — The student forgot to integrate the constant term 2.
- (D) 12x² − 6 + C — The student multiplied each coefficient by the original exponent instead of dividing by the new one.
23. Answer: B (sin(x²))
By the Fundamental Theorem of Calculus, Part 1: d/dx[∫₀ˣ f(t) dt] = f(x). Here, the upper limit is simply x, so the derivative is sin(x²). No chain rule adjustment is needed.
- (A) cos(x²) — The student differentiated sin(t²) with respect to t, getting cos(t²), but this is the integrand evaluated at x, and the integrand is sin(t²), not cos(t²).
- (C) 2x cos(x²) — The student incorrectly applied the chain rule, treating the upper limit as if it were x² rather than x.
- (D) −sin(x²) — The student introduced an extraneous negative sign.
24. Answer: B (6)
∫₀² (3x² − 1) dx = [x³ − x]₀² = (8 − 2) − (0 − 0) = 6. Apply the power rule to each term and evaluate at the bounds.
- (A) 4 — The student computed 8 − 2 = 6 but then subtracted 2 again, or made a similar arithmetic error.
- (C) 8 — The student evaluated only x³ at x = 2 and forgot to subtract x.
- (D) 10 — The student may have integrated −1 as −x² instead of −x, getting [x³ − x²]₀² = 8 − 4 = 4, then added errors.
25. Answer: A ((2/3)(x² + 1)^(3/2) + C)
Let u = x² + 1, so du = 2x dx. The integral becomes ∫ √u du = (2/3)u^(3/2) + C = (2/3)(x² + 1)^(3/2) + C.
- (B) (x² + 1)^(3/2) + C — The student forgot the factor of 2/3 from integrating u^(1/2).
- (C) (2/3)(x² + 1)^(1/2) + C — The student decreased the exponent instead of increasing it when integrating.
- (D) 2√(x² + 1) + C — The student treated √u as u^(−1/2) and integrated incorrectly.
26. Answer: A (1 + x)
With Δx = 2/n and right endpoints x_i = 2i/n, the Riemann sum is Σ f(2i/n) · (2/n) = Σ (1 + 2i/n)(2/n). So f(2i/n) = 1 + 2i/n, meaning f(x) = 1 + x.
- (B) 1 + 2x — The student confused the term 2i/n with f evaluated at the right endpoint.
- (C) x — The student ignored the constant 1 in the summand.
- (D) 2x + 1 — This is equivalent to choice (A), but the student should recognize the standard form; however, the function is the same, so this would also be technically correct. The best match is (A) since it directly corresponds.
Unit 7: Differential Equations
27. Answer: B (y² = x² + C)
Separate variables: y dy = x dx. Integrate both sides: y²/2 = x²/2 + C. Multiply by 2: y² = x² + 2C, and relabel the constant as C.
- (A) y = x² + C — The student did not separate variables correctly and integrated x/y as if it were just x.
- (C) y = x + C — The student treated the DE as if it were dy/dx = 1.
- (D) y² = 2x + C — The student integrated x dx incorrectly as x instead of x²/2.
28. Answer: B (16,000)
dP/dt = kP gives P(t) = P₀e^(kt). From P(2) = 500e^(2k) = 2000, we get e^(2k) = 4, so k = ln 2. Then P(5) = 500e^(5 ln 2) = 500 · 2⁵ = 500 · 32 = 16,000.
- (A) 8,000 — The student used k = ln 2 but computed P(5) = 500 · 2⁴ = 500 · 16.
- (C) 32,000 — The student used P(5) = 2000 · 2³ = 16,000 but then doubled it, or used 2⁵ = 32 without the factor of 500.
- (D) 64,000 — The student used P(5) = 2000 · 2⁵ = 64,000, forgetting that P₀ = 500, not 2000.
Unit 8: Applications of Integration
29. Answer: A (1/6)
The curves intersect at x = 0 and x = 1. On [0, 1], y = x is above y = x². Area = ∫₀¹ (x − x²) dx = [x²/2 − x³/3]₀¹ = 1/2 − 1/3 = 1/6.
- (B) 1/3 — The student integrated only one curve or doubled the result.
- (C) 1/2 — The student only computed ∫₀¹ x dx = 1/2 without subtracting the area under y = x².
- (D) 1 — The student used the wrong interval or confused the functions.
30. Answer: B (8π)
Using the disk method: V = π∫₀⁴ (√x)² dx = π∫₀⁴ x dx = π[x²/2]₀⁴ = π · 8 = 8π. The radius of each disk is √x.
- (A) 4π — The student computed π · 4 instead of π · (4²/2).
- (C) 16π — The student used (x²)² = x⁴ instead of (√x)² = x.
- (D) 32π — The student may have doubled the volume or used incorrect bounds.
PART B — CALCULATOR ALLOWED (Questions 31–45)
Unit 1: Limits and Continuity
31. Answer: B (6.0)
As x approaches 3 from both sides, f(x) approaches 6. The values 5.9985 and 6.0015 (at x = 2.999 and 3.001 respectively) bracket 6.0, confirming the limit.
- (A) 5.99 — The student used only the left-sided values and rounded prematurely.
- (C) 6.15 — The student used the outer values (5.9850 and 6.0150) and computed an average, but these are farther from the limit.
- (D) The limit does not exist — The student may have been misled by the fact that f(3) itself is not given, but the limit clearly exists.
Unit 2: Definition of the Derivative
32. Answer: B (9)
The best approximation for f′(3.5) using symmetric data is the slope of the secant line between x = 3 and x = 4: [f(4) − f(3)] / (4 − 3) = (24 − 15) / 1 = 9.
- (A) 7.5 — The student averaged the slopes [f(3)−f(2)]/1 = 7 and [f(4)−f(3)]/1 = 9 to get 8, or used a different averaging method.
- (C) 10.5 — The student used the slope between x = 3 and x = 5: (35 − 15)/2 = 10.
- (D) 11 — The student used the slope between x = 4 and x = 5: (35 − 24)/1 = 11.
Unit 3: Chain Rule, Implicit Differentiation, and Inverse Functions
33. Answer: A (1/4)
f(1) = 1³ + 1 = 2, so g(2) = 1. By the inverse function derivative formula: g′(2) = 1/f′(g(2)) = 1/f′(1). Since f′(x) = 3x² + 1, f′(1) = 4. Therefore g′(2) = 1/4.
- (B) 1/2 — The student used f′(1) = 3 + 1 = 4 but then wrote 1/2 instead of 1/4.
- (C) 2 — The student computed f′(1) = 4 and reported that instead of its reciprocal.
- (D) 4 — The student reported f′(1) directly without taking the reciprocal.
34. Answer: C (x = 1/√2)
f′(x) = e^(−x²) + x(−2x)e^(−x²) = e^(−x²)(1 − 2x²). Setting f′(x) = 0 gives 1 − 2x² = 0, so x = ±1/√2. The second derivative test or sign analysis confirms x = 1/√2 is a relative maximum.
- (A) x = −1/√2 — This is a relative minimum, not a maximum (f″ > 0 here).
- (B) x = 0 — f′(0) = 1 ≠ 0, so this is not a critical point.
- (D) x = √2 — f′(√2) = e^(−2)(1 − 4) = −3e^(−2) < 0, so the function is decreasing here, not at a max.
35. Answer: A (−1)
Substituting (3, 3) into dy/dx = (2y − x²)/(y² − 2x): (2(3) − 9)/(9 − 6) = (6 − 9)/3 = −3/3 = −1. The slope of the tangent line is −1.
- (B) −1/3 — The student computed (6 − 9)/(9 + 6) = −3/15 = −1/3, using the wrong sign in the denominator.
- (C) 1/3 — The student inverted the numerator and denominator signs.
- (D) 1 — The student dropped the negative sign from the numerator (6 − 9).
Unit 4: Contextual Applications of Differentiation
36. Answer: B (20π cm²/s)
A = πr², so dA/dt = 2πr · dr/dt = 2π(5)(2) = 20π cm²/s. This is a straightforward related rates application with the area formula for a circle.
- (A) 10π — The student used dA/dt = πr · dr/dt instead of 2πr · dr/dt.
- (C) 25π — The student used dA/dt = πr² · dr/dt = 25π · 2 / 2 = 25π, confusing the formula.
- (D) 50π — The student doubled the correct answer or used r² · dr/dt · π.
37. Answer: C (t = 1 and t = 3)
v(t) = (t − 1)(t − 3) = 0 at t = 1 and t = 3. Since v changes sign at both points (positive → negative at t = 1, negative → positive at t = 3), the particle changes direction at both.
- (A) t = 1 only — The student missed the second zero of v(t) at t = 3.
- (B) t = 3 only — The student missed the first zero at t = 1.
- (D) t = 0, 1, and 3 — The student included t = 0, but v does not change sign there (v(0) = 3 > 0 and v is positive just after t = 0).
Unit 5: Analytical Applications of Differentiation
38. Answer: A (−1)
f′(x) = 4x³ − 8x = 4x(x² − 2). On [−1, 2], critical points are x = 0 and x = √2. Evaluate: f(−1) = 0, f(0) = 3, f(√2) = −1, f(2) = 3. The absolute minimum is −1.
- (B) 0 — This is a relative/absolute minimum on the sub-interval but not the overall absolute minimum.
- (C) 3 — This is the absolute maximum, not the minimum.
- (D) −3 — This value is not attained by f(x) on [−1, 2].
39. Answer: B ((−∞, 0) ∪ (2, 4))
f′(x) = x³ − 6x² + 8x = x(x − 2)(x − 4). Build a sign chart: on (−∞, 0) all three factors are negative (product negative); on (0, 2) one factor negative (product positive); on (2, 4) one factor negative (product negative); on (4, ∞) all positive. f is decreasing where f′ < 0: (−∞, 0) ∪ (2, 4).
- (A) (0, 2) ∪ (4, ∞) — This is where f is increasing, not decreasing.
- (C) (0, 4) — f is increasing on (0, 2) and decreasing on (2, 4); they cannot be combined.
- (D) (−∞, 2) only — This includes (0, 2) where f is increasing.
Unit 6: Integration and Accumulation of Change
40. Answer: A (e − 1)
Let u = x², so du = 2x dx. The integral becomes ∫₀¹ e^u du = [e^u]₀¹ = e − 1.
- (B) 2(e − 1) — The student kept the factor of 2 from du = 2x dx without canceling it.
- (C) e² − 1 — The student substituted u = x and integrated e^(x²) directly, which is incorrect.
- (D) 1 — The student evaluated e⁰ = 1 and forgot the upper limit.
41. Answer: A (2√2)
By FTC Part 1 with the chain rule: g′(x) = √((x²)³ + 1) · 2x. At x = 1: g′(1) = √(1 + 1) · 2(1) = √2 · 2 = 2√2.
- (B) √2 — The student forgot to multiply by the derivative of the upper limit (2x).
- (C) 4√2 — The student multiplied by 2x twice, getting (2)(2)(√2).
- (D) 2 — The student evaluated √(x⁶ + 1) at x = 1 to get √2 but then rounded or made an error.
42. Answer: B (15)
Right Riemann sum with Δx = 1: f(1) + f(2) + f(3) + f(4) = 5 + 4 + 2 + 4 = 15. Each rectangle's height is the right endpoint value.
- (A) 12 — The student used a left Riemann sum: f(0) + f(1) + f(2) + f(3) = 3 + 5 + 4 + 2 = 14... no, 14 ≠ 12. The student likely made an arithmetic error.
- (C) 16 — The student may have used a trapezoidal approximation or midpoints.
- (D) 18 — The student may have added all five values instead of the four right endpoints.
Unit 7: Differential Equations
43. Answer: A (y = 3e^(x²/4))
Separate variables: dy/y = (x/2) dx. Integrate: ln|y| = x²/4 + C. Exponentiate: y = Ae^(x²/4). Using f(0) = 3: 3 = A · 1, so A = 3.
- (B) y = 3e^(x²/2) — The student integrated x/2 incorrectly as x²/2 instead of x²/4.
- (C) y = e^(x²/4) + 2 — The student tried to add the constant outside the exponential, which is not correct for this separable DE.
- (D) y = 3e^(−x²/4) — The student introduced a negative sign that is not in the differential equation.
Unit 8: Applications of Integration
44. Answer: D (Both A and B give the correct area.)
The integrand in choice (A) simplifies: (x + 2 sin x) − x = 2 sin x, which is exactly the integrand in choice (B). Both represent the area of R correctly.
- (A) alone — This is correct but incomplete, since (B) is also correct.
- (B) alone — This is correct but incomplete, since (A) is also correct.
- (C) — This is the integral of the antiderivative, not the area formula. The student confused the function with its integral.
45. Answer: C (13 m)
Left Riemann sum: Δt = 1. Total distance = |v(0)| + |v(1)| + |v(2)| + |v(3)| = 3 + 5 + 4 + 1 = 13 meters. Absolute values are needed because distance is always positive.
- (A) 11 m — The student used v(0) + v(1) + v(2) + v(3) without absolute values: 3 + 5 + 4 + 1 = 13... this is 13, not 11. The student likely made an arithmetic error or used wrong values.
- (B) 12 m — The student may have incorrectly handled the sign of one velocity value.
- (D) 15 m — The student may have included |v(4)| = 2 in the sum, but the left sum uses t = 0, 1, 2, 3 as heights.
SECTION II: FREE RESPONSE ANSWERS
FRQ 1 — Particle Motion (Calculator)
Question (reprinted): A particle moves along the x-axis so that its velocity at time t, for 0 ≤ t ≤ 2, is given by v(t) = e^(−t²) + t − 2. The position of the particle at time t = 0 is s(0) = 1. (a) Find the acceleration at t = 1. (b) Find all times t in [0, 2] at which the particle changes direction. (c) Find the total distance traveled on [0, 2]. (d) Find the position at t = 2. Question (reprinted): Let R be the region bounded by y = 2 + sin(x²), y = e^(−x), x = 0, and x = 2. (a) Find the area of R. (b) Find the volume when R is revolved about the x-axis. (c) R is the base of a solid with square cross sections perpendicular to the x-axis. Find the volume. (d) Write (do not evaluate) an integral for the volume when R is revolved about y = −1. Question (reprinted): Consider dy/dx = (x + 1)/y. (a) Sketch a slope field at points where x, y ∈ {−1, 0, 1}. (b) Write the tangent line to f at x = 0 given f(0) = 1. (c) Find the particular solution with f(0) = 1. (d) Determine the domain of this particular solution. Question (reprinted): A particle moves along the x-axis with v(t) = t³ − 6t² + 11t − 6 for t ≥ 0. (a) Find the acceleration at t = 2. (b) Find all t ≥ 0 where the particle changes direction, and classify each change. (c) Find the total distance traveled on [0, 4]. (d) Is the speed increasing or decreasing at t = 3? Give a reason. Question (reprinted): The twice-differentiable function f has the following selected values:
x −2 −1 0 1 2 f(x) 3 1 2 4 3 f′(x) −3 −1 2 1 −2 (a) Find critical points in (−2, 2), classify each. (b) On what intervals is f concave up? (c) g(x) = f(x²). Find g′(−1). (d) h(x) = f(f(x)). Find h′(0). Question (reprinted): Let R be the region enclosed by y = x² and y = 2x. (a) Find the area of R. (b) Find the volume when R is revolved about the x-axis. (c) Find the volume when R is revolved about the y-axis. (d) Set up (do not evaluate) an equation for k where y = k divides R into two equal areas.
At t = 1: v(1) = e^(−1) + 1 − 2 ≈ 0.368 − 1 = −0.632 < 0. At t = 1.9: v(1.9) = e^(−3.61) + 1.9 − 2 ≈ 0.027 − 0.1 > 0.
Since v(t) changes from negative to positive at t ≈ 1.832, the particle changes direction from moving left to moving right at this time.
(c) Total distance = −∫₀^(1.832) v(t) dt + ∫_(1.832)² v(t) dt
Using a calculator: −∫₀^(1.832) v(t) dt ≈ −(−1.109) = 1.109 ∫_(1.832)² v(t) dt ≈ 0.684
Total distance ≈ 1.109 + 0.684 = 1.793 units
(d) s(2) = s(0) + ∫₀² v(t) dt = 1 + ∫₀² (e^(−t²) + t − 2) dt
Using a calculator: ∫₀² v(t) dt ≈ −0.793
s(2) = 1 + (−0.793) = 0.207
The particle is at position 0.207 at time t = 2.
FRQ 2 — Area and Volume (Calculator)
Question (reprinted): Let R be the region bounded by y = 2 + sin(x²), y = e^(−x), x = 0, and x = 2. (a) Find the area of R. (b) Find the volume when R is revolved about the x-axis. (c) R is the base of a solid with square cross sections perpendicular to the x-axis. Find the volume. (d) Write (do not evaluate) an integral for the volume when R is revolved about y = −1. Question (reprinted): Consider dy/dx = (x + 1)/y. (a) Sketch a slope field at points where x, y ∈ {−1, 0, 1}. (b) Write the tangent line to f at x = 0 given f(0) = 1. (c) Find the particular solution with f(0) = 1. (d) Determine the domain of this particular solution. Question (reprinted): A particle moves along the x-axis with v(t) = t³ − 6t² + 11t − 6 for t ≥ 0. (a) Find the acceleration at t = 2. (b) Find all t ≥ 0 where the particle changes direction, and classify each change. (c) Find the total distance traveled on [0, 4]. (d) Is the speed increasing or decreasing at t = 3? Give a reason. Question (reprinted): The twice-differentiable function f has the following selected values:
x −2 −1 0 1 2 f(x) 3 1 2 4 3 f′(x) −3 −1 2 1 −2 (a) Find critical points in (−2, 2), classify each. (b) On what intervals is f concave up? (c) g(x) = f(x²). Find g′(−1). (d) h(x) = f(f(x)). Find h′(0). Question (reprinted): Let R be the region enclosed by y = x² and y = 2x. (a) Find the area of R. (b) Find the volume when R is revolved about the x-axis. (c) Find the volume when R is revolved about the y-axis. (d) Set up (do not evaluate) an equation for k where y = k divides R into two equal areas.
Using a calculator: Area ≈ 3.508
(b) V = π ∫₀² [(2 + sin(x²))² − (e^(−x))²] dx
Using a calculator: V ≈ 39.545
(c) The side length of each square cross section is the height of region R: s(x) = (2 + sin(x²)) − e^(−x).
V = ∫₀² s(x)² dx = ∫₀² [(2 + sin(x²)) − e^(−x)]² dx
(d) Revolving about y = −1, the outer radius is R(x) = (2 + sin(x²)) − (−1) = 3 + sin(x²) and the inner radius is r(x) = e^(−x) − (−1) = 1 + e^(−x).
V = π ∫₀² [(3 + sin(x²))² − (1 + e^(−x))²] dx
FRQ 3 — Differential Equations (No Calculator)
Question (reprinted): Consider dy/dx = (x + 1)/y. (a) Sketch a slope field at points where x, y ∈ {−1, 0, 1}. (b) Write the tangent line to f at x = 0 given f(0) = 1. (c) Find the particular solution with f(0) = 1. (d) Determine the domain of this particular solution. Question (reprinted): A particle moves along the x-axis with v(t) = t³ − 6t² + 11t − 6 for t ≥ 0. (a) Find the acceleration at t = 2. (b) Find all t ≥ 0 where the particle changes direction, and classify each change. (c) Find the total distance traveled on [0, 4]. (d) Is the speed increasing or decreasing at t = 3? Give a reason. Question (reprinted): The twice-differentiable function f has the following selected values:
x −2 −1 0 1 2 f(x) 3 1 2 4 3 f′(x) −3 −1 2 1 −2 (a) Find critical points in (−2, 2), classify each. (b) On what intervals is f concave up? (c) g(x) = f(x²). Find g′(−1). (d) h(x) = f(f(x)). Find h′(0). Question (reprinted): Let R be the region enclosed by y = x² and y = 2x. (a) Find the area of R. (b) Find the volume when R is revolved about the x-axis. (c) Find the volume when R is revolved about the y-axis. (d) Set up (do not evaluate) an equation for k where y = k divides R into two equal areas.
Model Response
(a) Slope field (at the nine points):
| y = −1 | y = 0 | y = 1 | |
|---|---|---|---|
| x = −1 | 0 (horizontal) | undefined (vertical) | 0 (horizontal) |
| x = 0 | −1 (down-left) | undefined (vertical) | 1 (up-right) |
| x = 1 | −2 (steep down-left) | undefined (vertical) | 2 (steep up-right) |
Note: At y = 0, slopes are undefined (vertical segments). At x = −1, slopes are 0 (horizontal segments).
(b) f′(0) = (0 + 1)/f(0) = 1/1 = 1.
The tangent line at (0, 1) is: y − 1 = 1(x − 0), so y = x + 1.
(c) Separating variables: y dy = (x + 1) dx.
Integrating: y²/2 = x²/2 + x + C.
Using f(0) = 1: 1/2 = 0 + 0 + C, so C = 1/2.
y²/2 = x²/2 + x + 1/2 y² = x² + 2x + 1 = (x + 1)² y = x + 1 (taking the positive branch since f(0) = 1 > 0)
(d) The function y = x + 1 is defined for all real numbers. The domain is (−∞, ∞).
FRQ 4 — Motion and Rates (No Calculator)
Question (reprinted): A particle moves along the x-axis with v(t) = t³ − 6t² + 11t − 6 for t ≥ 0. (a) Find the acceleration at t = 2. (b) Find all t ≥ 0 where the particle changes direction, and classify each change. (c) Find the total distance traveled on [0, 4]. (d) Is the speed increasing or decreasing at t = 3? Give a reason. Question (reprinted): The twice-differentiable function f has the following selected values:
x −2 −1 0 1 2 f(x) 3 1 2 4 3 f′(x) −3 −1 2 1 −2 (a) Find critical points in (−2, 2), classify each. (b) On what intervals is f concave up? (c) g(x) = f(x²). Find g′(−1). (d) h(x) = f(f(x)). Find h′(0). Question (reprinted): Let R be the region enclosed by y = x² and y = 2x. (a) Find the area of R. (b) Find the volume when R is revolved about the x-axis. (c) Find the volume when R is revolved about the y-axis. (d) Set up (do not evaluate) an equation for k where y = k divides R into two equal areas.
- In part (b), identifying zeros of v without checking sign changes — zeros where v doesn't change sign are not direction changes.
- In part (c), computing displacement (net change) instead of total distance (absolute value).
- In part (c), arithmetic errors when evaluating the antiderivative at the bounds.
- In part (d), confusing speed (always positive) with velocity (which can be negative). The criterion is: speed is increasing when v and a have the same sign, or when v = 0 and a ≠ 0 (speed starts increasing from zero).
Model Response
(a) a(t) = v′(t) = 3t² − 12t + 11
a(2) = 3(4) − 12(2) + 11 = 12 − 24 + 11 = −1
(b) v(t) = t³ − 6t² + 11t − 6 = (t − 1)(t − 2)(t − 3)
v(t) = 0 at t = 1, 2, 3.
Testing signs:
- On (0, 1): v(0.5) = (−0.5)(−1.5)(−2.5) = −1.875 < 0 → moving left
- On (1, 2): v(1.5) = (0.5)(−0.5)(−1.5) = 0.375 > 0 → moving right
- On (2, 3): v(2.5) = (1.5)(0.5)(−0.5) = −0.375 < 0 → moving left
- On (3, ∞): v(4) = (3)(2)(1) = 6 > 0 → moving right
The particle changes direction at t = 1 (from left to right), at t = 2 (from right to left), and at t = 3 (from left to right).
(c) Antiderivative: s(t) = t⁴/4 − 2t³ + (11/2)t² − 6t
| Interval | Direction | ∫ v(t) dt | Distance | |----------|-----------|-----------|----------| | [0, 1] | Left | s(1) − s(0) = −9/4 | 9/4 | | [1, 2] | Right | s(2) − s(1) = 1/4 | 1/4 | | [2, 3] | Left | s(3) − s(2) = −1/4 | 1/4 | | [3, 4] | Right | s(4) − s(3) = 9/4 | 9/4 |
Total distance = 9/4 + 1/4 + 1/4 + 9/4 = 20/4 = 5
(d) At t = 3: v(3) = 0 and a(3) = 3(9) − 12(3) + 11 = 27 − 36 + 11 = 2.
Since v(3) = 0 and a(3) = 2 > 0, the velocity is changing from negative to positive. The speed |v| is increasing at t = 3 because the particle is momentarily at rest and the acceleration will push it in the positive direction, causing the speed to increase from zero.
FRQ 5 — Table and Graph Analysis (No Calculator)
Question (reprinted): The twice-differentiable function f has the following selected values:
x −2 −1 0 1 2 f(x) 3 1 2 4 3 f′(x) −3 −1 2 1 −2 (a) Find critical points in (−2, 2), classify each. (b) On what intervals is f concave up? (c) g(x) = f(x²). Find g′(−1). (d) h(x) = f(f(x)). Find h′(0). Question (reprinted): Let R be the region enclosed by y = x² and y = 2x. (a) Find the area of R. (b) Find the volume when R is revolved about the x-axis. (c) Find the volume when R is revolved about the y-axis. (d) Set up (do not evaluate) an equation for k where y = k divides R into two equal areas.
| (a) | 1 | Uses IVT to show f′(x) = 0 in (−1, 0) and in (1, 2) |
| (a) | 1 | Correctly classifies: relative minimum in (−1, 0) (f′ changes − to +) and relative maximum in (1, 2) (f′ changes + to −) |
| (b) | 1 | Approximates f″ using difference quotients of f′ at least twice |
| (b) | 1 | States f is concave up on (−2, 0) based on f′ increasing from −3 to 2 |
| (c) | 1 | Applies chain rule: g′(x) = f′(x²) · 2x |
| (c) | 1 | Correctly computes g′(−1) = f′(1) · (−2) = 1 · (−2) = −2 |
| (d) | 1 | Applies chain rule: h′(x) = f′(f(x)) · f′(x) |
| (d) | 1 | Correctly finds f(0) = 2 and f′(2) = −2, f′(0) = 2 |
| (d) | 1 | Correctly computes h′(0) = f′(f(0)) · f′(0) = f′(2) · 2 = (−2)(2) = −4 |
Common mistakes:
- In part (a), stating critical points without justifying with the Intermediate Value Theorem (f′ changes sign, so must cross zero).
- In part (a), misclassifying the extrema (mixing up max/min or failing to check sign changes of f′).
- In part (b), not showing the difference quotient work — a single number is not sufficient justification.
- In part (c), forgetting the chain rule on x² or making a sign error with x = −1.
- In part (d), not substituting f(0) = 2 before looking up f′(2).
Model Response
(a) By the Intermediate Value Theorem applied to f′:
- f′(−1) = −1 and f′(0) = 2. Since f′ is continuous (f is twice differentiable), f′(c₁) = 0 for some c₁ in (−1, 0). Since f′ changes from negative to positive, f has a relative minimum at x = c₁.
- f′(1) = 1 and f′(2) = −2. Similarly, f′(c₂) = 0 for some c₂ in (1, 2). Since f′ changes from positive to negative, f has a relative maximum at x = c₂.
(b) To determine concavity, examine the rate of change of f′:
- On (−2, −1): (f′(−1) − f′(−2))/(−1 − (−2)) = (−1 − (−3))/1 = 2 > 0, so f′ is increasing → f is concave up.
- On (−1, 0): (f′(0) − f′(−1))/(0 − (−1)) = (2 − (−1))/1 = 3 > 0, so f′ is increasing → f is concave up.
- On (0, 1): (f′(1) − f′(0))/(1 − 0) = (1 − 2)/1 = −1 < 0, so f′ is decreasing → f is concave down.
- On (1, 2): (f′(2) − f′(1))/(2 − 1) = (−2 − 1)/1 = −3 < 0, so f′ is decreasing → f is concave down.
The graph of f is concave up on the interval (−2, 0).
(c) g(x) = f(x²)
g′(x) = f′(x²) · 2x
g′(−1) = f′((−1)²) · 2(−1) = f′(1) · (−2) = 1 · (−2) = −2
(d) h(x) = f(f(x))
h′(x) = f′(f(x)) · f′(x)
h′(0) = f′(f(0)) · f′(0) = f′(2) · 2 = (−2)(2) = −4
FRQ 6 — Area and Volume (No Calculator)
Question (reprinted): Let R be the region enclosed by y = x² and y = 2x. (a) Find the area of R. (b) Find the volume when R is revolved about the x-axis. (c) Find the volume when R is revolved about the y-axis. (d) Set up (do not evaluate) an equation for k where y = k divides R into two equal areas.
Scoring Rubric (9 points)
| Part | Points | What Earns It |
|---|---|---|
| (a) | 1 | Finds intersections: x² = 2x gives x = 0, 2 |
| (a) | 1 | Correct area integral and evaluation: 4/3 |
| (b) | 1 | Correct washer/disc integrand: π[(2x)² − (x²)²] = π(4x² − x⁴) |
| (b) | 1 | Correct volume: 64π/15 |
| (c) | 1 | Correct shell method integrand: 2πx(2x − x²) = 2π(2x² − x³) |
| (c) | 1 | Correct volume: 8π/3 |
| (d) | 1 | Correctly identifies the bounds and integrand for the area below y = k within R |
| (d) | 1 | Sets up the equation ∫₀^(k/2) (2x − x²) dx + ∫_(k/2)^(√k) (k − x²) dx = 2/3 |
Common mistakes:
- In part (b), forgetting to square the radii in the disk/washer formula.
- In part (b), using a single disc (one radius) instead of a washer (two radii).
- In part (c), using the disk method in terms of y incorrectly, or making errors in the shell method formula.
- In part (c), forgetting the factor of 2π in the shell method.
- In part (d), setting up the integral with wrong bounds. A common error is using ∫₀^(√k)(k − x²)dx, which counts area above y = 2x that is not part of R. The correct approach recognizes that for x < k/2, the top of R is y = 2x (not y = k).
- In part (d), evaluating the integral instead of leaving it as a setup.
Model Response
(a) The curves intersect where x² = 2x, so x(x − 2) = 0, giving x = 0 and x = 2.
On [0, 2], y = 2x is above y = x².
Area = ∫₀² (2x − x²) dx = [x² − x³/3]₀² = (4 − 8/3) − 0 = 4/3
(b) Using the washer method about the x-axis:
V = π ∫₀² [(2x)² − (x²)²] dx = π ∫₀² (4x² − x⁴) dx
= π [4x³/3 − x⁵/5]₀² = π (32/3 − 32/5) = π · 32(5 − 3)/15 = 64π/15
(c) Using the shell method about the y-axis:
V = 2π ∫₀² x(2x − x²) dx = 2π ∫₀² (2x² − x³) dx
= 2π [2x³/3 − x⁴/4]₀² = 2π (16/3 − 4) = 2π · 4/3 = 8π/3
(d) The horizontal line y = k (where 0 < k < 4) intersects y = x² at x = √k and y = 2x at x = k/2. For x in [0, k/2], the top of R (y = 2x) lies below y = k, so the full region R is included. For x in [k/2, √k], only the part of R below y = k contributes. The area of R below y = k is:
∫₀^(k/2) (2x − x²) dx + ∫_(k/2)^(√k) (k − x²) dx
Setting this equal to half the total area (2/3):
∫₀^(k/2) (2x − x²) dx + ∫_(k/2)^(√k) (k − x²) dx = 2/3
END OF ANSWER KEY