Getting ready for the AP Calculus AB exam? These 20 free practice questions cover the big ideas from every unit of the course, each with a clear explanation. Try each question on your own first, then tap Show answer to check your thinking.
The questions are written in the style of the multiple-choice section of the exam. Almost all of them can be done without a calculator, and the one that needs a calculator is marked. They’re a quick way to find out which units you’ve mastered and which ones need more review.
What the AP Calculus AB exam covers
- Limits and Continuity: evaluating limits, limits at infinity, continuity and the Intermediate Value Theorem
- Differentiation: Definition and Fundamental Properties: the derivative as a limit, tangent lines and the power, product and quotient rules
- Differentiation: Composite, Implicit, and Inverse Functions: the chain rule, implicit differentiation and derivatives of inverse functions
- Contextual Applications of Differentiation: rates of change, motion, related rates, linear approximation and L’Hôpital’s Rule
- Analytical Applications of Differentiation: the Mean Value Theorem, extreme values, increasing and decreasing behavior, concavity and optimization
- Integration and Accumulation of Change: Riemann sums, definite integrals, the Fundamental Theorem of Calculus and antiderivatives
- Differential Equations: slope fields, separation of variables and exponential models
- Applications of Integration: average value, motion, area between curves and volumes of solids
How to use these questions
Work through the questions without notes, the way you would on test day. Keep track of the ones you miss, and note which unit each one comes from. Two or three misses in the same unit is a sign to review that unit before moving on.
Don’t just check whether you got the right letter. Read every explanation, even for questions you answered correctly. AP Calculus questions often reward knowing why a rule applies, not just how to use it, and the explanations point out the most common traps.
Practice questions
Question 1
limx→∞ (3x² − 4x) / (6x² + 1) is
A. 0
B. 3
C. 1/2
D. nonexistent
Show answer
Answer: C. The numerator and denominator have the same degree, so the limit is the ratio of the leading coefficients: 3/6 = 1/2. Dividing every term by x² shows why: the −4/x and 1/x² terms go to 0.
Question 2
A function f is continuous on [1, 4], with f(1) = −2 and f(4) = 5. Which statement must be true?
A. f(c) = 0 for at least one c in (1, 4).
B. f′(c) = 7/3 for at least one c in (1, 4).
C. f has an absolute maximum at x = 4.
D. f is increasing on [1, 4].
Show answer
Answer: A. Because f is continuous and 0 is between −2 and 5, the Intermediate Value Theorem guarantees some c in (1, 4) with f(c) = 0. Choice B would need the Mean Value Theorem, which also requires f to be differentiable, and that is not given.
Question 3
If f(x) = x² sin x, then f′(x) =
A. 2x cos x
B. 2x sin x + x² cos x
C. x² cos x
D. 2x sin x − x² cos x
Show answer
Answer: B. Use the product rule: (first)′(second) + (first)(second)′. That gives (2x)(sin x) + (x²)(cos x). Choice A is the common trap of multiplying the two derivatives together.
Question 4
Which is an equation of the line tangent to the graph of f(x) = x³ − 2x at x = 1?
A. y = x + 1
B. y = −x
C. y = 3x − 4
D. y = x − 2
Show answer
Answer: D. First find the point: f(1) = 1 − 2 = −1, so the point is (1, −1). Then find the slope: f′(x) = 3x² − 2, so f′(1) = 1. The tangent line is y − (−1) = 1(x − 1), which simplifies to y = x − 2.
Question 5
If y = ln(3x² + 1), then dy/dx =
A. 6x / (3x² + 1)
B. 1 / (3x² + 1)
C. 1 / (6x)
D. 6x ln(3x² + 1)
Show answer
Answer: A. The derivative of ln(u) is u′/u. Here u = 3x² + 1 and u′ = 6x, so dy/dx = 6x/(3x² + 1). Choice B forgets the chain rule factor.
Question 6
A particle moves along a line with velocity v(t) = t² − 4t + 3. At time t = 1.5, the particle is
A. slowing down, because its velocity is negative
B. speeding up, because its velocity and acceleration are both negative
C. slowing down, because its acceleration is negative
D. at rest
Show answer
Answer: B. At t = 1.5, v = 2.25 − 6 + 3 = −0.75, and a(t) = v′(t) = 2t − 4, so a = −1. When velocity and acceleration have the same sign, the particle is speeding up. Looking at the sign of velocity or acceleration alone is not enough.
Question 7
At what value of x does f(x) = x³ − 3x² − 9x + 5 have a relative maximum?
A. x = 3
B. x = 1
C. x = −1
D. x = −3
Show answer
Answer: C. f′(x) = 3x² − 6x − 9 = 3(x − 3)(x + 1), so the critical points are x = −1 and x = 3. f′ is positive for x < −1 and negative between −1 and 3, so f′ changes from positive to negative at x = −1. That makes x = −1 a relative maximum, and x = 3 is a relative minimum.
Question 8
∫ cos(2x) dx =
A. 2 sin(2x) + C
B. (1/2) sin(2x) + C
C. −(1/2) sin(2x) + C
D. sin(2x) + C
Show answer
Answer: B. An antiderivative of cos(u) is sin(u). Because the inside function is 2x, you divide by 2 to undo the chain rule: (1/2) sin(2x) + C. Check by differentiating: (1/2)(cos(2x))(2) = cos(2x).
Question 9
Which function is the solution to dy/dx = 2xy with y(0) = 3?
A. y = 3e2x
B. y = x² + 3
C. y = 3 + ex²
D. y = 3ex²
Show answer
Answer: D. Separate the variables: dy/y = 2x dx. Integrating gives ln|y| = x² + C, so y = Aex². Using y(0) = 3 gives A = 3, so y = 3ex².
Question 10
The region bounded by y = 2x, the x-axis and the line x = 3 is revolved about the x-axis. What is the volume of the solid?
A. 36π
B. 18π
C. 9π
D. 108π
Show answer
Answer: A. Each cross section is a disk with radius 2x, so V = π ∫03 (2x)² dx = π ∫03 4x² dx = π(4/3)(27) = 36π. A common mistake is forgetting to square the radius.
More practice from the AP+ workbook
These next 10 questions come straight from our AP+ Review Workbook, with the same explanations you’ll find in its answer key.
Question 11
Let f(x) = kx + 1 if x < 3, and f(x) = x² − 2 if x ≥ 3. For what value of k is f continuous at x = 3?
A. 2
B. 7/3
C. 3
D. 8/3
Show answer
Answer: A. For f to be continuous at x = 3, the two pieces must meet there. The right piece gives 3² − 2 = 7, so the left piece must also equal 7 at x = 3: 3k + 1 = 7, which gives k = 2.
Question 12
If f(x) = 3x⁴ − 5x² + 2, then f′(x) =
A. 12x³ − 5x
B. 3x³ − 10x
C. 12x⁴ − 10x²
D. 12x³ − 10x
Show answer
Answer: D. Apply the power rule to each term: the derivative of 3x⁴ is 12x³, the derivative of −5x² is −10x, and the derivative of the constant 2 is 0. So f′(x) = 12x³ − 10x.
Question 13
If f(1) = 4 and f′(1) = 3, then (f−1)′(4) =
A. 3
B. 1/3
C. 4
D. 1/4
Show answer
Answer: B. Since f(1) = 4, the inverse function sends 4 back to 1. The derivative of the inverse at 4 is the reciprocal of f′ at the matching point: 1/f′(1) = 1/3.
Question 14
A sphere’s radius increases at 2 cm/s. How fast is its volume increasing when r = 3 cm?
A. 24π cm³/s
B. 36π cm³/s
C. 72π cm³/s
D. 108π cm³/s
Show answer
Answer: C. Differentiate V = (4/3)πr³ with respect to time: dV/dt = 4πr² (dr/dt). Substituting r = 3 and dr/dt = 2 gives 4π(9)(2) = 72π cm³/s.
Question 15
For f(x) = x³ on [0, 3], the Mean Value Theorem guarantees a c with f′(c) equal to the average rate of change. c =
A. 1
B. 3/2
C. √3
D. 2
Show answer
Answer: C. The average rate of change on [0, 3] is (27 − 0)/3 = 9. Setting f′(c) = 3c² equal to 9 gives c² = 3, so c = √3, which lies inside the interval.
Question 16
∫12 (6x² − 2x) dx =
A. 9
B. 11
C. 13
D. 14
Show answer
Answer: B. An antiderivative is 2x³ − x². Evaluating from 1 to 2 gives (16 − 4) − (2 − 1) = 12 − 1 = 11.
Question 17
A rate is measured as r(0) = 2, r(1) = 5, r(3) = 4, r(4) = 1. A trapezoidal sum for ∫04 r(t) dt using these values is
A. 12.0
B. 13.5
C. 15.0
D. 16.5
Show answer
Answer: C. The subintervals have different widths, so find each trapezoid separately: on [0, 1], 1(2 + 5)/2 = 3.5; on [1, 3], 2(5 + 4)/2 = 9; on [3, 4], 1(4 + 1)/2 = 2.5. The total is 3.5 + 9 + 2.5 = 15.0.
Question 18
For dy/dx = x − y, the slope of the solution curve through (2, 1) is
A. −1
B. 0
C. 3
D. 1
Show answer
Answer: D. The differential equation gives the slope directly at any point. Substituting x = 2 and y = 1 gives dy/dx = 2 − 1 = 1.
Question 19
A population satisfies dP/dt = 0.06P with P(0) = 250, t in years. When does the population reach 1000? (Calculator allowed.)
A. t ≈ 23.1
B. t ≈ 16.7
C. t ≈ 11.6
D. t ≈ 50.0
Show answer
Answer: A. The solution is P = 250e0.06t. Setting it equal to 1000 gives e0.06t = 4, so t = (ln 4)/0.06 ≈ 23.1 years.
Question 20
The area of the region enclosed by y = x and y = x² is
A. 1/2
B. 1/6
C. 1/3
D. 5/6
Show answer
Answer: B. The curves meet at x = 0 and x = 1, and y = x is on top between them. The area is ∫01 (x − x²) dx = 1/2 − 1/3 = 1/6.
How did you do?
If you got 17 or more right, you have a strong foundation. Focus your review on the explanations for any you missed. If you got 12 to 16 right, you’re on track, but a few units need more practice. If you got 11 or fewer right, start with the unit reviews before doing more practice questions.
Our AP+ Calculus AB Review Workbook covers every unit with clear lessons, worked examples, calculator and no-calculator practice, FRQ practice with self-scoring checklists and a full-length practice exam with answers and worked solutions. It’s available as an instant PDF download or a paperback.
See the AP+ Calculus AB Review Workbook →
Studying for more than one AP exam? Browse the full AP+ Exam Prep series.
Frequently asked questions
Are these questions like the real AP Calculus AB exam?
They’re written in the style of the multiple-choice section and cover the same units and skills. Official practice materials from the College Board are also worth using alongside them.
What’s the best way to study for AP Calculus AB?
Review one unit at a time, practice both calculator and no-calculator problems, and write out your setup and reasoning in full. Mixed practice from different units, plus reading every explanation, builds the understanding the exam rewards.
When is the AP Calculus AB exam?
AP exams are given each May. Check the College Board’s AP website for the current exam schedule.
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