A curve is given by x = t^2 + 1 and y = t^3 - 3t. Which statements are true?
Select all that apply.
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This is a free 15-question sampler in the style of the AP Calculus BC multiple-choice section, written by the StudyPDF team. It focuses on what BC adds to AB: parametric, polar and vector-valued functions, infinite series and Taylor polynomials, Euler's method and logistic growth, integration by parts, partial fractions, improper integrals and arc length. It mixes multiple choice, select all that apply and fill in the blank, and every question can be done without a calculator. Answer all 15, then check your score and read the worked explanation for every question.
All 15 questions at a glance
Prefer to read before you play? Every question in the sampler is listed below. Answers and explanations stay collapsed until you open them.
1.A curve is given by x = t^2 + 1 and y = t^3 - 3t. Which statements are true?
Select all that apply.
- A. dy/dx = 9/4 when t = 2
- B. The curve has a horizontal tangent when t = 1
- C. The curve has a vertical tangent when t = 0
- D. The curve has a horizontal tangent at exactly one value of t
- E. dy/dx = 3t^2 - 3 for all t
Answer: A. dy/dx = 9/4 when t = 2, B. The curve has a horizontal tangent when t = 1, C. The curve has a vertical tangent when t = 0
For a parametric curve, dy/dx = (dy/dt)/(dx/dt) = (3t^2 - 3)/(2t). At t = 2 that is 9/4. A horizontal tangent needs dy/dt = 0 with dx/dt not 0: dy/dt = 0 at t = 1 and t = -1, and dx/dt is 2 and -2 there, so there are two such values, not one. A vertical tangent needs dx/dt = 0 with dy/dt not 0: at t = 0, dx/dt = 0 and dy/dt = -3. The expression 3t^2 - 3 is dy/dt, not dy/dx.
2.A particle moves in the plane with position (x(t), y(t)) = (6t, 4t^2). What is its speed at t = 1?
- A. 8
- B. 10
- C. 14
- D. 100
Answer: B. 10
The velocity vector is (x'(t), y'(t)) = (6, 8t), which is (6, 8) at t = 1. Speed is the magnitude of the velocity vector: sqrt(6^2 + 8^2) = sqrt(100) = 10. The answer 14 adds the components, 100 forgets the square root, and 8 is only the vertical component of the velocity.
3.What is the area enclosed by the polar curve r = 1 + cos(theta) for 0 <= theta <= 2 pi?
- A. pi
- B. 2 pi
- C. 3 pi / 2
- D. 3 pi
Answer: C. 3 pi / 2
The area in polar coordinates is (1/2) times the integral of r^2 d(theta). Here r^2 = 1 + 2 cos(theta) + cos^2(theta). Over a full turn, the integral of 1 is 2 pi, the integral of 2 cos(theta) is 0 and the integral of cos^2(theta) is pi, so the total is 3 pi. Multiply by 1/2 to get 3 pi / 2. The answer 3 pi forgets the factor of 1/2, which is the most common slip on polar area.
4.A curve is given by x = t^2 and y = t^3 for t > 0. What is d^2y/dx^2 at t = 1?
- A. 3/4
- B. 3/2
- C. 3
- D. 3/8
Answer: A. 3/4
First dy/dx = (3t^2)/(2t) = 3t/2. For the second derivative, differentiate dy/dx with respect to t and then divide by dx/dt again: d^2y/dx^2 = (3/2)/(2t) = 3/(4t). At t = 1 that is 3/4. The answer 3/2 is d/dt of dy/dx, which skips the second division by dx/dt, and that skipped division is the trap in this topic.
5.Which of the following series converge?
Select all that apply.
- A. The sum from n = 1 to infinity of 1/n
- B. The sum from n = 1 to infinity of 1/n^2
- C. The sum from n = 1 to infinity of (-1)^n / n
- D. The sum from n = 1 to infinity of n/(n + 1)
- E. The sum from n = 0 to infinity of (2/3)^n
Answer: B. The sum from n = 1 to infinity of 1/n^2, C. The sum from n = 1 to infinity of (-1)^n / n, E. The sum from n = 0 to infinity of (2/3)^n
The sum of 1/n^2 is a p-series with p = 2, which is greater than 1, so it converges. The sum of (-1)^n / n passes the alternating series test, because the terms shrink to 0 and decrease in size, so it converges, but only conditionally. The geometric series with ratio 2/3 converges because the ratio is between -1 and 1. The harmonic series 1/n is a p-series with p = 1 and diverges. The terms n/(n + 1) approach 1, not 0, so the series diverges by the nth term test.
6.What is the interval of convergence of the series: the sum from n = 1 to infinity of (x - 2)^n / (n * 3^n)?
- A. [-1, 5)
- B. (-1, 5)
- C. (-1, 5]
- D. [-1, 5]
Answer: A. [-1, 5)
The ratio test gives |x - 2|/3 < 1, so the radius is 3 and the series is centered at 2, which makes the open interval (-1, 5). Now check each endpoint separately. At x = 5 the terms are 3^n / (n * 3^n) = 1/n, the harmonic series, which diverges. At x = -1 the terms are (-3)^n / (n * 3^n) = (-1)^n / n, the alternating harmonic series, which converges. So the interval includes -1 and excludes 5.
7.What is the coefficient of x^3 in the Maclaurin series for e^(2x)?
- A. 1/6
- B. 8
- C. 4/3
- D. 2/3
Answer: C. 4/3
Start from e^u = 1 + u + u^2/2! + u^3/3! + ... and substitute u = 2x. The x^3 term is (2x)^3 / 3! = 8x^3 / 6 = (4/3) x^3. The coefficient 1/6 forgets to cube the 2, and 8 is f'''(0), the third derivative at 0, before dividing by 3 factorial. Maclaurin coefficients are f^(n)(0) / n!.
8.Complete the sentence.
Substituting -x^2 into the geometric series gives 1/(1 + x^2) = 1 - x^2 + _____ - x^6 + ... , and integrating term by term gives the series arctan x = x - _____ + x^5/5 - ... .
Options for blank 1: x^3, x^4, x^6, x^2
Options for blank 2: x^3/2, x^3, x^3/3, x^2/2
Answer: x^4, x^3/3
The geometric series 1/(1 - u) = 1 + u + u^2 + u^3 + ... with u = -x^2 gives 1 - x^2 + x^4 - x^6 + ..., so the missing term is x^4. Integrating term by term raises each power by one and divides by the new power: x - x^3/3 + x^5/5 - ... . That is the series for arctan x, because the derivative of arctan x is 1/(1 + x^2) and arctan 0 = 0 fixes the constant of integration.
9.Complete the sentence.
The alternating series for the sum from n = 1 to infinity of (-1)^(n+1) / n^2 is approximated by its first 3 terms. The alternating series error bound says the error is at most _____, the size of the _____.
Options for blank 1: 1/9, 1/16, 1/4, 1/3
Options for blank 2: last term used, first omitted term, second omitted term, sum of all omitted terms
Answer: 1/16, first omitted term
The terms 1/n^2 decrease to 0, so the alternating series error bound applies: the error from stopping after a term is no larger than the absolute value of the first omitted term. After 3 terms, the first omitted term is the fourth, 1/4^2 = 1/16. The value 1/9 is the last term used, not the first one left out. The true error is the sum of all omitted terms, and the bound says that sum is no bigger than the first one in size.
10.Euler's method with a step size of 0.5 is applied to dy/dx = x + y with y(0) = 1. What is the approximation of y(1)?
- A. 2
- B. 2.25
- C. about 3.44
- D. 2.5
Answer: D. 2.5
Step 1: the slope at (0, 1) is 0 + 1 = 1, so y(0.5) is about 1 + 0.5(1) = 1.5. Step 2: the slope at (0.5, 1.5) is 0.5 + 1.5 = 2, so y(1) is about 1.5 + 0.5(2) = 2.5. The answer 2 is a single step of size 1, and 2.25 reuses the old y-value for the second slope. The exact solution, y = 2e^x - x - 1, gives about 3.44 at x = 1. Euler's method underestimates here because the solution is concave up, so each tangent line lies below the curve.
11.A population satisfies dP/dt = 0.2 P (1 - P/500) with P(0) = 50. Which statements are true?
Select all that apply.
- A. The carrying capacity is 500
- B. The population grows fastest when P = 250
- C. P(t) is increasing for all t >= 0
- D. The graph of P is concave up for all t > 0
- E. dP/dt is largest at t = 0
Answer: A. The carrying capacity is 500, B. The population grows fastest when P = 250, C. P(t) is increasing for all t >= 0
This is a logistic equation, dP/dt = kP(1 - P/K), so K = 500 is the carrying capacity. As a function of P, the rate 0.2P(1 - P/500) is a downward parabola with its peak halfway to K, at P = 250, where the rate is 25. The population stays between 50 and 500, so dP/dt is positive and P is always increasing. The graph is concave up while P is below 250 and concave down after that, so it is not concave up for all t. At t = 0 the rate is only 0.2(50)(0.9) = 9, far below the maximum.
12.What is the value of the integral of x ln x dx from x = 1 to x = e?
- A. (e^2 - 1)/4
- B. e^2 / 2
- C. (e^2 + 1)/2
- D. (e^2 + 1)/4
Answer: D. (e^2 + 1)/4
Use integration by parts with u = ln x and dv = x dx, so du = dx/x and v = x^2/2. The antiderivative is (x^2/2) ln x - the integral of x/2 dx = (x^2/2) ln x - x^2/4. At x = e that is e^2/2 - e^2/4 = e^2/4. At x = 1 it is 0 - 1/4 = -1/4. Subtracting the lower value gives e^2/4 + 1/4 = (e^2 + 1)/4. The answer (e^2 - 1)/4 comes from dropping the minus sign when subtracting -1/4.
13.What is the integral of 1/((x - 1)(x + 2)) dx?
- A. (1/3) ln|(x - 1)/(x + 2)| + C
- B. ln|(x - 1)(x + 2)| + C
- C. (1/3) ln|(x - 1)(x + 2)| + C
- D. 3 ln|(x - 1)/(x + 2)| + C
Answer: A. (1/3) ln|(x - 1)/(x + 2)| + C
Split the fraction: 1/((x - 1)(x + 2)) = A/(x - 1) + B/(x + 2), so 1 = A(x + 2) + B(x - 1). Setting x = 1 gives 1 = 3A, so A = 1/3. Setting x = -2 gives 1 = -3B, so B = -1/3. Integrating gives (1/3) ln|x - 1| - (1/3) ln|x + 2| + C = (1/3) ln|(x - 1)/(x + 2)| + C. The log of a product would come from adding the two terms, but the second term has a minus sign, so the logs combine as a quotient.
14.Complete the sentence.
The improper integral of 1/x^2 dx from x = 1 to infinity equals _____, and the improper integral of 1/x^p dx from x = 1 to infinity converges exactly when p is _____ 1.
Options for blank 1: 0, 1, 2, The integral diverges
Options for blank 2: less than, at most, equal to, greater than
Answer: 1, greater than
Write the improper integral as a limit: the limit as b approaches infinity of the integral from 1 to b, which is the limit of (1 - 1/b) = 1. For general p, the antiderivative x^(1-p)/(1-p) goes to 0 as b grows only when 1 - p is negative, that is, when p is greater than 1. At p = 1 the antiderivative is ln b, which grows without bound, so the integral diverges, just like the harmonic series.
15.What is the length of the curve y = (2/3) x^(3/2) from x = 0 to x = 3?
- A. 3
- B. 14/3
- C. 7/3
- D. 16/3
Answer: B. 14/3
Arc length is the integral of sqrt(1 + (dy/dx)^2) dx. Here dy/dx = x^(1/2), so 1 + (dy/dx)^2 = 1 + x. The integral of sqrt(1 + x) from 0 to 3 is (2/3)(1 + x)^(3/2) evaluated from 0 to 3, which is (2/3)(8 - 1) = 14/3. The curve starts at the origin and moves 3 units to the right, so the length has to be more than 3, and 14/3 is about 4.67.
What the AP Calculus BC exam looks like and what is BC-only
The AP Calculus BC exam is scored 1 to 5 and has two sections of equal weight. According to the College Board, Section I is 42 multiple-choice questions in 1 hour 40 minutes: Part A has 29 questions in 62 minutes with no calculator, and Part B has 13 questions in 38 minutes where a graphing calculator is required. Section II is 6 free-response questions in 1 hour 30 minutes: Part A has 2 problems in 30 minutes that need a graphing calculator and Part B has 4 problems in 60 minutes that do not. The exam is hybrid digital, so you answer the multiple-choice questions in the Bluebook app and handwrite your free-response answers in a paper booklet. The May 2027 exam is on Monday, May 10, in Session 1. Older PDFs and prep books describe 45 multiple-choice questions in 1 hour 45 minutes, which is the format before the update, so trust the current College Board page.
BC covers the same ground as AB and goes further, in ten units. Units 1 to 8 share their names with AB, but BC adds topics inside them: integration by parts, partial fractions and improper integrals in Unit 6, Euler's method and logistic models in Unit 7, and arc length in Unit 8. Then it adds two whole units, Unit 9 on parametric equations, polar coordinates and vector-valued functions, and Unit 10 on infinite sequences and series. The College Board gives Unit 9 roughly 10 to 15 percent of the multiple-choice section, and Units 6 and 10 roughly 15 to 20 percent each, so these are the units where a weak spot costs the most.
A 5 is the top score, but what a score is worth is up to each college. Some give credit for a 3, many ask for a 4 or a 5, and policies differ by course. Look up the schools you care about before you decide how hard to push.
How to use this sampler, and how to take it on paper
This page leaves limits, derivative rules and basic integrals to the AP Calculus AB practice test, because BC students need those too and they are not what sets BC apart. The 15 questions here are BC-only: four on parametric, polar and vector-valued functions, five on series and Taylor polynomials, three on integration techniques and improper integrals in Unit 6, two on Euler's method and logistic growth, and one on arc length. Series gets a third of the sampler, more than its share of the real exam, because it holds the most separate ideas: convergence tests, intervals of convergence, Taylor series and error bounds.
Take it in one sitting with scratch paper and no calculator, then read the explanation for every question, including the ones you got right. Treat a miss as a topic. If you missed the interval of convergence and the alternating series error question, the gap is probably which test to reach for and how to check endpoints, not two separate facts. Math is written in plain text here, so x^2 means x squared, e^(2x) means e raised to the power 2x and ln x is the natural logarithm.
If you would rather work on paper, use the Download test as PDF button above the questions. The file has all 15 questions with an answer key on the last page, and it needs a free StudyPDF account, while playing online is free. The questions are original and written to the published course outline. They are not released exam questions, and this sampler does not practice the free-response section, so pair it with past free-response questions from the College Board.
How StudyPDF builds AP Calculus BC practice tests from your class notes
Fifteen questions can show you which unit is shaky. They cannot get you ready on their own, and a generic bank rarely matches the problem types and notation your teacher actually uses.
StudyPDF works from your material instead. Upload your textbook chapters, your teacher's slides, a worked problem set or your review packet, press New material, and StudyPDF builds practice tests from those exact pages. Every question is grounded in your material and every explanation cites where the answer comes from, so you can go straight back to the page when a step does not make sense.
You do not need to upload anything to start. Name a topic instead, for example the ratio test or Taylor series error bounds, and Bo sets it up with its key concepts so you can build quizzes and flashcards on it. Your results track which concepts you keep missing, so the next practice test can target them. Starting is free.
Written by the StudyPDF team. Last updated August 19, 2026.
Good to know
Is this AP Calculus BC practice test free?
Yes. All 15 questions, the score screen and every worked explanation are free, and the quiz runs in your browser. A free StudyPDF account is needed to download the test as a PDF with an answer key or to build more practice tests from your own notes.
Who wrote these questions?
The StudyPDF team wrote every question and explanation, following the BC-only topics of the published AP Calculus BC course outline. They are original items, not real or released exam questions, and StudyPDF is not affiliated with the College Board, which does not endorse this page.
How is this different from the real AP Calculus BC exam?
The real multiple-choice section has one correct answer per question and is split into a part without a graphing calculator and a part with one. This sampler also uses select all that apply and fill in the blank, and every question can be done by hand. It covers only what BC adds to AB, so for limits, derivative rules and basic integrals use the AP Calculus AB practice test, and it does not practice the six free-response questions.
Can I get practice tests built from my own class materials?
Yes. Upload your textbook chapters, slides, problem sets or review packet to StudyPDF and you can build practice tests from those exact pages, with explanations tied to your sources. If you have nothing to upload, name a topic or a unit instead.