How to Use This BC Multiple-Choice Practice Set
This set contains twelve multiple-choice questions written in the style of the AP Calculus BC exam. The first six questions are no-calculator questions and the last six are calculator-active. The mix mirrors the exam's two multiple-choice parts, though the real exam is longer. Treat this page as a timed checkpoint: set a timer for about twenty minutes, work every question on paper, and only then open the solutions.
Each solution states the correct choice, shows the method, and explains why a tempting wrong choice fails. Reading the explanations of questions you answered correctly is worthwhile too: they show which wrong answers the exam writers expect students to choose.
Part A: No Calculator (Questions 1–6)
Question 1
- \(0\)
- \(\dfrac12\)
- \(1\)
- The limit does not exist
Show answer and solution to Question 1
Answer: B.
The form is \(0/0\), so apply L'Hôpital's Rule: \(\lim\dfrac{e^x-1}{2x}\), which is again \(0/0\). Apply the rule once more: \(\lim\dfrac{e^x}{2}=\dfrac12\). A series check agrees: \(e^x-1-x=\tfrac{x^2}{2}+\tfrac{x^3}{6}+\cdots\), so dividing by \(x^2\) leaves \(\tfrac12\) in the limit.
Question 2
- \([-1,5]\)
- \((-1,5]\)
- \([-1,5)\)
- \((-1,5)\)
Show answer and solution to Question 2
Answer: C.
By the ratio test, \(\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{|x-2|}{3}\cdot\dfrac{n}{n+1}\to\dfrac{|x-2|}{3}\), so the series converges absolutely for \(|x-2|<3\), that is, on \((-1,5)\). Test the endpoints separately. At \(x=5\), the series is \(\sum\frac1n\), the divergent harmonic series. At \(x=-1\), it is \(\sum\frac{(-1)^n}{n}\), which converges by the Alternating Series Test. The interval is \([-1,5)\).
Question 3
- \(\ln 2\)
- \(\dfrac12\)
- \(1\)
- The integral diverges
Show answer and solution to Question 3
Answer: A.
Use partial fractions: \(\dfrac{1}{x(x+1)}=\dfrac1x-\dfrac1{x+1}\). Then \(\displaystyle\int_1^b\left(\frac1x-\frac1{x+1}\right)dx=\left[\ln\frac{x}{x+1}\right]_1^b=\ln\frac{b}{b+1}-\ln\frac12\). As \(b\to\infty\), \(\ln\frac{b}{b+1}\to0\), so the value is \(-\ln\frac12=\ln2\). The integral converges even though each separate piece, \(\int\frac1x\) and \(\int\frac1{x+1}\), diverges on its own.
Question 4
- \(\dfrac49\)
- \(\dfrac32\)
- \(9\)
- \(\dfrac94\)
Show answer and solution to Question 4
Answer: D.
For a parametric curve, \(\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{3t^2-3}{2t}\). At \(t=2\), \(\dfrac{dy}{dt}=9\) and \(\dfrac{dx}{dt}=4\), so \(\dfrac{dy}{dx}=\dfrac94\). Choice A inverts the ratio, and choice C reports only \(dy/dt\).
Question 5
- \(\dfrac{e^2}{4}\)
- \(\dfrac{e^2+1}{4}\)
- \(\dfrac{e^2-1}{4}\)
- \(\dfrac{e^2+1}{2}\)
Show answer and solution to Question 5
Answer: B.
Use integration by parts with \(u=\ln x\) and \(dv=x\,dx\), so \(du=\dfrac{dx}{x}\) and \(v=\dfrac{x^2}{2}\). Then \(\displaystyle\int x\ln x\,dx=\frac{x^2}{2}\ln x-\int\frac x2\,dx=\frac{x^2}{2}\ln x-\frac{x^2}{4}\). Evaluating from \(1\) to \(e\): \(\left(\frac{e^2}{2}-\frac{e^2}{4}\right)-\left(0-\frac14\right)=\frac{e^2}{4}+\frac14=\frac{e^2+1}{4}\).
Question 6
- \(-30\)
- \(\dfrac1{24}\)
- \(30\)
- \(720\)
Show answer and solution to Question 6
Answer: C.
Multiply the cosine series by \(x^2\): \(x^2\cos x=x^2-\dfrac{x^4}{2}+\dfrac{x^6}{24}-\cdots\). The coefficient of \(x^6\) equals \(\dfrac{f^{(6)}(0)}{6!}\), so \(f^{(6)}(0)=720\cdot\dfrac1{24}=30\). Choice B is the coefficient itself, not the derivative.
Part B: Calculator Active (Questions 7–12)
Question 7
- \(1.000\)
- \(1.148\)
- \(2.000\)
- \(1.479\)
Show answer and solution to Question 7
Answer: D.
The arc length is \(\displaystyle\int_0^1\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx=\int_0^1\sqrt{1+4x^2}\,dx\approx1.479\) by numerical integration. Plausibility: the straight segment from \((0,0)\) to \((1,1)\) has length \(\sqrt2\approx1.414\), and the curve must be longer. Choice B, \(1.148\), results from forgetting the factor \(2\) in \(y'=2x\).
Question 8
- \(4.935\)
- \(5.168\)
- \(9.870\)
- \(10.335\)
Show answer and solution to Question 8
Answer: B.
The polar area formula is \(A=\tfrac12\displaystyle\int_0^\pi r^2\,d\theta=\tfrac12\int_0^\pi\theta^2\,d\theta=\tfrac12\cdot\tfrac{\pi^3}{3}=\tfrac{\pi^3}{6}\approx5.168\). Choice A, \(\pi^2/2\), integrates \(\theta\) instead of \(\theta^2\), and choice D omits the factor \(\tfrac12\).
Question 9
- \(280.0\)
- \(300.0\)
- \(324.4\)
- \(400.0\)
Show answer and solution to Question 9
Answer: C.
The logistic solution is \(P(t)=\dfrac{500}{1+Ae^{-0.2t}}\), where \(A=\dfrac{500-100}{100}=4\). So \(P(10)=\dfrac{500}{1+4e^{-2}}\approx\dfrac{500}{1.5413}\approx324.4\). Check: the population must stay below the carrying capacity \(500\) and above the starting value \(100\), and since \(P(0)=100\) is below \(250\), the growth is still accelerating at first.
Question 10
- \(\dfrac1{192}\)
- \(\dfrac1{24}\)
- \(\dfrac1{1920}\)
- \(\dfrac1{48}\)
Show answer and solution to Question 10
Answer: A.
For a degree-\(3\) polynomial the error satisfies \(|R_3(0.5)|\le\dfrac{M}{4!}(0.5)^4\), where \(M\) bounds \(|f^{(4)}|\) on \([0,0.5]\). Every derivative of \(e^x\) is \(e^x\), so \(M=2\) works. Then \(|R_3|\le\dfrac{2}{24}\cdot\dfrac1{16}=\dfrac1{192}\approx0.0052\). The exponent and the factorial must both be one more than the degree, which rules out choices B and C.
Question 11
- \(4.000\)
- \(4.173\)
- \(16.173\)
- \(4.022\)
Show answer and solution to Question 11
Answer: D.
The velocity vector is \(\vec v(t)=\langle2t,\cos t\rangle\), so at \(t=2\) it is \(\langle4,\cos 2\rangle\approx\langle4,-0.416\rangle\). The speed is its magnitude: \(\sqrt{16+\cos^2 2}\approx\sqrt{16.173}\approx4.022\). Choice C forgets the square root, and choice A ignores the vertical component.
Question 12
- \(\dfrac1{16}\)
- \(\dfrac1{25}\)
- \(\dfrac1{36}\)
- \(\dfrac14\)
Show answer and solution to Question 12
Answer: B.
The terms \(\dfrac1{n^2}\) decrease to zero, so the error after four terms is at most the first omitted term, \(b_5=\dfrac1{5^2}=\dfrac1{25}\). The index matters: \(S_4\) uses terms \(1\) through \(4\), so the first omitted magnitude is \(b_5\), not \(b_4\).
Answer Key
Use the key to score yourself after you have attempted every question. Give yourself one point per correct answer; the exam does not deduct points for wrong answers, so never leave a question blank.
| Question | Answer |
|---|---|
| 1 | B |
| 2 | C |
| 3 | A |
| 4 | D |
| 5 | B |
| 6 | C |
| 7 | D |
| 8 | B |
| 9 | C |
| 10 | A |
| 11 | D |
| 12 | B |
Scoring and Next Steps
Out of twelve questions, ten or more suggests strong command of these topics. Seven to nine means the ideas are present but some details, such as units, signs, limits of integration or the wrong form of a rule, are costing points. Six or fewer is a signal to return to the underlying tutorials before taking another timed set. The BC free-response set that follows this page applies series, parametric and polar ideas in written form.
Sort each miss into one of three types. A concept miss means you did not know the method. A setup miss means you knew the method but built the wrong expression, for example by dividing by the wrong interval length. An execution miss means the setup was right and the arithmetic or calculator entry was wrong. Concept misses need re-study, setup misses need slower reading of the question, and execution misses need a habit of checking the answer's size and sign.
Check Your Understanding
- Which of your missed questions asked for something different from what you computed, such as net change instead of total distance?
- For each calculator question, what quick estimate could have confirmed the size of your answer before you chose a letter?
- Which wrong answer choices did you find tempting, and what mistake would produce each one?
