How to Use This AB Multiple-Choice Practice Set
This set contains twelve multiple-choice questions written in the style of the AP Calculus AB 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
- \(0\)
- \(1\)
- \(3\)
- The limit does not exist
Show answer and solution to Question 1
Answer: C.
Split the quotient as \(\dfrac{\sin 3x}{x}\cdot\dfrac{1}{\cos x}\). Because \(\dfrac{\sin 3x}{x}=3\cdot\dfrac{\sin 3x}{3x}\to 3\cdot 1=3\) and \(\cos x\to1\), the limit is \(3\). The common error is to stop at \(\sin(3x)/x\to 1\), which forgets that the argument is \(3x\), not \(x\).
Question 2
- \(-3\)
- \(-1\)
- \(1\)
- \(3\)
Show answer and solution to Question 2
Answer: A.
By the quotient rule, \(f'(x)=\dfrac{2(x-1)-(2x+1)(1)}{(x-1)^2}=\dfrac{-3}{(x-1)^2}\). Then \(f'(2)=\dfrac{-3}{1^2}=-3\). A quick check: \(f\) is a decreasing hyperbola on each side of its vertical asymptote, so a negative slope is expected.
Question 3
- \(e^4-1\)
- \(\dfrac{e^4-1}{2}\)
- \(\dfrac{e^4}{2}\)
- \(2(e^4-1)\)
Show answer and solution to Question 3
Answer: B.
Let \(u=x^2\), so \(du=2x\,dx\) and \(x\,dx=\tfrac12\,du\). The limits become \(u=0\) and \(u=4\). Then \(\displaystyle\int_0^2 xe^{x^2}dx=\tfrac12\int_0^4 e^u\,du=\tfrac12(e^4-1)\). Choice A forgets the factor \(\tfrac12\); choice C drops the lower-limit value \(e^0=1\).
Question 4
- \(-\dfrac54\)
- \(-\dfrac45\)
- \(\dfrac45\)
- \(\dfrac54\)
Show answer and solution to Question 4
Answer: B.
First confirm the point lies on the curve: \(1+2+4=7\). Differentiate implicitly: \(2x+y+x\dfrac{dy}{dx}+2y\dfrac{dy}{dx}=0\). At \((1,2)\) this gives \(2+2+\dfrac{dy}{dx}(1+4)=0\), so \(5\dfrac{dy}{dx}=-4\) and \(\dfrac{dy}{dx}=-\dfrac45\). Choice A is the reciprocal, a common slip when solving for the derivative.
Question 5
- \(1\)
- \(\dfrac32\)
- \(\dfrac92\)
- \(3\)
Show answer and solution to Question 5
Answer: D.
The average value is \(\dfrac{1}{3-0}\displaystyle\int_0^3 x^2\,dx=\dfrac13\cdot\dfrac{27}{3}=\dfrac13\cdot 9=3\). Choice C, \(9/2\), is the result of dividing the integral by \(2\) instead of by the interval length \(3\).
Question 6
- \(\dfrac83\)
- \(\dfrac43\)
- \(0\)
- \(4\)
Show answer and solution to Question 6
Answer: A.
Total distance requires \(\int|v|\), so find the sign changes. Since \(v(t)=(t-1)(t-3)\), velocity is positive on \([0,1)\) and negative on \((1,3)\). With \(F(t)=\tfrac{t^3}{3}-2t^2+3t\): \(\int_0^1 v\,dt=F(1)-F(0)=\tfrac43\) and \(\int_1^3 v\,dt=F(3)-F(1)=0-\tfrac43=-\tfrac43\). Total distance is \(\tfrac43+\tfrac43=\tfrac83\). Choice C is the net displacement, \(F(3)-F(0)=0\).
Part B: Calculator Active (Questions 7–12)
Question 7
- \(2.667\)
- \(2.828\)
- \(3.241\)
- \(4.000\)
Show answer and solution to Question 7
Answer: C.
This integrand has no elementary antiderivative, so use the calculator's numerical integration command: \(\int_0^2\sqrt{1+x^3}\,dx\approx3.241\). A plausibility check: the integrand rises from \(1\) at \(x=0\) to \(3\) at \(x=2\), so the average height lies between those values and the area on an interval of width \(2\) lies between \(2\) and \(6\).
Question 8
- \(2.445\)
- \(2.000\)
- \(1.585\)
- \(3.000\)
Show answer and solution to Question 8
Answer: A.
Define \(h(x)=2^x-x-3\) and solve \(h(x)=0\). Because \(h(2)=4-5=-1<0\) and \(h(3)=8-6=2>0\), a root lies in \((2,3)\). The calculator's solver (or the intersection of the two graphs) gives \(x\approx2.445\). Choice D fails the check, since \(h(3)=2\ne0\), and choice B gives \(h(2)=-1\).
Question 9
| \(t\) (min) | 0 | 2 | 5 | 9 |
|---|---|---|---|---|
| \(r(t)\) (gal/min) | 3 | 5 | 4 | 8 |
- \(41.0\)
- \(43.5\)
- \(49.0\)
- \(45.5\)
Show answer and solution to Question 9
Answer: D.
The subintervals have unequal widths, so apply the trapezoid formula to each one separately: \(\tfrac12(2)(3+5)=8\), \(\tfrac12(3)(5+4)=13.5\), and \(\tfrac12(4)(4+8)=24\). The sum is \(8+13.5+24=45.5\) gallons. Choice B comes from treating all widths as equal.
Question 10
- \(-0.416\)
- \(-1.033\)
- \(1.033\)
- \(2.483\)
Show answer and solution to Question 10
Answer: B.
By the chain rule, \(f'(x)=\cos x\cdot e^{\sin x}\). Evaluate in radian mode: \(f'(2)=\cos 2\cdot e^{\sin 2}\approx(-0.416)(2.483)\approx-1.033\). Choice A is only \(\cos 2\), choice D is \(f(2)\), and choice C has the wrong sign.
Question 11
- \(1.868\)
- \(2.511\)
- \(3.652\)
- \(4.180\)
Show answer and solution to Question 11
Answer: C.
Solve \(e^x=3-x^2\) with the calculator to get the intersections \(x\approx-1.677\) and \(x\approx0.834\). Test a point between them: at \(x=0\), \(3-x^2=3\) lies above \(e^0=1\). So the area is \(\displaystyle\int_{-1.677}^{0.834}\left(3-x^2-e^x\right)dx\approx3.652\). Store the two roots in the calculator rather than retyping rounded values, which would shift the third decimal place.
Question 12
- \(1.571\)
- \(4.935\)
- \(6.283\)
- \(9.870\)
Show answer and solution to Question 12
Answer: B.
Use the disk method: \(V=\pi\displaystyle\int_0^\pi\sin^2x\,dx\). Since \(\int_0^\pi\sin^2x\,dx=\tfrac{\pi}{2}\), \(V=\tfrac{\pi^2}{2}\approx4.935\). Choice C, \(2\pi\), equals \(\pi\int_0^\pi\sin x\,dx\), which uses the radius \(\sin x\) without squaring it.
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 | C |
| 2 | A |
| 3 | B |
| 4 | B |
| 5 | D |
| 6 | A |
| 7 | C |
| 8 | A |
| 9 | D |
| 10 | B |
| 11 | C |
| 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. Once you have sorted your misses, the free-response set that follows this page tests the same skills with written justification.
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?
