AP Calculus AB/BC — Master Problem Set

20 Essential Multiple-Choice Questions · All Major Units

Name: _________________________ Date: _____________ Score: ____/20

Unit 1 — Limits & Continuity

Q1–Q3

Key Concepts

Limit Laws: \(\lim_{x\to a}[f(x)\pm g(x)]=L\pm M\). Squeeze Theorem: If \(g(x)\le f(x)\le h(x)\) and \(\lim g=\lim h=L\), then \(\lim f=L\).
L'Hôpital's Rule: If \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then \(\lim\frac{f}{g}=\lim\frac{f'}{g'}\). Continuity: \(f\) continuous at \(a\) iff \(\lim_{x\to a}f(x)=f(a)\).

Must-Memorize

\(\displaystyle\lim_{x\to 0}\frac{\sin x}{x}=1\) · \(\displaystyle\lim_{x\to 0}\frac{1-\cos x}{x}=0\) · \(\displaystyle\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x=e\)

Q 01 Limits · L'Hôpital Medium

\(\displaystyle\lim_{x\to 0}\frac{e^{3x}-1-3x}{x^2} = \)

Q 02 Continuity Medium

Let \(f(x)=\begin{cases}kx+2 & x<3\\x^2-1 & x\ge 3\end{cases}\). For what value of \(k\) is \(f\) continuous at \(x=3\)?

Q 03 Limits at Infinity Medium

\(\displaystyle\lim_{x\to\infty}\frac{4x^3-7x+2}{6x^3+x^2-5} = \)

Unit 2 — Differentiation

Q4–Q6

Key Rules

Chain Rule: \(\dfrac{d}{dx}[f(g(x))]=f'(g(x))\cdot g'(x)\) Product: \((fg)'=f'g+fg'\) Quotient: \(\left(\dfrac{f}{g}\right)'=\dfrac{f'g-fg'}{g^2}\)

Derivatives to Memorize

\(\frac{d}{dx}[\ln x]=\frac{1}{x}\) · \(\frac{d}{dx}[e^x]=e^x\) · \(\frac{d}{dx}[\sin x]=\cos x\) · \(\frac{d}{dx}[\tan x]=\sec^2 x\) · \(\frac{d}{dx}[\arcsin x]=\frac{1}{\sqrt{1-x^2}}\) · \(\frac{d}{dx}[\arctan x]=\frac{1}{1+x^2}\)

Q 04 Chain Rule Medium

If \(f(x)=\sin^3(x^2)\), then \(f'(x) = \)

Q 05 Implicit Differentiation Hard

If \(x^2y + y^3 = 5\), find \(\dfrac{dy}{dx}\) at the point \((2, 1)\).

Q 06 Higher-Order Derivatives Medium

If \(f(x)=xe^x\), what is \(f^{(n)}(x)\)?

Unit 3 — Applications of Derivatives

Q7–Q9

Key Theorems

MVT: \(f'(c)=\dfrac{f(b)-f(a)}{b-a}\) for some \(c\in(a,b)\). Concavity: \(f''>0\Rightarrow\) concave up; \(f''<0\Rightarrow\) concave down. Related Rates: Differentiate an equation implicitly with respect to time.

Q 07 Mean Value Theorem Medium

Let \(f(x)=x^3-3x\) on \([0,3]\). What value of \(c\) satisfies the Mean Value Theorem?

Q 08 Optimization Hard

A farmer has 200 m of fencing to enclose a rectangular pen against a barn wall (one side needs no fence). What dimensions maximize the area?

Q 09 Related Rates Hard

A spherical balloon is being inflated so that its volume increases at \(8\pi\text{ cm}^3/\text{s}\). How fast is the radius increasing when \(r=2\text{ cm}\)?

Unit 4 — Integration & the FTC

Q10–Q13

Fundamental Theorem of Calculus

FTC Part 1: \(\dfrac{d}{dx}\displaystyle\int_a^{g(x)}f(t)\,dt = f(g(x))\cdot g'(x)\).
FTC Part 2: \(\displaystyle\int_a^b f(x)\,dx = F(b)-F(a)\) where \(F'=f\).

Integration Formulas

\(\int x^n dx=\dfrac{x^{n+1}}{n+1}+C\) · \(\int e^x dx=e^x+C\) · \(\int \dfrac{1}{x}dx=\ln|x|+C\) · \(\int \sin x\,dx=-\cos x+C\) · \(\int \sec^2 x\,dx=\tan x+C\)

Q 10 FTC Part 1 Medium

If \(F(x)=\displaystyle\int_1^{x^2}\sqrt{t^3+1}\,dt\), then \(F'(x)=\)

Q 11 U-Substitution Medium

\(\displaystyle\int_0^1 x\,e^{x^2}\,dx = \)

Q 12 Area Between Curves Hard

The area enclosed by \(y=x^2\) and \(y=2x\) is:

Q 13 Volume · Disk Method Hard

The region bounded by \(y=\sqrt{x}\), \(x=4\), and the \(x\)-axis is revolved about the \(x\)-axis. The volume is:

Unit 5 — Differential Equations

Q14–Q15

Key Concepts

Separable DE: Rewrite as \(f(y)\,dy = g(x)\,dx\), then integrate both sides.
Logistic Growth: \(\dfrac{dP}{dt}=kP\!\left(1-\dfrac{P}{M}\right)\), carrying capacity \(M\), inflection at \(P=M/2\).

Q 14 Separable DE Medium

Solve \(\dfrac{dy}{dx}=\dfrac{x}{y}\) with initial condition \(y(0)=3\). What is \(y(4)\)?

Q 15 Logistic Growth Hard

A population follows logistic growth with carrying capacity \(M=500\). The population grows fastest when \(P = \)

Unit 6 — Parametric & Polar (BC)

Q16–Q17

Key Formulas

Parametric slope: \(\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}\). Arc length (param): \(L=\displaystyle\int_a^b\!\sqrt{\left(\tfrac{dx}{dt}\right)^2+\left(\tfrac{dy}{dt}\right)^2}\,dt\). Polar area: \(A=\dfrac{1}{2}\displaystyle\int_\alpha^\beta r^2\,d\theta\).

Q 16 Parametric · dy/dx Medium

A curve is defined parametrically by \(x=t^2+1\), \(y=t^3-3t\). At \(t=2\), the slope \(\dfrac{dy}{dx}\) equals:

Q 17 Polar Area Hard

The area enclosed by the polar curve \(r=2\cos\theta\) is:

Unit 7 — Infinite Series (BC)

Q18–Q20

Convergence Tests

Ratio Test: \(L=\lim\left|\dfrac{a_{n+1}}{a_n}\right|\); converges if \(L<1\), diverges if \(L>1\).
Geometric Series: \(\sum_{n=0}^\infty ar^n=\dfrac{a}{1-r}\) for \(|r|<1\).
p-Series: \(\sum\dfrac{1}{n^p}\) converges iff \(p>1\).

Taylor Series (Memorize)

\(e^x=\sum_{n=0}^\infty\dfrac{x^n}{n!}\) · \(\sin x=\sum_{n=0}^\infty\dfrac{(-1)^n x^{2n+1}}{(2n+1)!}\) · \(\cos x=\sum_{n=0}^\infty\dfrac{(-1)^n x^{2n}}{(2n)!}\) · \(\dfrac{1}{1-x}=\sum_{n=0}^\infty x^n,\,|x|<1\)

Q 18 Convergence · Ratio Test Hard

For what values of \(x\) does \(\displaystyle\sum_{n=1}^{\infty}\frac{(x-2)^n}{n\cdot 3^n}\) converge?

Q 19 Taylor Series Hard

The coefficient of \(x^4\) in the Maclaurin series for \(f(x)=\cos(x^2)\) is:

Q 20 Integration via Series Hard

Using the Maclaurin series, the first three nonzero terms of \(\displaystyle\int_0^x \frac{\sin t}{t}\,dt\) are:

Complete Solutions & Explanations

Question 1 — Limits · L'Hôpital

Answer: D — 9/2

The limit is \(\frac{0}{0}\) at \(x=0\). Apply L'Hôpital twice: first derivative gives \(\frac{3e^{3x}-3}{2x}\), still \(\frac{0}{0}\). Second derivative gives \(\frac{9e^{3x}}{2}\to\frac{9}{2}\).
Alternatively, using the Taylor expansion \(e^{3x}=1+3x+\frac{(3x)^2}{2!}+\cdots\), so \(e^{3x}-1-3x=\frac{9x^2}{2}+\cdots\), and \(\frac{9x^2/2}{x^2}=\frac{9}{2}\).

Question 2 — Continuity

Answer: C — 8/3

For continuity at \(x=3\): \(\lim_{x\to3^-}f(x)=3k+2\) must equal \(f(3)=3^2-1=8\). So \(3k+2=8\Rightarrow 3k=6\Rightarrow k=2\).
Wait — re-check: \(3k+2=8\Rightarrow k=2\). But answer C says \(\frac{8}{3}\). Let me be precise: \(3k+2=8\Rightarrow 3k=6\Rightarrow k=2\). Answer: B — \(k=2\).

Question 3 — Limits at Infinity

Answer: C — 4/6 = 2/3

When the leading degrees are equal, the limit equals the ratio of leading coefficients: \(\dfrac{4}{6}=\dfrac{2}{3}\). Choices B and C both show \(\frac{2}{3}\); choice C writes it as \(\frac{4}{6}\) explicitly — same value. Answer is \(\dfrac{2}{3}\).

Question 4 — Chain Rule

Answer: B — 6x sin²(x²) cos(x²)

Apply the chain rule twice. Let \(u=\sin(x^2)\), so \(f=u^3\). Then \(f'=3u^2\cdot u'\). Now \(u'=\cos(x^2)\cdot 2x\). Therefore \(f'=3\sin^2(x^2)\cdot\cos(x^2)\cdot 2x=6x\sin^2(x^2)\cos(x^2)\).

Question 5 — Implicit Differentiation

Answer: A — −4/7

Differentiate \(x^2y+y^3=5\) implicitly: \(2xy + x^2\dfrac{dy}{dx}+3y^2\dfrac{dy}{dx}=0\). Factor: \(\dfrac{dy}{dx}(x^2+3y^2)=-2xy\). At \((2,1)\): \(\dfrac{dy}{dx}=\dfrac{-2(2)(1)}{4+3}=\dfrac{-4}{7}\).

Question 6 — Higher-Order Derivatives

Answer: C — (x + n)eˣ

By the pattern: \(f'=(x+1)e^x\), \(f''=(x+2)e^x\), \(f'''=(x+3)e^x\). In general \(f^{(n)}(x)=(x+n)e^x\). This follows by induction using the product rule each time.

Question 7 — Mean Value Theorem

Answer: B — √3

\(f(0)=0\), \(f(3)=27-9=18\). Average rate \(=\frac{18-0}{3-0}=6\). Set \(f'(c)=3c^2-3=6\Rightarrow 3c^2=9\Rightarrow c^2=3\Rightarrow c=\sqrt{3}\approx1.73\in(0,3)\). ✓

Question 8 — Optimization

Answer: A — 50 m × 100 m

Let the side parallel to the barn wall have length \(L\) and the two perpendicular sides have length \(w\). Constraint: \(L+2w=200\Rightarrow L=200-2w\). Area \(A=Lw=(200-2w)w=200w-2w^2\). \(A'=200-4w=0\Rightarrow w=50\), \(L=100\). Maximum area \(=5000\text{ m}^2\).

Question 9 — Related Rates

Answer: A — 1/2 cm/s

\(V=\frac{4}{3}\pi r^3\Rightarrow\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}\). Given \(\dfrac{dV}{dt}=8\pi\) and \(r=2\): \(8\pi=4\pi(4)\dfrac{dr}{dt}\Rightarrow\dfrac{dr}{dt}=\dfrac{8\pi}{16\pi}=\dfrac{1}{2}\text{ cm/s}\).

Question 10 — FTC Part 1

Answer: B — 2x√(x⁶ + 1)

By FTC Part 1 with upper limit \(g(x)=x^2\): \(F'(x)=\sqrt{(x^2)^3+1}\cdot\dfrac{d}{dx}(x^2)=\sqrt{x^6+1}\cdot 2x=2x\sqrt{x^6+1}\).

Question 11 — U-Substitution

Answer: B — (e − 1)/2

Let \(u=x^2\Rightarrow du=2x\,dx\Rightarrow x\,dx=\frac{du}{2}\). When \(x=0, u=0\); when \(x=1, u=1\). \(\displaystyle\int_0^1 xe^{x^2}dx=\frac{1}{2}\int_0^1 e^u\,du=\frac{1}{2}[e^u]_0^1=\frac{e-1}{2}\).

Question 12 — Area Between Curves

Answer: B — 4/3

Intersections: \(x^2=2x\Rightarrow x=0,2\). On \([0,2]\), \(2x\ge x^2\). Area \(=\displaystyle\int_0^2(2x-x^2)dx=\left[x^2-\tfrac{x^3}{3}\right]_0^2=4-\tfrac{8}{3}=\tfrac{4}{3}\).

Question 13 — Volume · Disk Method

Answer: B — 8π

\(V=\pi\displaystyle\int_0^4 (\sqrt{x})^2\,dx=\pi\int_0^4 x\,dx=\pi\left[\tfrac{x^2}{2}\right]_0^4=\pi\cdot 8=8\pi\).

Question 14 — Separable DE

Answer: A — 5

Separate: \(y\,dy=x\,dx\). Integrate: \(\tfrac{y^2}{2}=\tfrac{x^2}{2}+C\Rightarrow y^2=x^2+C\). At \((0,3)\): \(9=0+C\Rightarrow C=9\). So \(y^2=x^2+9\). At \(x=4\): \(y^2=16+9=25\Rightarrow y=5\).

Question 15 — Logistic Growth

Answer: B — 250

For logistic growth \(\frac{dP}{dt}=kP(1-P/M)\), the growth rate is maximized at the inflection point, which occurs at \(P=M/2=500/2=250\). This is where \(\frac{d^2P}{dt^2}=0\).

Question 16 — Parametric · dy/dx

Answer: A — 3/2

\(\dfrac{dx}{dt}=2t\), \(\dfrac{dy}{dt}=3t^2-3\). At \(t=2\): \(\dfrac{dx}{dt}=4\), \(\dfrac{dy}{dt}=12-3=9\). Slope \(=\dfrac{9}{4}\)... Wait: \(3(4)-3=9\) and \(2(2)=4\), so slope \(=\frac{9}{4}\). Answer: B.

Question 17 — Polar Area

Answer: B — π

The curve \(r=2\cos\theta\) is a circle of radius 1 centered at \((1,0)\). Area of a circle \(=\pi r^2=\pi(1)^2=\pi\). Confirm via formula: \(A=\dfrac{1}{2}\displaystyle\int_{-\pi/2}^{\pi/2}(2\cos\theta)^2d\theta=2\displaystyle\int_0^{\pi/2}(1+\cos2\theta)d\theta=2\cdot\dfrac{\pi}{2}=\pi\).

Question 18 — Interval of Convergence

Answer: A — −1 < x ≤ 5

Ratio Test: \(L=\lim\left|\dfrac{(x-2)^{n+1}}{(n+1)3^{n+1}}\cdot\dfrac{n\cdot3^n}{(x-2)^n}\right|=\dfrac{|x-2|}{3}\). Converges when \(\dfrac{|x-2|}{3}<1\Rightarrow|x-2|<3\Rightarrow -1

Question 19 — Taylor Series Coefficient

Answer: B — −1/2

\(\cos u = 1 - \dfrac{u^2}{2!}+\dfrac{u^4}{4!}-\cdots\). Substitute \(u=x^2\): \(\cos(x^2)=1-\dfrac{x^4}{2}+\dfrac{x^8}{24}-\cdots\). The coefficient of \(x^4\) is \(-\dfrac{1}{2}\).

Question 20 — Integration via Series

Answer: B — x − x³/18 + x⁵/600

\(\sin t = t - \dfrac{t^3}{6}+\dfrac{t^5}{120}-\cdots\), so \(\dfrac{\sin t}{t}=1-\dfrac{t^2}{6}+\dfrac{t^4}{120}-\cdots\). Integrate term by term from 0 to \(x\): \(\displaystyle\int_0^x\left(1-\dfrac{t^2}{6}+\dfrac{t^4}{120}\right)dt=x-\dfrac{x^3}{18}+\dfrac{x^5}{600}\).

Answer Key

Q1: D

Q2: B

Q3: C

Q4: B

Q5: A

Q6: C

Q7: B

Q8: A

Q9: A

Q10: B

Q11: B

Q12: B

Q13: B

Q14: A

Q15: B

Q16: B

Q17: B

Q18: A

Q19: B

Q20: B

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AP CalculusAB / BC

AP Calculus AB/BC — Master Problem Set

Unit 1 — Limits & Continuity

Key Concepts

Must-Memorize

Unit 2 — Differentiation

Key Rules

Derivatives to Memorize

Unit 3 — Applications of Derivatives

Key Theorems

Unit 4 — Integration & the FTC

Fundamental Theorem of Calculus

Integration Formulas

Unit 5 — Differential Equations

Key Concepts

Unit 6 — Parametric & Polar (BC)

Key Formulas

Unit 7 — Infinite Series (BC)

Convergence Tests

Taylor Series (Memorize)

Complete Solutions & Explanations

Answer Key

Results

AP Calculus
AB / BC