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AP Calculus
AB / BC

Master Problem Set — All Units

20 Problems 45 Minutes Multiple Choice

Core Concepts & Key Formulas

Review before the timed quiz. Tap any formula to memorize it.

Unit 1 · AB & BC
Limits & Continuity
Limit Laws: If both limits exist, sums/products/quotients of limits obey arithmetic.
$\lim_{x\to a}[f(x)\pm g(x)] = L \pm M$
L'Hôpital's Rule: For $0/0$ or $\infty/\infty$ indeterminate forms only.
$\lim_{x\to a}\dfrac{f(x)}{g(x)}=\lim_{x\to a}\dfrac{f'(x)}{g'(x)}$
Squeeze Theorem: $g(x)\le f(x)\le h(x)$ near $a$ and $g,h\to L$ $\Rightarrow$ $f\to L$.
Continuity requires: $f(a)$ exists, $\lim_{x\to a}f(x)$ exists, and both are equal.
Quick Example
Evaluate $\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x}$.
Answer: 1 (famous standard limit — memorize this!)
Unit 2 · AB & BC
Differentiation: Definition & Rules
Definition: $f'(x)=\displaystyle\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}$
Power Rule:
$\dfrac{d}{dx}[x^n]=nx^{n-1}$
Product Rule: $(uv)'=u'v+uv'$
Quotient Rule: $\left(\dfrac{u}{v}\right)'=\dfrac{u'v - uv'}{v^2}$
Chain Rule: $[f(g(x))]'=f'(g(x))\cdot g'(x)$
Key trig derivatives: $(\sin x)'=\cos x$, $(\cos x)'=-\sin x$, $(\tan x)'=\sec^2 x$
Quick Example
Find $\dfrac{d}{dx}[\sin(x^2)]$.
Chain rule: $\cos(x^2)\cdot 2x = 2x\cos(x^2)$
Unit 3 · AB & BC
Applications of Derivatives
Mean Value Theorem: If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then $\exists c$ such that $f'(c)=\dfrac{f(b)-f(a)}{b-a}$.
Critical Points: $f'(c)=0$ or $f'(c)$ undefined. Use 1st or 2nd derivative test.
Concavity: $f''>0$ → concave up; $f''<0$ → concave down. Inflection point where $f''$ changes sign.
Related Rates: Differentiate an equation with respect to $t$ implicitly.
Quick Example
$f(x)=x^3-3x$. Find local extrema.
$f'(x)=3x^2-3=3(x-1)(x+1)$. Critical pts: $x=\pm1$. Local max at $x=-1$, local min at $x=1$.
Unit 4 · AB & BC
Integration: Definition & Techniques
FTC Part 1: $\dfrac{d}{dx}\displaystyle\int_a^x f(t)\,dt = f(x)$
FTC Part 2: $\displaystyle\int_a^b f(x)\,dx = F(b)-F(a)$ where $F'=f$
$u$-substitution: Replace inner function; $du$ absorbs the chain-rule factor.
Integration by Parts (BC):
$\int u\,dv = uv - \int v\,du$
Key antiderivatives: $\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C$ ($n\ne-1$); $\int\dfrac{1}{x}\,dx=\ln|x|+C$
Quick Example
Find $\displaystyle\int_0^1 2x\,e^{x^2}\,dx$.
Let $u=x^2$, $du=2x\,dx$. Integral $=\left[e^u\right]_0^1 = e^1-e^0 = e-1$.
Unit 5 · AB & BC
Applications of Integration
Area between curves: $\displaystyle\int_a^b [f(x)-g(x)]\,dx$ when $f\ge g$.
Disk/Washer Method (about $x$-axis):
$V=\pi\displaystyle\int_a^b\left[R(x)^2-r(x)^2\right]dx$
Average value: $f_{\text{avg}}=\dfrac{1}{b-a}\displaystyle\int_a^b f(x)\,dx$
Accumulation: Net change $=\displaystyle\int_a^b f'(x)\,dx = f(b)-f(a)$
Quick Example
Average value of $f(x)=x^2$ on $[0,3]$.
$\dfrac{1}{3}\displaystyle\int_0^3 x^2\,dx = \dfrac{1}{3}\cdot\dfrac{27}{3}=3$
Unit 6 · BC Only
Differential Equations & Parametric/Polar
Separable DE: Separate variables, integrate both sides.
$\dfrac{dy}{dx}=ky \;\Rightarrow\; y=Ce^{kx}$
Euler's Method: $y_{n+1}=y_n + h\cdot f'(x_n,y_n)$
Parametric arc length (BC):
$L=\displaystyle\int_a^b\sqrt{\left(\tfrac{dx}{dt}\right)^2+\left(\tfrac{dy}{dt}\right)^2}\,dt$
Polar area (BC):
$A=\dfrac{1}{2}\displaystyle\int_\alpha^\beta r^2\,d\theta$
Quick Example
Solve $\dfrac{dy}{dx}=2y$, $y(0)=3$.
$y=3e^{2x}$
Unit 7 · BC Only
Infinite Series
Geometric series: $\displaystyle\sum_{n=0}^\infty ar^n = \dfrac{a}{1-r}$, $|r|<1$
Ratio Test: Converges if $\displaystyle\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|<1$
Taylor Series:
$f(x)=\displaystyle\sum_{n=0}^\infty\dfrac{f^{(n)}(a)}{n!}(x-a)^n$
Key Maclaurin series (must memorize):
$e^x=\displaystyle\sum_{n=0}^\infty\dfrac{x^n}{n!}$,  $\sin x=\displaystyle\sum_{n=0}^\infty\dfrac{(-1)^n x^{2n+1}}{(2n+1)!}$,  $\cos x=\displaystyle\sum_{n=0}^\infty\dfrac{(-1)^n x^{2n}}{(2n)!}$
Alternating Series Error: $|$error$|\le |a_{n+1}|$ (next unused term)
Quick Example
Sum of $\displaystyle\sum_{n=0}^\infty \left(\dfrac{1}{3}\right)^n$.
Geometric: $a=1$, $r=\tfrac{1}{3}$. Sum $=\dfrac{1}{1-1/3}=\dfrac{3}{2}$.
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