Unit 1 · Limits & Continuity
01
Easy
Limits · L'Hôpital's Rule
0/0 or ∞/∞ → differentiate top & bottom separately
Quick Example
\(\displaystyle\lim_{x\to 0}\frac{\sin x}{x} = \frac{\cos 0}{1} = 1\)
Evaluate: \(\displaystyle\lim_{x \to 0} \frac{e^x - 1}{x}\)
This is an indeterminate form. Apply L'Hôpital's Rule.
해설 (Korean Explanation)
\(\frac{e^x-1}{x}\)은 \(x\to 0\)일 때 \(\frac{0}{0}\) 꼴입니다.로피탈 법칙: 분자 미분 \(\to e^x\), 분모 미분 \(\to 1\)
따라서 \(\displaystyle\lim_{x\to 0}\frac{e^x}{1} = e^0 = \mathbf{1}\)
02
Medium
Limits · Squeeze Theorem Trap
SQUEEZE: if g ≤ f ≤ h and g,h → L, then f → L
What is \(\displaystyle\lim_{x \to \infty} \frac{3x^2 - 5x + 1}{2x^2 + 7}\)?
Divide every term by the highest power of \(x\) in the denominator.
해설
분자·분모를 \(x^2\)으로 나누면:\(\dfrac{3 - \frac{5}{x} + \frac{1}{x^2}}{2 + \frac{7}{x^2}} \xrightarrow{x\to\infty} \dfrac{3-0+0}{2+0} = \mathbf{\dfrac{3}{2}}\)
⚠️ 오답 함정: 분자 최고차 계수(3)만 보고 3이라고 쓰면 틀립니다. 분모 계수(2)로 나눠야 합니다.
Unit 2 · Differentiation
03
Easy
Chain Rule
CHAIN: d/dx f(g(x)) = f'(g(x)) · g'(x) → "outside × inside prime"
Quick Example
\(\frac{d}{dx}\sin(x^2) = \cos(x^2)\cdot 2x\)
Find \(\dfrac{d}{dx}\left[e^{3x^2+1}\right]\)
해설
체인룰: 바깥 함수 \(e^u\)를 미분하면 그대로 \(e^u\), 안쪽 \(u=3x^2+1\)을 미분하면 \(6x\).\(\therefore\; 6x\cdot e^{3x^2+1}\)
⚠️ 함정: 안쪽 함수를 미분하지 않으면 (A)를 고르게 됩니다.
04
Medium
Implicit Differentiation
IMPLICIT: d/dx of y-terms → attach dy/dx every time you differentiate y
Given \(x^2 + y^2 = 25\), find \(\dfrac{dy}{dx}\).
해설
양변을 \(x\)에 대해 미분:\(2x + 2y\dfrac{dy}{dx} = 0\)
\(2y\dfrac{dy}{dx} = -2x\)
\(\dfrac{dy}{dx} = -\dfrac{x}{y}\)
⚠️ 부호 실수 주의: \(+2x\)를 이항하면 \(-2x\)입니다.
05
Hard
Related Rates
RELATED RATES: write equation → differentiate both sides w.r.t. t → plug in
A ladder 10 ft long leans against a wall. The bottom slides away at 2 ft/s. How fast is the top sliding down when the bottom is 6 ft from the wall?
Let \(x\) = distance of bottom from wall, \(y\) = height of top. Use Pythagorean theorem.
해설
\(x^2 + y^2 = 100\), 시간 미분: \(2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0\)\(x=6\)이면 \(y=\sqrt{100-36}=8\)
\(\dfrac{dx}{dt}=2\) ft/s 대입:
\(2(6)(2)+2(8)\dfrac{dy}{dt}=0\)
\(24+16\dfrac{dy}{dt}=0 \Rightarrow \dfrac{dy}{dt}=-\dfrac{24}{16}=-\dfrac{3}{2}\) ft/s
속도의 크기는 \(\mathbf{\dfrac{3}{2}}\) ft/s (음수는 아래로 내려감을 의미)
Unit 3 · Integration
06
Easy
U-Substitution
U-SUB: let u = inside → find du → rewrite → integrate → back-substitute
Quick Example
\(\int 2x\cos(x^2)\,dx\): let \(u=x^2\), \(du=2x\,dx\) → \(\int\cos u\,du = \sin u+C = \sin(x^2)+C\)
Evaluate \(\displaystyle\int 3x^2\cos(x^3)\,dx\)
해설
\(u=x^3 \Rightarrow du=3x^2\,dx\)\(\int\cos u\,du = \sin u + C = \mathbf{\sin(x^3)+C}\)
⚠️ (D) 함정: \(3x^2\)를 앞에 그대로 남겨두면 안 됩니다.
07
Medium
Integration by Parts
IBP: ∫u dv = uv − ∫v du · pick u with "LIATE" (Log, Inverse trig, Algebraic, Trig, Exp)
Evaluate \(\displaystyle\int x e^x\,dx\)
해설
LIATE 규칙: \(u=x\), \(dv=e^x dx\)\(du=dx\), \(v=e^x\)
\(\int xe^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C = \mathbf{e^x(x-1)+C}\)
⚠️ (A) 함정: 뒤의 적분 \(-\int e^x dx\) 부분을 빼먹으면 (A)가 됩니다.
08
Medium
FTC · Part 2
FTC2: d/dx ∫ₐˣ f(t)dt = f(x) · (derivative of upper limit)
If \(\displaystyle F(x) = \int_0^{x^2} \sin(t)\,dt\), find \(F'(x)\).
해설
FTC Part 2 + Chain Rule:상한이 \(x^2\)이므로 체인룰 적용:
\(F'(x) = \sin(x^2) \cdot \dfrac{d}{dx}(x^2) = \sin(x^2)\cdot 2x = \mathbf{2x\sin(x^2)}\)
⚠️ (A) 함정: 상한의 미분값 \(2x\)를 곱하지 않으면 틀립니다.
09
Hard
Partial Fractions
PARTIAL FRACTIONS: factor denominator → A/(x−a) + B/(x−b) → solve system
Evaluate \(\displaystyle\int \frac{1}{x^2-1}\,dx\)
해설
\(\dfrac{1}{x^2-1}=\dfrac{1}{(x-1)(x+1)}=\dfrac{A}{x-1}+\dfrac{B}{x+1}\)풀면 \(A=\frac{1}{2}\), \(B=-\frac{1}{2}\)
\(\int\left(\dfrac{1/2}{x-1}-\dfrac{1/2}{x+1}\right)dx = \dfrac{1}{2}\ln|x-1|-\dfrac{1}{2}\ln|x+1|+C\)
\(=\mathbf{\dfrac{1}{2}\ln\left|\dfrac{x-1}{x+1}\right|+C}\)
⚠️ (A): \(\int\frac{1}{x^2-1}dx \neq \frac{\ln|x^2-1|}{2x}\). 부분분수 분해를 꼭 써야 합니다.
Unit 4 · Infinite Series (BC Only)
10
Easy
Geometric Series
GEO SERIES: sum = a/(1−r), converges only if |r| < 1
Quick Example
\(\sum_{n=0}^{\infty}\left(\frac{1}{3}\right)^n = \frac{1}{1-\frac{1}{3}} = \frac{3}{2}\)
Find the sum: \(\displaystyle\sum_{n=0}^{\infty} \left(\frac{2}{3}\right)^n\)
해설
\(a=1\), \(r=\dfrac{2}{3}\), \(|r|<1\)이므로 수렴:\(S=\dfrac{1}{1-\frac{2}{3}}=\dfrac{1}{\frac{1}{3}}=\mathbf{3}\)
⚠️ (A) 함정: \(r=\frac{2}{3}\)를 \(1-r\)에 대입하면 \(\frac{1}{3}\)이므로 역수가 3입니다.
11
Medium
Ratio Test
RATIO TEST: L = lim|aₙ₊₁/aₙ| → L<1 converges, L>1 diverges, L=1 inconclusive
Use the Ratio Test to determine if \(\displaystyle\sum_{n=1}^{\infty} \frac{n!}{2^n}\) converges or diverges.
해설
\(\left|\dfrac{a_{n+1}}{a_n}\right|=\left|\dfrac{(n+1)!}{2^{n+1}}\cdot\dfrac{2^n}{n!}\right|=\dfrac{n+1}{2}\)\(\displaystyle\lim_{n\to\infty}\dfrac{n+1}{2}=\infty > 1\)
\(\therefore\) 비율 판정법에 의해 발산
팩토리얼은 지수보다 훨씬 빠르게 커집니다.
12
Hard
Taylor Series · Most Missed
TAYLOR: f(x) = Σ f⁽ⁿ⁾(a)/n! · (x−a)ⁿ · Memorize: eˣ, sinx, cosx series!
Must-Know Series
\(e^x=\sum_{n=0}^{\infty}\dfrac{x^n}{n!},\quad \sin x=\sum_{n=0}^{\infty}\dfrac{(-1)^n x^{2n+1}}{(2n+1)!}\)
The Maclaurin series for \(\sin x\) begins \(x - \dfrac{x^3}{6} + \cdots\) What is the Maclaurin series for \(\sin(x^2)\)?
해설
\(\sin u = u - \dfrac{u^3}{6}+\cdots\)에서 \(u = x^2\) 대입:\(\sin(x^2) = x^2 - \dfrac{(x^2)^3}{6}+\cdots = \mathbf{x^2 - \dfrac{x^6}{6}+\cdots}\)
⚠️ (A) 함정: \(\dfrac{x^6}{3}\)으로 계수를 바꾸는 오류. 분모는 \(6\)이 그대로입니다.
Unit 5 · Parametric, Polar & Vector
13
Medium
Parametric · dy/dx
PARAMETRIC SLOPE: dy/dx = (dy/dt) ÷ (dx/dt) — never forget to divide!
If \(x(t)=t^2\) and \(y(t)=t^3\), find \(\dfrac{dy}{dx}\) in terms of \(t\).
해설
\(\dfrac{dx}{dt}=2t,\quad \dfrac{dy}{dt}=3t^2\)\(\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{3t^2}{2t}=\mathbf{\dfrac{3t}{2}}\)
⚠️ (A) 함정: dy/dt만 쓰면 안 됩니다. 반드시 dx/dt로 나눠야 합니다.
14
Hard
Polar Area
POLAR AREA: A = ½∫r² dθ — the ½ is mandatory, don't drop it!
Find the area enclosed by the polar curve \(r = 2\cos\theta\) for \(0\le\theta\le\pi\).
해설
\(A=\dfrac{1}{2}\int_0^{\pi}(2\cos\theta)^2\,d\theta=\dfrac{1}{2}\int_0^{\pi}4\cos^2\theta\,d\theta\)\(=2\int_0^{\pi}\dfrac{1+\cos 2\theta}{2}\,d\theta=\int_0^{\pi}(1+\cos 2\theta)\,d\theta\)
\(=\left[\theta+\dfrac{\sin 2\theta}{2}\right]_0^{\pi}=\pi+0=\mathbf{\pi}\)
⚠️ \(r=2\cos\theta\)는 반지름 1인 원입니다. 넓이 \(\pi r^2=\pi\)로 검산!
Unit 6 · Applications of Calculus
15
Medium
Area Between Curves
AREA: ∫(top − bottom)dx → always subtract the lower curve from the upper!
Find the area between \(y = x^2\) and \(y = x\) on \([0,1]\).
해설
\([0,1]\)에서 \(x \ge x^2\)이므로 윗 곡선은 \(y=x\):\(A=\int_0^1(x-x^2)\,dx=\left[\dfrac{x^2}{2}-\dfrac{x^3}{3}\right]_0^1=\dfrac{1}{2}-\dfrac{1}{3}=\mathbf{\dfrac{1}{6}}\)
⚠️ (A) 함정: \(\int_0^1 x^2\,dx=\frac{1}{3}\)만 계산하면 틀립니다.
16
Hard
Volume · Disk Method
DISK: V = π∫[f(x)]²dx · WASHER: V = π∫([R]²−[r]²)dx
The region bounded by \(y=\sqrt{x}\), \(x=4\), and the \(x\)-axis is revolved about the \(x\)-axis. Find the volume.
해설
디스크 방법: \(V=\pi\int_0^4(\sqrt{x})^2\,dx=\pi\int_0^4 x\,dx\)\(=\pi\left[\dfrac{x^2}{2}\right]_0^4=\pi\cdot\dfrac{16}{2}=\mathbf{8\pi}\)
⚠️ (B) 함정: \(\pi\cdot 4^2=16\pi\)는 반지름을 \(\sqrt{x}\)가 아니라 4로 쓴 실수입니다.
Unit 7 · Differential Equations
17
Medium
Separable Differential Equations
SEPARABLE: get all y on one side, all x on other → integrate both sides → solve for y
Solve the differential equation \(\dfrac{dy}{dx} = 2xy\) with \(y(0)=3\).
해설
\(\dfrac{dy}{y}=2x\,dx\)양변 적분: \(\ln|y|=x^2+C\)
\(y=Ae^{x^2}\), 초기조건 \(y(0)=3\): \(A=3\)
\(\therefore y=\mathbf{3e^{x^2}}\)
⚠️ (C) 함정: \(2x\)를 적분하면 \(x^2\)인데 \(2x\)로 그냥 두면 틀립니다.
18
Hard
Logistic Growth · BC Exclusive
LOGISTIC: dP/dt = kP(1−P/M) → max growth rate at P = M/2
A population follows logistic growth with carrying capacity \(M=1000\) and growth constant \(k=0.3\). At what population size is the growth rate \(\dfrac{dP}{dt}\) maximized?
해설
로지스틱 성장: \(\dfrac{dP}{dt}=0.3P\!\left(1-\dfrac{P}{1000}\right)\)이를 P로 미분하여 0으로 놓거나, 대칭성 이용:
성장률은 \(P=\dfrac{M}{2}=\dfrac{1000}{2}=\mathbf{500}\)에서 최대
⚠️ (A) 함정: \(P=M=1000\)이면 \(dP/dt=0\)입니다 (성장 멈춤).
Unit 8 · Convergence Tests & Intervals
19
Hard
Interval of Convergence
IOC: use Ratio Test → find |x−a|<R → test endpoints separately!
Find the interval of convergence of \(\displaystyle\sum_{n=1}^{\infty} \frac{(x-2)^n}{n}\).
Remember: always check the endpoints by plugging them in separately.
해설
비율 판정법: \(\lim\left|\dfrac{(x-2)^{n+1}}{n+1}\cdot\dfrac{n}{(x-2)^n}\right|=|x-2|<1\)\(\Rightarrow 1<x<3\)
끝점 검사:
• \(x=1\): \(\sum\dfrac{(-1)^n}{n}\) — 교대급수, 수렴 ✔
• \(x=3\): \(\sum\dfrac{1}{n}\) — 조화급수, 발산 ✗
\(\therefore\) 수렴 구간 \(\mathbf{[1,\,3)}\)
20
Hard
Arc Length · Most Missed
ARC LENGTH: L = ∫√(1+[f'(x)]²)dx — don't forget the 1 inside the root!
Set up (do not evaluate) the arc length of \(y=x^2\) from \(x=0\) to \(x=2\).
해설
\(f(x)=x^2 \Rightarrow f'(x)=2x\)호의 길이 공식: \(L=\int_0^2\sqrt{1+(2x)^2}\,dx=\mathbf{\int_0^2\sqrt{1+4x^2}\,dx}\)
⚠️ (A) 함정: \([f'(x)]^2=(2x)^2=4x^2\)인데 \(x^2\)으로 쓰면 도함수를 제곱하지 않은 오류.
⚠️ (C) 함정: 루트 안의 \(1\)을 빠뜨리면 틀립니다.
Final Score
—Well done!
Keep practicing to master AP Calculus BC.