2025 HAFS 미적분학 I — 1학기 중간고사 완전 분석Updated 2026-04-21
시행일: 2025.04.23 | 2학년 1A, 1B반 (국제과정) | 22문항 선택형 100점
먼저 풀이 → 정답 보기 펼치기
시험 형식 요약
| 항목 | 내용 |
|---|
| 과목명 | 미적분학 I (Calculus I) |
| 문항 수 | 선택형 22문항 (논술형 0) |
| 총점 | 100점 |
| 배점 | 4.2 ~ 5.1 점진 증가 |
| 출제 언어 | 전체 영어 |
유형별 분포
| 유형 | 문항 수 | 비율 |
|---|
| Limits | 7 | 32% |
| Derivatives | 5 | 23% |
| Tangent line | 4 | 18% |
| Inverse trig | 4 | 18% |
| ε-δ | 1 | 4.5% |
| Orthogonal trajectories | 1 | 4.5% |
22문항 전체 분석
Find \(\displaystyle\lim_{x \to 0} \frac{(2+x)^3 - 2^3}{x}\).
풀이
\(f(x)=x^3\)의 \(x=2\)에서의 도함수 정의.
\((2+x)^3 - 8 = 12x + 6x^2 + x^3\). 나누기 \(x\) 후 \(x\to 0\) → \(12\).
정답: ④ 12
Find \(\displaystyle\lim_{x \to 0} \frac{\dfrac{1}{(2+x)^2} - \dfrac{1}{2^2}}{x}\).
- ① \(-1\)
- ② \(-\dfrac{1}{2}\)
- ③ \(-\dfrac{1}{4}\)
- ④ \(-\dfrac{1}{8}\)
- ⑤ \(-\dfrac{1}{16}\)
풀이
\(f(x)=(2+x)^{-2}\)의 도함수. \(f'(0) = -2 \cdot 2^{-3} = -\dfrac{1}{4}\).
정답: ③ \(-\dfrac{1}{4}\)
Find \(\displaystyle\lim_{x \to 0} \frac{\dfrac{1}{\sqrt{4+x}} - \dfrac{1}{\sqrt{4}}}{x}\).
- ① \(-\dfrac{1}{16}\)
- ② \(-\dfrac{1}{8}\)
- ③ \(-\dfrac{1}{4}\)
- ④ \(-\dfrac{1}{2}\)
- ⑤ \(-1\)
풀이
\(f(x)=(4+x)^{-1/2}\)의 도함수. \(f'(0) = -\dfrac{1}{2} \cdot 4^{-3/2} = -\dfrac{1}{16}\).
정답: ① \(-\dfrac{1}{16}\)
Find \(\displaystyle\lim_{x \to \infty}\left(\sqrt{4x^2+9x}-2x\right)\).
- ① 0
- ② 3
- ③ \(\dfrac{3}{2}\)
- ④ \(\dfrac{9}{2}\)
- ⑤ \(\dfrac{9}{4}\)
풀이
유리화: \(\dfrac{9x}{\sqrt{4x^2+9x}+2x} \to \dfrac{9}{2+2} = \dfrac{9}{4}\).
정답: ⑤ \(\dfrac{9}{4}\)
Find the maximum number of \(\delta\) such that if \(|x-2|<\delta\) then \(|x^2-4|<1\).
- ① \(2-\sqrt{3}\)
- ② \(\sqrt{5}-2\)
- ③ \(\dfrac{\sqrt{5}-\sqrt{3}}{2}\)
- ④ 1
- ⑤ 2
풀이
\(|x^2-4|<1 \Rightarrow 3<x^2<5 \Rightarrow \sqrt{3}<x<\sqrt{5}\).
\(\delta = \min(2-\sqrt{3},\;\sqrt{5}-2) = \sqrt{5}-2\).
정답: ② \(\sqrt{5}-2\)
Find \(\displaystyle\lim_{x \to 0^+} e^{-1/x}\).
- ① \(\dfrac{1}{e}\)
- ② \(e\)
- ③ 1
- ④ 0
- ⑤ \(\infty\)
풀이
\(x \to 0^+ \Rightarrow -1/x \to -\infty \Rightarrow e^{-\infty} = 0\).
정답: ④ 0
Find \(f'(x)\) when \(f(x)=\dfrac{2}{\sqrt{x}}\).
- ① \(-\dfrac{1}{x\sqrt{x}}\)
- ② \(-\dfrac{1}{x}\)
- ③ \(-\dfrac{1}{\sqrt{x}}\)
- ④ \(-\sqrt{x}\)
- ⑤ \(-x\sqrt{x}\)
풀이
\(f(x)=2x^{-1/2}\), \(f'(x)=-x^{-3/2}=-\dfrac{1}{x\sqrt{x}}\).
정답: ① \(-\dfrac{1}{x\sqrt{x}}\)
Find \(f'(\ln 2)\) when \(f(x)=2e^{2x}\).
풀이
\(f'(x) = 4e^{2x}\). \(f'(\ln 2) = 4e^{2\ln 2} = 4 \cdot 4 = 16\).
정답: ③ 16
Find \(f'(\ln 2)\) when \(f(x)=xe^{2x}\).
- ① \(2+2\ln 2\)
- ② \(2+4\ln 2\)
- ③ \(4+2\ln 2\)
- ④ \(4+4\ln 2\)
- ⑤ \(4+8\ln 2\)
풀이
Product rule: \(f'(x) = e^{2x}(1+2x)\). \(f'(\ln 2) = 4(1+2\ln 2) = 4+8\ln 2\).
정답: ⑤ \(4+8\ln 2\)
Find the slope of tangent line to the curve \(x^2+2xy+2y^2=5\) at the point \((1,1)\).
- ① \(-\dfrac{3}{2}\)
- ② \(-\dfrac{2}{3}\)
- ③ \(-\dfrac{5}{3}\)
- ④ \(-\dfrac{3}{5}\)
- ⑤ \(-\dfrac{4}{5}\)
풀이
암묵적 미분: \(2x + 2y + 2xy' + 4yy' = 0\).
\((1,1)\) 대입: \(4 + 6y' = 0 \Rightarrow y' = -\dfrac{2}{3}\).
정답: ② \(-\dfrac{2}{3}\)
Choose every true statement:
Ⓐ \(\displaystyle\lim_{x\to 0} x = 0\)
Ⓑ \(\displaystyle\lim_{x\to 0} \frac{1}{x} = \infty\)
Ⓒ \(\displaystyle\lim_{x\to 0} \frac{1}{x^2} = \infty\)
- ① Ⓐ only
- ② Ⓐ and Ⓑ only
- ③ Ⓐ and Ⓒ only
- ④ Ⓑ and Ⓒ only
- ⑤ Ⓐ, Ⓑ and Ⓒ
풀이
Ⓐ TRUE. Ⓑ FALSE (좌극한 \(=-\infty\), 우극한 \(=+\infty\) — 극한 존재하지 않음). Ⓒ TRUE (양쪽 모두 \(+\infty\)).
정답: ③ Ⓐ and Ⓒ only
Choose every true statement:
Ⓐ \(\displaystyle\lim_{x\to\infty}\left(\sqrt{x^2+2}-x\right)=0\)
Ⓑ \(\displaystyle\lim_{x\to\infty}\left(\sqrt{4x^2+6x}-2x\right)=3\)
Ⓒ \(\displaystyle\lim_{x\to\infty}\left(\sqrt{4x^2+6x}-3x\right)=2\)
- ① Ⓐ only
- ② Ⓐ and Ⓑ only
- ③ Ⓐ and Ⓒ only
- ④ Ⓑ and Ⓒ only
- ⑤ Ⓐ, Ⓑ and Ⓒ
풀이
Ⓐ TRUE (\(\to 0\)). Ⓑ FALSE (\(\to \frac{3}{2} \neq 3\)). Ⓒ FALSE (\(\to -\infty \neq 2\)).
정답: ① Ⓐ only
Find the \(y\)-intercept of the tangent line to the curve \(y=-x^2\) at \(x=2\).
풀이
\(f(2) = -4\), \(f'(2) = -4\). 접선: \(y = -4(x-2) - 4 = -4x + 4\). \(y\)-절편 \(= 4\).
정답: ② 4
Find the \(y\)-intercept of the tangent line to the curve \(xy=8\) at \(x=2\).
풀이
\(y(2) = 4\), \(y' = -y/x = -2\). 접선: \(y = -2(x-2)+4 = -2x+8\). \(y\)-절편 \(= 8\).
정답: ③ 8
Find the slope of the tangent line to the curve \(x^{2/3}+y^{2/3}=13\) at the point \((8,27)\).
- ① \(-\dfrac{3}{2}\)
- ② \(-\dfrac{2}{3}\)
- ③ \(-\dfrac{4}{3}\)
- ④ \(-\dfrac{4}{9}\)
- ⑤ \(-\dfrac{9}{4}\)
풀이
검증: \(8^{2/3}+27^{2/3} = 4+9 = 13\) ✓
\(y' = -\left(\dfrac{y}{x}\right)^{1/3} = -\left(\dfrac{27}{8}\right)^{1/3} = -\dfrac{3}{2}\).
정답: ① \(-\dfrac{3}{2}\)
Find \(\dfrac{d}{dx}(\arcsin x)\) at \(x=\dfrac{4}{5}\).
- ① \(\dfrac{4}{3}\)
- ② \(\dfrac{5}{4}\)
- ③ \(\dfrac{3}{4}\)
- ④ \(\dfrac{5}{3}\)
- ⑤ \(\dfrac{3}{5}\)
풀이
\(\dfrac{d}{dx}(\arcsin x) = \dfrac{1}{\sqrt{1-x^2}}\). \(x = \frac{4}{5}\) 대입: \(\dfrac{1}{\sqrt{1-16/25}} = \dfrac{1}{\sqrt{9/25}} = \dfrac{5}{3}\).
정답: ④ \(\dfrac{5}{3}\)
Find \(\dfrac{d}{dx}\left\{e^{x^2}\right\}\) at \(x=2\).
- ① \(e^2\)
- ② \(2e^2\)
- ③ \(2e^4\)
- ④ \(4e^2\)
- ⑤ \(4e^4\)
풀이
Chain rule: \(\dfrac{d}{dx}\left[e^{x^2}\right] = 2x \cdot e^{x^2}\). \(x=2\) 대입: \(4e^4\).
정답: ⑤ \(4e^4\)
Find \(\dfrac{d}{dx}\left\{e^{e^x}\right\}\) at \(x=\ln 2\).
- ① \(e^2\)
- ② \(2e^2\)
- ③ \(2e^4\)
- ④ \(4e^2\)
- ⑤ \(4e^4\)
풀이
Chain rule: \(\dfrac{d}{dx}\left[e^{e^x}\right] = e^{e^x} \cdot e^x\).
\(x = \ln 2\): \(e^{e^{\ln 2}} \cdot e^{\ln 2} = e^2 \cdot 2 = 2e^2\).
정답: ② \(2e^2\)
Find \(\dfrac{d}{dx}\left(e^x \cos x\right)\) at \(x=\arctan\!\left(\dfrac{3}{4}\right)\).
- ① \(\dfrac{7}{25}e^{\arctan(3/4)}\)
- ② \(-\dfrac{7}{5}e^{\arctan(3/4)}\)
- ③ \(\dfrac{7}{5}e^{\arctan(3/4)}\)
- ④ \(-\dfrac{1}{5}e^{\arctan(3/4)}\)
- ⑤ \(\dfrac{1}{5}e^{\arctan(3/4)}\)
풀이
Product rule: \(\dfrac{d}{dx}(e^x \cos x) = e^x(\cos x - \sin x)\).
\(\tan\theta = \frac{3}{4}\) → \(\sin\theta = \frac{3}{5}\), \(\cos\theta = \frac{4}{5}\) (3-4-5 삼각형).
\(\cos\theta - \sin\theta = \frac{4}{5} - \frac{3}{5} = \frac{1}{5}\). 결과: \(\dfrac{1}{5}e^{\arctan(3/4)}\).
정답: ⑤ \(\dfrac{1}{5}e^{\arctan(3/4)}\)
Find \(\dfrac{d}{dx}\left(e^{\sin x}\right)\) at \(x=\arctan\!\left(\dfrac{3}{4}\right)\).
- ① \(\dfrac{3}{5}e^{4/5}\)
- ② \(-\dfrac{3}{5}e^{4/5}\)
- ③ \(\dfrac{4}{5}e^{3/5}\)
- ④ \(-\dfrac{4}{5}e^{3/5}\)
- ⑤ \(\dfrac{3}{4}e^{3/4}\)
풀이
Chain rule: \(\dfrac{d}{dx}\left[e^{\sin x}\right] = e^{\sin x} \cdot \cos x\).
\(\sin\theta = \frac{3}{5}\), \(\cos\theta = \frac{4}{5}\) → \(\dfrac{4}{5} \cdot e^{3/5}\).
정답: ③ \(\dfrac{4}{5}e^{3/5}\)
The box below is the process to find \(\dfrac{d}{dx}(\arctan 2x)\). Choose the correct expression for Ⓐ.
Let \(y = \arctan 2x\)
By definition: \(x = \tfrac{1}{2}\tan y\)
\(\dfrac{dx}{dy} = \tfrac{1}{2}\sec^2 y\)
By trig identities: \(\dfrac{dx}{dy} =\) Ⓐ
By inverse relations: \(\dfrac{dy}{dx} = \cdots\)
- ① \(\dfrac{1+x^2}{2}\)
- ② \(\dfrac{1+2x^2}{2}\)
- ③ \(\dfrac{2+x^2}{2}\)
- ④ \(\dfrac{1+4x^2}{2}\)
- ⑤ \(\dfrac{4+x^2}{2}\)
풀이
\(\sec^2 y = 1 + \tan^2 y = 1 + (2x)^2 = 1 + 4x^2\).
\(\dfrac{dx}{dy} = \dfrac{1+4x^2}{2}\).
정답: ④ \(\dfrac{1+4x^2}{2}\)
Choose the curves that are orthogonal to the curves \(\dfrac{x^2}{4}+y^2=C\). (\(C\) and \(K\) are constants.)
- ① \(y = Kx^4\)
- ② \(y = Kx^2\)
- ③ \(y = Kx^{1/2}\)
- ④ \(y = Kx^{1/4}\)
- ⑤ \(y = Kx^{-1/4}\)
풀이
미분: \(\dfrac{x}{2} + 2yy' = 0 \Rightarrow y' = -\dfrac{x}{4y}\).
직교 궤적: \(\dfrac{dy}{dx} = \dfrac{4y}{x}\). 분리: \(\dfrac{dy}{y} = \dfrac{4\,dx}{x}\).
적분: \(\ln|y| = 4\ln|x| + C' \Rightarrow y = Kx^4\).
정답: ① \(y = Kx^4\)
정답표
| 번호 | 배점 | 유형 | 정답 |
|---|
| 1 | 4.2 | Limits | ④ 12 |
| 2 | 4.2 | Limits | ③ −1/4 |
| 3 | 4.2 | Limits | ① −1/16 |
| 4 | 4.3 | Limits | ⑤ 9/4 |
| 5 | 4.3 | ε-δ | ② √5−2 |
| 6 | 4.3 | Limits | ④ 0 |
| 7 | 4.4 | Derivatives | ① −1/(x√x) |
| 8 | 4.4 | Derivatives | ③ 16 |
| 9 | 4.4 | Derivatives | ⑤ 4+8ln2 |
| 10 | 4.5 | Tangent line | ② −2/3 |
| 11 | 4.5 | Limits | ③ Ⓐ and Ⓒ |
| 12 | 4.5 | Limits | ① Ⓐ only |
| 13 | 4.6 | Tangent line | ② 4 |
| 14 | 4.6 | Tangent line | ③ 8 |
| 15 | 4.6 | Tangent line | ① −3/2 |
| 16 | 4.7 | Inverse trig | ④ 5/3 |
| 17 | 4.7 | Derivatives | ⑤ 4e⁴ |
| 18 | 4.8 | Derivatives | ② 2e² |
| 19 | 4.8 | Derivatives | ⑤ (1/5)e^arctan(3/4) |
| 20 | 4.9 | Inverse trig | ③ (4/5)e^(3/5) |
| 21 | 5.0 | Inverse trig | ④ (1+4x²)/2 |
| 22 | 5.1 | Orthogonal trajectories | ① y=Kx⁴ |
출제 패턴 핵심 요약
- 난이도 설계: Q1~6 워밍업 → Q7~12 중간 → Q13~18 상 → Q19~22 최상
- 미분의 정의를 극한으로: Q1~3은 도함수 정의를 극한 형태로 출제
- ε-δ 출제: Q5 — 대학수학 수준의 정밀한 극한 정의
- 역삼각함수 4문항 (18%): Q16, Q20, Q21 + Q19. Q19는 태그상 Derivatives지만
arctan(3/4) 대입으로 역삼각 개념을 활용하는 문항이라 분포 집계에는 역삼각 범주에 포함
- arctan(3/4) 삼각비 대입: 3-4-5 삼각형을 활용한 Q19, Q20
- 증명 과정 빈칸: Q21 — arctan 미분 유도 과정의 핵심 단계 선택
- 직교 궤적: Q22 — 최고 배점 5.1점, 미분방정식 연계
- 참/거짓 복합: Q11, Q12 — 여러 명제의 참/거짓을 동시에 판별