한양대 에리카캠퍼스 › 2021학년도 수시 오후2 한양대 에리카캠퍼스 2021학년도 수시 오후2 자연계열 1번 · mathesis.kr/archive/hanyang-erica/2021su_pm2/jayeon_1
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한양대 에리카캠퍼스 2021학년도 수시 오후2 자연계열 1번 · mathesis.kr/archive/hanyang-erica/2021su_pm2/jayeon_1
[15점] · 1-1 1-1. [풀이] C : x 2 + y 2 = 1 \displaystyle \mathrm{C}:x^2+y^2=1 C : x 2 + y 2 = 1 y = a x + 1 \displaystyle y=ax+1 y = a x + 1 x 2 + ( a x + 1 ) 2 = 1 \displaystyle x^2+(ax+1)^2=1 x 2 + ( a x + 1 ) 2 = 1 ( a 2 + 1 ) x 2 + 2 a x = 0 \displaystyle (a^2+1)x^2+2ax=0 ( a 2 + 1 ) x 2 + 2 a x = 0 x ≠ 0 \displaystyle x\ne0 x = 0 x = − 2 a a 2 + 1 \displaystyle x=-\frac{2a}{a^2+1} x = − a 2 + 1 2 a y = − a 2 + 1 a 2 + 1 \displaystyle y=\frac{-a^2+1}{a^2+1} y = a 2 + 1 − a 2 + 1
따라서 R ( − 2 a a 2 + 1 , − a 2 + 1 a 2 + 1 ) \displaystyle \mathrm{R}\left(-\frac{2a}{a^2+1},\frac{-a^2+1}{a^2+1}\right) R ( − a 2 + 1 2 a , a 2 + 1 − a 2 + 1 ) R \displaystyle \mathrm{R} R − 2 a a 2 + 1 x + − a 2 + 1 a 2 + 1 y = 1 \displaystyle -\frac{2a}{a^2+1}x+\frac{-a^2+1}{a^2+1}y=1 − a 2 + 1 2 a x + a 2 + 1 − a 2 + 1 y = 1 − 2 a x + ( − a 2 + 1 ) y = a 2 + 1 \displaystyle -2ax+(-a^2+1)y=a^2+1 − 2 a x + ( − a 2 + 1 ) y = a 2 + 1
직선 y = a x + 1 \displaystyle y=ax+1 y = a x + 1 y = − 2 \displaystyle y=-2 y = − 2 y = − 2 \displaystyle y=-2 y = − 2 x = − 3 a \displaystyle x=-\frac3a x = − a 3 ( − 3 a , − 2 ) \displaystyle \left(-\frac3a,-2\right) ( − a 3 , − 2 ) 접선 − 2 a x + ( − a 2 + 1 ) y = a 2 + 1 \displaystyle -2ax+(-a^2+1)y=a^2+1 − 2 a x + ( − a 2 + 1 ) y = a 2 + 1 y = − 2 \displaystyle y=-2 y = − 2 y = − 2 \displaystyle y=-2 y = − 2 x = a 2 − 3 2 a \displaystyle x=\frac{a^2-3}{2a} x = 2 a a 2 − 3 ( a 2 − 3 2 a , − 2 ) \displaystyle \left(\frac{a^2-3}{2a},-2\right) ( 2 a a 2 − 3 , − 2 ) 이 때, 제시문 <나>를 이용하면, ∣ a 2 − 3 2 a + 3 a ∣ = a 2 + 3 2 a = 1 2 ( a + 3 a ) ≥ a ⋅ 3 a = 3 \displaystyle \left|\frac{a^2-3}{2a}+\frac3a\right|=\frac{a^2+3}{2a}=\frac12\left(a+\frac3a\right)\ge\sqrt{a\cdot\frac3a}=\sqrt3 2 a a 2 − 3 + a 3 = 2 a a 2 + 3 = 2 1 ( a + a 3 ) ≥ a ⋅ a 3 = 3 이 때, 등호가 성립하는 것은 a = 3 a \displaystyle a=\frac3a a = a 3 a = 3 \displaystyle a=\sqrt3 a = 3 따라서 a = 3 \displaystyle a=\sqrt3 a = 3 3 \displaystyle \sqrt3 3
[10점] · 1-2 1-2. [풀이] f ( x ) = x 2 + 2 m x \displaystyle f(x)=x^2+2mx f ( x ) = x 2 + 2 m x P ( cos α , sin α ) \displaystyle \mathrm{P}(\cos\alpha,\sin\alpha) P ( cos α , sin α ) y = f ( x ) \displaystyle y=f(x) y = f ( x ) sin α = cos 2 α + 2 m cos α \displaystyle \sin\alpha=\cos^2\alpha+2m\cos\alpha sin α = cos 2 α + 2 m cos α 따라서, m = sin α − cos 2 α 2 cos α = 1 2 ( tan α − cos α ) \displaystyle m=\frac{\sin\alpha-\cos^2\alpha}{2\cos\alpha}=\frac12(\tan\alpha-\cos\alpha) m = 2 cos α sin α − cos 2 α = 2 1 ( tan α − cos α )
[20점] · 1-3 1-3. [풀이] S 1 = 1 2 ⋅ 1 2 ( π 2 − α + π 2 + β ) = 1 2 ( π − α + β ) \displaystyle S_1=\frac12\cdot1^2\left(\frac\pi2-\alpha+\frac\pi2+\beta\right)=\frac12(\pi-\alpha+\beta) S 1 = 2 1 ⋅ 1 2 ( 2 π − α + 2 π + β ) = 2 1 ( π − α + β )
선분 O P : y = x tan α \displaystyle \mathrm{OP}:y=x\tan\alpha OP : y = x tan α y = f ( x ) \displaystyle y=f(x) y = f ( x ) S 2 \displaystyle S_2 S 2 S 2 = ∫ 0 cos α { x tan α − f ( x ) } d x = ∫ 0 cos α ( − x 2 + x cos α ) d x = 1 6 cos 3 α \displaystyle S_2=\int_0^{\cos\alpha}\{x\tan\alpha-f(x)\}dx=\int_0^{\cos\alpha}(-x^2+x\cos\alpha)dx=\frac16\cos^3\alpha S 2 = ∫ 0 c o s α { x tan α − f ( x )} d x = ∫ 0 c o s α ( − x 2 + x cos α ) d x = 6 1 cos 3 α
Q ( − cos β , − sin β ) \displaystyle \mathrm{Q}(-\cos\beta,-\sin\beta) Q ( − cos β , − sin β ) O Q : y = x tan β \displaystyle \mathrm{OQ}:y=x\tan\beta OQ : y = x tan β y = f ( x ) \displaystyle y=f(x) y = f ( x ) S 3 \displaystyle S_3 S 3 S 3 = ∫ − cos β 0 { x tan β − f ( x ) } d x = ∫ − cos β 0 ( − x 2 − x cos β ) d x = 1 6 cos 3 β \displaystyle S_3=\int_{-\cos\beta}^0\{x\tan\beta-f(x)\}dx=\int_{-\cos\beta}^0(-x^2-x\cos\beta)dx=\frac16\cos^3\beta S 3 = ∫ − c o s β 0 { x tan β − f ( x )} d x = ∫ − c o s β 0 ( − x 2 − x cos β ) d x = 6 1 cos 3 β S = S 1 + S 2 + S 3 = 1 2 ( π − α + β ) + 1 6 ( cos 3 α + cos 3 β ) . \displaystyle S=S_1+S_2+S_3=\frac12(\pi-\alpha+\beta)+\frac16(\cos^3\alpha+\cos^3\beta). S = S 1 + S 2 + S 3 = 2 1 ( π − α + β ) + 6 1 ( cos 3 α + cos 3 β ) .
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