+20점 (3-1) A \displaystyle \mathrm A A B \displaystyle \mathrm B B A ( 2 2 cos ( θ + π 4 ) , 2 2 sin ( θ + π 4 ) ) \displaystyle \mathrm A(2\sqrt2\cos(\theta+\frac\pi4),2\sqrt2\sin(\theta+\frac\pi4)) A ( 2 2 cos ( θ + 4 π ) , 2 2 sin ( θ + 4 π ))
+20점 , B ( 2 cos θ , 2 sin θ ) \displaystyle \mathrm B(2\cos\theta,2\sin\theta) B ( 2 cos θ , 2 sin θ )
+40점 · π 6 ≤ θ < π 2 \displaystyle \frac\pi6\le\theta<\frac\pi2 6 π ≤ θ < 2 π θ ≥ π 6 \displaystyle \theta\ge\frac\pi6 θ ≥ 6 π D \displaystyle \mathrm D D O B \displaystyle \mathrm{OB} OB 2 sin θ ≥ 1 \displaystyle 2\sin\theta\ge1 2 sin θ ≥ 1 π 6 ≤ θ < π 2 \displaystyle \frac\pi6\le\theta<\frac\pi2 6 π ≤ θ < 2 π
+10점 · C \displaystyle \mathrm C C ( cot ( θ + π 4 ) , 1 ) \displaystyle (\cot(\theta+\frac\pi4),1) ( cot ( θ + 4 π ) , 1 ) +10점 · D \displaystyle \mathrm D D ( cot θ , 1 ) \displaystyle (\cot\theta,1) ( cot θ , 1 ) (3-2) π 6 ≤ θ < π 2 \displaystyle \frac\pi6\le\theta<\frac\pi2 6 π ≤ θ < 2 π C \displaystyle \mathrm C C ( cot ( θ + π 4 ) , 1 ) \displaystyle (\cot(\theta+\frac\pi4),1) ( cot ( θ + 4 π ) , 1 ) D \displaystyle \mathrm D D ( cot θ , 1 ) \displaystyle (\cot\theta,1) ( cot θ , 1 )
+20점 따라서 선분 C D \displaystyle \mathrm{CD} CD cot θ − cot ( θ + π 4 ) \displaystyle \cot\theta-\cot(\theta+\frac\pi4) cot θ − cot ( θ + 4 π ) t = tan θ \displaystyle t=\tan\theta t = tan θ π 6 ≤ θ < π 2 \displaystyle \frac\pi6\le\theta<\frac\pi2 6 π ≤ θ < 2 π t ≥ 1 3 \displaystyle t\ge\frac1{\sqrt3} t ≥ 3 1 cot θ − cot ( θ + π 4 ) = 1 tan θ − 1 tan ( θ + π 4 ) = 1 tan θ − 1 1 + tan θ 1 − tan θ \displaystyle \cot\theta-\cot(\theta+\frac\pi4)=\frac1{\tan\theta}-\frac1{\tan(\theta+\frac\pi4)}=\frac1{\tan\theta}-\frac1{\frac{1+\tan\theta}{1-\tan\theta}} cot θ − cot ( θ + 4 π ) = tan θ 1 − tan ( θ + 4 π ) 1 = tan θ 1 − 1 − t a n θ 1 + t a n θ 1 이므로, 선분 C D \displaystyle \mathrm{CD} CD f ( t ) \displaystyle f(t) f ( t ) f ( t ) = 1 t − 1 − t 1 + t \displaystyle f(t)=\frac1t-\frac{1-t}{1+t} f ( t ) = t 1 − 1 + t 1 − t t ≥ 1 3 \displaystyle t\ge\frac1{\sqrt3} t ≥ 3 1 f ′ ( t ) = t 2 − 2 t − 1 t 2 ( 1 + t ) 2 = 0 \displaystyle f\prime(t)=\frac{t^2-2t-1}{t^2(1+t)^2}=0 f ′ ( t ) = t 2 ( 1 + t ) 2 t 2 − 2 t − 1 = 0 t = 1 + 2 \displaystyle t=1+\sqrt2 t = 1 + 2 1 3 ≤ t < 1 + 2 \displaystyle \frac1{\sqrt3}\le t<1+\sqrt2 3 1 ≤ t < 1 + 2 f ′ ( t ) < 0 \displaystyle f\prime(t)<0 f ′ ( t ) < 0 t > 1 + 2 \displaystyle t>1+\sqrt2 t > 1 + 2 f ′ ( t ) > 0 \displaystyle f\prime(t)>0 f ′ ( t ) > 0
+20점 · f ( t ) \displaystyle f(t) f ( t ) f ( 1 + 2 ) = 2 2 − 2 \displaystyle f(1+\sqrt2)=2\sqrt2-2 f ( 1 + 2 ) = 2 2 − 2 +20점 · f ( t ) \displaystyle f(t) f ( t ) f ( 1 3 ) = 2 3 − 2 \displaystyle f(\frac1{\sqrt3})=2\sqrt3-2 f ( 3 1 ) = 2 3 − 2 f ( 1 3 ) = 2 3 − 2 > 1 \displaystyle f(\frac1{\sqrt3})=2\sqrt3-2>1 f ( 3 1 ) = 2 3 − 2 > 1 f ( 1 + 2 ) = 2 2 − 2 \displaystyle f(1+\sqrt2)=2\sqrt2-2 f ( 1 + 2 ) = 2 2 − 2 lim t → ∞ f ( t ) = 1 \displaystyle \lim_{t\to\infty}f(t)=1 t → ∞ lim f ( t ) = 1 t ≥ 1 3 \displaystyle t\ge\frac1{\sqrt3} t ≥ 3 1 f ( t ) \displaystyle f(t) f ( t ) f ( 1 + 2 ) = 2 2 − 2 \displaystyle f(1+\sqrt2)=2\sqrt2-2 f ( 1 + 2 ) = 2 2 − 2 f ( 1 3 ) = 2 3 − 2 \displaystyle f(\frac1{\sqrt3})=2\sqrt3-2 f ( 3 1 ) = 2 3 − 2
(3-3) a = 2 − b \displaystyle a=2-b a = 2 − b g ( b ) = a × b = b ( 2 − b ) \displaystyle g(b)=a\times b=b(2-b) g ( b ) = a × b = b ( 2 − b ) 0 ≤ θ ≤ π 6 \displaystyle 0\le\theta\le\frac\pi6 0 ≤ θ ≤ 6 π θ \displaystyle \theta θ a \displaystyle a a b = 2 − a \displaystyle b=2-a b = 2 − a
+30점 · 0 ≤ θ ≤ π 6 \displaystyle 0\le\theta\le\frac\pi6 0 ≤ θ ≤ 6 π 3 − 1 ≤ b ≤ 3 2 \displaystyle \sqrt3-1\le b\le\frac32 3 − 1 ≤ b ≤ 2 3 θ = 0 \displaystyle \theta=0 θ = 0 b = 3 2 \displaystyle b=\frac32 b = 2 3 θ = π 6 \displaystyle \theta=\frac\pi6 θ = 6 π b = 3 − 1 \displaystyle b=\sqrt3-1 b = 3 − 1 0 ≤ θ ≤ π 6 \displaystyle 0\le\theta\le\frac\pi6 0 ≤ θ ≤ 6 π 3 − 1 ≤ b ≤ 3 2 \displaystyle \sqrt3-1\le b\le\frac32 3 − 1 ≤ b ≤ 2 3
+30점 · π 6 ≤ θ < π 2 \displaystyle \frac\pi6\le\theta<\frac\pi2 6 π ≤ θ < 2 π 2 − 1 ≤ b ≤ 3 − 1 \displaystyle \sqrt2-1\le b\le\sqrt3-1 2 − 1 ≤ b ≤ 3 − 1 또한 π 6 ≤ θ < π 2 \displaystyle \frac\pi6\le\theta<\frac\pi2 6 π ≤ θ < 2 π C D \displaystyle \mathrm{CD} CD 2 2 − 2 \displaystyle 2\sqrt2-2 2 2 − 2 2 3 − 2 \displaystyle 2\sqrt3-2 2 3 − 2 2 − 1 ≤ b ≤ 3 − 1 \displaystyle \sqrt2-1\le b\le\sqrt3-1 2 − 1 ≤ b ≤ 3 − 1 0 ≤ θ < π 2 \displaystyle 0\le\theta<\frac\pi2 0 ≤ θ < 2 π 2 − 1 ≤ b ≤ 3 2 \displaystyle \sqrt2-1\le b\le\frac32 2 − 1 ≤ b ≤ 2 3
+20점 · 최댓값은 1 \displaystyle 1 1 4 2 − 5 \displaystyle 4\sqrt2-5 4 2 − 5 1 − ( 2 − 1 ) > 3 2 − 1 \displaystyle 1-(\sqrt2-1)>\frac32-1 1 − ( 2 − 1 ) > 2 3 − 1 2 − 1 ≤ b ≤ 3 2 \displaystyle \sqrt2-1\le b\le\frac32 2 − 1 ≤ b ≤ 2 3 g ( b ) \displaystyle g(b) g ( b ) g ( 1 ) = 1 \displaystyle g(1)=1 g ( 1 ) = 1 g ( 2 − 1 ) = ( 2 − 1 ) ( 3 − 2 ) = 4 2 − 5 \displaystyle g(\sqrt2-1)=(\sqrt2-1)(3-\sqrt2)=4\sqrt2-5 g ( 2 − 1 ) = ( 2 − 1 ) ( 3 − 2 ) = 4 2 − 5
수리논술, 이게 다다.
