+20점 · ∠ P O 2 O 1 = π 6 − α \displaystyle \angle\mathrm{PO_2O_1}=\frac\pi6-\alpha ∠ P O 2 O 1 = 6 π − α +20점 · 0 < α < π 6 \displaystyle 0<\alpha<\frac\pi6 0 < α < 6 π +40점 · f ( α ) = 4 ( π − α ) sin ( π 6 − α ) + 4 ( 5 π 6 + α ) sin α \displaystyle f(\alpha)=4(\pi-\alpha)\sin(\frac\pi6-\alpha)+4(\frac{5\pi}6+\alpha)\sin\alpha f ( α ) = 4 ( π − α ) sin ( 6 π − α ) + 4 ( 6 5 π + α ) sin α (3-1) ∠ P O 2 O 1 = β \displaystyle \angle\mathrm{PO_2O_1}=\beta ∠ P O 2 O 1 = β C 1 \displaystyle C_1 C 1 P \displaystyle \mathrm P P P O 1 \displaystyle \mathrm{PO_1} P O 1 C 2 \displaystyle C_2 C 2 P \displaystyle \mathrm P P P O 2 \displaystyle \mathrm{PO_2} P O 2 ∠ O 1 P O 2 = 5 π 6 \displaystyle \angle\mathrm{O_1PO_2}=\frac{5\pi}6 ∠ O 1 P O 2 = 6 5 π α + β = π 6 \displaystyle \alpha+\beta=\frac\pi6 α + β = 6 π ∠ P O 2 O 1 = β = π 6 − α \displaystyle \angle\mathrm{PO_2O_1}=\beta=\frac\pi6-\alpha ∠ P O 2 O 1 = β = 6 π − α α > 0 \displaystyle \alpha>0 α > 0 β > 0 \displaystyle \beta>0 β > 0 0 < α < π 6 \displaystyle 0<\alpha<\frac\pi6 0 < α < 6 π 또한, 삼각형 O 1 O 2 P \displaystyle \mathrm{O_1O_2P} O 1 O 2 P s sin α = r sin β = 1 sin 5 π 6 = 2 \displaystyle \frac s{\sin\alpha}=\frac r{\sin\beta}=\frac1{\sin\frac{5\pi}6}=2 sin α s = sin β r = sin 6 5 π 1 = 2 이므로 s = 2 sin α \displaystyle s=2\sin\alpha s = 2 sin α r = 2 sin β \displaystyle r=2\sin\beta r = 2 sin β 그러므로 구하는 길이 중 C 1 \displaystyle C_1 C 1 2 ( π − α ) r = 4 ( π − α ) sin β \displaystyle 2(\pi-\alpha)r=4(\pi-\alpha)\sin\beta 2 ( π − α ) r = 4 ( π − α ) sin β C 2 \displaystyle C_2 C 2 2 ( π − β ) s = 4 ( π − β ) sin α \displaystyle 2(\pi-\beta)s=4(\pi-\beta)\sin\alpha 2 ( π − β ) s = 4 ( π − β ) sin α f ( α ) = 4 ( π − α ) sin β + 4 ( π − β ) sin α = 4 ( π − α ) sin ( π 6 − α ) + 4 ( 5 π 6 + α ) sin α \displaystyle f(\alpha)=4(\pi-\alpha)\sin\beta+4(\pi-\beta)\sin\alpha=4(\pi-\alpha)\sin(\frac\pi6-\alpha)+4(\frac{5\pi}6+\alpha)\sin\alpha f ( α ) = 4 ( π − α ) sin β + 4 ( π − β ) sin α = 4 ( π − α ) sin ( 6 π − α ) + 4 ( 6 5 π + α ) sin α
+10점 · g ′ ( α ) = 4 { sin α + ( 5 π 6 + α ) cos α } \displaystyle g\prime(\alpha)=4\{\sin\alpha+(\frac{5\pi}6+\alpha)\cos\alpha\} g ′ ( α ) = 4 { sin α + ( 6 5 π + α ) cos α } +10점 · g ′ ′ ( α ) = 4 { 2 cos α − ( 5 π 6 + α ) sin α } \displaystyle g\prime\prime(\alpha)=4\{2\cos\alpha-(\frac{5\pi}6+\alpha)\sin\alpha\} g ′′ ( α ) = 4 { 2 cos α − ( 6 5 π + α ) sin α } +40점 · g ′ ′ ( α ) > 0 \displaystyle g\prime\prime(\alpha)>0 g ′′ ( α ) > 0 +10점 · a \displaystyle a a 10 π 3 \displaystyle \frac{10\pi}3 3 10 π +10점 · b \displaystyle b b 2 + 2 3 π \displaystyle 2+2\sqrt3\pi 2 + 2 3 π (3-2) g ( α ) = 4 ( π − β ) sin α = 4 ( 5 π 6 + α ) sin α \displaystyle g(\alpha)=4(\pi-\beta)\sin\alpha=4(\frac{5\pi}6+\alpha)\sin\alpha g ( α ) = 4 ( π − β ) sin α = 4 ( 6 5 π + α ) sin α g ′ ( α ) = 4 { sin α + ( 5 π 6 + α ) cos α } \displaystyle g\prime(\alpha)=4\{\sin\alpha+(\frac{5\pi}6+\alpha)\cos\alpha\} g ′ ( α ) = 4 { sin α + ( 6 5 π + α ) cos α } g ′ ′ ( α ) = 4 { 2 cos α − ( 5 π 6 + α ) sin α } \displaystyle g\prime\prime(\alpha)=4\{2\cos\alpha-(\frac{5\pi}6+\alpha)\sin\alpha\} g ′′ ( α ) = 4 { 2 cos α − ( 6 5 π + α ) sin α } 0 < α < π 6 \displaystyle 0<\alpha<\frac\pi6 0 < α < 6 π 2 cos α > 3 \displaystyle 2\cos\alpha>\sqrt3 2 cos α > 3 ( 5 π 6 + α ) sin α < π 2 \displaystyle (\frac{5\pi}6+\alpha)\sin\alpha<\frac\pi2 ( 6 5 π + α ) sin α < 2 π g ′ ′ ( α ) = 4 { 2 cos α − ( 5 π 6 + α ) sin α } > 4 ( 3 − π 2 ) > 0 \displaystyle g\prime\prime(\alpha)=4\{2\cos\alpha-(\frac{5\pi}6+\alpha)\sin\alpha\}>4(\sqrt3-\frac\pi2)>0 g ′′ ( α ) = 4 { 2 cos α − ( 6 5 π + α ) sin α } > 4 ( 3 − 2 π ) > 0 이고, g ′ ( α ) \displaystyle g\prime(\alpha) g ′ ( α ) lim α → 0 + g ′ ( α ) = 10 π 3 \displaystyle \lim_{\alpha\to0+}g\prime(\alpha)=\frac{10\pi}3 α → 0 + lim g ′ ( α ) = 3 10 π lim α → π 6 − g ′ ( α ) = 2 + 2 3 π \displaystyle \lim_{\alpha\to\frac\pi6-}g\prime(\alpha)=2+2\sqrt3\pi α → 6 π − lim g ′ ( α ) = 2 + 2 3 π a \displaystyle a a 10 π 3 \displaystyle \frac{10\pi}3 3 10 π b \displaystyle b b 2 + 2 3 π \displaystyle 2+2\sqrt3\pi 2 + 2 3 π
+30점 · f ( α ) = g ( π 6 − α ) + g ( α ) \displaystyle f(\alpha)=g(\frac\pi6-\alpha)+g(\alpha) f ( α ) = g ( 6 π − α ) + g ( α ) +10점 · g ′ ( α ) \displaystyle g\prime(\alpha) g ′ ( α ) +30점 · f ′ ( α ) = g ′ ( α ) − g ′ ( π 6 − α ) = 0 \displaystyle f\prime(\alpha)=g\prime(\alpha)-g\prime(\frac\pi6-\alpha)=0 f ′ ( α ) = g ′ ( α ) − g ′ ( 6 π − α ) = 0 α = π 6 − α \displaystyle \alpha=\frac\pi6-\alpha α = 6 π − α +10점 · α = π 12 \displaystyle \alpha=\frac\pi{12} α = 12 π f ( α ) \displaystyle f(\alpha) f ( α ) (3-3) f ( α ) = 4 ( π − α ) sin ( π 6 − α ) + 4 ( 5 π 6 + α ) sin α = g ( π 6 − α ) + g ( α ) \displaystyle f(\alpha)=4(\pi-\alpha)\sin(\frac\pi6-\alpha)+4(\frac{5\pi}6+\alpha)\sin\alpha=g(\frac\pi6-\alpha)+g(\alpha) f ( α ) = 4 ( π − α ) sin ( 6 π − α ) + 4 ( 6 5 π + α ) sin α = g ( 6 π − α ) + g ( α ) f ′ ( α ) = − g ′ ( π 6 − α ) + g ′ ( α ) \displaystyle f\prime(\alpha)=-g\prime(\frac\pi6-\alpha)+g\prime(\alpha) f ′ ( α ) = − g ′ ( 6 π − α ) + g ′ ( α ) g ′ ( α ) \displaystyle g\prime(\alpha) g ′ ( α ) 0 < x < π 6 \displaystyle 0<x<\frac\pi6 0 < x < 6 π 0 < y < π 6 \displaystyle 0<y<\frac\pi6 0 < y < 6 π g ′ ( x ) = g ′ ( y ) \displaystyle g\prime(x)=g\prime(y) g ′ ( x ) = g ′ ( y ) x = y \displaystyle x=y x = y f ′ ( α ) = g ′ ( α ) − g ′ ( π 6 − α ) = 0 \displaystyle f\prime(\alpha)=g\prime(\alpha)-g\prime(\frac\pi6-\alpha)=0 f ′ ( α ) = g ′ ( α ) − g ′ ( 6 π − α ) = 0 α = π 6 − α \displaystyle \alpha=\frac\pi6-\alpha α = 6 π − α 0 < α < π 6 \displaystyle 0<\alpha<\frac\pi6 0 < α < 6 π f ′ ( α ) = 0 \displaystyle f\prime(\alpha)=0 f ′ ( α ) = 0 α = π 12 \displaystyle \alpha=\frac\pi{12} α = 12 π g ′ ′ ( α ) > 0 \displaystyle g\prime\prime(\alpha)>0 g ′′ ( α ) > 0 f ′ ′ ( π 12 ) = g ′ ′ ( π 12 ) + g ′ ′ ( π 12 ) > 0 \displaystyle f\prime\prime(\frac\pi{12})=g\prime\prime(\frac\pi{12})+g\prime\prime(\frac\pi{12})>0 f ′′ ( 12 π ) = g ′′ ( 12 π ) + g ′′ ( 12 π ) > 0 α = π 12 \displaystyle \alpha=\frac\pi{12} α = 12 π f ( α ) \displaystyle f(\alpha) f ( α )
수리논술, 배움에서 논증의 완성까지.