+10점 · ( t − f ( a ) ) 2 + ( t 2 − g ( t ) ) 2 = g ( t ) 2 \displaystyle (t-f(a))^2+(t^2-g(t))^2=g(t)^2 ( t − f ( a ) ) 2 + ( t 2 − g ( t ) ) 2 = g ( t ) 2 Correction. 1-1 첫 채점기준의 f ( a ) \displaystyle f(a) f ( a ) f ( t ) \displaystyle f(t) f ( t ) C ( f ( t ) , g ( t ) ) \displaystyle C(f(t),g(t)) C ( f ( t ) , g ( t )) P ( t , t 2 ) \displaystyle P(t,t^2) P ( t , t 2 ) ( t − f ( t ) ) 2 + ( t 2 − g ( t ) ) 2 = g ( t ) 2 \displaystyle (t-f(t))^2+(t^2-g(t))^2=g(t)^2 ( t − f ( t ) ) 2 + ( t 2 − g ( t ) ) 2 = g ( t ) 2 예시답안에는 f ( t ) \displaystyle f(t) f ( t )
【1-1】 원 R \displaystyle R R y = x 2 \displaystyle y=x^2 y = x 2 x \displaystyle x x R \displaystyle R R ( x − f ( t ) ) 2 + ( y − g ( t ) ) 2 = g ( t ) 2 \displaystyle (x-f(t))^2+(y-g(t))^2=g(t)^2 ( x − f ( t ) ) 2 + ( y − g ( t ) ) 2 = g ( t ) 2 을 만족시킨다. 점 P ( t , t 2 ) \displaystyle \mathrm P(t,t^2) P ( t , t 2 ) ( t − f ( t ) ) 2 + ( t 2 − g ( t ) ) 2 = g ( t ) 2 \displaystyle (t-f(t))^2+(t^2-g(t))^2=g(t)^2 ( t − f ( t ) ) 2 + ( t 2 − g ( t ) ) 2 = g ( t ) 2
+10점 · 2 t g ( t ) − t 2 f ( t ) − t = − 1 \displaystyle 2t\dfrac{g(t)-t^2}{f(t)-t}=-1 2 t f ( t ) − t g ( t ) − t 2 = − 1 을 만족시키고 선분 P C ‾ \displaystyle \overline{\mathrm{PC}} PC P ( t , t 2 ) \displaystyle \mathrm P(t,t^2) P ( t , t 2 ) y = x 2 \displaystyle y=x^2 y = x 2 2 t g ( t ) − t 2 f ( t ) − t = − 1 \displaystyle 2t\dfrac{g(t)-t^2}{f(t)-t}=-1 2 t f ( t ) − t g ( t ) − t 2 = − 1 을 만족시킨다. 따라서
+10점 · g ( t ) = 1 4 4 t 2 + 1 ( 4 t 2 + 1 − 1 ) \displaystyle g(t)=\dfrac14\sqrt{4t^2+1}\left(\sqrt{4t^2+1}-1\right) g ( t ) = 4 1 4 t 2 + 1 ( 4 t 2 + 1 − 1 ) +10점 · f ( t ) = t 2 ( 4 t 2 + 1 + 1 ) \displaystyle f(t)=\dfrac t2\left(\sqrt{4t^2+1}+1\right) f ( t ) = 2 t ( 4 t 2 + 1 + 1 ) g ( t ) = 1 4 4 t 2 + 1 ( 4 t 2 + 1 − 1 ) , \displaystyle g(t)=\dfrac14\sqrt{4t^2+1}\left(\sqrt{4t^2+1}-1\right), g ( t ) = 4 1 4 t 2 + 1 ( 4 t 2 + 1 − 1 ) , f ( t ) = t 2 ( 4 t 2 + 1 + 1 ) \displaystyle f(t)=\dfrac t2\left(\sqrt{4t^2+1}+1\right) f ( t ) = 2 t ( 4 t 2 + 1 + 1 )
+10점 · lim t → 0 + f ( t ) g ( t ) t 3 = 1 2 \displaystyle \lim_{t\to0+}\dfrac{f(t)g(t)}{t^3}=\dfrac12 t → 0 + lim t 3 f ( t ) g ( t ) = 2 1 이고 f ( t ) g ( t ) = t 3 2 4 t 2 + 1 \displaystyle f(t)g(t)=\dfrac{t^3}2\sqrt{4t^2+1} f ( t ) g ( t ) = 2 t 3 4 t 2 + 1 lim t → 0 + f ( t ) g ( t ) t 3 = 1 2 \displaystyle \lim_{t\to0+}\dfrac{f(t)g(t)}{t^3}=\dfrac12 t → 0 + lim t 3 f ( t ) g ( t ) = 2 1 이다.
+15점 · g ( t ) + g ( t ) sin ( θ 2 ) = t 2 \displaystyle g(t)+g(t)\sin\left(\dfrac\theta2\right)=t^2 g ( t ) + g ( t ) sin ( 2 θ ) = t 2 【1-2】 점 P ( t , t 2 ) \displaystyle \mathrm P(t,t^2) P ( t , t 2 ) y \displaystyle y y R \displaystyle R R g ( t ) sin ( θ 2 ) \displaystyle g(t)\sin\left(\dfrac\theta2\right) g ( t ) sin ( 2 θ ) b + b sin ( θ 2 ) = t 2 \displaystyle b+b\sin\left(\dfrac\theta2\right)=t^2 b + b sin ( 2 θ ) = t 2 Correction. 1-2 예시답안의 b + b sin ( θ / 2 ) = t 2 \displaystyle b+b\sin(\theta/2)=t^2 b + b sin ( θ /2 ) = t 2 b \displaystyle b b g ( t ) \displaystyle g(t) g ( t ) g ( t ) + g ( t ) sin θ 2 = t 2 \displaystyle g(t)+g(t)\sin\frac{\theta}{2}=t^2 g ( t ) + g ( t ) sin 2 θ = t 2 원문 표기는 그대로 실었다. 을 만족시킨다. 따라서
+15점 · d d t 1 sin 2 ( θ 2 ) = d d t ( 4 t 2 + 1 ) = 8 t \displaystyle \dfrac d{dt}\dfrac1{\sin^2\left(\frac\theta2\right)}=\dfrac d{dt}(4t^2+1)=8t d t d sin 2 ( 2 θ ) 1 = d t d ( 4 t 2 + 1 ) = 8 t t = 1 \displaystyle t=1 t = 1 8 \displaystyle 8 8 d d t 1 sin 2 ( θ 2 ) = d d t ( 4 t 2 + 1 ) = 8 t \displaystyle \dfrac d{dt}\dfrac1{\sin^2\left(\frac\theta2\right)}=\dfrac d{dt}(4t^2+1)=8t d t d sin 2 ( 2 θ ) 1 = d t d ( 4 t 2 + 1 ) = 8 t 이고 t = 1 \displaystyle t=1 t = 1 8 \displaystyle 8 8
+10점 · S ( t ) = t 3 3 + 1 2 ( t 2 + g ( t ) ) ( f ( t ) − t ) − g ( t ) 2 4 ( π + θ ) \displaystyle S(t)=\dfrac{t^3}3+\dfrac12(t^2+g(t))(f(t)-t)-\dfrac{g(t)^2}4(\pi+\theta) S ( t ) = 3 t 3 + 2 1 ( t 2 + g ( t )) ( f ( t ) − t ) − 4 g ( t ) 2 ( π + θ ) 【1-3】 넓이 S \displaystyle S S y = x 2 \displaystyle y=x^2 y = x 2 x \displaystyle x x P \displaystyle \mathrm P P B ( t , 0 ) \displaystyle \mathrm B(t,0) B ( t , 0 ) P B A C \displaystyle \mathrm{PBAC} PBAC C P A \displaystyle \mathrm{CPA} CPA S ( t ) \displaystyle S(t) S ( t ) S ( t ) = ∫ 0 t x 2 d x + 1 2 ( t 2 + g ( t ) ) ( f ( t ) − t ) − g ( t ) 2 4 ( π + θ ) = t 3 3 + 1 2 ( t 2 + g ( t ) ) ( f ( t ) − t ) − g ( t ) 2 4 ( π + θ ) \displaystyle \begin{aligned}S(t)&=\int_0^t x^2\,dx+\dfrac12(t^2+g(t))(f(t)-t)-\dfrac{g(t)^2}4(\pi+\theta)\\&=\dfrac{t^3}3+\dfrac12(t^2+g(t))(f(t)-t)-\dfrac{g(t)^2}4(\pi+\theta)\end{aligned} S ( t ) = ∫ 0 t x 2 d x + 2 1 ( t 2 + g ( t )) ( f ( t ) − t ) − 4 g ( t ) 2 ( π + θ ) = 3 t 3 + 2 1 ( t 2 + g ( t )) ( f ( t ) − t ) − 4 g ( t ) 2 ( π + θ ) 이다.
+5점 · lim t → 0 + g ( t ) t 2 = 1 2 \displaystyle \lim_{t\to0+}\dfrac{g(t)}{t^2}=\dfrac12 t → 0 + lim t 2 g ( t ) = 2 1 +5점 · lim t → 0 + f ( t ) t = 1 \displaystyle \lim_{t\to0+}\dfrac{f(t)}t=1 t → 0 + lim t f ( t ) = 1 lim t → 0 + g ( t ) t 2 = lim t → 0 + 4 t 2 + 1 ( 4 t 2 + 1 − 1 ) 4 t 2 = lim t → 0 + 4 t 2 + 1 4 t 2 + 1 + 1 = 1 2 , \displaystyle \lim_{t\to0+}\dfrac{g(t)}{t^2}=\lim_{t\to0+}\dfrac{\sqrt{4t^2+1}\left(\sqrt{4t^2+1}-1\right)}{4t^2}=\lim_{t\to0+}\dfrac{\sqrt{4t^2+1}}{\sqrt{4t^2+1}+1}=\dfrac12, t → 0 + lim t 2 g ( t ) = t → 0 + lim 4 t 2 4 t 2 + 1 ( 4 t 2 + 1 − 1 ) = t → 0 + lim 4 t 2 + 1 + 1 4 t 2 + 1 = 2 1 , lim t → 0 + f ( t ) t = lim t → 0 + 1 2 ( 4 t 2 + 1 + 1 ) = 1 , \displaystyle \lim_{t\to0+}\dfrac{f(t)}t=\lim_{t\to0+}\dfrac12\left(\sqrt{4t^2+1}+1\right)=1, t → 0 + lim t f ( t ) = t → 0 + lim 2 1 ( 4 t 2 + 1 + 1 ) = 1 , π ≤ π + θ ≤ 2 π \displaystyle \pi\le\pi+\theta\le2\pi π ≤ π + θ ≤ 2 π 이므로
+10점 · lim t → 0 + S ( t ) t 3 = 1 3 \displaystyle \lim_{t\to0+}\dfrac{S(t)}{t^3}=\dfrac13 t → 0 + lim t 3 S ( t ) = 3 1 lim t → 0 + S ( t ) t 3 = lim t → 0 + ( 1 3 + 1 2 ( 1 + g ( t ) t 2 ) ( f ( t ) t − 1 ) − g ( t ) 2 4 t 3 ( π + θ ) ) = 1 3 + 1 2 ( 1 + 1 2 ) ( 1 − 1 ) = 1 3 \displaystyle \begin{aligned}\lim_{t\to0+}\dfrac{S(t)}{t^3}&=\lim_{t\to0+}\left(\dfrac13+\dfrac12\left(1+\dfrac{g(t)}{t^2}\right)\left(\dfrac{f(t)}t-1\right)-\dfrac{g(t)^2}{4t^3}(\pi+\theta)\right)\\&=\dfrac13+\dfrac12\left(1+\dfrac12\right)(1-1)=\dfrac13\end{aligned} t → 0 + lim t 3 S ( t ) = t → 0 + lim ( 3 1 + 2 1 ( 1 + t 2 g ( t ) ) ( t f ( t ) − 1 ) − 4 t 3 g ( t ) 2 ( π + θ ) ) = 3 1 + 2 1 ( 1 + 2 1 ) ( 1 − 1 ) = 3 1 이다.
수리논술, 배움에서 논증의 완성까지.