(3-1) t = π 4 \displaystyle t=\frac\pi4 t = 4 π x \displaystyle x x π 4 \displaystyle \frac\pi4 4 π x \displaystyle x x 1 − 2 \displaystyle 1-\sqrt2 1 − 2
50점 · 답 2 − 1 \displaystyle \sqrt2-1 2 − 1 g ( π 4 ) = 2 − 1 \displaystyle g\left(\frac\pi4\right)=\sqrt2-1 g ( 4 π ) = 2 − 1
(3-2) 두 반원에 공통으로 접하는 직선이 x \displaystyle x x t = π 6 \displaystyle t=\frac\pi6 t = 6 π 0 < t < π 6 \displaystyle 0<t<\frac\pi6 0 < t < 6 π π 6 < t < π 2 \displaystyle \frac\pi6<t<\frac\pi2 6 π < t < 2 π g ( t ) \displaystyle g(t) g ( t )
+30점 · g ( t ) = { 3 sin t − 5 ( 0 < t ≤ π 6 ) 1 sin t − 1 ( π 6 < t < π 2 ) \displaystyle g(t)=\begin{cases}\frac3{\sin t}-5&\left(0<t\le\frac\pi6\right)\\\frac1{\sin t}-1&\left(\frac\pi6<t<\frac\pi2\right)\end{cases} g ( t ) = { s i n t 3 − 5 s i n t 1 − 1 ( 0 < t ≤ 6 π ) ( 6 π < t < 2 π ) g ( t ) = { 3 sin t − 5 ( 0 < t ≤ π 6 ) 1 sin t − 1 ( π 6 < t < π 2 ) \displaystyle g(t)=\begin{cases}\frac3{\sin t}-5&\left(0<t\le\frac\pi6\right)\\\frac1{\sin t}-1&\left(\frac\pi6<t<\frac\pi2\right)\end{cases} g ( t ) = { s i n t 3 − 5 s i n t 1 − 1 ( 0 < t ≤ 6 π ) ( 6 π < t < 2 π )
+20점 · lim t → π / 6 + 0 g ( t ) − 1 t − π / 6 = − 2 3 \displaystyle \lim_{t\to\pi/6+0}\frac{g(t)-1}{t-\pi/6}=-2\sqrt3 t → π /6 + 0 lim t − π /6 g ( t ) − 1 = − 2 3 lim t → π / 6 − 0 g ( t ) − 1 t − π / 6 = − 6 3 \displaystyle \lim_{t\to\pi/6-0}\frac{g(t)-1}{t-\pi/6}=-6\sqrt3 t → π /6 − 0 lim t − π /6 g ( t ) − 1 = − 6 3 t = π 6 \displaystyle t=\frac\pi6 t = 6 π 또한 t = π 6 \displaystyle t=\frac\pi6 t = 6 π lim t → π / 6 + 0 g ( t ) − 1 t − π / 6 = − 2 3 \displaystyle \lim_{t\to\pi/6+0}\frac{g(t)-1}{t-\pi/6}=-2\sqrt3 t → π /6 + 0 lim t − π /6 g ( t ) − 1 = − 2 3 lim t → π / 6 − 0 g ( t ) − 1 t − π / 6 = − 6 3 \displaystyle \lim_{t\to\pi/6-0}\frac{g(t)-1}{t-\pi/6}=-6\sqrt3 t → π /6 − 0 lim t − π /6 g ( t ) − 1 = − 6 3 t = π 6 \displaystyle t=\frac\pi6 t = 6 π
(3-3) 그림자의 길이는 t \displaystyle t t t = π 6 \displaystyle t=\frac\pi6 t = 6 π s ≥ 1 \displaystyle s\ge1 s ≥ 1 s = g ( h ( s ) ) = 3 sin h ( s ) − 5 \displaystyle s=g(h(s))=\frac3{\sin h(s)}-5 s = g ( h ( s )) = sin h ( s ) 3 − 5 s \displaystyle s s
1 = − 3 cos h ( s ) sin 2 h ( s ) h ′ ( s ) . \displaystyle 1=-\frac{3\cos h(s)}{\sin^2h(s)}h\prime(s). 1 = − sin 2 h ( s ) 3 cos h ( s ) h ′ ( s ) .
이제 4 = 3 sin ( h ( 4 ) ) − 5 \displaystyle 4=\frac3{\sin(h(4))}-5 4 = sin ( h ( 4 )) 3 − 5 sin ( h ( 4 ) ) = 1 3 \displaystyle \sin(h(4))=\frac13 sin ( h ( 4 )) = 3 1 cos ( h ( 4 ) ) = 2 2 3 \displaystyle \cos(h(4))=\frac{2\sqrt2}3 cos ( h ( 4 )) = 3 2 2
50점 · 답 h ′ ( 4 ) = − 1 18 2 \displaystyle h\prime(4)=-\frac1{18\sqrt2} h ′ ( 4 ) = − 18 2 1 h ′ ( 4 ) = − 1 18 2 \displaystyle h\prime(4)=-\frac1{18\sqrt2} h ′ ( 4 ) = − 18 2 1
(3-1) 반지름 1의 반원에 접하는 직선을 그렸거나 세 변의 길이가 1 , 1 , 2 \displaystyle 1,1,\sqrt2 1 , 1 , 2
(3-2) (합산으로 채점)
(3-2) 함수표현은 맞고 구간이 틀리면 (15점)
(3-2) 증명 없이 t = π 6 \displaystyle t=\frac\pi6 t = 6 π
(3-3) 답이 틀린 경우: 다음 두 부분점수의 합산으로 채점
(1) 1 = − 3 cos h ( s ) sin 2 h ( s ) h ′ ( s ) \displaystyle 1=-\frac{3\cos h(s)}{\sin^2h(s)}h\prime(s) 1 = − sin 2 h ( s ) 3 cos h ( s ) h ′ ( s ) h ′ ( s ) = − sin 2 h ( s ) 3 cos h ( s ) \displaystyle h\prime(s)=-\frac{\sin^2h(s)}{3\cos h(s)} h ′ ( s ) = − 3 cos h ( s ) sin 2 h ( s )
(별해)에서는 1 = h ′ ( g ( t ) ) ⋅ ( − 3 cos t sin 2 t ) \displaystyle 1=h\prime(g(t))\cdot\left(-\frac{3\cos t}{\sin^2t}\right) 1 = h ′ ( g ( t )) ⋅ ( − sin 2 t 3 cos t )
(2) sin ( h ( 4 ) ) = 1 3 \displaystyle \sin(h(4))=\frac13 sin ( h ( 4 )) = 3 1 cos ( h ( 4 ) ) = 2 2 3 \displaystyle \cos(h(4))=\frac{2\sqrt2}3 cos ( h ( 4 )) = 3 2 2
수리논술, 배움에서 논증의 완성까지.
