+2점 · g ( t ) = sin ( n l t ) \displaystyle g(t)=\sin\left(\frac nl t\right) g ( t ) = sin ( l n t ) g ′ ( t ) = n l cos ( n l t ) \displaystyle g^{\prime}(t)=\frac nl\cos\left(\frac nl t\right) g ′ ( t ) = l n cos ( l n t ) [문제 1-1] (25점) (1) g ( t ) = sin ( n l t ) \displaystyle g(t)=\sin\left(\frac nl t\right) g ( t ) = sin ( l n t ) g ′ ( t ) = n l cos ( n l t ) \displaystyle g^{\prime}(t)=\frac nl\cos\left(\frac nl t\right) g ′ ( t ) = l n cos ( l n t )
+3점 · ∣ g ′ ( t ) ∣ 2 = 1 2 ( n l ) 2 ( 1 + cos ( 2 n l t ) ) \displaystyle |g^{\prime}(t)|^2=\frac12\left(\frac nl\right)^2\left(1+\cos\left(\frac{2n}l t\right)\right) ∣ g ′ ( t ) ∣ 2 = 2 1 ( l n ) 2 ( 1 + cos ( l 2 n t ) ) 이것을 제곱하고 cos 2 θ = 1 + cos ( 2 θ ) 2 \displaystyle \cos^2\theta=\frac{1+\cos(2\theta)}2 cos 2 θ = 2 1 + cos ( 2 θ ) ∣ g ′ ( t ) ∣ 2 = 1 2 ( n l ) 2 ( 1 + cos ( 2 n l t ) ) \displaystyle |g^{\prime}(t)|^2=\frac12\left(\frac nl\right)^2\left(1+\cos\left(\frac{2n}l t\right)\right) ∣ g ′ ( t ) ∣ 2 = 2 1 ( l n ) 2 ( 1 + cos ( l 2 n t ) ) 이다.
+5점 · E = ∫ 0 2 π ∣ g ′ ( t ) ∣ 2 d t \displaystyle E=\int_0^{2\pi}|g^{\prime}(t)|^2\,dt E = ∫ 0 2 π ∣ g ′ ( t ) ∣ 2 d t = ( n l ) 2 π + n 4 l sin ( 4 n π l ) \displaystyle =\left(\frac nl\right)^2\pi+\frac n{4l}\sin\left(\frac{4n\pi}l\right) = ( l n ) 2 π + 4 l n sin ( l 4 nπ ) 따라서 E = ∫ 0 2 π ∣ g ′ ( t ) ∣ 2 d t = 1 2 ( n l ) 2 { 2 π + l 2 n sin ( 2 n l 2 π ) } = ( n l ) 2 π + n 4 l sin ( 4 n π l ) \displaystyle E=\int_0^{2\pi}|g^{\prime}(t)|^2\,dt=\frac12\left(\frac nl\right)^2\left\{2\pi+\frac l{2n}\sin\left(\frac{2n}l2\pi\right)\right\}=\left(\frac nl\right)^2\pi+\frac n{4l}\sin\left(\frac{4n\pi}l\right) E = ∫ 0 2 π ∣ g ′ ( t ) ∣ 2 d t = 2 1 ( l n ) 2 { 2 π + 2 n l sin ( l 2 n 2 π ) } = ( l n ) 2 π + 4 l n sin ( l 4 nπ ) 이다.
+5점 · n l < 1 \displaystyle \frac nl<1 l n < 1 ( n , l ) \displaystyle (n,l) ( n , l ) E < π \displaystyle E<\pi E < π (2) 다음과 같은 세 가지 경우를 고려하면 된다. 1) n l < 1 \displaystyle \frac nl<1 l n < 1 sin θ ≤ 1 \displaystyle \sin\theta\le1 sin θ ≤ 1 E = ( n l ) 2 π + n 4 l sin ( 4 n π l ) ≤ ( n l ) 2 π + n 4 l < { ( n l ) 2 + 1 4 n l } π \displaystyle E=\left(\frac nl\right)^2\pi+\frac n{4l}\sin\left(\frac{4n\pi}l\right)\le\left(\frac nl\right)^2\pi+\frac n{4l}<\left\{\left(\frac nl\right)^2+\frac14\frac nl\right\}\pi E = ( l n ) 2 π + 4 l n sin ( l 4 nπ ) ≤ ( l n ) 2 π + 4 l n < { ( l n ) 2 + 4 1 l n } π 를 관찰할 수 있다. 이와 관련하여 함수 h ( t ) = t 2 + 1 4 t \displaystyle h(t)=t^2+\frac14t h ( t ) = t 2 + 4 1 t h ( t ) \displaystyle h(t) h ( t ) t > 0 \displaystyle t>0 t > 0 n l < 1 \displaystyle \frac nl<1 l n < 1 n l \displaystyle \frac nl l n ( n , l ) = ( 5 , 6 ) \displaystyle (n,l)=(5,6) ( n , l ) = ( 5 , 6 ) h ( 5 6 ) < 1 \displaystyle h\left(\frac56\right)<1 h ( 6 5 ) < 1 n l < 1 \displaystyle \frac nl<1 l n < 1 ( n , l ) \displaystyle (n,l) ( n , l ) E < π \displaystyle E<\pi E < π
+5점 · E = ( n l ) 2 π + n 4 l sin 4 n π l = ( n l ) 2 π ≥ π \displaystyle E=\left(\frac nl\right)^2\pi+\frac n{4l}\sin\frac{4n\pi}l=\left(\frac nl\right)^2\pi\ge\pi E = ( l n ) 2 π + 4 l n sin l 4 nπ = ( l n ) 2 π ≥ π 2) n l ≥ 1 \displaystyle \frac nl\ge1 l n ≥ 1 n l \displaystyle \frac nl l n a 2 \displaystyle \frac a2 2 a a 4 \displaystyle \frac a4 4 a sin ( 4 n π l ) = 0 \displaystyle \sin\left(\frac{4n\pi}l\right)=0 sin ( l 4 nπ ) = 0 E = ( n l ) 2 π + n 4 l sin 4 n π l = ( n l ) 2 π ≥ π \displaystyle E=\left(\frac nl\right)^2\pi+\frac n{4l}\sin\frac{4n\pi}l=\left(\frac nl\right)^2\pi\ge\pi E = ( l n ) 2 π + 4 l n sin l 4 nπ = ( l n ) 2 π ≥ π 이 성립한다.
+5점 · 답은 n l ≥ 1 \displaystyle \frac nl\ge1 l n ≥ 1 ( n , l ) \displaystyle (n,l) ( n , l ) 3) n l ≥ 1 \displaystyle \frac nl\ge1 l n ≥ 1 n l \displaystyle \frac nl l n a 3 \displaystyle \frac a3 3 a a 5 \displaystyle \frac a5 5 a 이 경우에는 가능한 순서쌍이 다음 세 가지뿐이다. ( n , l ) = ( 4 , 3 ) , ( 5 , 3 ) , ( 6 , 5 ) \displaystyle (n,l)=(4,3),(5,3),(6,5) ( n , l ) = ( 4 , 3 ) , ( 5 , 3 ) , ( 6 , 5 ) − 1 ≤ sin θ \displaystyle -1\le\sin\theta − 1 ≤ sin θ E \displaystyle E E ( n , l ) = ( 4 , 3 ) : E = ( 4 3 ) 2 π + 1 3 sin 16 π 3 > 16 9 π − 1 3 = π + 7 9 π − 1 3 > π \displaystyle (n,l)=(4,3):\ E=\left(\frac43\right)^2\pi+\frac13\sin\frac{16\pi}3>\frac{16}9\pi-\frac13=\pi+\frac79\pi-\frac13>\pi ( n , l ) = ( 4 , 3 ) : E = ( 3 4 ) 2 π + 3 1 sin 3 16 π > 9 16 π − 3 1 = π + 9 7 π − 3 1 > π ( n , l ) = ( 5 , 3 ) : E = ( 5 3 ) 2 π + 5 12 sin 20 π 3 > 25 9 π − 5 12 = π + 16 9 π − 5 12 > π \displaystyle (n,l)=(5,3):\ E=\left(\frac53\right)^2\pi+\frac5{12}\sin\frac{20\pi}3>\frac{25}9\pi-\frac5{12}=\pi+\frac{16}9\pi-\frac5{12}>\pi ( n , l ) = ( 5 , 3 ) : E = ( 3 5 ) 2 π + 12 5 sin 3 20 π > 9 25 π − 12 5 = π + 9 16 π − 12 5 > π ( n , l ) = ( 6 , 5 ) : E = ( 6 5 ) 2 π + 3 10 sin 24 π 5 > 36 25 π − 3 10 = π + 11 25 π − 3 10 > π \displaystyle (n,l)=(6,5):\ E=\left(\frac65\right)^2\pi+\frac3{10}\sin\frac{24\pi}5>\frac{36}{25}\pi-\frac3{10}=\pi+\frac{11}{25}\pi-\frac3{10}>\pi ( n , l ) = ( 6 , 5 ) : E = ( 5 6 ) 2 π + 10 3 sin 5 24 π > 25 36 π − 10 3 = π + 25 11 π − 10 3 > π 그러므로 답은 n l ≥ 1 \displaystyle \frac nl\ge1 l n ≥ 1 ( n , l ) \displaystyle (n,l) ( n , l )
+5점 · 곡선의 길이는 모두 e − e − 1 2 \displaystyle \frac{e-e^{-1}}2 2 e − e − 1 [문제 1-2] (25점) (1) A ( 0 , f ( 0 ) ) \displaystyle \mathrm{A}(0,f(0)) A ( 0 , f ( 0 )) ( s , f ( s ) ) \displaystyle (s,f(s)) ( s , f ( s )) G ( s ) \displaystyle G(s) G ( s ) s ≥ 0 \displaystyle s\ge0 s ≥ 0 G ( s ) = ∫ 0 s 1 + f ′ ( x ) 2 d x = ∫ 0 s 1 + ( e x − e − x 2 ) 2 d x = ∫ 0 s e x + e − x 2 d x = e s − e − s 2 \displaystyle \begin{aligned}G(s)&=\int_0^s\sqrt{1+f^{\prime}(x)^2}\,dx=\int_0^s\sqrt{1+\left(\frac{e^x-e^{-x}}2\right)^2}\,dx\\&=\int_0^s\frac{e^x+e^{-x}}2\,dx=\frac{e^s-e^{-s}}2\end{aligned} G ( s ) = ∫ 0 s 1 + f ′ ( x ) 2 d x = ∫ 0 s 1 + ( 2 e x − e − x ) 2 d x = ∫ 0 s 2 e x + e − x d x = 2 e s − e − s 이고, s ≤ 0 \displaystyle s\le0 s ≤ 0 G ( s ) = e − s − e s 2 \displaystyle G(s)=\frac{e^{-s}-e^s}2 G ( s ) = 2 e − s − e s 이다. 특히 A ( 0 , f ( 0 ) ) \displaystyle \mathrm{A}(0,f(0)) A ( 0 , f ( 0 )) B ( 1 , f ( 1 ) ) \displaystyle \mathrm{B}(1,f(1)) B ( 1 , f ( 1 )) A ( 0 , f ( 0 ) ) \displaystyle \mathrm{A}(0,f(0)) A ( 0 , f ( 0 )) C ( − 1 , f ( − 1 ) ) \displaystyle \mathrm{C}(-1,f(-1)) C ( − 1 , f ( − 1 )) e − e − 1 2 \displaystyle \frac{e-e^{-1}}2 2 e − e − 1
+5점 · k \displaystyle k k a \displaystyle a a a = 0 \displaystyle a=0 a = 0 a = 1 \displaystyle a=1 a = 1 a = 2 \displaystyle a=2 a = 2 a = 3 \displaystyle a=3 a = 3 0 ≤ t ≤ 2 π \displaystyle 0\le t\le2\pi 0 ≤ t ≤ 2 π 0 ≤ n l t ≤ 2 n π l \displaystyle 0\le\frac nl t\le\frac{2n\pi}l 0 ≤ l n t ≤ l 2 nπ n l t \displaystyle \frac nl t l n t π 2 \displaystyle \frac\pi2 2 π n l 2 π π 2 = 4 n l \displaystyle \frac{\frac nl2\pi}{\frac\pi2}=\frac{4n}l 2 π l n 2 π = l 4 n k ≤ 4 n l < k + 1 \displaystyle k\le\frac{4n}l<k+1 k ≤ l 4 n < k + 1 k \displaystyle k k k \displaystyle k k k \displaystyle k k a \displaystyle a a - a = 0 \displaystyle a=0 a = 0 L = e − e − 1 2 k + e sin 2 n π l − e − sin 2 n π l 2 \displaystyle L=\frac{e-e^{-1}}2k+\frac{e^{\sin\frac{2n\pi}l}-e^{-\sin\frac{2n\pi}l}}2 L = 2 e − e − 1 k + 2 e s i n l 2 nπ − e − s i n l 2 nπ - a = 1 \displaystyle a=1 a = 1 L = e − e − 1 2 k + e − e − 1 2 − e sin 2 n π l − e − sin 2 n π l 2 = e − e − 1 2 ( k + 1 ) − e sin 2 n π l − e − sin 2 n π l 2 \displaystyle L=\frac{e-e^{-1}}2k+\frac{e-e^{-1}}2-\frac{e^{\sin\frac{2n\pi}l}-e^{-\sin\frac{2n\pi}l}}2=\frac{e-e^{-1}}2(k+1)-\frac{e^{\sin\frac{2n\pi}l}-e^{-\sin\frac{2n\pi}l}}2 L = 2 e − e − 1 k + 2 e − e − 1 − 2 e s i n l 2 nπ − e − s i n l 2 nπ = 2 e − e − 1 ( k + 1 ) − 2 e s i n l 2 nπ − e − s i n l 2 nπ - a = 2 \displaystyle a=2 a = 2 L = e − e − 1 2 k + e − sin 2 n π l − e sin 2 n π l 2 = e − e − 1 2 k − e sin 2 n π l − e − sin 2 n π l 2 \displaystyle L=\frac{e-e^{-1}}2k+\frac{e^{-\sin\frac{2n\pi}l}-e^{\sin\frac{2n\pi}l}}2=\frac{e-e^{-1}}2k-\frac{e^{\sin\frac{2n\pi}l}-e^{-\sin\frac{2n\pi}l}}2 L = 2 e − e − 1 k + 2 e − s i n l 2 nπ − e s i n l 2 nπ = 2 e − e − 1 k − 2 e s i n l 2 nπ − e − s i n l 2 nπ - a = 3 \displaystyle a=3 a = 3 L = e − e − 1 2 k + e − e − 1 2 − e − sin 2 n π l − e sin 2 n π l 2 = e − e − 1 2 ( k + 1 ) + e sin 2 n π l − e − sin 2 n π l 2 \displaystyle L=\frac{e-e^{-1}}2k+\frac{e-e^{-1}}2-\frac{e^{-\sin\frac{2n\pi}l}-e^{\sin\frac{2n\pi}l}}2=\frac{e-e^{-1}}2(k+1)+\frac{e^{\sin\frac{2n\pi}l}-e^{-\sin\frac{2n\pi}l}}2 L = 2 e − e − 1 k + 2 e − e − 1 − 2 e − s i n l 2 nπ − e s i n l 2 nπ = 2 e − e − 1 ( k + 1 ) + 2 e s i n l 2 nπ − e − s i n l 2 nπ
+2점 · k = 0 , 1 , 2 , 3 , 4 , 5 , 6 , 8 , 10 , 12 , 16 , 20 , 24 \displaystyle k=0,1,2,3,4,5,6,8,10,12,16,20,24 k = 0 , 1 , 2 , 3 , 4 , 5 , 6 , 8 , 10 , 12 , 16 , 20 , 24 한편 n , l \displaystyle n,l n , l k \displaystyle k k k = 0 , 1 , 2 , 3 , 4 , 5 , 6 , 8 , 10 , 12 , 16 , 20 , 24 \displaystyle k=0,1,2,3,4,5,6,8,10,12,16,20,24 k = 0 , 1 , 2 , 3 , 4 , 5 , 6 , 8 , 10 , 12 , 16 , 20 , 24 임을 알 수 있다.
+3점 · L = { e − e − 1 2 k + ( − 1 ) k 2 e sin 2 n π l − e − sin 2 n π l 2 ( k = 0 , 2 , 4 , 6 , 8 , 10 , 12 , 16 , 20 , 24 ) e − e − 1 2 ( k + 1 ) + ( − 1 ) k + 1 2 e sin 2 n π l − e − sin 2 n π l 2 ( k = 1 , 3 , 5 ) \displaystyle L=\begin{cases}\frac{e-e^{-1}}2k+(-1)^{\frac k2}\frac{e^{\sin\frac{2n\pi}l}-e^{-\sin\frac{2n\pi}l}}2&(k=0,2,4,6,8,10,12,16,20,24)\\\frac{e-e^{-1}}2(k+1)+(-1)^{\frac{k+1}2}\frac{e^{\sin\frac{2n\pi}l}-e^{-\sin\frac{2n\pi}l}}2&(k=1,3,5)\end{cases} L = ⎩ ⎨ ⎧ 2 e − e − 1 k + ( − 1 ) 2 k 2 e s i n l 2 nπ − e − s i n l 2 nπ 2 e − e − 1 ( k + 1 ) + ( − 1 ) 2 k + 1 2 e s i n l 2 nπ − e − s i n l 2 nπ ( k = 0 , 2 , 4 , 6 , 8 , 10 , 12 , 16 , 20 , 24 ) ( k = 1 , 3 , 5 ) 따라서 답은 다음과 같다. L = { e − e − 1 2 k + ( − 1 ) k 2 e sin 2 n π l − e − sin 2 n π l 2 ( k = 0 , 2 , 4 , 6 , 8 , 10 , 12 , 16 , 20 , 24 ) e − e − 1 2 ( k + 1 ) + ( − 1 ) k + 1 2 e sin 2 n π l − e − sin 2 n π l 2 ( k = 1 , 3 , 5 ) \displaystyle L=\begin{cases}\frac{e-e^{-1}}2k+(-1)^{\frac k2}\frac{e^{\sin\frac{2n\pi}l}-e^{-\sin\frac{2n\pi}l}}2&(k=0,2,4,6,8,10,12,16,20,24)\\\frac{e-e^{-1}}2(k+1)+(-1)^{\frac{k+1}2}\frac{e^{\sin\frac{2n\pi}l}-e^{-\sin\frac{2n\pi}l}}2&(k=1,3,5)\end{cases} L = ⎩ ⎨ ⎧ 2 e − e − 1 k + ( − 1 ) 2 k 2 e s i n l 2 nπ − e − s i n l 2 nπ 2 e − e − 1 ( k + 1 ) + ( − 1 ) 2 k + 1 2 e s i n l 2 nπ − e − s i n l 2 nπ ( k = 0 , 2 , 4 , 6 , 8 , 10 , 12 , 16 , 20 , 24 ) ( k = 1 , 3 , 5 )
+5점 · n l \displaystyle \frac nl l n L \displaystyle L L (2) n l \displaystyle \frac nl l n L \displaystyle L L
+3점 · 이때 L = e − e − 1 \displaystyle L=e-e^{-1} L = e − e − 1 n l = 1 2 \displaystyle \frac nl=\frac12 l n = 2 1 A ( 0 , 1 ) \displaystyle \mathrm{A}(0,1) A ( 0 , 1 ) B ( 1 , f ( 1 ) ) \displaystyle \mathrm{B}(1,f(1)) B ( 1 , f ( 1 )) A ( 0 , 1 ) \displaystyle \mathrm{A}(0,1) A ( 0 , 1 ) L = e − e − 1 \displaystyle L=e-e^{-1} L = e − e − 1
+2점 · L ≥ e − e − 1 \displaystyle L\ge e-e^{-1} L ≥ e − e − 1 ( n , l ) \displaystyle (n,l) ( n , l ) 2 n ≥ l \displaystyle 2n\ge l 2 n ≥ l ( n , l ) \displaystyle (n,l) ( n , l ) 따라서, L ≥ e − e − 1 \displaystyle L\ge e-e^{-1} L ≥ e − e − 1 ( n , l ) \displaystyle (n,l) ( n , l ) 2 n ≥ l \displaystyle 2n\ge l 2 n ≥ l ( n , l ) \displaystyle (n,l) ( n , l )
수리논술, 배움에서 논증의 완성까지.