+20점 · 현 PQ의 길이 f ( t ) \displaystyle f(t) f ( t ) g ( t ) \displaystyle g(t) g ( t ) +10점 · 삼각함수의 극한에 대한 성질을 활용해서 lim t → 0 + g ( t ) f ( t ) \displaystyle \lim_{t\to0+}\frac{g(t)}{f(t)} t → 0 + lim f ( t ) g ( t ) 1. 오른쪽 그림에서 원의 중심 O에서 현 PQ에 내린 수선의 발을 H라 하자. f ( t ) = 2 Q H ‾ = 2 sin t 2 \displaystyle f(t)=2\overline{\mathrm{QH}}=2\sin\frac{t}{2} f ( t ) = 2 QH = 2 sin 2 t g ( t ) = 1 2 × 1 2 × t − 1 2 × 1 2 × sin t = 1 2 ( t − sin t ) \displaystyle g(t)=\frac{1}{2}\times1^2\times t-\frac{1}{2}\times1^2\times\sin t=\frac{1}{2}(t-\sin t) g ( t ) = 2 1 × 1 2 × t − 2 1 × 1 2 × sin t = 2 1 ( t − sin t ) 이므로
lim t → 0 + g ( t ) f ( t ) = lim t → 0 + 1 2 ( t − sin t ) 2 sin t 2 = lim t → 0 + 1 2 t 2 sin t 2 ( 1 − sin t t ) = 1 2 × 1 × ( 1 − 1 ) = 0 \displaystyle \lim_{t\to0+}\frac{g(t)}{f(t)}=\lim_{t\to0+}\frac{\frac{1}{2}(t-\sin t)}{2\sin\frac{t}{2}}=\lim_{t\to0+}\frac{1}{2}\frac{\frac{t}{2}}{\sin\frac{t}{2}}\left(1-\frac{\sin t}{t}\right)=\frac{1}{2}\times1\times(1-1)=0 t → 0 + lim f ( t ) g ( t ) = t → 0 + lim 2 sin 2 t 2 1 ( t − sin t ) = t → 0 + lim 2 1 sin 2 t 2 t ( 1 − t sin t ) = 2 1 × 1 × ( 1 − 1 ) = 0
+20점 · 무한급수 lim n → ∞ P 0 P 1 ‾ + P 0 P 2 ‾ + ⋯ + P 0 P n − 1 ‾ n \displaystyle \lim_{n\to\infty}\frac{\overline{\mathrm{P}_0\mathrm{P}_1}+\overline{\mathrm{P}_0\mathrm{P}_2}+\cdots+\overline{\mathrm{P}_0\mathrm{P}_{n-1}}}{n} n → ∞ lim n P 0 P 1 + P 0 P 2 + ⋯ + P 0 P n − 1 +10점 · 정적분의 값을 구했는가? 2. 1 \displaystyle 1 1 f ( t ) \displaystyle f(t) f ( t ) P 0 P 1 ‾ = f ( 2 π n ) \displaystyle \overline{P_0P_1}=f\left(\frac{2\pi}{n}\right) P 0 P 1 = f ( n 2 π ) P 0 P 2 ‾ = f ( 2 π n ⋅ 2 ) \displaystyle \overline{P_0P_2}=f\left(\frac{2\pi}{n}\cdot2\right) P 0 P 2 = f ( n 2 π ⋅ 2 ) ⋯ \displaystyle \cdots ⋯ P 0 P n − 1 ‾ = f ( 2 π n ⋅ ( n − 1 ) ) \displaystyle \overline{P_0P_{n-1}}=f\left(\frac{2\pi}{n}\cdot(n-1)\right) P 0 P n − 1 = f ( n 2 π ⋅ ( n − 1 ) ) f ( 2 π n ⋅ n ) = f ( 2 π ) = 0 \displaystyle f\left(\frac{2\pi}{n}\cdot n\right)=f(2\pi)=0 f ( n 2 π ⋅ n ) = f ( 2 π ) = 0
lim n → ∞ P 0 P 1 ‾ + ⋯ + P 0 P n − 1 ‾ n = lim n → ∞ ∑ k = 1 n − 1 f ( 2 π n ⋅ k ) 1 n = lim n → ∞ ( ∑ k = 1 n − 1 f ( 2 π n ⋅ k ) 1 n + f ( 2 π n ⋅ n ) 1 n ) = lim n → ∞ ∑ k = 1 n f ( 2 π n ⋅ k ) 1 n = lim n → ∞ ∑ k = 1 n f ( 0 + 2 π − 0 n ⋅ k ) 2 π − 0 n ⋅ 1 2 π = 1 2 π ∫ 0 2 π f ( t ) d t \displaystyle \begin{aligned}\lim_{n\to\infty}\frac{\overline{P_0P_1}+\cdots+\overline{P_0P_{n-1}}}{n}&=\lim_{n\to\infty}\sum_{k=1}^{n-1}f\left(\frac{2\pi}{n}\cdot k\right)\frac{1}{n}\\&=\lim_{n\to\infty}\left(\sum_{k=1}^{n-1}f\left(\frac{2\pi}{n}\cdot k\right)\frac{1}{n}+f\left(\frac{2\pi}{n}\cdot n\right)\frac{1}{n}\right)\\&=\lim_{n\to\infty}\sum_{k=1}^n f\left(\frac{2\pi}{n}\cdot k\right)\frac{1}{n}\\&=\lim_{n\to\infty}\sum_{k=1}^n f\left(0+\frac{2\pi-0}{n}\cdot k\right)\frac{2\pi-0}{n}\cdot\frac{1}{2\pi}=\frac{1}{2\pi}\int_0^{2\pi}f(t)\,dt\end{aligned} n → ∞ lim n P 0 P 1 + ⋯ + P 0 P n − 1 = n → ∞ lim k = 1 ∑ n − 1 f ( n 2 π ⋅ k ) n 1 = n → ∞ lim ( k = 1 ∑ n − 1 f ( n 2 π ⋅ k ) n 1 + f ( n 2 π ⋅ n ) n 1 ) = n → ∞ lim k = 1 ∑ n f ( n 2 π ⋅ k ) n 1 = n → ∞ lim k = 1 ∑ n f ( 0 + n 2 π − 0 ⋅ k ) n 2 π − 0 ⋅ 2 π 1 = 2 π 1 ∫ 0 2 π f ( t ) d t 이고, f ( t ) = 2 sin t 2 \displaystyle f(t)=2\sin\frac{t}{2} f ( t ) = 2 sin 2 t 1 2 π ∫ 0 2 π f ( t ) d t = 4 π \displaystyle \frac{1}{2\pi}\int_0^{2\pi}f(t)\,dt=\frac{4}{\pi} 2 π 1 ∫ 0 2 π f ( t ) d t = π 4
+20점 · 무한급수 lim n → ∞ S 1 2 + S 2 2 + ⋯ + S n − 1 2 n \displaystyle \lim_{n\to\infty}\frac{S_1^2+S_2^2+\cdots+S_{n-1}^2}{n} n → ∞ lim n S 1 2 + S 2 2 + ⋯ + S n − 1 2 +20점 · 정적분의 값을 구했는가? 3. 1 \displaystyle 1 1 g ( t ) \displaystyle g(t) g ( t ) S 1 = g ( 2 π n ) \displaystyle S_1=g\left(\frac{2\pi}{n}\right) S 1 = g ( n 2 π ) S 2 = g ( 2 π n ⋅ 2 ) \displaystyle S_2=g\left(\frac{2\pi}{n}\cdot2\right) S 2 = g ( n 2 π ⋅ 2 ) ⋯ \displaystyle \cdots ⋯ S n − 1 = g ( 2 π n ⋅ ( n − 1 ) ) \displaystyle S_{n-1}=g\left(\frac{2\pi}{n}\cdot(n-1)\right) S n − 1 = g ( n 2 π ⋅ ( n − 1 ) ) g ( 2 π n ⋅ n ) = g ( 2 π ) = π \displaystyle g\left(\frac{2\pi}{n}\cdot n\right)=g(2\pi)=\pi g ( n 2 π ⋅ n ) = g ( 2 π ) = π
lim n → ∞ S 1 2 + ⋯ + S n − 1 2 n = lim n → ∞ ∑ k = 1 n − 1 ( g ( 2 π n ⋅ k ) ) 2 1 n \displaystyle \begin{aligned}\lim_{n\to\infty}\frac{S_1^2+\cdots+S_{n-1}^2}{n}&=\lim_{n\to\infty}\sum_{k=1}^{n-1}\left(g\left(\frac{2\pi}{n}\cdot k\right)\right)^2\frac{1}{n}\end{aligned} n → ∞ lim n S 1 2 + ⋯ + S n − 1 2 = n → ∞ lim k = 1 ∑ n − 1 ( g ( n 2 π ⋅ k ) ) 2 n 1 = lim n → ∞ ( ∑ k = 1 n ( g ( 2 π n ⋅ k ) ) 2 1 n − ( g ( 2 π n ⋅ n ) ) 2 1 n ) \displaystyle \begin{aligned}&=\lim_{n\to\infty}\left(\sum_{k=1}^n\left(g\left(\frac{2\pi}{n}\cdot k\right)\right)^2\frac{1}{n}-\left(g\left(\frac{2\pi}{n}\cdot n\right)\right)^2\frac{1}{n}\right)\end{aligned} = n → ∞ lim ( k = 1 ∑ n ( g ( n 2 π ⋅ k ) ) 2 n 1 − ( g ( n 2 π ⋅ n ) ) 2 n 1 ) = lim n → ∞ ( ∑ k = 1 n ( g ( 0 + 2 π − 0 n ⋅ k ) ) 2 2 π − 0 n ⋅ 1 2 π ) − lim n → ∞ π 2 n \displaystyle \begin{aligned}&=\lim_{n\to\infty}\left(\sum_{k=1}^n\left(g\left(0+\frac{2\pi-0}{n}\cdot k\right)\right)^2\frac{2\pi-0}{n}\cdot\frac{1}{2\pi}\right)-\lim_{n\to\infty}\frac{\pi^2}{n}\end{aligned} = n → ∞ lim ( k = 1 ∑ n ( g ( 0 + n 2 π − 0 ⋅ k ) ) 2 n 2 π − 0 ⋅ 2 π 1 ) − n → ∞ lim n π 2 = 1 2 π ∫ 0 2 π ( g ( t ) ) 2 d t − 0 = 1 2 π ∫ 0 2 π ( 1 2 ( t − sin t ) ) 2 d t \displaystyle \begin{aligned}&=\frac{1}{2\pi}\int_0^{2\pi}(g(t))^2\,dt-0=\frac{1}{2\pi}\int_0^{2\pi}\left(\frac{1}{2}(t-\sin t)\right)^2\,dt\end{aligned} = 2 π 1 ∫ 0 2 π ( g ( t ) ) 2 d t − 0 = 2 π 1 ∫ 0 2 π ( 2 1 ( t − sin t ) ) 2 d t = 1 8 π ∫ 0 2 π ( t 2 − 2 t sin t + sin 2 t ) d t \displaystyle \begin{aligned}&=\frac{1}{8\pi}\int_0^{2\pi}(t^2-2t\sin t+\sin^2t)\,dt\end{aligned} = 8 π 1 ∫ 0 2 π ( t 2 − 2 t sin t + sin 2 t ) d t 부분적분에 의해 ∫ t sin t d t = − t cos t + sin t + C 1 \displaystyle \int t\sin t\,dt=-t\cos t+\sin t+C_1 ∫ t sin t d t = − t cos t + sin t + C 1 ∫ sin 2 t d t = 1 2 ( t − cos t sin t ) + C 2 \displaystyle \int\sin^2t\,dt=\frac{1}{2}(t-\cos t\sin t)+C_2 ∫ sin 2 t d t = 2 1 ( t − cos t sin t ) + C 2 를 구하고, 따라서 1 8 π ∫ 0 2 π ( t 2 − 2 t sin t + sin 2 t ) d t = 1 8 π [ 1 3 t 3 − 2 ( − t cos t + 2 sin t ) + 1 2 ( t − cos t sin t ) ] 0 2 π \displaystyle \begin{aligned}\frac{1}{8\pi}\int_0^{2\pi}(t^2-2t\sin t+\sin^2t)\,dt&=\frac{1}{8\pi}\Biggl[\frac{1}{3}t^3-2(-t\cos t+2\sin t)+\frac{1}{2}(t-\cos t\sin t)\Biggr]_0^{2\pi}\end{aligned} 8 π 1 ∫ 0 2 π ( t 2 − 2 t sin t + sin 2 t ) d t = 8 π 1 [ 3 1 t 3 − 2 ( − t cos t + 2 sin t ) + 2 1 ( t − cos t sin t ) ] 0 2 π Correction. 대괄호 안의 2 sin t \displaystyle 2\sin t 2 sin t sin t \displaystyle \sin t sin t = 1 8 π ( 8 3 π 3 + 5 π ) = π 2 3 + 5 8 \displaystyle \begin{aligned}&=\frac{1}{8\pi}\left(\frac{8}{3}\pi^3+5\pi\right)=\frac{\pi^2}{3}+\frac{5}{8}\end{aligned} = 8 π 1 ( 3 8 π 3 + 5 π ) = 3 π 2 + 8 5
수리논술, 배움에서 논증의 완성까지.
