【3-1】 d x d t = − n cos n − 1 t sin t \displaystyle \dfrac{dx}{dt}=-n\cos^{n-1}t\sin t d t d x = − n cos n − 1 t sin t d y d t = n sin n − 1 t cos t \displaystyle \dfrac{dy}{dt}=n\sin^{n-1}t\cos t d t d y = n sin n − 1 t cos t s n = ∫ 0 π / 2 ( − n cos n − 1 t sin t ) 2 + ( n sin n − 1 t cos t ) 2 d t = n ∫ 0 π / 2 sin t cos t cos 2 n − 4 t + sin 2 n − 4 t d t \displaystyle s_n=\int_0^{\pi/2}\sqrt{(-n\cos^{n-1}t\sin t)^2+(n\sin^{n-1}t\cos t)^2}\,dt=n\int_0^{\pi/2}\sin t\cos t\sqrt{\cos^{2n-4}t+\sin^{2n-4}t}\,dt s n = ∫ 0 π /2 ( − n cos n − 1 t sin t ) 2 + ( n sin n − 1 t cos t ) 2 d t = n ∫ 0 π /2 sin t cos t cos 2 n − 4 t + sin 2 n − 4 t d t 임을 알 수 있다. 따라서,
+15점 · s 2 = 2 \displaystyle s_2=\sqrt2 s 2 = 2 +15점 · s 3 = 3 2 \displaystyle s_3=\dfrac32 s 3 = 2 3 s 2 = 2 ∫ 0 π / 2 sin t cos t 1 + 1 d t = 2 2 ∫ 0 π / 2 sin t cos t d t = 2 , s 3 = 3 ∫ 0 π / 2 sin t cos t cos 2 t + sin 2 t d t = 3 ∫ 0 π / 2 sin t cos t d t = 3 2 \displaystyle \begin{aligned}s_2&=2\int_0^{\pi/2}\sin t\cos t\sqrt{1+1}\,dt=2\sqrt2\int_0^{\pi/2}\sin t\cos t\,dt=\sqrt2,\\s_3&=3\int_0^{\pi/2}\sin t\cos t\sqrt{\cos^2t+\sin^2t}\,dt=3\int_0^{\pi/2}\sin t\cos t\,dt=\dfrac32\end{aligned} s 2 s 3 = 2 ∫ 0 π /2 sin t cos t 1 + 1 d t = 2 2 ∫ 0 π /2 sin t cos t d t = 2 , = 3 ∫ 0 π /2 sin t cos t cos 2 t + sin 2 t d t = 3 ∫ 0 π /2 sin t cos t d t = 2 3 이다.
+15점 · π 2 − t = u \displaystyle \dfrac\pi2-t=u 2 π − t = u +15점 · a n = b n \displaystyle a_n=b_n a n = b n 【3-2】 임의의 실수 t \displaystyle t t sin ( π 2 − t ) = cos t \displaystyle \sin\left(\dfrac\pi2-t\right)=\cos t sin ( 2 π − t ) = cos t cos ( π 2 − t ) = sin t \displaystyle \cos\left(\dfrac\pi2-t\right)=\sin t cos ( 2 π − t ) = sin t π 2 − t = u \displaystyle \dfrac\pi2-t=u 2 π − t = u a n = n ∫ 0 π / 4 sin t cos t cos 2 n − 4 t + sin 2 n − 4 t d t = n ∫ π / 4 π / 2 cos u sin u sin 2 n − 4 u + cos 2 n − 4 u d u = b n \displaystyle a_n=n\int_0^{\pi/4}\sin t\cos t\sqrt{\cos^{2n-4}t+\sin^{2n-4}t}\,dt=n\int_{\pi/4}^{\pi/2}\cos u\sin u\sqrt{\sin^{2n-4}u+\cos^{2n-4}u}\,du=b_n a n = n ∫ 0 π /4 sin t cos t cos 2 n − 4 t + sin 2 n − 4 t d t = n ∫ π /4 π /2 cos u sin u sin 2 n − 4 u + cos 2 n − 4 u d u = b n 임을 알 수 있다.
【3-3】 s n = n ∫ 0 π / 2 sin t cos t cos 2 n − 4 t + sin 2 n − 4 t d t = n ∫ 0 π / 4 sin t cos t cos 2 n − 4 t + sin 2 n − 4 t d t + n ∫ π / 4 π / 2 sin t cos t cos 2 n − 4 t + sin 2 n − 4 t d t = a n + b n \displaystyle \begin{aligned}s_n&=n\int_0^{\pi/2}\sin t\cos t\sqrt{\cos^{2n-4}t+\sin^{2n-4}t}\,dt\\&=n\int_0^{\pi/4}\sin t\cos t\sqrt{\cos^{2n-4}t+\sin^{2n-4}t}\,dt+n\int_{\pi/4}^{\pi/2}\sin t\cos t\sqrt{\cos^{2n-4}t+\sin^{2n-4}t}\,dt=a_n+b_n\end{aligned} s n = n ∫ 0 π /2 sin t cos t cos 2 n − 4 t + sin 2 n − 4 t d t = n ∫ 0 π /4 sin t cos t cos 2 n − 4 t + sin 2 n − 4 t d t + n ∫ π /4 π /2 sin t cos t cos 2 n − 4 t + sin 2 n − 4 t d t = a n + b n 이고 a n = b n \displaystyle a_n=b_n a n = b n s n = 2 a n \displaystyle s_n=2a_n s n = 2 a n
+20점 · 부등식 n ∫ 0 π / 4 sin t cos t ( cos n − 2 t − sin n − 2 t ) d t ≤ a n = n ∫ 0 π / 4 sin t cos t cos 2 n − 4 t + sin 2 n − 4 t d t ≤ n ∫ 0 π / 4 sin t cos t ( cos n − 2 t + sin n − 2 t ) d t \displaystyle n\int_0^{\pi/4}\sin t\cos t(\cos^{n-2}t-\sin^{n-2}t)\,dt\le a_n=n\int_0^{\pi/4}\sin t\cos t\sqrt{\cos^{2n-4}t+\sin^{2n-4}t}\,dt\le n\int_0^{\pi/4}\sin t\cos t(\cos^{n-2}t+\sin^{n-2}t)\,dt n ∫ 0 π /4 sin t cos t ( cos n − 2 t − sin n − 2 t ) d t ≤ a n = n ∫ 0 π /4 sin t cos t cos 2 n − 4 t + sin 2 n − 4 t d t ≤ n ∫ 0 π /4 sin t cos t ( cos n − 2 t + sin n − 2 t ) d t 한편, 0 ≤ t ≤ π 4 \displaystyle 0\le t\le\dfrac\pi4 0 ≤ t ≤ 4 π t \displaystyle t t cos n − 2 t − sin n − 2 t ≤ cos 2 n − 4 t + sin 2 n − 4 t ≤ cos n − 2 t + sin n − 2 t \displaystyle \cos^{n-2}t-\sin^{n-2}t\le\sqrt{\cos^{2n-4}t+\sin^{2n-4}t}\le\cos^{n-2}t+\sin^{n-2}t cos n − 2 t − sin n − 2 t ≤ cos 2 n − 4 t + sin 2 n − 4 t ≤ cos n − 2 t + sin n − 2 t 가 성립하므로, n ∫ 0 π / 4 sin t cos t ( cos n − 2 t − sin n − 2 t ) d t ≤ a n = n ∫ 0 π / 4 sin t cos t cos 2 n − 4 t + sin 2 n − 4 t d t ≤ n ∫ 0 π / 4 sin t cos t ( cos n − 2 t + sin n − 2 t ) d t \displaystyle \begin{aligned}n\int_0^{\pi/4}\sin t\cos t(\cos^{n-2}t-\sin^{n-2}t)\,dt&\le a_n=n\int_0^{\pi/4}\sin t\cos t\sqrt{\cos^{2n-4}t+\sin^{2n-4}t}\,dt\\&\le n\int_0^{\pi/4}\sin t\cos t(\cos^{n-2}t+\sin^{n-2}t)\,dt\end{aligned} n ∫ 0 π /4 sin t cos t ( cos n − 2 t − sin n − 2 t ) d t ≤ a n = n ∫ 0 π /4 sin t cos t cos 2 n − 4 t + sin 2 n − 4 t d t ≤ n ∫ 0 π /4 sin t cos t ( cos n − 2 t + sin n − 2 t ) d t 임을 알 수 있다.
+20점 · 부등식 2 − 4 ( 2 2 ) n ≤ s n ≤ 2 \displaystyle 2-4\left(\dfrac{\sqrt2}2\right)^n\le s_n\le2 2 − 4 ( 2 2 ) n ≤ s n ≤ 2 여기서 n ∫ 0 π / 4 sin t cos t ( cos n − 2 t − sin n − 2 t ) d t = [ − cos n t − sin n t ] 0 π / 4 = 1 − 2 ( 2 2 ) n , \displaystyle n\int_0^{\pi/4}\sin t\cos t(\cos^{n-2}t-\sin^{n-2}t)\,dt=\Biggl[-\cos^nt-\sin^nt\Biggr]_0^{\pi/4}=1-2\left(\dfrac{\sqrt2}2\right)^n, n ∫ 0 π /4 sin t cos t ( cos n − 2 t − sin n − 2 t ) d t = [ − cos n t − sin n t ] 0 π /4 = 1 − 2 ( 2 2 ) n , n ∫ 0 π / 4 sin t cos t ( cos n − 2 t + sin n − 2 t ) d t = [ − cos n t + sin n t ] 0 π / 4 = 1 \displaystyle n\int_0^{\pi/4}\sin t\cos t(\cos^{n-2}t+\sin^{n-2}t)\,dt=\Biggl[-\cos^nt+\sin^nt\Biggr]_0^{\pi/4}=1 n ∫ 0 π /4 sin t cos t ( cos n − 2 t + sin n − 2 t ) d t = [ − cos n t + sin n t ] 0 π /4 = 1 이므로, 1 − 2 ( 2 2 ) n ≤ a n ≤ 1 \displaystyle 1-2\left(\dfrac{\sqrt2}2\right)^n\le a_n\le1 1 − 2 ( 2 2 ) n ≤ a n ≤ 1
+20점 · lim n → ∞ s n = 2 \displaystyle \lim_{n\to\infty}s_n=2 n → ∞ lim s n = 2 따라서, 2 − 4 ( 2 2 ) n ≤ s n ≤ 2 \displaystyle 2-4\left(\dfrac{\sqrt2}2\right)^n\le s_n\le2 2 − 4 ( 2 2 ) n ≤ s n ≤ 2 lim n → ∞ s n = 2 \displaystyle \lim_{n\to\infty}s_n=2 n → ∞ lim s n = 2
수리논술, 배움에서 논증의 완성까지.