1-(1) θ \displaystyle \theta θ M \displaystyle M M
+2점 [풀이] 함수 f ( x ) = e x sin x \displaystyle f(x)=e^x\sin x f ( x ) = e x sin x f ′ ( x ) = e x sin x + e x cos x = e x ( sin x + cos x ) \displaystyle f^{\prime}(x)=e^x\sin x+e^x\cos x=e^x(\sin x+\cos x) f ′ ( x ) = e x sin x + e x cos x = e x ( sin x + cos x )
+1점 · x = 3 4 π \displaystyle x=\frac34\pi x = 4 3 π f ′ ( x ) = 0 \displaystyle f^{\prime}(x)=0 f ′ ( x ) = 0 x = 3 4 π \displaystyle x=\frac34\pi x = 4 3 π f ′ ( x ) = 0 \displaystyle f^{\prime}(x)=0 f ′ ( x ) = 0
+2점 · x < 3 4 π \displaystyle x<\frac34\pi x < 4 3 π f ′ ( x ) > 0 \displaystyle f^{\prime}(x)>0 f ′ ( x ) > 0 f ( x ) \displaystyle f(x) f ( x ) x < 3 4 π \displaystyle x<\frac34\pi x < 4 3 π f ′ ( x ) > 0 \displaystyle f^{\prime}(x)>0 f ′ ( x ) > 0 f ( x ) \displaystyle f(x) f ( x )
+2점 · x > 3 4 π \displaystyle x>\frac34\pi x > 4 3 π f ′ ( x ) < 0 \displaystyle f^{\prime}(x)<0 f ′ ( x ) < 0 f ( x ) \displaystyle f(x) f ( x ) x > 3 4 π \displaystyle x>\frac34\pi x > 4 3 π f ′ ( x ) < 0 \displaystyle f^{\prime}(x)<0 f ′ ( x ) < 0 f ( x ) \displaystyle f(x) f ( x )
+2점 · 최댓값 f ( 3 4 π ) = e 3 4 π sin 3 4 π = e 3 4 π 1 2 \displaystyle f\left(\frac34\pi\right)=e^{\frac34\pi}\sin\frac34\pi=e^{\frac34\pi}\frac1{\sqrt2} f ( 4 3 π ) = e 4 3 π sin 4 3 π = e 4 3 π 2 1 그러므로 f ( x ) \displaystyle f(x) f ( x ) x = 3 4 π \displaystyle x=\frac34\pi x = 4 3 π f ( 3 4 π ) = e 3 4 π sin 3 4 π = e 3 4 π 1 2 \displaystyle f\left(\frac34\pi\right)=e^{\frac34\pi}\sin\frac34\pi=e^{\frac34\pi}\frac1{\sqrt2} f ( 4 3 π ) = e 4 3 π sin 4 3 π = e 4 3 π 2 1
+1점 따라서 θ = 3 4 π \displaystyle \theta=\frac34\pi θ = 4 3 π M = e 3 4 π 1 2 \displaystyle M=e^{\frac34\pi}\frac1{\sqrt2} M = e 4 3 π 2 1
1-(2) n ≥ 4 \displaystyle n\ge4 n ≥ 4 n \displaystyle n n f ( θ + 1 n π ) = M e 1 n π ( cos 1 n π − sin 1 n π ) \displaystyle f\left(\theta+\frac1n\pi\right)=Me^{\frac1n\pi}\left(\cos\frac1n\pi-\sin\frac1n\pi\right) f ( θ + n 1 π ) = M e n 1 π ( cos n 1 π − sin n 1 π )
+3점 · f ( θ + 1 n π ) = e 3 4 π ⋅ e 1 n π { 1 2 cos 1 n π − 1 2 sin 1 n π } \displaystyle f\left(\theta+\frac1n\pi\right)=e^{\frac34\pi}\cdot e^{\frac1n\pi}\left\{\frac1{\sqrt2}\cos\frac1n\pi-\frac1{\sqrt2}\sin\frac1n\pi\right\} f ( θ + n 1 π ) = e 4 3 π ⋅ e n 1 π { 2 1 cos n 1 π − 2 1 sin n 1 π } +2점 · e 3 4 π ⋅ e 1 n π { 1 2 cos 1 n π − 1 2 sin 1 n π } = e 3 4 π 1 2 ⋅ e 1 n π { cos 1 n π − sin 1 n π } \displaystyle e^{\frac34\pi}\cdot e^{\frac1n\pi}\left\{\frac1{\sqrt2}\cos\frac1n\pi-\frac1{\sqrt2}\sin\frac1n\pi\right\}=e^{\frac34\pi}\frac1{\sqrt2}\cdot e^{\frac1n\pi}\left\{\cos\frac1n\pi-\sin\frac1n\pi\right\} e 4 3 π ⋅ e n 1 π { 2 1 cos n 1 π − 2 1 sin n 1 π } = e 4 3 π 2 1 ⋅ e n 1 π { cos n 1 π − sin n 1 π } [풀이]지수함수의 성질과 삼각함수의 덧셈정리에 의하여 아래의 등식을 얻는다. f ( θ + 1 n π ) = e θ + 1 n π sin ( θ + 1 n π ) = e θ + 1 n π { sin θ cos 1 n π + cos θ sin 1 n π } = e 3 4 π ⋅ e 1 n π { 1 2 cos 1 n π − 1 2 sin 1 n π } = e 3 4 π 1 2 ⋅ e 1 n π { cos 1 n π − sin 1 n π } = M ⋅ e 1 n π ( cos 1 n π − sin 1 n π ) \displaystyle \begin{aligned}f\left(\theta+\frac1n\pi\right)&=e^{\theta+\frac1n\pi}\sin\left(\theta+\frac1n\pi\right)\\&=e^{\theta+\frac1n\pi}\left\{\sin\theta\cos\frac1n\pi+\cos\theta\sin\frac1n\pi\right\}\\&=e^{\frac34\pi}\cdot e^{\frac1n\pi}\left\{\frac1{\sqrt2}\cos\frac1n\pi-\frac1{\sqrt2}\sin\frac1n\pi\right\}\\&=e^{\frac34\pi}\frac1{\sqrt2}\cdot e^{\frac1n\pi}\left\{\cos\frac1n\pi-\sin\frac1n\pi\right\}\\&=M\cdot e^{\frac1n\pi}\left(\cos\frac1n\pi-\sin\frac1n\pi\right)\end{aligned} f ( θ + n 1 π ) = e θ + n 1 π sin ( θ + n 1 π ) = e θ + n 1 π { sin θ cos n 1 π + cos θ sin n 1 π } = e 4 3 π ⋅ e n 1 π { 2 1 cos n 1 π − 2 1 sin n 1 π } = e 4 3 π 2 1 ⋅ e n 1 π { cos n 1 π − sin n 1 π } = M ⋅ e n 1 π ( cos n 1 π − sin n 1 π )
1-(3) n ≥ 4 \displaystyle n\ge4 n ≥ 4 n \displaystyle n n M n e π ( cos 1 n π − sin 1 n π ) n 1 n π < ∫ 0 π { f ( x ) } n d x < M n π \displaystyle M^ne^\pi\left(\cos\frac1n\pi-\sin\frac1n\pi\right)^n\frac1n\pi<\int_0^\pi\{f(x)\}^n dx<M^n\pi M n e π ( cos n 1 π − sin n 1 π ) n n 1 π < ∫ 0 π { f ( x ) } n d x < M n π
+2점 · 부등식의 오른쪽 부분:
x ∈ [ 0 , π ] \displaystyle x\in[0,\pi] x ∈ [ 0 , π ] 0 ≤ f ( x ) ≤ M \displaystyle 0\le f(x)\le M 0 ≤ f ( x ) ≤ M 0 ≤ { f ( x ) } n ≤ M n \displaystyle 0\le\{f(x)\}^n\le M^n 0 ≤ { f ( x ) } n ≤ M n [풀이] 오른쪽 부등식: x ∈ [ 0 , π ] \displaystyle x\in[0,\pi] x ∈ [ 0 , π ] 0 ≤ f ( x ) ≤ M \displaystyle 0\le f(x)\le M 0 ≤ f ( x ) ≤ M 0 ≤ { f ( x ) } n ≤ M n \displaystyle 0\le\{f(x)\}^n\le M^n 0 ≤ { f ( x ) } n ≤ M n
+2점 f ( x ) \displaystyle f(x) f ( x ) [ 0 , π ] \displaystyle [0,\pi] [ 0 , π ] ∫ 0 π { f ( x ) } n d x < ∫ 0 π M n d x = M n π \displaystyle \int_0^\pi\{f(x)\}^n dx<\int_0^\pi M^n dx=M^n\pi ∫ 0 π { f ( x ) } n d x < ∫ 0 π M n d x = M n π 이다.
+2점 · 부등식의 왼쪽 부분:
[ 3 4 π , 3 4 π + 1 n π ] \displaystyle \left[\frac34\pi,\frac34\pi+\frac1n\pi\right] [ 4 3 π , 4 3 π + n 1 π ] M n e π ( cos 1 n π − sin 1 n π ) n ≤ { f ( x ) } n \displaystyle M^ne^\pi\left(\cos\frac1n\pi-\sin\frac1n\pi\right)^n\le\{f(x)\}^n M n e π ( cos n 1 π − sin n 1 π ) n ≤ { f ( x ) } n 왼쪽 부등식: n ≥ 4 \displaystyle n\ge4 n ≥ 4 3 4 π < 3 4 π + 1 n π ≤ π \displaystyle \frac34\pi<\frac34\pi+\frac1n\pi\le\pi 4 3 π < 4 3 π + n 1 π ≤ π f ( x ) \displaystyle f(x) f ( x ) [ 3 4 π , π ] \displaystyle \left[\frac34\pi,\pi\right] [ 4 3 π , π ] [ 3 4 π , 3 4 π + 1 n π ] \displaystyle \left[\frac34\pi,\frac34\pi+\frac1n\pi\right] [ 4 3 π , 4 3 π + n 1 π ] x \displaystyle x x f ( π ) ≤ f ( 3 4 π + 1 n π ) ≤ f ( x ) \displaystyle f(\pi)\le f\left(\frac34\pi+\frac1n\pi\right)\le f(x) f ( π ) ≤ f ( 4 3 π + n 1 π ) ≤ f ( x ) 이다. 따라서 0 ≤ M e 1 n π ( cos 1 n π − sin 1 n π ) ≤ f ( x ) \displaystyle 0\le Me^{\frac1n\pi}\left(\cos\frac1n\pi-\sin\frac1n\pi\right)\le f(x) 0 ≤ M e n 1 π ( cos n 1 π − sin n 1 π ) ≤ f ( x ) 가 성립한다.
+2점 · M n e π ( cos 1 n π − sin 1 n π ) n 1 n π = ∫ 3 4 π 3 4 π + 1 n π M n e π ( cos 1 n π − sin 1 n π ) n d x \displaystyle M^ne^\pi\left(\cos\frac1n\pi-\sin\frac1n\pi\right)^n\frac1n\pi=\int_{\frac34\pi}^{\frac34\pi+\frac1n\pi}M^ne^\pi\left(\cos\frac1n\pi-\sin\frac1n\pi\right)^n dx M n e π ( cos n 1 π − sin n 1 π ) n n 1 π = ∫ 4 3 π 4 3 π + n 1 π M n e π ( cos n 1 π − sin n 1 π ) n d x 구간 [ 3 4 π , 3 4 π + 1 n π ] \displaystyle \left[\frac34\pi,\frac34\pi+\frac1n\pi\right] [ 4 3 π , 4 3 π + n 1 π ] { M e 1 n π ( cos 1 n π − sin 1 n π ) } n = M n e π ( cos 1 n π − sin 1 n π ) n ≤ { f ( x ) } n \displaystyle \left\{Me^{\frac1n\pi}\left(\cos\frac1n\pi-\sin\frac1n\pi\right)\right\}^n=M^ne^\pi\left(\cos\frac1n\pi-\sin\frac1n\pi\right)^n\le\{f(x)\}^n { M e n 1 π ( cos n 1 π − sin n 1 π ) } n = M n e π ( cos n 1 π − sin n 1 π ) n ≤ { f ( x ) } n 이므로 위의 부등식을 적분하면 M n e π ( cos 1 n π − sin 1 n π ) n 1 n π = ∫ 3 4 π 3 4 π + 1 n π M n e π ( cos 1 n π − sin 1 n π ) n d x ≤ ∫ 3 4 π 3 4 π + 1 n π { f ( x ) } n d x \displaystyle M^ne^\pi\left(\cos\frac1n\pi-\sin\frac1n\pi\right)^n\frac1n\pi=\int_{\frac34\pi}^{\frac34\pi+\frac1n\pi}M^ne^\pi\left(\cos\frac1n\pi-\sin\frac1n\pi\right)^n dx\le\int_{\frac34\pi}^{\frac34\pi+\frac1n\pi}\{f(x)\}^n dx M n e π ( cos n 1 π − sin n 1 π ) n n 1 π = ∫ 4 3 π 4 3 π + n 1 π M n e π ( cos n 1 π − sin n 1 π ) n d x ≤ ∫ 4 3 π 4 3 π + n 1 π { f ( x ) } n d x 가 성립한다.
+2점 · ∫ 3 4 π 3 4 π + 1 n π M n e π ( cos 1 n π − sin 1 n π ) n d x ≤ ∫ 3 4 π 3 4 π + 1 n π { f ( x ) } n d x \displaystyle \int_{\frac34\pi}^{\frac34\pi+\frac1n\pi}M^ne^\pi\left(\cos\frac1n\pi-\sin\frac1n\pi\right)^n dx\le\int_{\frac34\pi}^{\frac34\pi+\frac1n\pi}\{f(x)\}^n dx ∫ 4 3 π 4 3 π + n 1 π M n e π ( cos n 1 π − sin n 1 π ) n d x ≤ ∫ 4 3 π 4 3 π + n 1 π { f ( x ) } n d x 그리고 구간 [ 3 4 π , 3 4 π + 1 n π ] \displaystyle \left[\frac34\pi,\frac34\pi+\frac1n\pi\right] [ 4 3 π , 4 3 π + n 1 π ] [ 0 , π ] \displaystyle [0,\pi] [ 0 , π ] f ( x ) ≥ 0 \displaystyle f(x)\ge0 f ( x ) ≥ 0 ∫ 3 4 π 3 4 π + 1 n π { f ( x ) } n d x < ∫ 0 π { f ( x ) } n d x \displaystyle \int_{\frac34\pi}^{\frac34\pi+\frac1n\pi}\{f(x)\}^n dx<\int_0^\pi\{f(x)\}^n dx ∫ 4 3 π 4 3 π + n 1 π { f ( x ) } n d x < ∫ 0 π { f ( x ) } n d x 가 성립한다. 위의 두 부등식으로부터 M n e π ( cos 1 n π − sin 1 n π ) n 1 n π < ∫ 0 π { f ( x ) } n d x \displaystyle M^ne^\pi\left(\cos\frac1n\pi-\sin\frac1n\pi\right)^n\frac1n\pi<\int_0^\pi\{f(x)\}^n dx M n e π ( cos n 1 π − sin n 1 π ) n n 1 π < ∫ 0 π { f ( x ) } n d x 를 얻는다. 그러므로 n ≥ 4 \displaystyle n\ge4 n ≥ 4 M n e π ( cos 1 n π − sin 1 n π ) n 1 n π < ∫ 0 π { f ( x ) } n d x < M n π \displaystyle M^ne^\pi\left(\cos\frac1n\pi-\sin\frac1n\pi\right)^n\frac1n\pi<\int_0^\pi\{f(x)\}^n dx<M^n\pi M n e π ( cos n 1 π − sin n 1 π ) n n 1 π < ∫ 0 π { f ( x ) } n d x < M n π 가 성립한다.
1-(4) lim n → ∞ ( ∫ 0 π { f ( x ) } n d x ) 1 n = M \displaystyle \lim_{n\to\infty}\left(\int_0^\pi\{f(x)\}^n dx\right)^{\frac1n}=M n → ∞ lim ( ∫ 0 π { f ( x ) } n d x ) n 1 = M n \displaystyle n n 4 \displaystyle 4 4 lim n → ∞ n 1 n = 1 \displaystyle \lim_{n\to\infty}n^{\frac1n}=1 n → ∞ lim n n 1 = 1
[풀이] 문항 1-(3)의 부등식 M n e π ( cos 1 n π − sin 1 n π ) n 1 n π < ∫ 0 π { f ( x ) } n d x < M n π \displaystyle M^ne^\pi\left(\cos\frac1n\pi-\sin\frac1n\pi\right)^n\frac1n\pi<\int_0^\pi\{f(x)\}^n dx<M^n\pi M n e π ( cos n 1 π − sin n 1 π ) n n 1 π < ∫ 0 π { f ( x ) } n d x < M n π 에 x n \displaystyle \sqrt[n]{\phantom{x}} n x ( M n e π ( cos 1 n π − sin 1 n π ) n 1 n π ) 1 n < ( ∫ 0 π { f ( x ) } n d x ) 1 n < ( M n π ) 1 n = M ⋅ π 1 n \displaystyle \left(M^ne^\pi\left(\cos\frac1n\pi-\sin\frac1n\pi\right)^n\frac1n\pi\right)^{\frac1n}<\left(\int_0^\pi\{f(x)\}^n dx\right)^{\frac1n}<(M^n\pi)^{\frac1n}=M\cdot\pi^{\frac1n} ( M n e π ( cos n 1 π − sin n 1 π ) n n 1 π ) n 1 < ( ∫ 0 π { f ( x ) } n d x ) n 1 < ( M n π ) n 1 = M ⋅ π n 1 이다. 위 부등식의 왼쪽 부분을 정리하면 ( M n e π ( cos 1 n π − sin 1 n π ) n 1 n π ) 1 n = M e π n ( cos 1 n π − sin 1 n π ) 1 n 1 n π 1 n \displaystyle \left(M^ne^\pi\left(\cos\frac1n\pi-\sin\frac1n\pi\right)^n\frac1n\pi\right)^{\frac1n}=Me^{\frac\pi n}\left(\cos\frac1n\pi-\sin\frac1n\pi\right)\frac1{n^{\frac1n}}\pi^{\frac1n} ( M n e π ( cos n 1 π − sin n 1 π ) n n 1 π ) n 1 = M e n π ( cos n 1 π − sin n 1 π ) n n 1 1 π n 1
+2점 이므로, 위의 부등식을 다시 쓰면 M e π n ( cos 1 n π − sin 1 n π ) 1 n 1 n π 1 n < ( ∫ 0 π { f ( x ) } n d x ) 1 n < M ⋅ π 1 n \displaystyle Me^{\frac\pi n}\left(\cos\frac1n\pi-\sin\frac1n\pi\right)\frac1{n^{\frac1n}}\pi^{\frac1n}<\left(\int_0^\pi\{f(x)\}^n dx\right)^{\frac1n}<M\cdot\pi^{\frac1n} M e n π ( cos n 1 π − sin n 1 π ) n n 1 1 π n 1 < ( ∫ 0 π { f ( x ) } n d x ) n 1 < M ⋅ π n 1
+3점 이고, 극한값의 대소관계에 의하여 lim n → ∞ M e π n ( cos 1 n π − sin 1 n π ) 1 n 1 n π 1 n ≤ lim n → ∞ ( ∫ 0 π { f ( x ) } n d x ) 1 n ≤ lim n → ∞ M ⋅ π 1 n \displaystyle \lim_{n\to\infty}Me^{\frac\pi n}\left(\cos\frac1n\pi-\sin\frac1n\pi\right)\frac1{n^{\frac1n}}\pi^{\frac1n}\le\lim_{n\to\infty}\left(\int_0^\pi\{f(x)\}^n dx\right)^{\frac1n}\le\lim_{n\to\infty}M\cdot\pi^{\frac1n} n → ∞ lim M e n π ( cos n 1 π − sin n 1 π ) n n 1 1 π n 1 ≤ n → ∞ lim ( ∫ 0 π { f ( x ) } n d x ) n 1 ≤ n → ∞ lim M ⋅ π n 1 이 된다.
+1점 여기에서 π \displaystyle \pi π e \displaystyle e e lim n → ∞ e π n = 1 \displaystyle \lim_{n\to\infty}e^{\frac\pi n}=1 n → ∞ lim e n π = 1
+1점 lim n → ∞ π 1 n = 1 \displaystyle \lim_{n\to\infty}\pi^{\frac1n}=1 n → ∞ lim π n 1 = 1
+1점 을 얻고, 또한 lim n → ∞ ( cos 1 n π − sin 1 n π ) = cos 0 − sin 0 = 1 \displaystyle \lim_{n\to\infty}\left(\cos\frac1n\pi-\sin\frac1n\pi\right)=\cos0-\sin0=1 n → ∞ lim ( cos n 1 π − sin n 1 π ) = cos 0 − sin 0 = 1
이며, 주어진 조건에 의하여 lim n → ∞ n 1 n = 1 \displaystyle \lim_{n\to\infty}n^{\frac1n}=1 n → ∞ lim n n 1 = 1
+1점 이다. 그러므로 lim n → ∞ M e π n ( cos 1 n π − sin 1 n π ) 1 n 1 n π 1 n = M \displaystyle \lim_{n\to\infty}Me^{\frac\pi n}\left(\cos\frac1n\pi-\sin\frac1n\pi\right)\frac1{n^{\frac1n}}\pi^{\frac1n}=M n → ∞ lim M e n π ( cos n 1 π − sin n 1 π ) n n 1 1 π n 1 = M
+1점 이고 lim n → ∞ M ⋅ π 1 n = M \displaystyle \lim_{n\to\infty}M\cdot\pi^{\frac1n}=M n → ∞ lim M ⋅ π n 1 = M
이다. 따라서 위의 부등식은 M ≤ lim n → ∞ ( ∫ 0 π { f ( x ) } n d x ) 1 n ≤ M \displaystyle M\le\lim_{n\to\infty}\left(\int_0^\pi\{f(x)\}^n dx\right)^{\frac1n}\le M M ≤ n → ∞ lim ( ∫ 0 π { f ( x ) } n d x ) n 1 ≤ M 이다. 그러므로 lim n → ∞ ( ∫ 0 π { f ( x ) } n d x ) 1 n = M \displaystyle \lim_{n\to\infty}\left(\int_0^\pi\{f(x)\}^n dx\right)^{\frac1n}=M n → ∞ lim ( ∫ 0 π { f ( x ) } n d x ) n 1 = M
수리논술, 이게 다다.