+30점 · 그래프를 그리면 (2-1) 구간 [ − π , π ] \displaystyle [-\pi,\pi] [ − π , π ] x \displaystyle x x sin x ≥ 3 2 \displaystyle \sin x\ge\frac{\sqrt3}{2} sin x ≥ 2 3 π 3 ≤ x ≤ 2 π 3 \displaystyle \frac\pi3\le x\le\frac{2\pi}3 3 π ≤ x ≤ 3 2 π g ( x ) = { sin x − 3 ( π 3 ≤ x ≤ 2 π 3 ) − sin x ( 그 외의 경우 ) \displaystyle g(x)=\begin{cases}\sin x-\sqrt3&(\frac\pi3\le x\le\frac{2\pi}3)\\-\sin x&(\text{그 외의 경우})\end{cases} g ( x ) = { sin x − 3 − sin x ( 3 π ≤ x ≤ 3 2 π ) ( 그 외의 경우 ) y = g ( x ) \displaystyle y=g(x) y = g ( x )
+20점 · ∫ − π 0 g ( x ) d x = 2 \displaystyle \int_{-\pi}^0g(x)dx=2 ∫ − π 0 g ( x ) d x = 2 그러므로 ∫ − π 0 g ( x ) d x = ∫ − π 0 ( − sin x ) d x = [ cos x ] − π 0 = 2 \displaystyle \int_{-\pi}^0g(x)dx=\int_{-\pi}^0(-\sin x)dx=\Biggl[\cos x\Biggr]_{-\pi}^0=2 ∫ − π 0 g ( x ) d x = ∫ − π 0 ( − sin x ) d x = [ cos x ] − π 0 = 2 이고, 그래프의 대칭성을 이용하여 계산하면 다음을 얻는다.
+20점 · ∫ 0 π g ( x ) d x = − 3 π 3 \displaystyle \int_0^\pi g(x)dx=-\frac{\sqrt3\pi}{3} ∫ 0 π g ( x ) d x = − 3 3 π ∫ 0 π g ( x ) d x = 2 { ∫ 0 π 3 ( − sin x ) d x + ∫ π 3 π 2 ( sin x − 3 ) d x } = 2 [ cos x ] 0 π 3 + 2 [ − cos x − 3 x ] π 3 π 2 = − 3 π 3 \displaystyle \begin{aligned}\int_0^\pi g(x)dx&=2\left\{\int_0^{\frac\pi3}(-\sin x)dx+\int_{\frac\pi3}^{\frac\pi2}(\sin x-\sqrt3)dx\right\}\\&=2\Biggl[\cos x\Biggr]_0^{\frac\pi3}+2\Biggl[-\cos x-\sqrt3x\Biggr]_{\frac\pi3}^{\frac\pi2}=-\frac{\sqrt3\pi}{3}\end{aligned} ∫ 0 π g ( x ) d x = 2 { ∫ 0 3 π ( − sin x ) d x + ∫ 3 π 2 π ( sin x − 3 ) d x } = 2 [ cos x ] 0 3 π + 2 [ − cos x − 3 x ] 3 π 2 π = − 3 3 π
+30점 · α = 1 3 \displaystyle \alpha=\frac13 α = 3 1 (2-2) α = 1 3 \displaystyle \alpha=\frac13 α = 3 1 h ( x ) = − sin x 3 \displaystyle h(x)=-\sin\frac x3 h ( x ) = − sin 3 x ( − π , π ) \displaystyle (-\pi,\pi) ( − π , π )
+30점 · α > 1 3 \displaystyle \alpha>\frac13 α > 3 1 x = π 3 α \displaystyle x=\frac\pi{3\alpha} x = 3 α π 그렇지만 α > 1 3 \displaystyle \alpha>\frac13 α > 3 1 p ( x ) = sin α x − 3 2 \displaystyle p(x)=\sin\alpha x-\frac{\sqrt3}{2} p ( x ) = sin α x − 2 3 ( 0 , π 2 α ) \displaystyle (0,\frac\pi{2\alpha}) ( 0 , 2 α π ) p ′ ( x ) = α cos α x > 0 \displaystyle p^{\prime}(x)=\alpha\cos\alpha x>0 p ′ ( x ) = α cos α x > 0 p ( x ) \displaystyle p(x) p ( x ) p ( π 3 α ) = 0 \displaystyle p(\frac\pi{3\alpha})=0 p ( 3 α π ) = 0 ( 0 , π 3 α ) \displaystyle (0,\frac\pi{3\alpha}) ( 0 , 3 α π ) p ( x ) < 0 \displaystyle p(x)<0 p ( x ) < 0 ( π 3 α , π 2 α ) \displaystyle (\frac\pi{3\alpha},\frac\pi{2\alpha}) ( 3 α π , 2 α π ) p ( x ) > 0 \displaystyle p(x)>0 p ( x ) > 0 h ( x ) = f ( sin α x ) = ∣ sin α x − 3 2 ∣ − 3 2 = ∣ p ( x ) ∣ − 3 2 \displaystyle h(x)=f(\sin\alpha x)=\left|\sin\alpha x-\frac{\sqrt3}{2}\right|-\frac{\sqrt3}{2}=|p(x)|-\frac{\sqrt3}{2} h ( x ) = f ( sin α x ) = sin α x − 2 3 − 2 3 = ∣ p ( x ) ∣ − 2 3 은 구간 ( 0 , π 3 α ) \displaystyle (0,\frac\pi{3\alpha}) ( 0 , 3 α π ) ( π 3 α , π 2 α ) \displaystyle (\frac\pi{3\alpha},\frac\pi{2\alpha}) ( 3 α π , 2 α π ) h ( x ) \displaystyle h(x) h ( x ) x = π 3 α \displaystyle x=\frac\pi{3\alpha} x = 3 α π − 3 2 \displaystyle -\frac{\sqrt3}{2} − 2 3 sin α x = 3 2 \displaystyle \sin\alpha x=\frac{\sqrt3}{2} sin α x = 2 3
+20점 · h ( x ) \displaystyle h(x) h ( x ) α \displaystyle \alpha α 1 3 \displaystyle \frac13 3 1 따라서 h ( x ) \displaystyle h(x) h ( x ) α \displaystyle \alpha α 1 3 \displaystyle \frac13 3 1
(2-3) (2-1)에서 구한 곡선 y = f ( sin x ) \displaystyle y=f(\sin x) y = f ( sin x ) y = f ( sin 6 x ) \displaystyle y=f(\sin6x) y = f ( sin 6 x )
그림에서 ∣ ∫ 0 π 6 f ( sin 6 t ) d t ∣ < ∫ π 6 π 3 f ( sin u ) d u \displaystyle \left|\int_0^{\frac\pi6}f(\sin6t)dt\right|<\int_{\frac\pi6}^{\frac\pi3}f(\sin u)du ∫ 0 6 π f ( sin 6 t ) d t < ∫ 6 π 3 π f ( sin u ) d u f ( sin x ) \displaystyle f(\sin x) f ( sin x ) [ 0 , π ] \displaystyle [0,\pi] [ 0 , π ] F ( x ) = ∫ 0 x f ( sin 6 t ) d t \displaystyle F(x)=\int_0^xf(\sin6t)dt F ( x ) = ∫ 0 x f ( sin 6 t ) d t F ( π 6 ) \displaystyle F(\frac\pi6) F ( 6 π ) F ( π ) \displaystyle F(\pi) F ( π ) Correction. 넓이를 비교하는 부등식의 오른쪽 적분에서 f ( sin u ) \displaystyle f(\sin u) f ( sin u ) f ( sin 6 u ) \displaystyle f(\sin6u) f ( sin 6 u ) ∫ − π 0 f ( sin x ) d x = 2 \displaystyle \int_{-\pi}^0f(\sin x)dx=2 ∫ − π 0 f ( sin x ) d x = 2 ∫ 0 π f ( sin x ) d x = − 3 π 3 \displaystyle \int_0^\pi f(\sin x)dx=-\frac{\sqrt3\pi}{3} ∫ 0 π f ( sin x ) d x = − 3 3 π 및 치환적분과 함수의 주기성을 이용하면 다음과 같이 각각 계산된다.
+40점 · 최솟값 F ( π 6 ) = − 3 π 18 \displaystyle F(\frac\pi6)=-\frac{\sqrt3\pi}{18} F ( 6 π ) = − 18 3 π 최솟값: F ( π 6 ) = ∫ 0 π 6 f ( sin 6 t ) d t = 1 6 ∫ 0 π f ( sin u ) d u = − 3 π 18 \displaystyle F(\frac\pi6)=\int_0^{\frac\pi6}f(\sin6t)dt=\frac16\int_0^\pi f(\sin u)du=-\frac{\sqrt3\pi}{18} F ( 6 π ) = ∫ 0 6 π f ( sin 6 t ) d t = 6 1 ∫ 0 π f ( sin u ) d u = − 18 3 π
+40점 · 최댓값 F ( π ) = 1 − 3 π 6 \displaystyle F(\pi)=1-\frac{\sqrt3\pi}{6} F ( π ) = 1 − 6 3 π 최댓값: F ( π ) = ∫ 0 π f ( sin 6 t ) d t = 3 ∫ 0 π / 3 f ( sin 6 t ) d t = 3 6 ∫ 0 2 π f ( sin u ) d u = 1 2 ∫ − π π f ( sin u ) d u = 1 − 3 π 6 \displaystyle \begin{aligned}F(\pi)&=\int_0^\pi f(\sin6t)dt=3\int_0^{\pi/3}f(\sin6t)dt\\&=\frac36\int_0^{2\pi}f(\sin u)du=\frac12\int_{-\pi}^\pi f(\sin u)du=1-\frac{\sqrt3\pi}{6}\end{aligned} F ( π ) = ∫ 0 π f ( sin 6 t ) d t = 3 ∫ 0 π /3 f ( sin 6 t ) d t = 6 3 ∫ 0 2 π f ( sin u ) d u = 2 1 ∫ − π π f ( sin u ) d u = 1 − 6 3 π
참고: 구간 [ 0 , π ] \displaystyle [0,\pi] [ 0 , π ] y = F ( x ) = ∫ 0 x f ( sin 6 t ) d t \displaystyle y=F(x)=\int_0^xf(\sin6t)dt y = F ( x ) = ∫ 0 x f ( sin 6 t ) d t
수리논술, 배움에서 논증의 완성까지.![구간 [-π,π]에서 함수 g의 그래프](/figures/sejong-2026mo-2-g.png)
