+40점 · f ( π 2 ) = − 4 − 2 2 \displaystyle f(\frac\pi2)=-4-2\sqrt2 f ( 2 π ) = − 4 − 2 2 (1-1) x = π 2 \displaystyle x=\frac\pi2 x = 2 π f ( π 2 ) = ∫ π 2 2 π ( t − π 2 ) cos t 2 d t = [ 2 ( t − π 2 ) sin t 2 ] π 2 2 π − ∫ π 2 2 π 2 sin t 2 d t = − [ − 4 cos t 2 ] π 2 2 π = − 4 − 2 2 \displaystyle \begin{aligned}f\left(\frac\pi2\right)&=\int_{\frac\pi2}^{2\pi}\left(t-\frac\pi2\right)\cos\frac t2dt=\Biggl[2\left(t-\frac\pi2\right)\sin\frac t2\Biggr]_{\frac\pi2}^{2\pi}-\int_{\frac\pi2}^{2\pi}2\sin\frac t2dt\\&=-\Biggl[-4\cos\frac t2\Biggr]_{\frac\pi2}^{2\pi}=-4-2\sqrt2\end{aligned} f ( 2 π ) = ∫ 2 π 2 π ( t − 2 π ) cos 2 t d t = [ 2 ( t − 2 π ) sin 2 t ] 2 π 2 π − ∫ 2 π 2 π 2 sin 2 t d t = − [ − 4 cos 2 t ] 2 π 2 π = − 4 − 2 2 이다.
+30점 · x < π 2 \displaystyle x<\frac\pi2 x < 2 π f ′ ( x ) = 2 \displaystyle f\prime(x)=\sqrt2 f ′ ( x ) = 2 한편, x < π 2 \displaystyle x<\frac\pi2 x < 2 π f ( x ) = ∫ π 2 2 π ( t − x ) cos t 2 d t = ∫ π 2 2 π t cos t 2 d t − x ∫ π 2 2 π cos t 2 d t \displaystyle f(x)=\int_{\frac\pi2}^{2\pi}(t-x)\cos\frac t2dt=\int_{\frac\pi2}^{2\pi}t\cos\frac t2dt-x\int_{\frac\pi2}^{2\pi}\cos\frac t2dt f ( x ) = ∫ 2 π 2 π ( t − x ) cos 2 t d t = ∫ 2 π 2 π t cos 2 t d t − x ∫ 2 π 2 π cos 2 t d t 이므로 f ′ ( x ) = − ∫ π 2 2 π cos t 2 d t = [ − 2 sin t 2 ] π 2 2 π = 2 \displaystyle f\prime(x)=-\int_{\frac\pi2}^{2\pi}\cos\frac t2dt=\Biggl[-2\sin\frac t2\Biggr]_{\frac\pi2}^{2\pi}=\sqrt2 f ′ ( x ) = − ∫ 2 π 2 π cos 2 t d t = [ − 2 sin 2 t ] 2 π 2 π = 2
+10점 · x > 2 π \displaystyle x>2\pi x > 2 π f ′ ( x ) = − 2 \displaystyle f\prime(x)=-\sqrt2 f ′ ( x ) = − 2 (1-2) (1-1)에서 x < π 2 \displaystyle x<\frac\pi2 x < 2 π f ′ ( x ) = 2 \displaystyle f\prime(x)=\sqrt2 f ′ ( x ) = 2 x > 2 π \displaystyle x>2\pi x > 2 π f ( x ) = ∫ π 2 2 π ( x − t ) cos t 2 d t = x ∫ π 2 2 π cos t 2 d t − ∫ π 2 2 π t cos t 2 d t \displaystyle f(x)=\int_{\frac\pi2}^{2\pi}(x-t)\cos\frac t2dt=x\int_{\frac\pi2}^{2\pi}\cos\frac t2dt-\int_{\frac\pi2}^{2\pi}t\cos\frac t2dt f ( x ) = ∫ 2 π 2 π ( x − t ) cos 2 t d t = x ∫ 2 π 2 π cos 2 t d t − ∫ 2 π 2 π t cos 2 t d t 이므로 f ′ ( x ) = ∫ π 2 2 π cos t 2 d t = [ 2 sin t 2 ] π 2 2 π = − 2 \displaystyle f\prime(x)=\int_{\frac\pi2}^{2\pi}\cos\frac t2dt=\Biggl[2\sin\frac t2\Biggr]_{\frac\pi2}^{2\pi}=-\sqrt2 f ′ ( x ) = ∫ 2 π 2 π cos 2 t d t = [ 2 sin 2 t ] 2 π 2 π = − 2
+30점 · π 2 ≤ x ≤ 2 π \displaystyle \frac\pi2\le x\le2\pi 2 π ≤ x ≤ 2 π f ′ ( x ) = 4 sin x 2 − 2 \displaystyle f\prime(x)=4\sin\frac x2-\sqrt2 f ′ ( x ) = 4 sin 2 x − 2 한편 π 2 ≤ x ≤ 2 π \displaystyle \frac\pi2\le x\le2\pi 2 π ≤ x ≤ 2 π f ( x ) = ∫ π 2 x ( x − t ) cos t 2 d t + ∫ x 2 π ( t − x ) cos t 2 d t = x ∫ π 2 x cos t 2 d t − ∫ π 2 x t cos t 2 d t + ∫ x 2 π t cos t 2 d t − x ∫ x 2 π cos t 2 d t \displaystyle \begin{aligned}f(x)&=\int_{\frac\pi2}^x(x-t)\cos\frac t2dt+\int_x^{2\pi}(t-x)\cos\frac t2dt\\&=x\int_{\frac\pi2}^x\cos\frac t2dt-\int_{\frac\pi2}^xt\cos\frac t2dt+\int_x^{2\pi}t\cos\frac t2dt-x\int_x^{2\pi}\cos\frac t2dt\end{aligned} f ( x ) = ∫ 2 π x ( x − t ) cos 2 t d t + ∫ x 2 π ( t − x ) cos 2 t d t = x ∫ 2 π x cos 2 t d t − ∫ 2 π x t cos 2 t d t + ∫ x 2 π t cos 2 t d t − x ∫ x 2 π cos 2 t d t 이다. 따라서 f ′ ( x ) = ∫ π 2 x cos t 2 d t + x cos x 2 − x cos x 2 − x cos x 2 − ∫ x 2 π cos t 2 d t + x cos x 2 = ∫ π 2 x cos t 2 d t − ∫ x 2 π cos t 2 d t = [ 2 sin t 2 ] π 2 x − [ 2 sin t 2 ] x 2 π = 4 sin x 2 − 2 \displaystyle \begin{aligned}f\prime(x)&=\int_{\frac\pi2}^x\cos\frac t2dt+x\cos\frac x2-x\cos\frac x2-x\cos\frac x2-\int_x^{2\pi}\cos\frac t2dt+x\cos\frac x2\\&=\int_{\frac\pi2}^x\cos\frac t2dt-\int_x^{2\pi}\cos\frac t2dt\\&=\Biggl[2\sin\frac t2\Biggr]_{\frac\pi2}^x-\Biggl[2\sin\frac t2\Biggr]_x^{2\pi}=4\sin\frac x2-\sqrt2\end{aligned} f ′ ( x ) = ∫ 2 π x cos 2 t d t + x cos 2 x − x cos 2 x − x cos 2 x − ∫ x 2 π cos 2 t d t + x cos 2 x = ∫ 2 π x cos 2 t d t − ∫ x 2 π cos 2 t d t = [ 2 sin 2 t ] 2 π x − [ 2 sin 2 t ] x 2 π = 4 sin 2 x − 2 이다. 즉 f ′ ( x ) = { 2 ( x < π 2 ) 4 sin x 2 − 2 ( π 2 ≤ x ≤ 2 π ) − 2 ( x > 2 π ) \displaystyle f\prime(x)=\begin{cases}\sqrt2&(x<\frac\pi2)\\4\sin\frac x2-\sqrt2&(\frac\pi2\le x\le2\pi)\\-\sqrt2&(x>2\pi)\end{cases} f ′ ( x ) = ⎩ ⎨ ⎧ 2 4 sin 2 x − 2 − 2 ( x < 2 π ) ( 2 π ≤ x ≤ 2 π ) ( x > 2 π ) 이다.
+20점 · sin α 2 = 2 4 \displaystyle \sin\frac\alpha2=\frac{\sqrt2}4 sin 2 α = 4 2 따라서 f ′ ( x ) \displaystyle f\prime(x) f ′ ( x ) sin α 2 = 2 4 \displaystyle \sin\frac\alpha2=\frac{\sqrt2}4 sin 2 α = 4 2 α \displaystyle \alpha α x < α \displaystyle x<\alpha x < α x > α \displaystyle x>\alpha x > α f ( x ) \displaystyle f(x) f ( x ) x = α \displaystyle x=\alpha x = α
+20점 · cos α 2 = − 14 4 \displaystyle \cos\frac\alpha2=-\frac{\sqrt{14}}4 cos 2 α = − 4 14 한편 π < α < 2 π \displaystyle \pi<\alpha<2\pi π < α < 2 π cos α 2 = − 14 4 \displaystyle \cos\frac\alpha2=-\frac{\sqrt{14}}4 cos 2 α = − 4 14
(1-3) π 2 < x < 2 π \displaystyle \frac\pi2<x<2\pi 2 π < x < 2 π f ′ ′ ( x ) = 2 cos x 2 \displaystyle f\prime\prime(x)=2\cos\frac x2 f ′′ ( x ) = 2 cos 2 x f ( x ) \displaystyle f(x) f ( x )
x \displaystyle x x ⋯ \displaystyle \cdots ⋯ π 2 \displaystyle \frac\pi2 2 π ⋯ \displaystyle \cdots ⋯ π \displaystyle \pi π ⋯ \displaystyle \cdots ⋯ α \displaystyle \alpha α ⋯ \displaystyle \cdots ⋯ 2 π \displaystyle 2\pi 2 π ⋯ \displaystyle \cdots ⋯ f ′ ( x ) \displaystyle f\prime(x) f ′ ( x ) + \displaystyle + + + \displaystyle + + + \displaystyle + + + \displaystyle + + + \displaystyle + + 0 \displaystyle 0 0 − \displaystyle - − − \displaystyle - − − \displaystyle - − f ′ ′ ( x ) \displaystyle f\prime\prime(x) f ′′ ( x ) 0 \displaystyle 0 0 + \displaystyle + + 0 \displaystyle 0 0 − \displaystyle - − − \displaystyle - − − \displaystyle - − 0 \displaystyle 0 0 f ( x ) \displaystyle f(x) f ( x ) ↗ \displaystyle \nearrow ↗ ↗ \displaystyle \nearrow ↗ 변곡점 ↗ \displaystyle \nearrow ↗ 극댓값 ↘ \displaystyle \searrow ↘ ↘ \displaystyle \searrow ↘
이로부터 곡선 y = f ( x ) − f ( π 2 ) \displaystyle y=f(x)-f(\frac\pi2) y = f ( x ) − f ( 2 π )
+30점 · 직선 y = 2 x + k \displaystyle y=\sqrt2x+k y = 2 x + k y = f ( x ) − f ( π 2 ) \displaystyle y=f(x)-f(\frac\pi2) y = f ( x ) − f ( 2 π ) 2 \displaystyle \sqrt2 2 ( π 2 , 0 ) \displaystyle (\frac\pi2,0) ( 2 π , 0 ) y \displaystyle y y 따라서 직선 y = 2 x + k \displaystyle y=\sqrt2x+k y = 2 x + k y = f ( x ) − f ( π 2 ) \displaystyle y=f(x)-f(\frac\pi2) y = f ( x ) − f ( 2 π ) y = f ( x ) − f ( π 2 ) \displaystyle y=f(x)-f(\frac\pi2) y = f ( x ) − f ( 2 π ) 2 \displaystyle \sqrt2 2 ( π 2 , 0 ) \displaystyle (\frac\pi2,0) ( 2 π , 0 ) y \displaystyle y y
+30점 · 직선 y = 2 x + k \displaystyle y=\sqrt2x+k y = 2 x + k y = f ( x ) − f ( π 2 ) \displaystyle y=f(x)-f(\frac\pi2) y = f ( x ) − f ( 2 π ) k = ( 8 − 5 2 π ) 2 \displaystyle k=(8-\frac52\pi)\sqrt2 k = ( 8 − 2 5 π ) 2 직선 y = 2 x + k \displaystyle y=\sqrt2x+k y = 2 x + k y = f ( x ) − f ( π 2 ) \displaystyle y=f(x)-f(\frac\pi2) y = f ( x ) − f ( 2 π ) x \displaystyle x x β \displaystyle \beta β 4 sin β 2 − 2 = 2 \displaystyle 4\sin\frac\beta2-\sqrt2=\sqrt2 4 sin 2 β − 2 = 2 π 2 < β < 2 π \displaystyle \frac\pi2<\beta<2\pi 2 π < β < 2 π sin β 2 = 2 2 \displaystyle \sin\frac\beta2=\frac{\sqrt2}2 sin 2 β = 2 2 β = 3 2 π \displaystyle \beta=\frac32\pi β = 2 3 π f ( 3 2 π ) − f ( 1 2 π ) = ∫ π 2 3 2 π ( 4 sin t 2 − 2 ) d t = [ − 8 cos t 2 − 2 t ] π 2 3 2 π = 8 2 − 2 π \displaystyle f\left(\frac32\pi\right)-f\left(\frac12\pi\right)=\int_{\frac\pi2}^{\frac32\pi}\left(4\sin\frac t2-\sqrt2\right)dt=\Biggl[-8\cos\frac t2-\sqrt2t\Biggr]_{\frac\pi2}^{\frac32\pi}=8\sqrt2-\sqrt2\pi f ( 2 3 π ) − f ( 2 1 π ) = ∫ 2 π 2 3 π ( 4 sin 2 t − 2 ) d t = [ − 8 cos 2 t − 2 t ] 2 π 2 3 π = 8 2 − 2 π 이므로, 직선 y = 2 x + k \displaystyle y=\sqrt2x+k y = 2 x + k ( 3 2 π , f ( 3 2 π ) − f ( π 2 ) ) \displaystyle (\frac32\pi,f(\frac32\pi)-f(\frac\pi2)) ( 2 3 π , f ( 2 3 π ) − f ( 2 π )) k = ( 8 − 5 2 π ) 2 \displaystyle k=(8-\frac52\pi)\sqrt2 k = ( 8 − 2 5 π ) 2
+20점 · k < − 2 π 2 \displaystyle k<-\frac{\sqrt2\pi}2 k < − 2 2 π 또한 직선 y = 2 x + k \displaystyle y=\sqrt2x+k y = 2 x + k ( π 2 , 0 ) \displaystyle (\frac\pi2,0) ( 2 π , 0 ) k = − 2 2 π \displaystyle k=-\frac{\sqrt2}2\pi k = − 2 2 π k = ( 8 − 5 2 π ) 2 \displaystyle k=\left(8-\frac52\pi\right)\sqrt2 k = ( 8 − 2 5 π ) 2 k < − 2 2 π \displaystyle k<-\frac{\sqrt2}2\pi k < − 2 2 π 이다.
수리논술, 이게 다다.