+12점 · 정적분을 해결한 형태(별해1, 별해2 및 다른 방법 포함)를 제시 +5점 · ln 1 2 f ( a ) = 1 \displaystyle \ln\frac1{2f(a)}=1 ln 2 f ( a ) 1 = 1 [논제1] 치환적분법에 의하여 1 = ∫ 0 a 2 sin x + 1 2 cos x + sin x cos x d x = ∫ 0 a 2 sin x + 1 ( 2 + sin x ) 2 × 2 + sin x cos x d x = − ∫ 0 a f ′ ( x ) f ( x ) d x = [ − ln ∣ f ( x ) ∣ ] 0 a = ln ∣ f ( 0 ) ∣ ∣ f ( a ) ∣ \displaystyle \begin{aligned}1&=\int_0^a\frac{2\sin x+1}{2\cos x+\sin x\cos x}\,dx\\&=\int_0^a\frac{2\sin x+1}{(2+\sin x)^2}\times\frac{2+\sin x}{\cos x}\,dx=-\int_0^a\frac{f^{\prime}(x)}{f(x)}\,dx\\&=\Biggl[-\ln|f(x)|\Biggr]_0^a=\ln\frac{|f(0)|}{|f(a)|}\end{aligned} 1 = ∫ 0 a 2 cos x + sin x cos x 2 sin x + 1 d x = ∫ 0 a ( 2 + sin x ) 2 2 sin x + 1 × cos x 2 + sin x d x = − ∫ 0 a f ( x ) f ′ ( x ) d x = [ − ln ∣ f ( x ) ∣ ] 0 a = ln ∣ f ( a ) ∣ ∣ f ( 0 ) ∣ 이다.
+3점 · f ( a ) = 1 2 e \displaystyle f(a)=\frac1{2e} f ( a ) = 2 e 1 따라서 f ( a ) = 1 2 e \displaystyle f(a)=\frac1{2e} f ( a ) = 2 e 1
(별해1) 1 = ∫ 0 a 2 sin x + 1 2 cos x + sin x cos x d x = ∫ 0 a sin x ( 2 + sin x ) + 1 − sin 2 x 2 cos x + sin x cos x d x = ∫ 0 a tan x d x + ∫ 0 a cos x 2 + sin x d x = ln 1 2 f ( a ) \displaystyle \begin{aligned}1&=\int_0^a\frac{2\sin x+1}{2\cos x+\sin x\cos x}\,dx\\&=\int_0^a\frac{\sin x(2+\sin x)+1-\sin^2x}{2\cos x+\sin x\cos x}\,dx\\&=\int_0^a\tan x\,dx+\int_0^a\frac{\cos x}{2+\sin x}\,dx=\ln\frac1{2f(a)}\end{aligned} 1 = ∫ 0 a 2 cos x + sin x cos x 2 sin x + 1 d x = ∫ 0 a 2 cos x + sin x cos x sin x ( 2 + sin x ) + 1 − sin 2 x d x = ∫ 0 a tan x d x + ∫ 0 a 2 + sin x cos x d x = ln 2 f ( a ) 1
(별해2) 1 = ∫ 0 a 2 sin x + 1 2 cos x + sin x cos x d x = ∫ 0 a ( 2 sin x + 1 ) cos x cos 2 x ( 2 + sin x ) d x \displaystyle 1=\int_0^a\frac{2\sin x+1}{2\cos x+\sin x\cos x}\,dx=\int_0^a\frac{(2\sin x+1)\cos x}{\cos^2x(2+\sin x)}\,dx 1 = ∫ 0 a 2 cos x + sin x cos x 2 sin x + 1 d x = ∫ 0 a cos 2 x ( 2 + sin x ) ( 2 sin x + 1 ) cos x d x sin x = t \displaystyle \sin x=t sin x = t ln 1 2 f ( a ) = 1 \displaystyle \ln\frac1{2f(a)}=1 ln 2 f ( a ) 1 = 1
+5점 · 함수 f ( x ) \displaystyle f(x) f ( x ) 3 3 \displaystyle \frac{\sqrt3}3 3 3 − 3 3 \displaystyle -\frac{\sqrt3}3 − 3 3 [논제2] 구간 [ π 2 , 2 π ] \displaystyle \left[\frac\pi2,2\pi\right] [ 2 π , 2 π ] y = f ( x ) \displaystyle y=f(x) y = f ( x ) f ′ ( x ) = − ( 2 + sin x ) sin x − cos 2 x ( 2 + sin x ) 2 = − 2 sin x + 1 ( 2 + sin x ) 2 = 0 \displaystyle f^{\prime}(x)=\frac{-(2+\sin x)\sin x-\cos^2x}{(2+\sin x)^2}=-\frac{2\sin x+1}{(2+\sin x)^2}=0 f ′ ( x ) = ( 2 + sin x ) 2 − ( 2 + sin x ) sin x − cos 2 x = − ( 2 + sin x ) 2 2 sin x + 1 = 0 으로부터 x = 7 6 π \displaystyle x=\frac76\pi x = 6 7 π − 3 3 \displaystyle -\frac{\sqrt3}3 − 3 3 x = 11 6 π \displaystyle x=\frac{11}6\pi x = 6 11 π 3 3 \displaystyle \frac{\sqrt3}3 3 3
+8점 · 구간 [ π 2 , 2 π ] \displaystyle \left[\frac\pi2,2\pi\right] [ 2 π , 2 π ] f ( x ) \displaystyle f(x) f ( x ) 또한 ∙f ′ ( x ) > 0 \displaystyle f^{\prime}(x)>0 f ′ ( x ) > 0 2 sin x + 1 < 0 \displaystyle 2\sin x+1<0 2 sin x + 1 < 0 7 6 π < x < 11 6 π \displaystyle \frac76\pi<x<\frac{11}6\pi 6 7 π < x < 6 11 π f ( x ) \displaystyle f(x) f ( x ) ∙f ′ ( x ) < 0 \displaystyle f^{\prime}(x)<0 f ′ ( x ) < 0 2 sin x + 1 > 0 \displaystyle 2\sin x+1>0 2 sin x + 1 > 0 π 2 ≤ x < 7 6 π \displaystyle \frac\pi2\le x<\frac76\pi 2 π ≤ x < 6 7 π 11 6 π < x ≤ 2 π \displaystyle \frac{11}6\pi<x\le2\pi 6 11 π < x ≤ 2 π f ( x ) \displaystyle f(x) f ( x ) 한다. 이를 이용하여 y = f ( x ) \displaystyle y=f(x) y = f ( x )
곡선 y = f ( x ) \displaystyle y=f(x) y = f ( x )
+12점 · − 1 − 3 < k ≤ − 1 , 1 − 3 , − 1 + 3 < k ≤ 1 , 5 2 ≤ k < 1 + 3 \displaystyle -1-\sqrt3<k\le-1,\quad1-\sqrt3,\quad-1+\sqrt3<k\le1,\quad\frac52\le k<1+\sqrt3 − 1 − 3 < k ≤ − 1 , 1 − 3 , − 1 + 3 < k ≤ 1 , 2 5 ≤ k < 1 + 3 ∣ 3 f ( x ) − k ∣ = 1 \displaystyle |3f(x)-k|=1 ∣3 f ( x ) − k ∣ = 1 f ( x ) = k − 1 3 \displaystyle f(x)=\frac{k-1}3 f ( x ) = 3 k − 1 f ( x ) = k + 1 3 \displaystyle f(x)=\frac{k+1}3 f ( x ) = 3 k + 1 y = ∣ 3 f ( x ) − k ∣ \displaystyle y=|3f(x)-k| y = ∣3 f ( x ) − k ∣ y = 1 \displaystyle y=1 y = 1 2 \displaystyle 2 2 y = f ( x ) \displaystyle y=f(x) y = f ( x ) y = k − 1 3 \displaystyle y=\frac{k-1}3 y = 3 k − 1 y = f ( x ) \displaystyle y=f(x) y = f ( x ) y = k + 1 3 \displaystyle y=\frac{k+1}3 y = 3 k + 1 2 \displaystyle 2 2 두 직선 y = k − 1 3 \displaystyle y=\frac{k-1}3 y = 3 k − 1 y = k + 1 3 \displaystyle y=\frac{k+1}3 y = 3 k + 1 d \displaystyle d d 2 3 \displaystyle \frac23 3 2 d = 2 3 > 3 3 \displaystyle d=\frac23>\frac{\sqrt3}3 d = 3 2 > 3 3 d = 2 3 > 1 2 \displaystyle d=\frac23>\frac12 d = 3 2 > 2 1 y = k − 1 3 \displaystyle y=\frac{k-1}3 y = 3 k − 1 y = k + 1 3 \displaystyle y=\frac{k+1}3 y = 3 k + 1 2 \displaystyle 2 2 k \displaystyle k k (ⅰ) − 3 3 < k + 1 3 ≤ 0 \displaystyle -\frac{\sqrt3}3<\frac{k+1}3\le0 − 3 3 < 3 k + 1 ≤ 0 − 1 − 3 < k ≤ − 1 \displaystyle -1-\sqrt3<k\le-1 − 1 − 3 < k ≤ − 1 (ⅱ) k − 1 3 = − 3 3 \displaystyle \frac{k-1}3=-\frac{\sqrt3}3 3 k − 1 = − 3 3 k = 1 − 3 \displaystyle k=1-\sqrt3 k = 1 − 3 (ⅲ) 3 3 − 2 3 < k − 1 3 ≤ 0 \displaystyle \frac{\sqrt3}3-\frac23<\frac{k-1}3\le0 3 3 − 3 2 < 3 k − 1 ≤ 0 − 1 + 3 < k ≤ 1 \displaystyle -1+\sqrt3<k\le1 − 1 + 3 < k ≤ 1 (ⅳ) 1 2 ≤ k − 1 3 < 3 3 \displaystyle \frac12\le\frac{k-1}3<\frac{\sqrt3}3 2 1 ≤ 3 k − 1 < 3 3 5 2 ≤ k < 1 + 3 \displaystyle \frac52\le k<1+\sqrt3 2 5 ≤ k < 1 + 3 이다. 요약하면 − 1 − 3 < k ≤ − 1 , 1 − 3 , − 1 + 3 < k ≤ 1 , 5 2 ≤ k < 1 + 3 \displaystyle -1-\sqrt3<k\le-1,\quad1-\sqrt3,\quad-1+\sqrt3<k\le1,\quad\frac52\le k<1+\sqrt3 − 1 − 3 < k ≤ − 1 , 1 − 3 , − 1 + 3 < k ≤ 1 , 2 5 ≤ k < 1 + 3
예시 답안의 내용 및 평가 기준의 내용과는 일치하지 않더라도 기술한 내용이 논리적으로 타당한 경우에는 평가위원의 판단에 따라 부분 점수를 부여할 수 있음.
수리논술, 배움에서 논증의 완성까지.
