[문제 5-1] 예시 답안
+3점 · f ′ ( x ) = e x + π 3 ( a sin x + b cos x ) + e x + π 3 ( a cos x − b sin x ) \displaystyle f^{\prime}(x)=e^{x+\frac{\pi}{3}}(a\sin x+b\cos x)+e^{x+\frac{\pi}{3}}(a\cos x-b\sin x) f ′ ( x ) = e x + 3 π ( a sin x + b cos x ) + e x + 3 π ( a cos x − b sin x ) 함수 f ( x ) = e x + π 3 ( a sin x + b cos x ) \displaystyle f(x)=e^{x+\frac{\pi}{3}}(a\sin x+b\cos x) f ( x ) = e x + 3 π ( a sin x + b cos x ) f ′ ( x ) = e x + π 3 ( a sin x + b cos x ) + e x + π 3 ( a cos x − b sin x ) \displaystyle f^{\prime}(x)=e^{x+\frac{\pi}{3}}(a\sin x+b\cos x)+e^{x+\frac{\pi}{3}}(a\cos x-b\sin x) f ′ ( x ) = e x + 3 π ( a sin x + b cos x ) + e x + 3 π ( a cos x − b sin x ) 를 얻는다.
+4점 · 연립방정식 a + 3 b = 2 \displaystyle a+\sqrt{3}b=2 a + 3 b = 2 b = 3 a \displaystyle b=\sqrt{3}a b = 3 a 주어진 조건에 의해 f ( π 6 ) = e π 2 ( a 2 + 3 b 2 ) = e π 2 ⇒ a + 3 b = 2 \displaystyle f\left(\frac{\pi}{6}\right)=e^{\frac{\pi}{2}}\left(\frac{a}{2}+\frac{\sqrt{3}b}{2}\right)=e^{\frac{\pi}{2}}\ \Rightarrow\ a+\sqrt{3}b=2 f ( 6 π ) = e 2 π ( 2 a + 2 3 b ) = e 2 π ⇒ a + 3 b = 2 이고 f ′ ( π 6 ) = e π 2 ( a 2 + 3 b 2 ) + e π 2 ( 3 a 2 − b 2 ) = e π 2 + e π 2 ( 3 a 2 − b 2 ) = e π 2 ⇒ b = 3 a \displaystyle f^{\prime}\left(\frac{\pi}{6}\right)=e^{\frac{\pi}{2}}\left(\frac{a}{2}+\frac{\sqrt{3}b}{2}\right)+e^{\frac{\pi}{2}}\left(\frac{\sqrt{3}a}{2}-\frac{b}{2}\right)=e^{\frac{\pi}{2}}+e^{\frac{\pi}{2}}\left(\frac{\sqrt{3}a}{2}-\frac{b}{2}\right)=e^{\frac{\pi}{2}}\ \Rightarrow\ b=\sqrt{3}a f ′ ( 6 π ) = e 2 π ( 2 a + 2 3 b ) + e 2 π ( 2 3 a − 2 b ) = e 2 π + e 2 π ( 2 3 a − 2 b ) = e 2 π ⇒ b = 3 a
+3점 · a = 1 2 \displaystyle a=\frac{1}{2} a = 2 1 b = 3 2 \displaystyle b=\frac{\sqrt{3}}{2} b = 2 3 이므로, 연립방정식을 풀어 a = 1 2 \displaystyle a=\frac{1}{2} a = 2 1 b = 3 2 \displaystyle b=\frac{\sqrt{3}}{2} b = 2 3
[문제 5-2] 예시 답안
+4점 · f ( x ) = e x + π 3 sin ( x + π 3 ) \displaystyle f(x)=e^{x+\frac{\pi}{3}}\sin\left(x+\frac{\pi}{3}\right) f ( x ) = e x + 3 π sin ( x + 3 π ) a = cos π 3 \displaystyle a=\cos\frac{\pi}{3} a = cos 3 π b = sin π 3 \displaystyle b=\sin\frac{\pi}{3} b = sin 3 π f ( x ) = e x + π 3 { sin x cos ( π 3 ) + sin ( π 3 ) cos x } = e x + π 3 sin ( x + π 3 ) \displaystyle f(x)=e^{x+\frac{\pi}{3}}\left\{\sin x\cos\left(\frac{\pi}{3}\right)+\sin\left(\frac{\pi}{3}\right)\cos x\right\}=e^{x+\frac{\pi}{3}}\sin\left(x+\frac{\pi}{3}\right) f ( x ) = e x + 3 π { sin x cos ( 3 π ) + sin ( 3 π ) cos x } = e x + 3 π sin ( x + 3 π ) 를 얻을 수 있다.
+3점 · x = 5 π 12 \displaystyle x=\frac{5\pi}{12} x = 12 5 π 함수를 미분하여 f ′ ( x ) = e x + π 3 { sin ( x + π 3 ) + cos ( x + π 3 ) } = 0 ⇒ tan ( x + π 3 ) = − 1 \displaystyle f^{\prime}(x)=e^{x+\frac{\pi}{3}}\left\{\sin\left(x+\frac{\pi}{3}\right)+\cos\left(x+\frac{\pi}{3}\right)\right\}=0\ \Rightarrow\ \tan\left(x+\frac{\pi}{3}\right)=-1 f ′ ( x ) = e x + 3 π { sin ( x + 3 π ) + cos ( x + 3 π ) } = 0 ⇒ tan ( x + 3 π ) = − 1 를 풀면, 구간 [ 0 , π ] \displaystyle [0,\pi] [ 0 , π ] x + π 3 = 3 π 4 \displaystyle x+\frac{\pi}{3}=\frac{3\pi}{4} x + 3 π = 4 3 π
+3점 · 극값과 구간의 끝점을 비교하여 최댓값 f ( 5 π 12 ) = 1 2 e 3 π 4 \displaystyle f\left(\frac{5\pi}{12}\right)=\frac{1}{\sqrt{2}}e^{\frac{3\pi}{4}} f ( 12 5 π ) = 2 1 e 4 3 π 따라서 f ( x ) \displaystyle f(x) f ( x ) f ( 0 ) = e π 3 sin ( π 3 ) = 3 2 e π 3 , f ( 5 π 12 ) = e 3 π 4 sin ( 3 π 4 ) = 1 2 e 3 π 4 , f ( π ) = e 4 π 3 sin ( 4 π 3 ) = − 3 2 e 4 π 3 \displaystyle f(0)=e^{\frac{\pi}{3}}\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}e^{\frac{\pi}{3}},\quad f\left(\frac{5\pi}{12}\right)=e^{\frac{3\pi}{4}}\sin\left(\frac{3\pi}{4}\right)=\frac{1}{\sqrt{2}}e^{\frac{3\pi}{4}},\quad f(\pi)=e^{\frac{4\pi}{3}}\sin\left(\frac{4\pi}{3}\right)=-\frac{\sqrt{3}}{2}e^{\frac{4\pi}{3}} f ( 0 ) = e 3 π sin ( 3 π ) = 2 3 e 3 π , f ( 12 5 π ) = e 4 3 π sin ( 4 3 π ) = 2 1 e 4 3 π , f ( π ) = e 3 4 π sin ( 3 4 π ) = − 2 3 e 3 4 π 중 하나이다. 마지막 값은 음수이므로 제외하고, 앞의 두 값을 비교하여 f ( 5 π 12 ) = 1 2 e 3 π 4 \displaystyle f\left(\frac{5\pi}{12}\right)=\frac{1}{\sqrt{2}}e^{\frac{3\pi}{4}} f ( 12 5 π ) = 2 1 e 4 3 π 3 2 e π 3 < 1 2 e 3 π 4 ⇔ 3 2 < e 5 π 12 ⇔ 3 2 < 2 5 2 < e 5 π 6 = ( 2.718 ⋯ ) 5 6 × 3.14 ⋯ , \displaystyle \frac{\sqrt{3}}{2}e^{\frac{\pi}{3}}<\frac{1}{\sqrt{2}}e^{\frac{3\pi}{4}}\ \Leftrightarrow\ \sqrt{\frac{3}{2}}<e^{\frac{5\pi}{12}}\ \Leftrightarrow\ \frac{3}{2}<2^{\frac{5}{2}}<e^{\frac{5\pi}{6}}=(2.718\cdots)^{\frac{5}{6}\times 3.14\cdots}, 2 3 e 3 π < 2 1 e 4 3 π ⇔ 2 3 < e 12 5 π ⇔ 2 3 < 2 2 5 < e 6 5 π = ( 2.718 ⋯ ) 6 5 × 3.14 ⋯ ,
※ 논리 전개 과정이 맞으면 답이 틀리더라도 1 \displaystyle 1 1 2 \displaystyle 2 2
※ 채점자는 답안의 완성도에 따라 − 0.5 \displaystyle -0.5 − 0.5 + 0.5 \displaystyle +0.5 + 0.5
수리논술, 배움에서 논증의 완성까지.