+10점 · f ( 1 ) = ∫ ln 2 1 ( 1 − t ) e − t ln t d t + ∫ 1 ln 3 ( t − 1 ) e − t ln t d t \displaystyle f(1)=\int_{\ln2}^1(1-t)e^{-t}\ln t\,dt+\int_1^{\ln3}(t-1)e^{-t}\ln t\,dt f ( 1 ) = ∫ l n 2 1 ( 1 − t ) e − t ln t d t + ∫ 1 l n 3 ( t − 1 ) e − t ln t d t (3-1) f ( 1 ) = ∫ ln 2 ln 3 ∣ 1 − t ∣ e − t ln t d t = ∫ ln 2 1 ( 1 − t ) e − t ln t d t + ∫ 1 ln 3 ( t − 1 ) e − t ln t d t \displaystyle f(1)=\int_{\ln2}^{\ln3}|1-t|e^{-t}\ln t\,dt=\int_{\ln2}^1(1-t)e^{-t}\ln t\,dt+\int_1^{\ln3}(t-1)e^{-t}\ln t\,dt f ( 1 ) = ∫ l n 2 l n 3 ∣1 − t ∣ e − t ln t d t = ∫ l n 2 1 ( 1 − t ) e − t ln t d t + ∫ 1 l n 3 ( t − 1 ) e − t ln t d t
+30점 · ∫ ( t − 1 ) e − t ln t d t = − t e − t ln t − e − t + C \displaystyle \int(t-1)e^{-t}\ln t\,dt=-te^{-t}\ln t-e^{-t}+C ∫ ( t − 1 ) e − t ln t d t = − t e − t ln t − e − t + C +20점 · ∫ 1 ln 3 ( t − 1 ) e − t ln t d t = − ln 3 3 ln ( ln 3 ) − 1 3 + e − 1 \displaystyle \int_1^{\ln3}(t-1)e^{-t}\ln t\,dt=-\frac{\ln3}{3}\ln(\ln3)-\frac13+e^{-1} ∫ 1 l n 3 ( t − 1 ) e − t ln t d t = − 3 ln 3 ln ( ln 3 ) − 3 1 + e − 1 ∫ ( t − 1 ) e − t d t = − ( t − 1 ) e − t + ∫ e − t d t = − t e − t + C \displaystyle \int(t-1)e^{-t}dt=-(t-1)e^{-t}+\int e^{-t}dt=-te^{-t}+C ∫ ( t − 1 ) e − t d t = − ( t − 1 ) e − t + ∫ e − t d t = − t e − t + C 이므로 ∫ 1 ln 3 ( t − 1 ) e − t ln t d t = [ − t e − t × ln t ] 1 ln 3 + ∫ 1 ln 3 e − t d t = [ − t e − t × ln t ] 1 ln 3 + [ − e − t ] 1 ln 3 = − ln 3 3 ln ( ln 3 ) − 1 3 + e − 1 \displaystyle \begin{aligned}\int_1^{\ln3}(t-1)e^{-t}\ln t\,dt&=\Biggl[-te^{-t}\times\ln t\Biggr]_1^{\ln3}+\int_1^{\ln3}e^{-t}dt\\&=\Biggl[-te^{-t}\times\ln t\Biggr]_1^{\ln3}+\Biggl[-e^{-t}\Biggr]_1^{\ln3}=-\frac{\ln3}{3}\ln(\ln3)-\frac13+e^{-1}\end{aligned} ∫ 1 l n 3 ( t − 1 ) e − t ln t d t = [ − t e − t × ln t ] 1 l n 3 + ∫ 1 l n 3 e − t d t = [ − t e − t × ln t ] 1 l n 3 + [ − e − t ] 1 l n 3 = − 3 ln 3 ln ( ln 3 ) − 3 1 + e − 1
+20점 · 답을 구하면 ∫ ln 2 1 ( 1 − t ) e − t ln t d t = ∫ 1 ln 2 ( t − 1 ) e − t ln t d t = [ − t e − t × ln t ] 1 ln 2 + ∫ 1 ln 2 e − t d t = [ − t e − t × ln t ] 1 ln 2 + [ − e − t ] 1 ln 2 = − ln 2 2 ln ( ln 2 ) − 1 2 + e − 1 \displaystyle \begin{aligned}\int_{\ln2}^1(1-t)e^{-t}\ln t\,dt&=\int_1^{\ln2}(t-1)e^{-t}\ln t\,dt=\Biggl[-te^{-t}\times\ln t\Biggr]_1^{\ln2}+\int_1^{\ln2}e^{-t}dt\\&=\Biggl[-te^{-t}\times\ln t\Biggr]_1^{\ln2}+\Biggl[-e^{-t}\Biggr]_1^{\ln2}=-\frac{\ln2}{2}\ln(\ln2)-\frac12+e^{-1}\end{aligned} ∫ l n 2 1 ( 1 − t ) e − t ln t d t = ∫ 1 l n 2 ( t − 1 ) e − t ln t d t = [ − t e − t × ln t ] 1 l n 2 + ∫ 1 l n 2 e − t d t = [ − t e − t × ln t ] 1 l n 2 + [ − e − t ] 1 l n 2 = − 2 ln 2 ln ( ln 2 ) − 2 1 + e − 1 이다. 따라서 f ( 1 ) = − ln 3 3 ln ( ln 3 ) − ln 2 2 ln ( ln 2 ) − 5 6 + 2 e \displaystyle f(1)=-\frac{\ln3}{3}\ln(\ln3)-\frac{\ln2}{2}\ln(\ln2)-\frac56+\frac2e f ( 1 ) = − 3 ln 3 ln ( ln 3 ) − 2 ln 2 ln ( ln 2 ) − 6 5 + e 2 이다.
+30점 · ln x ≤ x − 1 \displaystyle \ln x\le x-1 ln x ≤ x − 1 (3-2) h ( x ) = ln x − x + 1 \displaystyle h(x)=\ln x-x+1 h ( x ) = ln x − x + 1 h ′ ( x ) = 1 x − 1 = 1 − x x \displaystyle h^{\prime}(x)=\frac1x-1=\frac{1-x}{x} h ′ ( x ) = x 1 − 1 = x 1 − x x = 1 \displaystyle x=1 x = 1 h ( 1 ) = 0 \displaystyle h(1)=0 h ( 1 ) = 0 Correction. h ( x ) = ln x − x + 1 \displaystyle h(x)=\ln x-x+1 h ( x ) = ln x − x + 1 x = 1 \displaystyle x=1 x = 1 0 \displaystyle 0 0 따라서 h ( x ) ≤ 0 \displaystyle h(x)\le0 h ( x ) ≤ 0 ln x ≤ x − 1 \displaystyle \ln x\le x-1 ln x ≤ x − 1
+20점 · ∫ ln 2 ln 3 e − t ln t d t ≤ ∫ ln 2 ln 3 e − t ( t − 1 ) d t \displaystyle \int_{\ln2}^{\ln3}e^{-t}\ln t\,dt\le\int_{\ln2}^{\ln3}e^{-t}(t-1)dt ∫ l n 2 l n 3 e − t ln t d t ≤ ∫ l n 2 l n 3 e − t ( t − 1 ) d t +30점 · g ( ln 3 ) < 0 \displaystyle g(\ln3)<0 g ( ln 3 ) < 0 g ( ln 3 ) = ∫ ln 2 ln 3 e − t ln t d t ≤ ∫ ln 2 ln 3 e − t ( t − 1 ) d t = [ − t e − t ] ln 2 ln 3 = ln 2 2 − ln 3 3 = 3 ln 2 − 2 ln 3 6 = 1 6 ln 8 9 < 0 \displaystyle \begin{aligned}g(\ln3)&=\int_{\ln2}^{\ln3}e^{-t}\ln t\,dt\le\int_{\ln2}^{\ln3}e^{-t}(t-1)dt=\Biggl[-te^{-t}\Biggr]_{\ln2}^{\ln3}\\&=\frac{\ln2}{2}-\frac{\ln3}{3}=\frac{3\ln2-2\ln3}{6}=\frac16\ln\frac89<0\end{aligned} g ( ln 3 ) = ∫ l n 2 l n 3 e − t ln t d t ≤ ∫ l n 2 l n 3 e − t ( t − 1 ) d t = [ − t e − t ] l n 2 l n 3 = 2 ln 2 − 3 ln 3 = 6 3 ln 2 − 2 ln 3 = 6 1 ln 9 8 < 0 이다.
(3-3) ln 2 ≤ x ≤ ln 3 \displaystyle \ln2\le x\le\ln3 ln 2 ≤ x ≤ ln 3 f ( x ) = ∫ ln 2 x ( x − t ) e − t ln t d t + ∫ x ln 3 ( t − x ) e − t ln t d t = x ∫ ln 2 x e − t ln t d t − ∫ ln 2 x t e − t ln t d t + x ∫ ln 3 x e − t ln t d t − ∫ ln 3 x t e − t ln t d t \displaystyle \begin{aligned}f(x)&=\int_{\ln2}^x(x-t)e^{-t}\ln t\,dt+\int_x^{\ln3}(t-x)e^{-t}\ln t\,dt\\&=x\int_{\ln2}^xe^{-t}\ln t\,dt-\int_{\ln2}^xte^{-t}\ln t\,dt+x\int_{\ln3}^xe^{-t}\ln t\,dt-\int_{\ln3}^xte^{-t}\ln t\,dt\end{aligned} f ( x ) = ∫ l n 2 x ( x − t ) e − t ln t d t + ∫ x l n 3 ( t − x ) e − t ln t d t = x ∫ l n 2 x e − t ln t d t − ∫ l n 2 x t e − t ln t d t + x ∫ l n 3 x e − t ln t d t − ∫ l n 3 x t e − t ln t d t 이므로 f ( ln 2 ) = ∫ ln 2 ln 3 ( t − ln 2 ) e − t ln t d t \displaystyle f(\ln2)=\int_{\ln2}^{\ln3}(t-\ln2)e^{-t}\ln t\,dt f ( ln 2 ) = ∫ l n 2 l n 3 ( t − ln 2 ) e − t ln t d t f ( ln 3 ) = ∫ ln 2 ln 3 ( ln 3 − t ) e − t ln t d t \displaystyle f(\ln3)=\int_{\ln2}^{\ln3}(\ln3-t)e^{-t}\ln t\,dt f ( ln 3 ) = ∫ l n 2 l n 3 ( ln 3 − t ) e − t ln t d t
+30점 · ln 2 < x < ln 3 \displaystyle \ln2<x<\ln3 ln 2 < x < ln 3 f ′ ( x ) = ∫ ln 2 x e − t ln t d t + ∫ ln 3 x e − t ln t d t \displaystyle f^{\prime}(x)=\int_{\ln2}^xe^{-t}\ln t\,dt+\int_{\ln3}^xe^{-t}\ln t\,dt f ′ ( x ) = ∫ l n 2 x e − t ln t d t + ∫ l n 3 x e − t ln t d t 따라서 ln 2 < x < ln 3 \displaystyle \ln2<x<\ln3 ln 2 < x < ln 3 f ′ ( x ) = ∫ ln 2 x e − t ln t d t + x e − x ln x − x e − x ln x + ∫ ln 3 x e − t ln t d t + x e − x ln x − x e − x ln x = ∫ ln 2 x e − t ln t d t + ∫ ln 3 x e − t ln t d t \displaystyle \begin{aligned}f^{\prime}(x)&=\int_{\ln2}^xe^{-t}\ln t\,dt+xe^{-x}\ln x-xe^{-x}\ln x+\int_{\ln3}^xe^{-t}\ln t\,dt+xe^{-x}\ln x-xe^{-x}\ln x\\&=\int_{\ln2}^xe^{-t}\ln t\,dt+\int_{\ln3}^xe^{-t}\ln t\,dt\end{aligned} f ′ ( x ) = ∫ l n 2 x e − t ln t d t + x e − x ln x − x e − x ln x + ∫ l n 3 x e − t ln t d t + x e − x ln x − x e − x ln x = ∫ l n 2 x e − t ln t d t + ∫ l n 3 x e − t ln t d t 이다.
+30점 · f ′ ( a ) = 0 \displaystyle f^{\prime}(a)=0 f ′ ( a ) = 0 f ′ ′ ( x ) = 2 e − x ln x \displaystyle f^{\prime\prime}(x)=2e^{-x}\ln x f ′′ ( x ) = 2 e − x ln x x < 1 \displaystyle x<1 x < 1 f ′ ′ ( x ) < 0 \displaystyle f^{\prime\prime}(x)<0 f ′′ ( x ) < 0 x > 1 \displaystyle x>1 x > 1 f ′ ′ ( x ) > 0 \displaystyle f^{\prime\prime}(x)>0 f ′′ ( x ) > 0 f ′ ( x ) \displaystyle f^{\prime}(x) f ′ ( x ) x < 1 \displaystyle x<1 x < 1 x > 1 \displaystyle x>1 x > 1 f ′ ( ln 2 ) = ∫ ln 3 ln 2 e − t ln t d t = − g ( ln 3 ) > 0 \displaystyle f^{\prime}(\ln2)=\int_{\ln3}^{\ln2}e^{-t}\ln t\,dt=-g(\ln3)>0 f ′ ( ln 2 ) = ∫ l n 3 l n 2 e − t ln t d t = − g ( ln 3 ) > 0 f ′ ( ln 3 ) = ∫ ln 2 ln 3 e − t ln t d t = g ( ln 3 ) < 0 \displaystyle f^{\prime}(\ln3)=\int_{\ln2}^{\ln3}e^{-t}\ln t\,dt=g(\ln3)<0 f ′ ( ln 3 ) = ∫ l n 2 l n 3 e − t ln t d t = g ( ln 3 ) < 0 이므로 f ′ ( x ) \displaystyle f^{\prime}(x) f ′ ( x )
즉 함수 f ( x ) \displaystyle f(x) f ( x ) f ′ ( a ) = 0 \displaystyle f^{\prime}(a)=0 f ′ ( a ) = 0 a ∈ [ ln 2 , ln 3 ] \displaystyle a\in[\ln2,\ln3] a ∈ [ ln 2 , ln 3 ] x < a \displaystyle x<a x < a x > a \displaystyle x>a x > a [ ln 2 , ln 3 ] \displaystyle [\ln2,\ln3] [ ln 2 , ln 3 ] x = a \displaystyle x=a x = a a \displaystyle a a f ′ ( a ) = ∫ ln 2 a e − t ln t d t − ∫ a ln 3 e − t ln t d t = 0 \displaystyle f^{\prime}(a)=\int_{\ln2}^ae^{-t}\ln t\,dt-\int_a^{\ln3}e^{-t}\ln t\,dt=0 f ′ ( a ) = ∫ l n 2 a e − t ln t d t − ∫ a l n 3 e − t ln t d t = 0 을 만족한다.
+20점 · 답을 구하면 한편 g ( ln 3 ) = ∫ ln 2 a e − t ln t d t + ∫ a ln 3 e − t ln t d t = 2 g ( a ) \displaystyle g(\ln3)=\int_{\ln2}^ae^{-t}\ln t\,dt+\int_a^{\ln3}e^{-t}\ln t\,dt=2g(a) g ( ln 3 ) = ∫ l n 2 a e − t ln t d t + ∫ a l n 3 e − t ln t d t = 2 g ( a ) g ( a ) g ( ln 3 ) = 1 2 \displaystyle \frac{g(a)}{g(\ln3)}=\frac12 g ( ln 3 ) g ( a ) = 2 1
수리논술, 이게 다다.