논제 1.
+5점 · F ( t ) = t e t \displaystyle F(t)=\frac{t}{e^t} F ( t ) = e t t 급수 ∑ n = 1 ∞ t n e t n \displaystyle \sum_{n=1}^{\infty}\frac{t^n}{e^{tn}} n = 1 ∑ ∞ e t n t n t e t \displaystyle \frac{t}{e^t} e t t 함수 F ( t ) = t e t \displaystyle F(t)=\frac{t}{e^t} F ( t ) = e t t F ′ ( t ) = 1 − t e t \displaystyle F^{\prime}(t)=\frac{1-t}{e^t} F ′ ( t ) = e t 1 − t F ( t ) = t e t \displaystyle F(t)=\frac{t}{e^t} F ( t ) = e t t ( − ∞ , 1 ] \displaystyle (-\infty,1] ( − ∞ , 1 ] [ 1 , ∞ ) \displaystyle [1,\infty) [ 1 , ∞ ) t = 1 \displaystyle t=1 t = 1 1 e \displaystyle \frac1e e 1 또한 lim t → ∞ F ( t ) = 0 , lim t → − ∞ F ( t ) = − ∞ , F ( 0 ) = 0 \displaystyle \lim_{t\to\infty}F(t)=0,\;\lim_{t\to-\infty}F(t)=-\infty,\;F(0)=0 t → ∞ lim F ( t ) = 0 , t → − ∞ lim F ( t ) = − ∞ , F ( 0 ) = 0 a < 0 \displaystyle a<0 a < 0 F ( a ) = − 1 \displaystyle F(a)=-1 F ( a ) = − 1
+5점 · M = 1 e − 1 \displaystyle M=\frac1{e-1} M = e − 1 1 따라서 t > a \displaystyle t>a t > a t \displaystyle t t r = t e t \displaystyle r=\frac{t}{e^t} r = e t t − 1 < r ≤ 1 e \displaystyle -1<r\leq\frac1e − 1 < r ≤ e 1 ∑ n = 1 ∞ t n e t n \displaystyle \sum_{n=1}^{\infty}\frac{t^n}{e^{tn}} n = 1 ∑ ∞ e t n t n r 1 − r \displaystyle \frac{r}{1-r} 1 − r r 그러므로 급수 ∑ n = 1 ∞ t n e t n \displaystyle \sum_{n=1}^{\infty}\frac{t^n}{e^{tn}} n = 1 ∑ ∞ e t n t n r = 1 e \displaystyle r=\frac1e r = e 1 t = 1 \displaystyle t=1 t = 1 ∑ n = 1 ∞ 1 e n = 1 e − 1 \displaystyle \sum_{n=1}^{\infty}\frac1{e^n}=\frac1{e-1} n = 1 ∑ ∞ e n 1 = e − 1 1 t > a \displaystyle t>a t > a t \displaystyle t t f ( x ) \displaystyle f(x) f ( x ) x = t \displaystyle x=t x = t ∑ n = 1 ∞ t n e t n \displaystyle \sum_{n=1}^{\infty}\frac{t^n}{e^{tn}} n = 1 ∑ ∞ e t n t n M = 1 e − 1 \displaystyle M=\frac1{e-1} M = e − 1 1
논제 2.
+8점 · 부등식의 필요충분조건은 cos α ≤ sin α \displaystyle \cos\alpha\leq\sin\alpha cos α ≤ sin α 함수 F ( t ) = t e t \displaystyle F(t)=\frac{t}{e^t} F ( t ) = e t t ( − ∞ , 1 ] \displaystyle (-\infty,1] ( − ∞ , 1 ] ∣ cos α ∣ ≤ 1 , ∣ sin α ∣ ≤ 1 \displaystyle |\cos\alpha|\leq1,\;|\sin\alpha|\leq1 ∣ cos α ∣ ≤ 1 , ∣ sin α ∣ ≤ 1 F ( cos α ) = cos α e cos α ≤ sin α e sin α = F ( sin α ) \displaystyle F(\cos\alpha)=\frac{\cos\alpha}{e^{\cos\alpha}}\leq\frac{\sin\alpha}{e^{\sin\alpha}}=F(\sin\alpha) F ( cos α ) = e c o s α cos α ≤ e s i n α sin α = F ( sin α ) cos α ≤ sin α \displaystyle \cos\alpha\leq\sin\alpha cos α ≤ sin α
+2점 · A = [ π 4 , 5 π 4 ] \displaystyle A=\left[\frac\pi4,\frac{5\pi}4\right] A = [ 4 π , 4 5 π ] 0 ≤ α ≤ 2 π \displaystyle 0\leq\alpha\leq2\pi 0 ≤ α ≤ 2 π cos α ≤ sin α \displaystyle \cos\alpha\leq\sin\alpha cos α ≤ sin α π 4 ≤ α ≤ 5 π 4 \displaystyle \frac\pi4\leq\alpha\leq\frac{5\pi}4 4 π ≤ α ≤ 4 5 π 따라서 A = [ π 4 , 5 π 4 ] \displaystyle A=\left[\frac\pi4,\frac{5\pi}4\right] A = [ 4 π , 4 5 π ]
+2점 · S = ∫ 0 2 π ∣ q ( x ) ∣ d x \displaystyle S=\int_0^{2\pi}|q(x)|dx S = ∫ 0 2 π ∣ q ( x ) ∣ d x q ( x ) = g ( x ) − h ( x ) = e cos x sin x − e sin x cos x \displaystyle q(x)=g(x)-h(x)=e^{\cos x}\sin x-e^{\sin x}\cos x q ( x ) = g ( x ) − h ( x ) = e c o s x sin x − e s i n x cos x 구하고자 하는 도형의 넓이는 S = ∫ 0 2 π ∣ q ( x ) ∣ d x \displaystyle S=\int_0^{2\pi}|q(x)|dx S = ∫ 0 2 π ∣ q ( x ) ∣ d x
+5점 · q ( x ) \displaystyle q(x) q ( x ) e cos x > 0 , e sin x > 0 \displaystyle e^{\cos x}>0,\;e^{\sin x}>0 e c o s x > 0 , e s i n x > 0 q ( x ) ≥ 0 \displaystyle q(x)\geq0 q ( x ) ≥ 0 cos x e cos x ≤ sin x e sin x \displaystyle \frac{\cos x}{e^{\cos x}}\leq\frac{\sin x}{e^{\sin x}} e c o s x cos x ≤ e s i n x sin x S = ∫ 0 2 π ∣ q ( x ) ∣ d x = − ∫ 0 π 4 q ( x ) d x + ∫ π 4 5 π 4 q ( x ) d x − ∫ 5 π 4 2 π q ( x ) d x \displaystyle S=\int_0^{2\pi}|q(x)|dx=-\int_0^{\frac\pi4}q(x)dx+\int_{\frac\pi4}^{\frac{5\pi}4}q(x)dx-\int_{\frac{5\pi}4}^{2\pi}q(x)dx S = ∫ 0 2 π ∣ q ( x ) ∣ d x = − ∫ 0 4 π q ( x ) d x + ∫ 4 π 4 5 π q ( x ) d x − ∫ 4 5 π 2 π q ( x ) d x
+8점 · Q ( x ) = − e cos x − e sin x \displaystyle Q(x)=-e^{\cos x}-e^{\sin x} Q ( x ) = − e c o s x − e s i n x q ( x ) \displaystyle q(x) q ( x ) u = cos x \displaystyle u=\cos x u = cos x d u d x = − sin x \displaystyle \frac{du}{dx}=-\sin x d x d u = − sin x ∫ g ( x ) d x = ∫ e cos x sin x d x = − ∫ e u d u = − e u + C = − e cos x + C \displaystyle \int g(x)dx=\int e^{\cos x}\sin x\,dx=-\int e^u du=-e^u+C=-e^{\cos x}+C ∫ g ( x ) d x = ∫ e c o s x sin x d x = − ∫ e u d u = − e u + C = − e c o s x + C u = sin x \displaystyle u=\sin x u = sin x d u d x = cos x \displaystyle \frac{du}{dx}=\cos x d x d u = cos x ∫ h ( x ) d x = ∫ e sin x cos x d x = ∫ e u d u = e u + C = e sin x + C \displaystyle \int h(x)dx=\int e^{\sin x}\cos x\,dx=\int e^u du=e^u+C=e^{\sin x}+C ∫ h ( x ) d x = ∫ e s i n x cos x d x = ∫ e u d u = e u + C = e s i n x + C 따라서 Q ( x ) = − e cos x − e sin x \displaystyle Q(x)=-e^{\cos x}-e^{\sin x} Q ( x ) = − e c o s x − e s i n x q ( x ) \displaystyle q(x) q ( x )
+5점 · S = 4 ( e 2 2 − e − 2 2 ) \displaystyle S=4\left(e^{\frac{\sqrt2}{2}}-e^{-\frac{\sqrt2}{2}}\right) S = 4 ( e 2 2 − e − 2 2 ) 그러므로 S = ∫ 0 2 π ∣ q ( x ) ∣ d x = − ∫ 0 π 4 q ( x ) d x + ∫ π 4 5 π 4 q ( x ) d x − ∫ 5 π 4 2 π q ( x ) d x = − ( Q ( π 4 ) − Q ( 0 ) ) + ( Q ( 5 π 4 ) − Q ( π 4 ) ) − ( Q ( 2 π ) − Q ( 5 π 4 ) ) = 2 ( Q ( 5 π 4 ) − Q ( π 4 ) ) = 4 ( e 2 2 − e − 2 2 ) \displaystyle \begin{aligned}S&=\int_0^{2\pi}|q(x)|dx=-\int_0^{\frac\pi4}q(x)dx+\int_{\frac\pi4}^{\frac{5\pi}4}q(x)dx-\int_{\frac{5\pi}4}^{2\pi}q(x)dx\\&=-\left(Q\left(\frac\pi4\right)-Q(0)\right)+\left(Q\left(\frac{5\pi}4\right)-Q\left(\frac\pi4\right)\right)-\left(Q(2\pi)-Q\left(\frac{5\pi}4\right)\right)\\&=2\left(Q\left(\frac{5\pi}4\right)-Q\left(\frac\pi4\right)\right)=4\left(e^{\frac{\sqrt2}{2}}-e^{-\frac{\sqrt2}{2}}\right)\end{aligned} S = ∫ 0 2 π ∣ q ( x ) ∣ d x = − ∫ 0 4 π q ( x ) d x + ∫ 4 π 4 5 π q ( x ) d x − ∫ 4 5 π 2 π q ( x ) d x = − ( Q ( 4 π ) − Q ( 0 ) ) + ( Q ( 4 5 π ) − Q ( 4 π ) ) − ( Q ( 2 π ) − Q ( 4 5 π ) ) = 2 ( Q ( 4 5 π ) − Q ( 4 π ) ) = 4 ( e 2 2 − e − 2 2 )
수리논술, 배움에서 논증의 완성까지.