[문제 3-1]
(풀이) 플라스틱 쓰레기 1500톤을 재활용하게 되는 플라스틱의 양을 모두 구하면 1500 ( 0.65 × 0.8 ) + 1500 ( 0.65 × 0.8 ) 2 + 1500 ( 0.65 × 0.8 ) 3 + ⋯ = ∑ n = 1 ∞ 1500 × 0.52 n = 1500 × 0.52 1 − 0.52 = 1625 \displaystyle 1500(0.65\times0.8)+1500(0.65\times0.8)^2+1500(0.65\times0.8)^3+\cdots=\sum_{n=1}^{\infty}1500\times0.52^n=\frac{1500\times0.52}{1-0.52}=1625 1500 ( 0.65 × 0.8 ) + 1500 ( 0.65 × 0.8 ) 2 + 1500 ( 0.65 × 0.8 ) 3 + ⋯ = n = 1 ∑ ∞ 1500 × 0.5 2 n = 1 − 0.52 1500 × 0.52 = 1625
이다. 그러므로 답은 1625톤 이다. (끝)
[문제 3-2]
(풀이) (i) A ( x ) = ∫ 1.5 x ( 10 t 2 − 9 t − 9 ) e − t d t \displaystyle A(x)=\int_{1.5}^x(10t^2-9t-9)e^{-t}\,dt A ( x ) = ∫ 1.5 x ( 10 t 2 − 9 t − 9 ) e − t d t
α ( t ) = 10 t 2 − 9 t − 9 \displaystyle \alpha(t)=10t^2-9t-9 α ( t ) = 10 t 2 − 9 t − 9 β ′ ( t ) = e − t \displaystyle \beta^{\prime}(t)=e^{-t} β ′ ( t ) = e − t α ′ ( t ) = 20 t − 9 \displaystyle \alpha^{\prime}(t)=20t-9 α ′ ( t ) = 20 t − 9 β ( t ) = − e − t \displaystyle \beta(t)=-e^{-t} β ( t ) = − e − t
A ( x ) = [ − ( 10 t 2 − 9 t − 9 ) e − t ] 1.5 x − ∫ 1.5 x ( 20 t − 9 ) ( − e − t ) d t = − ( 10 x 2 − 9 x − 9 ) e − x + ∫ 1.5 x ( 20 t − 9 ) e − t d t \displaystyle \begin{aligned}A(x)&=\Biggl[-(10t^2-9t-9)e^{-t}\Biggr]_{1.5}^x-\int_{1.5}^x(20t-9)(-e^{-t})\,dt\\&=-(10x^2-9x-9)e^{-x}+\int_{1.5}^x(20t-9)e^{-t}\,dt\end{aligned} A ( x ) = [ − ( 10 t 2 − 9 t − 9 ) e − t ] 1.5 x − ∫ 1.5 x ( 20 t − 9 ) ( − e − t ) d t = − ( 10 x 2 − 9 x − 9 ) e − x + ∫ 1.5 x ( 20 t − 9 ) e − t d t
이다. 같은 방법으로 ∫ 1.5 x ( 20 t − 9 ) e − t d t \displaystyle \int_{1.5}^x(20t-9)e^{-t}\,dt ∫ 1.5 x ( 20 t − 9 ) e − t d t α ( t ) = 20 t − 9 \displaystyle \alpha(t)=20t-9 α ( t ) = 20 t − 9 β ′ ( t ) = e − t \displaystyle \beta^{\prime}(t)=e^{-t} β ′ ( t ) = e − t α ′ ( t ) = 20 \displaystyle \alpha^{\prime}(t)=20 α ′ ( t ) = 20 β ( t ) = − e − t \displaystyle \beta(t)=-e^{-t} β ( t ) = − e − t
∫ 1.5 x ( 20 t − 9 ) e − t d t = [ − ( 20 t − 9 ) e − t ] 1.5 x − ∫ 1.5 x 20 ( − e − t ) d t = − { ( 20 x − 9 ) e − x − 21 e − 1.5 } + [ − 20 e − t ] 1.5 x = − ( 20 x + 11 ) e − x + 41 e − 1.5 \displaystyle \begin{aligned}\int_{1.5}^x(20t-9)e^{-t}\,dt&=\Biggl[-(20t-9)e^{-t}\Biggr]_{1.5}^x-\int_{1.5}^x20(-e^{-t})\,dt\\&=-\{(20x-9)e^{-x}-21e^{-1.5}\}+\Biggl[-20e^{-t}\Biggr]_{1.5}^x\\&=-(20x+11)e^{-x}+41e^{-1.5}\end{aligned} ∫ 1.5 x ( 20 t − 9 ) e − t d t = [ − ( 20 t − 9 ) e − t ] 1.5 x − ∫ 1.5 x 20 ( − e − t ) d t = − {( 20 x − 9 ) e − x − 21 e − 1.5 } + [ − 20 e − t ] 1.5 x = − ( 20 x + 11 ) e − x + 41 e − 1.5
이다. 따라서
A ( x ) = ∫ 1.5 x ( 10 t 2 − 9 t − 9 ) e − t d t = ( − 10 x 2 + 9 x + 9 ) e − x + ∫ 1.5 x ( 20 t − 9 ) e − t d t = ( − 10 x 2 − 11 x − 2 ) e − x + 41 e − 1.5 \displaystyle \begin{aligned}A(x)&=\int_{1.5}^x(10t^2-9t-9)e^{-t}\,dt\\&=(-10x^2+9x+9)e^{-x}+\int_{1.5}^x(20t-9)e^{-t}\,dt\\&=(-10x^2-11x-2)e^{-x}+41e^{-1.5}\end{aligned} A ( x ) = ∫ 1.5 x ( 10 t 2 − 9 t − 9 ) e − t d t = ( − 10 x 2 + 9 x + 9 ) e − x + ∫ 1.5 x ( 20 t − 9 ) e − t d t = ( − 10 x 2 − 11 x − 2 ) e − x + 41 e − 1.5
(ii) lim x → ∞ A ( x ) \displaystyle \lim_{x\to\infty}A(x) x → ∞ lim A ( x ) lim x → ∞ { ( − 10 x 2 − 11 x − 2 ) e − x + 41 e − 1.5 } = 41 e − 1.5 \displaystyle \lim_{x\to\infty}\{(-10x^2-11x-2)e^{-x}+41e^{-1.5}\}=41e^{-1.5} x → ∞ lim {( − 10 x 2 − 11 x − 2 ) e − x + 41 e − 1.5 } = 41 e − 1.5
[문제 3-3]
(풀이) 함수 f ( t ) = ( 10 t 2 − 9 t − 9 ) e − t ( t ≥ 1.5 ) \displaystyle f(t)=(10t^2-9t-9)e^{-t}\ (t\ge1.5) f ( t ) = ( 10 t 2 − 9 t − 9 ) e − t ( t ≥ 1.5 )
f ′ ( t ) = ( 20 t − 9 ) e − t − ( 10 t 2 − 9 t − 9 ) e − t = ( − 10 t 2 + 29 t ) e − t f ′ ′ ( t ) = ( − 20 t + 29 ) e − t − ( − 10 t 2 + 29 t ) e − t = ( 10 t 2 − 49 t + 29 ) e − t \displaystyle \begin{aligned}f^{\prime}(t)&=(20t-9)e^{-t}-(10t^2-9t-9)e^{-t}=(-10t^2+29t)e^{-t}\\f^{\prime\prime}(t)&=(-20t+29)e^{-t}-(-10t^2+29t)e^{-t}=(10t^2-49t+29)e^{-t}\end{aligned} f ′ ( t ) f ′′ ( t ) = ( 20 t − 9 ) e − t − ( 10 t 2 − 9 t − 9 ) e − t = ( − 10 t 2 + 29 t ) e − t = ( − 20 t + 29 ) e − t − ( − 10 t 2 + 29 t ) e − t = ( 10 t 2 − 49 t + 29 ) e − t
f ′ ′ ( t ) = 0 \displaystyle f^{\prime\prime}(t)=0 f ′′ ( t ) = 0 t = 49 − 49 2 − 40 × 29 20 = 49 − 1241 20 \displaystyle t=\frac{49-\sqrt{49^2-40\times29}}{20}=\frac{49-\sqrt{1241}}{20} t = 20 49 − 4 9 2 − 40 × 29 = 20 49 − 1241 t = 49 + 1241 20 \displaystyle t=\frac{49+\sqrt{1241}}{20} t = 20 49 + 1241
t > 1.5 \displaystyle t>1.5 t > 1.5 t = 49 + 1241 20 \displaystyle t=\frac{49+\sqrt{1241}}{20} t = 20 49 + 1241 f ′ ′ ( t ) \displaystyle f^{\prime\prime}(t) f ′′ ( t )
t \displaystyle t t 1.5 \displaystyle 1.5 1.5 ⋯ \displaystyle \cdots ⋯ 49 + 1241 20 \displaystyle \frac{49+\sqrt{1241}}{20} 20 49 + 1241 ⋯ \displaystyle \cdots ⋯ f ′ ′ ( t ) \displaystyle f^{\prime\prime}(t) f ′′ ( t ) − 22 e 1.5 \displaystyle -22e^{1.5} − 22 e 1.5 − \displaystyle - − 0 \displaystyle 0 0 + \displaystyle + +
Correction. 해당 함숫값의 − 22 e 1.5 \displaystyle -22e^{1.5} − 22 e 1.5 − 22 e − 1.5 \displaystyle -22e^{-1.5} − 22 e − 1.5
f ′ ′ ( t ) = 0 \displaystyle f^{\prime\prime}(t)=0 f ′′ ( t ) = 0 t = 49 + 1241 20 \displaystyle t=\frac{49+\sqrt{1241}}{20} t = 20 49 + 1241 f ′ ′ ( t ) \displaystyle f^{\prime\prime}(t) f ′′ ( t ) a = 49 + 1241 20 \displaystyle a=\frac{49+\sqrt{1241}}{20} a = 20 49 + 1241
[문제 3-1] (10점) (상) 답이 맞고 그 이유를 명확히 제시한 경우 (중) 답이 틀렸지만 접선의 등비급수 식이 맞는데 답이 틀린 경우 (하) 답이 틀린 경우
[문제 3-2] (15점) (상) 답이 맞고 그 이유를 명확히 제시한 경우 (중) 답이 틀렸지만 정적분 식의 부분적분법에서 계산 실수한 경우 (하) 정적분 식을 제시하지 못한 경우
[문제 3-3] (15점) (상) 답이 맞고 그 이유를 명확히 제시한 경우 (중상) 답이 맞지만 그 이유를 명확히 제시하지 못한 경우 (중하) 답이 틀렸지만 이계도함수를 구한 경우 (하) 이계도함수를 구하지 못한 경우
수리논술, 이게 다다.