+10점 · x ≠ 0 \displaystyle x\ne0 x = 0 f ′ ( x ) \displaystyle f\prime(x) f ′ ( x ) (2-1) f ( x ) = { ( 1 + x 3 ) − 2 ( x ≥ 0 ) ( 1 − x 3 ) − 2 ( x < 0 ) \displaystyle f(x)=\begin{cases}(1+x^3)^{-2}&(x\ge0)\\(1-x^3)^{-2}&(x<0)\end{cases} f ( x ) = { ( 1 + x 3 ) − 2 ( 1 − x 3 ) − 2 ( x ≥ 0 ) ( x < 0 ) f ′ ( x ) = { − 6 x 2 ( 1 + x 3 ) − 3 ( x > 0 ) 6 x 2 ( 1 − x 3 ) − 3 ( x < 0 ) \displaystyle f\prime(x)=\begin{cases}-6x^2(1+x^3)^{-3}&(x>0)\\6x^2(1-x^3)^{-3}&(x<0)\end{cases} f ′ ( x ) = { − 6 x 2 ( 1 + x 3 ) − 3 6 x 2 ( 1 − x 3 ) − 3 ( x > 0 ) ( x < 0 )
+10점 · f ′ ( 0 ) = 0 \displaystyle f\prime(0)=0 f ′ ( 0 ) = 0 한편 lim x → 0 + f ( x ) − f ( 0 ) x = lim x → 0 + 1 x { 1 ( 1 + x 3 ) 2 − 1 } = lim x → 0 + 1 x { − 2 x 3 − x 6 ( 1 + x 3 ) 2 } = 0 \displaystyle \lim_{x\to0+}\frac{f(x)-f(0)}x=\lim_{x\to0+}\frac1x\{\frac1{(1+x^3)^2}-1\}=\lim_{x\to0+}\frac1x\{\frac{-2x^3-x^6}{(1+x^3)^2}\}=0 x → 0 + lim x f ( x ) − f ( 0 ) = x → 0 + lim x 1 { ( 1 + x 3 ) 2 1 − 1 } = x → 0 + lim x 1 { ( 1 + x 3 ) 2 − 2 x 3 − x 6 } = 0 lim x → 0 − f ( x ) − f ( 0 ) x = lim x → 0 − 1 x { 1 ( 1 − x 3 ) 2 − 1 } = lim x → 0 − 1 x { 2 x 3 − x 6 ( 1 − x 3 ) 2 } = 0 \displaystyle \lim_{x\to0-}\frac{f(x)-f(0)}x=\lim_{x\to0-}\frac1x\{\frac1{(1-x^3)^2}-1\}=\lim_{x\to0-}\frac1x\{\frac{2x^3-x^6}{(1-x^3)^2}\}=0 x → 0 − lim x f ( x ) − f ( 0 ) = x → 0 − lim x 1 { ( 1 − x 3 ) 2 1 − 1 } = x → 0 − lim x 1 { ( 1 − x 3 ) 2 2 x 3 − x 6 } = 0 이므로 f ′ ( 0 ) = 0 \displaystyle f\prime(0)=0 f ′ ( 0 ) = 0
+10점 · f ′ ( x ) < 0 \displaystyle f\prime(x)<0 f ′ ( x ) < 0 x > 0 \displaystyle x>0 x > 0 따라서 f ′ ( x ) < 0 \displaystyle f\prime(x)<0 f ′ ( x ) < 0 x \displaystyle x x x > 0 \displaystyle x>0 x > 0
+20점 · x ≠ 0 \displaystyle x\ne0 x = 0 f ′ ′ ( x ) \displaystyle f\prime\prime(x) f ′′ ( x ) 한편 x > 0 \displaystyle x>0 x > 0 f ′ ′ ( x ) = − 12 x ( 1 + x 3 ) − 3 + 54 x 4 ( 1 + x 3 ) − 4 = 6 x ( 1 + x 3 ) − 4 ( − 2 + 7 x 3 ) \displaystyle f\prime\prime(x)=-12x(1+x^3)^{-3}+54x^4(1+x^3)^{-4}=6x(1+x^3)^{-4}(-2+7x^3) f ′′ ( x ) = − 12 x ( 1 + x 3 ) − 3 + 54 x 4 ( 1 + x 3 ) − 4 = 6 x ( 1 + x 3 ) − 4 ( − 2 + 7 x 3 ) 이고, x < 0 \displaystyle x<0 x < 0 f ′ ′ ( x ) = 12 x ( 1 − x 3 ) − 3 + 54 x 4 ( 1 − x 3 ) − 4 = 6 x ( 1 − x 3 ) − 4 ( 2 + 7 x 3 ) \displaystyle f\prime\prime(x)=12x(1-x^3)^{-3}+54x^4(1-x^3)^{-4}=6x(1-x^3)^{-4}(2+7x^3) f ′′ ( x ) = 12 x ( 1 − x 3 ) − 3 + 54 x 4 ( 1 − x 3 ) − 4 = 6 x ( 1 − x 3 ) − 4 ( 2 + 7 x 3 )
+10점 · f ′ ′ ( 0 ) = 0 \displaystyle f\prime\prime(0)=0 f ′′ ( 0 ) = 0 이며, lim x → 0 + f ′ ( x ) − f ′ ( 0 ) x = lim x → 0 + ( − 1 x ) × 6 x 2 ( 1 + x 3 ) − 3 = 0 \displaystyle \lim_{x\to0+}\frac{f\prime(x)-f\prime(0)}x=\lim_{x\to0+}(-\frac1x)\times6x^2(1+x^3)^{-3}=0 x → 0 + lim x f ′ ( x ) − f ′ ( 0 ) = x → 0 + lim ( − x 1 ) × 6 x 2 ( 1 + x 3 ) − 3 = 0 lim x → 0 − f ′ ( x ) − f ′ ( 0 ) x = lim x → 0 − 1 x × 6 x 2 ( 1 − x 3 ) − 3 = 0 \displaystyle \lim_{x\to0-}\frac{f\prime(x)-f\prime(0)}x=\lim_{x\to0-}\frac1x\times6x^2(1-x^3)^{-3}=0 x → 0 − lim x f ′ ( x ) − f ′ ( 0 ) = x → 0 − lim x 1 × 6 x 2 ( 1 − x 3 ) − 3 = 0 이므로 f ′ ′ ( 0 ) = 0 \displaystyle f\prime\prime(0)=0 f ′′ ( 0 ) = 0
+10점 · f ′ ′ ( x ) < 0 \displaystyle f\prime\prime(x)<0 f ′′ ( x ) < 0 − 2 7 3 < x < 0 \displaystyle -\sqrt[3]{\frac27}<x<0 − 3 7 2 < x < 0 0 < x < 2 7 3 \displaystyle 0<x<\sqrt[3]{\frac27} 0 < x < 3 7 2 따라서 f ′ ′ ( x ) < 0 ⇔ 0 < x 3 < 2 7 \displaystyle f\prime\prime(x)<0\Leftrightarrow0<x^3<\frac27 f ′′ ( x ) < 0 ⇔ 0 < x 3 < 7 2 − 2 7 < x 3 < 0 \displaystyle -\frac27<x^3<0 − 7 2 < x 3 < 0 이다. 그러므로 f ′ ′ ( x ) < 0 \displaystyle f\prime\prime(x)<0 f ′′ ( x ) < 0 x \displaystyle x x − 2 7 3 < x < 0 \displaystyle -\sqrt[3]{\frac27}<x<0 − 3 7 2 < x < 0 0 < x < 2 7 3 \displaystyle 0<x<\sqrt[3]{\frac27} 0 < x < 3 7 2
(2-2) 함수 f ( x ) \displaystyle f(x) f ( x ) f ( x ) > 0 \displaystyle f(x)>0 f ( x ) > 0 x \displaystyle x x f ( x ) = f ( − x ) \displaystyle f(x)=f(-x) f ( x ) = f ( − x ) y \displaystyle y y f ( 0 ) = 1 \displaystyle f(0)=1 f ( 0 ) = 1 y \displaystyle y y ( 0 , 1 ) \displaystyle (0,1) ( 0 , 1 )
x \displaystyle x x ⋯ \displaystyle \cdots ⋯ − 2 7 3 \displaystyle -\sqrt[3]{\frac27} − 3 7 2 ⋯ \displaystyle \cdots ⋯ 0 \displaystyle 0 0 ⋯ \displaystyle \cdots ⋯ 2 7 3 \displaystyle \sqrt[3]{\frac27} 3 7 2 ⋯ \displaystyle \cdots ⋯ f ′ ( x ) \displaystyle f\prime(x) f ′ ( x ) + \displaystyle + + + \displaystyle + + + \displaystyle + + 0 \displaystyle 0 0 − \displaystyle - − − \displaystyle - − − \displaystyle - − f ′ ′ ( x ) \displaystyle f\prime\prime(x) f ′′ ( x ) + \displaystyle + + 0 \displaystyle 0 0 − \displaystyle - − 0 \displaystyle 0 0 − \displaystyle - − 0 \displaystyle 0 0 + \displaystyle + + f ( x ) \displaystyle f(x) f ( x ) 변곡점 극대 변곡점
+30점 · 증감, 볼록성, 점근선을 설명하고 y = f ( x ) \displaystyle y=f(x) y = f ( x ) 즉, 함수 f ( x ) \displaystyle f(x) f ( x ) x < 0 \displaystyle x<0 x < 0 x > 0 \displaystyle x>0 x > 0 ( − 2 7 3 , 2 7 3 ) \displaystyle (-\sqrt[3]{\frac27},\sqrt[3]{\frac27}) ( − 3 7 2 , 3 7 2 ) lim x → ∞ f ( x ) = 0 \displaystyle \lim_{x\to\infty}f(x)=0 x → ∞ lim f ( x ) = 0 lim x → − ∞ f ( x ) = 0 \displaystyle \lim_{x\to-\infty}f(x)=0 x → − ∞ lim f ( x ) = 0 x \displaystyle x x f ( x ) \displaystyle f(x) f ( x )
한편 함수 g ( x ) = ∫ x x + 4 f ( t ) d t \displaystyle g(x)=\int_x^{x+4}f(t)dt g ( x ) = ∫ x x + 4 f ( t ) d t g ′ ( x ) = f ( x + 4 ) − f ( x ) \displaystyle g\prime(x)=f(x+4)-f(x) g ′ ( x ) = f ( x + 4 ) − f ( x ) g ′ ( x ) = 0 ⇔ f ( x + 4 ) = f ( x ) ⇔ 1 + ∣ x ∣ 3 = 1 + ∣ x + 4 ∣ 3 ⇔ ∣ x ∣ = ∣ x + 4 ∣ ⇔ x = − 2 \displaystyle g\prime(x)=0\Leftrightarrow f(x+4)=f(x)\Leftrightarrow1+|x|^3=1+|x+4|^3\Leftrightarrow|x|=|x+4|\Leftrightarrow x=-2 g ′ ( x ) = 0 ⇔ f ( x + 4 ) = f ( x ) ⇔ 1 + ∣ x ∣ 3 = 1 + ∣ x + 4 ∣ 3 ⇔ ∣ x ∣ = ∣ x + 4∣ ⇔ x = − 2 에서, x < − 2 \displaystyle x<-2 x < − 2 f ( x + 4 ) > f ( x ) \displaystyle f(x+4)>f(x) f ( x + 4 ) > f ( x ) x > − 2 \displaystyle x>-2 x > − 2 f ( x + 4 ) < f ( x ) \displaystyle f(x+4)<f(x) f ( x + 4 ) < f ( x )
+30점 · g ′ ( x ) \displaystyle g\prime(x) g ′ ( x ) x < − 2 \displaystyle x<-2 x < − 2 x > − 2 \displaystyle x>-2 x > − 2 그러므로 g ′ ( x ) \displaystyle g\prime(x) g ′ ( x ) x < − 2 \displaystyle x<-2 x < − 2 x > − 2 \displaystyle x>-2 x > − 2 g ( x ) \displaystyle g(x) g ( x ) x < − 2 \displaystyle x<-2 x < − 2 x > − 2 \displaystyle x>-2 x > − 2 g ( x ) \displaystyle g(x) g ( x ) x = − 2 \displaystyle x=-2 x = − 2
+20점 · a = − 2 \displaystyle a=-2 a = − 2 즉 a = − 2 \displaystyle a=-2 a = − 2
(참고) 함수 g ( x ) = ∫ x x + 4 f ( t ) d t \displaystyle g(x)=\int_x^{x+4}f(t)dt g ( x ) = ∫ x x + 4 f ( t ) d t g ′ ( x ) = f ( x + 4 ) − f ( x ) \displaystyle g\prime(x)=f(x+4)-f(x) g ′ ( x ) = f ( x + 4 ) − f ( x ) f ( x + 4 ) − f ( x ) = 1 ( 1 + ∣ x + 4 ∣ 3 ) 2 − 1 ( 1 + ∣ x ∣ 3 ) 2 = ( 1 + ∣ x ∣ 3 ) 2 − ( 1 + ∣ x + 4 ∣ 3 ) 2 ( 1 + ∣ x ∣ 3 ) 2 ( 1 + ∣ x + 4 ∣ 3 ) 2 = ( ∣ x ∣ 3 − ∣ x + 4 ∣ 3 ) ( 2 + ∣ x ∣ 3 + ∣ x + 4 ∣ 3 ) ( 1 + ∣ x ∣ 3 ) 2 ( 1 + ∣ x + 4 ∣ 3 ) 2 \displaystyle \begin{aligned}f(x+4)-f(x)&=\frac1{(1+|x+4|^3)^2}-\frac1{(1+|x|^3)^2}=\frac{(1+|x|^3)^2-(1+|x+4|^3)^2}{(1+|x|^3)^2(1+|x+4|^3)^2}\\&=\frac{(|x|^3-|x+4|^3)(2+|x|^3+|x+4|^3)}{(1+|x|^3)^2(1+|x+4|^3)^2}\end{aligned} f ( x + 4 ) − f ( x ) = ( 1 + ∣ x + 4 ∣ 3 ) 2 1 − ( 1 + ∣ x ∣ 3 ) 2 1 = ( 1 + ∣ x ∣ 3 ) 2 ( 1 + ∣ x + 4 ∣ 3 ) 2 ( 1 + ∣ x ∣ 3 ) 2 − ( 1 + ∣ x + 4 ∣ 3 ) 2 = ( 1 + ∣ x ∣ 3 ) 2 ( 1 + ∣ x + 4 ∣ 3 ) 2 ( ∣ x ∣ 3 − ∣ x + 4 ∣ 3 ) ( 2 + ∣ x ∣ 3 + ∣ x + 4 ∣ 3 ) 이고, 2 + ∣ x ∣ 3 + ∣ x + 4 ∣ 3 ( 1 + ∣ x ∣ 3 ) 2 ( 1 + ∣ x + 4 ∣ 3 ) 2 > 0 \displaystyle \frac{2+|x|^3+|x+4|^3}{(1+|x|^3)^2(1+|x+4|^3)^2}>0 ( 1 + ∣ x ∣ 3 ) 2 ( 1 + ∣ x + 4 ∣ 3 ) 2 2 + ∣ x ∣ 3 + ∣ x + 4 ∣ 3 > 0 g ′ ( x ) > 0 ⇔ ∣ x ∣ 3 > ∣ x + 4 ∣ 3 ⇔ x < − 2 \displaystyle g\prime(x)>0\Leftrightarrow|x|^3>|x+4|^3\Leftrightarrow x<-2 g ′ ( x ) > 0 ⇔ ∣ x ∣ 3 > ∣ x + 4 ∣ 3 ⇔ x < − 2 이다. 즉 g ( x ) \displaystyle g(x) g ( x ) x < − 2 \displaystyle x<-2 x < − 2 x > − 2 \displaystyle x>-2 x > − 2 g ( x ) \displaystyle g(x) g ( x ) x = − 2 \displaystyle x=-2 x = − 2 a = − 2 \displaystyle a=-2 a = − 2
+20점 · g ( a ) = − 2 ∫ 0 − 2 f ( t ) d t \displaystyle g(a)=-2\int_0^{-2}f(t)dt g ( a ) = − 2 ∫ 0 − 2 f ( t ) d t (2-3) f ( t ) = f ( − t ) \displaystyle f(t)=f(-t) f ( t ) = f ( − t ) g ( a ) = g ( − 2 ) = ∫ − 2 2 f ( t ) d t = 2 ∫ − 2 0 f ( t ) d t = − 2 ∫ 0 − 2 f ( t ) d t \displaystyle g(a)=g(-2)=\int_{-2}^2f(t)dt=2\int_{-2}^0f(t)dt=-2\int_0^{-2}f(t)dt g ( a ) = g ( − 2 ) = ∫ − 2 2 f ( t ) d t = 2 ∫ − 2 0 f ( t ) d t = − 2 ∫ 0 − 2 f ( t ) d t 이다. 또한
+30점 · S = − 2 27 − 1 3 ∫ 0 − 2 1 1 − t 3 d t \displaystyle S=-\frac2{27}-\frac13\int_0^{-2}\frac1{1-t^3}dt S = − 27 2 − 3 1 ∫ 0 − 2 1 − t 3 1 d t S = ∫ 0 − 2 t 3 ( 1 − t 3 ) 2 d t = ∫ 0 − 2 1 3 t × 3 t 2 ( 1 − t 3 ) 2 d t = [ 1 3 t ( 1 − t 3 ) − 1 ] 0 − 2 − 1 3 ∫ 0 − 2 1 1 − t 3 d t = − 2 27 − 1 3 ∫ 0 − 2 1 ( 1 − t 3 ) 2 d t + 1 3 ∫ 0 − 2 t 3 ( 1 − t 3 ) 2 d t = − 2 27 − 1 3 ∫ 0 − 2 f ( t ) d t + 1 3 S = − 2 27 + 1 6 g ( a ) + 1 3 S \displaystyle \begin{aligned}S&=\int_0^{-2}\frac{t^3}{(1-t^3)^2}dt=\int_0^{-2}\frac13t\times\frac{3t^2}{(1-t^3)^2}dt=\Biggl[\frac13t(1-t^3)^{-1}\Biggr]_0^{-2}-\frac13\int_0^{-2}\frac1{1-t^3}dt\\&=-\frac2{27}-\frac13\int_0^{-2}\frac1{(1-t^3)^2}dt+\frac13\int_0^{-2}\frac{t^3}{(1-t^3)^2}dt=-\frac2{27}-\frac13\int_0^{-2}f(t)dt+\frac13S\\&=-\frac2{27}+\frac16g(a)+\frac13S\end{aligned} S = ∫ 0 − 2 ( 1 − t 3 ) 2 t 3 d t = ∫ 0 − 2 3 1 t × ( 1 − t 3 ) 2 3 t 2 d t = [ 3 1 t ( 1 − t 3 ) − 1 ] 0 − 2 − 3 1 ∫ 0 − 2 1 − t 3 1 d t = − 27 2 − 3 1 ∫ 0 − 2 ( 1 − t 3 ) 2 1 d t + 3 1 ∫ 0 − 2 ( 1 − t 3 ) 2 t 3 d t = − 27 2 − 3 1 ∫ 0 − 2 f ( t ) d t + 3 1 S = − 27 2 + 6 1 g ( a ) + 3 1 S
+30점 · g ( a ) − 4 S = 4 9 \displaystyle g(a)-4S=\frac49 g ( a ) − 4 S = 9 4 이므로 g ( a ) − 4 S = 4 9 \displaystyle g(a)-4S=\frac49 g ( a ) − 4 S = 9 4
수리논술, 이게 다다.
