+20점 · f ( x ) = x ( x − 1 ) ( x 2 + a x + a ) \displaystyle f(x)=x(x-1)(x^2+ax+a) f ( x ) = x ( x − 1 ) ( x 2 + a x + a ) +40점 · 0 < a ≤ 4 \displaystyle 0<a\le4 0 < a ≤ 4 0 ≤ a ≤ 4 \displaystyle 0\le a\le4 0 ≤ a ≤ 4 +10점 · 답 − 4 ≤ f ′ ( 0 ) < 0 \displaystyle -4\le f\prime(0)<0 − 4 ≤ f ′ ( 0 ) < 0 (1-1) 조건 (나)로부터 f ( x ) = x ( x − 1 ) ( x 2 + a x + b ) \displaystyle f(x)=x(x-1)(x^2+ax+b) f ( x ) = x ( x − 1 ) ( x 2 + a x + b ) a = b \displaystyle a=b a = b f ( x ) = x ( x − 1 ) ( x 2 + a x + a ) \displaystyle f(x)=x(x-1)(x^2+ax+a) f ( x ) = x ( x − 1 ) ( x 2 + a x + a ) x 2 + a x + a = 0 \displaystyle x^2+ax+a=0 x 2 + a x + a = 0 a 2 − 4 a ≤ 0 \displaystyle a^2-4a\le0 a 2 − 4 a ≤ 0 0 ≤ a ≤ 4 \displaystyle 0\le a\le4 0 ≤ a ≤ 4 a = 0 \displaystyle a=0 a = 0 ∣ f ( x ) ∣ \displaystyle |f(x)| ∣ f ( x ) ∣ x = 0 \displaystyle x=0 x = 0 0 < a ≤ 4 \displaystyle 0<a\le4 0 < a ≤ 4 한편 f ′ ( 0 ) = − a \displaystyle f\prime(0)=-a f ′ ( 0 ) = − a − 4 ≤ f ′ ( 0 ) < 0 \displaystyle -4\le f\prime(0)<0 − 4 ≤ f ′ ( 0 ) < 0
+40점 · lim x → 0 + e ∣ f ( x ) ∣ − 1 x = 3 \displaystyle \lim_{x\to0+}\frac{e^{|f(x)|}-1}x=3 x → 0 + lim x e ∣ f ( x ) ∣ − 1 = 3 +40점 · lim x → 0 − e ∣ f ( x ) ∣ − 1 x = − 3 \displaystyle \lim_{x\to0-}\frac{e^{|f(x)|}-1}x=-3 x → 0 − lim x e ∣ f ( x ) ∣ − 1 = − 3 (1-2) lim x → 0 + e ∣ f ( x ) ∣ − 1 x = lim x → 0 + e − f ( x ) − 1 x = { e − f ( x ) } ′ ( 0 ) = − f ′ ( 0 ) = 3 \displaystyle \lim_{x\to0+}\frac{e^{|f(x)|}-1}x=\lim_{x\to0+}\frac{e^{-f(x)}-1}x=\{e^{-f(x)}\}\prime(0)=-f\prime(0)=3 x → 0 + lim x e ∣ f ( x ) ∣ − 1 = x → 0 + lim x e − f ( x ) − 1 = { e − f ( x ) } ′ ( 0 ) = − f ′ ( 0 ) = 3 lim x → 0 − e ∣ f ( x ) ∣ − 1 x = lim x → 0 − e f ( x ) − 1 x = { e f ( x ) } ′ ( 0 ) = f ′ ( 0 ) = − 3 \displaystyle \lim_{x\to0-}\frac{e^{|f(x)|}-1}x=\lim_{x\to0-}\frac{e^{f(x)}-1}x=\{e^{f(x)}\}\prime(0)=f\prime(0)=-3 x → 0 − lim x e ∣ f ( x ) ∣ − 1 = x → 0 − lim x e f ( x ) − 1 = { e f ( x ) } ′ ( 0 ) = f ′ ( 0 ) = − 3 이므로 e ∣ f ( x ) ∣ \displaystyle e^{|f(x)|} e ∣ f ( x ) ∣ x = 0 \displaystyle x=0 x = 0
+10점 · x ( x − 1 ) ( x 2 + a x + a ) = m x + n + ( x − α ) 2 ( x − β ) 2 \displaystyle x(x-1)(x^2+ax+a)=mx+n+(x-\alpha)^2(x-\beta)^2 x ( x − 1 ) ( x 2 + a x + a ) = m x + n + ( x − α ) 2 ( x − β ) 2 +40점 · m = ( a − 1 ) 3 8 − a \displaystyle m=\frac{(a-1)^3}8-a m = 8 ( a − 1 ) 3 − a (1-3) 조건으로부터 x ( x − 1 ) ( x 2 + a x + a ) = m x + n + ( x − α ) 2 ( x − β ) 2 \displaystyle x(x-1)(x^2+ax+a)=mx+n+(x-\alpha)^2(x-\beta)^2 x ( x − 1 ) ( x 2 + a x + a ) = m x + n + ( x − α ) 2 ( x − β ) 2 − 2 ( α + β ) = a − 1 , ( α + β ) 2 + 2 α β = 0 , m − 2 ( α + β ) α β = − a \displaystyle -2(\alpha+\beta)=a-1,\;(\alpha+\beta)^2+2\alpha\beta=0,\;m-2(\alpha+\beta)\alpha\beta=-a − 2 ( α + β ) = a − 1 , ( α + β ) 2 + 2 α β = 0 , m − 2 ( α + β ) α β = − a 를 얻는다. 따라서 m = 2 ( α + β ) α β − a = ( 1 − a ) ( − 1 2 ) ( 1 − a 2 ) 2 − a = ( a − 1 ) 3 8 − a \displaystyle m=2(\alpha+\beta)\alpha\beta-a=(1-a)(-\frac12)(\frac{1-a}2)^2-a=\frac{(a-1)^3}8-a m = 2 ( α + β ) α β − a = ( 1 − a ) ( − 2 1 ) ( 2 1 − a ) 2 − a = 8 ( a − 1 ) 3 − a 를 얻는다.
+30점 · 답 f ′ ( 0 ) = − 1 − 2 6 3 \displaystyle f\prime(0)=-1-\frac{2\sqrt6}3 f ′ ( 0 ) = − 1 − 3 2 6 또한 d m d a = 3 ( a − 1 ) 2 8 − 1 = 0 \displaystyle \frac{dm}{da}=\frac{3(a-1)^2}8-1=0 d a d m = 8 3 ( a − 1 ) 2 − 1 = 0 a = 1 ± 2 6 3 \displaystyle a=1\pm\frac{2\sqrt6}3 a = 1 ± 3 2 6 0 < a ≤ 4 \displaystyle 0<a\le4 0 < a ≤ 4 m \displaystyle m m a = 1 + 2 6 3 \displaystyle a=1+\frac{2\sqrt6}3 a = 1 + 3 2 6 f ′ ( 0 ) = − a = − 1 − 2 6 3 \displaystyle f\prime(0)=-a=-1-\frac{2\sqrt6}3 f ′ ( 0 ) = − a = − 1 − 3 2 6
수리논술, 배움에서 논증의 완성까지.