+4점 · 미분을 통하여 y = x 3 − x \displaystyle y=x^3-x y = x 3 − x y = x 3 − x − t \displaystyle y=x^3-x-t y = x 3 − x − t +4점 · t \displaystyle t t (1-1) x 3 − x − t = 0 \displaystyle x^3-x-t=0 x 3 − x − t = 0 y = x 3 − x \displaystyle y=x^3-x y = x 3 − x y = t \displaystyle y=t y = t y = x 3 − x \displaystyle y=x^3-x y = x 3 − x x = − 1 / 3 \displaystyle x=-1/\sqrt{3} x = − 1/ 3 2 / 3 3 \displaystyle 2/\sqrt{3}^3 2/ 3 3 x = 1 / 3 \displaystyle x=1/\sqrt{3} x = 1/ 3 − 2 / 3 3 \displaystyle -2/\sqrt{3}^3 − 2/ 3 3 − 2 / 3 3 < t < 2 / 3 3 \displaystyle -2/\sqrt{3}^3<t<2/\sqrt{3}^3 − 2/ 3 3 < t < 2/ 3 3 α < − 1 / 3 < β < 1 / 3 < γ \displaystyle \alpha<-1/\sqrt{3}<\beta<1/\sqrt{3}<\gamma α < − 1/ 3 < β < 1/ 3 < γ
+2점 · (1-1)에서 구한 두 극값을 갖는 점 사이의 범위를 인지함 +3점 · 서로 같은 두 실근을 인지함 (1-2)(a) (1-1)에서와 같이 구하고자 하는 방정식의 근을 y = x 3 − x \displaystyle y=x^3-x y = x 3 − x y = t \displaystyle y=t y = t − 2 / 3 3 < t < 2 / 3 3 \displaystyle -2/\sqrt{3}^3<t<2/\sqrt{3}^3 − 2/ 3 3 < t < 2/ 3 3 − 1 / 3 < β < 1 / 3 \displaystyle -1/\sqrt{3}<\beta<1/\sqrt{3} − 1/ 3 < β < 1/ 3 t = 2 / 3 3 \displaystyle t=2/\sqrt{3^3} t = 2/ 3 3 t = − 2 / 3 3 \displaystyle t=-2/\sqrt{3^3} t = − 2/ 3 3 x = − 1 / 3 \displaystyle x=-1/\sqrt{3} x = − 1/ 3 x = 1 / 3 \displaystyle x=1/\sqrt{3} x = 1/ 3 β \displaystyle \beta β − 1 / 3 ≤ β ≤ 1 / 3 \displaystyle -1/\sqrt{3}\le\beta\le1/\sqrt{3} − 1/ 3 ≤ β ≤ 1/ 3 t \displaystyle t t
+2점 · 근과 계수의 관계와 α , β , γ \displaystyle \alpha,\beta,\gamma α , β , γ x 3 − x − t = 0 \displaystyle x^3-x-t=0 x 3 − x − t = 0 (1-2)(b) 제시문 (나)에서 주어진 근과 계수의 관계를 이용하면 다음을 얻을 수 있다. α + β + γ = 0 , α β + β γ + γ α = − 1 , α β γ = t , α 3 − α = t , β 3 − β = t , γ 3 − γ = t \displaystyle \alpha+\beta+\gamma=0,\;\alpha\beta+\beta\gamma+\gamma\alpha=-1,\;\alpha\beta\gamma=t,\;\alpha^3-\alpha=t,\;\beta^3-\beta=t,\;\gamma^3-\gamma=t α + β + γ = 0 , α β + β γ + γ α = − 1 , α β γ = t , α 3 − α = t , β 3 − β = t , γ 3 − γ = t β \displaystyle \beta β
+5점 · 적분을 α , β , γ \displaystyle \alpha,\beta,\gamma α , β , γ +5점 · 적분을 β \displaystyle \beta β A = ∫ α γ ∣ x 3 − x − t ∣ d x = ∫ α β ( x 3 − x − t ) d x − ∫ β γ ( x 3 − x − t ) d x = β 4 / 2 − α 4 / 4 − γ 4 / 4 − β 2 + α 2 / 2 + γ 2 / 2 − t ( 2 β − α − γ ) = β ( β + t ) / 2 − α ( α + t ) / 4 − γ ( γ + t ) / 4 − β 2 + α 2 / 2 + γ 2 / 2 − t ( 2 β − α − γ ) = α 2 / 4 + γ 2 / 4 − β 2 / 2 − ( t − t / 4 ) ( 2 β − α − γ ) = 1 4 ( β 2 − 2 α γ ) − β 2 / 2 − 3 t 4 ( 3 β ) = 1 4 ( 2 − 3 β 2 − 9 t β ) = 1 4 ( − 9 β 4 + 6 β 2 + 2 ) = − 9 4 ( β 2 − 1 3 ) 2 + 3 4 \displaystyle \begin{aligned}A&=\int_\alpha^\gamma|x^3-x-t|dx=\int_\alpha^\beta(x^3-x-t)dx-\int_\beta^\gamma(x^3-x-t)dx\\&=\beta^4/2-\alpha^4/4-\gamma^4/4-\beta^2+\alpha^2/2+\gamma^2/2-t(2\beta-\alpha-\gamma)\\&=\beta(\beta+t)/2-\alpha(\alpha+t)/4-\gamma(\gamma+t)/4-\beta^2+\alpha^2/2+\gamma^2/2-t(2\beta-\alpha-\gamma)\\&=\alpha^2/4+\gamma^2/4-\beta^2/2-(t-t/4)(2\beta-\alpha-\gamma)\\&=\frac14(\beta^2-2\alpha\gamma)-\beta^2/2-\frac{3t}4(3\beta)\\&=\frac14(2-3\beta^2-9t\beta)=\frac14(-9\beta^4+6\beta^2+2)=-\frac94\left(\beta^2-\frac13\right)^2+\frac34\end{aligned} A = ∫ α γ ∣ x 3 − x − t ∣ d x = ∫ α β ( x 3 − x − t ) d x − ∫ β γ ( x 3 − x − t ) d x = β 4 /2 − α 4 /4 − γ 4 /4 − β 2 + α 2 /2 + γ 2 /2 − t ( 2 β − α − γ ) = β ( β + t ) /2 − α ( α + t ) /4 − γ ( γ + t ) /4 − β 2 + α 2 /2 + γ 2 /2 − t ( 2 β − α − γ ) = α 2 /4 + γ 2 /4 − β 2 /2 − ( t − t /4 ) ( 2 β − α − γ ) = 4 1 ( β 2 − 2 α γ ) − β 2 /2 − 4 3 t ( 3 β ) = 4 1 ( 2 − 3 β 2 − 9 tβ ) = 4 1 ( − 9 β 4 + 6 β 2 + 2 ) = − 4 9 ( β 2 − 3 1 ) 2 + 4 3
+5점 · 부등식 또는 이차함수의 최솟값을 이용하여 최솟값을 구함 문항 (1-2)(a)에서 − 1 / 3 ≤ β ≤ 1 / 3 \displaystyle -1/\sqrt{3}\le\beta\le1/\sqrt{3} − 1/ 3 ≤ β ≤ 1/ 3 0 ≤ β 2 ≤ 1 / 3 \displaystyle 0\le\beta^2\le1/3 0 ≤ β 2 ≤ 1/3 0 ≤ ( β 2 − 1 / 3 ) 2 ≤ 1 / 9 \displaystyle 0\le(\beta^2-1/3)^2\le1/9 0 ≤ ( β 2 − 1/3 ) 2 ≤ 1/9 β = 0 \displaystyle \beta=0 β = 0 1 / 2 \displaystyle 1/2 1/2
수리논술, 배움에서 논증의 완성까지.