[문제 1-ⅰ]
+6점 · 닫힌 구간 위에서 양수와 음수 값을 동시에 가지는 함수의 넓이를 구할 수 있다. g ( − 3 2 ) = ∫ − 3 2 − 1 ( 3 x 2 − 3 ) d x + ∫ − 1 − 1 2 ( 3 − 3 x 2 ) d x = [ x 3 − 3 x ] − 3 2 − 1 + [ 3 x − x 3 ] − 1 − 1 2 = 3 2 \displaystyle \begin{aligned} g\left(-\frac{3}{2}\right) &=\int_{-\frac{3}{2}}^{-1}(3x^{2}-3)dx+\int_{-1}^{-\frac{1}{2}}(3-3x^{2})dx \\ &=\Biggl[x^{3}-3x\Biggr]_{-\frac{3}{2}}^{-1}+\Biggl[3x-x^{3}\Biggr]_{-1}^{-\frac{1}{2}} \\ &=\frac{3}{2} \end{aligned} g ( − 2 3 ) = ∫ − 2 3 − 1 ( 3 x 2 − 3 ) d x + ∫ − 1 − 2 1 ( 3 − 3 x 2 ) d x = [ x 3 − 3 x ] − 2 3 − 1 + [ 3 x − x 3 ] − 1 − 2 1 = 2 3
[문제 1-ⅱ]
+12점 · 변수로 주어진 닫힌 구간 위에서 양수와 음수 값을 동시에 가지는 함수의 넓이를 함수로 나타낼 수 있다. 다음과 같이 세 가지 경우를 생각한다.
(1) − 3 ≤ t < − 2 \displaystyle -3\le t<-2 − 3 ≤ t < − 2 이때 t ≤ t + 1 < − 1 \displaystyle t\le t+1<-1 t ≤ t + 1 < − 1 t ≤ x ≤ t + 1 \displaystyle t\le x\le t+1 t ≤ x ≤ t + 1 ∣ 3 x 2 − 3 ∣ = 3 x 2 − 3 \displaystyle |3x^{2}-3|=3x^{2}-3 ∣3 x 2 − 3∣ = 3 x 2 − 3 따라서 g ( t ) = ∫ t t + 1 ( 3 x 2 − 3 ) d x = [ x 3 − 3 x ] t t + 1 = 3 t 2 + 3 t − 2 \displaystyle g(t)=\int_{t}^{t+1}(3x^{2}-3)dx=\Biggl[x^{3}-3x\Biggr]_{t}^{t+1}=3t^{2}+3t-2 g ( t ) = ∫ t t + 1 ( 3 x 2 − 3 ) d x = [ x 3 − 3 x ] t t + 1 = 3 t 2 + 3 t − 2
(2) − 2 ≤ t < − 1 \displaystyle -2\le t<-1 − 2 ≤ t < − 1 이때 t < − 1 ≤ t + 1 \displaystyle t<-1\le t+1 t < − 1 ≤ t + 1 t ≤ x < − 1 \displaystyle t\le x<-1 t ≤ x < − 1 ∣ 3 x 2 − 3 ∣ = 3 x 2 − 3 \displaystyle |3x^{2}-3|=3x^{2}-3 ∣3 x 2 − 3∣ = 3 x 2 − 3 − 1 ≤ x < t + 1 \displaystyle -1\le x<t+1 − 1 ≤ x < t + 1 ∣ 3 x 2 − 3 ∣ = 3 − 3 x 2 \displaystyle |3x^{2}-3|=3-3x^{2} ∣3 x 2 − 3∣ = 3 − 3 x 2 따라서 g ( t ) = ∫ t − 1 ( 3 x 2 − 3 ) d x + ∫ − 1 t + 1 ( 3 − 3 x 2 ) d x = [ x 3 − 3 x ] t − 1 + [ 3 x − x 3 ] − 1 t + 1 = ( − t 3 + 3 t + 2 ) + ( − t 3 − 3 t 2 + 4 ) = − 2 t 3 − 3 t 2 + 3 t + 6 \displaystyle \begin{aligned} g(t) &=\int_{t}^{-1}(3x^{2}-3)dx+\int_{-1}^{t+1}(3-3x^{2})dx \\ &=\Biggl[x^{3}-3x\Biggr]_{t}^{-1}+\Biggl[3x-x^{3}\Biggr]_{-1}^{t+1} \\ &=(-t^{3}+3t+2)+(-t^{3}-3t^{2}+4) \\ &=-2t^{3}-3t^{2}+3t+6 \end{aligned} g ( t ) = ∫ t − 1 ( 3 x 2 − 3 ) d x + ∫ − 1 t + 1 ( 3 − 3 x 2 ) d x = [ x 3 − 3 x ] t − 1 + [ 3 x − x 3 ] − 1 t + 1 = ( − t 3 + 3 t + 2 ) + ( − t 3 − 3 t 2 + 4 ) = − 2 t 3 − 3 t 2 + 3 t + 6
(3) − 1 ≤ t ≤ 0 \displaystyle -1\le t\le0 − 1 ≤ t ≤ 0 − 1 ≤ t < t + 1 ≤ 1 \displaystyle -1\le t<t+1\le1 − 1 ≤ t < t + 1 ≤ 1 t ≤ x ≤ t + 1 \displaystyle t\le x\le t+1 t ≤ x ≤ t + 1 ∣ 3 x 2 − 3 ∣ = 3 − 3 x 2 \displaystyle |3x^{2}-3|=3-3x^{2} ∣3 x 2 − 3∣ = 3 − 3 x 2 따라서 g ( t ) = ∫ t t + 1 ( 3 − 3 x 2 ) d x = [ 3 x − x 3 ] t t + 1 = − 3 t 2 − 3 t + 2 \displaystyle g(t)=\int_{t}^{t+1}(3-3x^{2})dx=\Biggl[3x-x^{3}\Biggr]_{t}^{t+1}=-3t^{2}-3t+2 g ( t ) = ∫ t t + 1 ( 3 − 3 x 2 ) d x = [ 3 x − x 3 ] t t + 1 = − 3 t 2 − 3 t + 2
위의 세 가지 경우를 종합하면 g ( t ) = { 3 t 2 + 3 t − 2 ( − 3 ≤ t < − 2 ) − 2 t 3 − 3 t 2 + 3 t + 6 ( − 2 ≤ t < − 1 ) − 3 t 2 − 3 t + 2 ( − 1 ≤ t ≤ 0 ) \displaystyle g(t)=\begin{cases}3t^{2}+3t-2 & (-3\le t<-2) \\ -2t^{3}-3t^{2}+3t+6 & (-2\le t<-1) \\ -3t^{2}-3t+2 & (-1\le t\le0)\end{cases} g ( t ) = ⎩ ⎨ ⎧ 3 t 2 + 3 t − 2 − 2 t 3 − 3 t 2 + 3 t + 6 − 3 t 2 − 3 t + 2 ( − 3 ≤ t < − 2 ) ( − 2 ≤ t < − 1 ) ( − 1 ≤ t ≤ 0 )
[문제 1-ⅲ]
+12점 · 문제는 이렇게 구해진 함수의 극값과 그래프의 개형을 조사해서 최솟값을 구할 수 있다. (1) − 3 ≤ t < − 2 \displaystyle -3\le t<-2 − 3 ≤ t < − 2 g ′ ( t ) = 6 t + 3 < 0 \displaystyle g^{\prime}(t)=6t+3<0 g ′ ( t ) = 6 t + 3 < 0 (2) − 2 ≤ t < − 1 \displaystyle -2\le t<-1 − 2 ≤ t < − 1 g ′ ( t ) = − 6 t 2 − 6 t + 3 = − 6 ( t + 1 + 3 2 ) ( t + 1 − 3 2 ) \displaystyle g^{\prime}(t)=-6t^{2}-6t+3=-6\left(t+\frac{1+\sqrt{3}}{2}\right)\left(t+\frac{1-\sqrt{3}}{2}\right) g ′ ( t ) = − 6 t 2 − 6 t + 3 = − 6 ( t + 2 1 + 3 ) ( t + 2 1 − 3 ) 따라서 − 2 ≤ t < − 1 + 3 2 \displaystyle -2\le t<-\frac{1+\sqrt{3}}{2} − 2 ≤ t < − 2 1 + 3 g ′ ( t ) < 0 \displaystyle g^{\prime}(t)<0 g ′ ( t ) < 0 − 1 + 3 2 < t < − 1 \displaystyle -\frac{1+\sqrt{3}}{2}<t<-1 − 2 1 + 3 < t < − 1 g ′ ( t ) > 0 \displaystyle g^{\prime}(t)>0 g ′ ( t ) > 0 (3) − 1 ≤ t ≤ 0 \displaystyle -1\le t\le0 − 1 ≤ t ≤ 0 g ′ ( t ) = − 6 t − 3 \displaystyle g^{\prime}(t)=-6t-3 g ′ ( t ) = − 6 t − 3 따라서 − 1 < t < − 1 2 \displaystyle -1<t<-\frac{1}{2} − 1 < t < − 2 1 g ′ ( t ) > 0 \displaystyle g^{\prime}(t)>0 g ′ ( t ) > 0 − 1 2 < t < − 1 \displaystyle -\frac{1}{2}<t<-1 − 2 1 < t < − 1 g ′ ( t ) < 0 \displaystyle g^{\prime}(t)<0 g ′ ( t ) < 0 Correction. g ′ ( t ) < 0 \displaystyle g^{\prime}(t)<0 g ′ ( t ) < 0 − 1 / 2 < t < − 1 \displaystyle -1/2<t<-1 − 1/2 < t < − 1 − 1 / 2 < t < 0 \displaystyle -1/2<t<0 − 1/2 < t < 0
이를 종합하여 닫힌구간 [ − 3 , 0 ] \displaystyle [-3,0] [ − 3 , 0 ] g ( t ) \displaystyle g(t) g ( t ) x − 3 − 2 − 1 + 3 2 − 1 − 1 2 0 g ′ ( t ) − − + + − g ( t ) 16 ↘ 4 ↘ 4 − 3 2 3 ↗ 2 ↗ 11 4 ↘ 2 \displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline x & -3 & & -2 & & -\frac{1+\sqrt{3}}{2} & & -1 & & -\frac{1}{2} & & 0 \\ \hline g^{\prime}(t) & & - & & - & & + & & + & & - & \\ \hline g(t) & 16 & \searrow & 4 & \searrow & 4-\frac{3}{2}\sqrt{3} & \nearrow & 2 & \nearrow & \frac{11}{4} & \searrow & 2 \\ \hline \end{array} x g ′ ( t ) g ( t ) − 3 16 − ↘ − 2 4 − ↘ − 2 1 + 3 4 − 2 3 3 + ↗ − 1 2 + ↗ − 2 1 4 11 − ↘ 0 2 따라서 함수 g \displaystyle g g t = − 1 + 3 2 \displaystyle t=-\frac{1+\sqrt{3}}{2} t = − 2 1 + 3 g ( − 1 + 3 2 ) = 4 − 3 2 3 \displaystyle g\left(-\frac{1+\sqrt{3}}{2}\right)=4-\frac{3}{2}\sqrt{3} g ( − 2 1 + 3 ) = 4 − 2 3 3
수리논술, 배움에서 논증의 완성까지.