+2점 · 곡선 C ′ \displaystyle C^{\prime} C ′ y = − 1 a ( x − 1 a ) 2 + a ( − a + 1 a ≤ x ≤ a + 1 a ) \displaystyle y=-\frac1a\left(x-\frac1a\right)^2+a\;\left(-a+\frac1a\le x\le a+\frac1a\right) y = − a 1 ( x − a 1 ) 2 + a ( − a + a 1 ≤ x ≤ a + a 1 ) [2-1] 곡선 C ′ \displaystyle C^{\prime} C ′ y = − 1 a ( x − 1 a ) 2 + a ( − a + 1 a ≤ x ≤ a + 1 a ) \displaystyle y=-\frac1a\left(x-\frac1a\right)^2+a\quad\left(-a+\frac1a\le x\le a+\frac1a\right) y = − a 1 ( x − a 1 ) 2 + a ( − a + a 1 ≤ x ≤ a + a 1 ) 의 그래프이다.
곡선 C \displaystyle C C C ′ \displaystyle C^{\prime} C ′ a \displaystyle a a
+3점 · 0 < a < 1 2 \displaystyle 0<a<\frac1{\sqrt2} 0 < a < 2 1 (i) 1 a > 2 a \displaystyle \frac1a>2a a 1 > 2 a 0 < a < 1 2 \displaystyle 0<a<\frac1{\sqrt2} 0 < a < 2 1
+5점 · a ≥ 1 2 \displaystyle a\ge\frac1{\sqrt2} a ≥ 2 1 ( 1 2 a , a − 1 4 a 3 ) \displaystyle \left(\frac1{2a},a-\frac1{4a^3}\right) ( 2 a 1 , a − 4 a 3 1 ) (ii) 1 a ≤ 2 a \displaystyle \frac1a\le2a a 1 ≤ 2 a a ≥ 1 2 \displaystyle a\ge\frac1{\sqrt2} a ≥ 2 1 곡선 C \displaystyle C C C ′ \displaystyle C^{\prime} C ′ x \displaystyle x x − 1 a x 2 + a = − 1 a ( x − 1 a ) 2 + a \displaystyle -\frac1a x^2+a=-\frac1a\left(x-\frac1a\right)^2+a − a 1 x 2 + a = − a 1 ( x − a 1 ) 2 + a x = 1 2 a \displaystyle x=\frac1{2a} x = 2 a 1 이다. 따라서 교점은 ( 1 2 a , a − 1 4 a 3 ) \displaystyle \left(\frac1{2a},a-\frac1{4a^3}\right) ( 2 a 1 , a − 4 a 3 1 )
+5점 · 곡선이 지나간 영역의 넓이를 a \displaystyle a a [2-2] 곡선 C \displaystyle C C S \displaystyle S S a \displaystyle a a
ⅰ) 1 a ≥ 2 a \displaystyle \frac1a\ge2a a 1 ≥ 2 a 0 < a ≤ 1 2 \displaystyle 0<a\le\frac1{\sqrt2} 0 < a ≤ 2 1 [2-1]에 의하여 곡선 C \displaystyle C C C ′ \displaystyle C^{\prime} C ′ 0 < a < 1 2 \displaystyle 0<a<\frac1{\sqrt2} 0 < a < 2 1 a = 1 2 \displaystyle a=\frac1{\sqrt2} a = 2 1 ( 1 2 , 0 ) \displaystyle \left(\frac1{\sqrt2},0\right) ( 2 1 , 0 ) 곡선 C \displaystyle C C
+7점 · 0 < a ≤ 1 2 \displaystyle 0<a\le\frac1{\sqrt2} 0 < a ≤ 2 1 S \displaystyle S S S = ∫ − a 0 ( − 1 a x 2 + a ) d x + ( a × 1 a ) + ∫ 1 a a + 1 a { − 1 a ( x − 1 a ) 2 + a } d x \displaystyle S=\int_{-a}^0\left(-\frac1a x^2+a\right)dx+\left(a\times\frac1a\right)+\int_{\frac1a}^{a+\frac1a}\left\{-\frac1a\left(x-\frac1a\right)^2+a\right\}dx S = ∫ − a 0 ( − a 1 x 2 + a ) d x + ( a × a 1 ) + ∫ a 1 a + a 1 { − a 1 ( x − a 1 ) 2 + a } d x 이 영역의 넓이 S \displaystyle S S [ − a , 0 ] \displaystyle [-a,0] [ − a , 0 ] y = − 1 a x 2 + a \displaystyle y=-\frac1a x^2+a y = − a 1 x 2 + a x \displaystyle x x 닫힌구간 [ 0 , 1 a ] \displaystyle \left[0,\frac1a\right] [ 0 , a 1 ] y = a \displaystyle y=a y = a x \displaystyle x x 닫힌구간 [ 1 a , a + 1 a ] \displaystyle \left[\frac1a,a+\frac1a\right] [ a 1 , a + a 1 ] y = − 1 a ( x − 1 a ) 2 + a \displaystyle y=-\frac1a\left(x-\frac1a\right)^2+a y = − a 1 ( x − a 1 ) 2 + a x \displaystyle x x 따라서 제시문 (나)에 의하여 S = ∫ − a 0 ( − 1 a x 2 + a ) d x + ( a × 1 a ) + ∫ 1 a a + 1 a { − 1 a ( x − 1 a ) 2 + a } d x \displaystyle S=\int_{-a}^0\left(-\frac1a x^2+a\right)dx+\left(a\times\frac1a\right)+\int_{\frac1a}^{a+\frac1a}\left\{-\frac1a\left(x-\frac1a\right)^2+a\right\}dx S = ∫ − a 0 ( − a 1 x 2 + a ) d x + ( a × a 1 ) + ∫ a 1 a + a 1 { − a 1 ( x − a 1 ) 2 + a } d x
+3점 · 0 < a ≤ 1 2 \displaystyle 0<a\le\frac1{\sqrt2} 0 < a ≤ 2 1 S = 1 + 4 3 a 2 \displaystyle S=1+\frac43a^2 S = 1 + 3 4 a 2 S = 1 + 2 ∫ 0 a ( − 1 a x 2 + a ) d x = 1 + 2 ( − a 3 3 a + a 2 ) = 1 + 4 3 a 2 \displaystyle \begin{aligned}S&=1+2\int_0^a\left(-\frac1a x^2+a\right)dx\\&=1+2\left(-\frac{a^3}{3a}+a^2\right)=1+\frac43a^2\end{aligned} S = 1 + 2 ∫ 0 a ( − a 1 x 2 + a ) d x = 1 + 2 ( − 3 a a 3 + a 2 ) = 1 + 3 4 a 2 이다.
ⅱ) 1 a < 2 a \displaystyle \frac1a<2a a 1 < 2 a a > 1 2 \displaystyle a>\frac1{\sqrt2} a > 2 1 [2-1]에 의하여, 곡선 C \displaystyle C C C ′ \displaystyle C^{\prime} C ′ ( 1 2 a , a − 1 4 a 3 ) \displaystyle \left(\frac1{2a},a-\frac1{4a^3}\right) ( 2 a 1 , a − 4 a 3 1 ) 곡선 C \displaystyle C C
+7점 · a > 1 2 \displaystyle a>\frac1{\sqrt2} a > 2 1 S \displaystyle S S S = 1 + 4 3 a 2 − ∫ − a + 1 a 1 2 a { − 1 a ( x − 1 a ) 2 + a } d x − ∫ 1 2 a a ( − 1 a x 2 + a ) d x \displaystyle S=1+\frac43a^2-\int_{-a+\frac1a}^{\frac1{2a}}\left\{-\frac1a\left(x-\frac1a\right)^2+a\right\}dx-\int_{\frac1{2a}}^a\left(-\frac1a x^2+a\right)dx S = 1 + 3 4 a 2 − ∫ − a + a 1 2 a 1 { − a 1 ( x − a 1 ) 2 + a } d x − ∫ 2 a 1 a ( − a 1 x 2 + a ) d x 이 영역의 넓이 S \displaystyle S S 닫힌구간 [ − a + 1 a , 1 2 a ] \displaystyle \left[-a+\frac1a,\frac1{2a}\right] [ − a + a 1 , 2 a 1 ] y = − 1 a ( x − 1 a ) 2 + a \displaystyle y=-\frac1a\left(x-\frac1a\right)^2+a y = − a 1 ( x − a 1 ) 2 + a x \displaystyle x x 닫힌구간 [ 1 2 a , a ] \displaystyle \left[\frac1{2a},a\right] [ 2 a 1 , a ] y = − 1 a x 2 + a \displaystyle y=-\frac1a x^2+a y = − a 1 x 2 + a x \displaystyle x x 따라서 S = 1 + 4 3 a 2 − ∫ − a + 1 a 1 2 a { − 1 a ( x − 1 a ) 2 + a } d x − ∫ 1 2 a a ( − 1 a x 2 + a ) d x \displaystyle S=1+\frac43a^2-\int_{-a+\frac1a}^{\frac1{2a}}\left\{-\frac1a\left(x-\frac1a\right)^2+a\right\}dx-\int_{\frac1{2a}}^a\left(-\frac1a x^2+a\right)dx S = 1 + 3 4 a 2 − ∫ − a + a 1 2 a 1 { − a 1 ( x − a 1 ) 2 + a } d x − ∫ 2 a 1 a ( − a 1 x 2 + a ) d x
+3점 · a > 1 2 \displaystyle a>\frac1{\sqrt2} a > 2 1 S = 2 − 1 12 a 4 \displaystyle S=2-\frac1{12a^4} S = 2 − 12 a 4 1 S = 1 + 4 3 a 2 − 2 ∫ 1 2 a a ( − 1 a x 2 + a ) d x = 1 + 4 3 a 2 − 2 [ − 1 3 a x 3 + a x ] 1 2 a a = 1 + 4 3 a 2 − 2 ( − a 2 3 + a 2 + 1 24 a 4 − 1 2 ) = 2 − 1 12 a 4 \displaystyle \begin{aligned}S&=1+\frac43a^2-2\int_{\frac1{2a}}^a\left(-\frac1a x^2+a\right)dx\\&=1+\frac43a^2-2\Biggl[-\frac1{3a}x^3+ax\Biggr]_{\frac1{2a}}^a\\&=1+\frac43a^2-2\left(-\frac{a^2}3+a^2+\frac1{24a^4}-\frac12\right)\\&=2-\frac1{12a^4}\end{aligned} S = 1 + 3 4 a 2 − 2 ∫ 2 a 1 a ( − a 1 x 2 + a ) d x = 1 + 3 4 a 2 − 2 [ − 3 a 1 x 3 + a x ] 2 a 1 a = 1 + 3 4 a 2 − 2 ( − 3 a 2 + a 2 + 24 a 4 1 − 2 1 ) = 2 − 12 a 4 1 이다.
수리논술, 배움에서 논증의 완성까지.
