[문제 2-ⅰ]
+8점 · 곡선과 직선으로 둘러싸인 영역의 넓이를 구할 수 있다. S P + S Q = ∫ − 1 1 ( 1 − x 2 ) d x = 2 [ x − x 3 3 ] 0 1 = 2 ( 1 − 1 3 ) = 4 3 \displaystyle \mathrm{S_{P}}+\mathrm{S_{Q}}=\int_{-1}^{1}(1-x^{2})dx=2\Biggl[x-\frac{x^{3}}{3}\Biggr]_{0}^{1}=2\left(1-\frac{1}{3}\right)=\frac{4}{3} S P + S Q = ∫ − 1 1 ( 1 − x 2 ) d x = 2 [ x − 3 x 3 ] 0 1 = 2 ( 1 − 3 1 ) = 3 4 S P = ∫ − 1 1 2 { ( 1 − x 2 ) − ( x 2 + 1 2 ) } d x = ∫ − 1 1 2 ( − x 2 − 1 2 x + 1 2 ) d x = 9 16 \displaystyle \mathrm{S_{P}}=\int_{-1}^{\frac{1}{2}}\left\{(1-x^{2})-\left(\frac{x}{2}+\frac{1}{2}\right)\right\}dx=\int_{-1}^{\frac{1}{2}}\left(-x^{2}-\frac{1}{2}x+\frac{1}{2}\right)dx=\frac{9}{16} S P = ∫ − 1 2 1 { ( 1 − x 2 ) − ( 2 x + 2 1 ) } d x = ∫ − 1 2 1 ( − x 2 − 2 1 x + 2 1 ) d x = 16 9 S Q = 37 48 \displaystyle \mathrm{S_{Q}}=\frac{37}{48} S Q = 48 37 S P S Q = 27 37 \displaystyle \frac{S_{P}}{S_{Q}}=\frac{27}{37} S Q S P = 37 27 답: 27 37 \displaystyle \frac{27}{37} 37 27
[문제 2-ⅱ]
+12점 · 주어진 조건을 만족하는 직사각형의 넓이를 변수로 표현할 수 있고, 미분을 통해 극점을 구할 수 있다. C ( q , 0 ) \displaystyle \mathrm{C}(q,0) C ( q , 0 ) B ( p , 0 ) \displaystyle \mathrm{B}(p,0) B ( p , 0 ) p + 1 2 = 1 − q 2 \displaystyle \frac{p+1}{2}=1-q^{2} 2 p + 1 = 1 − q 2 p = 1 − 2 q 2 \displaystyle p=1-2q^{2} p = 1 − 2 q 2 B C ‾ = q − p = q + 2 q 2 − 1 \displaystyle \overline{\mathrm{BC}}=q-p=q+2q^{2}-1 BC = q − p = q + 2 q 2 − 1 D C ‾ = 1 − q 2 \displaystyle \overline{\mathrm{DC}}=1-q^{2} DC = 1 − q 2 f ( q ) = ( 1 − q 2 ) ( 2 q 2 + q − 1 ) \displaystyle f(q)=(1-q^{2})(2q^{2}+q-1) f ( q ) = ( 1 − q 2 ) ( 2 q 2 + q − 1 ) f ′ ( q ) = − 8 q 3 − 3 q 2 + 6 q + 1 = 0 \displaystyle f^{\prime}(q)=-8q^{3}-3q^{2}+6q+1=0 f ′ ( q ) = − 8 q 3 − 3 q 2 + 6 q + 1 = 0 q = 1 2 + w \displaystyle q=\frac{1}{2}+w q = 2 1 + w 8 w 3 + 15 w 2 + 3 w = 9 4 \displaystyle 8w^{3}+15w^{2}+3w=\frac{9}{4} 8 w 3 + 15 w 2 + 3 w = 4 9 답: 9 4 \displaystyle \frac{9}{4} 4 9
(부연 설명) f ( q ) = ( 1 − q 2 ) ( 2 q 2 + q − 1 ) \displaystyle f(q)=(1-q^{2})(2q^{2}+q-1) f ( q ) = ( 1 − q 2 ) ( 2 q 2 + q − 1 ) f ( 1 2 ) = 0 \displaystyle f\left(\frac{1}{2}\right)=0 f ( 2 1 ) = 0 f ( 1 ) = 0 \displaystyle f(1)=0 f ( 1 ) = 0 f ′ ( 1 2 ) = − 1 − 3 4 + 3 + 1 = 9 4 > 0 \displaystyle f^{\prime}\left(\frac{1}{2}\right)=-1-\frac{3}{4}+3+1=\frac{9}{4}>0 f ′ ( 2 1 ) = − 1 − 4 3 + 3 + 1 = 4 9 > 0 f ′ ( 1 ) = − 11 + 7 = − 4 < 0 \displaystyle f^{\prime}(1)=-11+7=-4<0 f ′ ( 1 ) = − 11 + 7 = − 4 < 0 H ( q ) = f ′ ( q ) = − 8 q 3 − 3 q 2 + 6 q + 1 \displaystyle H(q)=f^{\prime}(q)=-8q^{3}-3q^{2}+6q+1 H ( q ) = f ′ ( q ) = − 8 q 3 − 3 q 2 + 6 q + 1 H ′ ( q ) = − 24 q 2 − 6 q + 6 \displaystyle H^{\prime}(q)=-24q^{2}-6q+6 H ′ ( q ) = − 24 q 2 − 6 q + 6 열린구간 ( 1 2 , 1 ) \displaystyle \left(\frac{1}{2},1\right) ( 2 1 , 1 ) H ′ ( q ) < 0 \displaystyle H^{\prime}(q)<0 H ′ ( q ) < 0 H ( 1 2 ) = f ′ ( 1 2 ) > 0 \displaystyle H\left(\frac{1}{2}\right)=f^{\prime}\left(\frac{1}{2}\right)>0 H ( 2 1 ) = f ′ ( 2 1 ) > 0 H ( 1 ) = f ′ ( 1 ) < 0 \displaystyle H(1)=f^{\prime}(1)<0 H ( 1 ) = f ′ ( 1 ) < 0 ( 1 2 , 1 ) \displaystyle \left(\frac{1}{2},1\right) ( 2 1 , 1 ) H ( q ) = 0 \displaystyle H(q)=0 H ( q ) = 0 q \displaystyle q q f ′ ( q ) = 0 \displaystyle f^{\prime}(q)=0 f ′ ( q ) = 0 q \displaystyle q q f ′ ( x ) \displaystyle f^{\prime}(x) f ′ ( x ) q \displaystyle q q f ( q ) \displaystyle f(q) f ( q )
[문제 2-ⅲ]
+10점 · 주어진 조건을 만족하는 직사각형의 좌표를 직선의 방정식을 활용하여 구할 수 있다. 점 A \displaystyle \mathrm{A} A ( w , w + 1 2 ) \displaystyle \left(w,\frac{w+1}{2}\right) ( w , 2 w + 1 ) A D ‾ = 17 4 \displaystyle \overline{\mathrm{AD}}=\frac{\sqrt{17}}{4} AD = 4 17 A D \displaystyle \mathrm{AD} AD = 1 4 \displaystyle =\frac{1}{4} = 4 1 ( D 의 x 좌표 ) − ( A 의 x 좌표 ) = A D ‾ × 4 17 = 1 \displaystyle (\mathrm{D}\text{의 }x\text{좌표})-(\mathrm{A}\text{의 }x\text{좌표})=\overline{\mathrm{AD}}\times\frac{4}{\sqrt{17}}=1 ( D 의 x 좌표 ) − ( A 의 x 좌표 ) = AD × 17 4 = 1 ∴ \displaystyle \therefore ∴ D \displaystyle \mathrm{D} D A D : y = 1 4 ( x − w ) + 1 2 ( w + 1 ) \displaystyle \mathrm{AD}:y=\frac{1}{4}(x-w)+\frac{1}{2}(w+1) AD : y = 4 1 ( x − w ) + 2 1 ( w + 1 ) y = 1 − x 2 \displaystyle y=1-x^{2} y = 1 − x 2 x \displaystyle x x Correction. 점 D \displaystyle D D x \displaystyle x x w + 1 \displaystyle w+1 w + 1 x = w + 1 \displaystyle x=w+1 x = w + 1 1 4 ( x − w ) + 1 2 ( w + 1 ) = 1 − x 2 \displaystyle \frac{1}{4}(x-w)+\frac{1}{2}(w+1)=1-x^{2} 4 1 ( x − w ) + 2 1 ( w + 1 ) = 1 − x 2 ∴ 4 w 2 + 10 w + 3 = 0 \displaystyle \therefore\ 4w^{2}+10w+3=0 ∴ 4 w 2 + 10 w + 3 = 0 w = − 5 + 13 4 \displaystyle w=\frac{-5+\sqrt{13}}{4} w = 4 − 5 + 13 B ( a , a + 1 4 ) \displaystyle \mathrm{B}\left(a,\frac{a+1}{4}\right) B ( a , 4 a + 1 ) 직선 B A \displaystyle \mathrm{BA} BA B C \displaystyle \mathrm{BC} BC − 4 \displaystyle -4 − 4 따라서, B A : y = − 4 ( x − a ) + a + 1 4 \displaystyle \mathrm{BA}:y=-4(x-a)+\frac{a+1}{4} BA : y = − 4 ( x − a ) + 4 a + 1 점 A \displaystyle \mathrm{A} A B A \displaystyle \mathrm{BA} BA y = x + 1 2 \displaystyle y=\frac{x+1}{2} y = 2 x + 1 점 A \displaystyle \mathrm{A} A x \displaystyle x x w \displaystyle w w − 4 ( w − a ) + a + 1 4 = w + 1 2 \displaystyle -4(w-a)+\frac{a+1}{4}=\frac{w+1}{2} − 4 ( w − a ) + 4 a + 1 = 2 w + 1 w = 1 18 ( − 1 + 17 a ) \displaystyle w=\frac{1}{18}(-1+17a) w = 18 1 ( − 1 + 17 a ) 1 18 ( − 1 + 17 a ) = − 5 + 13 4 \displaystyle \frac{1}{18}(-1+17a)=\frac{-5+\sqrt{13}}{4} 18 1 ( − 1 + 17 a ) = 4 − 5 + 13 ∴ a = − 43 + 9 13 34 \displaystyle \therefore a=\frac{-43+9\sqrt{13}}{34} ∴ a = 34 − 43 + 9 13 B ( − 43 + 9 13 34 , − 9 + 9 13 136 ) \displaystyle \mathrm{B}\left(\frac{-43+9\sqrt{13}}{34},\frac{-9+9\sqrt{13}}{136}\right) B ( 34 − 43 + 9 13 , 136 − 9 + 9 13 ) 답: ( − 43 + 9 13 34 , − 9 + 9 13 136 ) \displaystyle \left(\frac{-43+9\sqrt{13}}{34},\frac{-9+9\sqrt{13}}{136}\right) ( 34 − 43 + 9 13 , 136 − 9 + 9 13 )
수리논술, 배움에서 논증의 완성까지.