3-(1) 두 곡선 f ( x ) \displaystyle f(x) f ( x ) g ( x ) \displaystyle g(x) g ( x ) n \displaystyle n n
+5점 · 삼차식 h ( x ) = f ( x ) − g ( x ) = 0 \displaystyle h(x)=f(x)-g(x)=0 h ( x ) = f ( x ) − g ( x ) = 0 h ( x ) \displaystyle h(x) h ( x ) (풀이) h ( x ) = f ( x ) − g ( x ) = 2 x 3 − 3 ( n + 1 ) x 2 + 6 n x − 4 n 2 + 20 n − 28 \displaystyle h(x)=f(x)-g(x)=2x^3-3(n+1)x^2+6nx-4n^2+20n-28 h ( x ) = f ( x ) − g ( x ) = 2 x 3 − 3 ( n + 1 ) x 2 + 6 n x − 4 n 2 + 20 n − 28 라 하면, 삼차식 h ( x ) = 0 \displaystyle h(x)=0 h ( x ) = 0 h ( x ) \displaystyle h(x) h ( x )
+2점 · h ′ ( x ) = 6 x 2 − 6 ( n + 1 ) x + 6 n = 6 ( x − 1 ) ( x − n ) = 0 \displaystyle h^{\prime}(x)=6x^2-6(n+1)x+6n=6(x-1)(x-n)=0 h ′ ( x ) = 6 x 2 − 6 ( n + 1 ) x + 6 n = 6 ( x − 1 ) ( x − n ) = 0 x = 1 , x = n \displaystyle x=1,x=n x = 1 , x = n h ′ ( x ) = 6 x 2 − 6 ( n + 1 ) x + 6 n = 6 ( x − 1 ) ( x − n ) = 0 \displaystyle h^{\prime}(x)=6x^2-6(n+1)x+6n=6(x-1)(x-n)=0 h ′ ( x ) = 6 x 2 − 6 ( n + 1 ) x + 6 n = 6 ( x − 1 ) ( x − n ) = 0 이므로, 극값을 갖는 점은 x = 1 \displaystyle x=1 x = 1 x = n \displaystyle x=n x = n
+2점 · n \displaystyle n n n ≥ 1 \displaystyle n\ge1 n ≥ 1 h ( 1 ) \displaystyle h(1) h ( 1 ) h ( n ) \displaystyle h(n) h ( n ) n \displaystyle n n n ≥ 1 \displaystyle n\ge1 n ≥ 1 h ( x ) = 0 \displaystyle h(x)=0 h ( x ) = 0 h ( x ) \displaystyle h(x) h ( x ) x = 1 \displaystyle x=1 x = 1 x = n \displaystyle x=n x = n
+2점 · 극댓값 조건 h ( 1 ) = − 4 n 2 + 23 n − 29 ≥ 0 \displaystyle h(1)=-4n^2+23n-29\ge0 h ( 1 ) = − 4 n 2 + 23 n − 29 ≥ 0 23 − 65 8 ≤ n ≤ 23 + 65 8 \displaystyle \frac{23-\sqrt{65}}8\le n\le\frac{23+\sqrt{65}}8 8 23 − 65 ≤ n ≤ 8 23 + 65 극댓값이 0보다 크거나 같아야 하므로 h ( 1 ) = 2 − 3 ( n + 1 ) + 6 n − 4 n 2 + 20 n − 28 = − 4 n 2 + 23 n − 29 ≥ 0 \displaystyle h(1)=2-3(n+1)+6n-4n^2+20n-28=-4n^2+23n-29\ge0 h ( 1 ) = 2 − 3 ( n + 1 ) + 6 n − 4 n 2 + 20 n − 28 = − 4 n 2 + 23 n − 29 ≥ 0 즉 4 n 2 − 23 n + 29 ≤ 0 \displaystyle 4n^2-23n+29\le0 4 n 2 − 23 n + 29 ≤ 0 23 − 65 8 ≤ n ≤ 23 + 65 8 \displaystyle \frac{23-\sqrt{65}}8\le n\le\frac{23+\sqrt{65}}8 8 23 − 65 ≤ n ≤ 8 23 + 65
+2점 · n \displaystyle n n n = 2 \displaystyle n=2 n = 2 n = 3 \displaystyle n=3 n = 3 +3점 · 극댓값 h ( 1 ) = { 1 , n = 2 4 , n = 3 \displaystyle h(1)=\begin{cases}1,n=2\\4,n=3\end{cases} h ( 1 ) = { 1 , n = 2 4 , n = 3 즉, n = 2 \displaystyle n=2 n = 2 n = 3 \displaystyle n=3 n = 3 h ( 1 ) = { 1 , n = 2 4 , n = 3 \displaystyle h(1)=\begin{cases}1,n=2\\4,n=3\end{cases} h ( 1 ) = { 1 , n = 2 4 , n = 3
+3점 · 극솟값 h ( n ) \displaystyle h(n) h ( n ) n = 2 \displaystyle n=2 n = 2 h ( n ) = 0 \displaystyle h(n)=0 h ( n ) = 0 n = 3 \displaystyle n=3 n = 3 h ( 3 ) = − 4 < 0 \displaystyle h(3)=-4<0 h ( 3 ) = − 4 < 0 Correction. 두 경우를 함께 설명한 “0보다 작음”은 “0보다 작거나 같음”의 오기이다. 그러나 원칙에 따라 원문 표기를 그대로 실었다.
+1점 · 결론적으로 n = 2 \displaystyle n=2 n = 2 n = 3 \displaystyle n=3 n = 3 f ( x ) \displaystyle f(x) f ( x ) g ( x ) \displaystyle g(x) g ( x ) 한편 극솟값은 h ( n ) = 2 n 3 − 3 ( n + 1 ) n 2 + 6 n 2 − 4 n 2 + 20 n − 28 = − n 3 − n 2 + 20 n − 28 = − ( n − 2 ) ( n 2 + 3 n − 14 ) \displaystyle \begin{aligned}h(n)&=2n^3-3(n+1)n^2+6n^2-4n^2+20n-28\\&=-n^3-n^2+20n-28\\&=-(n-2)(n^2+3n-14)\end{aligned} h ( n ) = 2 n 3 − 3 ( n + 1 ) n 2 + 6 n 2 − 4 n 2 + 20 n − 28 = − n 3 − n 2 + 20 n − 28 = − ( n − 2 ) ( n 2 + 3 n − 14 ) 이므로, n = 2 \displaystyle n=2 n = 2 h ( 2 ) = 0 \displaystyle h(2)=0 h ( 2 ) = 0 n = 3 \displaystyle n=3 n = 3 h ( 3 ) = − 4 < 0 \displaystyle h(3)=-4<0 h ( 3 ) = − 4 < 0 n = 2 \displaystyle n=2 n = 2 n = 3 \displaystyle n=3 n = 3 f ( x ) \displaystyle f(x) f ( x ) g ( x ) \displaystyle g(x) g ( x )
3-(2) 위 (1)의 조건을 만족하는 자연수 n \displaystyle n n f ( x ) \displaystyle f(x) f ( x ) g ( x ) \displaystyle g(x) g ( x )
+1점 · (1)의 풀이를 근거로 두 곡선 f ( x ) \displaystyle f(x) f ( x ) g ( x ) \displaystyle g(x) g ( x ) n \displaystyle n n n = 2 \displaystyle n=2 n = 2 n = 3 \displaystyle n=3 n = 3 (풀이) (1)의 풀이에 의해 두 곡선 f ( x ) \displaystyle f(x) f ( x ) g ( x ) \displaystyle g(x) g ( x ) n \displaystyle n n n = 2 \displaystyle n=2 n = 2 n = 3 \displaystyle n=3 n = 3 f ( x ) \displaystyle f(x) f ( x ) g ( x ) \displaystyle g(x) g ( x ) h ( x ) \displaystyle h(x) h ( x ) x \displaystyle x x
+2점 · n = 2 \displaystyle n=2 n = 2 h ( x ) = 2 x 3 − 9 x 2 + 12 x − 4 = 0 \displaystyle h(x)=2x^3-9x^2+12x-4=0 h ( x ) = 2 x 3 − 9 x 2 + 12 x − 4 = 0 x = 1 2 , 2 \displaystyle x=\frac12,2 x = 2 1 , 2 (i) n = 2 \displaystyle n=2 n = 2 h ( x ) = 2 x 3 − 9 x 2 + 12 x − 4 = ( 2 x − 1 ) ( x − 2 ) 2 = 0 \displaystyle h(x)=2x^3-9x^2+12x-4=(2x-1)(x-2)^2=0 h ( x ) = 2 x 3 − 9 x 2 + 12 x − 4 = ( 2 x − 1 ) ( x − 2 ) 2 = 0 이므로 근은 x = 1 2 , 2 \displaystyle x=\frac12,2 x = 2 1 , 2
+4점 · 둘러싸인 도형의 넓이가 아래와 같음을 보임
A ( 2 ) = ∫ 1 2 2 ( 2 x 3 − 9 x 2 + 12 x − 4 ) d x = [ 1 2 x 4 − 3 x 3 + 6 x 2 − 4 x ] 1 2 2 = 27 32 \displaystyle A(2)=\int_{\frac12}^2(2x^3-9x^2+12x-4)\,dx=\Biggl[\frac12x^4-3x^3+6x^2-4x\Biggr]_{\frac12}^2=\frac{27}{32} A ( 2 ) = ∫ 2 1 2 ( 2 x 3 − 9 x 2 + 12 x − 4 ) d x = [ 2 1 x 4 − 3 x 3 + 6 x 2 − 4 x ] 2 1 2 = 32 27 A ( 2 ) = ∫ 1 2 2 ( 2 x 3 − 9 x 2 + 12 x − 4 ) d x = [ 1 2 x 4 − 3 x 3 + 6 x 2 − 4 x ] 1 2 2 = 27 32 \displaystyle A(2)=\int_{\frac12}^2(2x^3-9x^2+12x-4)\,dx=\Biggl[\frac12x^4-3x^3+6x^2-4x\Biggr]_{\frac12}^2=\frac{27}{32} A ( 2 ) = ∫ 2 1 2 ( 2 x 3 − 9 x 2 + 12 x − 4 ) d x = [ 2 1 x 4 − 3 x 3 + 6 x 2 − 4 x ] 2 1 2 = 32 27 이다.
+1점 · 2 − 3 ≤ 2 ≤ 2 + 3 \displaystyle 2-\sqrt3\le2\le2+\sqrt3 2 − 3 ≤ 2 ≤ 2 + 3 +2점 · n = 3 \displaystyle n=3 n = 3 h ( x ) = 2 x 3 − 12 x 2 + 18 x − 4 = 0 \displaystyle h(x)=2x^3-12x^2+18x-4=0 h ( x ) = 2 x 3 − 12 x 2 + 18 x − 4 = 0 x = 2 , 2 ± 3 \displaystyle x=2,2\pm\sqrt3 x = 2 , 2 ± 3 (ii) n = 3 \displaystyle n=3 n = 3 h ( x ) = 2 x 3 − 12 x 2 + 18 x − 4 = 2 ( x − 2 ) ( x 2 − 4 x + 1 ) = 0 \displaystyle h(x)=2x^3-12x^2+18x-4=2(x-2)(x^2-4x+1)=0 h ( x ) = 2 x 3 − 12 x 2 + 18 x − 4 = 2 ( x − 2 ) ( x 2 − 4 x + 1 ) = 0 이므로 근은 x = 2 , 2 ± 3 \displaystyle x=2,2\pm\sqrt3 x = 2 , 2 ± 3 ( 2 , 0 ) \displaystyle (2,0) ( 2 , 0 )
+2점 · 둘러싸인 도형의 넓이가
A ( 3 ) = ∫ 2 − 3 2 ( 2 x 3 − 12 x 2 + 18 x − 4 ) d x + ∫ 2 2 + 3 ( − 2 x 3 + 12 x 2 − 18 x + 4 ) d x \displaystyle A(3)=\int_{2-\sqrt3}^2(2x^3-12x^2+18x-4)\,dx+\int_2^{2+\sqrt3}(-2x^3+12x^2-18x+4)\,dx A ( 3 ) = ∫ 2 − 3 2 ( 2 x 3 − 12 x 2 + 18 x − 4 ) d x + ∫ 2 2 + 3 ( − 2 x 3 + 12 x 2 − 18 x + 4 ) d x A ( 3 ) = 2 ∫ 2 − 3 2 ( 2 x 3 − 12 x 2 + 18 x − 4 ) d x \displaystyle A(3)=2\int_{2-\sqrt3}^2(2x^3-12x^2+18x-4)\,dx A ( 3 ) = 2 ∫ 2 − 3 2 ( 2 x 3 − 12 x 2 + 18 x − 4 ) d x +5점 · A ( 3 ) = 9 \displaystyle A(3)=9 A ( 3 ) = 9 +2점 · 최댓값은 9이고 최솟값은 27 32 \displaystyle \frac{27}{32} 32 27 A ( 3 ) = 2 ∫ 2 − 3 2 ( 2 x 3 − 12 x 2 + 18 x − 4 ) d x = 4 ∫ 2 − 3 2 ( x − 2 ) ( x 2 − 4 x + 1 ) d x = 4 ∫ 2 − 3 2 ( x − 2 ) ( ( x − 2 ) 2 − 3 ) d x = 4 ∫ − 3 0 x ( x 2 − 3 ) d x = 4 ∫ − 3 0 ( x 3 − 3 x ) d x = 4 [ 1 4 x 4 − 3 2 x 2 ] − 3 0 = 9 \displaystyle \begin{aligned}A(3)&=2\int_{2-\sqrt3}^2(2x^3-12x^2+18x-4)\,dx\\&=4\int_{2-\sqrt3}^2(x-2)(x^2-4x+1)\,dx\\&=4\int_{2-\sqrt3}^2(x-2)((x-2)^2-3)\,dx\\&=4\int_{-\sqrt3}^0x(x^2-3)\,dx\\&=4\int_{-\sqrt3}^0(x^3-3x)\,dx\\&=4\Biggl[\frac14x^4-\frac32x^2\Biggr]_{-\sqrt3}^0\\&=9\end{aligned} A ( 3 ) = 2 ∫ 2 − 3 2 ( 2 x 3 − 12 x 2 + 18 x − 4 ) d x = 4 ∫ 2 − 3 2 ( x − 2 ) ( x 2 − 4 x + 1 ) d x = 4 ∫ 2 − 3 2 ( x − 2 ) (( x − 2 ) 2 − 3 ) d x = 4 ∫ − 3 0 x ( x 2 − 3 ) d x = 4 ∫ − 3 0 ( x 3 − 3 x ) d x = 4 [ 4 1 x 4 − 2 3 x 2 ] − 3 0 = 9 이다. 따라서 최댓값은 9이고 최솟값은 27 32 \displaystyle \frac{27}{32} 32 27
수리논술, 배움에서 논증의 완성까지.