[수학2-ⅰ]
+3점 · g ( x ) = ( − 1 − 2 ) x \displaystyle g(x)=(-1-\sqrt{2})x g ( x ) = ( − 1 − 2 ) x b \displaystyle b b ( b , f ( b ) ) \displaystyle (b,f(b)) ( b , f ( b )) y = ( 2 b − 1 ) ( x − b ) + ( b 2 − b ) \displaystyle y=(2b-1)(x-b)+(b^{2}-b) y = ( 2 b − 1 ) ( x − b ) + ( b 2 − b ) y = − x 2 − x \displaystyle y=-x^{2}-x y = − x 2 − x x = − b ± 2 b \displaystyle x=-b\pm\sqrt{2}b x = − b ± 2 b g ( x ) \displaystyle g(x) g ( x ) b \displaystyle b b g ( b ) \displaystyle g(b) g ( b ) g ( b ) = ( − 1 − 2 ) b \displaystyle g(b)=(-1-\sqrt{2})b g ( b ) = ( − 1 − 2 ) b 비슷한 방법으로, b \displaystyle b b ( b , f ( b ) ) \displaystyle (b,f(b)) ( b , f ( b )) y = ( − 2 b − 1 ) ( x − b ) + ( − b 2 − b ) \displaystyle y=(-2b-1)(x-b)+(-b^{2}-b) y = ( − 2 b − 1 ) ( x − b ) + ( − b 2 − b ) y = x 2 − x \displaystyle y=x^{2}-x y = x 2 − x x = − b ± 2 b \displaystyle x=-b\pm\sqrt{2}b x = − b ± 2 b g ( x ) \displaystyle g(x) g ( x ) b \displaystyle b b g ( b ) \displaystyle g(b) g ( b ) g ( b ) = ( − 1 − 2 ) b \displaystyle g(b)=(-1-\sqrt{2})b g ( b ) = ( − 1 − 2 ) b
+2점 · h ( x ) = ( 1 − 2 ) x \displaystyle h(x)=(1-\sqrt{2})x h ( x ) = ( 1 − 2 ) x 따라서 모든 실수 x \displaystyle x x g ( x ) \displaystyle g(x) g ( x ) g ( x ) = ( − 1 − 2 ) x \displaystyle g(x)=(-1-\sqrt{2})x g ( x ) = ( − 1 − 2 ) x h ( x ) \displaystyle h(x) h ( x ) h ( x ) = ( 1 − 2 ) x \displaystyle h(x)=(1-\sqrt{2})x h ( x ) = ( 1 − 2 ) x 답: g ( x ) = ( − 1 − 2 ) x \displaystyle g(x)=(-1-\sqrt{2})x g ( x ) = ( − 1 − 2 ) x h ( x ) = ( 1 − 2 ) x \displaystyle h(x)=(1-\sqrt{2})x h ( x ) = ( 1 − 2 ) x
[수학2-ⅱ]
문제에 주어진 상황을 그림으로 나타내면 아래와 같다.
+5점 · L 2 m ( x ) \displaystyle L_{2m}(x) L 2 m ( x ) − 2 α 2 m + 1 − 1 \displaystyle -2\alpha^{2m+1}-1 − 2 α 2 m + 1 − 1 L 2 m ( x ) \displaystyle L_{2m}(x) L 2 m ( x ) α 4 m + 2 \displaystyle \alpha^{4m+2} α 4 m + 2 x 0 = 1 \displaystyle x_{0}=1 x 0 = 1 n \displaystyle n n x n + 1 = h ( x n ) = ( 1 − 2 ) x n \displaystyle x_{n+1}=h(x_{n})=(1-\sqrt{2})x_{n} x n + 1 = h ( x n ) = ( 1 − 2 ) x n x n = ( 1 − 2 ) n \displaystyle x_{n}=(1-\sqrt{2})^{n} x n = ( 1 − 2 ) n y = L 2 m ( x ) \displaystyle y=L_{2m}(x) y = L 2 m ( x ) P 2 m + 1 \displaystyle \mathrm{P}_{2m+1} P 2 m + 1 α = 1 − 2 \displaystyle \alpha=1-\sqrt{2} α = 1 − 2 b = x 2 m + 1 = α 2 m + 1 \displaystyle b=x_{2m+1}=\alpha^{2m+1} b = x 2 m + 1 = α 2 m + 1 L 2 m ( x ) = ( − 2 b − 1 ) x + b 2 = ( − 2 α 2 m + 1 − 1 ) x + α 4 m + 2 \displaystyle L_{2m}(x)=(-2b-1)x+b^{2}=(-2\alpha^{2m+1}-1)x+\alpha^{4m+2} L 2 m ( x ) = ( − 2 b − 1 ) x + b 2 = ( − 2 α 2 m + 1 − 1 ) x + α 4 m + 2
+5점 · L 2 m + 1 ( x ) \displaystyle L_{2m+1}(x) L 2 m + 1 ( x ) 2 α 2 m + 2 − 1 \displaystyle 2\alpha^{2m+2}-1 2 α 2 m + 2 − 1 L 2 m + 1 ( x ) \displaystyle L_{2m+1}(x) L 2 m + 1 ( x ) − α 4 m + 4 \displaystyle -\alpha^{4m+4} − α 4 m + 4 한편, 직선 y = L 2 m + 1 ( x ) \displaystyle y=L_{2m+1}(x) y = L 2 m + 1 ( x ) P 2 m + 2 \displaystyle \mathrm{P}_{2m+2} P 2 m + 2 c = x 2 m + 2 = α 2 m + 2 \displaystyle c=x_{2m+2}=\alpha^{2m+2} c = x 2 m + 2 = α 2 m + 2 L 2 m + 1 ( x ) = ( 2 c − 1 ) x − c 2 = ( 2 α 2 m + 2 − 1 ) x − α 4 m + 4 \displaystyle L_{2m+1}(x)=(2c-1)x-c^{2}=(2\alpha^{2m+2}-1)x-\alpha^{4m+4} L 2 m + 1 ( x ) = ( 2 c − 1 ) x − c 2 = ( 2 α 2 m + 2 − 1 ) x − α 4 m + 4 답: L 2 m ( x ) \displaystyle L_{2m}(x) L 2 m ( x ) − 2 α 2 m + 1 − 1 \displaystyle -2\alpha^{2m+1}-1 − 2 α 2 m + 1 − 1 L 2 m ( x ) \displaystyle L_{2m}(x) L 2 m ( x ) α 4 m + 2 \displaystyle \alpha^{4m+2} α 4 m + 2 L 2 m + 1 ( x ) \displaystyle L_{2m+1}(x) L 2 m + 1 ( x ) 2 α 2 m + 2 − 1 \displaystyle 2\alpha^{2m+2}-1 2 α 2 m + 2 − 1 L 2 m + 1 ( x ) \displaystyle L_{2m+1}(x) L 2 m + 1 ( x ) − α 4 m + 4 \displaystyle -\alpha^{4m+4} − α 4 m + 4
[별해] 점 P n = ( x n , y n ) \displaystyle \mathrm{P}_{n}=(x_{n},y_{n}) P n = ( x n , y n ) n \displaystyle n n α \displaystyle \alpha α n \displaystyle n n 즉, x n = α n \displaystyle x_{n}=\alpha^{n} x n = α n y n = { α 2 n − α n , n 은 짝수 − α 2 n − α n , n 은 홀수 \displaystyle y_{n}=\begin{cases}\alpha^{2n}-\alpha^{n}, & n\text{은 짝수} \\ -\alpha^{2n}-\alpha^{n}, & n\text{은 홀수}\end{cases} y n = { α 2 n − α n , − α 2 n − α n , n 은 짝수 n 은 홀수 점 P 2 m \displaystyle \mathrm{P}_{2m} P 2 m P 2 m + 1 \displaystyle \mathrm{P}_{2m+1} P 2 m + 1 L 2 m ( x ) \displaystyle L_{2m}(x) L 2 m ( x ) 점 P 2 m + 1 \displaystyle \mathrm{P}_{2m+1} P 2 m + 1 P 2 m + 2 \displaystyle \mathrm{P}_{2m+2} P 2 m + 2 L 2 m + 1 ( x ) \displaystyle L_{2m+1}(x) L 2 m + 1 ( x ) L 2 m ( x ) = ( y 2 m + 1 − y 2 m x 2 m + 1 − x 2 m ) ( x − x 2 m ) + y 2 m , \displaystyle L_{2m}(x)=\left(\frac{y_{2m+1}-y_{2m}}{x_{2m+1}-x_{2m}}\right)(x-x_{2m})+y_{2m}, L 2 m ( x ) = ( x 2 m + 1 − x 2 m y 2 m + 1 − y 2 m ) ( x − x 2 m ) + y 2 m , L 2 m + 1 ( x ) = ( y 2 m + 2 − y 2 m + 1 x 2 m + 2 − x 2 m + 1 ) ( x − x 2 m + 1 ) + y 2 m + 1 \displaystyle L_{2m+1}(x)=\left(\frac{y_{2m+2}-y_{2m+1}}{x_{2m+2}-x_{2m+1}}\right)(x-x_{2m+1})+y_{2m+1} L 2 m + 1 ( x ) = ( x 2 m + 2 − x 2 m + 1 y 2 m + 2 − y 2 m + 1 ) ( x − x 2 m + 1 ) + y 2 m + 1 을 구한 후, 이 식의 계수들을 α \displaystyle \alpha α m \displaystyle m m
[수학2-ⅲ]
+5점 · A 2 m = − 2 3 α 6 m + 1 \displaystyle A_{2m}=-\frac{2}{3}\alpha^{6m+1} A 2 m = − 3 2 α 6 m + 1 [수학2-ⅱ]의 풀이에서처럼 편의상 α = 1 − 2 \displaystyle \alpha=1-\sqrt{2} α = 1 − 2 b = x 2 m + 1 = α 2 m + 1 \displaystyle b=x_{2m+1}=\alpha^{2m+1} b = x 2 m + 1 = α 2 m + 1 c = x 2 m + 2 = α 2 m + 2 \displaystyle c=x_{2m+2}=\alpha^{2m+2} c = x 2 m + 2 = α 2 m + 2 A 2 m = ∫ x 2 m + 1 x 2 m ( L 2 m ( x ) − f ( x ) ) d x = ∫ b 0 ( ( − 2 b − 1 ) x + b 2 − ( − x 2 − x ) ) d x + ∫ 0 b / α ( ( − 2 b − 1 ) x + b 2 − ( x 2 − x ) ) d x = ∫ b 0 ( x 2 − 2 b x + b 2 ) d x + ∫ 0 b / α ( − x 2 − 2 b x + b 2 ) d x = [ x 3 3 − b x 2 + b 2 x ] b 0 + [ − x 3 3 − b x 2 + b 2 x ] 0 b / α = − b 3 3 + ( − b 3 3 α 3 − b 3 α 2 + b 3 α ) = − ( α 3 + 1 + 3 α − 3 α 2 3 α 3 ) b 3 = − ( 2 + 5 α + 1 + 3 α − 3 − 6 α 3 α 3 ) α 6 m + 3 = − 2 3 α 6 m + 1 \displaystyle \begin{aligned} A_{2m} &=\int_{x_{2m+1}}^{x_{2m}}(L_{2m}(x)-f(x))dx=\int_{b}^{0}((-2b-1)x+b^{2}-(-x^{2}-x))dx+\int_{0}^{b/\alpha}((-2b-1)x+b^{2}-(x^{2}-x))dx \\ &=\int_{b}^{0}(x^{2}-2bx+b^{2})dx+\int_{0}^{b/\alpha}(-x^{2}-2bx+b^{2})dx=\Biggl[\frac{x^{3}}{3}-bx^{2}+b^{2}x\Biggr]_{b}^{0}+\Biggl[-\frac{x^{3}}{3}-bx^{2}+b^{2}x\Biggr]_{0}^{b/\alpha} \\ &=-\frac{b^{3}}{3}+\left(-\frac{b^{3}}{3\alpha^{3}}-\frac{b^{3}}{\alpha^{2}}+\frac{b^{3}}{\alpha}\right)=-\left(\frac{\alpha^{3}+1+3\alpha-3\alpha^{2}}{3\alpha^{3}}\right)b^{3}=-\left(\frac{2+5\alpha+1+3\alpha-3-6\alpha}{3\alpha^{3}}\right)\alpha^{6m+3} \\ &=-\frac{2}{3}\alpha^{6m+1} \end{aligned} A 2 m = ∫ x 2 m + 1 x 2 m ( L 2 m ( x ) − f ( x )) d x = ∫ b 0 (( − 2 b − 1 ) x + b 2 − ( − x 2 − x )) d x + ∫ 0 b / α (( − 2 b − 1 ) x + b 2 − ( x 2 − x )) d x = ∫ b 0 ( x 2 − 2 b x + b 2 ) d x + ∫ 0 b / α ( − x 2 − 2 b x + b 2 ) d x = [ 3 x 3 − b x 2 + b 2 x ] b 0 + [ − 3 x 3 − b x 2 + b 2 x ] 0 b / α = − 3 b 3 + ( − 3 α 3 b 3 − α 2 b 3 + α b 3 ) = − ( 3 α 3 α 3 + 1 + 3 α − 3 α 2 ) b 3 = − ( 3 α 3 2 + 5 α + 1 + 3 α − 3 − 6 α ) α 6 m + 3 = − 3 2 α 6 m + 1 이 되고,
+5점 · A 2 m + 1 = 2 3 α 6 m + 4 \displaystyle A_{2m+1}=\frac{2}{3}\alpha^{6m+4} A 2 m + 1 = 3 2 α 6 m + 4 A 2 m + 1 = ∫ x 2 m + 1 x 2 m + 2 ( f ( x ) − L 2 m + 1 ( x ) ) d x = ∫ 0 c ( ( x 2 − x ) − ( ( 2 c − 1 ) x − c 2 ) ) d x + ∫ c / α 0 ( ( − x 2 − x ) − ( ( 2 c − 1 ) x − c 2 ) ) d x = ∫ 0 c ( x 2 − 2 c x + c 2 ) d x + ∫ c / α 0 ( − x 2 − 2 c x + c 2 ) d x = [ x 3 3 − c x 2 + c 2 x ] 0 c + [ − x 3 3 − c x 2 + c 2 x ] c / α 0 = c 3 3 − ( − c 3 3 α 3 − c 3 α 2 + c 3 α ) = ( α 3 + 1 + 3 α − 3 α 2 3 α 3 ) c 3 = ( 2 + 5 α + 1 + 3 α − 3 − 6 α 3 α 3 ) α 6 m + 6 = 2 3 α 6 m + 4 \displaystyle \begin{aligned} A_{2m+1} &=\int_{x_{2m+1}}^{x_{2m+2}}(f(x)-L_{2m+1}(x))dx=\int_{0}^{c}((x^{2}-x)-((2c-1)x-c^{2}))dx+\int_{c/\alpha}^{0}((-x^{2}-x)-((2c-1)x-c^{2}))dx \\ &=\int_{0}^{c}(x^{2}-2cx+c^{2})dx+\int_{c/\alpha}^{0}(-x^{2}-2cx+c^{2})dx=\Biggl[\frac{x^{3}}{3}-cx^{2}+c^{2}x\Biggr]_{0}^{c}+\Biggl[-\frac{x^{3}}{3}-cx^{2}+c^{2}x\Biggr]_{c/\alpha}^{0} \\ &=\frac{c^{3}}{3}-\left(-\frac{c^{3}}{3\alpha^{3}}-\frac{c^{3}}{\alpha^{2}}+\frac{c^{3}}{\alpha}\right)=\left(\frac{\alpha^{3}+1+3\alpha-3\alpha^{2}}{3\alpha^{3}}\right)c^{3}=\left(\frac{2+5\alpha+1+3\alpha-3-6\alpha}{3\alpha^{3}}\right)\alpha^{6m+6} \\ &=\frac{2}{3}\alpha^{6m+4} \end{aligned} A 2 m + 1 = ∫ x 2 m + 1 x 2 m + 2 ( f ( x ) − L 2 m + 1 ( x )) d x = ∫ 0 c (( x 2 − x ) − (( 2 c − 1 ) x − c 2 )) d x + ∫ c / α 0 (( − x 2 − x ) − (( 2 c − 1 ) x − c 2 )) d x = ∫ 0 c ( x 2 − 2 c x + c 2 ) d x + ∫ c / α 0 ( − x 2 − 2 c x + c 2 ) d x = [ 3 x 3 − c x 2 + c 2 x ] 0 c + [ − 3 x 3 − c x 2 + c 2 x ] c / α 0 = 3 c 3 − ( − 3 α 3 c 3 − α 2 c 3 + α c 3 ) = ( 3 α 3 α 3 + 1 + 3 α − 3 α 2 ) c 3 = ( 3 α 3 2 + 5 α + 1 + 3 α − 3 − 6 α ) α 6 m + 6 = 3 2 α 6 m + 4 이 된다. 답: A 2 m = − 2 3 α 6 m + 1 \displaystyle A_{2m}=-\frac{2}{3}\alpha^{6m+1} A 2 m = − 3 2 α 6 m + 1 A 2 m + 1 = 2 3 α 6 m + 4 \displaystyle A_{2m+1}=\frac{2}{3}\alpha^{6m+4} A 2 m + 1 = 3 2 α 6 m + 4 α = 1 − 2 \displaystyle \alpha=1-\sqrt{2} α = 1 − 2
[수학2-ⅳ]
+5점 · A 2 m + 1 A 2 m = − α 3 = ( 2 − 1 ) 3 \displaystyle \frac{A_{2m+1}}{A_{2m}}=-\alpha^{3}=(\sqrt{2}-1)^{3} A 2 m A 2 m + 1 = − α 3 = ( 2 − 1 ) 3 [수학2-ⅲ]의 풀이에 있는 결과로부터, A 2 m + 1 A 2 m = − α 3 = ( 2 − 1 ) 3 \displaystyle \frac{A_{2m+1}}{A_{2m}}=-\alpha^{3}=(\sqrt{2}-1)^{3} A 2 m A 2 m + 1 = − α 3 = ( 2 − 1 ) 3 이 되므로, m \displaystyle m m 답: A 2 m + 1 A 2 m = − α 3 = ( 2 − 1 ) 3 \displaystyle \frac{A_{2m+1}}{A_{2m}}=-\alpha^{3}=(\sqrt{2}-1)^{3} A 2 m A 2 m + 1 = − α 3 = ( 2 − 1 ) 3
수리논술, 배움에서 논증의 완성까지.