+10점 · 접선의 방정식 y − y n = ( ln x n + 3 2 ) ( x − x n ) \displaystyle y-y_n=\left(\ln x_n+\frac32\right)(x-x_n) y − y n = ( ln x n + 2 3 ) ( x − x n ) 【2-1】 (30점) ○ 예시 답안: f ( x ) = x ln x + x 2 \displaystyle f(x)=x\ln x+\frac x2 f ( x ) = x ln x + 2 x f ′ ( x ) = ln x + 3 2 \displaystyle f^{\prime}(x)=\ln x+\frac32 f ′ ( x ) = ln x + 2 3 C \displaystyle C C ( x n , y n ) \displaystyle (x_n,y_n) ( x n , y n ) y − y n = ( ln x n + 3 2 ) ( x − x n ) \displaystyle y-y_n=\left(\ln x_n+\frac32\right)(x-x_n) y − y n = ( ln x n + 2 3 ) ( x − x n ) 이다. 이 접선과 y \displaystyle y y ( 0 , a n ) \displaystyle (0,a_n) ( 0 , a n ) a n − y n = ( ln x n + 3 2 ) ( − x n ) \displaystyle a_n-y_n=\left(\ln x_n+\frac32\right)(-x_n) a n − y n = ( ln x n + 2 3 ) ( − x n )
+10점 · x n = r n − 1 \displaystyle x_n=r^{n-1} x n = r n − 1 한편, 수열 { a n } \displaystyle \{a_n\} { a n } r \displaystyle r r a n = − r n − 1 \displaystyle a_n=-r^{n-1} a n = − r n − 1 y n = f ( x n ) = x n ln x n + x n 2 \displaystyle y_n=f(x_n)=x_n\ln x_n+\frac{x_n}2 y n = f ( x n ) = x n ln x n + 2 x n x n = r n − 1 \displaystyle x_n=r^{n-1} x n = r n − 1
+10점 · ln x 21 ln x 20 = 20 19 \displaystyle \frac{\ln x_{21}}{\ln x_{20}}=\frac{20}{19} ln x 20 ln x 21 = 19 20 이로부터 ln x n = ( n − 1 ) ln r \displaystyle \ln x_n=(n-1)\ln r ln x n = ( n − 1 ) ln r ln x 21 ln x 20 = 20 ln r 19 ln r = 20 19 \displaystyle \frac{\ln x_{21}}{\ln x_{20}}=\frac{20\ln r}{19\ln r}=\frac{20}{19} ln x 20 ln x 21 = 19 ln r 20 ln r = 19 20
+10점 · A n = ∫ r n − 1 r n ( x ln x + x 2 ) d x \displaystyle A_n=\int_{r^{n-1}}^{r^n}\left(x\ln x+\frac x2\right)dx A n = ∫ r n − 1 r n ( x ln x + 2 x ) d x A 20 = ∫ r 19 r 20 ( x ln x + x 2 ) d x \displaystyle A_{20}=\int_{r^{19}}^{r^{20}}\left(x\ln x+\frac x2\right)dx A 20 = ∫ r 19 r 20 ( x ln x + 2 x ) d x 【2-2】 (30점) ○ 예시 답안: (2-1)의 풀이과정에서 얻은 결과를 바탕으로 A n \displaystyle A_n A n A n = ∫ x n x n + 1 ∣ f ( x ) ∣ d x = ∫ r n − 1 r n ∣ x ln x + x 2 ∣ d x = ∫ r n − 1 r n ( x ln x + x 2 ) d x \displaystyle A_n=\int_{x_n}^{x_{n+1}}|f(x)|\,dx=\int_{r^{n-1}}^{r^n}\left|x\ln x+\frac x2\right|dx=\int_{r^{n-1}}^{r^n}\left(x\ln x+\frac x2\right)dx A n = ∫ x n x n + 1 ∣ f ( x ) ∣ d x = ∫ r n − 1 r n x ln x + 2 x d x = ∫ r n − 1 r n ( x ln x + 2 x ) d x 이를 부분적분법을 이용해 계산하면 A n = ∫ r n − 1 r n ( x ln x + x 2 ) d x = [ 1 2 x 2 ln x ] r n − 1 r n = 1 2 ( r 2 n ln r n − r 2 n − 2 ln r n − 1 ) = 1 2 { n r 2 − ( n − 1 ) } r 2 n − 2 ln r \displaystyle \begin{aligned}A_n&=\int_{r^{n-1}}^{r^n}\left(x\ln x+\frac x2\right)dx=\Biggl[\frac12x^2\ln x\Biggr]_{r^{n-1}}^{r^n}\\&=\frac12(r^{2n}\ln r^n-r^{2n-2}\ln r^{n-1})=\frac12\{nr^2-(n-1)\}r^{2n-2}\ln r\end{aligned} A n = ∫ r n − 1 r n ( x ln x + 2 x ) d x = [ 2 1 x 2 ln x ] r n − 1 r n = 2 1 ( r 2 n ln r n − r 2 n − 2 ln r n − 1 ) = 2 1 { n r 2 − ( n − 1 )} r 2 n − 2 ln r 이다.
+10점 · A 20 = 1 2 ( 20 r 2 − 19 ) r 38 ln r \displaystyle A_{20}=\frac12(20r^2-19)r^{38}\ln r A 20 = 2 1 ( 20 r 2 − 19 ) r 38 ln r 이로부터 A 20 = 1 2 ( 20 r 2 − 19 ) r 38 ln r \displaystyle A_{20}=\frac12(20r^2-19)r^{38}\ln r A 20 = 2 1 ( 20 r 2 − 19 ) r 38 ln r
+10점 · 1 r 2 − 1 ( A 20 r 38 ln r − 1 2 ) = 10 \displaystyle \frac1{r^2-1}\left(\frac{A_{20}}{r^{38}\ln r}-\frac12\right)=10 r 2 − 1 1 ( r 38 ln r A 20 − 2 1 ) = 10 따라서 1 r 2 − 1 ( A 20 r 38 ln r − 1 2 ) = 1 r 2 − 1 { 1 2 ( 20 r 2 − 19 ) − 1 2 } = 10 ( r 2 − 1 ) r 2 − 1 = 10 \displaystyle \frac1{r^2-1}\left(\frac{A_{20}}{r^{38}\ln r}-\frac12\right)=\frac1{r^2-1}\left\{\frac12(20r^2-19)-\frac12\right\}=\frac{10(r^2-1)}{r^2-1}=10 r 2 − 1 1 ( r 38 ln r A 20 − 2 1 ) = r 2 − 1 1 { 2 1 ( 20 r 2 − 19 ) − 2 1 } = r 2 − 1 10 ( r 2 − 1 ) = 10
+15점 · (1) B 1 = 1 ln r ∫ 1 r 20 ( x ln x + x 2 ) d x \displaystyle B_1=\frac1{\ln r}\int_1^{r^{20}}\left(x\ln x+\frac x2\right)dx B 1 = ln r 1 ∫ 1 r 20 ( x ln x + 2 x ) d x 【2-3】 (60점) (1) (30점) ○ 예시 답안: (2-2)의 풀이과정에서 얻은 식 A n = ∫ r n − 1 r n ( x ln x + x 2 ) d x \displaystyle A_n=\int_{r^{n-1}}^{r^n}\left(x\ln x+\frac x2\right)dx A n = ∫ r n − 1 r n ( x ln x + 2 x ) d x B 1 = ∑ n = 1 20 A n ln r = 1 ln r ∑ n = 1 20 A n = 1 ln r ∑ n = 1 20 ∫ r n − 1 r n ( x ln x + x 2 ) d x = 1 ln r ∫ 1 r 20 ( x ln x + x 2 ) d x = 1 ln r [ 1 2 x 2 ln x ] 1 r 20 = 10 r 40 \displaystyle \begin{aligned}B_1&=\sum_{n=1}^{20}\frac{A_n}{\ln r}=\frac1{\ln r}\sum_{n=1}^{20}A_n=\frac1{\ln r}\sum_{n=1}^{20}\int_{r^{n-1}}^{r^n}\left(x\ln x+\frac x2\right)dx\\&=\frac1{\ln r}\int_1^{r^{20}}\left(x\ln x+\frac x2\right)dx=\frac1{\ln r}\Biggl[\frac12x^2\ln x\Biggr]_1^{r^{20}}=10r^{40}\end{aligned} B 1 = n = 1 ∑ 20 ln r A n = ln r 1 n = 1 ∑ 20 A n = ln r 1 n = 1 ∑ 20 ∫ r n − 1 r n ( x ln x + 2 x ) d x = ln r 1 ∫ 1 r 20 ( x ln x + 2 x ) d x = ln r 1 [ 2 1 x 2 ln x ] 1 r 20 = 10 r 40 이다.
+15점 · (1) B 1 2 r 40 = 5 \displaystyle \frac{B_1}{2r^{40}}=5 2 r 40 B 1 = 5 따라서 B 1 2 r 40 = 5 \displaystyle \frac{B_1}{2r^{40}}=5 2 r 40 B 1 = 5
+10점 · (2) B m = 10 r 40 − m − 1 2 r 2 m − 2 \displaystyle B_m=10r^{40}-\frac{m-1}2r^{2m-2} B m = 10 r 40 − 2 m − 1 r 2 m − 2 (2) (30점) ○ 예시 답안: (2-2)의 풀이과정에서 얻은 식 A n = ∫ r n − 1 r n ( x ln x + x 2 ) d x \displaystyle A_n=\int_{r^{n-1}}^{r^n}\left(x\ln x+\frac x2\right)dx A n = ∫ r n − 1 r n ( x ln x + 2 x ) d x B m = ∑ n = m 20 A n ln r = 1 ln r ∑ n = m 20 A n = 1 ln r ∑ n = m 20 ∫ r n − 1 r n ( x ln x + x 2 ) d x = 1 ln r ∫ r m − 1 r 20 ( x ln x + x 2 ) d x = 1 ln r [ 1 2 x 2 ln x ] r m − 1 r 20 = 10 r 40 − m − 1 2 r 2 m − 2 \displaystyle \begin{aligned}B_m&=\sum_{n=m}^{20}\frac{A_n}{\ln r}=\frac1{\ln r}\sum_{n=m}^{20}A_n=\frac1{\ln r}\sum_{n=m}^{20}\int_{r^{n-1}}^{r^n}\left(x\ln x+\frac x2\right)dx\\&=\frac1{\ln r}\int_{r^{m-1}}^{r^{20}}\left(x\ln x+\frac x2\right)dx=\frac1{\ln r}\Biggl[\frac12x^2\ln x\Biggr]_{r^{m-1}}^{r^{20}}=10r^{40}-\frac{m-1}2r^{2m-2}\end{aligned} B m = n = m ∑ 20 ln r A n = ln r 1 n = m ∑ 20 A n = ln r 1 n = m ∑ 20 ∫ r n − 1 r n ( x ln x + 2 x ) d x = ln r 1 ∫ r m − 1 r 20 ( x ln x + 2 x ) d x = ln r 1 [ 2 1 x 2 ln x ] r m − 1 r 20 = 10 r 40 − 2 m − 1 r 2 m − 2 이다.
+10점 · (2) B 1 − B m = m − 1 2 r 2 m − 2 \displaystyle B_1-B_m=\frac{m-1}2r^{2m-2} B 1 − B m = 2 m − 1 r 2 m − 2 (2-3) (1)에서 구한 B 1 = 10 r 40 \displaystyle B_1=10r^{40} B 1 = 10 r 40 B 1 − B m = m − 1 2 r 2 m − 2 \displaystyle B_1-B_m=\frac{m-1}2r^{2m-2} B 1 − B m = 2 m − 1 r 2 m − 2
+10점 · (2) ∑ m = 1 20 ( B 1 − B m r 2 m − 2 ) = 95 \displaystyle \sum_{m=1}^{20}\left(\frac{B_1-B_m}{r^{2m-2}}\right)=95 m = 1 ∑ 20 ( r 2 m − 2 B 1 − B m ) = 95 따라서 ∑ m = 1 20 ( B 1 − B m r 2 m − 2 ) = ∑ m = 1 20 ( m − 1 2 ) = 1 2 ∑ m = 1 20 ( m − 1 ) = 1 2 ( 20 × 21 2 − 20 ) = 95 \displaystyle \sum_{m=1}^{20}\left(\frac{B_1-B_m}{r^{2m-2}}\right)=\sum_{m=1}^{20}\left(\frac{m-1}2\right)=\frac12\sum_{m=1}^{20}(m-1)=\frac12\left(\frac{20\times21}2-20\right)=95 m = 1 ∑ 20 ( r 2 m − 2 B 1 − B m ) = m = 1 ∑ 20 ( 2 m − 1 ) = 2 1 m = 1 ∑ 20 ( m − 1 ) = 2 1 ( 2 20 × 21 − 20 ) = 95 이다.
수리논술, 배움에서 논증의 완성까지.