1-(1) 부등식 x 2 ≤ x + 1 \displaystyle x^2\le x+1 x 2 ≤ x + 1 α ≤ x ≤ β \displaystyle \alpha\le x\le\beta α ≤ x ≤ β α , β \displaystyle \alpha,\beta α , β [ 0 , β ] \displaystyle [0,\beta] [ 0 , β ] f ( x ) = x + 1 x + 1 \displaystyle f(x)=x+\frac1{x+1} f ( x ) = x + x + 1 1 2 \displaystyle 2 2
[풀이]
+2점 · x 2 − x − 1 = 0 \displaystyle x^2-x-1=0 x 2 − x − 1 = 0 x = 1 ± 5 2 \displaystyle x=\frac{1\pm\sqrt5}2 x = 2 1 ± 5 함수 g ( x ) = x 2 − x − 1 \displaystyle g(x)=x^2-x-1 g ( x ) = x 2 − x − 1 g ( x ) = 0 \displaystyle g(x)=0 g ( x ) = 0 x = 1 ± 5 2 \displaystyle x=\frac{1\pm\sqrt5}2 x = 2 1 ± 5
+2점 · x 2 ≤ x + 1 \displaystyle x^2\le x+1 x 2 ≤ x + 1 1 − 5 2 ≤ x ≤ 1 + 5 2 \displaystyle \frac{1-\sqrt5}2\le x\le\frac{1+\sqrt5}2 2 1 − 5 ≤ x ≤ 2 1 + 5 α = 1 − 5 2 , β = 1 + 5 2 \displaystyle \alpha=\frac{1-\sqrt5}2,\beta=\frac{1+\sqrt5}2 α = 2 1 − 5 , β = 2 1 + 5 따라서 g ( x ) ≤ 0 \displaystyle g(x)\le0 g ( x ) ≤ 0 x 2 ≤ x + 1 \displaystyle x^2\le x+1 x 2 ≤ x + 1 1 − 5 2 ≤ x ≤ 1 + 5 2 \displaystyle \frac{1-\sqrt5}2\le x\le\frac{1+\sqrt5}2 2 1 − 5 ≤ x ≤ 2 1 + 5 α = 1 − 5 2 \displaystyle \alpha=\frac{1-\sqrt5}2 α = 2 1 − 5 β = 1 + 5 2 \displaystyle \beta=\frac{1+\sqrt5}2 β = 2 1 + 5
+3점 · 구간 ( 0 , 1 + 5 2 ) \displaystyle \left(0,\frac{1+\sqrt5}2\right) ( 0 , 2 1 + 5 ) f ′ ( x ) = 1 − 1 ( x + 1 ) 2 > 0 \displaystyle f^{\prime}(x)=1-\frac1{(x+1)^2}>0 f ′ ( x ) = 1 − ( x + 1 ) 2 1 > 0 f ( x ) \displaystyle f(x) f ( x ) [ 0 , 1 + 5 2 ] \displaystyle \left[0,\frac{1+\sqrt5}2\right] [ 0 , 2 1 + 5 ] f ( x ) \displaystyle f(x) f ( x ) [ 0 , 1 + 5 2 ] \displaystyle \left[0,\frac{1+\sqrt5}2\right] [ 0 , 2 1 + 5 ] 구간 [ 0 , β ] = [ 0 , 1 + 5 2 ] \displaystyle [0,\beta]=\left[0,\frac{1+\sqrt5}2\right] [ 0 , β ] = [ 0 , 2 1 + 5 ] f ( x ) = x + 1 x + 1 \displaystyle f(x)=x+\frac1{x+1} f ( x ) = x + x + 1 1 ( 0 , 1 + 5 2 ) \displaystyle \left(0,\frac{1+\sqrt5}2\right) ( 0 , 2 1 + 5 ) f ′ ( x ) = 1 − 1 ( x + 1 ) 2 > 0 \displaystyle f^{\prime}(x)=1-\frac1{(x+1)^2}>0 f ′ ( x ) = 1 − ( x + 1 ) 2 1 > 0 이다. 따라서 함수 f ( x ) \displaystyle f(x) f ( x ) [ 0 , 1 + 5 2 ] \displaystyle \left[0,\frac{1+\sqrt5}2\right] [ 0 , 2 1 + 5 ] x = 1 + 5 2 \displaystyle x=\frac{1+\sqrt5}2 x = 2 1 + 5
+3점 · x = 1 + 5 2 \displaystyle x=\frac{1+\sqrt5}2 x = 2 1 + 5 f ( 1 + 5 2 ) = 2 \displaystyle f\left(\frac{1+\sqrt5}2\right)=2 f ( 2 1 + 5 ) = 2 f ( 1 + 5 2 ) = 1 + 5 2 + 1 1 + 5 2 + 1 = 2 \displaystyle f\left(\frac{1+\sqrt5}2\right)=\frac{1+\sqrt5}2+\frac1{\frac{1+\sqrt5}2+1}=2 f ( 2 1 + 5 ) = 2 1 + 5 + 2 1 + 5 + 1 1 = 2 f ( 1 + 5 2 ) = 2 \displaystyle f\left(\frac{1+\sqrt5}2\right)=2 f ( 2 1 + 5 ) = 2 f ( 1 + 5 2 ) = 1 + 5 2 + 1 1 + 5 2 + 1 = 2 \displaystyle f\left(\frac{1+\sqrt5}2\right)=\frac{1+\sqrt5}2+\frac1{\frac{1+\sqrt5}2+1}=2 f ( 2 1 + 5 ) = 2 1 + 5 + 2 1 + 5 + 1 1 = 2 이므로 구간 [ 0 , β ] \displaystyle [0,\beta] [ 0 , β ] f ( x ) \displaystyle f(x) f ( x ) 2 \displaystyle 2 2
1-(2) β \displaystyle \beta β n \displaystyle n n a n ≥ β \displaystyle a_n\ge\beta a n ≥ β b n ≤ β \displaystyle b_n\le\beta b n ≤ β
[풀이]
+2점 · 첫 항 a 1 = 2 ≥ 1 + 5 2 \displaystyle a_1=2\ge\frac{1+\sqrt5}2 a 1 = 2 ≥ 2 1 + 5 (i) 수열 { a n } \displaystyle \{a_n\} { a n } a 1 = 2 ≥ 1 + 5 2 \displaystyle a_1=2\ge\frac{1+\sqrt5}2 a 1 = 2 ≥ 2 1 + 5
+2점 · n = k \displaystyle n=k n = k a k ≥ 1 + 5 2 \displaystyle a_k\ge\frac{1+\sqrt5}2 a k ≥ 2 1 + 5 +3점 · ( a k + 1 ) 2 = a k + 1 ≥ 1 + 5 2 + 1 = ( 1 + 5 2 ) 2 \displaystyle (a_{k+1})^2=a_k+1\ge\frac{1+\sqrt5}2+1=\left(\frac{1+\sqrt5}2\right)^2 ( a k + 1 ) 2 = a k + 1 ≥ 2 1 + 5 + 1 = ( 2 1 + 5 ) 2 +1점 · a k + 1 \displaystyle a_{k+1} a k + 1 a k + 1 ≥ 1 + 5 2 \displaystyle a_{k+1}\ge\frac{1+\sqrt5}2 a k + 1 ≥ 2 1 + 5 n = k \displaystyle n=k n = k a k ≥ 1 + 5 2 \displaystyle a_k\ge\frac{1+\sqrt5}2 a k ≥ 2 1 + 5 ( a k + 1 ) 2 = a k + 1 ≥ 1 + 5 2 + 1 = ( 1 + 5 2 ) 2 \displaystyle (a_{k+1})^2=a_k+1\ge\frac{1+\sqrt5}2+1=\left(\frac{1+\sqrt5}2\right)^2 ( a k + 1 ) 2 = a k + 1 ≥ 2 1 + 5 + 1 = ( 2 1 + 5 ) 2 이 성립한다. a k + 1 \displaystyle a_{k+1} a k + 1 n = k + 1 \displaystyle n=k+1 n = k + 1 a k + 1 ≥ 1 + 5 2 \displaystyle a_{k+1}\ge\frac{1+\sqrt5}2 a k + 1 ≥ 2 1 + 5 따라서 수학적 귀납법에 의해서 모든 자연수 n \displaystyle n n a n ≥ β \displaystyle a_n\ge\beta a n ≥ β
+2점 · 첫 항 b 1 = 1 ≤ 1 + 5 2 \displaystyle b_1=1\le\frac{1+\sqrt5}2 b 1 = 1 ≤ 2 1 + 5 (ii) 수열 { b n } \displaystyle \{b_n\} { b n } b 1 = 1 ≤ 1 + 5 2 \displaystyle b_1=1\le\frac{1+\sqrt5}2 b 1 = 1 ≤ 2 1 + 5
+2점 · n = k \displaystyle n=k n = k b k ≤ 1 + 5 2 \displaystyle b_k\le\frac{1+\sqrt5}2 b k ≤ 2 1 + 5 +3점 · b k + 1 = 2 − 1 b k + 1 ≤ 2 − 1 1 + 5 2 + 1 = 1 + 5 2 \displaystyle b_{k+1}=2-\frac1{b_k+1}\le2-\frac1{\frac{1+\sqrt5}2+1}=\frac{1+\sqrt5}2 b k + 1 = 2 − b k + 1 1 ≤ 2 − 2 1 + 5 + 1 1 = 2 1 + 5 n = k \displaystyle n=k n = k b k ≤ 1 + 5 2 \displaystyle b_k\le\frac{1+\sqrt5}2 b k ≤ 2 1 + 5 b k + 1 = 2 − 1 b k + 1 ≤ 2 − 1 1 + 5 2 + 1 = 2 − 2 3 + 5 = 1 + 5 2 \displaystyle b_{k+1}=2-\frac1{b_k+1}\le2-\frac1{\frac{1+\sqrt5}2+1}=2-\frac2{3+\sqrt5}=\frac{1+\sqrt5}2 b k + 1 = 2 − b k + 1 1 ≤ 2 − 2 1 + 5 + 1 1 = 2 − 3 + 5 2 = 2 1 + 5 가 되어 n = k + 1 \displaystyle n=k+1 n = k + 1 b k + 1 ≤ 1 + 5 2 \displaystyle b_{k+1}\le\frac{1+\sqrt5}2 b k + 1 ≤ 2 1 + 5 따라서 수학적 귀납법에 의해서 모든 자연수 n \displaystyle n n b n ≤ β \displaystyle b_n\le\beta b n ≤ β
1-(3) 모든 자연수 n \displaystyle n n a n ≥ a n + 1 \displaystyle a_n\ge a_{n+1} a n ≥ a n + 1 b n ≤ b n + 1 \displaystyle b_n\le b_{n+1} b n ≤ b n + 1
[풀이]
+1점 · 문항 (1)의 결과를 인용하여 x ≥ 1 + 5 2 \displaystyle x\ge\frac{1+\sqrt5}2 x ≥ 2 1 + 5 x 2 ≥ x + 1 \displaystyle x^2\ge x+1 x 2 ≥ x + 1 (i) 문항 (1)의 결과로부터 x ≥ 1 + 5 2 \displaystyle x\ge\frac{1+\sqrt5}2 x ≥ 2 1 + 5 x 2 ≥ x + 1 \displaystyle x^2\ge x+1 x 2 ≥ x + 1
+1점 · 문항 (2)의 결과를 인용하여 a n ≥ 1 + 5 2 \displaystyle a_n\ge\frac{1+\sqrt5}2 a n ≥ 2 1 + 5 +2점 · ( a n + 1 ) 2 = a n + 1 ≤ ( a n ) 2 \displaystyle (a_{n+1})^2=a_n+1\le(a_n)^2 ( a n + 1 ) 2 = a n + 1 ≤ ( a n ) 2 +1점 · a n , a n + 1 \displaystyle a_n,a_{n+1} a n , a n + 1 a n + 1 ≤ a n \displaystyle a_{n+1}\le a_n a n + 1 ≤ a n 문항 (2)의 결과로부터 a n ≥ 1 + 5 2 \displaystyle a_n\ge\frac{1+\sqrt5}2 a n ≥ 2 1 + 5 ( a n + 1 ) 2 = a n + 1 ≤ ( a n ) 2 \displaystyle (a_{n+1})^2=a_n+1\le(a_n)^2 ( a n + 1 ) 2 = a n + 1 ≤ ( a n ) 2 이 성립한다. a n , a n + 1 \displaystyle a_n,a_{n+1} a n , a n + 1 a n + 1 ≤ a n \displaystyle a_{n+1}\le a_n a n + 1 ≤ a n
+1점 · 문항 (1)의 결과를 인용하여 0 ≤ x ≤ 1 + 5 2 \displaystyle 0\le x\le\frac{1+\sqrt5}2 0 ≤ x ≤ 2 1 + 5 x + 1 x + 1 ≤ 2 \displaystyle x+\frac1{x+1}\le2 x + x + 1 1 ≤ 2 (ii) 문항 (1)의 결과로부터 0 ≤ x ≤ 1 + 5 2 \displaystyle 0\le x\le\frac{1+\sqrt5}2 0 ≤ x ≤ 2 1 + 5 x + 1 x + 1 ≤ 2 \displaystyle x+\frac1{x+1}\le2 x + x + 1 1 ≤ 2
+3점 · 문항 (2)의 결과 b n ≤ 1 + 5 2 \displaystyle b_n\le\frac{1+\sqrt5}2 b n ≤ 2 1 + 5 b n \displaystyle b_n b n b n + 1 b n + 1 ≤ 2 \displaystyle b_n+\frac1{b_n+1}\le2 b n + b n + 1 1 ≤ 2 +1점 · b n + 1 = 2 − 1 b n + 1 ≥ b n \displaystyle b_{n+1}=2-\frac1{b_n+1}\ge b_n b n + 1 = 2 − b n + 1 1 ≥ b n 양수 b n \displaystyle b_n b n b n ≤ 1 + 5 2 \displaystyle b_n\le\frac{1+\sqrt5}2 b n ≤ 2 1 + 5 b n + 1 b n + 1 ≤ 2 \displaystyle b_n+\frac1{b_n+1}\le2 b n + b n + 1 1 ≤ 2 따라서 b n + 1 = 2 − 1 b n + 1 ≥ b n \displaystyle b_{n+1}=2-\frac1{b_n+1}\ge b_n b n + 1 = 2 − b n + 1 1 ≥ b n 이 성립한다.
수리논술, 배움에서 논증의 완성까지.