문제 2-1 예시 답안
+3점 · [2-1] log 1.01 365 = 1.5695 \displaystyle \log1.01^{365}=1.5695 log 1.0 1 365 = 1.5695 a = 1.01 365 \displaystyle a=1.01^{365} a = 1.0 1 365 log a = 365 log 1.01 = 365 × 0.0043 = 1.5695 \displaystyle \log a=365\log1.01=365\times0.0043=1.5695 log a = 365 log 1.01 = 365 × 0.0043 = 1.5695
+2점 · [2-1] log 3.71 = 0.5694 < 0.5695 < 0.5705 = log 3.72 \displaystyle \log3.71=0.5694<0.5695<0.5705=\log3.72 log 3.71 = 0.5694 < 0.5695 < 0.5705 = log 3.72 상용로그표에 의하면, log 3.71 = 0.5694 < 0.5695 < 0.5705 = log 3.72 \displaystyle \log3.71=0.5694<0.5695<0.5705=\log3.72 log 3.71 = 0.5694 < 0.5695 < 0.5705 = log 3.72 log 37.1 = 1 + log 3.71 = 1.5694 < 1.5695 < 1.5705 = 1 + log 3.72 = log 37.2 \displaystyle \log37.1=1+\log3.71=1.5694<1.5695<1.5705=1+\log3.72=\log37.2 log 37.1 = 1 + log 3.71 = 1.5694 < 1.5695 < 1.5705 = 1 + log 3.72 = log 37.2
+5점 · [2-1] m \displaystyle m m 따라서 37.1 < a < 37.2 \displaystyle 37.1<a<37.2 37.1 < a < 37.2 m = 37 \displaystyle m=37 m = 37
문제 2-2 예시 답안
+5점 · [2-2] f ( n ) = ∑ k = 1 n − 1 ( 3 ⋅ 4 k − 2 ⋅ 2 k + 1 ) \displaystyle f(n)=\sum_{k=1}^{n-1}(3\cdot4^k-2\cdot2^{k+1}) f ( n ) = k = 1 ∑ n − 1 ( 3 ⋅ 4 k − 2 ⋅ 2 k + 1 ) +5점 · [2-2] f ( n ) = 4 ⋅ 4 n − 1 − 8 ⋅ 2 n − 1 + 4 \displaystyle f(n)=4\cdot4^{n-1}-8\cdot2^{n-1}+4 f ( n ) = 4 ⋅ 4 n − 1 − 8 ⋅ 2 n − 1 + 4 f ( n ) = ( 2 n − 2 ) 2 \displaystyle f(n)=(2^n-2)^2 f ( n ) = ( 2 n − 2 ) 2 +5점 · [2-2] 자연수 n \displaystyle n n n = 6 \displaystyle n=6 n = 6 g ( x ) = 2 ⋅ 2 x \displaystyle g(x)=2\cdot2^x g ( x ) = 2 ⋅ 2 x h ( x ) = 3 ⋅ 4 x \displaystyle h(x)=3\cdot4^x h ( x ) = 3 ⋅ 4 x x ≥ 0 \displaystyle x\ge0 x ≥ 0 g ( x ) ≤ h ( x ) \displaystyle g(x)\le h(x) g ( x ) ≤ h ( x ) k \displaystyle k k g ( k + 1 ) = 2 ⋅ 2 k + 1 \displaystyle g(k+1)=2\cdot2^{k+1} g ( k + 1 ) = 2 ⋅ 2 k + 1 h ( k ) = 3 ⋅ 4 k \displaystyle h(k)=3\cdot4^k h ( k ) = 3 ⋅ 4 k { ( x , y ) ∣ k ≤ x ≤ k + 1 , g ( x ) ≤ y ≤ h ( x ) } \displaystyle \{(x,y)\mid k\le x\le k+1,g(x)\le y\le h(x)\} {( x , y ) ∣ k ≤ x ≤ k + 1 , g ( x ) ≤ y ≤ h ( x )} 3 ⋅ 4 k − 2 ⋅ 2 k + 1 \displaystyle 3\cdot4^k-2\cdot2^{k+1} 3 ⋅ 4 k − 2 ⋅ 2 k + 1 f ( n ) = ∑ k = 1 n − 1 ( 3 ⋅ 4 k − 2 ⋅ 2 k + 1 ) = 12 ( 4 n − 1 − 1 ) 4 − 1 − 8 ( 2 n − 1 − 1 ) 2 − 1 = ( 2 ⋅ 2 n − 1 − 2 ) 2 = ( 2 n − 2 ) 2 ≥ 2016 \displaystyle \begin{aligned}f(n)&=\sum_{k=1}^{n-1}(3\cdot4^k-2\cdot2^{k+1})=\frac{12(4^{n-1}-1)}{4-1}-\frac{8(2^{n-1}-1)}{2-1}\\&=(2\cdot2^{n-1}-2)^2=(2^n-2)^2\ge2016\end{aligned} f ( n ) = k = 1 ∑ n − 1 ( 3 ⋅ 4 k − 2 ⋅ 2 k + 1 ) = 4 − 1 12 ( 4 n − 1 − 1 ) − 2 − 1 8 ( 2 n − 1 − 1 ) = ( 2 ⋅ 2 n − 1 − 2 ) 2 = ( 2 n − 2 ) 2 ≥ 2016 이므로 ( 2 n − 2 ) 2 ≥ 2016 > 1936 = 44 2 \displaystyle (2^n-2)^2\ge2016>1936=44^2 ( 2 n − 2 ) 2 ≥ 2016 > 1936 = 4 4 2 2 n > 44 + 2 = 46 \displaystyle 2^n>44+2=46 2 n > 44 + 2 = 46 45 2 = 2025 > 2016 \displaystyle 45^2=2025>2016 4 5 2 = 2025 > 2016 ( 2 n − 2 ) 2 ≥ 2016 \displaystyle (2^n-2)^2\ge2016 ( 2 n − 2 ) 2 ≥ 2016 n \displaystyle n n n = 6 \displaystyle n=6 n = 6
수리논술, 배움에서 논증의 완성까지.