+4점 · f ( a ) = ln a + 1 a − 1 \displaystyle f(a)=\ln a+\frac1a-1 f ( a ) = ln a + a 1 − 1 +2점 · h ( a ) = ln a + 2 a − 2 \displaystyle h(a)=\ln a+\frac2a-2 h ( a ) = ln a + a 2 − 2 [수학2-ⅰ] 문제의 조건으로 부터 f ( a ) = ∫ 1 a ( 1 x − 1 a ) d x = ( ln a − 1 ) − ( ln 1 − 1 a ) = ln a + 1 a − 1 \displaystyle f(a)=\int_{1}^{a}\left(\frac1x-\frac1a\right)dx=(\ln a-1)-\left(\ln1-\frac1a\right)=\ln a+\frac1a-1 f ( a ) = ∫ 1 a ( x 1 − a 1 ) d x = ( ln a − 1 ) − ( ln 1 − a 1 ) = ln a + a 1 − 1 g ( a ) = ( a − 1 ) 1 a = 1 − 1 a \displaystyle g(a)=(a-1)\frac1a=1-\frac1a g ( a ) = ( a − 1 ) a 1 = 1 − a 1 따라서 h ( a ) = ln a + 2 a − 2 \displaystyle h(a)=\ln a+\frac2a-2 h ( a ) = ln a + a 2 − 2
+3점 · h ′ ( a ) = a − 2 a 2 \displaystyle h^{\prime}(a)=\frac{a-2}{a^{2}} h ′ ( a ) = a 2 a − 2 +3점 · 최솟값 h ( 2 ) = ln 2 − 1 \displaystyle h(2)=\ln2-1 h ( 2 ) = ln 2 − 1 [수학2-ⅱ] h ′ ( a ) = 1 a − 2 a 2 = a − 2 a 2 \displaystyle h^{\prime}(a)=\frac1a-\frac2{a^{2}}=\frac{a-2}{a^{2}} h ′ ( a ) = a 1 − a 2 2 = a 2 a − 2 h ′ ′ ( a ) = − 1 a 2 + 4 a 3 \displaystyle h^{\prime\prime}(a)=-\frac1{a^{2}}+\frac4{a^{3}} h ′′ ( a ) = − a 2 1 + a 3 4 이고 h ′ ′ ( 2 ) = 1 4 > 0 \displaystyle h^{\prime\prime}(2)=\frac14>0 h ′′ ( 2 ) = 4 1 > 0 h ( a ) \displaystyle h(a) h ( a ) h ( 2 ) = ln 2 − 1 \displaystyle h(2)=\ln2-1 h ( 2 ) = ln 2 − 1
+4점 · lim a → 1 ln a a − 1 = lim x → 0 ln ( 1 + x ) 1 x \displaystyle \lim_{a\to1}\frac{\ln a}{a-1}=\lim_{x\to0}\ln(1+x)^{\frac1x} a → 1 lim a − 1 ln a = x → 0 lim ln ( 1 + x ) x 1 [수학2-ⅲ] h ( a ) a − 1 = ln a a − 1 − 2 a \displaystyle \frac{h(a)}{a-1}=\frac{\ln a}{a-1}-\frac2a a − 1 h ( a ) = a − 1 ln a − a 2 lim a → 1 h ( a ) a − 1 = lim a → 1 ( ln a a − 1 − 2 a ) = lim a → 1 ln a a − 1 − 2 = lim x → 0 ln ( 1 + x ) x − 2 = lim x → 0 ln ( 1 + x ) 1 x − 2 \displaystyle \begin{aligned}\lim_{a\to1}\frac{h(a)}{a-1}&=\lim_{a\to1}\left(\frac{\ln a}{a-1}-\frac2a\right)=\lim_{a\to1}\frac{\ln a}{a-1}-2\\&=\lim_{x\to0}\frac{\ln(1+x)}{x}-2\\&=\lim_{x\to0}\ln(1+x)^{\frac1x}-2\end{aligned} a → 1 lim a − 1 h ( a ) = a → 1 lim ( a − 1 ln a − a 2 ) = a → 1 lim a − 1 ln a − 2 = x → 0 lim x ln ( 1 + x ) − 2 = x → 0 lim ln ( 1 + x ) x 1 − 2
+2점 · lim a → 1 h ( a ) a − 1 = − 1 \displaystyle \lim_{a\to1}\frac{h(a)}{a-1}=-1 a → 1 lim a − 1 h ( a ) = − 1 따라서 <제시문 2>에 의해서 lim a → 1 h ( a ) a − 1 = ln e − 2 = − 1 \displaystyle \lim_{a\to1}\frac{h(a)}{a-1}=\ln e-2=-1 a → 1 lim a − 1 h ( a ) = ln e − 2 = − 1 임을 알수 있다.
+2점 · k ′ ( a ) = a 3 / 2 − a + 2 a 2 \displaystyle k^{\prime}(a)=\frac{a^{3/2}-a+2}{a^{2}} k ′ ( a ) = a 2 a 3/2 − a + 2 +3점 · a > 1 \displaystyle a>1 a > 1 k ′ ( a ) = a ( a − 1 ) + 2 a 2 > 0 \displaystyle k^{\prime}(a)=\frac{a(\sqrt{a}-1)+2}{a^{2}}>0 k ′ ( a ) = a 2 a ( a − 1 ) + 2 > 0 +1점 · k ( 1 ) > 0 \displaystyle k(1)>0 k ( 1 ) > 0 [수학2-ⅳ] k ( a ) = 2 a − h ( a ) = 2 a − ln a − 2 a + 2 \displaystyle k(a)=2\sqrt{a}-h(a)=2\sqrt{a}-\ln a-\frac2a+2 k ( a ) = 2 a − h ( a ) = 2 a − ln a − a 2 + 2 라 두자. a > 1 \displaystyle a>1 a > 1 k ′ ( a ) = a 3 / 2 − a + 2 a 2 = a ( a − 1 ) + 2 a 2 > 0 \displaystyle k^{\prime}(a)=\frac{a^{3/2}-a+2}{a^{2}}=\frac{a(\sqrt{a}-1)+2}{a^{2}}>0 k ′ ( a ) = a 2 a 3/2 − a + 2 = a 2 a ( a − 1 ) + 2 > 0 그런데 k ( 1 ) = 2 > 0 \displaystyle k(1)=2>0 k ( 1 ) = 2 > 0 이므로 k ( a ) = 2 a − h ( a ) > 0 \displaystyle k(a)=2\sqrt{a}-h(a)>0 k ( a ) = 2 a − h ( a ) > 0 이다. 따라서 h ( a ) < 2 a \displaystyle h(a)<2\sqrt{a} h ( a ) < 2 a
+3점 · ln 2 − 1 a ≤ h ( a ) a < 2 a \displaystyle \frac{\ln2-1}{a}\le\frac{h(a)}{a}<\frac2{\sqrt{a}} a ln 2 − 1 ≤ a h ( a ) < a 2 +3점 · lim a → ∞ ln 2 − 1 a = lim a → ∞ 2 a = 0 \displaystyle \lim_{a\to\infty}\frac{\ln2-1}{a}=\lim_{a\to\infty}\frac2{\sqrt{a}}=0 a → ∞ lim a ln 2 − 1 = a → ∞ lim a 2 = 0 [수학2-ⅴ] [수학1-ⅱ]와 [수학1-ⅳ]의 결과로부터 Correction. 참조한 문항 번호는 [수학2-ⅱ]와 [수학2-ⅳ]다. 그러나 원칙에 따라 원문 표기를 그대로 실었다.ln 2 − 1 ≤ h ( a ) < 2 a \displaystyle \ln2-1\le h(a)<2\sqrt{a} ln 2 − 1 ≤ h ( a ) < 2 a 양변을 a \displaystyle a a ln 2 − 1 a ≤ h ( a ) a < 2 a \displaystyle \frac{\ln2-1}{a}\le\frac{h(a)}a<\frac2{\sqrt{a}} a ln 2 − 1 ≤ a h ( a ) < a 2 그런데 lim a → ∞ ln 2 − 1 a = lim a → ∞ 2 a = 0 \displaystyle \lim_{a\to\infty}\frac{\ln2-1}{a}=\lim_{a\to\infty}\frac2{\sqrt{a}}=0 a → ∞ lim a ln 2 − 1 = a → ∞ lim a 2 = 0 이므로 <제시문 3>에 의해서 lim a → ∞ h ( a ) a = 0 \displaystyle \lim_{a\to\infty}\frac{h(a)}a=0 a → ∞ lim a h ( a ) = 0
수리논술, 배움에서 논증의 완성까지.