자연계열 2번

[1]$\displaystyle \begin{aligned}&\lim_{n\to\infty}n\sum_{k=0}^{n}\ln\left(1-\frac{1}{(n+k)^2}\right)\\ &=\lim_{n\to\infty}n\left\{\ln\left(1-\frac{1}{n^2}\right)+\ln\left(1-\frac{1}{(n+1)^2}\right)+\ln\left(1-\frac{1}{(n+2)^2}\right)+\cdots\ln\left(1-\frac{1}{(2n)^2}\right)\right\}\\ &=\lim_{n\to\infty}n\ln\left\{\frac{(n-1)(n+1)}{n^2}\times\frac{n(n+2)}{(n+1)^2}\times\frac{(n+1)(n+3)}{(n+2)^2}\times\cdots\times\frac{(2n-1)(2n+1)}{(2n)^2}\right\}\\ &=\lim_{n\to\infty}\ln\left(\frac{n-1}{n}\times\frac{2n+1}{2n}\right)^n\\ &=\lim_{n\to\infty}\ln\left[\left\{1+\left(-\frac{1}{n}\right)\right\}^n\times\left(1+\frac{1}{2n}\right)^{2n\times\frac{1}{2}}\right]\end{aligned}$

$\displaystyle -\frac{1}{n}=t$라고 하면 $\displaystyle \lim_{n\to\infty}\left(1-\frac{1}{n}\right)^n=\lim_{t\to0}(1+t)^{\frac{1}{t}\times(-1)}=\frac{1}{e}$이므로

$\displaystyle \lim_{n\to\infty}n\sum_{k=0}^{n}\ln\left(1-\frac{1}{(n+k)^2}\right)=\ln\left(e^{-1}\times e^{\frac{1}{2}}\right)=-\frac{1}{2}$

[2](1) 조건 (나)로부터$\displaystyle \frac{1}{2}\ln\left(2xe^{g(x)}+1\right)=g(x)$이로부터 $\displaystyle e^{g(x)}$에 관한 이차방정식을 얻고 이를 풀면$\displaystyle e^{2g(x)}-2xe^{g(x)}-1=0$

$\displaystyle e^{g(x)}=x+\sqrt{x^2+1}$

따라서$\displaystyle g(x)=\ln\left(x+\sqrt{x^2+1}\right)$

(2) $\displaystyle \lim_{x\to\ln\sqrt{2}}\frac{f(2x)-a}{x-\ln\sqrt{2}}=b$에서 극한값이 존재하고 $\displaystyle \lim_{x\to\ln\sqrt{2}}(x-\ln\sqrt{2})=0$이므로$\displaystyle \lim_{x\to\ln\sqrt{2}}\{f(2x)-a\}=0$, 즉 $\displaystyle f(2\ln\sqrt{2})=f(\ln2)=a$$\displaystyle f(\ln2)=a\Leftrightarrow g(a)=\ln2\Leftrightarrow\ln\left(a+\sqrt{a^2+1}\right)=\ln2\Leftrightarrow\sqrt{a^2+1}=2-a$양변을 제곱하여 풀면$\displaystyle a=\frac{3}{4}$

함수 $\displaystyle g$를 미분하면$\displaystyle g^{\prime}(x)=\frac{1+\frac{x}{\sqrt{x^2+1}}}{x+\sqrt{x^2+1}}=\frac{1}{\sqrt{x^2+1}}$제시문 2로부터$\displaystyle b=2\lim_{x\to\ln\sqrt{2}}\frac{f(2x)-\frac{3}{4}}{2x-2\ln\sqrt{2}}=2f^{\prime}(\ln2)=\frac{2}{g^{\prime}(f(\ln2))}=\frac{2}{g^{\prime}\left(\frac{3}{4}\right)}=\frac{2}{\frac{1}{\sqrt{\left(\frac{3}{4}\right)^2+1}}}=\frac{5}{2}$따라서$\displaystyle a+b=\frac{3}{4}+\frac{5}{2}=\frac{13}{4}$

[3]$\displaystyle e^{1-\sqrt{x}}\left(\frac{1}{2\sqrt{x}}f(x)-f^{\prime}(x)\right)=\frac{e^{1-\sqrt{x}}}{2\sqrt{x}}f(x)-e^{1-\sqrt{x}}f^{\prime}(x)=-\frac{d}{dx}\left(e^{1-\sqrt{x}}f(x)\right)=-\frac{1}{2\sqrt{x}}$위 식을 적분하면$\displaystyle e^{1-\sqrt{x}}f(x)=\sqrt{x}+C$

조건(가)로부터 $\displaystyle f(1)=1$이므로 $\displaystyle C=0$이고 다음을 얻는다.$\displaystyle f(x)=e^{\sqrt{x}-1}\sqrt{x}$

$\displaystyle \sqrt{x}-1=t$라고 치환하고 부분적분법을 적용하면$\displaystyle \begin{aligned}\int_1^4f(x)dx&=\int_1^4e^{\sqrt{x}-1}\sqrt{x}\,dx\\ &=2\int_0^1e^t(t+1)^2dt\\ &=2\Biggl[e^t(t+1)^2\Biggr]_0^1-4\int_0^1e^t(t+1)dt\\ &=8e-2-4\Biggl[e^t(t+1)\Biggr]_0^1+4\Biggl[e^t\Biggr]_0^1\\ &=4e-2\end{aligned}$

[4]$\displaystyle \overline{\mathrm{OD}}$는 반원의 반지름이므로 $\displaystyle \overline{\mathrm{OD}}=\frac{a+b}{2}$

직각삼각형 $\displaystyle \mathrm{COD}$에서 $\displaystyle \overline{\mathrm{CO}}=\frac{b-a}{2}$이므로 피타고라스 정리를 이용하면$\displaystyle \overline{\mathrm{CD}}=\sqrt{\overline{\mathrm{OD}}^2-\overline{\mathrm{CO}}^2}=\sqrt{\frac{(a+b)^2}{4}-\frac{(b-a)^2}{4}}=\sqrt{ab}$

직각삼각형 $\displaystyle \mathrm{COD}$와 직각삼각형 $\displaystyle \mathrm{EDC}$에서 $\displaystyle \angle\mathrm{ECD}=\angle\mathrm{CDO}$이고 $\displaystyle \angle\mathrm{COD}=\angle\mathrm{EDC}\,(=\angle\mathrm{ACE})$이므로 직각삼각형 $\displaystyle \mathrm{COD}$와 직각삼각형 $\displaystyle \mathrm{EDC}$는 닮음이다.이로부터 다음 비례관계가 성립한다.$\displaystyle \overline{\mathrm{EC}}:\overline{\mathrm{CD}}=\overline{\mathrm{CD}}:\overline{\mathrm{DO}}$$\displaystyle \overline{\mathrm{EC}}=\frac{\overline{\mathrm{CD}}^2}{\overline{\mathrm{DO}}}=\frac{(\sqrt{ab})^2}{\frac{a+b}{2}}=\frac{2ab}{a+b}$

$\displaystyle \overline{\mathrm{OD}}>\overline{\mathrm{CD}}>\overline{\mathrm{EC}}$로부터 $\displaystyle \frac{a+b}{2}>\sqrt{ab}>\frac{2ab}{a+b}$