자연계열 2번

2-1.$\displaystyle \begin{aligned}f^{\prime}(x)=\frac{2x+1}{x^2+x+1}\Rightarrow\frac1{f^{\prime}(x)}&=\frac{x^2+x+1}{2x+1}=\frac12\frac{\left(x+\frac12\right)^2+\frac34}{x+\frac12}\\&=\frac12\left(x+\frac12\right)+\frac34\frac1{2x+1}\end{aligned}$즉, $\displaystyle \begin{aligned}\frac1{f^{\prime}(-n)}+\frac1{f^{\prime}(n)}-\frac12&=\frac12\left(-n+\frac12\right)+\frac34\frac1{-2n+1}+\frac12\left(n+\frac12\right)+\frac34\frac1{2n+1}-\frac12\\&=\frac34\left(\frac1{2n+1}-\frac1{2n-1}\right)\end{aligned}$따라서$\displaystyle \begin{aligned}\sum_{n=1}^{\infty}\left(\frac1{f^{\prime}(-n)}+\frac1{f^{\prime}(n)}-\frac12\right)&=\frac34\sum_{n=1}^{\infty}\left(\frac1{2n+1}-\frac1{2n-1}\right)\\&=\frac34\left\{\left(\frac13-1\right)+\left(\frac15-\frac13\right)+\left(\frac17-\frac15\right)+\cdots\right\}\\&=-\frac34\end{aligned}$

2-2.우선 $\displaystyle g(x)=(x-1)^2+a(x-1)+b$라 놓는다.$\displaystyle f^{\prime}(x)=\frac{2x+1}{x^2+x+1}$, $\displaystyle g^{\prime}(x)=2(x-1)+a$에서$\displaystyle f(1)=\ln3$, $\displaystyle f^{\prime}(1)=1$, $\displaystyle g(1)=b$, $\displaystyle g^{\prime}(1)=a$를 얻는다.$\displaystyle h(x)=(f(x)+g(x))^2$의 최솟값이 $\displaystyle 0$이 아니므로 $\displaystyle f(x)+g(x)$는 항상 양의 값을 가지던지 음의 값을 갖는다.그런데 $\displaystyle x\to\infty\Rightarrow f(x)+g(x)\to\infty$이므로 $\displaystyle f(x)+g(x)$는 항상 양의 값을 갖는다.$\displaystyle \therefore h(1)=1\Leftrightarrow(f(1)+g(1))^2=1$이므로 $\displaystyle f(1)+g(1)=1$이다.$\displaystyle \therefore b=g(1)=1-\ln3$

$\displaystyle h(x)$가 $\displaystyle x=1$에서 최솟값을 가지므로$\displaystyle h^{\prime}(1)=2(f(1)+g(1))(f^{\prime}(1)+g^{\prime}(1))=0$이므로 $\displaystyle f^{\prime}(1)+g^{\prime}(1)=0$$\displaystyle \therefore a=g^{\prime}(1)=-1$결국 $\displaystyle g(x)=(x-1)^2-(x-1)+(1-\ln3)=x^2-3x+3-\ln3$