자연계열 (오전) 2번

1.$\displaystyle \begin{aligned}f^{\prime}(x)&=(x-3)\cdots(x-2017)+(x-1)(x-5)\cdots(x-2017)\\&\quad+(x-1)(x-3)(x-7)\cdots(x-2017)+\cdots+(x-1)(x-3)\cdots(x-2015)\end{aligned}$이고,$\displaystyle \begin{aligned}f^{\prime}(2)&=(2-3)(2-5)\cdots(2-2017)+(2-1)(2-5)\cdots(2-2017)\\&\quad+(2-1)(2-3)(2-7)\cdots(2-2017)\\&\quad+\cdots+(2-1)(2-3)\cdots(2-2015)\end{aligned}$이다. 한편 우변의 첫 두항의 합은$\displaystyle \begin{aligned}&(2-3)(2-5)\cdots(2-2017)+(2-1)(2-5)\cdots(2-2017)\\&=(-1)(2-5)\cdots(2-2017)+(1)(2-5)\cdots(2-2017)=0\end{aligned}$이므로$\displaystyle f^{\prime}(2)=(2-1)(2-3)(2-7)\cdots(2-2017)+\cdots+(2-1)(2-3)\cdots(2-2015)$한편 우변의 각항은 음수 1007개의 곱이므로 역시 음수이고, $\displaystyle f^{\prime}(2)<0$이다.

2. 우선 임의의 $\displaystyle a=1,3,\ldots,2017$에 대하여 $\displaystyle f(a)=0$이고 $\displaystyle f^{\prime}(a)\ne0$이므로$\displaystyle f^{\prime\prime}(a)f(a)<(f^{\prime}(a))^2.$만약 $\displaystyle x$가 $\displaystyle 1,3,\ldots,2017$이 아닌 실수라면 $\displaystyle f(x)\ne0$이므로$\displaystyle \frac{f(x)f^{\prime\prime}(x)-f^{\prime}(x)^2}{f(x)^2}<0\quad------(*)$임을 보이면 된다. 한편$\displaystyle \frac{f^{\prime}(x)}{f(x)}=\frac1{x-1}+\cdots+\frac1{x-2017}$이고$\displaystyle \left(\frac{f^{\prime}(x)}{f(x)}\right)^{\prime}=\frac{f(x)f^{\prime\prime}(x)-f^{\prime}(x)^2}{f(x)^2}=-\left(\frac1{(x-1)^2}+\cdots+\frac1{(x-2017)^2}\right)<0.$

3. 우선$\displaystyle -1=g(\alpha)^3-1=p_1(\alpha)\cdots p_n(\alpha)\quad-----(*)$이고 $\displaystyle p_i(\alpha)\ (i=1,\ldots,n)$는 정수이므로 $\displaystyle p_i(\alpha)=\pm1$이고$\displaystyle h(\alpha)=p_1(\alpha)^2+\cdots+p_n(\alpha)^2-n=0.$한편$\displaystyle 3g(x)^2g^{\prime}(x)=p_1^{\prime}(x)p_2(x)\cdots p_n(x)+\cdots+p_1(x)\cdots p_{n-1}(x)p_n^{\prime}(x)$이므로$\displaystyle \begin{aligned}0&=p_1^{\prime}(\alpha)p_2(\alpha)\cdots p_n(\alpha)+\cdots+p_1(\alpha)\cdots p_{n-1}(\alpha)p_n^{\prime}(\alpha)\\&=-\left(\frac{p_1^{\prime}(\alpha)}{p_1(\alpha)}+\cdots+\frac{p_n^{\prime}(\alpha)}{p_n(\alpha)}\right)\quad-------------\text{ (* 적용)}\\&=-\left(p_1^{\prime}(\alpha)p_1(\alpha)+\cdots+p_n^{\prime}(\alpha)p_n(\alpha)\right)\quad-----------\left(p_i(\alpha)=\pm1=\frac1{p_i(\alpha)}\right)\end{aligned}$이고 $\displaystyle h^{\prime}(x)=2(p_1(x)p_1^{\prime}(x)+\cdots+p_n(x)p_n^{\prime}(x))$이므로 $\displaystyle h^{\prime}(\alpha)=0$.