삼차함수 극한 조건

[출제의도] 도함수를 이용하여 추론하기 조건 (가)에서 $\displaystyle \lim\limits _{x \rightarrow 1} {\frac{f \left( x \right)}{x - 1} = 3}$이고 $\displaystyle \lim\limits _{x \rightarrow 1} {\left( x - 1 \right)} = 0$이므로 $\displaystyle \lim\limits _{x \rightarrow 1} {f \left( x \right)} = 0$ 함수 $f \left( x \right)$는 $x = 1$에서 연속이므로 $f \left( 1 \right) = 0$ $\cdots \text{㉠}$ 조건 (나)에서 $\displaystyle \lim\limits _{x \rightarrow 2} {\frac{f \left( x \right)}{\left( x - 2 \right) f' \left( x \right)} = \alpha}$ $\left( \alpha \ne 1 \right)$이고 $\displaystyle \lim\limits _{x \rightarrow 2} {\left( x - 2 \right) f' \left( x \right) = 0}$이므로 $\displaystyle \lim\limits _{x \rightarrow 2} {f \left( x \right) = 0}$ 함수 $f \left( x \right)$는 $x = 2$에서 연속이므로 $f \left( 2 \right) = 0$ $\cdots \text{㉡}$ ㉠, ㉡에 의해 $f \left( x \right) = k \left( x - 1 \right) \left( x - 2 \right) \left( x + a \right)$ $( k , a$는 상수, $k \ne 0 )$ $f' \left( x \right) = k \left\{ \left( x - 2 \right) \left( x + a \right) + \left( x - 1 \right) \left( x + a \right) + \left( x - 1 \right) \left( x - 2 \right) \right\}$ $a \ne - 2$라 하면 $\displaystyle \lim\limits _{x \rightarrow 2} {\frac{f \left( x \right)}{\left( x - 2 \right) f' \left( x \right)}}$$\displaystyle = \lim\limits _{x \rightarrow 2} {\frac{k \left( x - 1 \right) \left( x + a \right)}{f' \left( x \right)}}$ $\displaystyle = \frac{2 + a}{2 + a}$ $= 1 \ne a$ 그러므로 $a = - 2$이며 $f \left( x \right) = k \left( x - 1 \right) \left( x - 2 \right) ^{2}$ $\alpha$$\displaystyle = \lim\limits _{x \rightarrow 2} {\frac{f \left( x \right)}{\left( x - 2 \right) f' \left( x \right)}}$ $\displaystyle = \lim\limits _{x \rightarrow 2} {\frac{k \left( x - 1 \right) \left( x - 2 \right) ^{2}}{\left( x - 2 \right) \left\{ k \left( x - 2 \right) ^{2} + 2 k \left( x - 1 \right) \left( x - 2 \right) \right\}}}$ $\displaystyle = \lim\limits _{x \rightarrow 2} {\frac{x - 1}{\left( x - 2 \right) + 2 \left( x - 1 \right)}}$ $\displaystyle = \frac{1}{0 + 2 \times 1} = \frac{1}{2}$ 조건 (가)에서 $\displaystyle \lim\limits _{x \rightarrow 1} {\frac{f \left( x \right)}{x - 1} = \lim\limits _{x \rightarrow 1} {\frac{f \left( x \right) - f \left( 1 \right)}{x - 1} = f' \left( 1 \right) = 3}}$이므로 $k = 3$ 따라서 $\displaystyle \alpha \times f \left( 4 \right) = \frac{1}{2} \times \left( 3 \times 3 \times 2 ^{2} \right) = 18$