분수꼴 함수의 연속

[출제의도] 함수의 연속을 이해하여 상수의 값을 구한다. 함수 $f ( x )$는 $f ( 1 ) = 0$, $f ( a ) = 0$이고, $\displaystyle \lim\limits _{x \rightarrow 2 -} {} f ( x )$$= \mathit{-} 1$, $\displaystyle \lim\limits _{x \rightarrow 2 +} {} f ( x )$$= \mathit{-} 4 + 2 a$에서 $\displaystyle \lim\limits _{x \rightarrow 2 -} {} f ( x )$$\displaystyle \ne \lim\limits _{x \rightarrow 2 +} {} f ( x )$이므로 $x = 2$에서 불연속이다. 함수 $h ( x )$가 실수 전체의 집합에서 연속이므로 함수 $h ( x )$는 $x = 1$, $x = a$, $x = 2$에서 연속이어야 한다. $\displaystyle \lim\limits _{x \rightarrow 1} {} \frac{g ( x )}{f ( x )} = h ( 1 )$, $\displaystyle \lim\limits _{x \rightarrow a} {} \frac{g ( x )}{f ( x )} = h ( a )$에서 $\displaystyle \lim\limits _{x \rightarrow 1} {} f ( x )$$= f ( 1 )$$= 0$, $\displaystyle \lim\limits _{x \rightarrow a} {} f ( x )$$= f ( a )$$= 0$이므로 $\displaystyle \lim\limits _{x \rightarrow 1} {} g ( x ) = 0$, $\displaystyle \lim\limits _{x \rightarrow a} {} g ( x ) = 0$ 즉, $g ( 1 ) = 0$, $g ( a ) = 0$ 또, $\displaystyle \lim\limits _{x \rightarrow 2 -} {\frac{g ( x )}{f ( x )}}$$\displaystyle = \lim\limits _{x \rightarrow 2 +} {\frac{g ( x )}{f ( x )}}$, $\displaystyle \frac{g ( 2 )}{- 1} = \frac{g ( 2 )}{- 4 + 2 a}$이므로 $g ( 2 ) = 0$이고 $g ( x )$$= \left( x - 1 \right) \left( x - 2 \right) \left( x - a \right)$이다. $\displaystyle \lim\limits _{x \rightarrow 1} {} h ( x )$$\displaystyle = \lim\limits _{x \rightarrow 1} {} \frac{\left( x - 1 \right) \left( x - 2 \right) \left( x - a \right)}{\left( x - 1 \right) \left( x - 3 \right)}$ $\displaystyle = \lim\limits _{x \rightarrow 1} {} \frac{\left( x - 2 \right) \left( x - a \right)}{x - 3}$ $\displaystyle = \frac{1 - a}{2}$ $\displaystyle \lim\limits _{x \rightarrow a} {} h ( x )$$\displaystyle = \lim\limits _{x \rightarrow a} {} \frac{\left( x - 1 \right) \left( x - 2 \right) \left( x - a \right)}{- x \left( x - a \right)}$ $\displaystyle = \lim\limits _{x \rightarrow a} {} \frac{\left( x - 1 \right) \left( x - 2 \right)}{- x}$ $\displaystyle = \mathit{-} \frac{\left( a - 1 \right) \left( a - 2 \right)}{a}$ $h ( 1 )$$= h ( a )$이므로 $\displaystyle \frac{1 - a}{2}$$\displaystyle = \mathit{-} \frac{\left( a - 1 \right) \left( a - 2 \right)}{a}$ $a > 2$이므로 $a = 4$ 따라서 $h ( x )$$\displaystyle = \frac{g ( x )}{f ( x )}$$\displaystyle = {\begin{cases} \frac{\left( x - 2 \right) \left( x - 4 \right)}{x - 3} & & \left( x {\leq} 2 \right) \\ - \frac{\left( x - 1 \right) \left( x - 2 \right)}{x} & & \left( x > 2 \right) \end{cases}}$ 이므로 $h ( 1 ) + h ( 3 )$$\displaystyle = \mathit{-} \frac{3}{2} + \left( - \frac{2}{3} \right)$$\displaystyle = \mathit{-} \frac{13}{6}$