극한으로 함수 결정

[출제의도] 함수의 극한을 활용하여 문제해결하기 두 함수 $f \left( x \right) g \left( x \right) , f \left( x \right) + g \left( x \right)$는 $x = \mathit{-} 3$에서 연속이다. $\displaystyle \lim\limits _{x \rightarrow - 3} {} \frac{f ( x ) g ( x )}{\left( x + 3 \right) ^{2}} = 4$에서 $\displaystyle \lim\limits _{x \rightarrow - 3} {} \left( x + 3 \right) ^{2} = 0$이므로 $\displaystyle \lim\limits _{x \rightarrow - 3} {} f ( x ) g ( x ) = f ( - 3 ) g ( - 3 ) = 0$ $\cdots \cdots$ ㉠ $\displaystyle \lim\limits _{x \rightarrow - 3} {} \frac{f ( x ) + g ( x )}{x + 3} = \mathit{-} 4$에서 $\displaystyle \lim\limits _{x \rightarrow - 3} {} ( x + 3 ) = 0$이므로 $\displaystyle \lim\limits _{x \rightarrow - 3} {} \left\{ f ( x ) + g ( x ) \right\} = f ( - 3 ) + g ( - 3 ) = 0$ $\cdots \cdots$ ㉡ ㉠, ㉡에 의하여 $f \left( - 3 \right) = g \left( - 3 \right) = 0$이므로 두 함수 $f \left( x \right) , g ( x )$는 각각 $x + 3$을 인수로 갖는다. $f ( x ) = a ( x + 3 ) , g ( x ) = ( x + 3 ) ( x + b )$ ($a , b$는 상수)라 하면 $\displaystyle \lim\limits _{x \rightarrow - 3} {} \frac{f ( x ) g ( x )}{( x + 3 ) ^{2}}$$\displaystyle = \lim\limits _{x \rightarrow - 3} {} \frac{a ( x + 3 ) ^{2} ( x + b )}{( x + 3 ) ^{2}}$$\displaystyle = \lim\limits _{x \rightarrow - 3} {} a ( x + b )$ $= a ( - 3 + b ) = 4$ $\cdots \cdots$ ㉢ $\displaystyle \lim\limits _{x \rightarrow - 3} {} \frac{f ( x ) + g ( x )}{x + 3}$$\displaystyle = \lim\limits _{x \rightarrow - 3} {} \frac{a ( x + 3 ) + ( x + 3 ) ( x + b )}{x + 3}$ $\displaystyle = \lim\limits _{x \rightarrow - 3} {} ( a + x + b )$ $= a - 3 + b = \mathit{-} 4$ $\cdots \cdots$ ㉣ ㉢, ㉣을 연립하면 $a ( - a - 4 ) = 4 , a ^{2} + 4 a + 4 = 0$ $( a + 2 ) ^{2} = 0$에서 $a = \mathit{-} 2$이고 $b = 1$ $f ( x ) = \mathit{-} 2 ( x + 3 ) , g ( x ) = ( x + 3 ) ( x + 1 )$ 따라서 $g ( 2 ) - f ( 2 ) = 5 \times 3 - ( - 2 ) \times 5 = 25$