다항함수 결정

[출제의도] 함수의 극한을 이용하여 문제 해결하기 $f \left( x \right)$가 다항함수이고 $\displaystyle \lim\limits _{x \rightarrow \infty} {\frac{f \left( x \right)}{2 x ^{2}} = 1}$이므로 $f \left( x \right) = 2 x ^{2} + ax + b$ ($a , b$는 실수) $\displaystyle \lim\limits _{x \rightarrow 1} {\frac{f \left( x \right) - 3}{\left( x - 1 \right) \left( x - 2 \right)} = 4}$이고 $\displaystyle \lim\limits _{x \rightarrow 1} {\left( x - 1 \right) \left( x - 2 \right) = 0}$이므로 $\displaystyle \lim\limits _{x \rightarrow 1} {\left\{ f \left( x \right) - 3 \right\}} = 0$ $\displaystyle \lim\limits _{x \rightarrow 1} {f \left( x \right)}$$\displaystyle = \lim\limits _{x \rightarrow 1} {\left[ \left\{ f \left( x \right) - 3 \right\} + 3 \right]}$ $\displaystyle = \lim\limits _{x \rightarrow 1} {\left\{ f \left( x \right) - 3 \right\}} + \lim\limits _{x \rightarrow 1} {3}$ $= 0 + 3 = 3$ $\displaystyle \lim\limits _{x \rightarrow 1} {f \left( x \right)} = 2 + a + b = 3$, $b = - a + 1$ $f \left( x \right) - 3$$= 2 x ^{2} + ax - a - 2$ $= \left( x - 1 \right) \left( 2 x + a + 2 \right)$ $\displaystyle \lim\limits _{x \rightarrow 1} {\frac{f \left( x \right) - 3}{\left( x - 1 \right) \left( x - 2 \right)}}$$\displaystyle = \lim\limits _{x \rightarrow 1} {\frac{\left( x - 1 \right) \left( 2 x + a + 2 \right)}{\left( x - 1 \right) \left( x - 2 \right)}}$ $\displaystyle = \frac{2 + a + 2}{- 1} = 4$ $a = - 8$, $b = 9$이므로 $f \left( x \right) = 2 x ^{2} - 8 x + 9$ 따라서 $f \left( 4 \right) = 9$