극한과 미분 조건의 다항함수

[출제의도] 도함수를 이용하여 추론하기 $f \left( x \right) = x ^{2} + bx + c$ ($b$, $c$는 상수)라 하자. 조건 (가)에서 $\displaystyle \lim\limits _{x \rightarrow 1} {\frac{g \left( x \right) - f \left( x \right)}{x - 1}}$$\displaystyle = \lim\limits _{x \rightarrow 1} {\frac{\left( x ^{2} - x + a \right) f \left( x \right) - f \left( x \right)}{x - 1}}$ $\displaystyle = \lim\limits _{x \rightarrow 1} {\frac{\left( x ^{2} - x + a - 1 \right) f \left( x \right)}{x - 1}} = 0$ $\displaystyle \lim\limits _{x \rightarrow 1} {\left( x - 1 \right)} = 0$이므로 $\displaystyle \lim\limits _{x \rightarrow 1} {\left( x ^{2} - x + a - 1 \right) f \left( x \right)} = \left( a - 1 \right) f \left( 1 \right) = 0$ 따라서 $a = 1$ 또는 $f \left( 1 \right) = 0$ $a \ne 1$이라 하면 $f \left( 1 \right) = 0$이어야 하므로 $c = \mathit{-} b - 1$, $f \left( x \right) = \left( x - 1 \right) \left( x + b + 1 \right)$ $\displaystyle \lim\limits _{x \rightarrow 1} {\frac{g \left( x \right) - f \left( x \right)}{x - 1}}$ $\displaystyle = \lim\limits _{x \rightarrow 1} {\frac{\left( x ^{2} - x + a - 1 \right) \left( x - 1 \right) \left( x + b + 1 \right)}{x - 1}}$ $= \left( a - 1 \right) \left( b + 2 \right) = 0$ 이므로 $b = \mathit{-} 2$, $f \left( x \right) = \left( x - 1 \right) ^{2}$ $g' \left( x \right) = \left( 2 x - 1 \right) \left( x - 1 \right) ^{2} + \left( x ^{2} - x + a \right) \left( 2 x - 2 \right)$에서 $g' \left( 1 \right) = 0$이므로 조건 (나)를 만족시키지 않는다. $f \left( 1 \right) \ne 0$이라 하면 $a = 1$이고 $\displaystyle \lim\limits _{x \rightarrow 1} {\frac{g \left( x \right) - f \left( x \right)}{x - 1}}$$\displaystyle = \lim\limits _{x \rightarrow 1} {\frac{x \left( x - 1 \right) f \left( x \right)}{x - 1}}$$= f \left( 1 \right)$$= 0$ 이므로 모순이다. 따라서 $a = 1$이고 $f \left( 1 \right) = 0$이며 $b \ne \mathit{-} 2$ $f \left( x \right) = \left( x - 1 \right) \left( x + b + 1 \right)$에서 $f' \left( x \right) = 2 x + b$ $g \left( x \right) = \left( x ^{2} - x + 1 \right) f \left( x \right)$에서 $g' \left( x \right) = \left( 2 x - 1 \right) f \left( x \right) + \left( x ^{2} - x + 1 \right) f' \left( x \right)$ 조건 (다)에서 $f \left( \alpha \right) = f' \left( \alpha \right)$이므로 $\left( \alpha - 1 \right) \left( \alpha + b + 1 \right) = 2 \alpha + b$ $\cdots$ ㉠ $g' \left( \alpha \right) = \left( 2 \alpha - 1 \right) f \left( \alpha \right) + \left( \alpha ^{2} - \alpha + 1 \right) f' \left( \alpha \right) = 2 f' \left( \alpha \right)$ $\left( 2 \alpha - 1 \right) f' \left( \alpha \right) + \left( \alpha ^{2} - \alpha + 1 \right) f' \left( \alpha \right) = 2 f' \left( \alpha \right)$ $\left( \alpha ^{2} + \alpha - 2 \right) f' \left( \alpha \right) = 0$ $\left( \alpha + 2 \right) \left( \alpha - 1 \right) \left( 2 \alpha + b \right) = 0$ 따라서 $\alpha = \mathit{-} 2$ 또는 $\alpha = 1$ 또는 $\displaystyle \alpha = \mathit{-} \frac{b}{2}$ $\alpha = 1$ 또는 $\displaystyle \alpha = \mathit{-} \frac{b}{2}$일 때 ㉠에서 $b = \mathit{-} 2$이므로 $b \ne \mathit{-} 2$인 것에 모순이다. $\alpha = \mathit{-} 2$이므로 ㉠에서 $\displaystyle b = \frac{7}{4}$ $\displaystyle g \left( x \right) = \left( x ^{2} - x + 1 \right) \left( x - 1 \right) \left( x + \frac{11}{4} \right)$ $\displaystyle g \left( \alpha + 4 \right) = g \left( 2 \right) = 3 \times 1 \times \frac{19}{4} = \frac{57}{4}$ 따라서 $p = 4$, $q = 57$이므로 $p + q = 61$