역함수 합성 미분가능

[출제의도] 역함수의 미분법을 활용하여 문제해결하기 함수 $h \left( x \right)$가 실수 전체의 집합에서 미분가능하므로 연속함수이다. 함수 $h \left( x \right)$는 $x$$= 0$에서 연속이므로 $h \left( 0 \right)$$\displaystyle = \lim\limits _{x \rightarrow 0 -} {} h \left( x \right)$$\displaystyle = \lim\limits _{x \rightarrow 0 +} {} h \left( x \right)$에서 $h \left( 0 \right)$$= 0$이고 $f \left( g ^{- 1} \left( 0 \right) \right)$$= 0$ $g ^{- 1} \left( 0 \right)$$= \alpha$라 하면 $f \left( \alpha \right)$$= 0$, $g \left( \alpha \right)$$= 0$ $f \left( \alpha \right)$$= 0$에서 $\alpha$$= \mathit{-} 1$ 또는 $\alpha$$= 0$ 또는 $\alpha$$= 1$ $\cdots \cdots$ ㉠ 함수 $h \left( x \right)$는 $x$$= 1$에서 연속이므로 $h \left( 1 \right)$$\displaystyle = \lim\limits _{x \rightarrow 1 -} {} h \left( x \right)$$\displaystyle = \lim\limits _{x \rightarrow 1 +} {} h \left( x \right)$에서 $h \left( 1 \right)$$= 0$이고 $f \left( g ^{- 1} \left( 1 \right) \right)$$= 0$ $g \left( 0 \right)$$= 1$이므로 $g ^{- 1} \left( 1 \right)$$= 0$이고 $f \left( 0 \right)$$= 0$이므로 $f \left( g ^{- 1} \left( 1 \right) \right)$$= 0$은 성립한다. 함수 $h \left( x \right)$는 $x$$= 0$에서 미분가능하므로 $\displaystyle \lim\limits _{x \rightarrow 0 -} {} \frac{h \left( x \right) - h \left( 0 \right)}{x - 0}$$=$$\displaystyle \lim\limits _{x \rightarrow 0 +} {} \frac{h \left( x \right) - h \left( 0 \right)}{x - 0}$에서 $\displaystyle \lim\limits _{x \rightarrow 0 -} {} \frac{f \left( g ^{- 1} \left( x \right) \right)}{x}$$=$$\displaystyle \lim\limits _{x \rightarrow 0 +} {} \frac{\frac{1}{\pi} \sin \pi x}{x}$$= 1$ $f' \left( g ^{- 1} \left( 0 \right) \right) \left( g ^{- 1} \right)' \left( 0 \right)$$= 1$ $g ^{- 1} \left( 0 \right)$$= \alpha$이고 $\left( g ^{- 1} \right)' \left( 0 \right)$$\displaystyle = \frac{\mathit{1}}{g' \left( \alpha \right)}$이므로 $\displaystyle f' \left( \alpha \right) \times \frac{1}{g' \left( \alpha \right)}$$= 1$ $f' \left( \alpha \right)$$= g' \left( \alpha \right)$ $3 \alpha ^{2} - 1$$= 3 a \alpha ^{2} + 2 \alpha + b$ $\cdots \cdots$ ㉡ 함수 $h \left( x \right)$는 $x = 1$에서 미분가능하므로 $\displaystyle \lim\limits _{x \rightarrow 1 -} {} \frac{h \left( x \right) - h \left( 1 \right)}{x - 1}$$=$$\displaystyle \lim\limits _{x \rightarrow 1 +} {} \frac{h \left( x \right) - h \left( 1 \right)}{x - 1}$에서 $\displaystyle \lim\limits _{x \rightarrow 1 -} {} \frac{\frac{1}{\pi} \sin \pi x}{x - 1}$$=$$\displaystyle \lim\limits _{x \rightarrow 1 +} {} \frac{f \left( g ^{- 1} \left( x \right) \right)}{x - 1}$ $\displaystyle \lim\limits _{x \rightarrow 1 -} {} \frac{\frac{1}{\pi} \sin \pi x}{x - 1}$에서 $x - 1$$= t$라 하면 $\displaystyle \lim\limits _{x \rightarrow 1 -} {} \frac{\frac{1}{\pi} \sin \pi x}{x - 1}$$\displaystyle = \lim\limits _{t \rightarrow 0 -} {} \frac{- \sin \pi t}{\pi t}$$= \mathit{-} 1$ $\displaystyle \lim\limits _{x \rightarrow 1 +} {} \frac{f \left( g ^{- 1} \left( x \right) \right)}{x - 1}$$= \mathit{-} 1$에서 $f' \left( g ^{- 1} \left( 1 \right) \right) \left( g ^{- 1} \right)' \left( 1 \right)$$= \mathit{-} 1$ $g ^{- 1} \left( 1 \right)$$= 0$이고 $\left( g ^{- 1} \right)' \left( 1 \right)$$\displaystyle = \frac{1}{g' \left( 0 \right)}$이므로 $\displaystyle f' \left( 0 \right) \times \frac{1}{g' \left( 0 \right)}$$= \mathit{-} 1$ $f' \left( 0 \right)$$= \mathit{-} 1$이므로 $g' \left( 0 \right)$$= b$$= 1$ 삼차함수 $g \left( x \right)$는 역함수 $g ^{- 1} \left( x \right)$를 가지고 $g' \left( 0 \right)$$= 1$$> 0$이므로 증가함수이다. $g \left( \alpha \right)$$= 0$, $g \left( 0 \right)$$= 1$이므로 $\alpha < 0$ ㉠에 의하여 $\alpha$$= \mathit{-} 1$ ㉡에 의하여 $a$$= 1$ $g \left( x \right)$$= x ^{3} + x ^{2} + x + 1$ 따라서 $g \left( a + b \right)$$= g \left( 2 \right)$$= 15$