합성함수 실근 극소

$f ( x ) = x ^{2} - ax + b$ $\displaystyle g ( x ) = x ^{2} e ^{- \frac{x}{2}}$

$g ( x )$는 극솟값 $g ( 0 ) = 0$, 극댓값 $\displaystyle g ( 4 ) = \frac{16}{e ^{2}}$ $\displaystyle \lim\limits _{x\rightarrow \infty} g ( x ) = 0$, $\displaystyle \lim\limits _{x\rightarrow -\infty} g ( x ) = \infty$ (가) $h ( 0 ) = f ( 0 ) = b$, $\displaystyle h ( 4 ) = f \left( \frac{16}{e ^{2}} \right)$$> b$ 이므로 $\displaystyle 0 < a < \frac{16}{e ^{2}}$ (단, $\displaystyle \frac{5}{2} < e < 3$) (나)를 만족시키려면 $h ( x ) = f ( g ( x ) )$의 그래프는 다음과 같아야 한다.

$f' ( x ) = 2 x - a$, $\displaystyle g' ( x ) = \frac{1}{2} x ( 4 - x ) e ^{- \frac{x}{2}}$ $h' ( x ) = g' ( x ) f' ( g ( x ) )$ $\displaystyle = \frac{1}{2} x ( 4 - x ) e ^{- \frac{x}{2}} \{ 2 g ( x ) - a \}$ $g ( x )$$\displaystyle = \frac{a}{2}$, ($\displaystyle 0 < \frac{a}{2} < \frac{8}{e ^{2}}$)에서 $h ( x )$는 극솟값인데, 그 값이 $- k = - f ( 0 ) = - b$$\displaystyle = f \left( \frac{a}{2} \right)$ $\displaystyle f ( x ) = \left( x - \frac{a}{2} \right) ^{2} + b - \frac{a ^{2}}{4}$에서 $\displaystyle \frac{a ^{2}}{4} = 2 b$, 즉 $\displaystyle b = \frac{a ^{2}}{8}$ 한편 $\displaystyle f ( 1 ) = 1 - a + b = 1 - a + \frac{a ^{2}}{8} = - \frac{7}{32}$이므로 $4 a ^{2} - 32 a + 39 = 0$, $( 2 a - 13 ) ( 2 a - 3 ) = 0$ $\displaystyle 0 < a < \frac{16}{e ^{2}}$이므로 $\displaystyle a = \frac{3}{2}$이고 $\displaystyle b = \frac{9}{32}$이다. $\therefore$ $\displaystyle a + 16 b = \frac{3}{2} + \frac{9}{2} = 6$