곱함수의 미분가능성

[출제의도] 함수의 미분가능성을 활용하여 문제해결하기 함수 $g \left( x \right)$는 실수 전체의 집합에서 미분가능하다. (ⅰ) $a \leq 10$일 때 $\displaystyle f \left( x \right) = {\begin{cases} x + 5 & \left( x < 5 \right) \\ 2 x - a & \left( x \geq 5 \right) \end{cases}}$이므로 함수 $f \left( x \right) g \left( x \right)$가 실수 전체의 집합에서 미분가능하기 위해서는 $x = 5$에서 미분가능하여야 한다. $\displaystyle \lim\limits _{x \rightarrow 5 -} {} \frac{f \left( x \right) g \left( x \right) - f \left( 5 \right) g \left( 5 \right)}{x - 5}$ $\displaystyle = \lim\limits _{x \rightarrow 5 -} {} \frac{\left( x + 5 \right) \left( x - 5 \right) \left( x - b \right)}{x - 5}$ $\displaystyle = \lim\limits _{x \rightarrow 5 -} {} \left( x + 5 \right) \left( x - b \right) = 10 \left( 5 - b \right)$ $\displaystyle \lim\limits _{x \rightarrow 5 +} {} \frac{f \left( x \right) g \left( x \right) - f \left( 5 \right) g \left( 5 \right)}{x - 5}$ $\displaystyle = \lim\limits _{x \rightarrow 5 +} {} \frac{\left( 2 x - a \right) \left( x - 5 \right) \left( x - b \right)}{x - 5}$ $\displaystyle = \lim\limits _{x \rightarrow 5 +} {} \left( 2 x - a \right) \left( x - b \right) = \left( 10 - a \right) \left( 5 - b \right)$ 에서 $10 \left( 5 - b \right) = \left( 10 - a \right) \left( 5 - b \right)$, $a \left( 5 - b \right) = 0$ $a$는 자연수이므로 $b = 5$ 그러므로 순서쌍 $\left( a , b \right)$는 $\left( 1 , 5 \right)$, $\left( 2 , 5 \right)$, $\cdots$, $\left( 10 , 5 \right)$ (ⅱ) $a \geq 11$일 때 $\displaystyle f \left( x \right) = {\begin{cases} x + 5 & \left( x < 5 \right) \\ - 2 x + a & \left( 5 \leq x < \frac{a}{2} \right) \\ 2 x - a & \left( x \geq \frac{a}{2} \right) \end{cases}}$ 이므로 함수 $f \left( x \right) g \left( x \right)$가 실수 전체의 집합에서 미분가능하기 위해서는 $x = 5$와 $\displaystyle x = \frac{a}{2}$에서 미분가능하여야 한다. $\displaystyle \lim\limits _{x \rightarrow 5 -} {} \frac{f \left( x \right) g \left( x \right) - f \left( 5 \right) g \left( 5 \right)}{x - 5} = 10 \left( 5 - b \right)$ $\displaystyle \lim\limits _{x \rightarrow 5 +} {} \frac{f \left( x \right) g \left( x \right) - f \left( 5 \right) g \left( 5 \right)}{x - 5}$ $\displaystyle = \lim\limits _{x \rightarrow 5 +} {} \frac{\left( - 2 x + a \right) \left( x - 5 \right) \left( x - b \right)}{x - 5}$ $\displaystyle = \lim\limits _{x \rightarrow 5 +} {} \left( - 2 x + a \right) \left( x - b \right)$$= \left( - 10 + a \right) \left( 5 - b \right)$ 에서 $10 \left( 5 - b \right) = \left( - 10 + a \right) \left( 5 - b \right)$, $\left( a - 20 \right) \left( 5 - b \right) = 0$ $a = 20$ 또는 $b = 5$ $\cdots \cdots$ ㉠ 또한 $\displaystyle \lim\limits _{x \rightarrow \frac{a}{2} -} {} \frac{f \left( x \right) g \left( x \right) - f \left( \frac{a}{2} \right) g \left( \frac{a}{2} \right)}{x - \frac{a}{2}}$ $\displaystyle = \lim\limits _{x \rightarrow \frac{a}{2} -} {} \frac{\left( - 2 x + a \right) \left( x - 5 \right) \left( x - b \right)}{x - \frac{a}{2}}$ $\displaystyle = \lim\limits _{x \rightarrow \frac{a}{2} -} {} \left\{ - 2 \left( x - 5 \right) \left( x - b \right) \right\} = \left( - a + 10 \right) \left( \frac{a}{2} - b \right)$ $\displaystyle \lim\limits _{x \rightarrow \frac{a}{2} +} {} \frac{f \left( x \right) g \left( x \right) - f \left( \frac{a}{2} \right) g \left( \frac{a}{2} \right)}{x - \frac{a}{2}}$ $\displaystyle = \lim\limits _{x \rightarrow \frac{a}{2} +} {} \frac{\left( 2 x - a \right) \left( x - 5 \right) \left( x - b \right)}{x - \frac{a}{2}}$ $\displaystyle = \lim\limits _{x \rightarrow \frac{a}{2} +} {} 2 \left( x - 5 \right) \left( x - b \right) = \left( a - 10 \right) \left( \frac{a}{2} - b \right)$ 에서 $\displaystyle \left( - a + 10 \right) \left( \frac{a}{2} - b \right) = \left( a - 10 \right) \left( \frac{a}{2} - b \right)$ $\left( a - 10 \right) \left( a - 2 b \right) = 0$ $a \geq 11$이므로 $a = 2 b$ $\cdots \cdots$ ㉡ 그러므로 ㉠, ㉡에 의하여 순서쌍 $\left( a , b \right)$는 $\left( 20 , 10 \right)$ (ⅰ), (ⅱ)에 의하여 모든 순서쌍 $\left( a , b \right)$의 개수는 $11$