f(x)+f(1-x)=1 함수

ㄱ. $\displaystyle f \left( \frac{1}{2} \right) = \frac{4 ^{\frac{1}{2}}}{4 ^{\frac{1}{2}} + 2} = \frac{\sqrt{4}}{\sqrt{4} + 2} = \frac{1}{2}$ (참) ㄴ. $f ( 1 - x )$$\displaystyle = \frac{4 ^{1-x}}{4 ^{1-x} + 2} = \frac{\frac{4}{4 ^{x}}}{\frac{4}{4 ^{x}} + 2}$ $\displaystyle = \frac{\frac{4}{4 ^{x}}}{\frac{4 + 2 \cdot 4 ^{x}}{4 ^{x}}}$$\displaystyle = \frac{4}{2 \cdot 4 ^{x} + 4}$$\displaystyle = \frac{2}{4 ^{x} + 2}$이므로 $\displaystyle f ( x ) + f ( 1 - x ) = \frac{4 ^{x}}{4 ^{x} + 2} + \frac{2}{4 ^{x} + 2} = \frac{4 ^{x} + 2}{4 ^{x} + 2} = 1$$= 1$ (참) ㄷ. $\displaystyle \sum\limits _{{k} = 1} ^{^{100}} f \left( \frac{k}{101} \right)$$\displaystyle = f \left( \frac{1}{101} \right) + f \left( \frac{2}{101} \right)$$\displaystyle + f \left( \frac{3}{101} \right) + \cdots + f \left( \frac{100}{101} \right)$ $\displaystyle = \left\{ f \left( \frac{1}{101} \right) + f \left( \frac{100}{101} \right) \right\} + \left\{ f \left( \frac{2}{101} \right) + f \left( \frac{99}{101} \right) \right\}$$\displaystyle \cdots + \left\{ f \left( \frac{50}{101} \right) + f \left( \frac{51}{101} \right) \right\}$ $= 1 \times 50$$= 50$ (참) 따라서 옳은 것은 ㄱ, ㄴ, ㄷ이다.