대칭 주기함수 보기

[출제의도] 정적분을 활용하여 함수 추론하기 ㄱ. $f \left( 2 + x \right)$$= f \left( 2 - x \right) = f \left( 1 + \left( 1 - x \right) \right)$$= f \left( 1 - \left( 1 - x \right) \right) = f \left( x \right)$ $f \left( x + 2 \right) = f \left( x \right)$ (참) ㄴ. $\displaystyle \int _{2} ^{5} {f' \left( x \right) dx}$${= \left[ f \left( x \right) \right] _{2} ^{5}}$$= f \left( 5 \right) - f \left( 2 \right) = 4$이고 ㄱ에 의하여 $f \left( 5 \right) = f \left( 1 \right)$이고 $f \left( 2 \right) = f \left( 0 \right)$이므로 $f \left( 1 \right) - f \left( 0 \right) = 4$ (참) ㄷ. $f \left( 0 \right) = a$라 하면 $f \left( 1 \right) = a + 4$ $f \left( x \right) = t$라 치환하면 $\displaystyle \frac{dt}{dx} = f' \left( x \right)$이므로 $\displaystyle \int _{0} ^{1} {f \left( f \left( x \right) \right) f' \left( x \right) dx = \int _{f \left( 0 \right)} ^{f \left( 1 \right)} {f \left( t \right) dt = 6}}$ ㄱ, ㄴ에 의하여 $\displaystyle \int _{a} ^{a + 4} {f \left( t \right) dt = 6}$$\displaystyle = 2 \int _{a} ^{a + 2} {f \left( t \right) dt}$에서 $\displaystyle \int _{0} ^{2} {f \left( t \right) dt} = 3$ $\displaystyle \int _{0} ^{10} {f ( x ) dx} = 5 \int _{0} ^{2} {f ( x ) dx} = 15$ $f \left( 1 + x \right) = f \left( 1 - x \right)$이므로 $\displaystyle \int _{0} ^{1} {f ( x ) dx} = \int _{1} ^{2} {f ( x ) dx} = \frac{3}{2}$ $\displaystyle \int _{1} ^{10} {f ( x ) dx}$$\displaystyle = \int _{0} ^{10} {f ( x ) dx} - \int _{0} ^{1} {f ( x ) dx}$ $\displaystyle = 15 - \frac{3}{2} = \frac{27}{2}$ (참) 따라서 옳은 것은 ㄱ, ㄴ, ㄷ