Given that the domain of function $f ( x )$ is $\mathbb { R }$, $f ( x + 2 )$ is an even function, and $f ( 2 x + 1 )$ is an odd function, then A. $f \left( - \frac { 1 } { 2 } \right) = 0$ B. $f ( - 1 ) = 0$ C. $f ( 2 ) = 0$ D. $f ( 4 ) = 0$
B Since $f ( x + 2 )$ is an even function, $f ( 2 + x ) = f ( 2 - x )$, which gives $f ( x + 3 ) = f ( 1 - x )$. Since $f ( 2 x + 1 )$ is an odd function, $f ( 1 - 2 x ) = - f ( 2 x + 1 )$, so $f ( 1 - x ) = - f ( x + 1 )$. Therefore, $f ( x + 3 ) = - f ( x + 1 ) = f ( x - 1 )$, i.e., $f ( x ) = f ( x + 4 )$, so $f$ has period 4. Since $F ( x ) = f ( 2 x + 1 )$ is odd, $F ( 0 ) = f ( 1 ) = 0$. Therefore, $f ( -1 ) = -f(1) = 0$.
Given that the domain of function $f ( x )$ is $\mathbb { R }$, $f ( x + 2 )$ is an even function, and $f ( 2 x + 1 )$ is an odd function, then
A. $f \left( - \frac { 1 } { 2 } \right) = 0$
B. $f ( - 1 ) = 0$
C. $f ( 2 ) = 0$
D. $f ( 4 ) = 0$