8. Given that the domain of function $f ( x )$ is $\mathbf { R }$, $f ( x + 2 )$ 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. $f ( 4 ) = 0$

【Answer】B\\
【Solution】\\
【Analysis】Derive that $f ( x )$ is a periodic function with period 4. From the given conditions, deduce that $f ( 1 ) = 0$, and combine with the given conditions to reach the conclusion.

【Detailed Solution】Since $f ( x + 2 )$ is an even function, we have $f ( 2 + x ) = f ( 2 - x )$, which gives $f ( x + 3 ) = f ( 1 - x )$. Since $f ( 2 x + 1 )$ is an odd function, we have $f ( 1 - 2 x ) = - f ( 2 x + 1 )$, so $f ( 1 - x ) = - f ( x + 1 )$. \\
Therefore, $f ( x + 3 ) = - f ( x + 1 ) = f ( x - 1 )$, that is, $f ( x ) = f ( x + 4 )$. \\
Thus, $f ( x )$ is a periodic function with period 4. \\
Since $F ( x ) = f ( 2 x + 1 )$ is an odd function, we have $F ( 0 ) = f ( 1 ) = 0$. \\
Therefore, $f ( - 1 ) = - f ( 1 ) = 0$, while the other three options are unknown.\\
Therefore, the answer is: B.

II. Multiple Choice Questions: This section contains 4 questions, each worth 5 points, for a total of 20 points. For each question, there may be multiple correct options. Full marks (5 points) are awarded for selecting all correct options, 2 points for partially correct selections, and 0 points for any incorrect selection.