2. Let $f(x)$ be a function defined on $\mathbf{R}$, for any $x \in \mathbf{R}$, we have $f(x+1)=\frac{1}{2}+$ $\sqrt{f(x)-[f(x)]^{2}}$, and $f(-1)=\frac{1}{2}$, then the value of $f(2007)$ is
A. -1
B. 1
C. 2007
D. $\frac{1}{2}$
$$
\begin{array}{l}
\text { 2. } \mathrm{D} \begin{array}{l} 
f(x+2)=f[(x+1)+1] \\
=\frac{1}{2}+\sqrt{f(x+1)-[f(x+1)]^{2}} \\
=\frac{1}{2}+\sqrt{\frac{1}{2} \sqrt{f(x)-[f(x)]^{2}}-\left[\frac{1}{2}+\sqrt{f(x)-[f(x)]^{2}}\right]^{2}} \\
=\frac{1}{2}+\sqrt{\frac{1}{4}-f(x)+[f(x)]^{2}}=\frac{1}{2}+\left|\frac{1}{2}-f(x)\right| . \\
\text { Also, } f(x)=f[(x-1)+1] \\
=\frac{1}{2}+\sqrt{f(x-1)-[f(x-1)]^{2}} \geqslant \frac{1}{2} .
\end{array}
\end{array}
$$

Therefore, $f(x+2)=\frac{1}{2}+\left[f(x)-\frac{1}{2}\right]=f(x)$, which means 2 is a period of $f(x)$, hence $f(2007)=f(2 \times 1004-1)=f(-1)=\frac{1}{2}$. Therefore, the answer is D.