8. If one edge length of a tetrahedron is $x$, and the other edges are all 1, the volume is $F(x)$, then the function $F(x)$ on its domain
A. is an increasing function but has no maximum value
B. is an increasing function and has a maximum value
C. is not an increasing function and has no maximum value
D. is not an increasing function but has a maximum value
In tetrahedron $ABCD$, $\triangle ABC$ and $\triangle ABD$ are equilateral triangles with side length 1, and $E$ is the midpoint of $AB$, $CD=x$.
$$
\begin{array}{l}
\text { Then in } \triangle DEC, \cos \theta=\frac{\frac{3}{4}+\frac{3}{4}-x^{2}}{2 \cdot \frac{3}{4}}=1-\frac{2}{3} x^{2}, \\
DO=DE \sin \theta=\frac{\sqrt{3}}{2} \cdot \sqrt{1-\left(1-\frac{2}{3} x^{2}\right)^{2}}=\frac{\sqrt{3}}{2} \cdot \sqrt{\frac{4}{3} x^{2}-\frac{4}{9} x^{4}}=\frac{\sqrt{3}}{3} \cdot \sqrt{3 x^{2}-x^{4}} \\
\Rightarrow V_{D-ABC}=\frac{1}{3} \cdot \frac{\sqrt{3}}{4} \cdot \frac{\sqrt{3}}{3} \cdot \sqrt{3 x^{2}-x^{4}}=\frac{1}{12} \sqrt{3 x^{2}-x^{4}} .
\end{array}
$$

Thus, when $x=\frac{\sqrt{6}}{2}$, $V_{\max }=\frac{1}{8}$. Therefore, the answer is $D$.