II. (50 points) The function $f(x)$ is defined on $[0,1]$, $\alpha \in\left(0, \frac{\pi}{2}\right)$. If for any $x \geqslant y, x, y \in[0,1]$, we have $f\left(\frac{x+y}{2}\right)=f(x) \sin \alpha+f(y)(1-\sin \alpha)$. Determine the expression for $f(x)$.
II. Solution: Given $f\left(\frac{1}{2}\right)=f\left(\frac{1+0}{2}\right)=\sin \alpha, f\left(\frac{1}{4}\right)=f\left(\frac{\frac{1}{2}+0}{2}\right)=\sin ^{2} \alpha$,
$$
f\left(\frac{3}{4}\right)=f\left(\frac{1+\frac{1}{2}}{2}\right)=-\sin ^{2} \alpha+2 \sin \alpha,
$$

Since $f\left(\frac{\frac{3}{4}+\frac{1}{4}}{2}\right)=f\left(\frac{1}{2}\right)$, we have
$$
\sin \alpha=\left(-\sin ^{2} \alpha+2 \sin \alpha\right) \sin \alpha+(1-\sin \alpha) \sin ^{2} \alpha,
$$

which simplifies to $2 \sin ^{3} \alpha-3 \sin ^{2} \alpha+\sin \alpha=0$,
$\because \sin \alpha \in(0,1), \therefore 2 \sin ^{2} \alpha-3 \sin \alpha+1=0$,
$\therefore \sin \alpha=\frac{1}{2}$, or $\sin \alpha=1$ (discard), hence $\alpha=\frac{\pi}{6}$.
Thus, we have $f\left(\frac{x+y}{2}\right)=\frac{1}{2} f(x)+\frac{1}{2} f(y)$.
In the Cartesian coordinate system $O-x y z$, let $A\left(x_{1}, f\left(x_{1}\right)\right), B\left(x_{2}, f\left(x_{2}\right)\right), C\left(\frac{x_{1}+x_{2}}{2}, f\left(\frac{x_{1}+x_{2}}{2}\right)\right)$, then $\overrightarrow{O C}=\frac{1}{2} \cdot \overrightarrow{O A}+\frac{1}{2} \cdot \overrightarrow{O B}$, by vector knowledge, we know: $A, B, C$ are collinear.
Noting that $x, y$ are any real numbers in $[0,1]$, and the property that two points determine a line, we have:
Any point $A(x, f(x))$ lies on a fixed line $l$,
Also, since $f\left(\frac{1}{2}\right)=\sin \frac{\pi}{6}=\frac{1}{2}, f(1)=\frac{1}{2} f(1)+\frac{1}{2} f(1)=1$,
we get that the equation of the fixed line $l$ is $y=x$. Therefore, $f(x)=x$.