3. Given the quadratic function about the parameter $a(a>0)$: $y=a x^{2}+\sqrt{1-a^{2}} x+a^{2}-3 a-\frac{1}{4}+\frac{1}{4 a}(x \in \mathbf{R})$, the minimum value of which is a function $f(a)$ of $a$. Then the minimum value of $f(a)$ is ( ).
(A) -2
(B) $-\frac{137}{64}$
(C) $-\frac{1}{4}$
(D) None of the above results is correct
3.A.

When $x=-\frac{\sqrt{1-a^{2}}}{2 a}$, the minimum value of $y$ is $f(a)=a^{2}-\frac{11}{4} a-\frac{1}{4}(0<a \leqslant 1)$.

Since the axis of symmetry is $a=\frac{11}{8}$, when $a=1$, the minimum value of $f(a)$ is -2.