$$
\begin{array}{ll} 
& \text { 3. Let } a b \neq 0 \text {, and the function } \\
& f_{1}(x)=x^{2}+2 a x+4 b \\
\text { and } & f_{2}(x)=x^{2}+4 a x+2 b
\end{array}
$$

have the same minimum value $u$, and the functions
$$
\begin{aligned}
f_{3}(x) & =-x^{2}+2 b x+4 a \\
\text { and } f_{4}(x) & =-x^{2}+4 b x+2 a
\end{aligned}
$$

have the same maximum value $v$. Then the value of $u+v$ ( ).
(A) must be a positive number
(B) must be a negative number
(C) must be 0
(D) the sign cannot be determined
$$
3. C.
$$
\begin{array}{c}
\text { Given } f_{1}(x)=(x+a)^{2}+4 b-a^{2} \geqslant 4 b-a^{2}, \\
f_{2}(x)=(x+2 a)^{2}+2 b-4 a^{2} \geqslant 2 b-4 a^{2},
\end{array}
$$

Also, $4 b-a^{2}=u=2 b-4 a^{2}$, then
$$
-2 b=3 a^{2} \text {. }
$$

Given $f_{3}(x)=-(x-b)^{2}+4 a+b^{2} \leqslant 4 a+b^{2}$,
$$
f_{4}(x)=-(x-2 b)^{2}+2 a+4 b^{2} \leqslant 2 a+4 b^{2} \text {, }
$$

Also, $4 a+b^{2}=v=2 a+4 b^{2}$, then
$$
2 a=3 b^{2} \text {. }
$$
(2) - (1) gives
$$
2(a+b)=3\left(b^{2}-a^{2}\right) \text {. }
$$

Solving, we get $a+b=0$ or $b-a=\frac{2}{3}$.
If $a+b=0$, then
$$
\begin{array}{l}
2(u+v)=\left(6 b-5 a^{2}\right)+\left(6 a+5 b^{2}\right) \\
=(a+b)[6+5(b-a)]=0 ;
\end{array}
$$

If $b-a=\frac{2}{3}$, combining with equation (2) we get
$$
2\left(b-\frac{2}{3}\right)=3 b^{2} \text {, }
$$

which simplifies to $(3 b-1)^{2}+3=0$,
a contradiction.