2. Given the inequality $2^{2 x}+a 2^{x}+b \geqslant 0$ has a minimum solution of 2. Then the range of values for $a$ is ( ).
(A) $(-\infty, 0)$
(B) $[0,+\infty)$
(C) $[2,+\infty)$
(D) $[-4,+\infty)$
2.D.
It is known that 2 should be a root of the equation $2^{2 x}+a 2^{x}+b=0$.
We can obtain $b=-4 a-16$.
Thus, $2^{2 x}+a 2^{x}+b=\left(2^{x}-4\right)\left(2^{x}+4+a\right) \geqslant 0$.
Then $a \geqslant-\left(4+2^{x}\right)$ for $x \in \mathbf{R}-\{2\}$ always holds.
Also, $-\left(4+2^{x}\right)<-4$, therefore, $a \geqslant-4$.