2. Given the function $y=2 x^{2}$ on $[a, b](a<b)$, its value range is $[0,2]$. Then the trajectory of the point $(a, b)$ is ( ).
(A) line segments $A B, B C$
(B) line segments $A B, O C$
(C) line segments $O A, B C$
(D) line segments $O A, O C$
2. (A).

From $y=2 x^{2} \geqslant 0$, we know that when $y=0$, $x=0$.
Therefore, $0 \in[a, b]$.
When $y=2$, $x=1$ or -1. There are two possibilities:
If $1 \in[a, b]$, then by $a<b$ and the monotonicity of the function on $R^{+}$, we have $b=1$ and $-1 \leqslant a \leqslant 0$
(see figure); the corresponding trajectory is the line segment $A B$.
If $-1 \in[a, b]$, then by $a<b$ and the monotonicity of the function on $R^{-}$, we have $a=-1$ and $0 \leqslant b \leqslant 1$; the corresponding trajectory is the line segment $B C$.