2. As shown in Figure 1-2, the graph of the function $y=f(x)$ is two arcs of an ellipse centered at the origin with foci on the $x$-axis. Then the solution set of the inequality $f(x)<f(-x)+x$ is ( )
A. $\{x \mid-\sqrt{2}<x<0$, or $\sqrt{2}<x \leqslant 2\}$
B. $\{x \mid-2 \leqslant x<-\sqrt{2}$, or $\sqrt{2}<x \leqslant 2\}$
C. $\left\{x \left\lvert\,-2 \leqslant x<-\frac{\sqrt{2}}{2}\right.\right.$. or $\left.\frac{\sqrt{2}}{2}<x \leqslant 2\right\}$
D. $\{x \mid-\sqrt{2}<x<\sqrt{2}$, and $x \neq 0\}$
2. A Prompt: From the left graph, we know that $f(x)$ is an odd function, $f(-x)=-f(x)$, so the original inequality can be transformed into $f(x)<\frac{x}{2}$. The geometric meaning of this inequality is the set of $x$ values for which the graph of $f(x)$ lies below the graph of $g(x)=\frac{x}{2}$. By combining the ellipse $\frac{x^{2}}{4}+y^{2}=1$ with the line $y=\frac{x}{2}$, we get $\frac{x^{2}}{4}+\frac{x^{2}}{4}=1$, so $x^{2}=2, x= \pm \sqrt{2}$.
Observing the graph, we find that $-\sqrt{2}<x<0$ or $\sqrt{2}<x \leqslant 2$. Therefore, the answer is A.