$\text{A.}$ $f^{\prime}(0)=0$. $\text{B.}$ $f^{\prime}(0) \neq 0$. $\text{C.}$ $f(0)=0$. $\text{D.}$ $f(0) \neq 0$.
【答案】 A

【解析】 $F(x)=f[g(x)]=\left\{\begin{array}{ll}f\left(x^2 \sin \frac{1}{|x|}+\frac{1}{|x|} \sin ^2 x\right), & x \neq 0 \\ f(0), & x=0\end{array}\right.$,

\begin{aligned} &=\lim _{x \rightarrow 0^{-}} \frac{f\left(-x^2 \sin \frac{1}{x}-\frac{1}{x} \sin ^2 x\right)-f(0)}{-x^2 \sin \frac{1}{x}-\frac{1}{x} \sin ^2 x} \cdot \lim _{x \rightarrow 0^{-}} \frac{-x^2 \sin \frac{1}{x}-\frac{1}{x} \sin ^2 x}{x} \\ &=f^{\prime}(0) \cdot \lim _{x \rightarrow 0^{-}}\left(-x \sin \frac{1}{x}-\frac{\sin ^2 x}{x^2}\right)=-f^{\prime}(0) . \\ F_{+}^{\prime}(0)=& \lim _{x \rightarrow 0^{+}} \frac{F(x)-F(0)}{x-0}=\lim _{x \rightarrow 0^{+}} \frac{f\left(x^2 \sin \frac{1}{x}+\frac{1}{x} \sin ^2 x\right)-f(0)}{x} \\ =& \lim _{x \rightarrow 0^{+}} \frac{f\left(x^2 \sin \frac{1}{x}+\frac{1}{x} \sin ^2 x\right)-f(0)}{x^2 \sin \frac{1}{x}+\frac{1}{x} \sin ^2 x} \cdot \lim _{x \rightarrow 0^{+}} \frac{x^2 \sin \frac{1}{x}+\frac{1}{x} \sin ^2 x}{x} \end{aligned}

$$=f^{\prime}(0) \cdot \lim _{x \rightarrow 0^{+}}\left(x \sin \frac{1}{x}+\frac{\sin ^2 x}{x^2}\right)=f^{\prime}(0) .$$