设函数 $f(x)$ 可导, $g(x)=\left\{\begin{array}{ll}x^2 \sin \frac{1}{|x|}+\frac{1}{|x|} \sin ^2 x, & x \neq 0 \\ 0, & x=0\end{array}, F(x)=f[g(x)]\right.$,
则 $F(x)$ 在 $x=0$ 点可导的充分必要条件是
$ \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】 $F(x)=f[g(x)]=\left\{\begin{array}{ll}f\left(x^2 \sin \frac{1}{|x|}+\frac{1}{|x|} \sin ^2 x\right), & x \neq 0 \\ f(0), & x=0\end{array}\right.$,
则 $F_{-}^{\prime}(0)=\lim _{x \rightarrow 0^{-}} \frac{F(x)-F(0)}{x-0}=\lim _{x \rightarrow 0^{-}} \frac{f\left(-x^2 \sin \frac{1}{x}-\frac{1}{x} \sin ^2 x\right)-f(0)}{x}$
$$
\begin{aligned}
&=\lim _{x \rightarrow 0^{-}} \frac{f\left(-x^2 \sin \frac{1}{x}-\frac{1}{x} \sin ^2 x\right)-f(0)}{-x^2 \sin \frac{1}{x}-\frac{1}{x} \sin ^2 x} \cdot \lim _{x \rightarrow 0^{-}} \frac{-x^2 \sin \frac{1}{x}-\frac{1}{x} \sin ^2 x}{x} \\
&=f^{\prime}(0) \cdot \lim _{x \rightarrow 0^{-}}\left(-x \sin \frac{1}{x}-\frac{\sin ^2 x}{x^2}\right)=-f^{\prime}(0) . \\
F_{+}^{\prime}(0)=& \lim _{x \rightarrow 0^{+}} \frac{F(x)-F(0)}{x-0}=\lim _{x \rightarrow 0^{+}} \frac{f\left(x^2 \sin \frac{1}{x}+\frac{1}{x} \sin ^2 x\right)-f(0)}{x} \\
=& \lim _{x \rightarrow 0^{+}} \frac{f\left(x^2 \sin \frac{1}{x}+\frac{1}{x} \sin ^2 x\right)-f(0)}{x^2 \sin \frac{1}{x}+\frac{1}{x} \sin ^2 x} \cdot \lim _{x \rightarrow 0^{+}} \frac{x^2 \sin \frac{1}{x}+\frac{1}{x} \sin ^2 x}{x}
\end{aligned}
$$

$$
=f^{\prime}(0) \cdot \lim _{x \rightarrow 0^{+}}\left(x \sin \frac{1}{x}+\frac{\sin ^2 x}{x^2}\right)=f^{\prime}(0) .
$$
则 $F(x)$ 在 $x=0$ 点可导的充要条件是 $F_{-}^{\prime}(0)=F_{+}^{\prime}(0)$, 即 $f^{\prime}(0)=0$. 故应选 (A).
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