( I ) 若 $f(x)=0$, 则存在 $\xi \in(-a, a)$ 使得 $f^{\prime \prime}(\xi)=\frac{1}{a^2}[f(a)+f(-a)]$;
( II ) 若 $f(x)$ 在 $(-a, a)$ 内取得极值, 则存在 $\eta \in(-a, a)$, 使得 $\left|f^{\prime \prime}(\eta)\right| \geq \frac{1}{a^2}|f(a)-f(-a)|$
【答案】 【解析】(1)证明: $f(x)=f(0)+f^{\prime}(0) x+\frac{f^{\prime \prime}(\eta)}{2 !} x^2=f^{\prime}(0) x+\frac{f^{\prime \prime}(\eta)}{2 !} x^2, \eta$ 介于 0 与 $x$ 之间, 则 $f(a)=f^{\prime}(0) a+\frac{f^{\prime \prime}\left(\eta_1\right)}{2 !} a^2, 0 < \eta_1 < a$ (1)
$$f(-a)=f^{\prime}(0)(-a)+\frac{f^{\prime \prime}\left(\eta_2\right)}{2 !} a^2,-a < \eta_2 < 0 \text { (2) }$$
(1)+2)得: $f(a)+f(-a)=\frac{a^2}{2}\left[f^{\prime \prime}\left(\eta_1\right)+f^{\prime \prime}\left(\eta_2\right)\right]$ (3)

$$m \leq f^{\prime \prime}\left(\eta_1\right) \leq M ; m \leq f^{\prime \prime}\left(\eta_2\right) \leq M \text {; 从而 } m \leq \frac{f^{\prime \prime}\left(\eta_1\right)+f^{\prime \prime}\left(\eta_2\right)}{2} \leq M \text {; }$$

$$f(a)+f(-a)=a^2 f^{\prime \prime}(\xi) \text {, 即 } f^{\prime \prime}(\xi)=\frac{f(a)+f(-a)}{a^2}$$
(2) 证明: 设 $f(x)$ 在 $x=x_0 \in(-a, a)$ 取极值, 且 $f(x)$ 在 $x=x_0$ 可导, 则 $f^{\prime}\left(x_0\right)=0$.

$f(a)=f\left(x_0\right)+\frac{f^{\prime \prime}\left(\gamma_2\right)}{2 !}\left(a-x_0\right)^2, 0 < \gamma_2 < a$

$$\leq \frac{1}{2}\left|\left(a-x_0\right)^2 f^{\prime \prime}\left(\gamma_2\right)\right|+\frac{1}{2}\left|\left(a+x_0\right)^2 f^{\prime \prime}\left(\gamma_1\right)\right|$$

$|f(a)-f(-a)| \leq \frac{1}{2} M\left(a+x_0\right)^2+\frac{1}{2} M\left(a-x_0\right)^2=M\left(a^2+x_0{ }^2\right)$

$M \geq \frac{1}{2 a^2}|f(a)-f(-a)|$, 即存在 $\eta=\gamma_1$ 或 $\eta=\gamma_2 \in(-a, a)$, 有
$$\left|f^{\prime \prime}(\eta)\right| \geq \frac{1}{2 a^2}|f(a)-f(-a)| \text {. }$$