$$
\begin{array}{l}
f(x) \text { 在 }(0,2 \pi) 2 \text { 阶可导 } ，f^{\prime \prime}(x) \neq f(x) & \\
\exists \xi \in(0,2 \pi) \text { 使得 } \tan \xi=\frac{2 \cdot f^{\prime}(\xi)}{f(\xi)-f^{\prime \prime}(\xi)} \text { 成立 }
\end{array}
$$
解：
$$\frac{sin(\xi)}{cos(\xi)}=\frac{2 \cdot f^{\prime}(\xi)}{f(\xi)-f^{\prime \prime}(\xi)}$$
$由于f^{\prime \prime}(x)\neq f(x),交叉相乘得$
$$
\begin{array}{l}
f(\xi)sin(\xi)-f^{\prime \prime}(\xi)sin(\xi)=2f’(\xi)cos(\xi)\\
f(\xi)sin(\xi)-f’(\xi)cos(\xi)=f^{\prime \prime}(\xi)sin(\xi)+f’(\xi)cos(\xi)\\
-\left(f(\xi)cos(\xi)\right)’=(f’(\xi)sin(\xi))’\\
\left(f(\xi)cos(\xi)+f’(\xi)sin(\xi)\right)’=0\\
\left(f(\xi)sin(\xi)\right)’’=0
\end{array}
$$
故令$F(x)=f(x)sin(x),F(0)=F(\pi)=F(2\pi)=0$
$\exists \xi_1 \in(0,\pi),\exists \xi_2 \in(\pi,2\pi)$,使得$F’(\xi_1)=F’(\xi_2)=0$
$\exists \xi_3 \in(0,2\pi)$,使得$F’’(\xi_3)=0$