📝 题目
例 22 设函数 $f\left( x\right)$ 在 $\left\lbrack {0,\pi }\right\rbrack$ 上连续,且
$$ {\int }_{0}^{\pi }f\left( x\right) \mathrm{d}x = 0,\;{\int }_{0}^{\pi }f\left( x\right) \cos x\mathrm{\;d}x = 0. $$
求证: 在 $\left( {0,\pi }\right)$ 内至少存在两个不同的点 ${\xi }_{1},{\xi }_{2}$ ,使
$$ f\left( {\xi }_{1}\right) = f\left( {\xi }_{2}\right) = 0. $$
💡 答案与解析
证 令 $F\left( x\right) = {\int }_{0}^{x}f\left( t\right) \mathrm{d}t\left( {x \in \left\lbrack {0,\pi }\right\rbrack }\right)$ ,则有 $F\left( 0\right) = F\left( \pi \right) = 0$ . 又因为
$$ 0 = {\int }_{0}^{\pi }f\left( x\right) \cos x\mathrm{\;d}x = {\int }_{0}^{\pi }\cos x\mathrm{\;d}F\left( x\right) $$
$$ = {\left. F\left( x\right) \cos x\right| }_{0}^{\pi } + {\int }_{0}^{\pi }F\left( x\right) \sin x\mathrm{\;d}x $$
$$ = {\int }_{0}^{\pi }F\left( x\right) \sin x\mathrm{\;d}x, $$
所以存在 $\xi \in \left( {0,\pi }\right)$ ,使得 $F\left( \xi \right) \sin \xi = 0$ . 因若不然,则在 $\left( {0,\pi }\right)$ 内或 $F\left( x\right) \sin x$ 恒为正,或 $F\left( x\right) \sin x$ 恒为负,都与 $\displaystyle{\int }_{0}^{\pi }F\left( x\right) \sin x\mathrm{\;d}x = 0$ 矛盾. 又当 $\xi \in \left( {0,\pi }\right)$ 时, $\sin \xi \neq 0$ ,故 $F\left( \xi \right) = 0$ . 于是 $F\left( x\right)$ 在 $\left\lbrack {0,\pi }\right\rbrack$ 上有三个不同零点, $0 < \xi < \pi$ . 再用罗尔定理,则存在 ${\xi }_{1} \in \left( {0,\xi }\right) ,{\xi }_{2} \in$ $\left( {\xi ,\pi }\right)$ ,使得 ${F}^{\prime }\left( {\xi }_{1}\right) = 0,{F}^{\prime }\left( {\xi }_{2}\right) = 0$ ,即 $f\left( {\xi }_{1}\right) = 0,f\left( {\xi }_{2}\right) = 0$ .