855

已知函数 ${\displaystyle f(x)={\mathrm{e}}^x+a\sin x-1(a<0)}$.

1. 若 ${\displaystyle a=-1}$, 求证: 当 ${\displaystyle x\ge 0}$ 时，${\displaystyle f(x)\ge 0}$;
2. 讨论函数 ${\displaystyle g(x)=f(x)+ax\cos x}$ 在区间 ${\displaystyle [0,\pi ]}$ 上的零点个数.

1. 证明 当 ${\displaystyle a=-1,x\ge 0}$，有 ${\displaystyle f^{\prime }(x)={\mathrm{e}}^x-\cos x\ge 1-\cos x\ge 0}$, 故 ${\displaystyle f(x)\ge f(0)=0}$.

2. 解 给定 ${\displaystyle a}$，记 ${\displaystyle g(x)=g_a(x)}$，是关于 ${\displaystyle a}$ 的一次函数.

   • 当 ${\displaystyle {\displaystyle -\frac{1}{2}\le a<0}}$,则 ${\displaystyle \mathrm{\forall }x\in [0,\pi ]}$: ${\displaystyle g(x)\ge min{\displaystyle \{g_0(x),g_{-\frac{1}{2}}(x)\}}}$.
     因为${\displaystyle g_0(x)={\mathrm{e}}^x-1\ge 0}$,而${\displaystyle [0,\pi ]}$上$$\begin{array}{r}{g}_{-\frac{1}{2}}^{\prime }(x)=({\mathrm{e}}^x-\cos x)+\frac{1}{2}x\cos x\ge 0,\end{array}$$
     故${\displaystyle {\displaystyle g_{-\frac{1}{2}}(x)\ge g(0)=0}}$.因此此时${\displaystyle g(x)}$只有唯一零点${\displaystyle x=0}$.
   • 当 ${\displaystyle a<-{\displaystyle \frac{1}{2}}}$,考察 $$g^{\prime }(x)={\mathrm{e}}^x+a(2\cos x-x\sin x),g^{″}(x)={\mathrm{e}}^x-3a(\sin x+x\cos x).$$ 则 ${\displaystyle x\in {\displaystyle [0,\frac{\pi }{2}]}}$ 时 ${\displaystyle g^{″}(x)>0}$，故此时 ${\displaystyle g^{\prime }(x)}$ 单增.又 $$g^{\prime }(0)=1+2a<0<{\displaystyle g^{\prime }\left(\frac{\pi }{2}\right)={\mathrm{e}}^{\frac{\pi }{2}}-\frac{\pi }{2}a,}$$ 故 ${\displaystyle g^{\prime }(x)}$ 在 ${\displaystyle {\displaystyle (0,\frac{\pi }{2})}}$ 上存在唯一零点 ${\displaystyle x_0}$. 当 ${\displaystyle x\in {\displaystyle [\frac{\pi }{2},\pi ]}}$ 时显然 ${\displaystyle g^{″}(x)>0}$，故 ${\displaystyle g^{\prime }(x)\ge g^{\prime }{\displaystyle \left(\frac{\pi }{2}\right)>0}}$.
     这样 ${\displaystyle x_0}$ 是 ${\displaystyle g(x)}$ 的唯一极小值点.又 $$g(0)=0,g(\pi )={\mathrm{e}}^{\pi }-a\pi -1>{\displaystyle {\mathrm{e}}^{\pi }+\frac{\pi }{2}-1>0,}$$ 故 ${\displaystyle g(x)}$ 存在两个零点.

   综上，${\displaystyle {\displaystyle -\frac{1}{2}\le a<0}}$ 时有 ${\displaystyle 1}$ 个零点，${\displaystyle {\displaystyle a<-\frac{1}{2}}}$ 时有两个零点.