简单的零点分析
已知\(f(x)=ae^x-\sin x-1\)
\((1)\) 当\(a=1\)证明：\(\forall x\in[0,+\infty),f(x)\geq 0\)
\((2)\) 若\(f(x)\)在区间\(\left(0,\dfrac{\pi}{2}\right)\)上存在极值，求实数\(a\)的取值范围.
\((1)\) \(a=1,f(x)=e^x-\sin x-1,f^{\prime}(x)=e^x-\cos x\geq 0\)
从而\(f(x)\)单调递增，进而\(f(x)\geq f(0)=0\)
得证.
\((2)\)
法一：
\(f^{\prime}(x)=ae^x-\cos x=0\)
即\(a=\dfrac{\cos x}{e^x}\)
记\(h(x)=\dfrac{\cos x}{e^x},h^{\prime}(x)=\dfrac{-\sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right)}{e^x}\)
不难得到\(h(x)\)在\(\left(0,\dfrac{\pi}{2}\right)\)上单调递减，而\(h(0)=1,h\left(\dfrac{\pi}{2}\right)=0\)
从而\(a\in(0,1)\)
法二:
\(f^{\prime}(x)=ae^x-\cos x\)
若\(a\geq 1\)，则\(f^{\prime}(x)\geq 0\)，没有零点，从而\(f(x)\)没有极值.
若\(0<a<1\)时，\(f^{\prime\prime}(x)=ae^x+\sin x>0\)
则\(f^{\prime}(x)\)单调递增\(f^{\prime}(0)=a-1<0,f\left(\dfrac{\pi}{2}\right)=ae^{\frac{\pi}{2}}>0\)
从而由零点存在定理，\(f^{\prime}(x)\)在\(\left(0,\dfrac{\pi}{2}\right)\)上存在零点，从而\(f(x)\)在\(\left(0,\dfrac{\pi}{2}\right)\)上存在极值.
若\(a\leq 0\)时，\(f^{\prime}(x)=ae^x-\cos x<0\)，没有零点，从而\(f(x)\)没有极值.
综上\(a\in(0,1)\)