26.已知函数f(x)=ex+acosxf(x)=e^{x}+a\cos xf(x)=ex+acosx.(1)若函数f(x)f(x)f(x)在区间(0,π2)(0,\frac{\pi }{2})(0,2π)上恰有两个极值点,求aaa的取值范围;(2)证明:当1⩽a⩽eπ2−2−π21\leqslant a\leqslant {e}^{\frac{\pi }{2}}-2-\frac{\pi }{2}1⩽a⩽e2π−2−2π时,在(0,+∞)(0,+\infty )(0,+∞)上,f(x)>2+xf(x)>2+xf(x)>2+x恒成立.