1)導(dǎo)數(shù)法可以,也可以利用三角變換的方法.
f(x)=2cos^x+√3*sin2x+a
=cos2x+1+√3sin2x+a
=cos2x+√3sin2x+a+1
=2*(1/2cos2x+√3/2sin2x)+a+1
=2*sin(2x+1/6π)+a+1
設(shè)Y=2x+1/6π,因?yàn)檎液瘮?shù)的遞增區(qū)間為(2Kπ-1/2π,2Kπ+1/2π),
即當(dāng)Y∈(2Kπ-π/2,2Kπ+π/2)時(shí)函數(shù)遞增,即x∈(Kπ-π/3,Kπ+π/6)為函數(shù)的遞增區(qū)間.
2)當(dāng)x∈[0,π/2]時(shí),Y=2x+1/6π∈[π/6,7π/6],此時(shí)只需研究函數(shù)f(Y)=2sinY+a+1,在[π/6,7π/6]區(qū)間,當(dāng)Y=π/2時(shí)函數(shù)有最大值=2+a+1,另2+a+1≤4,得到a≤1.得出結(jié)論.
如果有算錯(cuò)的地方,請(qǐng)見(jiàn)諒!