Question

已知函数f(x)=cos2x-8mcosx+1,x∈[0,π/2] 当m=0时,求f（x）的单调区间 求f（x）=cos

Answer 1

当m=0时,求f（x）=cos2x+1

x∈[0,π/2] ,f（x）单调递减

当x=π/2,f（x）=cos2x+1有最小值0

f（x）=cos2x-8mcosx+1=2(cosx)^2-8mcosx+1

=2(cosx-m^2/4)^2+1-m^2/2

当m∈[-2,2]时,cosx=m^2/4,f(x)有最小值1-m^2/2

当m∉[-2,2]时,cosx=1f(x)有最小值3-8m