设a属于R,f(x)=cosx(ashinx-cosx)+cos^(π/2-x)满足f(-π/3)=f(0),

题目

设a属于R,f(x)=cosx(ashinx-cosx)+cos^(π/2-x)满足f(-π/3)=f(0),
求函数f(x)在[π/4,11π/24]上的最大值

答案

f(x)=cosx(ashinx-cosx)+cos²(π/2-x)
=asinxcosx-cos²x+sin²x
=(a/2)sin2x-cos2x
因为满足f(-π/3)=f(0)
所以(a/2)sin-2π/3-cos2π/3
=-a√3/4+1/2=(a/2)sin0-cos0=-1
所以a=2√3
f(x)=(a/2)sin2x-cos2x=√3sin2x-cos2x
=2sin(2x-π/6)
x∈[π/4,11π/24]
2x-π/6∈[π/3,3π/4]
f(x)为最大值时
2x-π/6=π/2
此时f(x)最大值=2sin(π/2)=2