Question

设α∈R,函数f(x)=√2sin2xcosα+√2cos2xsinα-√2cos(2x+α)+cosα,

若a∈[π/4,π/2],求f(x)在区间[0,π/2]上的最大值

Answer 1

f(x)=√2sin2xcosα+√2cos2xsinα-√2cos(2x+α)+cosα
=√2sin(2x+α)-√2cos(2x+α)+cosα
=2(sin(2x+α)√2/2-cos(2x+α)√2/2)+cosα
=2sin(2x+α-π/4)+cosα

f'(x)=2*2cos(2x+α-π/4)=4cos(2x+α-π/4)
当x∈[0,π/2]，α∈[π/4,π/2]时，2x+α-π/4∈[0+π/4-π/4,π+π/2-π/4]
即2x+α-π/4∈[0,5π/4]

显然2x+α-π/4∈[0,π/2]，f'(x)≥0，函数单调增
2x+α-π/4∈[π/2,5π/4]，f'(x)≤0，函数单调减
则f(x)在2x+α-π/4=π/2时，即x=3π/8-α/2取得最大值，
f(3π/8-α/2)=2sin(2(3π/8-α/2)+α-π/4)+cosα
=2sin(3π/4-π/4)+cosα
=2+cosα