Question

f(x)=2sin^2(x+π/4)-√3cos2x+α在[∏／4，∏／2]上的最小值为5，求α的值

Answer 1

α=4

f(x)=2sin^2(x+π/4)-√3cos2x+α
=2(sinxcosπ/4+cosxsinπ/4)^2-√3cos2x+α
=2(√2/2sinx+√2/2cosx)^2-√3cos2x+α
=2((sinx)^2/2+(cosx)^2/2+sinxcosx)-√3cos2x+α
=1+2sinxcosx-√3cosx+α
=1+sin2x-√3cos2x+α
=1+2((1/2)sin2x-(√3/2)cos2x)+α
=1+2(cosπ/6sin2x-sinπ/6cos2x)+α
=1+2sin(2x-π/6)+α
x∈[π/4，π/2]，2x∈[π/2，π]，2x-π/6∈[π/3，5π/6]
2sin(2x-π/6)在[π/3，5π/6]的最小值为0
1+2sin(2x-π/6)+α的最小值为5
所以α=5-1=4