Question

设函数f(x)=(cosx)^2+根号3*sinxcosx+a若x∈【0,π/2】时,f(x)的最小值为2,求a的值

Answer 1

f(x)=(cos2x+1)/2+√3*sin2x/2+a

=sin(2x+π/6)+1/2+a

所以当x∈[0,π/2]

那么

2x+π/6∈[π/6,7π/6]

所以最小值

f(x)=sin(7π/6)+1/2+a

=-1/2+1/2+a

=a=2

所以a=2