Question

设函数f (x)＝√3（cosωx）平方＋sinωxcosωx＋a（其中ω＞0,a∈r）,

Answer 1

f (x)＝√3（cosωx）平方＋sinωxcosωx＋a

=√3(1+cos2ωx)/2+(sin2ωx)/2+a

=sin(2ωx+π/3)+a+√3/2

(1) 2ω*π/6 +π/3=π/2

所以 ω=1/2

(2) f(x)=sin(x+π/3)+a+√3/2

x+π/3∈[0,7π/6]

所以 x+π/3=7π/6

时,y有最小值 -1/2+a+√3/2=√3

a=1/2+√3/2