Question

求函数值域，

设函数fx＝｜√3sinx＋cosx｜＋√3sinxcosx－1/2cos2x，若x属于【－派/3，派/2，求函数的值域】

Answer 1

f(x)=|√3sinx＋cosx|+√3sinxcosx－1/2cos2x
=2|sinxcosπ/6+sinπ/6cosx|+√3/2 sin2x－1/2cos2x
=2|sin(x+π/6)|+sin2xsinπ/3-cosπ/3cos2x
=2|sin(x+π/6)|-cos(2x+π/3)
=2|sin(x+π/6)|-cos2(x+π/6)
=2|sin(x+π/6)|+2sin²(x+π/6)-1
=2(|sin(x+π/6)|+1/2)²-3/2
因为x∈[-π/3,π/2]，所以 - 1/2≤sin(x+π/6)≤1
0≤|sin(x+π/6)|≤1
令t=|sin(x+π/6)|，则0≤t≤1
g(t)=f(x)=2(t+1/2)²-3/2，在0≤t≤1单调递增
所以g(0)≤f(x)≤g(1)
g(0)=-1，g(1)=3
故f(x)的值域为[-1,3]

Answer 2