Question

已知函数f(x)=sin^2x+根号3sinxcosx+2cos^2x当x∈【0,π/4】时求值域

Answer 1

已知函数f（x）=sin平方x+根号3sinxcosx+2cos平方x，求函数f（x）的最小正周期和单调递增区间
f(x)=
(sinx)^2
+3
cosx*sinx
+
2（cosx)^2
=
(cosx)^2+1
+
1.5*2sinx*cosx
=
-cos2x/2
+1.5
+1.5
sin2x
故
最小正周期
为
π

Answer 2

答：
f(x)=(sinx)^2+√3sinxcosx+2(cosx)^2
=1+(cosx)^2+(√3/2)*2sinxcosx
=1+(1/2)(cos2x+1)+(√3/2)*sin2x
=(√3/2)sin2x+(1/2)cos2x+3/2
=sin(2x+π/6)+3/2
0<=x<=π/4
0<=2x<=π/2
π/6<=2x+π/6<=2π/3
所以：1/2<=sin(2x+π/6)<=1
所以：1/2+3/2<=f(x)<=1+3/2
所以：值域为[2,5/2]

Answer 3

f(x)=根号3sin2x+cos2x
=2sin(2x+π/6)
x∈【0，π/4】
2x+π/6∈【π/6，2π/3】
f(x)∈【1，2】