Question

已知函数f（x）=sin²x+根号3sinxcosx+2cos²x，x∈R

Answer 1

f(x)=sin^2x+√3sinxcosx+2cos^2x
=1+1/2+1/2*cos2x+√3/2*sin2x
=3/2+sin(2x+π/6)
(1)f(x)的最小正周期π
单调递增区间
2kπ-2π/2<=2x+π/6<=2kπ+π/2
即x属于[kπ-π/3,kπ+π/6]
(2)
x∈【0，π/2】时求函数的值域
2x+π/6属于[π/2,7π/6]
sin(2x+π/6)属于[-1/2,1]
f(x)属于[1,5/2]
（3）
f(x)单调递增区间为[kπ-π/3,kπ+π/6]
x的定义域为[0,π/2]
令k=0,则可得f(x)递增区间为[0,π/6]

Answer 2

f(x)=(1-cos2x)/2+√3/2sin2x+1+cos2x
=√3/2sin2x+1/2cos2x+3/2
=sin(2x+π/6)+3/2
的最小正周期
T=2π/2=π
2kπ-π/2<=2x+π/6<=2kπ+π/2
kπ-π/3<=x<=kπ+π/6
单调地增区间
[kπ-π/3，kπ+π/6]
k∈Z
x∈【0，π/2】
2x+π/6∈【π/6，5π/6】
函数的值域
[2,9/2]
函数的单调递增区间
[0,π/6]