Question

已知函数f（x）=sin²x+2sinxcosx+3cos²x，x∈R

求函数f（x）的最大值及取得最大值的自变量x的集合

求函数f（X）的单调增区间

Answer 1

解：
(2)
f（x）=sin²x+2sinxcosx+3cos²x
=sin²x+cos²x+2cos²x+sin2x
=1+2cos²x+sin2x
=1+con2x+1+sin2x
=2+√2sin（π/4+2x)
∵|sin（π/4+2x)|<=1
∴f（x）=2+√2sin（π/4+2x) >0
2nπ-π/2<= π/4+2x <=2nπ+π/2
2nπ-π3/4<= 2x <=2nπ+π/4
nπ-π3/8<= x <=nπ+π/8
f（x）的单调增区间为：[nπ-π3/8 , nπ+π/8] （n为整数）
（1）
π/4+2x=2nπ+π/2
X=nπ+π/8时 f(x)最大=2+√2 （n为整数）

Answer 2

f（x）=sin²x+2sinxcosx+3cos²x
=1+sin2x+2cos²x-1+1
=sin2x+cos2x+2
=√2sin(2x+π/4)+2
函数f（x）的最大值=2+√2
此时 2x+π/4=2kπ+π/2 x=kπ+π/8 自变量x的集合 {x|x=kπ+π/8}

单调增区间 2kπ-π/2<=2x+π/4<=2kπ+π/2
kπ-3π/8<=x<kπ+π/8
单调增区间[kπ-3π/8,kπ+π/8] k∈Z