Question

已知函数f(x)=√3cos2x-2sin²(π/4+x)+1,x∈[π/6,π/2].

(1)求f(x)的最大值，并求当f(x)取得最大值时x的值(2)求f(x)单调递增区间.

Answer 1

f(x)=√3cos2x-2sin²(π/4+x)+1
=√3cos2x-2[(√2/2)(sinx+cosx)]²+1
=√3cos2x-(1-sin2x)+1
=√3cos2x+sin2x
=2sin(2x+π/3)
当
sin(2x+π/3)=1时
f(x)取最大值为
2
这时
2x+π/3=2kπ+π/2
x=kπ+π/12
k∈z
单增:
2x+π/3∈[2kπ-π/2,2kπ+π/2]
x∈[kπ-5π/12,kπ+π/12]
k∈z
单调增区间为
[kπ-5π/12,kπ+π/12]
k∈z