Question

已知a＞0,函数f(x)＝-2asin(2x＋π/6)＋2a＋b,当x∈［0，π/2］时，-5≤f(

Answer 1

(1)因为，x∈［0，π/2］，

2x＋π/6∈［π/6，7π/6］，

sin(2x＋π/6)∈［-1/2，1］，

又
a＞0

所以，
-2a＋2a＋b=-5

a＋2a＋b=1

解得：
a=2,
b=-5
(2)
由(1)知，f(x)＝-4sin(2x＋π/6)-1

由题意
g(x)＝f(x＋π/2)

＝-4sin(2x+π+π/6)-1

=4sin(2x＋π/6)-1＞1

即
sin(2x＋π/6)＞1/2

所以
2x＋π/6∈(2kπ+π/6,2kπ+5π/6)

单调增区间满足
2x＋π/6∈(2kπ+π/6,2kπ+π/2]

单调减区间满足
2x＋π/6∈[2kπ+π/2,2kπ+5π/6)

解得
g(x)的单调增区间为
(kπ,kπ+π/6]

单调减区间为
[kπ+π/6,kπ+π/3]