已知a>0,函数f(x)=-2asin(2x+π/6)+2a+b,当x∈[0,π/2]时,-5≤f(x)≤1.(1)求常数a,b的值.(2)设g(x)=f(x+π/2)且lg[g(x)]>0,求g(x)单调区间

Question

已知a>0,函数f(x)=-2asin(2x+π/6)+2a+b,当x∈[0,π/2]时,-5≤f(x)≤1.(1)求常数a,b的值.(2)设g(x)=f(x+π/2)且lg[g(x)]>0,求g(x)单调区间

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]