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lgg(x)＞0,求g(x)的单调 我算了好几遍第一问,a+b

(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]
打字不易,如满意,望采纳.第4行，当x=6/派的时候，y不等于1呀对不起，打错了当x=0的时候，sin(2x+6/派）不等于0不等于1。。