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已知A>0,函数F(X)=-2ASIN(2X+π|6)+2A+B,当X∈2,π|2时,-5≤F(X)≤1,

题目详情
已知A>0,函数F(X)=-2ASIN(2X+π|6)+2A+B,当X∈【2,π|2】时,-5≤F(X)≤1,
▼优质解答
答案和解析
(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]