已知a∈R,函数f(X)=-2aSin(2x+π/6）+2a+b当x∈〔0,π/2〕,-5≤f(X）≤1. （1）求常数a,b的值
问题描述：
已知a∈R,函数f(X)=-2aSin(2x+π/6）+2a+b当x∈〔0,π/2〕,-5≤f(X）≤1. （1）求常数a,b的值
答
x∈〔0，π/2)
2x+π/6∈〔π/6，7π/6)
Sin(2x+π/6）∈[-1/2,1]
(1)-2a(-1/2)+2a+b=-5且-2a+2a+b=1
得a=-2,b=1
(2)-2a(-1/2)+2a+b=1且-2a+2a+b=-5
得a=2,b=-5
所以a=-2,b=1或a=2,b=-5
答
由x∈[0,π/2],可知2x+π/6∈[π/6,7π/6],Sin(2x+π/6)∈[-1/2,1],
(1)a>0时,有a+2a+b=1,-2a+2a+b=-5,解得a=2,b=-5
(2)a综上,a=2,b=-5或
a=-2,b=1