Question

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

Answer 1

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<0时，有,-2a+2a+b=1，a+2a+b=-5解得a=-2,b=1
综上，a=2,b=-5或
a=-2,b=1

Answer 2

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