Question

已知函数a>0,函数f(x)=-2asin(2x+π/6)+2a+b,当x∈[0,π/2]

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

Answer 1

当x∈[0,π/2]时, -1/2 <=sin(2x+π/6)<=1, b<=f(x)<=3a+b
而已知-5<=f(x)<=1 则b=-5 3a+b=1 解得 a=2 b=-5
所以函数f(x)=-4sin(2x+π/6)-1 所以g(x)=f(x+π/2)=4sin(2x+π/6)-1
由lg g(x)>0得g(x)>1 即sin(2x+π/6)>1/2
单调增时:π/6+2kπ<2x+π/6<=π/2+2kπ 即 kπ<x<=π/6+kπ
单调减时:π/2+2kπ<2x+π/6<5π/6+2kπ 即 π/6+kπx<=π/3+kπ k为整数