Question

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),求使g(x)＞1的x值的范围

Answer 1

f(x)=-2asin(2x+π/6)+2a+b,
当x∈[0,π/2]时，2x+π/6 ∈[π/6 ,7π/6 ]
sin(2x+π/6) ∈ [-1/2 ,1 ]
-2asin(2x+π/6)+2a+b ∈ [b ,3a+b ] 因为a>0
又因为-5<=f(x)<=1.
所以
b=-5
3a+b = 1
a = 2
f(x)=-4sin(2x+π/6)-1

g(x)=f(x+π/2)
= -4sin(2x+π/6+π/2)-1
= -4cos(2x+π/6)-1 >1
则cos(2x+π/6)<-1/2
2x+π/6 ∈[2kπ+2π/3 ,2kπ+4π/3]
x ∈[kπ+π/4 ,kπ+7π/12]

如果答案对您有帮助，真诚希望您的采纳和好评哦！！
祝：学习进步哦！！
*^_^* *^_^*

追问

3Q