若x∈[π/6,3/π]时,k+tan(2x-π/3)>0恒成立,则k的取值范围为

问题描述:

若x∈[π/6,3/π]时,k+tan(2x-π/3)>0恒成立,则k的取值范围为
(A)k>-2 (b)K>-1
(C)K>0 (D)K>1

当x∈[π/6,π/3]时,2x-π/3∈[0,π/3]
所以tan(2x-π/3)∈[0,√3〕
因为k+tan(2x-π/3)>0
所以k>0
选C