(3)由-π/2+kπ<2x<π/2+kπ(k∈Z),得-π/4+kπ/2

　　(4)函数y=tan 2x在区间[-π,π]内的图像如图所示.

B组　能力提升

1.函数t=tan(3x+π/3)的图像的对称中心不可能是0(　　)

A.("-" π/9 "," 0) B.(π/18 "," 0)

C.("-" π/18 "," 0) D.("-" 5π/18 "," 0)

解析因为正切函数y=tan x图像的对称中心是(kπ/2 "," 0),k∈Z.令3x+π/3=kπ/2(k∈Z),解得x=kπ/6-π/9(k∈Z),所以函数y=tan(3x+π/3)的图像的对称中心为(kπ/6 "-" π/9 "," 0),k∈Z.

　　当k=0,1,-1时,得kπ/6-π/9=-π/9,π/18,-5π/18.

　　所以A,B,D选项是函数图像的对称中心.故选C.

答案C

2.导学号93774024下列图形分别是①y=|tan x|;②y=tan x;③y=tan(-x);④y=tan|x|在x∈("-" 3π/2 "," 3π/2)内的大致图像,则由a到d对应的函数关系式应是(　　)

A.①②③④ B.①③④②

C.③②④① D.①②④③

解析y=tan(-x)=-tan x在("-" π/2 "," π/2)上是减少的,只有图像d符合,即d对应③.

答案D

3.已知函数y=tan ωx在区间("-" π/2 "," π/2)上是减少的,则ω的取值可能是(　　)

A.1 B.-1

C.2 D.-2

解析根据各选项ω的值在给定区间上取特殊值来进行验证.选B.

答案B

4.若不等式tan x>a在x∈("-" π/4 "," π/2)上恒成立,则a的取值范围为(　　)

A.a>1 B.a≤1