Question

急！急！！急！！！ 已知函数f(x)=sin(2ωx-π/6)-1/2的最小正周期为π/2

（1）求函数f（x）的图像的对称轴方程
（2）求函数f（x）的单调增区间
（3）求函数f（x）在区间[-π/12，π/2]上的值域

Answer 1

如图

Answer 2

最小正周期=2π/2=π/2 =2
1 当4x-π/6=kπ+π/2时 即x=kπ/4+π/6是对称轴
2 当4x-π/6∈[kπ-π/2，kπ+π/2]时 函数单调增
kπ-π/2<=4x-π/6<=kπ+π/2 kπ/4-π/12<=x<=kπ/4+π/6
即单调增区间为[kπ/4-π/12,kπ/4+π/6]
3 当x∈[-π/12，π/2]时 4x-π/6∈[-π/2,11π/6]
所以最大值为1-1/2=1/2 最小值为-1-1/2=-3/2
所以值域为[-3/2,1/2]

Answer 3

解：2ω=2π/T=4
所以ω=2
所以f(x)=sin(4x-π/6)-1/2
(1).由4x-π/6=π/2+kπ得x=π/6+kπ/4
(2). 由-π/2+2kπ<=4x-π/6<=π/2+2kπ得-π/12+kπ/2<=x<=π/6+kπ/2
所以单调增区间为[kπ/2-π/12,kπ/2+π/6]
(3).由-π/12<=x<=π/2得-π/2<=4x-π/6<=11π/6
所以-1<=sin(4x-π/6)<=1
所以-3/2<=f(x)<=1/2
所以值域为[-3/2,1/2]

Answer 4

最小正周期T=2π/w=π/2
w=4
f(x)=sin(8x-π/6)-1/2
对称轴方程为
8x-π/6=kπ+π/2
x=kπ/8+π/12(k属于整数)
(2)令8x-π/6=t
则t的增区间为[2kπ-π/2,2kπ+π/2]
x的增区间为[kπ/4-π/24,kπ/4+π/12](k属于整数)
(3)π/2-(-π/12)>最小正周期π/2
所以值域为一个周期内的值域
fmax=1-1/2=1/2
fmin=-1-1/2=-3/2
即[-3/2,1/2]