Question

已知函数f(x)=√3sin(wx+φ)-cos(wx+φ)(0<φ<pai,w>0)为偶函数,

且函数y=f(x)图象的两相邻对称轴间的距离为pai/2
1)求f(pai/8)的值
2)将函数y=f(x)的图象向右平移pai/6个单位后,再将得到的图象上各点的横坐标伸长到原来的4倍,纵坐标不变,得到函数y=g(x)的图象,求g(x)的单调递减区间

Answer 1

f(x)=√3sin(wx+φ)-cos(wx+φ)
=2[√3/2sin(wx+φ)-1/2cos(wx+φ)]
=2sin(wx+φ-π/6)
对称轴：令wx+φ-π/6=kπ+π/2（k∈Z)
解之：x=（kπ+2π/3-φ）/w
故两相邻对称轴间的距离为π/|w|=π/2
又w>0
故w=2
又y=f(x)为偶函数
故x=0是他的一条对称轴
故φ-π/6=kπ+π/2
即φ=kπ+2π/3（k∈Z)
又0<φ<π
故φ=2π/3
即f(x)=2sin(wx+φ-π/6)
=2sin(2x+2π/3-π/6)
=2cos2x
故f(π/8)=2cosπ/4=√2
f(x)向右平移π/6个单位：y=2cos(2x-π/3)
再将得到的图象上各点的横坐标伸长到原来的4倍,纵坐标不变：g(x)=2cos(x/2-π/3)
令2kπ≤x/2-π/3≤2kπ+π
解之：4kπ+2π/3≤x≤4kπ+8π/3
即单调递减区间是[4kπ+2π/3,4kπ+8π/3](k∈Z）

Answer 2

f(x)=√3sin(wx+φ)-cos(wx+φ)
=2[√3/2sin(wx+φ)-1/2cos(wx+φ)]
=2sin(wx+φ-π/6)
对称轴：令wx+φ-π/6=kπ+π/2（k∈Z)
解之：x=（kπ+2π/3-φ）/w
故两相邻对称轴间的距离为π/|w|=π/2
又w>0
故w=2
又y=f(x)为偶函数
故x=0是他的一条对称轴
故φ-π/6=kπ+π/2
即φ=kπ+2π/3（k∈Z)
又0<φ<π
故φ=2π/3
即f(x)=2sin(wx+φ-π/6)
=2sin(2x+2π/3-π/6)
=2cos2x
故f(π/8)=2cosπ/4=√2
f(x)向右平移π/6个单位：y=2cos(2x-π/3)
再将得到的图象上各点的横坐标伸长到原来的4倍,纵坐标不变：g(x)=2cos(x/2-π/3)
令2kπ≤x/2-π/3≤2kπ+π
解之：4kπ+2π/3≤x≤4kπ+8π/3
即单调递减区间是[4kπ+2π/3,4kπ+8π/3](k∈Z）

Answer 3

f(x)=√3sin(wx+φ)-cos(wx+φ)
=2[√3/2sin(wx+φ)-1/2cos(wx+φ)]
=2sin(wx+φ-π/6)
对称轴：令wx+φ-π/6=kπ+π/2（k∈Z)
解之：x=（kπ+2π/3-φ）/w
故两相邻对称轴间的距离为π/|w|=π/2
又w>0
故w=2
又y=f(x)为偶函数
故x=0是他的一条对称轴
故φ-π/6=kπ+π/2
即φ=kπ+2π/3（k∈Z)
又0<φ<π
故φ=2π/3
即f(x)=2sin(wx+φ-π/6)
=2sin(2x+2π/3-π/6)
=2cos2x
故f(π/8)=2cosπ/4=√2
f(x)向右平移π/6个单位：y=2cos(2x-π/3)
再将得到的图象上各点的横坐标伸长到原来的4倍,纵坐标不变：g(x)=2cos(x/2-π/3)
令2kπ≤x/2-π/3≤2kπ+π
解之：4kπ+2π/3≤x≤4kπ+8π/3
即单调递减区间是[4kπ+2π/3,4kπ+8π/3](k∈Z）

Answer 4

f(x)=√3sin(wx+φ)-cos(wx+φ)
=2[√3/2sin(wx+φ)-1/2cos(wx+φ)]
=2sin(wx+φ-π/6)
对称轴：令wx+φ-π/6=kπ+π/2（k∈Z)
解之：x=（kπ+2π/3-φ）/w
故两相邻对称轴间的距离为π/|w|=π/2
又w>0
故w=2
又y=f(x)为偶函数
故x=0是他的一条对称轴
故φ-π/6=kπ+π/2
即φ=kπ+2π/3（k∈Z)
又0<φ<π
故φ=2π/3
即f(x)=2sin(wx+φ-π/6)
=2sin(2x+2π/3-π/6)
=2cos2x
故f(π/8)=2cosπ/4=√2
f(x)向右平移π/6个单位：y=2cos(2x-π/3)
再将得到的图象上各点的横坐标伸长到原来的4倍,纵坐标不变：g(x)=2cos(x/2-π/3)
令2kπ≤x/2-π/3≤2kπ+π
解之：4kπ+2π/3≤x≤4kπ+8π/3
即单调递减区间是[4kπ+2π/3,4kπ+8π/3](k∈Z）

Answer 5

解:∵函数y=f(x)图像的两相邻对称轴的距离为π/2
∴f(x)的周期为π
即:w=2
化简得:f(x)=√3
sin(
wx+φ)-cos(wx+φ)=2sin(2x+φ+π/6)
又∵f(x)为偶函数,所以φ+π/6=π，解得φ=5π/6
所以函数f(x)=2sin(2x+π)=2cos2x(1)f(π/8)=2cos(2×π/8)=√2
(2)y=f(x)的图像向右平移π/6
个单位,f(x)=2cos(2(x-π/6)=2cos(2x-π/3)
得到的图像上各点的横坐标伸长到原来的4倍g(x)=2cos(x/2-π/3)
2kπ≤x/2-π/3≤2kπ+π
解得:4kπ+2π/3≤x≤4kπ+8π/3
所以，g(x)的单调递减区间[4kπ+2π/3,4kπ+8π/3]