已知函数y=sin(wx+Q)(w>0,绝对值Q>π/2)在同一周期内

问题描述:

已知函数y=sin(wx+Q)(w>0,绝对值Q>π/2)在同一周期内
已知函数f(x)=sin(wx+Q)(w>0,绝对值Q>π/2)在同一周期内,当x=pai/4时,y取最大值1,当7pai/12时,y取最小值-1
⑴求函数y=f(x)
⑵函数y=sinx的图像经过怎样的变换可得到y=f(x)
⑶若函数f(x)满足方程f(x)=(0<a<1),求在[0,2pai]内所有实数根之和

已知函数f(x)=sin(wx+Q)(w>0,绝对值Q>π/2)在同一周期内,当x=pai/4时,y取最大值1,当7pai/12时,y取最小值-1
⑴求函数y=f(x)
⑵函数y=sinx的图像经过怎样的变换可得到y=f(x)
⑶若函数f(x)满足方程f(x)=(0<a<1),求在[0,2pai]内所有实数根之和
(1)解析:∵函数f(x)=sin(wx+Q)(w>0,绝对值Q>π/2)在同一周期内,当x=pai/4时,y取最大值1,当7pai/12时,y取最小值-1
∴T/2=7π/12-π/4=π/3==>T=2π/3==>w=2π/T=3
∴f(x)=sin(3x+Q)==> f(π/4)=sin(3π/4+Q)=1==>3π/4+Q=π/2==>Q=-π/4
∴f(x)=sin(3x-π/4)
(2)解析:函数y=sinx
A:将函数y水平右移π/4单位得y=sin(x-π/4)
B:将X轴坐标单位水平压缩到原单位的1/3,得y=sin(3x-π/4)

A:将X轴坐标单位水平压缩到原单位的1/3,得y=sin(3x)
B:将函数y水平右移π/12单位得y=sin(3(x-π/12))= sin(3x-π/4)
(3)解析:方程f(x)=a(0<a<1)
sin(3x-π/4)= √2/2==>3x-π/4=2kπ+π/4==>x=2kπ/3+π/6;3x-π/4=2kπ+3π/4==>x=2kπ/3+π/3
sin(3x-π/4)=1/2==>3x-π/4=2kπ+π/6==>x=2kπ/3+5π/36;3x-π/4=2kπ+5π/6==>x=2kπ/3+13π/36
sin(3x-π/4)= √3/2==>3x-π/4=2kπ+π/3==>x=2kπ/3+7π/36;3x-π/4=2kπ+4π/6==>x=2kπ/3+11π/36
在[0,2pai]内所有实数根之和
π/6+2π/6+5π/6+6π/6+9π/6+10π/6=11π/2
5π/36+13π/36+29π/36+37π/36+53π/36+61π/36=11π/2
7π/36+11π/36+31π/36+35π/36+55π/36+59π/36=11π/2可见,无论a∈(0,1)取何值,在[0,2pai]内所有实数根之和均为11π/2