
函数f(x)=根号2sin(wx+y)(w>0,且y的绝对值<π/2)的最小正周期是π,且f(0)=1.
(1)求f(x)的解析式(2)令g(x)=f(x+π/8)-a,若g(x)在x属于[-(6/π),3/π]时有两个零点,求a的取值范围...
(1)求f(x)的解析式 (2)令g(x)=f(x+π/8)-a,若g(x)在x属于[-(6/π),3/π]时有两个零点,求a的取值范围
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函数f(x)=根号2sin(wx+y)(w>0,且y的绝对值<π/2)的最小正周期是π,且f(0)=1.
(1)求f(x)的解析式 (2)令g(x)=f(x+π/8)-a,若g(x)在x属于[-(6/π),3/π]时有两个零点,求a的取值范围
(1)解析:∵f(x)=√2sin(wx+y)(w>0,且|y|<π/2)的最小正周期是π,且f(0)=1
∴w=2π/π=2==> f(x)=√2sin(2x+y)==> f(0)=√2sin(0+y)=1==>y=π/4
∴f(x)=√2sin(2x+π/4)
(2)解析:∵g(x)=f(x+π/8)-a
∴g(x)=√2sin(2x+π/2)-a=√2cos(2x)-a
∵g(x)在x属于[-π/6,π/3]时有两个零点
令g(-π/6) =√2cos(-π/3)-a=√2/2-a<=0==>a>=√2/2
g(π/3)=√2cos(2π/3)-a=-√2/2-a<=0==>a>=-√2/2
∴√2/2<=a<√2
即要g(x)在x属于[-π/6,π/3]时有两个零点,a的取值范围√2/2<=a<√2
(1)求f(x)的解析式 (2)令g(x)=f(x+π/8)-a,若g(x)在x属于[-(6/π),3/π]时有两个零点,求a的取值范围
(1)解析:∵f(x)=√2sin(wx+y)(w>0,且|y|<π/2)的最小正周期是π,且f(0)=1
∴w=2π/π=2==> f(x)=√2sin(2x+y)==> f(0)=√2sin(0+y)=1==>y=π/4
∴f(x)=√2sin(2x+π/4)
(2)解析:∵g(x)=f(x+π/8)-a
∴g(x)=√2sin(2x+π/2)-a=√2cos(2x)-a
∵g(x)在x属于[-π/6,π/3]时有两个零点
令g(-π/6) =√2cos(-π/3)-a=√2/2-a<=0==>a>=√2/2
g(π/3)=√2cos(2π/3)-a=-√2/2-a<=0==>a>=-√2/2
∴√2/2<=a<√2
即要g(x)在x属于[-π/6,π/3]时有两个零点,a的取值范围√2/2<=a<√2
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