help~help~1:F(x)是以5为周期的奇函数,F(-3)=4且cosα=0.5,则F(4cos2α)=?2:已知复数Z1=√3 sin2x+λi,Z2=m+(m-cos2x)i (λ,m,x∈R),且Z1=Z2问:1)若λ=0,且0
来源:学生作业学帮网 编辑:学帮网 时间:2024/05/05 05:28:46
help~help~
1:F(x)是以5为周期的奇函数,F(-3)=4且cosα=0.5,则F(4cos2α)=?
2:已知复数Z1=√3 sin2x+λi,Z2=m+(m-cos2x)i (λ,m,x∈R),且Z1=Z2
问:1)若λ=0,且0
1、因为cosα=0.5,所以cos(2α)=2(cosα)^2-1=-0.5.则F[4cos(2α)]=F[4*(-0.5)]=F(-2)].因为F(x)为奇函数,所以F(3)=-F(-3)=-4.又F(x)周期为5,所以,F(-2)=F(3)=-4.
2、因为Z1=Z2,所以有√3sin(2x)=m,λ=m-cos(2x).(λ,m,x∈R)
(1)当λ=0时,m=cos(2x),代入第一式消去m化简得:√3sin(2x)=cos(2x).显然cos(2x)≠0,故有tan(2x)=√3/3.0<x<π,则0<2x<2π.
所以2x=π/6或7π/6,得x=π/12或7π/12.
(2)f(x)=m-cos(2x)=√3sin(2x)-cos(2x)=2sin(2x-π/6)
易知f(x)的最小正周期为π,单调递增区间为kπ-π/2≤2x-π/6≤kπ+π/2,化简得:(3k-1)π/6≤x≤(3k+2)π/6.
即单调递增区间为[(3k-1)π/6,(3k+2)π/6],(k∈Z).
help~help~1:F(x)是以5为周期的奇函数,F(-3)=4且cosα=0.5,则F(4cos2α)=?2:已知复数Z1=√3 sin2x+λi,Z2=m+(m-cos2x)i (λ,m,x∈R),且Z1=Z2问:1)若λ=0,且0
【help 已知二次函数f(x)=ax^2+bx(a,b为常数,且a不等于0)满足条件f(-x+5)=f(x-3),且方程f(x)=x有等根.1、求f(x)的解析式 2、是否存在实数m、n(m
f(x)是以4为周期的函数,且f(1)=a,则f(5)等于
f(x)=sin(x φ) cos(x φ)为奇函数的充分不必要条件HELP!
HELP!
Help!
Help
Help!
help,
help
help
help!.
help...
.help
help,
help
help
help