1、化简函数f(x)=(sin^4 x+cos^4 x+sin^2 xcos^2 x)/(2-sin2x)

1、化简函数f(x)=(sin^4 x+cos^4 x+sin^2 xcos^2 x)/(2-sin2x)
2、已知函数f(x)=4sinxsin^2(π/4 +x/2)+cos2x
1) 设ω>0为常数,若y=f(ωx)在区间[-π/2,2π/3]上是增函数,求ω的取值范围
2)设集合A={π/6≤x≤2π/3},B={f(x)-m
Ethan_liu 1年前 已收到1个回答 举报

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1、f(x)=(sin^4 x+cos^4 x+sin^2 xcos^2 x)/(2-sin2x)
=[(sin^2 x+cos^2 x)^2-2sin^2 xcos^2 x+sin^2 xcos^2 x]/(2-sin2x)
=(1-sin^2 xcos^2 x)/(2-sin2x)
=[(1-sinxcosx)(1+sinxcosx)]/[2(1-sinxcosx)]=(1+sinxcosx)/2
2、f(x)=4sinxsin^2(π/4 +x/2)+cos2x
=2sinx(cosx/2+sinx/2)^2+cos2x
=2sinx(1+sinx)+cos2x
=2sinx+2sin^2 x+1-2sin^2 x
=2sinx+1
(1)y=f(ωx)=2sinωx+1
要使y=f(ωx)在区间[-π/2,2π/3]上是增函数
则T≥4*(2π/3)=8π/3
T=2π/ω 解得:ω≤3/4
(2)f(x)的T为2π,A是B的子集
根据题意得,f(x)在π/6≤x≤2π/3之间的取值范围为[-1,3]
则 f(x)-m1
3、(1)f(x)=2cos^2 x+ 根号3 sin2x+α
=cos2x+1+ 根号3 sin2x+α
=2sin(x+π/6)+α+1 T=2π
则在[-π/6,π/6]上最大值与最小值之和分别在π/6和-π/6处取得
即2sin(π/6+π/6)+α+1+2sin(-π/6+π/6)+α+1=3
解得:α=(1-根号3)/2
(2)要使f(x)=0在[0,π]内有两相异实根x1 x2
则0

1年前

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