Question

求函数f(x)=（sin^4x+cos^4x+sin^2xcos^2x）/（2-sin2x）的最小正周期,值域及单调增区间

如题

Answer 1

f(x)=(sin^4 x+cos^4 x+sin^2 xcos^2 x)/(2-sin2 x）
=(sin^4 x+cos^4 x+2sin^2 xcos^2 x
-sin^2 xcos^2 x)/(2-sin2 x）
=[(sin^2 x+cos^2 x)^2-(sinxcosx)^2]/(2-sin2x)
=[1-(sinxcosx)^2]/(2-2sinxcosx)
=[(1+sinxcosx)(1-sinxcosx)]/[2(1-sinxcosx)]
=1/2+1/2*sinxcosx
=1/2+1/4*sin2x
=1/4*sin2x+1/2
则T=2pi/2=pi
由sin2x属于[-1,1]
则f(x)属于[1/4,3/4]
即值域为[1/4,3/4]
单调增区间:[kpi-pi/4,kpi+pi/4] (k属于Z)

Answer 2

f(x)
=[(sin^2x+cos^2x)^2-sin^2xcos^2x]/(2-2sinxcosx)
=1/2 *(1+sinxcosx)
=1/2 *(1+1/2 *sin2x)
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