已知函数f(x)=(sinx+cosx)눀+2cos눀x 1.求函数f(x)的单调递增区间 2.当x

2025-01-01 12:36:59
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回答1:

下面a为根号2,pi是圆周率
f(x)=(sinx+cosx)²+2cos²x
=(sinx)^2+(cosx)^2+2sinx*cosx+2cos²x
=1+sin2x+cos2x+1
=2+a(a/2sin2x+a\2cos2x)
=2+asin(2x+pi/4)
1.单调增 2kpi-pi/2<=2x+pi/4<=2kpi+pi/2
kpi-3pi/8<=x<=kpi+pi/8 k属于整
2.2x+pi/4在x∈[0,pi/2]时的取值[pi/4,5pi/4]
sint在这上面取最大值是t=pi/2
当x=pi/8时,sin(2x+pi/4)=1
此时f(x)=2+a