问题
解答题
设向量
(I)求函数F(x)=f(x)f′(x)+f2(x)的最大值和最小正周期; (II)若f(x)=2f′(x),求
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答案
(1)f(x)=sinx+cosx
∴f′(x)=cosx-sinx,
∴F(x)=f(x)f′(x)+f2(x)
=cos2x-sin2x+1+2sinxcosx
=1+sin2x+cos2x
=1+
sin(2x+2
)π 4
∴当2x+
=2kπ+π 4
⇒x=kπ+π 2
(k∈Z)时,π 8
F(x)max=1+2
最小正周期为T=
=π2π 2
(2)∵f(x)=2f′(x)⇒sinx+cosx=2cosx-2sinx
∴cosx=3sinx⇒tanx=1 3
∴
=1+2sin2x cos2x-sinxcosx
=3sin2x+cos2x cos2x-sinxcosx
=3tan2x+1 1-tanx
=2.2 2 3