霉到柱
幼苗
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解
f(x)=ab
=(cosx,sin2x)(2cosx,1)
=2cos²x+sin2x
=cos2x+sin2x+1
=√2sin(2x+π/4)+1
∴f(x)的最小正周期实事2π/2=π
当sin(2x+π/4)=1时
f(x)max=√2+1
当sin(2x+π/4)=-1时
f(x)min=-√2+1
x∈(0,π),2x+π/4∈(π/4,9π/4)
ab=√2sin(2x+π/4)+1>2即
sin(2x+π/4)>√2/2
∴π/4
1年前
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