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设函数f(x)=sinxcosx+cosx^2, 求f(x)的最小正周期, 当x属于【0,π/2】时,求函数f(x)的最大值和最小值。
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f(x)=sinxcosx + (cosx)^2
=(1/2)sin2x + (1/2)cos2x + 1/2
=(√2/2)[sin2x*cos(π/4) + cos2x*sin(π/4)] + 1/2
=(√渣拍迹2/2)sin(2x + π/如并4) + 1/2
最小正周期T=2π/2=π
当x∈[0,π/2]时,(2x + π/4)∈[π/4 ,5π/4]
f(x)最大值=(√2+1)/2
f(x)最小值贺卜=0
=(1/2)sin2x + (1/2)cos2x + 1/2
=(√2/2)[sin2x*cos(π/4) + cos2x*sin(π/4)] + 1/2
=(√渣拍迹2/2)sin(2x + π/如并4) + 1/2
最小正周期T=2π/2=π
当x∈[0,π/2]时,(2x + π/4)∈[π/4 ,5π/4]
f(x)最大值=(√2+1)/2
f(x)最小值贺卜=0
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