设函数f(x)=2sin(π/2x+π/5) 若对任意x∈R,都有f(x1)
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设函数f(x)=2sin(π/2x+π/5) 若对任意x∈R,都有f(x1)
设函数f(x)=2sin(π/2x+π/5) 若对任意x∈R,都有f(x1)
设函数f(x)=2sin(π/2x+π/5) 若对任意x∈R,都有f(x1)
因为当x∈R时,f(x)的值域为[-2,2]
所以只有当f(x1)=-2,f(x2)=2时对任意x∈R,都有f(x1)<=f(x)<=f(x2)才成立.
而要求(x1-x2)的绝对值的最小值,则f(x1),f(x2)为相邻的最大最小值,则(x1-x2)的绝对值为半周期
周期T=2π/(π/2)=4
所以(x1-x2)的绝对值的最小值为2
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