设函数f(x)=2sin(πx/2+π/5),若对任意x∈R都有f(x1)≤f(x)≤f(x2)成立,则|x1-x2|的最小值为
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![设函数f(x)=2sin(πx/2+π/5),若对任意x∈R都有f(x1)≤f(x)≤f(x2)成立,则|x1-x2|的最小值为](/uploads/image/z/3979020-12-0.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3D2sin%28%CF%80x%2F2%2B%CF%80%2F5%29%2C%E8%8B%A5%E5%AF%B9%E4%BB%BB%E6%84%8Fx%E2%88%88R%E9%83%BD%E6%9C%89f%28x1%29%E2%89%A4f%28x%29%E2%89%A4f%28x2%29%E6%88%90%E7%AB%8B%2C%E5%88%99%7Cx1-x2%7C%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%E4%B8%BA)
设函数f(x)=2sin(πx/2+π/5),若对任意x∈R都有f(x1)≤f(x)≤f(x2)成立,则|x1-x2|的最小值为
设函数f(x)=2sin(πx/2+π/5),若对任意x∈R都有f(x1)≤f(x)≤f(x2)成立,则|x1-x2|的最小值为
设函数f(x)=2sin(πx/2+π/5),若对任意x∈R都有f(x1)≤f(x)≤f(x2)成立,则|x1-x2|的最小值为
对任意x∈R都有f(x1)≤f(x)≤f(x2)成立
所以f(x1)是最小值,f(x2)是最大值
所以f(x1)=-2
f(x2)=2
所以πx1/2+π/5=2kπ-π/2
πx2/2+π/5=2mπ+π/2
x1=4k-7/5
x2=4m+3/5
|x1-x2|=|4(k-m)-2|
k-m是整数
所以|x1-x2|最小=2
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