已知函数f(x)=2cos(π/2x+π/5),若实数x1,x2满足f(x1)-f(x2)=4,则(x1-x2)的绝对值
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已知函数f(x)=2cos(π/2x+π/5),若实数x1,x2满足f(x1)-f(x2)=4,则(x1-x2)的绝对值的最小值是
![已知函数f(x)=2cos(π/2x+π/5),若实数x1,x2满足f(x1)-f(x2)=4,则(x1-x2)的绝对值](/uploads/image/z/18525313-1-3.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D2cos%28%CF%80%2F2x%2B%CF%80%2F5%29%2C%E8%8B%A5%E5%AE%9E%E6%95%B0x1%2Cx2%E6%BB%A1%E8%B6%B3f%28x1%29-f%28x2%29%3D4%2C%E5%88%99%EF%BC%88x1-x2%EF%BC%89%E7%9A%84%E7%BB%9D%E5%AF%B9%E5%80%BC)
f(x)=2cos(π/2x+π/5)
f(x1)-f(x2)=2cos(π/2x1+π/5)-2cos(π/2x2+π/5)= 4
cos(π/2x1+π/5)-cos(π/2x2+π/5) = 2
∵-1 ≤ cos(π/2x1+π/5) ≤ 1,-1 ≤ cos(π/2x2+π/5) ≤ 1
∴ cos(π/2x1+π/5) = 1,cos(π/2x2+π/5) = - 1
π/2x1+π/5 = 2kπ,π/2x2+π/5=2kπ±π
x1+2/5 = 4k,x2+2/5=4k±2
x1 = 4k-2/5,x2=4k-2/5±2
|x1- x2| =|±2| = 2
f(x1)-f(x2)=2cos(π/2x1+π/5)-2cos(π/2x2+π/5)= 4
cos(π/2x1+π/5)-cos(π/2x2+π/5) = 2
∵-1 ≤ cos(π/2x1+π/5) ≤ 1,-1 ≤ cos(π/2x2+π/5) ≤ 1
∴ cos(π/2x1+π/5) = 1,cos(π/2x2+π/5) = - 1
π/2x1+π/5 = 2kπ,π/2x2+π/5=2kπ±π
x1+2/5 = 4k,x2+2/5=4k±2
x1 = 4k-2/5,x2=4k-2/5±2
|x1- x2| =|±2| = 2
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