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_数学解答

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

人气：421 ℃　时间：2019-08-18 17:25:51

解答

f(x)=2cos(π/2x+π/5)f(x1)-f(x2)=2cos(π/2x1+π/5)-2cos(π/2x2+π/5)= 4cos(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+...