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

问题描述:

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

由f(x)的式子可知它的周期是4,f(x1)-f(x2)=4可知x1,x2最小相差半个周期,所以答案应该是2、、、应该是这么做的,嘻嘻、、

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+...