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)=-2f(x2)=2所以πx1/2+π/5=2kπ-π/2πx2/2+π/5=2mπ+π/2x1=4k-7/5x2=4m+3/5|x1-x2|=|4(k-m)-2|k-m是整数所以|x1-x2|最小=2...