已知以T=4为周期的函数f(x)=

x∈(4k+1,4k+3]时, f(x)=f(x-4k)=1-|x-4k-2|∈(0,1]∴3f(x)∈(0,3]∴3f(x)=x仅在(1,3]上有根x∈(1,2]时, f(x)=1-(2-x)=x-1, 3f(x)=3(x-1)=3x-3=x, x=3/2有一解x∈(2,3]时, f(x)=1-(x-2)=3-x, 3f(x)=3(3-x)=9-3x=x. x=9/4有一解∴x∈(4k-1,4k+1]上有3解,此时f(x)=f(x-4k)=m×√(1-(x-4k)?)∈(0,m]即3f(x)=x在(0,3m]∩(4k-1,4k+1]上∴在(0,1],(3,4]和(4,5]上各有一根,在(7,8]上无根,其中最大值为m,在x=4和x=8处取得稍微画下图可知 x=4时, 3f(x)=3m>x=4, m>4/3x∈(7,8]时, 3f(x)=3f(x-8)=3m√(1-(x-8)?)=x, 9m?[1-(x-8)?]=x?, (1+9m?)x?-16×9m?x+63×9m?=0在(7,8]上无解设g(x)=(1+9m?)x?-16×9m?x+63×9m?, 对称轴x=8×9m?/(1+9m?)∈(0,8)g(7)=49>0,g(9)=81>0若8×9m?/(1+9m?)>7, 即m?>7/9, 此时△=(16×9m?)?-4×63×9m?(1+9m?)