∵定义域为R,∴b﹥0
f′(x)=(ax²+ab-2ax²)/(x²+b)²=0∴ab-ax²=0∵x=1处取得极值2∴ab-a=0∴a≠0b=1
∵f(1)=a/(1+1)=2∴a=4,∴f(x)=4x/(x²+1)
(1)|f(x2)-f(x1)|≤|f(x2)|+|f(x1)=|4x2/(x2²+1)|+|4x1/(x1²+1）|
∵x2²+1≥2|x2|, x1²+1≥2|x1|
∴|4x2/(x2²+1)|≤4|x2|/2|x2|=2,|4x1/(x1²+1)|≤4|x1|/2|x1|=2
∴|f(x2)-f(x1)|≤2+2=4
(2)