f(x)=(a x²+1)/(bx+c)为奇函数,
则有f(-x)=-f(x).
(a x²+1)/(-bx+c)=- (a x²+1)/(bx+c)
-bx+c=-bx-c,c=0.
由f(1)=2得,(a +1)/b=2.2b=a+1.
由f(2)