Question

f(x)=(ax^2+1)/(bx+c) (a,b,c属于Z)为奇函数,且f1=(2),f(2)﹤3

Answer 1

1.f(x)为奇函数

则f(-x)=-f(x)

即(ax²+1)/(-bx+c)=-(ax²+1)/(bx+c)

解得c=0

f(1)=(a+1)/(b+c)=(a+1)/b=2

a+1=2b (1)

f(2)=(4a+1)/(2b)=(4a+1)/(a+1)=4-3/(a+1)1

解得-1