函数奇偶性的题要过程 急急急急急加分！！！！

已知函数f[x]=｛a乘以x的平方+1｝除以bx+c ,, [a.b.c属于z】是奇函数，又f[1]=2.f[2]《3,求 a，b，c 的值

f(x) = (ax² + 1)/(bx + c)

由f(-x) = -f(x)得

[a(-x)² + 1]/[b×(-x) + c] = -(ax² + 1)/(bx + c)

-bx + c = -bx - c

c = 0 …………………………<1>

由 f(1) = 2得

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

c = 0代入得

a = 2b - 1 ……………………<2>

由f(2) ≤ 3得

(ax² + 1)/(bx + c) ≤ 3

(4a + 1)(2b + c) ≤ 3

4a + 1 ≤ 6b +3c

c = 0和a = 2b - 1代入得

b ≤ 3/2

由于b ∈ Z，故

b = 1

代入<2>得

a = 1

即

a = 1

b = 1

c = 0

 

也即

f(x) = (x² + 1)/x