If $f(x)=(1+b^2)x^2+2bx+1$ and let m(b) be the minimum value of f(x). Then as b varies, the range of m(b) must be

[1 , 0] $\left(0,\frac{1}{2}\right]$ $\left[\frac{1}{2},1\right]$ (0, 1]

(0, 1]

$f(x)=(1+b^2)x^2+2bx+1$ Minimum value of $f(x)=1-\frac{4b^2}{4(1+b^2)}=\frac{1}{(1+b^2)}$ $⇒m(b)=\frac{1}{1+b^2}⇒m(b)∈(0, 1];∵b∈R$