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