Question

Let $f\left(x\right)=(1+b^2)x^2+2bx+1$ and let $m\left(b\right)$ be the minimum value of $f\left(x\right).$ As $b$ varies, the range of $m\left(b\right)$ is

Answer 1

Given, $f\left(x\right)$$=(1+b^2)x^2+2bx+1=(1+b^2)(x+\frac{b}{1+b^2}{)}^2+1-\frac{b^2}{1+b^2}=(1+b^2)(x+\frac{b}{1+b^2}{)}^2+\frac{1}{1+b^2}$

When $x=-\frac{b}{1+b^2},f\left(x\right)$ is minimum.
$\therefore m\left(b\right)=\frac{1}{1+b^2}$
Range of $m\left(b\right)\equiv (0,1]$