Question

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

• A. $[0,1]$
• B. $(0,\frac{1}{2}]$
• C. $[\frac{1}{2},1]$
• D. $(0,1]$

Answer 1

The correct option is D $(0,1]$

$f\left(x\right)=(1+b^2)x^2+2bx+\frac{b^2}{(1+b^2)}-\frac{b^2}{1+b^2}+1=(1+b^2)(x+\frac{b}{1+b^2}{)}^2+\frac{1}{1+b^2}\ge \frac{1}{1+b^2}$
$\therefore m\left(b\right)=\frac{1}{1+b^2}$,so range of m(b) = (0, 1].