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

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

Answer 1

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

$f\left(x\right)=(1+b^2)x^2+2bx+1$

Coefficient of $x^2$ is $(1+b^2)>0$

Minimum value of $f\left(x\right)$ is $\frac{-D}{4a}$

$\therefore m\left(b\right)=-\frac{\{4b^2-4(1+b^2\left)\right\}}{4(1+b^2)}$
$\Rightarrow m\left(b\right)=\frac{1}{1+b^2}$
$\because 1\le 1+b^2<\infty \forall b\in R\Rightarrow 0<\frac{1}{1+b^2}\le 1$
$\therefore$ Range of $m\left(b\right)$ is $(0,1]$