Question

Let $f\left(x\right)=\left(1+b^2\right)x^2+2bx+1$ 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. $\left[0,1\right]$
• B. $(0,\frac{1}{2}]$
• C. $\left[\frac{1}{2},1\right]$
• D. $(0,1]$

Answer 1

The correct option is D

$(0,1]$

Explanation for the correct option:

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

This is a quadratic equation, so its minimum value is$\frac{-D}{4a}$, $D$being the discriminant

$$\begin{gathered} m\left(b\right)=\frac{-\left(4b^2-4\left(1+b^2\right)\right)}{4\left(1+b^2\right)} \\ =\frac{1}{1+b^2} \end{gathered}$$

Finding range of $m\left(b\right)$

$$\begin{gathered} y=\frac{1}{1+b^2} \\ \Rightarrow b^2=\frac{1-y}{y} \end{gathered}$$

since $b^2\ge 0$

$\frac{1-y}{y}\ge 0$

This means $1-y\ge 0,y>0$

That is

$$\begin{gathered} y>0,y\le 1 \\ y\in (0,1] \end{gathered}$$

Hence, option (D) is the correct answer