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 $b$ caries, 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 $(0,1]$

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

$\because 1+b^2>$ so upward parabola.

$minf\left(x\right)=m\left(b\right)=-\frac{D}{4a}$

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

for $1+b^2\ge 1$ and $\frac{1}{1+b^2}>0$

$\frac{1}{1+b^2}\le 1$

So $0<\frac{1}{1+b^2}\le 1$

$\Rightarrow 0<m\left(b\right)\le 1$