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. $\left[\frac{1}{2},1\right]$
• D. $(0,1]$

Answer 1

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

$Weknowthatminimumvalueoff\left(x\right)=a{x}^2+bx+cism\left(b\right)=\frac{4ac-b^2}{4a}\therefore forf\left(x\right)=(1+b^2)x^2+2bx+1Wehave,a⟶1+b^{2}b⟶2bc⟶1\therefore m\left(b\right)=\frac{4(1+b^2)-4b^2}{4(1+b^2)}=\frac{1}{(1+b^2)}Now\frac{d}{db}\left\{m\right(b\left)\right\}=\frac{-2b}{{(1+b^2)}^2}=0forextremevalue\Rightarrow b=0\therefore m\left(b\right)=\frac{1}{1+0}=1m\left(b\right)=\frac{1}{(1+b^2)}isapositivequantity.Soitsminimumvalueiszero.\therefore Therangeofm\left(b\right)is(0,1]Ans-OptionD.$