Question

Let $f\left(x\right)=(1+b^2)x^2+2bx+1$ and $m\left(b\right)$ be the minimum value of $f\left(x\right)$. If $b$ can assume different values, then range of $m\left(b\right)$ is equal to

• A. $[0,1]$
• B. $(0,1/2]$
• C. $[1/2,1]$
• D. $(0,1]$

Answer 1

Answer: (D)

$(0,1]$

$f\left(x\right)=(1+b^2)x^2+2bx+1$
It is quadratic expression with coefficient of $x^2$ as $(1+b^2)>0$
Whose minimum value is $-\frac{D}{4a}$
$\therefore$ $m\left(b\right)=-\frac{\{4b^2-4(1+b^2\left)\right\}}{4.(1+b^2)}$
$m\left(b\right)=\frac{1}{1+b^2}$
For range of $m\left(b\right)$
Let $y=\frac{1}{1+b^2}$
$\Rightarrow b^2=\frac{1-y}{y}$
Since, $b^2\ge 0$
$\Rightarrow \frac{1-y}{y}\ge 0$
$\Rightarrow y>0and1-y\ge 0$
$\Rightarrow y>0andy\le 1$
Hence, $0<\frac{1}{1+b^2}\le 1$
$\Rightarrow$ range of $m\left(b\right)$ is $(0,1]$