Let f(x) = (1 + b^2)x^2 + 2bx + 1 and m(b) be | vedclass

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(D) The given function is a quadratic expression $f(x) = (1 + b^2)x^2 + 2bx + 1$. Since the coefficient of $x^2$ is $(1 + b^2) > 0$ for all real $b$,t

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$(0, 1]$

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(D) The given function is a quadratic expression $f(x) = (1 + b^2)x^2 + 2bx + 1$. Since the coefficient of $x^2$ is $(1 + b^2) > 0$ for all real $b$,the function has a minimum value. The minimum value of a quadratic $ax^2 + bx + c$ is given by $-\frac{D}{4a}$,where $D = b^2 - 4ac$. Here,$a = (1 + b^2)$,$B = 2b$,and $C = 1$. $D = (2b)^2 - 4(1 + b^2)(1) = 4b^2 - 4 - 4b^2 = -4$. Thus,$m(b) = -\frac{-4}{4(1 + b^2)} = \frac{4}{4(1 + b^2)} = \frac{1}{1 + b^2}$. Since $b^2 \ge 0$,we have $1 + b^2 \ge 1$. Therefore,$0 The range of $m(b)$ is $(0, 1]$.

Column-$I$	Column-$II$
$A$. Imaginary roots	$P$. $a \in (-1, 4)$
$B$. One root less than $3$ and other greater than $3$	$Q$. $a \in (-\infty, -1)$
$C$. One root less than $1$ and other greater than $3$	$R$. $a \in (-\infty, -4/3)$

Let $f(x) = (1 + b^2)x^2 + 2bx + 1$ and $m(b)$ be the minimum value of $f(x)$. If $b$ can take any real value,what is the range of $m(b)$?

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