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 in $x$ of the form $f(x) = Ax^2 + Bx + C$,where $A = 1 + b^2$,$B = 2b$,and $C = 1$.Since $A = 1 + b^2 > 0$ for a

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$(0, 1]$

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(D) The given function is a quadratic in $x$ of the form $f(x) = Ax^2 + Bx + C$,where $A = 1 + b^2$,$B = 2b$,and $C = 1$.Since $A = 1 + b^2 > 0$ for all real $b$,the function has a minimum value at $x = -\frac{B}{2A}$.The minimum value $m(b)$ is given by $f(-\frac{B}{2A}) = C - \frac{B^2}{4A}$.Substituting the values: $m(b) = 1 - \frac{(2b)^2}{4(1 + b^2)} = 1 - \frac{4b^2}{4(1 + b^2)} = 1 - \frac{b^2}{1 + b^2}$.Simplifying this,we get $m(b) = \frac{1 + b^2 - b^2}{1 + b^2} = \frac{1}{1 + b^2}$.Since $b^2 \ge 0$,we have $1 + b^2 \ge 1$,which implies $0 Thus,the range of $m(b)$ is $(0, 1]$.

Let $f(x) = (1 + b^2)x^2 + 2bx + 1$ and $m(b)$ be the minimum value of $f(x)$ for a given $b$. As $b$ varies,the range of $m(b)$ is

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