Let \[f(x)=(1+{{b}^{2}}){{x}^{2}}+2bx+1\] and \[m(b)\] the minimum value of \[f(x)\]for a given b. As b varies, the range of m is [IIT Screening 2001]

\[\left( 0,\ \frac{1}{2} \right]\]

\[\left[ \frac{1}{2},\ 1 \right]\]

Let \[f(x)=(1+{{b}^{2}}){{x}^{2}}+2bx+1\] and \[m(b)\] the minimum val

1.	Let \[f(x)=(1+{{b}^{2}}){{x}^{2}}+2bx+1\] and \[m(b)\] the minimum value of \[f(x)\]for a given b. As b varies, the range of m is [IIT Screening 2001]
A.	[0, 1]
B.	\[\left( 0,\ \frac{1}{2} \right]\]
C.	\[\left[ \frac{1}{2},\ 1 \right]\]
D.	\[(0,\ 1]\]
Answer» E.

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Let \[f(x)=(1+{{b}^{2}}){{x}^{2}}+2bx+1\] and \[m(b)\] the minimum value of \[f(x)\]for a given b. As b varies, the range of m is [IIT Screening 2001]

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