Letf(x)=(1+bº) x2 +2bx +1 and m(b) ...
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Letf(x)=(1+bº) x2 +2bx +1 and m(b) the minimum value of f(x) for a given b. As b váries, the range of m (b) is (a) [0, 1] (b) (0,1/2] (d) (0, 1)

JEE/Engineering Exams
Maths
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( f(x)=left(1+b^{2}right) x^{2}+2 b x+1 ) ( f^{prime}(x)=2left(1+b^{2}right) x+2 b=0 ) ( x=frac{-2 b}{2left(1+b^{2}right)}=frac{-b}{1+b^{2}} ) ( m=frac{left(1+b^{2}right) b^{2}-frac{2 b^{2}}{1+b^{2}}+1}{left(1+b^{2}right)^{2}} frac{b^{2}-2 b^{2}+1+b^{2}}{1+b^{2}}=frac{1}{1+b^{2}} ) ( =frac{1}{1+b^{2}} ) ( 13 leqslant b^{2}frac{1}{b^{2}+1}>0 ) ( m(b) in(0,1] ) ( begin{aligned}=) & text { option. }(d) end{aligned} )
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