Let f(x)=(1+b2)x2+2bx+1 n let m(b) be the minimum value of f(x).As b varies,the range of m(b) isa.[0,1]b.[0,1/2]c.[1/2,1]d.(0,1].

f'(x) = (1+b2).2.x + 2b = 0

f'(x) = 0 implies x = -b/(1+b2)

f''(x) = (1+b2).2 > 0 means f(x) has min. at above given x

on solving for min value for f(x) at x we have

m(b) = 1/(1+b2)

its range is [0,1]

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regards

Ramesh