FunctionHard

Question

Let f(x) = (1 + b2)x2 + 2bx + 1 and let m (b) the minimum value of f(x) As b varies, the renge of m (b) is

Options

A.[0, 1]
B.
C.
D.(0,1]

Solution

Given, f(x) = (1 + b2)x2 + 2bx + 1
= (1 + b2)
m (b) = minimum value of f(x)is positive and
m (b) varies from 1 to 0, so range = (0,1]

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