Question

If the minimum value of f(x) = (1 + b2) x2 + 2bx + 1 is m(b), then the maximum value of m(b) is

(A) 0

(B) −1

(C) 2

(D) 1

Answer 1

Correct option**(D) 1**

Therefore, f(x) has minimum value at x = -b/1 + b2

Minimum value of + f(x), is given by

Clearly, 0 < m (b) ≤ 1. Since b2 ≥ 0 maximum value of m(b) is 1.