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.