Let f (x) = (1+b 2 ) x 2 + 2bx + 1 and m (b ) the minimum value of f (x) for a given b…
Mathematics · JEE Main · NTA Exams — Limit, Continuity and Differentiability
Let f (x) = (1+b2) x2 + 2bx + 1 and m (b ) the minimum value of f (x) for a given b. As b varies, the range of m (b ) is
- [0, 1]
- (0, 1/2]
- \(\left[\frac{1}{2}, 1\right]\)
- (0, 1]
Answer
(D) (0, 1]
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