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If x2 + 2ax + b ≥ c, ∀x, ∈ R, then

(A)  b - c ≥ a2

(B) c − a ≥ b2

(C) c - a ≥ b2 

(D) None of these

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Best answer

Correct option  (A) b - c ≥ a2

x2 + 2ax + b ≥ c, ∀x, ∈ R,

⇒ x2 + 2ax + b - c ≥ 0

D = 4a2 - 4(b - c) ≤ 0

⇒ a2  ≤ b - c

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