Q : If ax^{2} + bx + 6 = 0 does not have two distinct real roots where a, b ∈ R, then the least value of 3a + b is

(A) 4

(B) -1

(C) 1

(D) -2

Ans: (D)

Sol. D ≤ 0 ⇒ f(x) ≥ 0 or, f(x) ≤ 0 for all real x, but f(0) = 6 > 0

⇒ f(x) ≥ 0 ∀ x ∈ R. In particular f(3) ≥ 0 ⇒ 9a + 3b + 6 ≥ 0

⇒ 3a + b ≥ – 2.

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