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If the definition field of function f (x) = 1ax2 + 4ax + 3 is r, then the value range of real number a is () A. [0，34）B. （0，34）C. （34，+∞）D. （-∞，0）
∵ f (x) = 1ax2 + 4ax + 3 is defined as R, ∵ AX2 + 4ax + 3 ≠ 0. If a = 0, the inequality is equivalent to 3 ≠ 0. If a ≠ 0, the inequality is equivalent to the discriminant △ = 16a2-12a ＜ 0, that is, 4a2-3a ＜ 0, and the solution is 0 ＜ a ＜ 34. To sum up, a ∈ [0, 34), so: a
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