若命题“彐x∈R,2x^2-3ax+9

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若命题“彐x∈R,2x^2-3ax+9

若命题“彐x∈R,2x^2-3ax+9
若命题“彐x∈R,2x^2-3ax+9

若命题“彐x∈R,2x^2-3ax+9
由2x^2-3ax+9<0,得:2[x^2-(3a/2)x+(3a/4)^2]-2×(3a/4)^2+9<0,
∴2(x-3a/4)^2-9a^2/8+9<0.
显然,当-9a^2/8+9≧0时,2x^2-3ax+9<0就是假命题.
由-9a^2/8+9≧0,得:a^2≦8,∴-2√2≦a≦2√2.
∴满足条件的a的取值范围是[-2√2,2√2]