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Let f (x) = x2 + BX + 1, and f (- 1) = f (3), then the solution set of F (x) > 0 is () A. （-∞，-1）∪（3，+∞）B. RC. {x∈R|x≠1}D. {x∈R|x=1}
∵ f (x) = x2 + BX + 1, and f (- 1) = f (3), ∵ B2 = − 1 + 321 − B + 1 = 9 + 3B + 1, the solution is b = - 2. ∵ f (x) = x2-2x + 1 = (x-1) 2, the solution set of ∵ f (x) > 0 is {x | x ≠ 1}
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