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Let m be a set of functions f (x) which satisfy the following properties: in the domain of definition, in x0, f (x0 + 1) = f (x0) + F (1) holds. The following functions are known: ① f (x) = 1x; ② f (x) = 2X; ③ f (x) = LG (x2 + 2); ④ f (x) = cos π x, where the function belonging to the set M is () A. ①③B. ②③C. ③④D. ②④
① If there is x, such that f (x + 1) = f (x) + F (1), then 1x + 1 = LX + 1, that is, X2 + X + 1 = 0, ∵ △ = 1-4 = - 3 ＜ 0, so the equation has no solution. That is, in F (x) = ∉ M 2, there is x = 1, such that f (x + 1) = 2x + 1 = f (x) + F (1) = 2x + 2, that is, f (x) = 2x ∈ m; in 3, if there is x, f (x + 1
RELATED INFORMATIONS
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