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If the maximum and minimum values of F (x) = x3-3x-a in the interval [0, 3] are m and N respectively, then the value of M-N is () A. 2B. 4C. 18D. 20
When 0 ≤ x ＜ 1, f ′ (x) ＜ 0; when 1 ≤ x ≤ 3, f ′ (x) ＞ 0. Then f (1) is the minimum, then n = f (1) and f (0) = - A, f (3) = 18-a, and F (3) ＞ f (0), the maximum is f (3), that is, M = f (3), so M-N = f (3
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