It is known that y = f (x) is an even function. When x > 0, f (x) = (x-1) 2. If n ≤ f (x) ≤ m is constant when x ∈ [- 2, - 12], then the minimum value of M-N is () A. 13B. 12C. 34D. 1

It is known that y = f (x) is an even function. When x > 0, f (x) = (x-1) 2. If n ≤ f (x) ≤ m is constant when x ∈ [- 2, - 12], then the minimum value of M-N is () A. 13B. 12C. 34D. 1

Let x ＜ 0, then - x ＞ 0, f (- x) = (- x-1) 2 = (x + 1) 2, the original function is even, so f (x) = f (- x) = (x + 1) 2, that is, f (x) = (x + 1) 2 when x ＜ 0. The maximum value of the function on [- 2, - 12] is 1, and the minimum value is 0. According to the meaning of the problem, n ≤ f (x) ≤ m, it is always true, n ≥ 0, m ≤ 1, that is, M-N ≥ 1