Given the function f (x), for X ∈ R, f (4-x) = f (x). If f (x) has exactly four unequal zeros x1, X2, X3, x4, then X1 + x2 + X3 + X4=

From F (4-x) = f (x), we can get f (x1) = f (4-x1). In this case, let 4-x1 = X2, and X1 is not equal to X2, then X1 + x2 = 4, similarly, X3 + X4 = 4, the total sum is equal to 8. Of course, X1 cannot be equal to 4-x1, otherwise X1 = 2, and the principle of symmetry, then X1 = 2 will be the axis of symmetry, resulting in only three zeros (odd), the answer is 8

Given the function f (x), for X ∈ R, f (4-x) = f (x). If f (x) has exactly four unequal zeros x1, X2, X3, x4, then X1 + x2 + X3 + X4=

Given the function f (x), for X ∈ R, f (4-x) = f (x). If f (x) has exactly four unequal zeros x1, X2, X3, x4, then X1 + x2 + X3 + X4=

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