Given that the function f (x) = x2 + BX + 1 is an even function on R, then the real number B= The solution set of inequality f (x-1) ＜ x | is ．

∵ function f (x) = x2 + BX + 1 is a even function on R, f (- x) = f (x) holds for any x ∈ R, that is (- x) 2-bx + 1 = x2 + BX + 1, the comparison coefficient is b = 0; therefore, f (x) = x2 + 1, f (x-1) = (x-1) 2 + 1 = x2-2x + 2, the inequality f (x-1) ＜| x | that is: x2-2x + 2 ＜| x | is reduced to X |