Given the function f (x) = (12x-1 + 12) SiNx & nbsp; (- π 2 < x < π 2 and X ≠ 0) (1) judge the parity of F (x); (2) prove that f (x) > 0

Given the function f (x) = (12x-1 + 12) SiNx & nbsp; (- π 2 < x < π 2 and X ≠ 0) (1) judge the parity of F (x); (2) prove that f (x) > 0

(1) ∵ f (- x) = (12-x-1 + 12) sin (- x) = - (112x-1 + 12) SiN x = - (2x1-2x + 12) SiN x = (2x2x-1-12) SiN x = [(1 + 12x-1) - 12] SiN x = (12x-1 + 12) SiN x = f (x), f (x) is even function. (2) when 0 ＜ x ＜ π 2, 2x ＞ 1, & nbsp; & nbsp; 2x-1 ＞ 0 & nbsp; & nbsp;, SiNx ＞ 0 & nbsp; & nbsp;, & nbsp; & nbsp; & nbsp; When - π 2 ＜ x ＜ 0, ∵ f (x) is an even function. From the above formula, we know that f (x) ＞ 0, so f (x) ＞ 0 holds. In conclusion, f (x) ＞ 0