Solution: because 2f (x)+xf' (x) > x 2. ...................................................................................................................................................

When (1)x= 0, substitute ① to get: f (0) > 0.

(2) when x > 0, ① both sides are multiplied by x: 2)x>0 (x)+x 2f' (x) > x 3, i.e.

[x 2f (x)]' > x 3 > 0, so the function y = x 2f (x) is the increasing function on R+, and x > 0,

So: x 2f (x) > 0 2f (0) = 0, so f (x) > 0.

(3) when x < 0, ① both sides are multiplied by x: 2xf (x)+x 2f' (x) < x 3, that is

[x 2f (x)]' < x 3 < 0, so the function y = x 2f (x) is the increasing function on R- and x < 0,

So: X2f (x) > 0 2F (x) > 0) = 0, so there is also F (x) > 0.

To sum up, when x∈R, there is always f (x) > 0, so choose a.

—————————————————————————————————————————————————————————

Multiple choice questions should be done like this! If f (0) > 0, options b and c are excluded.

Obviously, when f (x) = x 2+a (a > 0), it is known that the condition 2f (x)+xf' (x) > x 2 holds, but

F (x) > x is not necessarily true, so d is also wrong, so choose a.