a f(x)> 0 B f(x)& lt； 0 C f(x)> x D f(x)& lt； x

Answer a

solve

Write the original inequality as 1

Let x=0, the inequality can be changed to f (0) >: 0, and if x=0 in the option, then b and d can be excluded.

Then bring 1/x into the inequality, 2f (1/x)+(1/x) * (1/x)' * f' (1/x) > (1x).

Simplified as 2f (1/x)-(1/x) 3 * f' (1/x) > (1/x) 2.

Replace all 1/x:X2 with 2f(x)-x3 * f '(x of x >) as type 2.

Let x= 1 bring in 1, 2, and you get

2f (1)+f' (1) >1is recorded as type 3.

2f (1)-f' (1) >1is recorded as type 4.

Formula 3 plus formula 4 gives that 4f( 1)>2 is f( 1)> 1/2 >0.

You can push out a, but you can't push out c, so choose a.