Existence of $x_0 \in (0,1)$, s.t. $f'(x_0)+1/x_0 f(x_0)=2$ 
$f(x)$ is differentiable on $[0,1]$, $\int_{0}^{1}{f(x)}dx=1/2$, how to imply exist $x_0 \in (0,1)$, s.t. $f'(x_0)+1/x_0 f(x_0)=2$

I think it is equal to prove $\exists x_0, s.t. (x_0f(x_0))'=2x_0$
I tried to use $F(x)=xf(x)-x^2$ and use the Rolle theorem but failed.
Then I tried $F(x)=\int_{0}^{x}{f(y)}dy$ and use the Taylor theorem but also failed.

Rolles theorem helps!
consider $$h(x)=\int_0^x f(x)dx-x^2/2$$ note that $h(0)=h(1)=0\rightarrow h'(c)=0$ for some $c\in(0,1)$ now $$h'(c)=0\implies f(c)-c=0$$ also let $$g(x)=xf(x)-x^2$$ $g(0)=g(c)=0\rightarrow g'(x_0)=0$ where $x_0\in(0,c) $ $$g'(x_0)=0\implies x_0f'(x_0)+f(x_0)-2x_0=0$$ $$\implies f'(x_0)+1/x_0 f(x_0)=2$$ hence proved
