(1)
f(x)+f(1/x)
=arctan(1/x) +arctan(x)
=π/2
(2)
f(x)=x
arctan(1/x) =x
let
g(x) = arctan(1/x) -x
g'(x)
=(-1/x^2)/(1+1/x^2) -1
=-1/(1+x^2) -1
=(-2-x^2)/(1+x^2)
<0
g(4/π)= 1- π/4 >0
lim(x->+∞) g(x)
=lim(x->+∞) arctan(1/x) -x
->-∞
=> g(x) = 0， 只有1个实数根在(0,+∞)
=> f(x) = x， 只有1个实数根在(0,+∞)
