Let $$f\left( x \right) = \frac{x}{{1 + {x^2}}}$$   and $$g\left( x \right) = \frac{{{e^{ - x}}}}{{1 + \left[ x \right]}},$$    where $$\left[ x \right]$$ is the greatest integer less than or equal to $$x.$$ Then,

Solution :
$$\eqalign{ & D\left( f \right) = R;\,D\left( g \right) = R - \left[ { - 1,\,0} \right) \cr & \therefore \,\,D\left( {f + g} \right) \cr & = D\left( f \right) \cap D\left( g \right) \cr & = R \cap \left( {R - \left[ { - 1,\,0} \right)} \right) \cr & = R \cap \left[ { - 1,\,0} \right) \cr & R\left( f \right) = \left[ { - \frac{1}{2},\,\frac{1}{2}} \right];\,\,R\left( g \right) = R - \left\{ 0 \right\} \cr & \therefore \,\,R\left( f \right) \cap R\left( g \right) \cr & = \left[ { - \frac{1}{2},\,\frac{1}{2}} \right] \cap \left( {R - \left\{ 0 \right\}} \right) \cr & = \left[ { - \frac{1}{2},\,\frac{1}{2}} \right] - \left\{ 0 \right\} \cr} $$