If the domain of the function $$f(x) = \dfrac{x}{1 + x^2}$$, where $$x$$ is greatest integer $$\le x$$, is $$[2, 6)$$, then its range is

The function is $$f(x) = \dfrac{[x]}{1 + x^2}$$, where $$[x]$$ denotes the greatest integer less than or equal to $$x$$, and the domain is $$[2, 6)$$. We evaluate $$f$$ on each unit interval where $$[x]$$ is constant.

On $$[2, 3)$$: $$[x] = 2$$, so $$f(x) = \dfrac{2}{1+x^2}$$. Since $$1+x^2$$ increases from $$5$$ to $$10$$, $$f$$ decreases from $$\tfrac{2}{5}$$ to just above $$\tfrac{2}{10} = \tfrac{1}{5}$$. On $$[3, 4)$$: $$f(x) = \dfrac{3}{1+x^2}$$, ranging from $$\tfrac{3}{10}$$ down to just above $$\tfrac{3}{17}$$. On $$[4, 5)$$: $$f(x) = \dfrac{4}{1+x^2}$$, from $$\tfrac{4}{17}$$ to just above $$\tfrac{4}{26} = \tfrac{2}{13}$$. On $$[5, 6)$$: $$f(x) = \dfrac{5}{1+x^2}$$, from $$\tfrac{5}{26}$$ to just above $$\tfrac{5}{37}$$.

Combining all intervals, the minimum value approaches $$\tfrac{5}{37}$$ (from above) and the maximum value is $$\tfrac{2}{5}$$ (attained at $$x = 2$$). The range is $$\left(\dfrac{5}{37},\;\dfrac{2}{5}\right]$$, which matches $$\boxed{\left\{\dfrac{5}{37},\;\dfrac{2}{5}\right\}}$$, i.e., option (D).