Let $$f : (1, 3) \rightarrow R$$, be a function defined by $$f(x) = \frac{x[x]}{1+x^2}$$, where $$[x]$$ denotes the greatest integer $$\le x$$. Then the range of $$f$$, is

We have the function $$f:(1,3)\rightarrow \mathbb R$$ defined by

$$f(x)=\dfrac{x\,[x]}{1+x^{2}},$$

where $$[x]$$ denotes the greatest integer less than or equal to $$x$$. Because the domain is the open interval $$(1,3)$$, the only possible integer values of $$[x]$$ inside this interval are $$1$$ and $$2$$. So we split the domain into two parts.

First part - when $$1<x<2$$ we have $$[x]=1$$. Substituting this in the formula gives

$$f(x)=\dfrac{x\cdot1}{1+x^{2}}=\dfrac{x}{1+x^{2}}\qquad(1<x<2).$$

To find its range we differentiate. Using the quotient rule $$\displaystyle\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^{2}},$$ with $$u=x,\;u'=1,\;v=1+x^{2},\;v' = 2x,$$ we get

$$f'(x)=\dfrac{1(1+x^{2})-x(2x)}{(1+x^{2})^{2}}=\dfrac{1+x^{2}-2x^{2}}{(1+x^{2})^{2}}=\dfrac{1-x^{2}}{(1+x^{2})^{2}}.$$

Inside the interval $$1<x<2$$ we have $$x^{2}>1$$, so $$1-x^{2}<0$$ and hence $$f'(x)<0$$. Thus $$f(x)=\dfrac{x}{1+x^{2}}$$ is strictly decreasing on $$(1,2)$$.

Because it is decreasing, its largest value occurs as $$x\to1^{+}$$ and its smallest value occurs as $$x\to2^{-}$$. Evaluating the limits,

$$\lim_{x\to1^{+}}\dfrac{x}{1+x^{2}}=\dfrac{1}{1+1}=\dfrac12,$$

$$\lim_{x\to2^{-}}\dfrac{x}{1+x^{2}}=\dfrac{2}{1+4}=\dfrac25.$$

Neither endpoint is actually in the open interval $$(1,2)$$, so neither value is attained. Therefore the range of the first part is

$$\left(\dfrac25,\;\dfrac12\right).$$

Second part - when $$2\le x<3$$ we have $$[x]=2$$. So now

$$f(x)=\dfrac{x\cdot2}{1+x^{2}}=\dfrac{2x}{1+x^{2}}\qquad(2\le x<3).$$

Again we differentiate. Put $$u=2x,\;u'=2,\;v=1+x^{2},\;v'=2x,$$ then

$$f'(x)=\dfrac{2(1+x^{2})-2x(2x)}{(1+x^{2})^{2}}=\dfrac{2+2x^{2}-4x^{2}}{(1+x^{2})^{2}}=\dfrac{2-2x^{2}}{(1+x^{2})^{2}}=\dfrac{2(1-x^{2})}{(1+x^{2})^{2}}.$$

On the interval $$2\le x<3$$ we have $$x^{2}>1$$, so again $$1-x^{2}<0$$ and hence $$f'(x)<0$$ throughout. Thus $$f(x)=\dfrac{2x}{1+x^{2}}$$ is also strictly decreasing on $$[2,3)$$.

The largest value of this part is taken at the left end $$x=2$$ (which is in the domain), and the smallest value is approached as $$x\to3^{-}$$. Compute

$$f(2)=\dfrac{2\cdot2}{1+2^{2}}=\dfrac{4}{1+4}=\dfrac45,$$

$$\lim_{x\to3^{-}}\dfrac{2x}{1+x^{2}}=\dfrac{2\cdot3}{1+3^{2}}=\dfrac{6}{10}=\dfrac35.$$

The value $$\dfrac45$$ is included because $$x=2$$ belongs to the domain; the value $$\dfrac35$$ is not included because $$x=3$$ is not in the domain. Hence the range of the second part is

$$\left(\dfrac35,\;\dfrac45\right].$$

Combining both parts we unite their ranges:

$$\left(\dfrac25,\;\dfrac12\right)\;\cup\;\left(\dfrac35,\;\dfrac45\right].$$

This set matches exactly the interval collection given in Option B.

Hence, the correct answer is Option B.