Key Concepts and Formulas

• Greatest Integer Function $[x]$: The greatest integer less than or equal to $x$. For example, $[3.7] = 3$ and $[5] = 5$.
• Domain and Range of Functions: The domain is the set of all possible input values ($x$) for a function, and the range is the set of all possible output values ($f(x)$).
• Piecewise Functions: A function defined by multiple sub-functions, each applying to a certain interval of the main function's domain.

Step-by-Step Solution

The function is given by $f(x) = \frac{x[x]}{1 + x^2}$ and the domain is $f : (1, 3) \to \mathbb{R}$.

Step 1: Decompose the Domain based on the Greatest Integer Function. The domain of the function is the open interval $(1, 3)$. We need to identify the values of $x$ within this interval for which $[x]$ remains constant.

• For $1 < x < 2$, the greatest integer less than or equal to $x$ is $[x] = 1$.
• For $2 \le x < 3$, the greatest integer less than or equal to $x$ is $[x] = 2$.

Step 2: Rewrite the Function as a Piecewise Function. Now, we substitute the constant values of $[x]$ into the function definition for each interval.

• For $1 < x < 2$: Here, $[x] = 1$. So, the function becomes: $f(x) = \frac{x(1)}{1 + x^2} = \frac{x}{1 + x^2}$

• For $2 \le x < 3$: Here, $[x] = 2$. So, the function becomes: $f(x) = \frac{x(2)}{1 + x^2} = \frac{2x}{1 + x^2}$

Step 3: Analyze the Range of each Piece of the Piecewise Function.

• Case 1: $1 < x < 2$, where $f(x) = \frac{x}{1 + x^2}$ To find the range of this part, we can analyze the derivative of $g(x) = \frac{x}{1 + x^2}$. $g'(x) = \frac{(1)(1 + x^2) - x(2x)}{(1 + x^2)^2} = \frac{1 + x^2 - 2x^2}{(1 + x^2)^2} = \frac{1 - x^2}{(1 + x^2)^2}$ For $1 < x < 2$, $x^2 > 1$, so $1 - x^2 < 0$. Thus, $g'(x) < 0$ in this interval. This means the function $g(x)$ is decreasing for $1 < x < 2$. We evaluate the function at the boundaries of the interval $(1, 2)$:

  • As $x \to 1^+$, $f(x) \to \frac{1}{1 + 1^2} = \frac{1}{2}$.
  • As $x \to 2^-$, $f(x) \to \frac{2}{1 + 2^2} = \frac{2}{5}$. Since the function is decreasing on $(1, 2)$, the range for this interval is $\left(\frac{2}{5}, \frac{1}{2}\right)$.

• Case 2: $2 \le x < 3$, where $f(x) = \frac{2x}{1 + x^2}$ Let $h(x) = \frac{2x}{1 + x^2}$. We can analyze its derivative. $h'(x) = \frac{(2)(1 + x^2) - (2x)(2x)}{(1 + x^2)^2} = \frac{2 + 2x^2 - 4x^2}{(1 + x^2)^2} = \frac{2 - 2x^2}{(1 + x^2)^2} = \frac{2(1 - x^2)}{(1 + x^2)^2}$ For $2 \le x < 3$, $x^2 \ge 4$. Thus, $1 - x^2 < 0$. This means $h'(x) < 0$ in this interval. The function $h(x)$ is decreasing for $2 \le x < 3$. We evaluate the function at the boundaries of the interval $[2, 3)$:

  • At $x = 2$, $f(2) = \frac{2(2)}{1 + 2^2} = \frac{4}{5}$.
  • As $x \to 3^-$, $f(x) \to \frac{2(3)}{1 + 3^2} = \frac{6}{10} = \frac{3}{5}$. Since the function is decreasing on $[2, 3)$, the range for this interval is $\left(\frac{3}{5}, \frac{4}{5}\right]$. Note that the interval is closed at $x=2$ because the domain includes $x=2$, and open at $x=3$ because the domain does not include $x=3$.

Step 4: Combine the Ranges of the Two Pieces. The overall range of $f(x)$ is the union of the ranges from the two intervals. Range of $f$ = Range from Case 1 $\cup$ Range from Case 2 Range of $f$ = $\left(\frac{2}{5}, \frac{1}{2}\right) \cup \left(\frac{3}{5}, \frac{4}{5}\right]$

Common Mistakes & Tips

• Incorrectly handling interval endpoints: Pay close attention to whether the domain intervals are open or closed, as this affects whether the endpoints are included in the range. For example, if the domain is $(a, b)$, the function values at $a$ and $b$ are not attained. If the domain is $[a, b]$, the function values at $a$ and $b$ are attained.
• Assuming monotonicity without checking: Always verify if the function is increasing or decreasing on the interval by checking the sign of its derivative.
• Confusing the domain of $[x]$ with the domain of $f(x)$: The definition of $[x]$ partitions the domain of $f(x)$, but the domain of $f(x)$ itself is what you should work with.

Summary

We analyzed the function $f(x) = \frac{x[x]}{1 + x^2}$ over its domain $(1, 3)$. By considering the intervals where $[x]$ is constant, we broke the function into two pieces: $f(x) = \frac{x}{1+x^2}$ for $1 < x < 2$ and $f(x) = \frac{2x}{1+x^2}$ for $2 \le x < 3$. We then determined the range of each piece by examining the function's behavior (monotonicity) and its values at the interval boundaries. The union of these ranges gives the complete range of $f(x)$. For $1 < x < 2$, the range is $\left(\frac{2}{5}, \frac{1}{2}\right)$. For $2 \le x < 3$, the range is $\left(\frac{3}{5}, \frac{4}{5}\right]$. Combining these, the total range is $\left(\frac{2}{5}, \frac{1}{2}\right) \cup \left(\frac{3}{5}, \frac{4}{5}\right]$.

The final answer is \boxed{{\left( {{2 \over 5},{1 \over 2}} \right) \cup \left( {{3 \over 5},{4 \over 5}} \right]}}. This corresponds to option (A).