Let `f(x) = (x[x])/(x^2+1) : (1, 3) rarr R` then range of f(x) is (where [ . ] denotes greatest integer function

Options

A. `((3)/(5),(4)/(5))`
B. `((2)/(5),(3)/(5)]uu((3)/(4),(4)/(5))`
C. `((2)/(5),(1)/(2))uu((3)/(5),(4)/(5]]`
D. `((2)/(5),(4)/(5)]`

Answer

To find the range of the function \( f(x) = \frac{x \lfloor x \rfloor}{x^2 + 1} \) for \( x \) in the interval \( [1, 3] \), we will analyze the function by breaking it down into two parts based on the greatest integer function, \( \lfloor x \rfloor \).

### Step 1: Identify the intervals based on the greatest integer function
The greatest integer function \( \lfloor x \rfloor \) takes different constant values in the interval \( [1, 3] \):
- For \( x \in [1, 2) \), \( \lfloor x \rfloor = 1 \)
- For \( x \in [2, 3) \), \( \lfloor x \rfloor = 2 \)

### Step 2: Define the function in each interval
1. For \( x \in [1, 2) \):
\[ f(x) = \frac{x \cdot 1}{x^2 + 1} = \frac{x}{x^2 + 1} \]

2. For \( x \in [2, 3] \):
\[ f(x) = \frac{x \cdot 2}{x^2 + 1} = \frac{2x}{x^2 + 1} \]

### Step 3: Analyze \( f(x) \) in each interval

#### Interval 1: \( x \in [1, 2) \)
We need to find the range of \( f(x) = \frac{x}{x^2 + 1} \) in this interval.

- Calculate \( f(1) \):
\[ f(1) = \frac{1}{1^2 + 1} = \frac{1}{2} \]

- Calculate \( f(2) \) (as \( x \) approaches 2 from the left):
\[ f(2) = \frac{2}{2^2 + 1} = \frac{2}{5} \]

- Since \( f(x) \) is continuous and differentiable in \( [1, 2) \), we can find the derivative to check if it is increasing or decreasing.

The derivative \( f'(x) \) can be computed using the quotient rule:
\[ f'(x) = \frac{(x^2 + 1)(1) - x(2x)}{(x^2 + 1)^2} = \frac{1 - x^2}{(x^2 + 1)^2} \]

- For \( x \in [1, 2) \), \( 1 - x^2 < 0 \) (since \( x^2 \) is greater than 1), indicating that \( f(x) \) is decreasing in this interval.

Thus, the range in this interval is:
\[ \text{Range} = \left[ \frac{2}{5}, \frac{1}{2} \right) \]

#### Interval 2: \( x \in [2, 3] \)
Now we analyze \( f(x) = \frac{2x}{x^2 + 1} \).

- Calculate \( f(2) \):
\[ f(2) = \frac{2 \cdot 2}{2^2 + 1} = \frac{4}{5} \]

- Calculate \( f(3) \):
\[ f(3) = \frac{2 \cdot 3}{3^2 + 1} = \frac{6}{10} = \frac{3}{5} \]

- Again, we find the derivative:
\[ f'(x) = \frac{(x^2 + 1)(2) - 2x(2x)}{(x^2 + 1)^2} = \frac{2 - 2x^2}{(x^2 + 1)^2} \]

- For \( x \in [2, 3] \), \( 2 - 2x^2 < 0 \) (since \( x^2 \) is greater than 2), indicating that \( f(x) \) is decreasing in this interval.

Thus, the range in this interval is:
\[ \text{Range} = \left( \frac{4}{5}, \frac{3}{5} \right] \]

### Step 4: Combine the ranges
Combining the ranges from both intervals, we have:
\[ \text{Total Range} = \left[ \frac{2}{5}, \frac{1}{2} \right) \cup \left( \frac{3}{5}, \frac{4}{5} \right] \]

### Final Answer
The range of \( f(x) \) is:
\[ \left[ \frac{2}{5}, \frac{1}{2} \right) \cup \left( \frac{3}{5}, \frac{4}{5} \right] \]