Let `f (x) =(2 |x| -1)/(x-3)`
Range of `f (x):`

Options

A. `R- {3}`
B. `(-oo, (1)/(3) ] uu (2,oo)`
C. `(-2, (1)/(3)]` uu (2,oo)`
D. R

Answer

To find the range of the function \( f(x) = \frac{2|x| - 1}{x - 3} \), we will analyze the function by considering two cases based on the absolute value of \( x \).

### Step 1: Case 1 - When \( x \geq 0 \)

For \( x \geq 0 \), we have \( |x| = x \). Thus, the function becomes:
\[ f(x) = \frac{2x - 1}{x - 3} \]

Let \( y = f(x) \):
\[ y = \frac{2x - 1}{x - 3} \]

Cross-multiplying gives:
\[ yx - 3y = 2x - 1 \]
Rearranging this, we get:
\[ yx - 2x = 3y - 1 \]
Factoring out \( x \):
\[ x(y - 2) = 3y - 1 \]
Thus,
\[ x = \frac{3y - 1}{y - 2} \]

Since \( x \geq 0 \), we require:
\[ \frac{3y - 1}{y - 2} \geq 0 \]

### Step 2: Solve the Inequality

To solve \( \frac{3y - 1}{y - 2} \geq 0 \), we find the critical points by setting the numerator and denominator to zero:
- \( 3y - 1 = 0 \) gives \( y = \frac{1}{3} \)
- \( y - 2 = 0 \) gives \( y = 2 \)

Now we analyze the sign of the expression on the number line:
- The critical points divide the number line into intervals: \( (-\infty, \frac{1}{3}) \), \( (\frac{1}{3}, 2) \), and \( (2, \infty) \).

Testing each interval:
1. For \( y < \frac{1}{3} \): Choose \( y = 0 \) → \( \frac{3(0) - 1}{0 - 2} = \frac{-1}{-2} > 0 \) (positive)
2. For \( \frac{1}{3} < y < 2 \): Choose \( y = 1 \) → \( \frac{3(1) - 1}{1 - 2} = \frac{2}{-1} < 0 \) (negative)
3. For \( y > 2 \): Choose \( y = 3 \) → \( \frac{3(3) - 1}{3 - 2} = \frac{8}{1} > 0 \) (positive)

Thus, the solution to the inequality is:
\[ y \in (-\infty, \frac{1}{3}] \cup (2, \infty) \]

### Step 3: Case 2 - When \( x < 0 \)

For \( x < 0 \), we have \( |x| = -x \). Thus, the function becomes:
\[ f(x) = \frac{-2x - 1}{x - 3} \]

Let \( y = f(x) \):
\[ y = \frac{-2x - 1}{x - 3} \]

Cross-multiplying gives:
\[ yx - 3y = -2x - 1 \]
Rearranging this, we get:
\[ yx + 2x = 3y - 1 \]
Factoring out \( x \):
\[ x(y + 2) = 3y - 1 \]
Thus,
\[ x = \frac{3y - 1}{y + 2} \]

Since \( x < 0 \), we require:
\[ \frac{3y - 1}{y + 2} < 0 \]

### Step 4: Solve the Inequality

To solve \( \frac{3y - 1}{y + 2} < 0 \), we find the critical points:
- \( 3y - 1 = 0 \) gives \( y = \frac{1}{3} \)
- \( y + 2 = 0 \) gives \( y = -2 \)

Now we analyze the sign of the expression on the number line:
- The critical points divide the number line into intervals: \( (-\infty, -2) \), \( (-2, \frac{1}{3}) \), and \( (\frac{1}{3}, \infty) \).

Testing each interval:
1. For \( y < -2 \): Choose \( y = -3 \) → \( \frac{3(-3) - 1}{-3 + 2} = \frac{-10}{-1} > 0 \) (positive)
2. For \( -2 < y < \frac{1}{3} \): Choose \( y = 0 \) → \( \frac{3(0) - 1}{0 + 2} = \frac{-1}{2} < 0 \) (negative)
3. For \( y > \frac{1}{3} \): Choose \( y = 1 \) → \( \frac{3(1) - 1}{1 + 2} = \frac{2}{3} > 0 \) (positive)

Thus, the solution to the inequality is:
\[ y \in (-2, \frac{1}{3}) \]

### Step 5: Combine the Results

Now we combine the results from both cases:
1. From Case 1: \( y \in (-\infty, \frac{1}{3}] \cup (2, \infty) \)
2. From Case 2: \( y \in (-2, \frac{1}{3}) \)

Combining these intervals, we get:
\[ \text{Range of } f(x) = (-\infty, \frac{1}{3}) \cup (2, \infty) \]

### Final Answer
The range of the function \( f(x) \) is:
\[ (-\infty, \frac{1}{3}) \cup (2, \infty) \]