Let `f (x) =(2 |x| -1)/(x-3)`
Range of the values of 'k' for which `f (x) = k` has exactly two distinct solutions:

Options

A. `(-2 , (1)/(3))`
B. `(-2,1]`
C. `(0,(2)/(3)]`
D. `(-oo, -2)`

Answer

To find the range of values of \( k \) for which the equation \( f(x) = k \) has exactly two distinct solutions, we start with the function:

\[ f(x) = \frac{2|x| - 1}{x - 3} \]

### Step 1: Analyze the function
The function \( f(x) \) is defined for all \( x \) except \( x = 3 \) (where the denominator becomes zero). We need to analyze the behavior of \( f(x) \) as \( x \) approaches 3 and also as \( x \) approaches positive and negative infinity.

### Step 2: Find the limits
1. As \( x \to 3^+ \):
\[ f(x) \to +\infty \]
2. As \( x \to 3^- \):
\[ f(x) \to -\infty \]
3. As \( x \to +\infty \):
\[ f(x) \to 2 \]
4. As \( x \to -\infty \):
\[ f(x) \to -2 \]

### Step 3: Determine the critical points
Next, we need to find where the function changes behavior. We can find the derivative \( f'(x) \) to determine the intervals of increase and decrease.

#### Case 1: \( x \geq 0 \)
For \( x \geq 0 \):
\[ f(x) = \frac{2x - 1}{x - 3} \]
Calculating the derivative:
\[ f'(x) = \frac{(2)(x - 3) - (2x - 1)(1)}{(x - 3)^2} = \frac{2x - 6 - 2x + 1}{(x - 3)^2} = \frac{-5}{(x - 3)^2} \]
Since \( f'(x) < 0 \) for \( x > 3 \), \( f(x) \) is decreasing for \( x > 3 \).

#### Case 2: \( x < 0 \)
For \( x < 0 \):
\[ f(x) = \frac{-2x - 1}{x - 3} \]
Calculating the derivative:
\[ f'(x) = \frac{(-2)(x - 3) - (-2x - 1)(1)}{(x - 3)^2} = \frac{-2x + 6 + 2x + 1}{(x - 3)^2} = \frac{7}{(x - 3)^2} \]
Since \( f'(x) > 0 \) for \( x < 3 \), \( f(x) \) is increasing for \( x < 3 \).

### Step 4: Identify the range of \( f(x) \)
From the analysis:
- As \( x \to -\infty \), \( f(x) \to -2 \).
- As \( x \to 3^- \), \( f(x) \to -\infty \).
- As \( x \to 3^+ \), \( f(x) \to +\infty \).
- As \( x \to +\infty \), \( f(x) \to 2 \).

Thus, the range of \( f(x) \) is \( (-2, 2) \).

### Step 5: Determine the values of \( k \)
For \( f(x) = k \) to have exactly two distinct solutions, \( k \) must be within the range of \( f(x) \) but not equal to the endpoints, which means:

\[ -2 < k < 2 \]

### Final Answer
The range of values of \( k \) for which \( f(x) = k \) has exactly two distinct solutions is:

\[ (-2, 2) \]