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 defined as:

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

### Step 1: Analyze the function for different cases of \( x \)

Since \( f(x) \) involves the absolute value function, we need to consider two cases based on the sign of \( x \).

**Case 1:** \( x \geq 0 \)

For \( x \geq 0 \), we have \( |x| = x \). Thus, the function simplifies to:

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

**Case 2:** \( x < 0 \)

For \( x < 0 \), we have \( |x| = -x \). Thus, the function simplifies to:

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

### Step 2: Set up the equations for both cases

We need to find the values of \( k \) such that \( f(x) = k \) has exactly two distinct solutions.

**For Case 1:** Setting \( f(x) = k \):

\[ \frac{2x - 1}{x - 3} = k \]

Cross-multiplying gives:

\[ 2x - 1 = k(x - 3) \implies 2x - 1 = kx - 3k \]

Rearranging this leads to:

\[ (2 - k)x = -3k + 1 \implies x = \frac{-3k + 1}{2 - k} \]

**For Case 2:** Setting \( f(x) = k \):

\[ \frac{-2x - 1}{x - 3} = k \]

Cross-multiplying gives:

\[ -2x - 1 = k(x - 3) \implies -2x - 1 = kx - 3k \]

Rearranging this leads to:

\[ (-2 - k)x = -3k + 1 \implies x = \frac{-3k + 1}{-2 - k} \]

### Step 3: Determine the conditions for distinct solutions

For \( f(x) = k \) to have exactly two distinct solutions, one solution must come from each case. This means we need to ensure that the two expressions for \( x \) yield valid solutions.

1. **From Case 1:** \( x = \frac{-3k + 1}{2 - k} \) must be non-negative.
2. **From Case 2:** \( x = \frac{-3k + 1}{-2 - k} \) must be negative.

### Step 4: Solve the inequalities

**For Case 1:** \( \frac{-3k + 1}{2 - k} \geq 0 \)

This occurs when:

- The numerator and denominator are both positive, or both negative.

1. **Numerator:** \( -3k + 1 \geq 0 \implies k \leq \frac{1}{3} \)
2. **Denominator:** \( 2 - k > 0 \implies k < 2 \)

**For Case 2:** \( \frac{-3k + 1}{-2 - k} < 0 \)

This occurs when:

- The numerator is positive and the denominator is negative.

1. **Numerator:** \( -3k + 1 > 0 \implies k < \frac{1}{3} \)
2. **Denominator:** \( -2 - k < 0 \implies k > -2 \)

### Step 5: Combine the results

From the inequalities, we have:

1. From Case 1: \( k \leq \frac{1}{3} \) and \( k < 2 \)
2. From Case 2: \( k < \frac{1}{3} \) and \( k > -2 \)

Combining these gives:

\[ -2 < k < \frac{1}{3} \]

### Conclusion

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

\[ \boxed{(-2, \frac{1}{3})} \]