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 will analyze the function \( f(x) = \frac{2|x| - 1}{x - 3} \).

### Step 1: Identify the critical points and intervals
The function \( f(x) \) involves the absolute value \( |x| \), which changes its behavior at \( x = 0 \). Additionally, the function is undefined at \( x = 3 \). Therefore, we will consider three intervals:
1. \( x < 0 \)
2. \( 0 \leq x < 3 \)
3. \( x > 3 \)

### Step 2: Analyze the function in each interval

#### Interval 1: \( x < 0 \)
In this interval, \( |x| = -x \). Thus, the function becomes:
\[ f(x) = \frac{2(-x) - 1}{x - 3} = \frac{-2x - 1}{x - 3} \]
We can simplify this:
\[ f(x) = \frac{-2x + 6 - 7}{x - 3} = -2 + \frac{7}{x - 3} \]
As \( x \) approaches \( -\infty \), \( f(x) \) approaches \( -2 \). As \( x \) approaches \( 3 \) from the left, \( f(x) \) approaches \( -\infty \).

#### Interval 2: \( 0 \leq x < 3 \)
In this interval, \( |x| = x \). Thus, the function becomes:
\[ f(x) = \frac{2x - 1}{x - 3} \]
We can simplify this:
\[ f(x) = \frac{2x - 6 + 5}{x - 3} = 2 + \frac{5}{x - 3} \]
As \( x \) approaches \( 0 \), \( f(x) \) approaches \( \frac{5}{-3} = -\frac{5}{3} \). As \( x \) approaches \( 3 \) from the left, \( f(x) \) approaches \( -\infty \).

#### Interval 3: \( x > 3 \)
In this interval, \( |x| = x \). Thus, the function becomes:
\[ f(x) = \frac{2x - 1}{x - 3} \]
We can simplify this:
\[ f(x) = 2 + \frac{5}{x - 3} \]
As \( x \) approaches \( 3 \) from the right, \( f(x) \) approaches \( +\infty \). As \( x \) approaches \( +\infty \), \( f(x) \) approaches \( 2 \).

### Step 3: Determine the range of \( f(x) \)
Now we summarize the behavior of \( f(x) \) in each interval:
- For \( x < 0 \): \( f(x) \) decreases from \( -2 \) to \( -\infty \).
- For \( 0 \leq x < 3 \): \( f(x) \) decreases from \( -\frac{5}{3} \) to \( -\infty \).
- For \( x > 3 \): \( f(x) \) increases from \( +\infty \) to \( 2 \).

### Step 4: Find the values of \( k \) for two solutions
To have exactly two distinct solutions for \( f(x) = k \), the horizontal line \( y = k \) must intersect the graph of \( f(x) \) in two places.

From our analysis:
- The function decreases to \( -\infty \) as \( x \) approaches \( 3 \) from the left and increases from \( +\infty \) as \( x \) approaches \( 3 \) from the right.
- The maximum value of \( f(x) \) in the interval \( x < 0 \) is \( -2 \) and the minimum value in the interval \( x > 3 \) is \( 2 \).

Thus, for \( k \) to have exactly two solutions, it must lie in the range:
\[ -2 < k < \frac{5}{3} \]

### Conclusion
The range of values of \( k \) for which \( f(x) = k \) has exactly two distinct solutions is:
\[ \boxed{(-2, \frac{5}{3})} \]