The solution set of the inequation `|(2x-1)/(x-1)| gt 2`, is

Options

A. `(34, 1) cup (1, oo)`
B. `(3//4, oo)`
C. `(-oo, 3//4)`
D. none of these

Answer

To solve the inequality \(\left|\frac{2x-1}{x-1}\right| > 2\), we will break it down into two cases based on the definition of absolute value.

### Step 1: Set up the two cases for the absolute value
We can rewrite the inequality as two separate inequalities:
1. \(\frac{2x-1}{x-1} > 2\)
2. \(\frac{2x-1}{x-1} < -2\)

### Step 2: Solve the first inequality \(\frac{2x-1}{x-1} > 2\)
To solve this, we first rearrange the inequality:
\[ \frac{2x-1}{x-1} - 2 > 0 \]
This can be rewritten as:
\[ \frac{2x-1 - 2(x-1)}{x-1} > 0 \]
Simplifying the numerator:
\[ 2x - 1 - 2x + 2 = 1 \]
So, we have:
\[ \frac{1}{x-1} > 0 \]
This inequality holds when \(x - 1 > 0\) or \(x > 1\).

### Step 3: Solve the second inequality \(\frac{2x-1}{x-1} < -2\)
Rearranging this inequality gives:
\[ \frac{2x-1}{x-1} + 2 < 0 \]
This can be rewritten as:
\[ \frac{2x-1 + 2(x-1)}{x-1} < 0 \]
Simplifying the numerator:
\[ 2x - 1 + 2x - 2 = 4x - 3 \]
So, we have:
\[ \frac{4x - 3}{x - 1} < 0 \]
To find the intervals where this fraction is negative, we need to find the critical points by setting the numerator and denominator to zero:
1. \(4x - 3 = 0 \Rightarrow x = \frac{3}{4}\)
2. \(x - 1 = 0 \Rightarrow x = 1\)

### Step 4: Test intervals
We will test the intervals determined by the critical points \(\frac{3}{4}\) and \(1\):
1. For \(x < \frac{3}{4}\), choose \(x = 0\):
\[ \frac{4(0) - 3}{0 - 1} = \frac{-3}{-1} = 3 > 0 \quad \text{(not valid)} \]
2. For \(\frac{3}{4} < x < 1\), choose \(x = \frac{7}{8}\):
\[ \frac{4(\frac{7}{8}) - 3}{\frac{7}{8} - 1} = \frac{\frac{28}{8} - \frac{24}{8}}{-\frac{1}{8}} = \frac{\frac{4}{8}}{-\frac{1}{8}} = -4 < 0 \quad \text{(valid)} \]
3. For \(x > 1\), choose \(x = 2\):
\[ \frac{4(2) - 3}{2 - 1} = \frac{8 - 3}{1} = 5 > 0 \quad \text{(not valid)} \]

### Step 5: Combine the solution sets
From the first inequality, we have \(x > 1\). From the second inequality, we have \(\frac{3}{4} < x < 1\).

Thus, the complete solution set is:
\[ \left(\frac{3}{4}, 1\right) \cup (1, \infty) \]

### Final Answer
The solution set of the inequality \(\left|\frac{2x-1}{x-1}\right| > 2\) is:
\[ \left(\frac{3}{4}, 1\right) \cup (1, \infty) \]