The solution set of `(|x-1|)/(x)gt2` is

Options

A. `0ltxlt1/3`
B. `xlt1/3`
C. `xgt1/3`
D. `1/3ltxlt1`

Explanation

To solve the inequality \(\frac{|x-1|}{x} > 2\), we will follow these steps:

### Step 1: Identify the domain
The expression \(\frac{|x-1|}{x}\) is undefined when \(x = 0\). Therefore, we need to consider two cases: \(x < 0\) and \(x > 0\).

**Hint:** Always check for values that make the denominator zero, as they are not in the domain.

### Step 2: Analyze the case \(x < 0\)
For \(x < 0\), the absolute value can be simplified as follows:
\[ |x-1| = -(x-1) = -x + 1 \]
Thus, the inequality becomes:
\[ \frac{-x + 1}{x} > 2 \]
This simplifies to:
\[ \frac{1 - x}{x} > 2 \]
Multiplying both sides by \(x\) (note that \(x\) is negative, so we reverse the inequality):
\[ 1 - x < 2x \]
This leads to:
\[ 1 < 3x \quad \Rightarrow \quad x > \frac{1}{3} \]
However, since we are considering \(x < 0\), there are no valid solutions in this case.

**Hint:** When multiplying or dividing by a negative number, remember to reverse the inequality sign.

### Step 3: Analyze the case \(x > 0\)
For \(x > 0\), the absolute value remains the same:
\[ |x-1| = x - 1 \]
Thus, the inequality becomes:
\[ \frac{x - 1}{x} > 2 \]
This simplifies to:
\[ 1 - \frac{1}{x} > 2 \]
Rearranging gives:
\[ - \frac{1}{x} > 1 \quad \Rightarrow \quad \frac{1}{x} < -1 \]
Since \(x > 0\), this inequality cannot hold true. Therefore, we need to check the boundary condition where the left-hand side equals 2.

### Step 4: Set the boundary condition
Set the inequality to equality:
\[ \frac{|x-1|}{x} = 2 \]
For \(x > 0\):
\[ x - 1 = 2x \quad \Rightarrow \quad -1 = x \quad \text{(not valid since \(x > 0\))} \]
Now consider the case where \(x - 1 = -2x\):
\[ -x + 1 = 2x \quad \Rightarrow \quad 1 = 3x \quad \Rightarrow \quad x = \frac{1}{3} \]

### Step 5: Determine the solution set
We now need to check the intervals:
1. For \(0 < x < \frac{1}{3}\), check a value like \(x = \frac{1}{4}\):
\[ \frac{\frac{1}{4} - 1}{\frac{1}{4}} = \frac{-\frac{3}{4}}{\frac{1}{4}} = -3 \quad \text{(not valid)} \]
2. For \(x > \frac{1}{3}\), check a value like \(x = 1\):
\[ \frac{1 - 1}{1} = 0 \quad \text{(not valid)} \]

Thus, the solution set is:
\[ 0 < x < \frac{1}{3} \]

### Final Answer
The solution set is \(0 < x < \frac{1}{3}\).