The solution set of the inequlity `(|x-2|-x)/(x)lt2 ` is

Options

A. (0,1)
B. [0,2]
C. `(-oo,0) cup (1,oo)`
D. None of these

Explanation

To solve the inequality \(\frac{|x-2| - x}{x} < 2\), we will break it down step by step.

### Step 1: Rearranging the Inequality
We start with the given inequality:
\[ \frac{|x-2| - x}{x} < 2 \]
We can rearrange this by moving \(2\) to the left side:
\[ \frac{|x-2| - x}{x} - 2 < 0 \]
This can be rewritten as:
\[ \frac{|x-2| - x - 2x}{x} < 0 \]
or
\[ \frac{|x-2| - 3x}{x} < 0 \]

### Step 2: Considering Cases for the Absolute Value
Next, we need to consider the two cases for the absolute value \(|x-2|\).

**Case 1:** \(x - 2 \geq 0\) (i.e., \(x \geq 2\))

In this case, \(|x-2| = x - 2\). The inequality becomes:
\[ \frac{(x - 2) - 3x}{x} < 0 \]
This simplifies to:
\[ \frac{-2x - 2}{x} < 0 \]
or
\[ \frac{-2(x + 1)}{x} < 0 \]

**Case 2:** \(x - 2 < 0\) (i.e., \(x < 2\))

In this case, \(|x-2| = -(x - 2) = 2 - x\). The inequality becomes:
\[ \frac{(2 - x) - 3x}{x} < 0 \]
This simplifies to:
\[ \frac{2 - 4x}{x} < 0 \]
or
\[ \frac{2 - 4x}{x} < 0 \]

### Step 3: Solving Case 1
From Case 1:
\[ \frac{-2(x + 1)}{x} < 0 \]
We analyze the sign of the expression:
- The numerator \(-2(x + 1)\) is negative when \(x + 1 > 0\) (i.e., \(x > -1\)).
- The denominator \(x\) is positive when \(x > 0\).

Thus, the critical points are \(x = -1\) and \(x = 0\). We check the intervals:
- For \(x < -1\): both parts are negative, so the expression is positive.
- For \(-1 < x < 0\): the numerator is positive and the denominator is negative, so the expression is negative.
- For \(0 < x < 2\): both parts are positive, so the expression is positive.
- For \(x > 2\): both parts are negative, so the expression is negative.

Thus, for Case 1, the solution is:
\[ x \in (-1, 0) \cup (2, \infty) \]

### Step 4: Solving Case 2
From Case 2:
\[ \frac{2 - 4x}{x} < 0 \]
The critical points are \(x = 0\) and \(x = \frac{1}{2}\). We analyze the sign:
- For \(x < 0\): both parts are positive, so the expression is positive.
- For \(0 < x < \frac{1}{2}\): the numerator is positive and the denominator is positive, so the expression is positive.
- For \(x = \frac{1}{2}\): the expression is zero.
- For \(x > \frac{1}{2}\): the numerator is negative and the denominator is positive, so the expression is negative.

Thus, for Case 2, the solution is:
\[ x \in (0, \frac{1}{2}) \cup (2, \infty) \]

### Step 5: Combining Solutions
Combining the solutions from both cases, we have:
\[ x \in (-\infty, 0) \cup \left(\frac{1}{2}, 2\right) \cup (2, \infty) \]

### Final Solution
The final solution set for the inequality is:
\[ x \in (-\infty, 0) \cup \left(\frac{1}{2}, 2\right) \cup (2, \infty) \]