For `(|x-1|)/(x+2) lt 1` , solution set of x is given by :

Options

A. `(-oo,-2) cup (-1//2,oo)`
B. `(-oo,-1) cup [2,3]`
C. `(-oo,-4)`
D. `[-1//2,1]`

Answer

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

### Step 1: Rearrange the Inequality
We start with the inequality:
\[ \frac{|x-1|}{x+2} < 1 \]
Subtracting 1 from both sides gives:
\[ \frac{|x-1|}{x+2} - 1 < 0 \]
This simplifies to:
\[ \frac{|x-1| - (x+2)}{x+2} < 0 \]
Thus, we have:
\[ \frac{|x-1| - x - 2}{x+2} < 0 \]

### Step 2: Consider Cases for the Absolute Value
Since we have the absolute value \(|x-1|\), we need to consider two cases:

**Case 1: \(x < 1\)**
In this case, \(|x-1| = -(x-1) = -x + 1\). Substituting this into the inequality gives:
\[ \frac{-x + 1 - x - 2}{x+2} < 0 \]
This simplifies to:
\[ \frac{-2x - 1}{x+2} < 0 \]
Multiplying both sides by -1 (which reverses the inequality):
\[ \frac{2x + 1}{x + 2} > 0 \]

**Case 2: \(x \geq 1\)**
Here, \(|x-1| = x - 1\). Substituting this into the inequality gives:
\[ \frac{x - 1 - (x + 2)}{x + 2} < 0 \]
This simplifies to:
\[ \frac{-3}{x + 2} < 0 \]
Since \(-3\) is always negative, this inequality holds true when \(x + 2 > 0\) or \(x > -2\).

### Step 3: Solve Each Case
**For Case 1:**
We need to find when \(\frac{2x + 1}{x + 2} > 0\). The critical points are found by setting the numerator and denominator to zero:
- \(2x + 1 = 0 \Rightarrow x = -\frac{1}{2}\)
- \(x + 2 = 0 \Rightarrow x = -2\)

Now, we can test intervals based on these critical points:
- Interval 1: \( (-\infty, -2) \)
- Interval 2: \( (-2, -\frac{1}{2}) \)
- Interval 3: \( (-\frac{1}{2}, \infty) \)

Testing these intervals:
1. For \(x = -3\) (Interval 1): \(\frac{2(-3) + 1}{-3 + 2} = \frac{-6 + 1}{-1} = \frac{-5}{-1} > 0\) (True)
2. For \(x = -1\) (Interval 2): \(\frac{2(-1) + 1}{-1 + 2} = \frac{-2 + 1}{1} = \frac{-1}{1} < 0\) (False)
3. For \(x = 0\) (Interval 3): \(\frac{2(0) + 1}{0 + 2} = \frac{1}{2} > 0\) (True)

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

**For Case 2:**
From the previous analysis, we found that the inequality holds for \(x > -2\). Since we are in the case where \(x \geq 1\), the solution for Case 2 is:
\[ x \in [1, \infty) \]

### Step 4: Combine Solutions
Now we combine the solutions from both cases:
1. From Case 1: \(x \in (-\infty, -2) \cup \left(-\frac{1}{2}, 1\right)\)
2. From Case 2: \(x \in [1, \infty)\)

Thus, the complete solution set is:
\[ x \in (-\infty, -2) \cup \left(-\frac{1}{2}, 1\right) \cup [1, \infty) \]

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