The solution set of inequality `|x-1|+|x+1| lt 4` is

Options

A. `(-oo, -2)`
B. `(-2, 2)`
C. `(2, oo)`
D. `(-oo, oo)`

Answer

To solve the inequality \( |x-1| + |x+1| < 4 \), we will break it down into different cases based on the values of \( x \).

### Step 1: Identify critical points
The critical points for the absolute values are found by setting the expressions inside the absolute values to zero:
- \( x - 1 = 0 \) gives \( x = 1 \)
- \( x + 1 = 0 \) gives \( x = -1 \)

Thus, the critical points are \( x = -1 \) and \( x = 1 \). These points divide the number line into three intervals:
1. \( (-\infty, -1) \)
2. \( [-1, 1] \)
3. \( (1, \infty) \)

### Step 2: Analyze each interval

**Interval 1: \( x < -1 \)**
In this interval, both \( x - 1 \) and \( x + 1 \) are negative. Therefore:
\[ |x-1| = -(x-1) = -x + 1 \]
\[ |x+1| = -(x+1) = -x - 1 \]
Substituting these into the inequality:
\[ (-x + 1) + (-x - 1) < 4 \]
Simplifying:
\[ -2x < 4 \implies x > -2 \]
So, in this interval, we have \( -2 < x < -1 \).

**Interval 2: \( -1 \leq x \leq 1 \)**
In this interval, \( x - 1 \) is negative and \( x + 1 \) is positive. Therefore:
\[ |x-1| = -(x-1) = -x + 1 \]
\[ |x+1| = x + 1 \]
Substituting these into the inequality:
\[ (-x + 1) + (x + 1) < 4 \]
Simplifying:
\[ 2 < 4 \]
This is always true, so the solution for this interval is \( -1 \leq x \leq 1 \).

**Interval 3: \( x > 1 \)**
In this interval, both \( x - 1 \) and \( x + 1 \) are positive. Therefore:
\[ |x-1| = x - 1 \]
\[ |x+1| = x + 1 \]
Substituting these into the inequality:
\[ (x - 1) + (x + 1) < 4 \]
Simplifying:
\[ 2x < 4 \implies x < 2 \]
So, in this interval, we have \( 1 < x < 2 \).

### Step 3: Combine the results
From the three intervals, we have:
1. From Interval 1: \( -2 < x < -1 \)
2. From Interval 2: \( -1 \leq x \leq 1 \)
3. From Interval 3: \( 1 < x < 2 \)

Combining these, the solution set is:
\[ (-2, -1) \cup [-1, 1] \cup (1, 2) \]

### Final Answer
The solution set of the inequality \( |x-1| + |x+1| < 4 \) is:
\[ (-2, 2) \]