The solution of `| |x|-1|lt|1-x|,x in R` is

Options

A. `(-1,1)`
B. `(o,oo)`
C. `(-1,oo)`
D. None of these

Answer

To solve the inequality \( ||x| - 1| < |1 - x| \), we will break it down step by step.

### Step 1: Analyze the Absolute Values
We start with the given inequality:
\[ ||x| - 1| < |1 - x| \]
We need to consider the cases for \( |x| \) and \( |1 - x| \).

### Step 2: Case Analysis for \( |x| \)
The expression \( |x| \) can be split into two cases based on the value of \( x \):
1. **Case 1**: \( x \geq 0 \) (Here, \( |x| = x \))
2. **Case 2**: \( x < 0 \) (Here, \( |x| = -x \))

### Step 3: Solve for Case 1: \( x \geq 0 \)
In this case, we have:
\[ ||x| - 1| = |x - 1| \quad \text{and} \quad |1 - x| = |1 - x| \]
Thus, the inequality becomes:
\[ |x - 1| < |1 - x| \]
Since \( |1 - x| = |x - 1| \), the inequality simplifies to:
\[ |x - 1| < |x - 1| \]
This is never true, so there are no solutions for \( x \geq 0 \).

### Step 4: Solve for Case 2: \( x < 0 \)
In this case, we have:
\[ ||x| - 1| = |-x - 1| = |-(x + 1)| = |x + 1| \quad \text{and} \quad |1 - x| = |1 - x| \]
Thus, the inequality becomes:
\[ |x + 1| < |1 - x| \]
Now, we need to analyze \( |1 - x| \):
- Since \( x < 0 \), \( 1 - x > 1 \) and thus \( |1 - x| = 1 - x \).

So we rewrite the inequality:
\[ |x + 1| < 1 - x \]
Since \( x < 0 \), \( |x + 1| = x + 1 \) (because \( x + 1 > 0 \)):
\[ x + 1 < 1 - x \]
### Step 5: Solve the Inequality
Now, we solve the inequality:
\[ x + 1 < 1 - x \]
Adding \( x \) to both sides gives:
\[ 2x + 1 < 1 \]
Subtracting \( 1 \) from both sides:
\[ 2x < 0 \]
Dividing by \( 2 \):
\[ x < 0 \]
This is true for all \( x < 0 \).

### Step 6: Conclusion
Thus, the solution to the inequality \( ||x| - 1| < |1 - x| \) is:
\[ x < 0 \]
In interval notation, the solution is:
\[ (-\infty, 0) \]