Complete set of values of x satisfying inequality `||x-1|-5| lt 2x -5 ` is

Options

A. `(5//2, oo)`
B. `(11 //3 , oo)`
C. `(-1 ,oo)`
D. `(-oo ,1//3)`

Explanation

To solve the inequality \( ||x-1|-5| < 2x - 5 \), we will break it down step by step.

### Step 1: Identify the critical points
First, we need to find the points where the expressions inside the absolute values become zero.

1. Set \( |x - 1| - 5 = 0 \):
\[ |x - 1| = 5 \]
This gives us two equations:
\[ x - 1 = 5 \quad \Rightarrow \quad x = 6 \]
\[ x - 1 = -5 \quad \Rightarrow \quad x = -4 \]

2. Set \( |x - 1| = 0 \):
\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]

So, the critical points are \( x = -4, 1, 6 \).

### Step 2: Determine intervals
The critical points divide the real line into four intervals:
1. \( (-\infty, -4) \)
2. \( [-4, 1) \)
3. \( [1, 6) \)
4. \( [6, \infty) \)

### Step 3: Analyze each interval
We will analyze the inequality \( ||x-1|-5| < 2x - 5 \) in each interval.

#### Case 1: \( x < -4 \)
In this interval, \( |x - 1| = -(x - 1) = -x + 1 \). Thus,
\[ ||x - 1| - 5| = |-x + 1 - 5| = |-x - 4| = x + 4 \]
The inequality becomes:
\[ x + 4 < 2x - 5 \]
Solving this:
\[ 4 + 5 < 2x - x \quad \Rightarrow \quad 9 < x \quad \Rightarrow \quad x > 9 \]
Since \( x < -4 \) and \( x > 9 \) cannot be true simultaneously, there are no solutions in this interval.

#### Case 2: \( -4 \leq x < 1 \)
Here, \( |x - 1| = -(x - 1) = -x + 1 \). Thus,
\[ ||x - 1| - 5| = |-x + 1 - 5| = |-x - 4| = -x - 4 \]
The inequality becomes:
\[ -x - 4 < 2x - 5 \]
Solving this:
\[ -4 + 5 < 2x + x \quad \Rightarrow \quad 1 < 3x \quad \Rightarrow \quad x > \frac{1}{3} \]
Since \( -4 \leq x < 1 \), the solution in this interval is:
\[ \frac{1}{3} < x < 1 \]

#### Case 3: \( 1 \leq x < 6 \)
In this interval, \( |x - 1| = x - 1 \). Thus,
\[ ||x - 1| - 5| = |x - 1 - 5| = |x - 6| \]
The inequality becomes:
\[ |x - 6| < 2x - 5 \]
This gives us two subcases:
1. \( x - 6 < 2x - 5 \) which simplifies to \( -1 < x \) (always true in this interval).
2. \( -(x - 6) < 2x - 5 \) which simplifies to \( 6 < 3x - 5 \) or \( 11 < 3x \) or \( x > \frac{11}{3} \).

Thus, the solution in this interval is:
\[ \frac{11}{3} < x < 6 \]

#### Case 4: \( x \geq 6 \)
In this interval, \( |x - 1| = x - 1 \). Thus,
\[ ||x - 1| - 5| = |x - 6| \]
The inequality becomes:
\[ |x - 6| < 2x - 5 \]
This gives us two subcases:
1. \( x - 6 < 2x - 5 \) which simplifies to \( -1 < x \) (always true in this interval).
2. \( -(x - 6) < 2x - 5 \) which simplifies to \( 6 < 3x - 5 \) or \( 11 < 3x \) or \( x > \frac{11}{3} \) (always true in this interval).

Thus, the solution in this interval is:
\[ x \geq 6 \]

### Step 4: Combine the solutions
From the intervals, we have:
1. From Case 2: \( \frac{1}{3} < x < 1 \)
2. From Case 3: \( \frac{11}{3} < x < 6 \)
3. From Case 4: \( x \geq 6 \)

Combining these intervals gives us the complete solution:
\[ x \in \left( \frac{1}{3}, 1 \right) \cup \left( \frac{11}{3}, \infty \right) \]

### Final Answer
The complete set of values of \( x \) satisfying the inequality \( ||x-1|-5| < 2x - 5 \) is:
\[ \left( \frac{1}{3}, 1 \right) \cup \left( \frac{11}{3}, \infty \right) \]