Let `A={x in R : |x+1| = |x|+1}` and `B ={x in R : |x-1| = |x|-1}`, Then

Options

A. A = B
B. `B sube A`
C. `A sube B`
D. `A cup B = R`

Answer

To solve the problem, we need to find the sets \( A \) and \( B \) based on the given conditions and then analyze their relationship.

### Step 1: Determine Set A
Set \( A \) is defined as:
\[ A = \{ x \in \mathbb{R} : |x + 1| = |x| + 1 \} \]

To solve \( |x + 1| = |x| + 1 \), we consider two cases based on the definition of absolute value.

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

In this case, \( |x + 1| = x + 1 \) and \( |x| = x \) (since \( x \geq 0 \)). Thus, the equation becomes:
\[ x + 1 = x + 1 \]
This is true for all \( x \geq -1 \).

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

Here, \( |x + 1| = -(x + 1) = -x - 1 \) and \( |x| = -x \) (since \( x < 0 \)). The equation becomes:
\[ -x - 1 = -x + 1 \]
Simplifying gives:
\[ -1 = 1 \]
This is a contradiction, so there are no solutions in this case.

Thus, the solution set for \( A \) is:
\[ A = \{ x \in \mathbb{R} : x \geq -1 \} \]

### Step 2: Determine Set B
Set \( B \) is defined as:
\[ B = \{ x \in \mathbb{R} : |x - 1| = |x| - 1 \} \]

Again, we consider two cases.

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

In this case, \( |x - 1| = x - 1 \) and \( |x| = x \) (since \( x \geq 0 \)). Thus, the equation becomes:
\[ x - 1 = x - 1 \]
This is true for all \( x \geq 1 \).

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

Here, \( |x - 1| = -(x - 1) = -x + 1 \) and \( |x| = -x \) (since \( x < 0 \)). The equation becomes:
\[ -x + 1 = -x - 1 \]
Simplifying gives:
\[ 1 = -1 \]
This is also a contradiction, so there are no solutions in this case.

Thus, the solution set for \( B \) is:
\[ B = \{ x \in \mathbb{R} : x \geq 1 \} \]

### Step 3: Analyze the Relationship between Sets A and B
Now we have:
- \( A = \{ x \in \mathbb{R} : x \geq -1 \} \)
- \( B = \{ x \in \mathbb{R} : x \geq 1 \} \)

Since every element in \( B \) (which starts at 1) is also in \( A \) (which starts at -1), we can conclude that:
\[ B \subseteq A \]

### Final Conclusion
The correct relationship between the sets is that \( B \) is a subset of \( A \).