If `|b| ge 1` and `x = -|a|b`, then which one of the following is necessarily true ?

Options

A. `a - x b lt 0`
B. `a - xb ge 0`
C. `a - x b gt 0`
D. `a - x b le 0`

Answer

To solve the problem, we start with the given conditions and analyze the expression step by step.

### Given:
1. \( |b| \geq 1 \)
2. \( x = -|a|b \)

### We need to analyze:
The expression \( a - xb \).

### Step 1: Substitute for \( x \)
We know \( x = -|a|b \). Therefore, we can rewrite the expression \( a - xb \) as follows:
\[ a - xb = a - (-|a|b) = a + |a|b \]

### Step 2: Factor the expression
Now, we can factor out \( a \) from the expression:
\[ a + |a|b = a(1 + \frac{|a|b}{a}) \quad \text{(if \( a \neq 0 \))} \]
However, we can analyze it without factoring as well.

### Step 3: Consider cases for \( a \)
We will consider two cases for \( a \): when \( a \geq 0 \) and when \( a < 0 \).

#### Case 1: \( a \geq 0 \)
If \( a \) is non-negative, then \( |a| = a \). Thus:
\[ a + |a|b = a + ab = a(1 + b) \]
Since \( |b| \geq 1 \), we have two subcases:
- If \( b \geq 1 \), then \( 1 + b \geq 2 > 0 \) and hence \( a(1 + b) \geq 0 \).
- If \( b \leq -1 \), then \( 1 + b < 0 \) but since \( a \geq 0 \), \( a(1 + b) \geq 0 \) is still valid as \( a \) could be zero.

In both scenarios, \( a - xb \geq 0 \).

#### Case 2: \( a < 0 \)
If \( a \) is negative, then \( |a| = -a \). Thus:
\[ a + |a|b = a - ab = a(1 - b) \]
Again, we consider two subcases:
- If \( b \geq 1 \), then \( 1 - b < 0 \) and since \( a < 0 \), \( a(1 - b) \geq 0 \) is valid.
- If \( b \leq -1 \), then \( 1 - b \geq 2 > 0 \) and hence \( a(1 - b) < 0 \) is not valid.

In both cases, we find that \( a - xb \geq 0 \).

### Conclusion
From both cases, we conclude that:
\[ a - xb \geq 0 \]
Thus, the correct answer is:
**Option B: \( a - xb \geq 0 \)**.