The set of all values of `' x '` which satisfies the inequation `|1-(|x|)/(1+|x|)|geq1/2` is: `[-1,1]` (b) `(-oo,-1)` `(1,oo)` (d) `(0,1)`

Options

A. `[(3)/(2), oo)`
B. `(-oo, -1)uu[(3)/(2), oo)`
C. `(-1, 0)uu[(3)/(2), oo)`
D. `R - {0, -1}`

Explanation

To solve the inequation \( |1 - \frac{|x|}{1 + |x|}| \geq \frac{1}{2} \), we will break it down step by step.

### Step 1: Analyze the expression
The expression inside the absolute value is \( 1 - \frac{|x|}{1 + |x|} \). We need to find when this expression is greater than or equal to \( \frac{1}{2} \) or less than or equal to \( -\frac{1}{2} \).

### Step 2: Set up the two cases
We will consider two cases based on the definition of absolute value:
1. \( 1 - \frac{|x|}{1 + |x|} \geq \frac{1}{2} \)
2. \( 1 - \frac{|x|}{1 + |x|} \leq -\frac{1}{2} \)

### Step 3: Solve the first case
**Case 1:**
\[ 1 - \frac{|x|}{1 + |x|} \geq \frac{1}{2} \]
Subtract 1 from both sides:
\[ -\frac{|x|}{1 + |x|} \geq -\frac{1}{2} \]
Multiply through by -1 (remember to flip the inequality):
\[ \frac{|x|}{1 + |x|} \leq \frac{1}{2} \]
Cross-multiply:
\[ 2|x| \leq 1 + |x| \]
Rearranging gives:
\[ 2|x| - |x| \leq 1 \]
\[ |x| \leq 1 \]
This means:
\[ -1 \leq x \leq 1 \]

### Step 4: Solve the second case
**Case 2:**
\[ 1 - \frac{|x|}{1 + |x|} \leq -\frac{1}{2} \]
Subtract 1 from both sides:
\[ -\frac{|x|}{1 + |x|} \leq -\frac{3}{2} \]
Multiply through by -1 (flip the inequality):
\[ \frac{|x|}{1 + |x|} \geq \frac{3}{2} \]
Cross-multiply:
\[ 2|x| \geq 3 + 3|x| \]
Rearranging gives:
\[ 2|x| - 3|x| \geq 3 \]
\[ -|x| \geq 3 \]
This is not possible since \( |x| \) cannot be negative. Thus, there are no solutions from this case.

### Conclusion
The only valid solution comes from Case 1, which gives us:
\[ x \in [-1, 1] \]

### Final Answer
The set of all values of \( x \) which satisfies the inequation is:
\[ \boxed{[-1, 1]} \]