`if f(x) ={{:((1-|x|)/(1+x),xne-1),(1, x=-1):}`then f([2x]) , where [.] represents the greatest integer function , is

Options

A. discontinuous at x=-1
B. continuous at x=0
C. continuous at x=1/2
D. continuous at x=1

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined as:

\[ f(x) = \begin{cases} \frac{1 - |x|}{1 + x} & \text{if } x \neq -1 \\ 1 & \text{if } x = -1 \end{cases} \]

We need to find \( f([\cdot]) \) where \( [2x] \) represents the greatest integer function (GIF) applied to \( 2x \).

### Step 1: Analyze the function \( f(x) \)

1. **For \( x < 0 \)**:
- The function simplifies to \( f(x) = 1 \).

2. **For \( x \geq 0 \)**:
- The modulus function \( |x| = x \), so:
\[ f(x) = \frac{1 - x}{1 + x} \]

### Step 2: Determine \( f([\cdot]) \) for \( [2x] \)

We will now evaluate \( f([\cdot]) \) based on the value of \( [2x] \).

#### Case 1: \( x < 0 \)
- Here, \( [2x] \) will be negative or zero.
- Thus, \( f([2x]) = 1 \).

#### Case 2: \( 0 \leq x < \frac{1}{2} \)
- Here, \( 2x \) will be in the range \( [0, 1) \), so \( [2x] = 0 \).
- Thus, \( f([2x]) = f(0) = \frac{1 - 0}{1 + 0} = 1 \).

#### Case 3: \( \frac{1}{2} \leq x < 1 \)
- Here, \( 2x \) will be in the range \( [1, 2) \), so \( [2x] = 1 \).
- Thus, \( f([2x]) = f(1) = \frac{1 - 1}{1 + 1} = 0 \).

#### Case 4: \( 1 \leq x < \frac{3}{2} \)
- Here, \( 2x \) will be in the range \( [2, 3) \), so \( [2x] = 2 \).
- Thus, \( f([2x]) = f(2) = \frac{1 - 2}{1 + 2} = \frac{-1}{3} \).

### Step 3: Summary of Results

- For \( x < 0 \): \( f([\cdot]) = 1 \)
- For \( 0 \leq x < \frac{1}{2} \): \( f([\cdot]) = 1 \)
- For \( \frac{1}{2} \leq x < 1 \): \( f([\cdot]) = 0 \)
- For \( 1 \leq x < \frac{3}{2} \): \( f([\cdot]) = -\frac{1}{3} \)

### Conclusion

The function \( f([\cdot]) \) is piecewise defined as follows:

\[ f([\cdot]) = \begin{cases} 1 & \text{if } x < \frac{1}{2} \\ 0 & \text{if } \frac{1}{2} \leq x < 1 \\ -\frac{1}{3} & \text{if } 1 \leq x < \frac{3}{2} \end{cases} \]