`Let f(x) ={{:( [x],-2le xle -(1)/(2) ),( 2x^(2)-1,-(1)/(2)lt xle2):}and g(x) =f(|x|)+|f(x)|,` where [.] represents the greatest integer function .
the number of point where g(x) is discontinuous is

Options

A. 1
B. 2
C. 3
D. none of these

Answer

To determine the number of points where the function \( g(x) = f(|x|) + |f(x)| \) is discontinuous, we first need to analyze the function \( f(x) \).

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

The function \( f(x) \) is defined piecewise as follows:
- For \( -2 \leq x \leq -\frac{1}{2} \), \( f(x) = [x] \) (the greatest integer function).
- For \( -\frac{1}{2} < x \leq 2 \), \( f(x) = 2x^2 - 1 \).

### Step 2: Analyze \( f(|x|) \) and \( |f(x)| \)

1. **For \( |x| \)**:
- If \( x \geq 0 \), then \( |x| = x \).
- If \( x < 0 \), then \( |x| = -x \).

2. **For \( f(|x|) \)**:
- If \( x \geq 0 \): \( f(|x|) = f(x) = 2x^2 - 1 \) for \( 0 \leq x \leq 2 \).
- If \( x < 0 \): \( f(|x|) = f(-x) = [ -x ] \) for \( -2 \leq -x \leq -\frac{1}{2} \).

3. **For \( |f(x)| \)**:
- If \( -2 \leq x \leq -\frac{1}{2} \): \( f(x) = [x] \) which can be negative.
- If \( -\frac{1}{2} < x \leq 2 \): \( f(x) = 2x^2 - 1 \), which is non-negative for \( x \geq \frac{1}{\sqrt{2}} \).

### Step 3: Determine \( g(x) \)

Now we can express \( g(x) \) based on the above analysis:
- For \( -2 \leq x < -\frac{1}{2} \):
\[ g(x) = [|x|] + |[x]| = [ -x ] + |[ -x ]| = [ -x ] + [ -x ] = 2[-x] \]
- For \( -\frac{1}{2} < x \leq 2 \):
\[ g(x) = f(x) + |f(x)| = (2x^2 - 1) + |(2x^2 - 1)| \]

### Step 4: Identify discontinuities

To find points of discontinuity, we need to check where \( g(x) \) changes its definition:
1. **At \( x = -2 \)** and **\( x = -\frac{1}{2} \)**: Check limits.
2. **At \( x = 0 \)**: Check limits.
3. **At \( x = \frac{1}{\sqrt{2}} \)**: Check limits.
4. **At \( x = 2 \)**: Check limits.

### Step 5: Check continuity at critical points

1. **At \( x = -1 \)**:
- Left-hand limit: \( g(-1) = 3 \)
- Right-hand limit: \( g(-1) = 1 \)
- Discontinuous.

2. **At \( x = -\frac{1}{2} \)**:
- Left-hand limit: \( g(-\frac{1}{2}) = 1 \)
- Right-hand limit: \( g(-\frac{1}{2}) = 0 \)
- Discontinuous.

3. **At \( x = \frac{1}{\sqrt{2}} \)**:
- Left-hand limit: \( g(\frac{1}{\sqrt{2}}) = 0 \)
- Right-hand limit: \( g(\frac{1}{\sqrt{2}}) = 0 \)
- Continuous.

4. **At \( x = 2 \)**:
- Left-hand limit: \( g(2) = 4 \)
- Right-hand limit: \( g(2) = 4 \)
- Continuous.

### Conclusion

The function \( g(x) \) is discontinuous at two points: \( x = -1 \) and \( x = -\frac{1}{2} \).

Thus, the number of points where \( g(x) \) is discontinuous is **2**.

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