`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 non - differentiable is

Options

A. 4
B. 5
C. 2
D. 3

Answer

To solve the problem, we need to analyze the functions \( f(x) \) and \( g(x) \) given in the question.

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

The function \( f(x) \) is defined piecewise as follows:

\[ f(x) = \begin{cases} [x] & \text{for } -2 \leq x \leq -\frac{1}{2} \\ 2x^2 - 1 & \text{for } -\frac{1}{2} < x \leq 2 \end{cases} \]

### Step 2: Define the function \( g(x) \)

The function \( g(x) \) is defined as:

\[ g(x) = f(|x|) + |f(x)| \]

### Step 3: Analyze \( g(x) \) for different intervals

1. **For \( x < -2 \)**:
- \( f(|x|) = f(-x) = f(-(-x)) = f(x) \) is not defined since \( -x > 2 \).
- So, \( g(x) \) is not defined.

2. **For \( -2 \leq x < -\frac{1}{2} \)**:
- \( f(|x|) = f(-x) = [x] \) and \( |f(x)| = |[x]| = -[x] \) (since \( [x] \) is negative).
- Thus, \( g(x) = [x] - [x] = 0 \).

3. **For \( -\frac{1}{2} < x < 0 \)**:
- \( f(|x|) = f(-x) = 2(-x)^2 - 1 = 2x^2 - 1 \) and \( |f(x)| = |[x]| = 0 \).
- Thus, \( g(x) = 2x^2 - 1 \).

4. **For \( 0 \leq x < \frac{1}{2} \)**:
- \( f(|x|) = f(x) = 2x^2 - 1 \) and \( |f(x)| = |2x^2 - 1| \).
- Thus, \( g(x) = 2x^2 - 1 + |2x^2 - 1| \).

5. **For \( \frac{1}{2} \leq x < 2 \)**:
- \( f(|x|) = f(x) = 2x^2 - 1 \) and \( |f(x)| = 2x^2 - 1 \).
- Thus, \( g(x) = 2x^2 - 1 + (2x^2 - 1) = 4x^2 - 2 \).

### Step 4: Identify points of non-differentiability

To find points of non-differentiability, we check the points where \( g(x) \) changes its definition:

1. **At \( x = -1 \)**:
- Left-hand limit: \( g(-1) = 0 \)
- Right-hand limit: \( g(-1) = 2(-1)^2 - 1 = 1 \)
- Since the left-hand limit does not equal the right-hand limit, \( g(x) \) is discontinuous at \( x = -1 \).

2. **At \( x = -\frac{1}{2} \)**:
- Left-hand limit: \( g(-\frac{1}{2}) = 0 \)
- Right-hand limit: \( g(-\frac{1}{2}) = 2(-\frac{1}{2})^2 - 1 = 0 \)
- Since the left-hand limit equals the right-hand limit, we need to check differentiability.

3. **At \( x = \frac{1}{2} \)**:
- Left-hand limit: \( g(\frac{1}{2}) = 2(\frac{1}{2})^2 - 1 = 0 \)
- Right-hand limit: \( g(\frac{1}{2}) = 4(\frac{1}{2})^2 - 2 = 0 \)
- Since the left-hand limit equals the right-hand limit, we need to check differentiability.

4. **At \( x = 1 \)**:
- Left-hand limit: \( g(1) = 4(1)^2 - 2 = 2 \)
- Right-hand limit: \( g(1) = 4(1)^2 - 2 = 2 \)
- Since the left-hand limit equals the right-hand limit, we need to check differentiability.

### Step 5: Check differentiability at critical points

1. **At \( x = -1 \)**: Non-differentiable (discontinuous).
2. **At \( x = -\frac{1}{2} \)**: Non-differentiable (check derivatives).
3. **At \( x = \frac{1}{2} \)**: Non-differentiable (check derivatives).
4. **At \( x = 1 \)**: Non-differentiable (check derivatives).

### Conclusion

The points where \( g(x) \) is non-differentiable are \( x = -1 \), \( x = -\frac{1}{2} \), and \( x = \frac{1}{2} \).

Thus, the total number of points where \( g(x) \) is non-differentiable is **3**.