`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 `|f(x)|` is non - differentiable is

Options

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

Answer

To solve the problem, we need to analyze the function \( f(x) \) and determine the points where \( |f(x)| \) is non-differentiable.

### Step-by-step Solution:

1. **Define the function \( f(x) \)**:
The function \( f(x) \) is given as:
\[ f(x) = \begin{cases} [x] & \text{for } -2 \leq x \leq -\frac{1}{2} \\ 2x^2 - 1 & \text{for } -\frac{1}{2} < x \leq 2 \end{cases} \]
Here, \([x]\) is the greatest integer function.

2. **Evaluate \( f(x) \) at critical points**:
- For \( x = -2 \): \( f(-2) = [-2] = -2 \)
- For \( x = -\frac{1}{2} \): \( f(-\frac{1}{2}) = [-\frac{1}{2}] = -1 \)
- For \( x = 0 \): \( f(0) = 2(0)^2 - 1 = -1 \)
- For \( x = 1 \): \( f(1) = 2(1)^2 - 1 = 1 \)
- For \( x = 2 \): \( f(2) = 2(2)^2 - 1 = 7 \)

3. **Identify points of non-differentiability for \( |f(x)| \)**:
We need to check where \( |f(x)| \) is non-differentiable. The absolute value function can introduce non-differentiability at points where \( f(x) = 0 \) or where \( f(x) \) is not differentiable.

- **Find where \( f(x) = 0 \)**:
- For \( 2x^2 - 1 = 0 \):
\[ 2x^2 = 1 \implies x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}} \approx \pm 0.707 \]
- For \( -2 \leq x \leq -\frac{1}{2} \), \( f(x) = [x] \) is never zero since \([x]\) takes values -2, -1, and 0 in that interval.

4. **Check differentiability at critical points**:
- **At \( x = -1 \)**:
- \( f(x) \) transitions from \(-2\) to \(-1\), which is a jump discontinuity.
- **At \( x = -\frac{1}{2} \)**:
- \( f(x) \) transitions from \(-1\) to \(0\), which is also a jump discontinuity.
- **At \( x = \frac{1}{\sqrt{2}} \)**:
- Check left-hand and right-hand derivatives:
- Left-hand derivative: \( \lim_{x \to \frac{1}{\sqrt{2}}^-} \frac{d}{dx}(2x^2 - 1) = 4\left(\frac{1}{\sqrt{2}}\right) = 2\sqrt{2} \)
- Right-hand derivative: \( \lim_{x \to \frac{1}{\sqrt{2}}^+} \frac{d}{dx}(2x^2 - 1) = 4\left(\frac{1}{\sqrt{2}}\right) = 2\sqrt{2} \)
- Since both derivatives are equal, \( |f(x)| \) is differentiable at this point.

5. **Conclusion**:
The points where \( |f(x)| \) is non-differentiable are:
- \( x = -1 \)
- \( x = -\frac{1}{2} \)
- \( x = \frac{1}{\sqrt{2}} \)

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

### Final Answer:
The number of points where \( |f(x)| \) is non-differentiable is **3**.

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