`if f(x) ={{:( x-1, xlt 0),( x^(2) - 2x , x ge 0):}` , then

Options

A. f(|x|) is discontinuous at x=0
B. f(x) | is differerntiavble at x=0
C. |f(x)| is non - differentiable at =0,2
D. |f(x)| is comtinuous at x=0

Answer

To solve the problem step by step, we need to analyze the function \( f(x) \) given by:

\[ f(x) = \begin{cases} x - 1 & \text{if } x < 0 \\ x^2 - 2x & \text{if } x \geq 0 \end{cases} \]

We need to evaluate the continuity and differentiability of \( f(\lvert x \rvert) \) and \( \lvert f(x) \rvert \) at specific points, particularly at \( x = 0 \) and \( x = 2 \).

### Step 1: Check Continuity of \( f(\lvert x \rvert) \) at \( x = 0 \)

1. **Evaluate \( f(\lvert x \rvert) \)**:
- For \( x < 0 \): \( \lvert x \rvert = -x \), so \( f(\lvert x \rvert) = f(-x) = -x - 1 \).
- For \( x \geq 0 \): \( \lvert x \rvert = x \), so \( f(\lvert x \rvert) = f(x) = x^2 - 2x \).

2. **Calculate the left-hand limit as \( x \to 0^- \)**:
\[ \lim_{x \to 0^-} f(\lvert x \rvert) = \lim_{x \to 0^-} (-x - 1) = -1 \]

3. **Calculate the right-hand limit as \( x \to 0^+ \)**:
\[ \lim_{x \to 0^+} f(\lvert x \rvert) = \lim_{x \to 0^+} (x^2 - 2x) = 0 \]

4. **Check continuity**:
Since the left-hand limit \(-1\) is not equal to the right-hand limit \(0\), \( f(\lvert x \rvert) \) is **not continuous** at \( x = 0 \).

### Step 2: Check Differentiability of \( f(\lvert x \rvert) \) at \( x = 0 \)

Since \( f(\lvert x \rvert) \) is not continuous at \( x = 0 \), it cannot be differentiable at that point.

### Step 3: Check Continuity of \( \lvert f(x) \rvert \) at \( x = 0 \)

1. **Evaluate \( \lvert f(x) \rvert \)**:
- For \( x < 0 \): \( f(x) = x - 1 \), thus \( \lvert f(x) \rvert = \lvert x - 1 \rvert = 1 - x \) (since \( x - 1 < 0 \)).
- For \( x \geq 0 \): \( f(x) = x^2 - 2x \), thus \( \lvert f(x) \rvert = x^2 - 2x \) (since \( x^2 - 2x \geq 0 \) at \( x = 0 \)).

2. **Calculate the left-hand limit as \( x \to 0^- \)**:
\[ \lim_{x \to 0^-} \lvert f(x) \rvert = \lim_{x \to 0^-} (1 - x) = 1 \]

3. **Calculate the right-hand limit as \( x \to 0^+ \)**:
\[ \lim_{x \to 0^+} \lvert f(x) \rvert = \lim_{x \to 0^+} (x^2 - 2x) = 0 \]

4. **Check continuity**:
Since the left-hand limit \(1\) is not equal to the right-hand limit \(0\), \( \lvert f(x) \rvert \) is **not continuous** at \( x = 0 \).

### Step 4: Check Differentiability of \( \lvert f(x) \rvert \) at \( x = 2 \)

1. **Evaluate \( \lvert f(x) \rvert \)** at \( x = 2 \):
- For \( x < 2 \): \( f(x) = x^2 - 2x \).
- For \( x \geq 2 \): \( f(x) = x^2 - 2x \).

2. **Calculate the left-hand derivative as \( x \to 2^- \)**:
\[ \lim_{h \to 0^-} \frac{f(2 + h) - f(2)}{h} = \lim_{h \to 0^-} \frac{(2 + h)^2 - 2(2 + h) - (2^2 - 2 \cdot 2)}{h} \]
Simplifying gives \( 2 \).

3. **Calculate the right-hand derivative as \( x \to 2^+ \)**:
\[ \lim_{h \to 0^+} \frac{f(2 + h) - f(2)}{h} = \lim_{h \to 0^+} \frac{(2 + h)^2 - 2(2 + h) - (2^2 - 2 \cdot 2)}{h} \]
Simplifying gives \( 2 \).

4. **Check differentiability**:
Since the left-hand and right-hand derivatives are equal, \( \lvert f(x) \rvert \) is **differentiable** at \( x = 2 \).

### Conclusion

- \( f(\lvert x \rvert) \) is not continuous at \( x = 0 \) and therefore not differentiable at \( x = 0 \).
- \( \lvert f(x) \rvert \) is not continuous at \( x = 0 \) and is differentiable at \( x = 2 \. ### Final Answer The correct option is that \( \lvert f(x) \rvert \) is non-differentiable at \( x = 0 \) and \( x = 2 \).