If `f(x)=sgn(x)={(|x|)/x,x!=0, 0, x=0 and g(x)=f(f(x)),`then at `x=0,g(x)` is

Options

A. continuous and differentiable
B. continuous but not differentiable
C. differentiable but not continuous
D. neither continuous nor differentiable

Explanation

To solve the problem, we need to analyze the functions \( f(x) \) and \( g(x) \) step by step.

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

The function \( f(x) \) is defined as follows:
\[ f(x) = \begin{cases} 1 & \text{if } x > 0 \\ 0 & \text{if } x = 0 \\ -1 & \text{if } x < 0 \end{cases} \]

### Step 2: Determine \( g(x) = f(f(x)) \)

Now, we need to find \( g(x) = f(f(x)) \). We will evaluate \( f(f(x)) \) for three cases: \( x > 0 \), \( x = 0 \), and \( x < 0 \).

1. **For \( x > 0 \)**:
- \( f(x) = 1 \)
- Therefore, \( g(x) = f(f(x)) = f(1) = 1 \)

2. **For \( x = 0 \)**:
- \( f(x) = 0 \)
- Therefore, \( g(x) = f(f(x)) = f(0) = 0 \)

3. **For \( x < 0 \)**:
- \( f(x) = -1 \)
- Therefore, \( g(x) = f(f(x)) = f(-1) = -1 \)

### Step 3: Summarize the function \( g(x) \)

Now we can summarize \( g(x) \):
\[ g(x) = \begin{cases} 1 & \text{if } x > 0 \\ 0 & \text{if } x = 0 \\ -1 & \text{if } x < 0 \end{cases} \]

### Step 4: Check continuity at \( x = 0 \)

To check if \( g(x) \) is continuous at \( x = 0 \), we need to see if:
\[ \lim_{x \to 0^-} g(x) = g(0) = \lim_{x \to 0^+} g(x) \]

- \( \lim_{x \to 0^-} g(x) = -1 \)
- \( g(0) = 0 \)
- \( \lim_{x \to 0^+} g(x) = 1 \)

Since \( -1 \neq 0 \neq 1 \), \( g(x) \) is not continuous at \( x = 0 \).

### Step 5: Check differentiability at \( x = 0 \)

A function must be continuous at a point to be differentiable there. Since \( g(x) \) is not continuous at \( x = 0 \), it cannot be differentiable at that point.

### Conclusion

Thus, at \( x = 0 \), \( g(x) \) is neither continuous nor differentiable.

### Final Answer
At \( x = 0 \), \( g(x) \) is neither continuous nor differentiable.

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