If `f(x) = |sinx|`, then

Options

A. f is everywhere differentiable
B. f is everywhere continuous but not differentiable at `x = npi, n in Z`
C. f is everywhere continuous but not differentiable at `x = (2n+1)""(pi)/(2), n in Z`
D. None of the above

Explanation

To solve the problem, we need to analyze the function \( f(x) = |\sin x| \) in terms of its continuity and differentiability.

### Step-by-step Solution:

1. **Understanding the Function**:
The function \( f(x) = |\sin x| \) takes the sine function and reflects any negative values above the x-axis. This means where \( \sin x \) is negative, \( |\sin x| \) will be positive.

2. **Graphing the Function**:
To visualize \( f(x) \), we can sketch the graph of \( \sin x \) and then reflect the portions below the x-axis:
- From \( 0 \) to \( \pi \), \( \sin x \) is non-negative, so \( f(x) = \sin x \).
- From \( \pi \) to \( 2\pi \), \( \sin x \) is negative, so \( f(x) = -\sin x \).
- This pattern continues for each interval of \( [n\pi, (n+1)\pi] \).

3. **Identifying Points of Non-Differentiability**:
The points where \( \sin x \) crosses the x-axis are \( x = n\pi \) for \( n \in \mathbb{Z} \). At these points, the function \( f(x) \) has a sharp corner (or cusp), which makes it non-differentiable.

4. **Checking Continuity**:
The function \( f(x) \) is continuous everywhere because there are no breaks in the graph. The limit from the left and right at any point \( x \) matches the function value at that point.

5. **Conclusion**:
Therefore, we conclude that \( f(x) = |\sin x| \) is continuous everywhere but not differentiable at points \( x = n\pi \).

### Final Answer:
The function \( f(x) = |\sin x| \) is continuous everywhere but not differentiable at \( x = n\pi \) for \( n \in \mathbb{Z} \).