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 determine the continuity and differentiability of the function \( f(x) = |\sin x| \), we will analyze the function step by step.

### Step 1: Understanding the Function
The function \( f(x) = |\sin x| \) takes the sine of \( x \) and applies the absolute value. This means that wherever \( \sin x \) is negative, \( f(x) \) will reflect that value above the x-axis.

**Hint:** Consider the behavior of \( \sin x \) over its period to understand where it is positive and negative.

### Step 2: Analyzing the Graph
The graph of \( \sin x \) oscillates between -1 and 1. The points where \( \sin x = 0 \) (i.e., \( x = n\pi \), where \( n \) is an integer) will be the points where \( f(x) \) changes direction, creating a "corner" in the graph.

**Hint:** Identify the key points of \( \sin x \) where it crosses the x-axis.

### Step 3: Continuity of \( f(x) \)
To check for continuity, we need to see if there are any breaks in the function. Since \( |\sin x| \) is defined for all \( x \) and does not have any jumps or breaks, it is continuous everywhere.

**Hint:** Use the definition of continuity: a function is continuous at a point if the limit as you approach that point equals the function's value at that point.

### Step 4: Differentiability of \( f(x) \)
Next, we check for differentiability. A function is differentiable at a point if it has a defined derivative there. The function \( f(x) = |\sin x| \) has corners (non-differentiable points) at \( x = n\pi \) because the slope changes abruptly at these points.

**Hint:** Look for points where the function has sharp turns or corners, as these indicate non-differentiability.

### Step 5: Conclusion
1. **Continuity:** \( f(x) = |\sin x| \) is continuous for all \( x \in \mathbb{R} \).
2. **Differentiability:** \( f(x) \) is not differentiable at \( x = n\pi \) (where \( n \) is an integer) because of the corners.

Thus, we conclude that \( f(x) \) is continuous everywhere but not differentiable at integer multiples of \( \pi \).

**Final Answer:**
- \( f(x) \) is continuous for all \( x \in \mathbb{R} \).
- \( f(x) \) is not differentiable at \( x = n\pi \) (where \( n \) is an integer).