Let f(x) = |sin x| .Then

Options

A. f is everywhere differentiable
B. f is everywhere continuous but no differentiable at x = n`pi,n epsilon`Z
C. f is everywhere continuous but not differentiable at x=`(2n+1) pi/2,n epsilon Z`
D. None of these

Answer

To solve the problem, we need to analyze the function \( f(x) = |\sin x| \) and determine its properties, particularly its continuity and differentiability.

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

### Step 2: Graphing the Function
To visualize \( f(x) \), we can sketch the graph of \( \sin x \) and then reflect the negative portions above the x-axis:
- The sine function oscillates between -1 and 1.
- The points where \( \sin x = 0 \) (i.e., \( x = n\pi \) for \( n \in \mathbb{Z} \)) will remain at 0.
- The points where \( \sin x < 0 \) will be reflected to positive values.

### Step 3: Identifying Points of Non-Differentiability
The function \( f(x) \) will not be differentiable at points where \( \sin x \) changes from negative to positive, which occurs at:
- \( x = n\pi \) for \( n \in \mathbb{Z} \) (where \( \sin x \) crosses the x-axis).

At these points, the left-hand derivative and the right-hand derivative will not be equal, indicating non-differentiability.

### Step 4: Conclusion
Thus, we conclude that:
- The function \( f(x) = |\sin x| \) is continuous everywhere because the sine function is continuous and the absolute value function does not introduce any discontinuities.
- The function is not differentiable at \( x = n\pi \) for \( n \in \mathbb{Z} \).

### Summary of Results
- **Continuity**: \( f(x) \) is continuous for all \( x \).
- **Differentiability**: \( f(x) \) is not differentiable at \( x = n\pi \) for \( n \in \mathbb{Z} \).