Let `f(x)=|x|+|sin x|, x in (-pi//2,pi//2).` Then, f is

Options

A. nowhere continuous
B. continuous and differentiable everywhere
C. nowhere differentiable
D. differentiable everywhere except at x=0

Explanation

To determine the properties of the function \( f(x) = |x| + |\sin x| \) over the interval \( x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \), we will analyze its continuity and differentiability step by step.

### Step 1: Analyze the function
The function \( f(x) \) is composed of two parts: \( |x| \) and \( |\sin x| \). We need to consider the behavior of these two components separately.

### Step 2: Break down the intervals
1. For \( x < 0 \):
\[ f(x) = -x + (-\sin x) = -x - \sin x \]
2. For \( x = 0 \):
\[ f(0) = |0| + |\sin(0)| = 0 + 0 = 0 \]
3. For \( x > 0 \):
\[ f(x) = x + \sin x \]

### Step 3: Check continuity at \( x = 0 \)
To check continuity at \( x = 0 \), we need to evaluate the left-hand limit and right-hand limit as \( x \) approaches 0.

- **Left-hand limit** as \( x \to 0^- \):
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-x - \sin x) = -0 - \sin(0) = 0 \]

- **Right-hand limit** as \( x \to 0^+ \):
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x + \sin x) = 0 + \sin(0) = 0 \]

Since both limits equal \( f(0) \):
\[ \lim_{x \to 0} f(x) = f(0) = 0 \]
Thus, \( f(x) \) is continuous at \( x = 0 \).

### Step 4: Check differentiability at \( x = 0 \)
To check differentiability, we need to find the left-hand derivative and right-hand derivative at \( x = 0 \).

- **Left-hand derivative**:
\[ f'(x) = \frac{d}{dx}(-x - \sin x) = -1 - \cos x \]
Evaluating at \( x = 0 \):
\[ f'_-(0) = -1 - \cos(0) = -1 - 1 = -2 \]

- **Right-hand derivative**:
\[ f'(x) = \frac{d}{dx}(x + \sin x) = 1 + \cos x \]
Evaluating at \( x = 0 \):
\[ f'_+(0) = 1 + \cos(0) = 1 + 1 = 2 \]

Since the left-hand derivative \( f'_-(0) = -2 \) and the right-hand derivative \( f'_+(0) = 2 \) are not equal, \( f(x) \) is not differentiable at \( x = 0 \).

### Conclusion
The function \( f(x) \) is continuous everywhere in the interval \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \) but not differentiable at \( x = 0 \). Therefore, the final answer is:

**Answer: \( f \) is continuous everywhere but not differentiable at \( x = 0 \).**