Assertion (A) `|sinx|` is continuous at x=0 .
Reason (R ) : |sinx| is differentiable at x=0.

Options

A. Both A and R are true and R is the correct explanation of A
B. Both A and R are true but R is NOT the correct explanation of A
C. A is true but R is false
D. A is false and R is true

Answer

To solve the problem, we need to analyze the assertion and the reason provided.

**Assertion (A)**: \(|\sin x|\) is continuous at \(x = 0\).
**Reason (R)**: \(|\sin x|\) is differentiable at \(x = 0\).

### Step-by-Step Solution:

1. **Check Continuity of \(|\sin x|\) at \(x = 0\)**:
- A function \(f(x)\) is continuous at a point \(c\) if:
\[ \lim_{x \to c} f(x) = f(c) \]
- Here, we need to check if \(\lim_{x \to 0} |\sin x| = |\sin(0)|\).
- We know that \(\sin(0) = 0\), so \(|\sin(0)| = 0\).
- Now, calculate the limit:
\[ \lim_{x \to 0} |\sin x| = |\sin(0)| = 0 \]
- Since both sides are equal, \(|\sin x|\) is continuous at \(x = 0\).

2. **Check Differentiability of \(|\sin x|\) at \(x = 0\)**:
- A function is differentiable at a point if the derivative exists at that point.
- We need to find the left-hand and right-hand derivatives at \(x = 0\).
- For \(x \geq 0\), \(|\sin x| = \sin x\).
\[ f'(0^+) = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{\sin h - 0}{h} = \lim_{h \to 0^+} \frac{\sin h}{h} = 1 \]
- For \(x < 0\), \(|\sin x| = -\sin x\).
\[ f'(0^-) = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{-\sin h - 0}{h} = \lim_{h \to 0^-} \frac{-\sin h}{h} = -1 \]
- Since the left-hand derivative \(f'(0^-)\) and the right-hand derivative \(f'(0^+)\) are not equal:
\[ f'(0^+) = 1 \quad \text{and} \quad f'(0^-) = -1 \]
- Thus, \(|\sin x|\) is not differentiable at \(x = 0\).

3. **Conclusion**:
- The assertion (A) is true: \(|\sin x|\) is continuous at \(x = 0\).
- The reason (R) is false: \(|\sin x|\) is not differentiable at \(x = 0\).

### Final Answer:
- Assertion (A) is true, and Reason (R) is false.