Statement I f(x) = |x| sin x is differentiable at x = 0.
Statement II If g(x) is not differentiable at x = a and h(x) is differentiable at x = a, then g(x).h(x) cannot be differentiable at x = a

Options

A. A. Statement I is correct, Statement II is also correct, Statement II is the correct explanation of Statement I
B. B. Statement I is correct, Statement II is also correct, Statement II is not the correct explanation of Statement I
C. C. Statement I is correct, Statement II is incorrect
D. D. Statement I is incorrect, Statement II is correct.

Answer

To solve the given problem, we need to analyze both statements one by one.

### Step 1: Analyze Statement I
The first statement is about the function \( f(x) = |x| \sin x \) and whether it is differentiable at \( x = 0 \).

1. **Definition of the Function**:
- For \( x \geq 0 \): \( f(x) = x \sin x \)
- For \( x < 0 \): \( f(x) = -x \sin x \)

2. **Finding the Derivative**:
We need to find the left-hand derivative and the right-hand derivative at \( x = 0 \).

- **Right-hand derivative** (\( f'(0^+) \)):
\[ f'(0^+) = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h \sin h - 0}{h} = \lim_{h \to 0^+} \sin h = \sin(0) = 0 \]

- **Left-hand derivative** (\( f'(0^-) \)):
\[ f'(0^-) = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{-h \sin h - 0}{h} = \lim_{h \to 0^-} -\sin h = -\sin(0) = 0 \]

3. **Conclusion for Statement I**:
Since both the left-hand derivative and the right-hand derivative at \( x = 0 \) are equal (both are 0), we conclude that \( f(x) \) is differentiable at \( x = 0 \). Thus, Statement I is **True**.

### Step 2: Analyze Statement II
The second statement claims that if \( g(x) \) is not differentiable at \( x = a \) and \( h(x) \) is differentiable at \( x = a \), then \( g(x) h(x) \) cannot be differentiable at \( x = a \).

1. **Understanding the Statement**:
- Let \( g(x) \) be a function that is not differentiable at \( x = a \).
- Let \( h(x) \) be a function that is differentiable at \( x = a \).

2. **Product of Functions**:
The product \( g(x) h(x) \) can potentially be differentiable even if one of the functions is not differentiable. A classic example is:
- Let \( g(x) = |x| \) (not differentiable at \( x = 0 \)).
- Let \( h(x) = x \) (differentiable everywhere).

Then:
\[ g(x) h(x) = |x| \cdot x = x^2 \]
which is differentiable at \( x = 0 \).

3. **Conclusion for Statement II**:
Since we found a counterexample where \( g(x) h(x) \) is differentiable even though \( g(x) \) is not, we conclude that Statement II is **False**.

### Final Conclusion
- Statement I is **True**.
- Statement II is **False**.