Let `f(x)""=""x|x|""and""g(x)""=""sinx in x`

Statement 1 : gof is differentiable at `x""=""0` and its derivative is continuous at that point
Statement 2: gof is twice differentiable at `x""=""0`

(1) Statement1 is true, Statement2 is true, Statement2 is a correct explanation for statement1
(2) Statement1 is true, Statement2 is true; Statement2 is not a correct explanation for statement1.
(3) Statement1 is true, statement2 is false.
(4) Statement1 is false, Statement2 is true

Options

A. Statement-1 is True, Statement -2 is True, Statement -2 is True and Statement -2 is a correct explanation for statement -1.
B. Statement -1 is True, Statement -2 is True and Statement -2 is Not is a correct explanation for Statement -1.
C. Statement -1 is True, Statement -2 is False.
D. Statement -1 is False, Statement -2 is True.

Answer

To solve the problem, we need to analyze the functions \( f(x) = x |x| \) and \( g(x) = \sin x \), and evaluate the statements regarding the differentiability of the composition \( g(f(x)) \) at \( x = 0 \).

### Step 1: Define the functions
We start with the definitions of the functions:
- \( f(x) = x |x| \)
- \( g(x) = \sin x \)

### Step 2: Analyze \( f(x) \)
The function \( f(x) = x |x| \) can be rewritten as:
- \( f(x) = x^2 \) for \( x \geq 0 \)
- \( f(x) = -x^2 \) for \( x < 0 \)

### Step 3: Find \( g(f(x)) \)
Now we compute \( g(f(x)) \):
- For \( x \geq 0 \):
\[ g(f(x)) = g(x^2) = \sin(x^2) \]
- For \( x < 0 \):
\[ g(f(x)) = g(-x^2) = \sin(-x^2) = -\sin(x^2) \]

Thus, we can express \( g(f(x)) \) as:
\[ g(f(x)) = \begin{cases} \sin(x^2) & \text{if } x \geq 0 \\ -\sin(x^2) & \text{if } x < 0 \end{cases} \]

### Step 4: Check differentiability at \( x = 0 \)
To check if \( g(f(x)) \) is differentiable at \( x = 0 \), we need to find the derivative from both sides.

**For \( x \geq 0 \)**:
\[ g(f'(x)) = \frac{d}{dx}(\sin(x^2)) = 2x \cos(x^2) \]

**For \( x < 0 \)**:
\[ g(f'(x)) = \frac{d}{dx}(-\sin(x^2)) = -2x \cos(x^2) \]

### Step 5: Evaluate the derivatives at \( x = 0 \)
Now we evaluate the derivatives at \( x = 0 \):
- From the right (as \( x \to 0^+ \)):
\[ \lim_{x \to 0^+} 2x \cos(x^2) = 0 \]
- From the left (as \( x \to 0^- \)):
\[ \lim_{x \to 0^-} -2x \cos(x^2) = 0 \]

Since both limits are equal, \( g(f(x)) \) is differentiable at \( x = 0 \).

### Step 6: Check continuity of the derivative
Next, we check if the derivative \( g(f'(x)) \) is continuous at \( x = 0 \):
- As \( x \to 0^+ \):
\[ g(f'(0^+)) = 0 \]
- As \( x \to 0^- \):
\[ g(f'(0^-)) = 0 \]

Since both limits are equal and equal to \( g(f'(0)) \), the derivative is continuous at \( x = 0 \).

### Step 7: Check if \( g(f(x)) \) is twice differentiable at \( x = 0 \)
Now we need to check the second derivative:
- The second derivative for \( x \geq 0 \):
\[ g(f''(x)) = \frac{d}{dx}(2x \cos(x^2)) = 2\cos(x^2) - 4x^2\sin(x^2) \]
- The second derivative for \( x < 0 \):
\[ g(f''(x)) = \frac{d}{dx}(-2x \cos(x^2)) = -2\cos(x^2) + 4x^2\sin(x^2) \]

### Step 8: Evaluate the second derivatives at \( x = 0 \)
- As \( x \to 0^+ \):
\[ \lim_{x \to 0^+} (2\cos(x^2) - 4x^2\sin(x^2)) = 2 \]
- As \( x \to 0^- \):
\[ \lim_{x \to 0^-} (-2\cos(x^2) + 4x^2\sin(x^2)) = -2 \]

Since the left-hand limit and right-hand limit are not equal, \( g(f(x)) \) is not twice differentiable at \( x = 0 \).

### Conclusion
- Statement 1 is true: \( g(f(x)) \) is differentiable at \( x = 0 \) and its derivative is continuous.
- Statement 2 is false: \( g(f(x)) \) is not twice differentiable at \( x = 0 \).

Thus, the correct answer is (3): Statement 1 is true, Statement 2 is false.