Let `fandg` be real valued functions defined on interval `(-1,1)` such that `g''(x)` is constinous, `g(0)=0`, `g'(0)=0,g''(0)=0andf(x)=g(x)sinx`.
Statement I `lim_(xrarr0)(g(x)cotx-g(0)cosecx)=f''(0)`
Statement II `f'(0)=g'(0)`

Options

A. Both statement I and Statement II are correct and Statement II is the correct explanation of Statement I
B. Both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I
C. Statement I is correct but Statement II is incorrect
D. Statement II is correct but Statement I is incorrect.

Answer

To solve the problem step by step, we will analyze both statements provided and verify their correctness.

### Given:
- Functions \( f \) and \( g \) are defined on the interval \((-1, 1)\).
- \( g''(x) \) is continuous.
- \( g(0) = 0 \), \( g'(0) = 0 \), and \( g''(0) = 0 \).
- \( f(x) = g(x) \sin x \).

### Statement I:
We need to prove that:
\[ \lim_{x \to 0} \left( g(x) \cot x - g(0) \csc x \right) = f''(0) \]

#### Step 1: Simplify the left-hand side
We know \( g(0) = 0 \), so we can rewrite the limit:
\[ \lim_{x \to 0} g(x) \cot x = \lim_{x \to 0} g(x) \frac{\cos x}{\sin x} \]

#### Step 2: Combine terms
The expression becomes:
\[ \lim_{x \to 0} \frac{g(x) \cos x}{\sin x} \]
Since both \( g(x) \) and \( \sin x \) approach 0 as \( x \to 0 \), we have a \( \frac{0}{0} \) form.

#### Step 3: Apply L'Hôpital's Rule
Using L'Hôpital's Rule:
\[ \lim_{x \to 0} \frac{g(x) \cos x}{\sin x} = \lim_{x \to 0} \frac{g'(x) \cos x - g(x) \sin x}{\cos x} \]

#### Step 4: Evaluate the limit
Substituting \( x = 0 \):
- \( g'(0) = 0 \)
- \( g(0) = 0 \)
- \( \cos(0) = 1 \)
- \( \sin(0) = 0 \)

Thus:
\[ \lim_{x \to 0} \frac{0 \cdot 1 - 0 \cdot 0}{1} = 0 \]

#### Step 5: Find \( f''(0) \)
Now we need to find \( f''(0) \):
1. First derivative:
\[ f'(x) = g'(x) \sin x + g(x) \cos x \]
2. Second derivative:
\[ f''(x) = g''(x) \sin x + 2g'(x) \cos x - g(x) \sin x \]

Substituting \( x = 0 \):
- \( g''(0) = 0 \)
- \( g'(0) = 0 \)
- \( g(0) = 0 \)

Thus:
\[ f''(0) = 0 + 0 - 0 = 0 \]

### Conclusion for Statement I:
Both sides equal 0, hence:
\[ \lim_{x \to 0} \left( g(x) \cot x - g(0) \csc x \right) = f''(0) \]
**Statement I is true.**

### Statement II:
We need to verify if:
\[ f'(0) = g'(0) \]

#### Step 1: Find \( f'(0) \)
From our earlier calculation:
\[ f'(x) = g'(x) \sin x + g(x) \cos x \]
Substituting \( x = 0 \):
\[ f'(0) = g'(0) \sin(0) + g(0) \cos(0) = 0 + 0 = 0 \]

#### Step 2: Compare with \( g'(0) \)
Since \( g'(0) = 0 \):
\[ f'(0) = 0 = g'(0) \]

### Conclusion for Statement II:
**Statement II is true.**

### Final Answer:
Both statements are correct.