Limit and Derivative Relations for Product Function Involving Sine

To determine the correctness of the statements, we evaluate each statement systematically.

Evaluation of STATEMENT-2

Given the function:
$f(x) = g(x) \sin x$

Differentiating $f(x)$ with respect to $x$ using the product rule:
$f'(x) = g'(x) \sin x + g(x) \cos x$

Evaluating at $x = 0$:
$f'(0) = g'(0) \sin(0) + g(0) \cos(0)$

Since $g'(0) = 0$ and $\cos(0) = 1$:
$f'(0) = 0 \cdot 0 + g(0) \cdot 1 = g(0)$

Thus, STATEMENT-2 is True.

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Evaluation of STATEMENT-1

We need to evaluate the limit:
$L = \lim_{x \to 0} \left[ g(x) \cot x - g(0) \csc x \right]$

Rewriting in terms of $\sin x$ and $\cos x$:
$L = \lim_{x \to 0} \frac{g(x) \cos x - g(0)}{\sin x}$

From STATEMENT-2, we have $g(0) = f'(0)$. Furthermore, from $f'(x) = g'(x) \sin x + g(x) \cos x$, we get:
$g(x) \cos x = f'(x) - g'(x) \sin x$

Substituting $g(0) = f'(0)$ and $g(x) \cos x = f'(x) - g'(x) \sin x$ into the limit:
$L = \lim_{x \to 0} \frac{f'(x) - g'(x) \sin x - f'(0)}{\sin x}$

Split the fraction into two terms:
$L = \lim_{x \to 0} \left[ \frac{f'(x) - f'(0)}{\sin x} - \frac{g'(x) \sin x}{\sin x} \right]$
$L = \lim_{x \to 0} \left[ \frac{f'(x) - f'(0)}{x} \cdot \frac{x}{\sin x} - g'(x) \right]$

Using standard limit results:

1. $\lim_{x \to 0} \frac{f'(x) - f'(0)}{x} = f''(0)$ (by definition of the second derivative)
2. $\lim_{x \to 0} \frac{x}{\sin x} = 1$
3. $\lim_{x \to 0} g'(x) = g'(0) = 0$ (since $g''(x)$ is continuous)

Substituting these values:
$L = f''(0) \cdot 1 - 0 = f''(0)$

Thus, STATEMENT-1 is True.

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Conclusion

STATEMENT-1 and STATEMENT-2 are both True, and STATEMENT-2 ($f'(0) = g(0)$) is directly used to simplify the expression and prove STATEMENT-1. Hence, STATEMENT-2 is the correct explanation for STATEMENT-1.

Correct Answer: (A)