Analysis of Statement - 2

1. We are given the function $f(x) = g(x) \sin x$.
2. To find $f'(x)$, we use the product rule for differentiation: $f'(x) = \frac{d}{dx}(g(x) \sin x) = g'(x) \sin x + g(x) \cos x$
3. Now, we evaluate $f'(x)$ at $x=0$: $f'(0) = g'(0) \sin(0) + g(0) \cos(0)$
4. We are given that $g'(0) = 0$. Also, we know $\sin(0) = 0$ and $\cos(0) = 1$. $f'(0) = (0)(0) + g(0)(1) = g(0)$
5. Thus, the statement $f'(0) = g(0)$ is True.

Analysis of Statement - 1

1. Statement - 1 is the equality: $\mathop {\lim }\limits_{x \to 0} \left[ {g\left( x \right)\cot x - g\left( 0 \right)\csc\,x} \right] = f''\left( 0 \right)$.

2. Let's first evaluate the limit on the left-hand side (LHS). Let's call it $L$. $L = \lim_{x \to 0} \left[ g(x) \frac{\cos x}{\sin x} - g(0) \frac{1}{\sin x} \right] = \lim_{x \to 0} \frac{g(x)\cos x - g(0)}{\sin x}$

3. As $x \to 0$, the numerator becomes $g(0)\cos(0) - g(0) = g(0) - g(0) = 0$, and the denominator becomes $\sin(0) = 0$. This is a $\frac{0}{0}$ indeterminate form, so we can apply L'Hôpital's Rule.

4. Differentiating the numerator and the denominator with respect to $x$:

   • Numerator derivative: $\frac{d}{dx}(g(x)\cos x - g(0)) = g'(x)\cos x - g(x)\sin x$.
   • Denominator derivative: $\frac{d}{dx}(\sin x) = \cos x$.

5. The limit becomes: $L = \lim_{x \to 0} \frac{g'(x)\cos x - g(x)\sin x}{\cos x}$

6. Substituting $x=0$ and using the given condition $g'(0)=0$: $L = \frac{g'(0)\cos(0) - g(0)\sin(0)}{\cos(0)} = \frac{(0)(1) - g(0)(0)}{1} = 0$ So, the LHS is 0.

7. Now, let's evaluate the right-hand side (RHS), which is $f''(0)$. We have $f'(x) = g'(x) \sin x + g(x) \cos x$. Differentiating again with respect to $x$: $f''(x) = \frac{d}{dx}(g'(x) \sin x) + \frac{d}{dx}(g(x) \cos x)$ $f''(x) = [g''(x)\sin x + g'(x)\cos x] + [g'(x)\cos x - g(x)\sin x]$ $f''(x) = g''(x)\sin x + 2g'(x)\cos x - g(x)\sin x$

8. Evaluating at $x=0$ and using $g'(0)=0$: $f''(0) = g''(0)\sin(0) + 2g'(0)\cos(0) - g(0)\sin(0)$ $f''(0) = g''(0)(0) + 2(0)(1) - g(0)(0) = 0$ So, the RHS is 0.

9. Since LHS = 0 and RHS = 0, Statement - 1 is True.

Analysis of the Explanation

1. We need to determine if Statement - 2 is the correct explanation for Statement - 1.
2. Let's establish a direct connection between the LHS and RHS of Statement - 1. Start with the definition of the second derivative: $f''(0) = \lim_{x \to 0} \frac{f'(x) - f'(0)}{x}$
3. From Statement - 2, we know $f'(0) = g(0)$. Substituting this into the definition of $f''(0)$: $f''(0) = \lim_{x \to 0} \frac{f'(x) - g(0)}{x}$
4. Now, substitute the expression for $f'(x) = g'(x) \sin x + g(x) \cos x$: $f''(0) = \lim_{x \to 0} \frac{g'(x) \sin x + g(x) \cos x - g(0)}{x}$
5. We can split this limit into two parts: $f''(0) = \lim_{x \to 0} \left( g'(x) \frac{\sin x}{x} \right) + \lim_{x \to 0} \left( \frac{g(x) \cos x - g(0)}{x} \right)$
6. The first term evaluates to $g'(0) \cdot 1 = g'(0)$.
7. The second term can be related to the limit $L$ from Statement - 1: $L = \lim_{x \to 0} \frac{g(x)\cos x - g(0)}{\sin x} = \lim_{x \to 0} \left( \frac{g(x)\cos x - g(0)}{x} \cdot \frac{x}{\sin x} \right)$ Since $\lim_{x \to 0} \frac{x}{\sin x} = 1$, the second term in the expression for $f''(0)$ is equal to $L$.
8. So, we have established the relationship: $f''(0) = g'(0) + L$.
9. Statement - 1 asserts that $L = f''(0)$. This is true if and only if $g'(0)=0$.
10. Since we are given that $g'(0)=0$, Statement - 1 is true.
11. The derivation of the crucial relationship $f''(0) = g'(0) + L$ depends on using Statement - 2 (substituting $f'(0)$ with $g(0)$). This relationship explains why Statement - 1 holds true under the given condition $g'(0)=0$. Therefore, Statement - 2 is a correct explanation for Statement - 1.

Both statements are true, and Statement - 2 is the correct explanation for Statement - 1.