Let $f$ and $g$ be real valued functions defined on interval $(-1, 1)$ such that $g''(x)$ is continuous, $g(0) \neq 0$, $g'(0) = 0$, $g''(0) \neq 0$, and $f(x) = g(x)\sin x$ Statement I : $\lim_{x \to 0} [g(x)\cot x - g(0)\operatorname{cosec} x] = f''(0)$ and Statement II : $f'(0) = g(0)$

$g'' \text{ continuous},\quad g(0)\neq 0,\quad g'(0)=0,\quad g''(0)\neq 0$

$f(x) = g(x)\sin x$

$\text{I: } \lim_{x\to 0}\left[g(x)\cot x - g(0)\operatorname{cosec} x\right] = f''(0)$

$\text{II: } f'(0) = g(0)$

$1.\ \text{check II: } f'(0)$

$2.\ \text{evaluate the limit in I}$

$3.\ \text{compare with } f''(0)$

$f'(x) = g'(x)\sin x + g(x)\cos x$

$f'(0) = g'(0)\cdot 0 + g(0)\cdot 1 = g(0) \Rightarrow \text{II TRUE}$

$g(x)\cot x - g(0)\operatorname{cosec} x = \frac{g(x)\cos x - g(0)}{\sin x}$

$\lim_{x\to 0}\frac{g(x)\cos x - g(0)}{\sin x} = \frac{g'(0)\cos 0 - g(0)\sin 0}{\cos 0} = \frac{0}{1} = 0$

$f''(x) = g''(x)\sin x + 2g'(x)\cos x - g(x)\sin x$

$f''(0) = 0$

$\lim = 0 = f''(0) \Rightarrow \text{I TRUE}$

$\text{II } (f'(0)=g(0)) \text{ does not explain I}$

$\boxed{\text{Both true; II not a correct explanation} \Rightarrow (2)}$