Let $y = f(x) = \begin{cases} e^{-\frac{1}{x^2}}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}$. Then which of the following can best represent the graph of $y = f(x)$?

Let $f_1: R \rightarrow R, f_2:\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \rightarrow R, f_3:\left(-1, e^{\frac{\pi}{2}}-2\right) \rightarrow R$ and $f_4: R \rightarrow R$ be functions defined by:
$(i)$ $f_1(x)=\sin \left(\sqrt{1-e^{-x^2}}\right)$
$(ii)$ $f_2(x)=\begin{cases} \frac{|\sin x|}{\tan^{-1} x} & \text{if } x \neq 0 \\ 1 & \text{if } x=0 \end{cases}$,where the inverse trigonometric function $\tan^{-1} x$ assumes values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
$(iii)$ $f_3(x)=\left[\sin \left(\log_e(x+2)\right)\right]$,where,for $t \in R, [t]$ denotes the greatest integer less than or equal to $t$.
$(iv)$ $f_4(x)=\begin{cases} x^2 \sin \left(\frac{1}{x}\right) & \text{if } x \neq 0 \\ 0 & \text{if } x=0 \end{cases}$

$LIST-I$	$LIST-II$
$P$. The function $f_1$ is	$1$. $NOT$ continuous at $x=0$
$Q$. The function $f_2$ is	$2$. Continuous at $x=0$ and $NOT$ differentiable at $x=0$
$R$. The function $f_3$ is	$3$. Differentiable at $x=0$ and its derivative is $NOT$ continuous at $x=0$
$S$. The function $f_4$ is	$4$. Differentiable at $x=0$ and its derivative is continuous at $x=0$

The correct option is:

Which of the following is not true?

The number of polynomials $p: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $p(0)=0$,$p(x) > x^2$ for all $x \neq 0$,and $p^{\prime \prime}(0) = \frac{1}{2}$ is

If $f$ is defined by $f(x) = \begin{cases} x & \text{for } 0 \leq x < 1 \\ 2-x & \text{for } x \geq 1 \end{cases}$,then at $x=1$,$f(x)$ is

Let $f$ and $g$ be real-valued functions defined on the interval $(-1, 1)$ such that $g^{\prime \prime}(x)$ is continuous,$g(0) \neq 0$,$g^{\prime}(0) = 0$,$g^{\prime \prime}(0) \neq 0$,and $f(x) = g(x) \sin x$.
$STATEMENT-1$: $\lim_{x \rightarrow 0} [g(x) \cot x - g(0) \operatorname{cosec} x] = f^{\prime \prime}(0)$.
$STATEMENT-2$: $f^{\prime}(0) = g(0)$.