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 $y = f(x)$?

$\frac{1}{e^{3x}}(e^x + e^{5x}) = a_0 + a_1x + a_2x^2 + \ldots$
$\Rightarrow 2a_1 + 2^3a_3 + 2^5a_5 + \ldots$ is equal to

If $f(x) = \begin{cases} x, & x \le 0 \\ 0, & x > 0 \end{cases}$ then $f(x)$ at $x = 0$ is

The value of $f(4)-f(3)$ is

Let $f: R \rightarrow R$ be given by
$f(x) = \begin{cases} x^5+5x^4+10x^3+10x^2+3x+1, & x < 0 \\ x^2-x+1, & 0 \leq x < 1 \\ \frac{2}{3}x^3-4x^2+7x-\frac{8}{3}, & 1 \leq x < 3 \\ (x-2)\log_e(x-2)-x+\frac{10}{3}, & x \geq 3 \end{cases}$
Then which of the following options is/are correct?
$(1)$ $f^{\prime}$ has a local maximum at $x = 1$
$(2)$ $f$ is onto
$(3)$ $f$ is increasing on $(-\infty, 0)$
$(4)$ $f^{\prime}$ is $NOT$ differentiable at $x = 1$

Let $f(x) = \begin{cases} 3-x & \text{if } x < -3 \\ 6 & \text{if } -3 \leq x \leq 3 \\ 3+x & \text{if } x > 3 \end{cases}$. Let $\alpha$ be the number of points of discontinuity of $f$ and $\beta$ be the number of points where $f$ is not differentiable. Then $\alpha+\beta=$