$\mathop {\lim }\limits_{x \to 0} \frac{d}{{dx}}\left( {\frac{{{e^{{e^{{x^2}}}}} - e}}{x}} \right)$ is

Consider the following three statements for the function $f : (0, \infty) \rightarrow \mathbb{R}$ defined by $f(x) = |\log_{e} x| - |x - 1|$:
$(I)$ $f$ is differentiable at all $x > 0$.
$(II)$ $f$ is increasing in $(0, 1)$.
$(III)$ $f$ is decreasing in $(1, \infty)$.
Then:

Consider $f(x) = \begin{cases} \tan^{-1}(\frac{\alpha x + \beta}{\gamma}) & x \in (0, \frac{1}{2}) \\ 0 & x = \frac{1}{2} \\ \ln(\beta x^2 + 2) & x \in (\frac{1}{2}, 1) \end{cases}$. If $f(x)$ is continuous and differentiable in its domain,then the value of $\alpha + \beta + \gamma$ is:

Let $f(x) = \begin{cases} 3x, & x < 0 \\ \min \{1+x+[x], x+2[x]\}, & 0 \leq x \leq 2 \\ 5, & x > 2 \end{cases}$ where $[.]$ denotes the greatest integer function. If $\alpha$ and $\beta$ are the number of points where $f$ is not continuous and is not differentiable,respectively,then $\alpha + \beta$ equals.......

Let $f(x) = \begin{cases} e^{x-1}; x < 0 \\ x^2-5x+6; x \ge 0 \end{cases}$ and $g(x) = f(|x|) + |f(x)|$. If the number of points where $g$ is not continuous and is not differentiable are $\alpha$ and $\beta$ respectively, then $\alpha + \beta$ is equal to ————

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 $y = f(x)$?