If the function $f(\alpha) = \begin{cases} \frac{1-\cos 6 \alpha}{36 \alpha^2}, & \alpha \neq 0 \\ k, & \alpha=0 \end{cases}$ is continuous at $\alpha=0$,then $k$ is equal to . . . . . . .

If the function $f(x) = \frac{\log(1 + ax) - \log(1 - bx)}{x}$,$x \neq 0$ is continuous at $x = 0$,then $f(0) = $ . . . . . .

If $f(x) = \begin{cases} \frac{\sin x}{x} + \cos x, & x \ne 0 \\ 2, & x = 0 \end{cases}$,then which of the following is true?

The values of $a$ and $b$,so that the function $f(x) = \begin{cases} x+a \sqrt{2} \sin x, & 0 \leq x \leq \frac{\pi}{4} \\ 2 x \cot x+b, & \frac{\pi}{4} < x \leq \frac{\pi}{2} \\ a \cos 2 x-b \sin x, & \frac{\pi}{2} < x \leq \pi \end{cases}$ is continuous for $0 \leq x \leq \pi$,are respectively given by

Let $f(x) = \begin{cases} \frac{(1 + \tan x)^{\frac{1}{x}} - e}{x} & x \neq 0 \\ k & x = 0 \end{cases}$ be continuous at $x = 0$,then the value of $k$ is:

Consider the function $f(x) = [x] + |1 - x|$ for $-1 \le x \le 3$,where $[x]$ is the greatest integer function.
Statement $1$: $f$ is not continuous at $x = 0, 1, 2$ and $3$.
Statement $2$: $f(x) = \begin{cases} -1 - x, & -1 \le x < 0 \\ 1 - x, & 0 \le x < 1 \\ 1 - x, & 1 \le x < 2 \\ 2 + x - 2, & 2 \le x < 3 \\ 3, & x = 3 \end{cases}$ (Note: The provided Statement $2$ in the prompt is incorrect).