The value of $a$ for which the function $f(x) = \begin{cases} \frac{1-\cos 4 x}{x^2}, & x < 0 \\ a, & x=0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x>0 \end{cases}$ is continuous at $x=0$, is

Consider $f(x) = \left[ \frac{2(\sin x - \sin^3 x) + |\sin x - \sin^3 x|}{2(\sin x - \sin^3 x) - |\sin x - \sin^3 x|} \right]$ for $x \in (0, \pi), x \neq \frac{\pi}{2}$,and $f(\frac{\pi}{2}) = 3$,where $[ \cdot ]$ denotes the greatest integer function. Then:

If $f(x) = \frac{x}{2} - 1$,then on the interval $[0, \pi]$,where $[.]$ represents the greatest integer function,which of the following is true?

Let $f :[0,1] \rightarrow \mathbb{R}$ and $g :[0,1] \rightarrow \mathbb{R}$ be defined as follows:
$f(x) = \begin{cases} 1 & \text{if } x \text{ is rational} \\ 0 & \text{if } x \text{ is irrational} \end{cases}$
$g(x) = \begin{cases} 0 & \text{if } x \text{ is rational} \\ 1 & \text{if } x \text{ is irrational} \end{cases}$
Then:

If $f: [0, 2) \to R$ is defined by $f(x) = \begin{cases} 1 + 2x^k, & 0 \le x < 1 \\ kx, & 1 \le x < 2 \end{cases}$ where $k > 0$ and $f$ is such that $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)$,then the value of $k^2$ is:

If the function $f(x)$ is continuous on its domain $[-2, 2]$,where $f(x) = \begin{cases} \frac{\sin ax}{x} + 3, & -2 \leq x < 0 \\ x + 5, & 0 \leq x \leq 1 \\ \sqrt{x^2 + 8} - b, & 1 < x \leq 2 \end{cases}$,then $7a + b + 1$ is equal to: