If $f(x) = \begin{cases} \frac{x^2 - 4x + 3}{x^2 - 1}, & x \ne 1 \\ 2, & x = 1 \end{cases}$,then:

Let $f(x) = \begin{cases} \frac{\tan^2 \{x\}}{x^2 - [x]^2} & \text{for } x > 0 \\ 1 & \text{for } x = 0 \\ \sqrt{\{x\} \cot \{x\}} & \text{for } x < 0 \end{cases}$ where $[x]$ is the greatest integer function and $\{x\}$ is the fractional part function of $x$,then:

Let $[x]$ be the greatest integer less than or equal to $x$. At which of the following point$(s)$ is the function $f(x) = x \cos(\pi(x + [x]))$ discontinuous?
$[A]$ $x = -1$
$[B]$ $x = 0$
$[C]$ $x = 2$
$[D]$ $x = 1$

If the function $f(x)$ is defined as:
$f(x) = \begin{cases} 1 + \sin \frac{\pi x}{2}, & -\infty < x \leq 1 \\ ax + b, & 1 < x < 3 \\ 6 \tan \frac{x \pi}{12}, & 3 \leq x < 6 \end{cases}$
and is continuous in $(-\infty, 6)$,then the values of $a$ and $b$ are respectively.

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:

If $f(x) = \begin{cases} \frac{\sqrt{\pi} - \sqrt{\cos^{-1} x}}{\sqrt{x+1}}, & x \neq -1 \\ \frac{1}{\sqrt{\lambda \pi}}, & x = -1 \end{cases}$ is right continuous at $x = -1$,then $\lambda = $