If $f(x) = [x]$ for $x \in (-1, 2)$,then $f$ is discontinuous at (where $[x]$ represents the floor function).

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

If the function defined by $f(x) = \begin{cases} (x^2 + e^{\frac{1}{2-x}})^{-1}, & x \neq 2 \\ k, & x = 2 \end{cases}$ is right continuous at $x = 2$, then $k =$

Let $f(x) = \begin{cases} \frac{(x - 1)(6x - 1)}{2x - 1}, & \text{if } x \neq \frac{1}{2} \\ 0, & \text{if } x = \frac{1}{2} \end{cases}$. Then at $x = \frac{1}{2}$,

Let $f(x) = \begin{cases} (x - 1)^{\frac{1}{2 - x}}, & x > 1, x \neq 2 \\ k, & x = 2 \end{cases}$. The value of $k$ for which $f$ is continuous at $x = 2$ is

Let $f(x) = \begin{cases} -2 \sin x, & \text{if } x \leq -\frac{\pi}{2} \\ A \sin x + B, & \text{if } -\frac{\pi}{2} < x < \frac{\pi}{2} \\ \cos x, & \text{if } x \geq \frac{\pi}{2} \end{cases}$. For what values of $A$ and $B$ is $f$ continuous?