Find the values of $k$ so that the function $f$ is continuous at the indicated point. $f(x) = \begin{cases} kx + 1, & \text{if } x \le 5 \\ 3x - 5, & \text{if } x > 5 \end{cases}$ at $x = 5$. (in $/5$)

If the function $f(x) = \begin{cases} \frac{x^2 - 1}{x - 1}, & x \ne 1 \\ k, & x = 1 \end{cases}$ is continuous at $x = 1$,then the value of $k$ is:

$f$ is continuous at $x=\frac{\pi}{2}$ where,
$f(x)=\begin{cases}\frac{2 k \cos x}{\pi-2 x}, & x \neq \frac{\pi}{2} \\ 2024, & x=\frac{\pi}{2}\end{cases}$ then,the value of $k$ is . . . . . .

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.

Let $f(x) = \begin{cases} 0, & \text{if } -1 \leq x < 0 \\ 1, & \text{if } x = 0 \\ 2, & \text{if } 0 < x \leq 1 \end{cases}$ and let $F(x) = \int_{-1}^{x} f(t) \, dt, -1 \leq x \leq 1$. Then:

Let $\alpha \in R$ be such that the function $f(x) = \begin{cases} \frac{\cos^{-1}(1-\{x\}^2) \sin^{-1}(1-\{x\})}{\{x\}-\{x\}^3}, & x \neq 0 \\ \alpha, & x=0 \end{cases}$ is continuous at $x=0$,where $\{x\} = x - [x]$ and $[x]$ is the greatest integer less than or equal to $x$. Then: