The value of $k$ so that the function $f(x) = \begin{cases} k(2x - x^2), & x < 0 \\ \cos x, & x \ge 0 \end{cases}$ is continuous at $x = 0$,is

Assertion $(A)$: $f(x)=|x-a|+|x-b|$ is continuous on $R$. Reason $(R)$: $\frac{|x-\alpha|}{x-\alpha}$ is continuous at $x \in R-\{\alpha\}$. The correct option among the following is:

Given that $f(x) = \begin{cases} \frac{1-\cos 4x}{x^2}, & x < 0 \\ a, & x = 0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x > 0 \end{cases}$ is continuous at $x = 0$,then $a = $

Find the values of $k$ so that the function $f$ is continuous at the indicated point. $f(x) = \begin{cases} kx^2, & \text{if } x \le 2 \\ 3, & \text{if } x > 2 \end{cases}$ at $x=2$.

Show that the function defined by $f(x)=|\cos x|$ is a continuous function.

Let $f(x) = \begin{cases} \alpha + \frac{\sin [x]}{x}, & \text{if } x > 0 \\ 2, & \text{if } x = 0 \\ \beta + \left[ \frac{\sin x - x}{x^3} \right], & \text{if } x < 0 \end{cases}$ where $[x]$ denotes the greatest integer function. If $f$ is continuous at $x = 0$, then $\beta - \alpha$ is equal to