$f(x) = \begin{cases} \frac{1-\cos kx}{x^2}, & x \le 0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x > 0 \end{cases}$ is continuous at $x = 0$,then the value of $k$ is:

Let $f: R \rightarrow R$ be defined by $f(x)=\begin{cases} \alpha+\frac{\sin [x]}{x}, & x>0 \\ 2, & x=0 \\ \beta+\left[\frac{\sin x-x}{x^3}\right], & x < 0 \end{cases}$. If $f$ is continuous at $x=0$,find the value of $\alpha + \beta$.

If $f(x) = \begin{cases} x, & x > 1 \\ x^2, & x < 1 \end{cases}$,then $\lim_{x \to 1} f(x) = $

Let $f(x) = \begin{cases} x, & x \in \mathbb{Q} \\ 1 - x, & x \notin \mathbb{Q} \end{cases}$. Then at $x = \frac{1}{2}$,$f(x)$ is:

If $f(x) = \frac{\ln(e^{x^2} + 2\sqrt{x})}{\sqrt{x}}$ is continuous at $x = 0$,then $f(0)$ must be equal to:

If the function $f(x) = x^2[\sin^{-1}x]$ is discontinuous at $x = \alpha$ and $x = \beta$,where $\alpha, \beta \in R - \{0\}$ and $[.]$ denotes the greatest integer function,then the value of $\alpha + \beta$ is: