If $f(x) = \begin{cases} x - 1, & x < 0 \\ \frac{1}{4}, & x = 0 \\ x^2, & x > 0 \end{cases}$,then

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

If $f(x) = \begin{cases} 1 + x, & \text{when } x \le 2 \\ 5 - x, & \text{when } x > 2 \end{cases}$,then which of the following is true?

If a real valued function $f(x) = \begin{cases} \log(1+[x]), & x \geq 0 \\ \sin^{-1}[x], & -1 \leq x < 0 \\ k([x]+|x|), & x < -1 \end{cases}$ is continuous at $x = -1$,then $k =$

If the function $f: R \rightarrow R$ defined by $f(x) = \begin{cases} \frac{a(1-\cos 2x)}{x^2}, & x < 0 \\ b, & x = 0 \\ \frac{\sqrt{x}}{\sqrt{4+\sqrt{x}}-2}, & x > 0 \end{cases}$ is continuous at $x = 0$, then $a+b=$

Let $f(x) = \begin{cases} \operatorname{sgn}(x^2 - 3x + 2) & x \in \mathbb{Q} \\ 0 & x \notin \mathbb{Q} \end{cases}$,then the number of points where $f(x)$ is continuous is (where $\operatorname{sgn}(x)$ denotes the signum function of $x$).