If $f(x) = \begin{cases} \frac{x^3 + x^2 - 16x + 20}{(x-2)^2}, x \neq 2 \\ k, x = 2 \end{cases}$ is continuous at $x = 2$,then $k = \rule{1cm}{0.15mm}$

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:

Let $f(x) = \begin{cases} \frac{x^3 + x^2 - 16x + 20}{(x - 2)^2}, & \text{if } x \neq 2 \\ k, & \text{if } x = 2 \end{cases}$. If $f(x)$ is continuous for all $x$,then $k =$

Let $[\bullet]$ denote the greatest integer function,and let $f(x) = \min \{\sqrt{2}x, x^2\}$. Let $S = \{x \in (-2, 2) : \text{the function } g(x) = |x|[x^2] \text{ is discontinuous at } x\}$. Then $\sum_{x \in S} f(x)$ equals:

Let $f(x) = x \left[ \frac{x}{2} \right]$,for $-10 < x < 10$,where $[t]$ denotes the greatest integer function. Then the number of points of discontinuity of $f$ is equal to

If $f(x) = \begin{cases} \frac{x^4 - 16}{x - 2}, & x \neq 2 \\ 16, & x = 2 \end{cases}$,then: