If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} \frac{x + 2}{x^2 + 3 x + 2}, & x \in R - \{-1, -2\} \\ -1, & x = -2 \\ 0, & x = -1 \end{cases}$ then $f$ is continuous on the set

In order that the function $f(x) = (x + 1)^{1/x}$ is continuous at $x = 0$,$f(0)$ must be defined as

Let $[t]$ denote the greatest integer less than or equal to $t$. Let $f:[0, \infty) \rightarrow \mathbb{R}$ be a function defined by $f(x) = [\frac{x}{2} + 3] - [\sqrt{x}]$. Let $S$ be the set of all points in the interval $[0, 8]$ at which $f$ is not continuous. Then $\sum_{a \in S} a$ is equal to:

Let $f: R \rightarrow R$ be a differentiable function such that $f(3)=3$ and $f^{\prime}(3)=\frac{1}{27}$. If $g(x)=\begin{cases} \int_3^{f(x)} \frac{3t^2}{x-3} dt & \text{if } x \neq 3 \\ K & \text{if } x=3 \end{cases}$ is continuous at $x=3$,then $K=$

Let $f(x) = [x^2 - x] + |-x + [x]|$,where $x \in R$ and $[t]$ denotes the greatest integer less than or equal to $t$. Then,$f$ is

If $f(x) = \begin{cases} 6 \beta - 3 \alpha x, & \text{if } -4 \leq x < -2 \\ 4x + 1, & \text{if } -2 \leq x \leq 2 \end{cases}$ is continuous on $[-4, 2]$,then $\alpha + \beta = $