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

$A$ function is $f(x) = \begin{cases} \frac{e^{1/x}-1}{e^{1/x}+1}, & \text{if } x \neq 0 \\ 0, & \text{if } x=0 \end{cases}$

The function $f(x) = \frac{\log(1 + ax) - \log(1 - bx)}{x}$ is not defined at $x = 0$. The value which should be assigned to $f$ at $x = 0$ so that it is continuous at $x = 0$ is:

$f$ is continuous at $x=\frac{\pi}{2}$ where,
$f(x)=\begin{cases}\frac{2 k \cos x}{\pi-2 x}, & x \neq \frac{\pi}{2} \\ 2024, & x=\frac{\pi}{2}\end{cases}$ then,the value of $k$ is . . . . . .

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

If $f(x) = \begin{cases} x, & x \le 0 \\ 0, & x > 0 \end{cases}$ then the function $f(x)$ at $x = 0$ is: