Let $f(x) = \begin{cases} \frac{(x - 1)(6x - 1)}{2x - 1}, & \text{if } x \neq \frac{1}{2} \\ 0, & \text{if } x = \frac{1}{2} \end{cases}$. Then at $x = \frac{1}{2}$,

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

Which of the following functions is discontinuous at $x = 0$?

Let $f(x) = [2x^3 - 5]$,where $[\cdot]$ denotes the Greatest Integer Function. Find the number of points in the interval $(1, 2)$ where the function $f(x)$ is discontinuous.

Let $f: R \rightarrow R$ be defined by $f(x)=\begin{cases} \alpha+\frac{\sin [x]}{x}, & \text{if } x>0 \\ 2, & \text{if } x=0 \\ \beta+\left[\frac{\sin x-x}{x^3}\right], & \text{if } x < 0 \end{cases}$ where $[x]$ denotes the greatest integer function. If $f$ is continuous at $x=0$,then $\beta-\alpha$ is equal to

If $f(x) = \begin{cases} x, & 0 < x < 1/2 \\ 1, & x = 1/2 \\ 1 - x, & 1/2 < x < 1 \end{cases}$,then which of the following is true?