If $f(x) = \frac{x+x^2+x^3+\ldots+x^{n}-n}{x-1}$ for $x \neq 1$ is continuous at $x=1$,then $f(1) =$

Let $f(x) = \begin{cases} \frac{1 + \cos 2\pi x}{1 - \sin \pi x}, & x < \frac{1}{2} \\ p, & x = \frac{1}{2} \\ \frac{\sqrt{2x - 1}}{\sqrt{4 + \sqrt{2x - 1}} - 2}, & x > \frac{1}{2} \end{cases}$. If $f(x)$ is discontinuous at $x = \frac{1}{2}$,then:

Let $f(x)=\lim _{\theta \rightarrow 0}\left(\frac{\cos \pi x-x^{\left(\frac{2}{\theta}\right)} \sin (x-1)}{1+x^{\left(\frac{2}{\theta}\right)}(x-1)}\right), x \in R$. Consider the following two statements: $(I)$ $f(x)$ is discontinuous at $x=1$. $(II)$ $f(x)$ is continuous at $x=-1$. Then,

The set of points of discontinuity of the function $f(x) = x - [x]$, where $x \in R$, is:

$f(x) = \begin{cases} \frac{e^{\alpha x} - e^{x} - x}{x^{2}}, & x \neq 0 \\ \frac{3}{2}, & x = 0 \end{cases}$ Find the value of $\alpha$ for which the function $f$ is continuous.

If $f: R \rightarrow R$ is defined by $f(x) = x - [x]$, where $[x]$ is the greatest integer not exceeding $x$, then the set of points of discontinuity of $f$ is