The function $f(x) = [x] \cos \left( \frac{2x - 1}{2} \pi \right)$,where $[.]$ denotes the greatest integer function,is discontinuous at

Let $f: R \rightarrow R$ be a function given by $f(x) = \begin{cases} \frac{1-\cos 2x}{x^2} & , x < 0 \\ \alpha & , x = 0 \\ \frac{\beta \sqrt{1-\cos x}}{x} & , x > 0 \end{cases}$. If $f$ is continuous at $x = 0$,then $\alpha^2 + \beta^2$ is equal to:

The function defined by $f(x) = \begin{cases} \frac{x-4}{|x-4|} + a, & x < 4 \\ a + b, & x = 4 \\ \frac{x-4}{|x-4|} + b, & x > 4 \end{cases}$ is continuous at $x = 4$,then:

Determine if $f$ defined by $f(x) = \begin{cases} x^2 \sin \frac{1}{x}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}$ is a continuous function.

If $f: [0, 2) \to R$ is defined by $f(x) = \begin{cases} 1 + 2x^k, & 0 \le x < 1 \\ kx, & 1 \le x < 2 \end{cases}$ where $k > 0$ and $f$ is such that $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)$,then the value of $k^2$ is:

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