If $\begin{aligned} f(x) &= \frac{4 \sin \pi x}{5 x} \text{ for } x \neq 0 \\ &= 2k \text{ for } x = 0 \end{aligned}$ is continuous at $x = 0$,then the value of $k$ is

If $f(x) = \begin{cases} \frac{\sin [x]}{[x] + 1}, & \text{for } x > 0 \\ \frac{\cos \frac{\pi }{2}[x]}{[x]}, & \text{for } x < 0 \\ k, & \text{at } x = 0 \end{cases}$; where $[x]$ denotes the greatest integer less than or equal to $x$,then in order that $f$ be continuous at $x = 0$,the value of $k$ is

If $f(x) = \begin{cases} 1 + \cos x, & x \le 0 \\ a - x, & 0 < x < 2 \\ (x - b)^2, & x \ge 2 \end{cases}$ is continuous at $x=0$ and $x=2$, then find the value of $a^2+b^2$.

Let $f: R \rightarrow R$ be such that $f(2x-1) = f(x)$ for all $x \in R$. If $f$ is continuous at $x = 1$ and $f(1) = 1$, then:

Let the function $f$ be defined by the equation $f(x) = \begin{cases} 3x & \text{if } 0 \le x \le 1 \\ 5 - 3x & \text{if } 1 < x \le 2 \end{cases}$,then:

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